Nirmala

CBSE previous Year Solved  Papers  Class 12 Biology Delhi 2015

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5.  Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SET-I

SECTION-A

Question.1. A geneticist interested in studying variations and patterns of inheritance in living beings prefers to choose organisms
for experiments with shorter life cycle. Provide a reason. 5.
Answer : A geneticist interested in studying variations & pattern of inheritance in Living beings. Prefers organisms having short life cycle as then he will be able to study to detect variations that occur & inherit from one generation to another.

Question.2. Name the transcriptionally active region of chromatin in a nucleus.
Answer : The transcriptionally active region of chromatin in
a nucleus is Euchromatin.

Question.3. 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.
Answer: ‘Deposition of soot and smoke causes the tree trunks to become darker (after industrialization); hence the number of dark moths increased as they were not easily visible to their predators while the white-winged ones were easily picked up by the predators. Thus, light coloured ones fail to survive and dark ones were selected by nature (natural selection) .

Question.4. Indiscriminate diagnostic practices using X-rays etc., should be avoided. Give one reason.
Answer: X-rays are ionizing radiations that may cause adversed effects in children in the form of mutations. The mutations can alter the genetic make-up of an organisation so the usage of X-rays causing these mutations should be avoided.

Question.5.What is Biopiracy ?
Answer: Biopiracy is defined as the illegal removal of biological material of a country by organizations or multinational companies without proper authorization from the concerned countries.

SECTION – B

Question.6.After a brief medical examination a healthy couple came to know that both of them are unable to produce functional gametes and should look for an ART’ (Assisted Reproductive Technique). Name the ART’ and the procedure involved that you can suggest to them to help them bear a child.
Answer: The assisted reproductive technique (ART) for such couples is ZIFT, which stands for Zygote Intra Fallopian transfer.
In this technique, the sperm and ovum are collected from the donor male and the donor female respectively. The sperm and the ovum are fused in laboratory conditions and developed till the 8-blastomere stage. This early embryo or zygote is then transferred to the fallopian tube of the mother for further development.

Question.7.Differentiate between male and female heterogamety.
Answer:

SECTION – C

Question.12. Describe the process of Parturition in humans.
Answer : Parturition is the process of expelling the fully developed foetus from mother’s uterus at the end of the gestation period. Parturition involves forceful muscular contraction of uterine wall called labour. Expel thebaby from the uterus.
Signal of parturition is controlled by a complex neuroendocrine mechanism. Signals originate from fully formed foetus & secreting certain hormones which diffuse into mother’s blood & cause secretion of oxytocin. Oxytocin stimulate uterine contractions. These are called foetal ejection reflexes. These reflexes become stronger & more frequent to push the foetus out of the uterus. These contractions called labour are induced by placental & foetal hormones Labour period is divided into three stages.

Question.13. A teacher wants his/her students to find the genotype of pea plants bearing purple coloured flowers in their school garden. Name and explain the cross that will make it possible.
Answer : Purple colour of flowers is a dominant phenotype in pea plants. The genotype of such plants can be determined by a test cross.
Test Cross
When an F, hybrid is crossed with its homolygous recessive parent, it is known as test cross. Test cross is used to determine that the genotype of a plant with the dominant phenotype is homozygous or heterozygous, e.g. purple flower coming from PP or Pw.
In this cross, plant with purple-coloured flowers is crossed with plant with white-coloured flowers, which will always have homozygous recessive genotype.
If the offspring produced is all purple-flowered plants, then the genotype of the parent purple-coloured plant is PP (homozygous). But if the offspring produced are purple & white flowered plants in equal proportions, then genotype of parent purple flowered plant is Pw (heterozygous).
Thus, the genotype of pea plant bearing purple-coloured flowers in the school garden can be Pw or PP.

Question.14. (a) 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?
(b) Draw a diagrammatic sketch of a portion of DNA segment to support your answer.
Answer : (a) Cytosine and thymine are pyrimidines. According to Chargaff’s rule, ratio of purines to pyrimidines is equal.
Thus, the number of adenine (A) will be equal to the number of thymine (T). .
Number of adenine (A) containing nucleotides = 240
Thus, A = T = 240
Therefore, A + T = 240+240 = 480
The number of cytosine (C) will be equal to the number of guanine (G).
Thus, G + C = Total number of nucleotides — Nucleotides containing A and T nitrogenous bases = 1000 — 480
= 520
Therefore, G = 260, C = 260
Number of guanine will be equal to number of cytosine, which will be 260.
Therefore, the number of pyrimidines that the segment possess = C + T
= 260 + 240
= 500
(b) A diagrammatic sketch of a portion of DNA segment is given below:

Question.15. Explain adaptive radiation with the help of a suitable example.
Answer : Adaptive radiation also referred as divergent evolution is the formation of a number of divergent species from a common ancestor, with new species moving out of the native habit at & adapting to new ecological niches.
The best example of adaptive radiation is the finches of Galapagos islands. Finches are sparrow-like small black- coloured birds. There are around 14 species of these finches 13 of which are on the galapagos island & 1 species is found on the nearby cocos island. The species on the cocus island resemble the mainland finches in plumge body structure & short tails. But these show differences amongst themselves & those on the mainland finches in shape & size of beaks, food habits, body size & feather colour.

Question.16. A team of students are preparing to participate in the interschool sports meet. During a practice session you find some vials with labels of certain cannabinoids.
(a) Will you report to the authorities ? Why ?
(b) Name of a plant from which such chemicals are obtained.
(c) Write the effect of these chemicals on human body.
Answer: (a) Yes, I would report the matter to the authorities because cannabinoids are classified under drugs and drug abuse1 is an illegal practice.
(b) Cannabinoids can be obtained from a plant called Cannabis sativa.
(c) The cannabinoids bind to the brain which has more cannabinoid receptors and they affect the cardiovascular system.

Question.17. Enlist the steps involved in inbreeding of cattle. Suggest two disadvantages of this practice.
Answer: In breeding involves mating of closely related individuals within the same breed for 4-6 generations.
The process of inbreeding of cattle is done when, Superior males and superior females of the same breed are identified and are made to mate in pairs. Then, evaluation of obtained progeny is done to identify the superior males and females for further mating. Following are the two disadvantages of the process of inbreeding:
(i) Inbreeding depression used by continuous inbreeding among cattle.
(ii) It decreases the fertility and also the productivity of an animal.

Question.18. Choose the 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.
Mvcorrhiza ; Monascus ; Anabaena; Rhizobium: Methanobacterium: Trichoderma.
Answer: Following three microbes can be chosen for organic farming.
Mycorrhiza : Absorb phosphorous from soil.
. Anabaena : It is involved in the process of atmospheric nitrogen fixation.
Rhizobium : It also plays a role in nitrogen fixation in legumhour plants.
Monascus : It produces a group of drugs called statins that are used to lower the cholesterol in body.
Methanobacterium : It is used in the biological generation of methane by anaerobic processes.
Trichoderma : It produces cyclosporin A which is used as an immunosuppressive agent.

Question.19. Recombination DNA – technology is of great importance in
the field of medicine. With the help of a flow chart, show how this technology has been used in preparing genetically engineered human insulins.
Answer : Preparation of Human Insulin Using Recombinant DNA Technology

Question.20. Draw a labelled sketch of sparged-stirred-tank bioreactor. Write its application.
Answer:

A bioreactor is an apparatus in which a biological process is carried out. Following are the applications of
bioreactors :

1.  They are also used to produce beverages like alcohol.
2.  They are used to produce large quantities of proteins.

Question.21. 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 bodies to their relatives. Name a modern scientific method and * write the procedure that would help in the identification of kinship. 
Answer: DNA fingerprinting is the modern scientific method used for the identification of kinship.
DNA Fingerprinting process can be carried out by follpwing point:

1.  Extract and purify DNA from cells (A drop of blood or semen or a piece of hair root or any tissues can be used to isolate DNA.)
2. DNA is digested or restricted with enzymes (restriction endonucleases) without causing cut in the mini satellite region.
3. DNA fragments are separated by electrophoresis.
4.  Denature DNA.
5.  Separated DNA fragments are transferred to synthetic membranes like nitrocellulose or nylon.
6. Add labelled VNTR (Variable Number of Tandem Repeats or Mini Satellite) probe for hybridization to take place.
7.  Wash off unbound probe.
8. Ffybridised DNA fragments are detected by autora-diography.
9.  Matching banding pattern of DNA of the passengers killed and that of relatives.

Question.22. Many plants and animals species are on the verge of their extinction because of loss of forest land by indiscriminates use by the humans. As biology student what method would you suggest along with its advantages that can protect such threatened species from getting extinct ?
OR
“Determination of Biological oxygen Demand (BOD) can help in suggesting the quality of a water body”. Explain.
Answer: Ex-situ conservation is the preservation of components of biological diversity outside their natural habitats. This involves conservation of genetic resources, as well as wild and cultivated or species and draws on a diverse body of techniques and facilities.
Some of these include :

1. Gene banks, e.g. seed banks, sperm and ova banks, field banks.
2. In vitro plant tissue and microbial culture collections.
3. Captive breeding of animals and artificial propagation of plants, with possible reintroduction into the wild, and
4.  Collecting living organisms for zoos, aquaria, and botanical gardens for research and public awareness.

OR
Biochemical oxygen demand or B.O.D. is a chemical procedure for determining the amount of dissolved oxygen needed by aerobic biological organisms in a body of water to break down organic material present in a given water sample at certain temperature over a specific time period. It is widely used as an indication of the organic quality of water. The greater the BOD, the more rapidly oxygen is depleted in the stream. This . means less oxygen is available to higher forms of aquatic life. Lower BOD of water body indicates less number of micro organisms in water, quality of water, less polluting potential and aquatic life fluorishes.

SECTION-D.

Question.23 . Since October 02, 2014 “Swachh Bharat Abhiyan” has been launched in our country.
(a) Write your views on this initiative giving justification.
(b) As a biologist name two problems that you may face while implementing the programme in your locality.
(c) Suggest two remedial methods to overcome these problems.
Answer:
(a) “Swachh Bharat Abhiyan” is an initiative started on 2nd October, 2014 by India’s Prime Minister, Narendra Modi. It is India’s biggest cleanliness drive since independence. On a personal note, I am completely in support for this movement. It is our main duty to clean our nation as wastes are the biggest evil that hinders the development and progress of a country. An unclean surrounding will lead to a lot of problems, starting from the health of citizens to the shortage of land. On a large scale, it is responsible for the air and water pollution, which creates problem both on economic aspects and development of the country.
(b) As a biologist, two problem faced while implementing the programme in my locality are as follows :

1. Problem of proper sanitation.
2. Separation of biodegradable and non-biodegradable wastes.

(c) Remedies to overcome the problems :

1.  To overcome sanitation problems, we should make people aware regarding the benefits of proper sanitation and encourage the local people to make proper toilets.
2.  There should be separate bins for both biodegradable and non-biodegradable wastes so that they can be disposed and recycled accordingly.

SECTION-E

Question.24. A flower of tomato plant following the process of sexual reproduction produces 240 viable seeds.
Answer the following questions giving reasons :
(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 ?
OR
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 ?
Answer : The number of viable seeds produced by the tomato plant through sexual reproduction = 240
(a) The minimum number of pollen grains that must have been involved in the pollination of its pistil are 240 because each pollen grain contains two male gametes. Out of theses two gametes, one fuses with polar nuclei and forms endosperm. While, the other male gamete fuses with the egg cell to form the zygote that eventually give rise to seeds. Therefore, in order to obtain 240 seeds, number of pollen grains needed would be 240.
(b) The number of ovules minimally involved in this process would be 240, as the number of viable seeds are 240. After fertilization, the ovary turns into fruit and the ovules turn into seeds. Therefore, the number of ovules are corresponding to the number of seeds formed.
(c) During the process of gametogenesis, 240 megaspore mother cells are involved as only one megaspore of the tetrad becomes functional and develops further and the rest three megaspores get degenerated.
(d) In the above case, 60 microspore mother cells must have undergone reduction division prior to dehiscence of anther, as each microspore mother cell would give rise to 4 microspores. Since 1 microspore mother cell would produce 4 microspores, therefore, to obtain 240 microspores, there must be 60 microspore mother cells.
(e) The number of male gametes involved in seed formation would be 240 as each male gamete will fuse with one egg nuclei to form zygote, which will further give rise to the seed.
OR
Placenta is a foetal maternal connective that develops during pregnancy and forms a temporar y association between foetal and mother tissues. It supports the foetus during its development. The foetus is connected to the placenta by a long, flexible string called umbilical cord. It is made up of allonto is after fertilisation, the zygote divides and leads to the formation of blastocyst. The blastocyst comes into contact with the endometrium. The outer layer of blastocyst (trophoblast) secrete lytic enzymes which cause wearing away of endometrial lining. These give rise to finger-like projections called chorionic villi. This forms the foetal part of placenta. These villi extend into the material part of placenta called- Decidua. The villi are immersed into the blood sinuses found in decidua region which is divided into chree distinct regions.
(A) Decidua Basalis
(B) Decidua capsularies
(C) Decidua Parietalis
Decidua Basalis is the part of decidua underlying the chorionic „ villi and overlying the myometrium. Decidua capsularis is the part lying between the embryo and lumen of uterus and decidua parietalis line the uterus at places other than the site of attachment of embryo.

Functions of placenta during pregnancy and embryonic development are as follows :

1.  Placenta serves as a medium to provide nutrition required by the foetus. The nutrients like amino acids, minerals,sugars, lipids, vitamins etc. are transferred from mothers blood to foetal blood.
2.  Placenta aids in exchange of respiratory gases-oxygen & carbon dioxide (oxygen from mother & carbon dioxide from foetus).
3. Placenta acts as an important endocrine gland that produces hormones-human chorionic gonadotropin (hCG) chorionic thyrotropin, chorionic corticotropin, human placental lactogen (hPL), progesterone, estrogen & relaxin. Relaxin, hCG & hPL are secreted only during . pregnancy.

Question.25. Explain the genetic basis of blood grouping in human population.
OR
How did Hershey and Chase established that DNA is transferred from virus to bacteria ?
Answer : Blood groups in human beings : Blood groups in human beings is a character that is controlled by three different alleles, namely, IA, IB and Ic or i. The letter I is related to their is agglutination. The phenomenon of more than two alleles controlling a single character is referred as multiple allelism and the alleles are called multiple alleles. The three different alleles combine differently to form four blood groups. A, B, AB&O.

1.  A person will have blood group as A if he has the presence of allele IA in homozygous condition (IAIA) or alleles IA & 1° (i) in heterozygous condition (IAi),
2.  A person will have blood group as B, if he has the presence of allele IB in homozygous condition (IBIB) or alleles P & i in heterozygous condition (Pi).
3.  Blood group AB is characterised by thje presence of
   both IA & P alleles. This indicated that IA & IB are co-dominant when these occur together.
4. Blood group O is present in an individual when allele ‘i’ occurs in homozygous condition (ii or I°I°). It indicates that ‘i’ allele can express only in the absence of either IA or IB allele. It is thus recessive in nature.
   

OR
Alfred Hershey & Martha Chase in 1952 conducted
Transduction experiments (Bacteriophage infecting bacteria
to prove that DNA is the genetic material.

1.  They first took T2 bacteriophages and made them infect two separate bacterial colonies. One bacterial colony was having radioactive phosphorus 32P & the other was impregnated with radioactive sulphur 35S.
2. When T2 bacteriophages infected these two colonies, radioactive sulphur (35S) got incorporated in the capsid proteins of bacteriophage. While radioactive phosphorus (32P) became a component of phage DNA.
3.  The two types of bacteriophages were then introduces to infect two different bacterial colonies of E.coli.
4.  The two culture were then centrifuged independence. The phage capsids got separated from the bacterial. The bacterial cells were present at the bottom of centrifuge tube as pellets.
5. Hershey & Chase observed that in experiment using 35S the supernatant continuing capsids showed presence of radioactivity. The bacterial cells in this case, were without radioactive whereas in the experiment using 32P. the supernatant showed no signs of radioactivity. It was present in the bacteria & the phages, which multiplied in them.
   Thus, it was concluded that the DNA which is able to enter the bacteria is the genetic material. It aided in phage multiplication. The protein part had no such function.
   

Question.26. “Analysis of age-pyramids for human population can provide important inputs for long-term planning strategies”. Explain.
OR
Describe the advantages for keeping the ecosystems healthy.
Answer : Analysis of age pyramids for human population can provide important inputs for long-term planning strategies. The different age groups present in a population determine its reproductive status. Distribution of age groups highly influences the growth of the population. Each population displays following three ecological ages or age groups :

1.  Pre-reproductive
2.  Reproductive
3.  Post-reproductive

Population having large number of young members grow rapidly, while, the population bearing more number of post-reproductive members tends to be declining. There are basically three types of age pyramids found to be present in human population. These are as follows :

Therefore, through the analysis of age pyramids of a particular population, the distribution of resources can be done more efficiently. A better fanning strategy can be adopted considering the demand of the resource;/thus, long-term management of resources can be done in Such a way that the population can derive maximum bendfhfvvith minimum effects on nature, leading the population to flourish efficiently.
The advantages of keeping the ecosystems healthy are as follows :

1.  The products of ecosystem processes are named as ecosystem services, as they are of great help to the organisms living within an ecosystem.
2.  Healthy ecosystem is the base for a wide range of economic, environmental and aesthetic goods and services.
3.  It also mitigates droughts and floods and cycle nutrients.
4.  Healthy forest ecosystem purify air and water.
5. It also provides aesthetic, cultural and spiritual values.
6. Maintenance of biodiversity is also an important aspect of healthy ecosystem.
7. Healthy ecosystem generates fertile soil and provides wildlife habitat.

SET-II

SECTION-B

Question.7. Name any two common Indian millet crops. State one characteristic of millets that has been improved as a result of hybrid breeding so as to produce high yielding millet crops.
Answer : The common Indian millet crops : Maize and Jowar Hybrid breeding has resulted in the production of high yielding millet varieties that are resistant to water stress.

Question.9. Explain mechanism of sex-determination in birds.
Answer : In birds, the method of sex determination is s ZZ-ZW type. In this system, both the sexes have two sex chromosomes. The females are heteromorphic with two different sex chromosomes (Z & W) whereas the males are homomorphic with two identifical sex chromosomes (Z). The female produces two types of eggs, one with Z chromosome. Thus, in birds the egg determines the sex of the offspririg. Fusion of a sperm with egg having z chromosome, will give rise to male offspring while fusion of egg with w’ chromosome, will give rise to female offspring. It is called female heterogamety.

Question.10. After a brief medical examination a healthy couple come to know that both of them are unable to produce functional gametes and should look for an ART’ (Assisted Reproductive Technique). Name the ART’ and the procedure involved that you can suggest them to help them bear a child.
Answer: The Assisted Reproductive Technique (ART) for such couples is ZIFT, which stands for zygote intrafallopian transfer.
In this technique, The sperms and the eggs are collected from the donor male and donor female respectively. The sperm and the ovum are fused in laboratory conditions to make embryos. These zygotes or early embryos are then transferred to the fallopian tube of the mothers for further development.

SECTION – C

Question.13. (a) A DNA segment has a total of 1,500 nucleotides, out of which 410 are Guanine containing nucleotides. How many pyrimidine bases this segment possesses ? (b) Draw a diagrammatic sketch of a portion of DNA segment to support your answer.
Answer : Cytosine and thymine are pyrimidines.
According to ChargafFs rule, the ratio of purines to pyrimidines is equal.
Thus, the number of cytosine (C) will be equal to the number of guanine (G).
Number of guanine-containing nucleotides = 410
Thus,
G = C = 410 Now,
G + C = 410 + 410 = 820
The number of adenine (A) will be equal to the number of thymine (T).
Thus,
A + T= Total number of nucleotides — Nucleotides containing G and C nitrogenous bases = 1500-820 = 680
Therefore,
A = 340 and T = 340
The number of adenine will be equal to number of thymine, which is 340.
Therefore,
Number of pyrimidines that the segment possess = C + T = 410 + 340 = 750

Question.14. Name the stage of human embryo at which it gets implanted. Explain the process of implantation.
Answer : Blastocyst is that stage of the human embryo in which it gets implanted in the uterus.
The process of implantation can be explained as follows :

1. Fertilization results in the formation of a diploid zygote.
2.  Mitotic divisions occur in the zygote as it moves through the uterus and 2, 4, 8, 16 daughter cells are formed by this divisions which are called as blastomers.
3.  Embryo with 8 to 16 blastomers is known as morula.
4.  Morula undergoes further cleavage and develops into blastocyst, blastomeres of blastocyst get arranged into an inner cell mass where as outer layer called trophoblast.
5.  The trophoblast does not take part in the formation of the embryo & gives rise to extra-embryonic membranes-Chorion & Amnion for nourishment of embryo & protection. The inner cell mass is referred as ‘precursor of embryo’.
6.  The blastocyst comos in contact with the endometrium
   in the region of embryonal knob. The surface cells of trophoblast secrete lytic enzymes which cause corrosion of endometrial lining. These give rise to finger-like projections called chorionic villi, which assists in fixation & absorption of nutrients.
   

Question.15. A non biology person is quite shocked to know that apple is a false fruit, mango is a true fruit and banana is a seedless fruit. As a biology student how would you satisfy this person ?
Answer : Fruits that are derived from matured ovaries of flowers are called true fruits and fruits that develop from other floral parts of plants are called false fruits. For example apple is a false fruit as in apple, the thalamus produces the fleshy edible part. Mango, on the other hand is a true fruit as it develops from the ovary of the flower, Fruits that develop without fertilization are called parthenocarpic fruits. Since no fertilization takes place, such fruits are seedless. Banana is an example of a seedless fruit.

SECTION -E

Question.25. A flower of brinjal plant following the process of sexual reproduction produces 360 viable seeds. Answer the following questions giving reasons.
(a) How many ovules are minimally involved ?
(b) How many megaspore mother cells are involved ?
(c) What is the minimum number of pollen grains that must land on stigma for pollination ?
(d) How many male gametes are involved in the above case ?
(e) How many microspore mother cells must have undergone reduction division prior to dehiscence of anther in the above case ?
Or
Describe the changes that occur in ovaries and uterus in human female during the reproductive cycle.
Answer: The number of viable seeds produced by the brinjal plant through sexual reproduction = 360
(a) The number of ovules minimally involved in this process would be 360, as the number of viable seeds are 360. After fertilization, the ovary turns into fruit and the ovules turn into seeds. Therefore, the number of ovules are corresponding to the number of seeds formed.
(b) During the process of garnetogenesis, 360 megaspore mother cells are involved as only one megaspore of the tetrad becomes functional and develops further and the rest three megaspores get degenerated.
(c) The minimum number of pollen grains that must land on stigma for pollination are 360 becauseeach pollen grain contains two male gametes. Out of these two gametes,
one fuses with polar nuclei and forms endosperm, while, the other male gamete fuses with the egg cell to form the zygote that eventually give rise to seeds.
(d) 720, each pollen grain carries two male gametes which participate in double fertilization.
(e) In the above case, 90 microspore mother cells must have undergone reduction division prior to dehiscence of anther, as each microspore mother cell would give rise to 4 microspores. Since 1 microspore mother cell would produce 4 microspores, therefore, to obtain 360 microspores, there must be 90 microspore mother cells.
OR
Menstrual cycle is the reproductive cycle in all primates , and begins at puberty (menarche). In human females, menstruation occurs once in 28 to 29 days. The cycle of 1 events starting from one menstruation till the next one is
called the menstrual cycle. These changes are brought about by ovarian and pituitary hormones.

Changes occurring in the Ovary :

1. During menstrual phase the levels of estrogen and progesterone falls considerably.
2.  This induces adenohypophysis to secrete FSH and LH.
3.  Increased levels of FSH stimulates the graafian follicle to mature and secrete estrogens. The rising level of estrogen, causes the endometrium to become thicker and more richly supplied with blood vessels. The level of estrogens in blood increases gradually for a few days and is at the peak on the 12th day of the cycle.
4. The estrogen surge reduces FSH secretion and this in turn introduces LH surge within 12 hours i.e. on the 13th day of the cycle.
5.  LH causes ovulation and formation of corpus futeum.
6.  During post ovulatory phase corpus luteum secretes progesterone.
7.  Corpus luteum secretes progesterone, which facilitates the preparation of the endometrium of the uterus for receiving the blastocyst and its implantation and it also inhibits the contraction of the uterus and any further development of a new follicle.

Change occurring in the uterus :

1.  If fertilization does not occur, the rising progesterone level inhibits the release of Gonadotropin releasing hormone (GnRH), which in turn inhibits the production of FSH, LH and progesterone.
2.  Once the progesterone level drops, the corpus luteum begins to degenerate resulting in its transformation into a white body called the corpus albicans.
3.  These hormonal changes, further, cause the breakdown of the endometrium, inhibition of uterine contraction ceases and the menstrual bleeding begins.
4. The low level of estrogens and progesterone stimulates secretion of FSH and LH from anterior pituitary initiating the next ovarian cycle.
5.  If fertilization occurs, corpus luteum persists and secretes progesterone and estrogens during pregnancy. Fertilised egg starts developing and simultaneously travels down and gets implanted in the uterus. Ovarian cycle comes to a temporary halt.

SET-III

SECTION-B

Question.6. Differentiate between ‘ZZ’ and ‘XY’ type of sex-determination mechanisms
Answer : The differences between ZZ and XY type of sex- determination mechanisms are listed below :

Question.7. An infertile couple is advised to adopt test-tube baby programme. Describe two principle procedures adopted for such technologies.
Answer : In-Vitro fertilization (IVF): In this process, the fertilization takes place outside the body (test tube baby). The ‘ following techniques are included in IVF :

1.  Zygote Intrafallopian Transfer (ZIFT) – In ZIFT, the sperm front a male donor and the ovum from a female donor are fused in the laboratory. The zygote so formed is transferred into the fallopian tube at the 8 blastomeres stage.
2.  Gamete Intrafallopian Transfer (GIFT) – In GIFT, females who cannot produce ovum, but can provide suitable conditions for the fertilization of ovum, are provided with ovum from a donor, which is transferred to the follopian tube for fertilisation.

Question.9. Enumerate four objectives for improving the nutritional quality of different crops for the health benefits of the human population by the process of “Biofortification”.
Answer : Biofortification involves the breeding of crops to increase their nutritional value. The objectives for biofortification are as follows :

1.  To improve protein content and quality
2.  To improve oil content and quality
3.  To improve vitamin content
4.  To improve micronutrient and mineral content

SECTION-C

Question.12. Describe the development of endosperm after double fertilization in an angiosperm. Why does endosperm development preceeds that of zygote ?
Answer : Endosperm is the nutritive tissue formed as a result of triple fusion in the angiosperms Endosperm is generally triploid meant for nourishing the embryo. The formation of endosperm starts with degeneration of the unclear tissue. Based on the mode of development there are three types of . endosperms, (i) Nuclear (ii) Cellular (iii) Helobial.

1. Nuclear type : Primary endosperm nucleus divides ‘ repeatedly to form a large number of free nuclei. No cell plate formation takes place at this stage. A central vacuole appears later.
2. Cellular type : In this case, there is cytokinesis after each nuclear division of endosperm nucleus. The endosperm, thus, has a cellular form, from the very beginning because first and subsequent divisions are all accompanied by wall formation e.g. Petunia, Datura, Adoxa, etc.
3.  Helobial type : It is an intermediate type between the nuclear and cellular types. The first division is accompanied by cytokinesis but the subsequent ones are free nuclear. The chamber towards micropylar end_ of embryo sac is usually much larger than the chamber towards chalazal and.
   The endosperm develops before embryo because the cells of endosperm provide nutrition to the developing embryos.

CBSE Previous Year Solved  Papers  Class 12 Physics Delhi 2010

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  All questions are compulsory. There are 26 questions in all.
2.  This question paper has five sections : Section A, Section B, Section C, Section D and Section E.
3. Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each.
4.  There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions.
5. You may use the following values of physical constants wherever necessary:

SET I

Note s Except for the following questions, all the remaining questions have been asked in Previous Set.

Question.1.In which orientation, a dipole placed in a uniform electric
field is in

1.  stable,
2. unstable equilibrium?

Answer :

1.  The dipole is in stable equilibrium when electric dipole is in the direction of electric field.
2. The dipole is in unstable equilibrium when electric dipole is in the opposite direction of electric field.

Question.2.Which part of electromagnetic spectrum has largest penetrating power?
Answer: Gamma rays

Question.3. A plot of magnetic flux (∅) versus current (I) is shown in the figure for two inductors A and B. Which of the two has larger value of self inductance.

Answer: Line A

Question.4.Figure shows three point charges, +2q, -q and +3q. Two charges +2q and —q are enclosed within a surface ‘S’. What is the electric flux due to this configuration through the surface ‘S’?

Answer : Total charge within a surface S = +2q + (-q) = +q

Question.5. A glass lens of refractive index 1.45 disappears when immersed in a liquid. What is the value of refractive index of the liquid?
Answer : The value of refractive index of the liquid is 1.45.

Question.6. What is the ratio of radii of the orbits corresponding to first excited state and ground state in a hydrogen atom?

Question.7. A wire of resistance’ 8R is bent in the form of a circle. What is the effective resistance between the ends of a diameter AB?
Answer : Resistance of each semi-circular part of circle is 4R.

Question.8. State the conditions for the phenomenon of total internal reflection to occur.
Answer : (i) Light should travel from denser to rarer medium (ii) Angle of incidence in denser medium should be greater than the critical angle for the pair of media in contact.

Question.9. Explain the function of a repeater in a communication system.
Answer : A repeater, picks up the signal from the transmitter, amplifies and retransmits it to the receiver sometimes with a change in carrier frequency. Repeaters are used to extend the range of a communication system.

Question.10.(i) Write two characteristics of a material used for making permanent magnets.
(ii) Why is core of an electromagnet made of ferromagnetic materials?
OR
Draw magnetic field lines when a (i) diamagnetic, (ii) paramagnetic substance is placed in an external magnetic field. WTiich magnetic property distinguishes this behaviour of the field lines due to the two substances?
Answer:
(i) Two characteristics of a material used for making permanent magnets are:
(a) High Coercivity
(b) High Retentivity
(ii) Core of an electromagnet made of ferromagnetic materials because of high permeability and low retentivity.
OR
(i) Behaviour of magnetic field lines when a diamagnetic substance is placed in-an external magnetic field.

Question.11. Draw the circuit diagram of an illuminated photodiode in reverse bias. How is photodiode used to measure light intensity?
Answer : Circuit diagram of an illuminated photodiode :

Explanation: The magnitude of the photocurrent depends on the intensity of incident light (photocurrent is proportional to incident light intensity). Thus photodiode can be used to measure light intensity.

Question.12. An electric lamp having coil of negligible inductance connected in series with a capacitor and an AC source is glowing with certain brightness. How does the brightness of the lamp change on reducing the (i) capacitance, and (ii) the frequency? Justify your answer.

The current flows in the circuit and the lamp glows.
(i) On reducing the capacitance C, X_c. increases. Therefore, the brightness of the bulb decreases.
(ii)On reducing the frequency u,X_c increases. Therefore, the brightness of the bulb decreases.

Question.13. Arrange the following electromagnetic radiations in ascending order of their frequencies :

1.  Microwave
2. Radiowave
3. X-rays
4. Gamma rays

Write two uses of any one of these.
Answer : Radiowave <Microwaves < X-rays < Gamma rays Uses of microwaves are :

1.  Microwaves are used in radar systems for aircraft navigations.
2. Microwaves are used in microwave ovens for cooking purposes.

Question.14. The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. If focal length of the lens is 12 cm, find the refractive index of the material of the lens.

Question.15.An electron is accelerated through a potential difference of 100 volts. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond?

Question.16. A heavy nucleus X of mass number 240 and binding energy per nucleon 7.6 MeV is split into two fragments Y and Z of mass numbers 110 and 130. The binding energy of nucleons in Y and Z is 8.5 MeV per nucleon. Calculate the energy Q released per fission in MeV.

Question.17. (a) The bluish colour predominates in clear shy.
(b) Violet colour is seen at the bottom of the spectrum when white light is dispersed by a prism.
State reasons to explain these observations. 
Answer .
(a) As per Rayleighs law (scattering oc 1/λ⁴), lights of shorter wavelengths scattered more by an atmospheric particles. This results in a dominance of bluish colour in the scattered light.
(b) In the visible spectrum, violet light having its shortest wavelength, has the highest refractive index. Hence it is deviated the most.

Question.18. Plot a graph showing the variation of stopping potential
with the frequency of incident radiation for two different photosensitive materials having work functions W₁ and W₂  (W₁>W₂). On what factors does the (i) slope and

Question.19. A parallel plate capacitor is charged by a battery. After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. How will (i) the capacitance of the capacitor, (ii) potential difference between the plates and
(iii) the energy stored in the capacitor be affected?
Justify your answer in each case.
Answer : As battery in dissconnected, charge will remain constant.

Question.20. Write the principle of working of a potentiometer. Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a given cell.
Answer : Working principle :When constant current flows through a wire of uniform cross-section, then potential difference across the wire is directly proportional to the length v00l

Question.21. Write the expression for the magnetic moment due to a planar square loop of side 7’ carrying a steady current I in a vector form.
In the given figure this loop is placed in a horizontal plane near a long straight conductor carrying a steady current Ii at a distance l as shown.
Give reasons to explain that the loop will experience a net force but no torque. Write the expression for this force acting on the loop.

Question.22.(a) Depict the equipotential surfaces for a system of two identical positive point charges placed a distance ‘d apart.
(b) Deduce the expression for the potential energy of a system of two point charges q₁and q₂ brought from
infinity to the points respectively in the presence of external electric field
Answer : (a) Equipotential surfaces of two identical positive point charges placed at a distance ‘d ’ apart.

Question.23. (i) Define‘activity7 of a radioactive material and write its S.I.unit.
(ii)Plot a graph showing variation of activity of a given radioactive sample with time.
(iii)The sequence of stepwise decay of a radioactive nucleus is

If the atomic number and mass number of D2 are 71 and 176 respectively, what are their corresponding values for D?
Answer : (i) The total decay rate (of a sample) at the given instant, i. e., the number of radio nuclides disintegrating per unit time is called the activity of that sample. The SI unit for activity is becquerel (Bq).

.^.. Atomic number of D = 74
Mass number of D = 180

Question.25. A long straight wire of a circular cross-section of radius V carries a steady current T. The current is uniformly distributed across the cross-section. Apply Amperes circuital law to calculate the magnetic field at a point V in the region for (i) r< a (ii) r> a.
OR
State the underlying principle of working of a moving coil galvanometer. Write two reasons why a galvanometer cannot be used as such to measure current in a given circuit. Name any two factors on which the current sensitivity of a galvanometer depends.



Two reasons:
(i) Galvanometer is a very sensitive device, it gives a full-scale deflection for a current of the order of pA.
(ii) For measuring currents, the galvanometer has to be connected in series, and it has a finite resistance, this will change the value of the current in the circuit.
Two factors : The current sensitivity of a moving coil galvanometer can be increased by
(i) increasing the number of turns , (ii) increasing cross-section area of the loop

Question.26. What is space wave propagation? Give two examples of communication system which use space wave mode.
A TV tower is 80 m tall. Calculate the maximum distance upto which the signal transmitted from the lower can be received.
Answer : When waves travel in space in a straight line from the transmitting antenna to the receiving antenna, this mode of propagation is called the space wave propagation. Examples: Television broadcast, satellite communication. Here

Question.27. In a meter bridge, the null point iHound at a distance of 40 cm from A. If a resistance of 12 Q is connected in parallel with S, the null point occurs at 50.0 cm front A. Determine the values of R and S.

Question.28. Describe briefly, with the help of a labelled diagram, the basic elements of an A.C. generator. State its underlying principle. Show diagrammatically how an alternating emf is generated by a loop of wire rotating in a magnetic field. Write the expression for the instantaneous value of the emf induced in the rotating loop.
OR ,
A series LCR circuit is connected to an ac source having voltage v = vm sin ϖt. Derive the expression for the instantaneous current I and its phase relationship to the applied voltage.
Obtain the condition for resonance to occur. Define ‘power factor’. Sate the conditions under which it is (i) maximum and (ii) minimum.
Answer: It consists of a coil mounted on a rotor shaft. The axis of rotation of the coil is perpendicular to the direction of the magnetic field. The coil (called armature) is mechanically rotated in the uniform magnetic field by some external means.
The ends of the coil are connected to an external circuit by means of slip rings and brushes.

Underlying Principle : As the coil rotates in a magnetic field B, the effective area of the loop (the face perpendicular to the field) which is AcosO, where 0 is the angle between area (A) and magnetic field (B) changes continuously. Hence, magnetic flux linked with the coil keeps on changing with time and an induced emf is produced.
The instantaneous value of the emf is e = NBA® sin cor

Question.29. State Huygens principle. Show, with the help of a, suitable diagram, how this principle is used to obtain the diffraction pattern by a single slit.
Draw a plot of intensity distribution and explain clearly why the secondary maxima become weaker with increasing order (n) of the secondary maxima.
OR
Draw a ray diagram to show the working of a compound microscope. Deduce an expression for the total magnification . when the final image is formed at the near point.
In a compound microscope, an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eye-piece has a focal length of 5 cm and the final image is formed at the near point, estimate the magnifying power of the microscope.
Answer : Huygens principle : Each point of wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave. The common tangent/forward envelope, to all these secondary wavelets gives the new wavefront at later time. Application to diffraction pattern: All the points of incoming wavefront (parallel to the plane of slit) are in phase with plane of slit. However the contributions of the secondary wavelets from different points, at any point, on the screen. Total contribution, at any point, may add up to give a maxima or minima dependent on the phase differences.


The central point is a maxima as the contribution of all secondary wavelet pairs are in phase here. Consider next a point on the screen where an angle θ = 3λ/2a. Divide the slit into three equal parts. Here the first two-thirds of the slit can be divided into two halves which have a λ/2 path difference. The contributions of these two halves cancel. Only the remaining one-third of the slit contributes to the intensity at a point between the two minima. Hence, this will be much weaker than the central maximum (where the entire slit contributes in phase). We can similarly show that there are maxima at θ = (n + 1/2) λ/ a with n = 2, 3, etc. These become weaker with increasing n, since only one- fifth, one-seventh, etc. of the slit contributes in these cases.

Question.30.(a) Explain the formation of depletion layer and potential barrier in a p-n junction.
Output
(b) In the figure given  below the input waveform is converted into the output waveform by a device ‘X’. Name the device and draw its circuit diagram.

(a) With the help of the circuit diagram explain the working principle of a transistor amplifier as an oscillator.
(b) Distinguish between a conductor, a semiconductor and an insulator on the basis of energy band diagrams.
Answer : (a) Depletion region : Due to the concentration gradient across p-side and n-sides, holes diffuse from p-side to n-side (p→n) and electrons diffuse from n-side to p-side (n→p).As the electrons diffuse from n→p,a layer of positive charge (or positive space-charge region) is developed on n-side of the junction. Similarly as the holes diffuse, a layer of negative charge (or negative space-charge region) is developed on the p-side of the junction. The space-charge region on either side of the junction together is known as depletion region.
Barrier potential : The loss of electrons from the n-region and the gain of electron by the p-region cause a difference of potential across the junction of the two regions. The polarity of this potential is such as to oppose further flow of carriers.
>

SET II

Note: Except for the following questions, all the remaining questions have been asked in Previous Set.

Question.3. The radius of innermost electron orbit of a hydrogen atom is 5.3 x 10-11m. What is the radius of orbit in the second excited state?

Question.6.Which part of the electromagnetic spectrum is absorbed from sunlight by ozone layer?
Answer: Ultravoilet rays

Question.9. (i) When primary coil is moved towards secondary coil S (as shown in figure below) the galvanometer shows momentary deflection. What can be done to have larger deflection in the galvanometer with the same battery?
(ii) State the related law.

Answer: (i) To get the larger deflection in the galvanometer with the same battery, the coil P should be moved faster.
(ii) Electromagnetic Induction : It is the process in which an emf is induced in a circuit placed in a magnetic field when the magnetic flux linked with the circuit changes.

Question.10.What is the range of frequencies used for TV transmission? What is common between these waves and light wave?
Answer : Range of frequencies used for TV transmission is 54-72 MHz (VHF-Very High Frequencies).
The ionosphere is unable to reflect back these waves to earth. Both waves are electromagnetic waves.

Question.11.A bioconvex lens has a focal length 2/3 times the radius of curvature of either surface. Calculate the refractive index of lens material.

Question.14.

(i)Why does the Sun appear reddish at Sunset or Sunrise?
(ii) For which colour the refractive index of prism material is maximum and minimum?
Answer:

1.  At Sunset or Sunrise, the sun’s rays have to pass through a larger distance in the atmosphere. Most of the blue and other shorter wavelengths are removed by scattering. The least scattered light reaching our eyes, therefore, the sun looks reddish.
2.  Refractive index of prism material is
   (a) maximum of violet colour
   (b) minimum of red colour

Question.20.(i) Why is communication line of sight mode limited to frequencies above 40 MHz?
(ii) A transmitting antenna at the top of a tower has a height 32m and the height of the receiving antenna is 50m. What is the maximum distance between them for satisfactory communication in line of sight mode?
Answer: (i) At frequencies above 40 MHz, communication is essentially limited to line-of-sight paths. At these frequencies, the antennas are relatively smaller and can be placed at height of many wavelengths above the ground. Because of the line- of-sight nature of propagation, direct waves get blocked at some point by the curvature of earth.

SET III

Note: Except for the following questions, all the remaining questions have been asked in Set-I and Previous Set.

Question.4.Which part of electromagnetic spectrum is used in radar systems?
Answer: Microwave

Question.5.Calculate the speed of light in a medium whose critical angle is 30°.

Question.7. Write the expression for Bohr’s radius in hydrogen atom.

Question.11. What is the range of frequencies used in satellite communication? What is common between these waves and light waves?
Answer: Range of frequencies used in satellite communication is
Uplink = 5.925 — 6.425 GHz Downlink = 3.7 – 4.2 GHz
Both space waves and light waves are electromagnetic waves.

Question.12. A coil Q is connected to low voltage bulb B and placed near another coil P as shown in the figure. Give reason to explain the following observations :
(a) The bulb ‘B’ lights
(b) Bulb gets dimmer if the coil Q is moved towards left.

Answer: (i) The bulb ‘B’ lights up due to an emf induced in the coil Q.
(ii) When the coil ,Q is moved towards left, the emf induced in the coil Q becomes less. Hence the bulb gets dimmer.

Question.13. Find the radius of curvature of the convex surface of the plano-convex lens, whose focal length is 0.3m and the refractive index of the material of the lens is 1.5.

Question.14. An electron is accelerated through a potential difference of 64 volts. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond?

This wavelength is associated with X-rays (range : lnm to 10^-3 nm)

Question.15. (i) Out of the blue and red light which is deviated more by a prism? Give reason.
(ii) Give the formula that can be used to determine refractive index of material of a prism in minimum deviation condition.

Question.20. In a meter bridge, the null point is found at a distance of l₁ cm from A. If a resistance of X is connected in parallel with S, the null point occurs at l₂ cm. Obtain a formula for X in terms of
l₁,l₂ h and S.

Question.27. A parallel plate capacitor is charged to a potential difference V by a dc source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following will change;
(i) electric field between the plates
(ii) capacitance, and
(iii) energy stored in the capacitor

CBSE Previous Year Solved  Papers  Class 12 Chemistry Delhi 2010

Time allowed: 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  All questions are compulsory. 
2.  Questions number 1 to 5 are very short-answer questions and carry 1 mark each.
3.  Questions number 6 to 10 are short-answer questions and carry 2 marks each.
4.  Questions number 11 to 22 are also short-answer questions and carry 3 marks each.
5.  Questions number 23 is a value based question and carry 4 marks.
6.  Questions number 24 to 26 are long-answer questions and carry 5 marks each. 
7.  Use log tables, if necessary. Use of calculators is not allowed.

SET I

Note : Except for the following questions, all the remaining questions have been asked in previous sets.

Question.1. Write a feature which will distinguish a metallic solid from
an ionic solid.
Answer : A metallic solid does not ionize in water.

Question.2. Define ‘order of a reaction’.
Answer : The sum of power of the concentration terms on which rate of reaction depend^ according to the rate law is called order of the reaction.

Question.3. What is an emulsion?
Answer : A colloid in which both dispersed phase and dispersing medium are liquid is called an emulsion, e. g. fat or oil in milk.

Question.4. Why does N02 dimerise?
Answer : Due to the presence of unpaired electrons in N02 it dimerises to form  N₂O₄.

Question.5. Give an example of linkage isomerism.
Answer : The example of linkage isomerism are : [Cr(SCN) (H₂O)₅]³⁺ and [Cr(NCS)(H₂O)₅]³⁺

Question.6. A solution KOH hydrolyses CH₃CHClCH₂CH₃and CH₃CH₂CH₂CH₂Cl which one of these in more easily hydrolysed?
Answer : CH₃CHCICH₂CH₃ is easily hydrolysed due to more stability of secondary carbocation.

Question.7. Draw the structural formula of 1-phenyl Propan-l-one molecule.
Answer:

Question.8.Give the IUPAC name of
H₂N – CH₂ – CH₂ – CH = CH₂.
Answer: But-3-ene-l-amine

Question.9. Non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused ? Explain with one example for each type.
Answer : Negative deviation from Raoult’s law : For any composition of the non-ideal solution the partial vapour pressure of each component and total vapour pressure of the solution is less than expected from Raoult’s law. Such solutions show negative deviation.
Example : Mixture of CHCl₃ and acetone, water and nitric acid etc.
Non-ideal solutions show positive deviations from Raoult’s law on mixing of two volatile components of the solution the partial pressure of each component and total vapour pressure of the solution is more than expected from Raoult’s Law, such solution show positive deviation.
Example : Mixture of acetone and benzene solutions show., positive deviation, water and ethanol, etc.

Question.10. A reaction is of first order in reactant A and of second order in reactants B. How is the rate of this reaction affected when
(i) the concentration of B alone is increased to three times
(ii)the concentrations of A as well as B are doubed?

Question.11. The rate constant for a reaction of zero order in A is 0.0030 mol L^-1. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M?

Question.12. Draw the structures of white phosphorus and red phosphorus. Which one of these two types of phosphorus is more reactive and why?
Answer:

Question.13. Explain the following observations:
(i) Generally there is an increase in density of elements from titanium (Z = 22) to copper (Z = 29) in the first series of transition elements.
(ii)Transition elements and their compounds are generally found to be good catalysts in chemical reactions.
Answer :
(i) Due to decrease in atomic radii from titanium to copper, the corresponding density increases.
(ii) The catalytic activity of transition elements is attributed to the following reasons.

(1) Because of their variable valencies transition metal
sometimes form unstable intermediate compound and provide a new path with lower activation energy for the reaction.
(2) In some cases, transition metals provide a suitable surface ‘ for the reaction to take place. The reactants are adsorbed on
the surface of the catalysts where the reaction occurs.

Question.14. Name the following coordination compounds according to IUPAC system of nomenclature:

Answer: (i) Tetraamineaquachloridocobalt (III) chloride (ii) Dichlorido-bi$(ethane-l, 2-diamine)-Chromium(III) chloride

Question.15. Illustrate the following reactions giving a chemical equation for each:
(i) Kolbe’s reaction,
(ii)Williamson synthesis..
Answer: (i) Kolbe’s reaction :

Question.16. How are the following conversions carried out?
(i) Benzyl chloride to benzyl alcohol.
(ii)Methyl magnesium bromide to 2-Methylpropan-2-ol.
Answer:(i)

Question.17. Explain the following terms :
(i) Invert sugar (ii) Polypeptides .
OR
Name the products of hydrolysis of sucrose. Why is sucrose not a reducing sugar ?
Answer : (i) Sucrose is dextrorotatory but after hydrolysis it gives an equimolar mixture of Q (+) -glucose and q- (—)—fructose which is laevorotatory. This change in specific rotation from dextrorotation to laevorotation is called inversion of sugar and the mixture obtained is called invert sugar.
(ii)Polypeptide is a polyamide formed by large number of ‘a-amino acids joined together by peptide bonds.

The hydrolysis product of sucrose are D-(+)-glucose D-(-)-fructose. It is not considered as a reducing sugar , because the anomeric carbon of both units are involved in glycoside formation. Also it is unable to reduce Tollen’s reagent or Fehling solution.

Question.18. What are essential and non-essential Amino acids in human food? Give one example of each.
Answer : Amino acids which the body cannot synthesise like phenylalanine, valine are essential amino acids and those which can be synthesised by the body are non-essential like glucose and serine.

Question.19. The well known mineral fluorite is chemically calcium fluoride. It is known that in one unit cell of this mineral there are 4 Ca²⁺ ions and 8 F ions and that Ca²⁺ ions are arranged in a fee lattice. The F^–ions fill all the tetrahedral holes in the face centred cubic lattice ofCa²⁺ jions. The edge of the unit cell is 5.46 x 10^-8 cm in length. The density of the solid is 3.18 g cm 3. Use this information to calculate Avogadro’s number (Molar mass of CaF2 = 78.08 g mol-1)

Question.20.A solution prepared by dissolving 1.25g of oil of winter green (methyle salicylate) in 99.0g of benzene has a boiling point of 80.31°C. Determine the molar mass of this compound. (B.P. of pure benzene = 80.10°C and K_b for benzene = 2.53°C kg mol^-1)

Question.21. What is the difference between multimolecular and macromolecular colloids? Give one example of each type. How are associated colloids different from these two types of colloids?
Answer : Multimolecular colloids : In this type of colloids, colloidal particles are aggregates of atoms or molecules each having size less than 1 am. e. g. Sulphur sol, gold sol. Macromolecular colloids : In this type of colloids, colloidal particles are themselves large molecules of colloidal dimensions, e. g. Starch, proteins, polythene etc.
Associated colloids : There are certain substances which at low concentration behave as normal electrolyte, but at higher concentration exhibit colloidal behaviour due to the formation of aggregates. Such colloids are known as associated colloids, e. g. Soaps and detergents.

Question.22. Describe how the following changes are brought about:

1. Pig iron into steel.
2. Zinc oxide into metallic zinc.
3. Impure titanium into pure titanium.

OR
Describe the role of

1.  NaCN in the extraction of gold from gold ore.
2. SiO₂ in the extraction of copper from copper matter.
3. Iodine in the refining of zirconium.

Answer :

1. For the conversion of pig iron into steel, basic oxygen process (BOP) is used. In this process, the furnance is charged with molten pig iron and lime. The pure O₂ is blown over the surface of the metal at a great speed through water-cooled retractable lances. The O₂ penetrates through the metal and oxidizes the impurities rapidly. When all the impurities are removed, this required elements (Cr, Ni etc.) are added to produce steel of desirable properties.
   
2. Calcination of zinc oxide reduces it to metallic zinc using coke this can be explained as follows – above 1270 k Δ_{f  G}° for ZnO is higher than that of CO₂ and CO from carbon therefore, above 1270 K Δ₂G° ΔlG° for reduction of ZnO by carbon is negative and hence ZnO is easily reduced by coke. The ZnO is made into brickettes with coke and clay ‘ and heated above 1270 K so that the reduction process goes to complete
   
3. The impure titanium is converted to pure titanium by , Van Arkel method. In this method, the impure titanium
   is heated in an evacuated vessel with iodine at 870 K. The volatile titanium tetraiodide thus formed is separated which is then decomposed by heating over a tungsten filament at 2075 * K to give pure titanium.
   

(i) NaCN is used to leach the metal present in the ore, when an ore of gold is leached with NaCN in a current of air, gold particle pass into solution as Au+, which combines with CN’ ions, to form soluble complex of aurocyanides. The gold metal is then recovered from the complex by precipitating it with electropositive reducing agent i.e., zinc.

To remove the basic impurities i. e. FeO, an acidic flux silica is added during smelting to remove it as iron silicate slag (FeSiO₃) which floats over molten matte and hence can be easily removed.
(iii) Iodine is used as a reducing agent in refining of zirconium to form a volatile iodide which on decomposition at high temperature gives pure metal.

Question.23. How would you account for the following?

1. The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series.
2. The E° value for the _Mn³⁺_/Mn²⁺ couple is much more positive than that of _Cr³⁺_/Cr²⁺ couple or _Fe³⁺_/Fe²⁺ couple.
3.  The highest oxidation state of a metal is exhibited in its oxide of fluoride.

Answer :

1.  Due to lanthanide contraction in second series i. e. Ad after lanthanum, and poor shielding of electrons, the atomic radii of elements of second (Ad) and third (5d) transition series become almost same.
2.  This is mainly due to much greater third ionization energy of Mn when d⁵ configuration changes to d⁴. 
3. Due to small size, low ionization energy and higher electronegativity the highest oxidation state of a metal is exhibited in its oxide or fluoride.

Question.24. (i) State one use of DDT and iodoform.
(ii) Which compound in the following couples will react fester in S_N2 displacement and why?
(a) 1-Bromopentane or 2-bromopentane
(b) l-Bromo-2 methylbutane or 2-Bromo-2 methyl- butance.
Answer : (i) DDT is used as an insecticide to prevent sugarcane and fodder crops by killing lice and mosquitoes which carry pathogens.
Iodoform dissolved in alcohol is used as an antiseptic for dressing wounds.
(ii) (a) 1-bromopentane being primary alkyl halide have less steric hindrance than 2-bromopentane, therefore, it will be more reactive and hence undergoes S_N2 reaction faster.
(b) l-bromo-2-methylbutane, experience less steric hindrance as it has -Br at primary carbon as compared to 2-bromo-2- methylbutane which has -Br at secondary carbon. Therefore, l-bromo-2methyl butane will be more reactive and hence undergoes S_N2 reaction faster. –

Question.25. Arrange the following in order of property indicated for each set.

Question.26. Give one example each of
(i) Addition polymers
(ii)Condensation polymers
(iii)Copolymers.

Question.27. What are analgesic medicines? How are they classified and when are they commonly recommended for use?
Answer: Analgesics are drugs, which-relieve or decrease pain without causing unconsciousness, paralysis or coordination and mental confusion. They are classified into the following two categories:
(i) Non-Narcotic (non-addictive) drugs, e. g. aspirin, ibuprofen etc. These drug are used to relief skeletal pain.
(ii)Narcotic (addictive) drugs, e. g. Heroin, morphine etc. These are used to relieve postoperative pain, cardiac pain.

Question.28. (a) State Kohlrausch law of independent migration of ions. Write an expression for the molar conductivity of acetic add at infinite dilution according to Kohlrauch law.

(a) Write the anode and cathode reactions and the overall reaction occurring in a lead storage battery.
(b) A copper silver cell is set up. The copper ion concentration is 0.10 M. The concentration of silver ion is not known. The cell potential when measured was 0.422 V. Determine the concentration of silver ions in the cell.

Answer : (a) Kohlrausch law : It states that “at infinite dilution, the molar conductance of an electrolyte is equal to the sum of the molar conductances of the two ions i. e. cation and anions.
Mathematically, A°_m= A°₊ + A°_.
Expression for the molar conductivity of acetic acid.

Question.29. (a) Complete the following chemical equations :

(b) How would you account for the following?
(i) The value of electron gain enthalpy with negative . sign for sulphur is higher than that for oxygen.
(ii)NF₃ is an exothermic compound but NCl₃   is endothermic compound.
(iii)ClF₃   molecule has a T-shaped structure and not a trigonal planar one.
OR
(a) Complete the following chemical reaction equations:

(b) Explain the following observations giving appropriate reasons:
(i) The stability of +5 oxidation state decreases down the group in group 15 of the periodic table.
(ii)Solid phosphorus pentachloride behaves as an ionic compound.
(iii)Halogens are strong oxidizing agents.
Answer: (a) (i)

(b) (i) Because enthalpy of S-S bond is higher then 0-0 bond and the hydration energy of S^2- is less than that of O^2- ion.
(ii) N F₃   is an exothermic compound whereas NCl₃   is an endothermic compound because in case of NF₃ , N-F bond strength is greater than the F-F bond strength while in case of NCl₃  , N-Cl bond strength is lower than the Cl-Cl bond strength. Thus, the formation NF₃ is spontaneous while energy has to be supplied during the formation of NCl₃. In other words, the formation of NF₃ is an exothermic reaction while formation of NCl₃ is an endothermic reaction.
(iii) ClF₃ molecule has a T-shaped structure. This is due to the presence of two lone pairs in the outer shell of chlorine in ClF₃ molecule which repel the bond pairs.

state becomes more and more common-on moving down the group from N to Bi. This is because of inert pair effect.
(ii) Solid PCl₅  behaves as ionic compound because it is a salt containing the tetrahedral cation [PCl4 ]⁺ and octahedral anion [PCl₅ ]^–
(iii) Halogens are strong oxidizing agents because they have high electron affinities, so they easily accept electrons from other substances and change into negative ions i.e. they undergo reduction. Therefore, they are strong oxidizing agents.

Question.30. (a) Explain the mechanism of a nucleophilic attack on the carbonyl group of an aldehyde or a ketone.
(b) An organic compound (A) (molecular formulaC₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (G). Oxidation of (C) with chromic add also produced (B). Oh dehydration (C) gives but-l-ene. Write the equations for the reactions involved.
OR
(a) Give chemical tests to distinguish between the following pairs of compounds:
(i) Ethanal and Propanal
(ii)Phenol and Benzoic add
(b) How will you bring about the following conversions?
(i) Benzoic add to benzaldehyde 
(ii)Ethanal to but-2-enal
(iii)Propanone to propene
Give complete reaction in each case.
Answer: (a)




(b) The hydrolysis of given compound (A) with molecular formula C₈H₁₆O₂ upon hydrolysis with dil H₂SO₄ gives an alcohol (C) and carboxylic acid (B) it suggests that the compound is an ester further since oxidation of (C) with chronic acid produces the acid (B), therefore, both the carboxylic acid (B) and the alcohol must contain the same number of carbon atom.

SET II

Note : Except for the following questions, all the remaining question have been asked in previous sets.

Question.1.Which point defect in the crystals of a solid does not change the density of the solid?
Answer: Frenkel defect.

Question.4. What is the oxidation number of phosphorus in H₃PO₂ molecule?

Question.5. Give an example of coordination isomerism.
Answer:

Question.12. Draw the structural formulae of molecules of following compounds:
Answer:

Question.14.Describe the shapes and magnetic behaviour of following complexes:

Question.15. Explain the following reactions with an example for each:

1.  Reimer-Tiemann reaction
2.  Friedel-Crafts reaction

Answer:

1.  Reimer-Tiemann reaction: Phenol on treatment with chloroform in aqueous sodium hydroxide at 340 K give salicyldehyde.
   
   
   Friedel-crafts reaction : This reaction is used for introducing an alkyl or an acyl group into an aromatic compound in presence of Lewis acid catalyst (AlCl₃).
   

Question.16. How are the following conversions carried out?

1. Propane to propan-2-ol
2. Etylmagnesium chloride to propan- l-ol

Answer:

Question.22. A solution of glycerol (C₃H₈O₃; molar mass = 92 g mol^-1) in water was prepared by dissolving some glycerol 500 g of water. This solution has a boiling point of 100.42 C. , What mass of glycerol was dissolved to make this solution?
K_b for water = 0.512K kg mol^-1.

Question.24. Complete the following chemical equations

Question.25. Write the name and structure of the monomer of each of the following polymers :
(i) Neoprene
(ii)Buna-S
(iii)Teflon 3.

SET III

Note: Except for the following questions, all the remaining question have been asked in previous sets.
Question.1. Which point defect in the crystals of a solid decreases the density of the solid.
Answer: Schottky defect

Question.2. Define‘rate of reaction’.
Answer : The rate of decrease of reactant concentration or increase in product concentration per unit time is called rate of reation.

Question.3. Give an example of‘shape-selective catalyst’.
Answer : Zeolites with its honey-comb structure is an example of shape selective catalysts.

Question.4. Draw the structure of 03 molecule.
Answer:

Question.5. Give an example of ionization isomerism.
Answer :[CO(NH₃)₅Br]SO₄ and [CO(NH₃)₅SO₄]Br

Question.13. Explain the following observations :

1. Transition elements generally form coloured compounds.
2. Zinc is not regarded as a transition element.

Answer:

1. Because of the presence of incomplete d-orbitals, transition elements, undergo d-d transition by four coloured absorption of energy from visible region and then they emit the light of complementary colour compounds.
2.  Zinc has completely filled d-subsheil in its common oxidation state. Hence, it is considered as a non-transition element.

Question.18. State clearly what are known as nucleotides and nucleosides?
Answer : Nucleoside : A nucleoside contains only two ’ basic components of nucleic acid i.e. a pentose sugar and a nitrogenous base. During their formation 1-position of the pyrimidine or 1-position of the purine moiety is linked to C₁ of the sugar (ribose or deoxyribose) by a β-linkage

Nucleotides : A nucleotide contrain all the three basic components of nucleic acid a phosphoric acid group, at 5′-position of pentose sugar and a nitrogenous base

Question.19. The density of copper metal is 8.95 g cm^-3. If the radius of copper is 127.8 pm, is the copper unit cell a simple cubic, a body centred cubic or a face centred cubic structure. (Givens At. Mass of Cu = 63.54 g mol^-1and N_A = 6.02 x 10²³ mol^-1)
Answer : If copper atom were simple cubic:

Question.20. How are the following colloids different from each other in respect of their dispersion medium and dispersed phase? Give one example of each.

1.  Aerosol
2. Emulsion
   (iii)Hydrosol
   Answer:
   

Question.26. Differentiate between thermoplastic and thermosetting polymers. Give one example of each.
Answer:

Question.27. Explain the following terms with one suitable example in each case.

1. Cationic detergents
2. Enzymes
3. Antifertility drugs

Answer :

1.  Cationic detergents are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions and the cationic part possess a long hydrocarbon chain and a positive charge on nitrogen atom. Therefore, they are called as cationic detergents. e.g. Cetyltrimethyl ammonium bromide.
2. Enzymes are biological catalysts which are chemically made up of globular proteins. They have high molecular mass and are highly specific in their actions due to presence of active sites of definite shape and size on their surfaces so that only specific substrate can fit in them. e.g. Pepsin, Trypsin etc.
3. Antifertility drugs are oral contraceptives which are used to check pregnancy in women. It control the female menstrual cycle and ovulation, e.g. N.ovestrol, Norethindrone etc.

CBSE Previous Year Solved  Papers  Class 12 Chemistry Delhi 2009

Time allowed: 3 hours                                                                                            Maximum Marks: 70

General Instructions:

1.  All questions are compulsory. 
2.  Questions number 1 to 5 are very short-answer questions and carry 1 mark each.
3.  Questions number 6 to 10 are short-answer questions and carry 2 marks each.
4.  Questions number 11 to 22 are also short-answer questions and carry 3 marks each.
5.  Questions number 23 is a value based question and carry 4 marks.
6.  Questions number 24 to 26 are long-answer questions and carry 5 marks each. 
7.  Use log tables, if necessary. Use of calculators is not allowed.

SET I

Note : Except for the following questions, all the remaining questions have been asked in previous sets.

Question.1. Which point defect in crystals does not affect the density of the relevant solid?
Answer : Frenkel defect does not alter the density of the relevant solids ?

Question.2. Define the term ‘Tyndall effect’.
Answer : Tyndall effect is the phenomenon due to which the yaath perpendicular to the path of the incident light becomes visible, when a beam pf light , is passed through a colloidal solution.

Question.3. Why is the froth flotation method selected for the concentration of sulphide ores?
Answer : The surface of sulphide ores is preferentially wetted by heavy oils like turpentine or pine oil while the gangue is wetted by water hence, froth floatation method is selected for concentration of sulphide ores.

Question.4. Why is Br(V) a stronger oxidant than Sb (V)?
Answer : Br (V) has a strong metallic character compared to Sb(V) which is a metalloid. The inert pair effect also contributes to Br (V) being a stronger oxidizing agent compound to Sb(V).

Question.5. Give the IUPAC name of the following compound:

Answer: 2-Bromo-3-methyl-but-2-en-l-ol

Question.6. Write t! s e structure of 3-oxopentanal.
Answer:

Question.7. Why is an alkylamine more basic than ammonia?
Answer : Since alkyl groups are electron donating groups, the electron density on nitrogen»,atom in case of alkylamines increases making it to donate the lone pair more easily than ammonia. Hence, alkylamines are more basic than ammonia.

Question.8.Given an example of elastomers.
Answer : The example of elastomers are Buna-S and Buna-N.

Question.9. A reacti on is of second order with respect to a reactant. How will the rate of reaction be affected if the concentration of this reactant is
(i) Doubled,
(ii) Reduced to half ?
Answer : For second order reaction,

So rate becomes 4 times.
If [A] is reduced to half i.e, a/2

Question.10. Explain the role of

1. Carbon electrolytic reduction of alumina.
2. Carbon monoxide in the purification of nickel.

Answer :

1. In the electrolytic reduction of alumina, cryolite is used to make it a good conductor of electricity and to lower the melting point of the mixture.
2.  Carbon monoxide acts as a reducing agent in the process of purification of nickel. When nickel is heated in a stream of CO gas, it forms a volatile complex, nickel tetracarbonyl, which when subjected to higher temperature decomposes to give pure nickel.
   

Question.11. Draw the structures of the following molecules :
Answer.

Question.12. Complete the following chemical reaction equations :

Question.13. Differentiate between molality and molarity of a solution. What is the effect of change in temperature of a solution on its molality and molarity?
Answer : Molality is the number of moles of the solute dissolve per kg of the solvent.
Molarity is the number of moles of the solute dissolved per litre of the solution.
Molality does not change with temperature. On the other hand, molarity increases with rise in temperature but only to a certain limit.

Question.14. Which ones in the following pairs of substances undergoes Sn₂ substitution reaction faster and why?

Question.15. Complete the following reaction equations :

Question.16. Explain what is meant by

1. A peptide linkage
2. A glycosidic linkage

Answer :

1. Peptide linkage is the amide linkage formed by the condensation of an amino group of an a-amino acid with the carboxyl group of another molecule of the same or a different a-amino acid with dehydration.
2.  The linkage between two monosaccharides through oxygen atom in an oligosaccharide or a polysaccharide is known as glycosidic linkage.

Question.17. Name two water soluble vitamins, their sources and the disease caused due to their deficiency in diet.
Answer:

Question.18. Draw the structures of the monomers of the following polymers:
(i) Teflon
(ii)Polyethene

Question.19.Iron has a body-centred cubic unit cell with a cell edge of 286.65 pm.The density of iron is 7.87 g cm”3. Use this information to calculate Avogadro’s number.(At. Mass of Fe = 56 g mol^-1)

Question.20. 100 mg of a protein is dissolved in just enough water to make 10.0 mL of solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of the protein?

Question.21. A first-order reaction has a rate constant of0.0051 min-1. If we begin with 0.10 M concentration of the reactant, what concentration of reactant will remain in the solution after 3 hours?

Question.22. How are the following colloids differ from each other with respect to dispersion medium and dispersed phase ? Give on example of each type. (i) An aerosol (ii) A hydrosol (iii) An emulsion
Answer :

1.  An aerosol is a colloidal solution in which dispersion medium is gas whereas dispersed phase can be liquid or solid eg. Perfume, smoke, dust particles in air.
2. A hydrosol is a colloidal solution in which the dispersion medium is water and dispersed phase is mostly solid, e.g. Carbonated drinks, starch sol.
3. An emulsion is a colloidal solution in which both the dispersed phase and dispersing medium are in liquid state. e.g. Cold cream, milk.

Question.23. Account for the following

1. NH3 is a stronger base than PH₃.
2. Sulphur has a greater tendency for catenation than
   oxygen.
3. Bond dissociation energy of F₂ is less than that of Cl₂.

OR
Explain the following situations:

1.  In the structure of HNO₃molecule, the N-O bond (121 pm) is shorter than the N-OH bond   (140 pm).
2. SF₄ is easily hydrolysed whereas SF₆ is not easily hydrolysed.
3. XeF₂ has a straight linear structure and not a bent angular structure.

Answer:

1.  Due to the small size and greater electronegativity of nitrogen compared to phosphorus, it has greater electron density on its surface and hence NH₃ is a stronger base than PH₃.
2. Due to presence of d-orbitals in sulphur and lower electronegativity of sulphur compared to oxygen, it has greater tendency for catenation as compared to oxygen.
3. Due to small size and high electronegativity of F compared to Cl, the inter-electronic repulsion between lone pair of electrons is very large, hence bond dissociation energy of F₂ is less than Cl₂.

OR

1.  In gaseous state the HNO₃ molecule has a planar structure which is a resonance hybrid in which nitrogen atom is sp2 hybridized in nitrate ions. Hence, N-O bond is shorter than N-OH bond in HNO₃.
   
2. SF₄ has sp³d hybridization with a lone pair of electrons in equatorial position which makes it unstable hence, it is easily hydrolyzed.
   SF₆ does not have a lone pair and is therefore, exceptionally stable.
3.  XeF₂ has three lone pairs of electrons and two bond pairs and thus show sp³d hybridization. But according to VSEPR theory due to presence of l.p-l.p and l.p-b p. repulsion its geometry will in order to minimize repulsion and to have maximum stability.
   

Question.24. For the complex [Fe(en)₂Cl₂] Cl, (en = enthyiene diamine) identify
(i) The oxidation number of iron,
(ii)The hybrid orbitals and the shape of the complex,
(iii)The magnetic behaviour of the complex,
(iv)The number of geometrical isomers,
(v) Whether there is an optical isomer also, and
(vi)Name of the complex. (At. No. of Fe = 26)

Question.25. Explain the mechanism of the following reactions :
(i) Addition of Grignard’s reagent to the carboxyl group of a compound forming an adduct followed by hydrolysis.
(ii)Acid catalysed dehydration of an alcohol forming an alkene.
(iii)Acid catalysed hydration of an alkene forming an alcohol.
Answer :
(i) Carboynl group under goes nucleophilic addi¬tion reaction with Grignard reagent to form an adduct which undergoes hydrolysis to give alcohol in the following manner

Question.26. Giving an example for each, describe the following reactions

1.  Hofmann’s bromamide reaction
2. Gatterman reaction
3. A coupling reaction

Answer :

1.  Hofmann’s bromamide reaction : This reaction involves the conversion of a primary amide to a primary amine on heating with a mixture of bromine in presence of
   
2.  Gattermann Reaction : When Benzene diazonium chloride is treated with halo-acid in presence of Cu powder, it form arylhalide.
   
3.  Coupling reaction : when benzene diazonium chloride reacts with phenol at 0.5°C, it form an azo compound i. e.,p -hydroxyazobenzene (orange dye).
   

Question.27. Explain the following types of substances with one suitable example, for each case s

1. Cationic detergents
2. Food preservatives
3. Analgesics

Answer :

1.  Cationic detergents are chloride, bromides or acetates of quarternary ammonium salts where the cationic part possesses a long hydrocarbon chain e.g. (Cetyltrimethyl ammonium bromide.)
   
2. Food preservatives are substances used to preveht food spoilage due to microbial growth and help extend the life span of an edible substance without any decline in its nutritional value, e.g. Sugar, salt, sodium benzoate.
3. The drugs that are used to rexluce or abolish pain without causing impairment of the consciousness mental confusion in co ordination or paralysis or some, other disturbance of nervous system, e.g. Novalgin, Butazolidine etc.

Question.28. (a) Define molar conductivity of a substance and describe how for weak and strong electrolytes, molar conductivity changes with concentration of solute. How is such change explained?
(b) A voltaic cell is set up at 25°C with the following half cells:

(a) State the relationship amongst the cell constant of a cell, the resistance of the solution in the cell and the conductivity of the solution. How is molar conductivity of a solute related to the conductivity of its solution?
(b) A voltaic cell is set up at 25°C with the following half-cells :

Answer : (a) Molar conductivity of a solution at a given concentration is the conductance of a volume ‘V’ of the solution containing a mole of electrolyte placed in between two electrodes with area of cross-section ‘A’ at distance of unit length Y, that is

Effect of change of concentrations on molar conductivity : Molar conductivity increases with decrease in concentration. This is because both number of ions as well as mobility of ions increases with dilution.
In case of strong electrolyte number of ions do not increase appreciably only mobility of ions increases therefore, Km increases as shown in graph as straight line.
Ir» case of weak electrolytes both no. of ions as well as mobility of ions increases therefore, Km increases sharply as shown by curve in the figure.

Question.29. (a) Complete the following equations :

(b) Explain the fallowing observations about the transition/ inner transition elements:
(i) There is in general an increase in density of element from titanium (Z = 22) to copper (Z = 29)
(ii)There occurs much more frequent metal-metal bonding in compounds of heavy transition elements (3rd series).
(iii)The members of the actinoid series exhibit a larger number of oxidation states than the corresponding members in the lanthanoid series.
OR
(a) Complete the following equations :

(b) Give an explanation for each of the following observations:
(i) The gradual decrease in sire (actinoid contraction) from element to element is greater among the actinoids than among the lanthanoids (lanthanoid contraction).
(ii)The greatest number of oxidation states are exhibited by the members in the middle of a transition series.
(iii)With the same d-orbital configuration  (d⁴) Cr²⁺ ion is a reducing agent but Mn³⁺ ion is an oxidizing agent.
Answer: (a) (i)

(b) (i) The density increases from titanium to copper due to decrease in atomic radii with increase in nuclear charge Hence the atomic volume decreases. At the same time, atomic mass increases, therefore, density increases.
(ii) It is due to more number of unpaired electrons and smaller size due to poor shielding effect of/orbitals and high enthalpy of atomization of heavy transition elements.
(iii)Actinoids are larger in size and have lower ionization energy, therefore it can show greater range of oxidation states. Secondary, 5/ 6dand7s orbitals have comparable energies in actinoid series.

(i) This is due to poor shielding of 4f and 5f electrons in actions from element to element whereas in lanthanoids, there is poor shielding effect of 4f electrons only, that is why nuclear charge increases from element to element in actinoids than lanthanoids.
(ii)The numbers in the middle of transition series elements posses one unpaired electron per d-orbital and hence exhibit greater number of oxidation states.
(iii) Mn³⁺ is strongly oxidizing as it gets reduced to much more stable Mn²⁺ ion. This involves change from 3d⁴to 3d⁵ configurations which is very stable being half filled.
Cr²⁺ is a reducing agent as it gets oxidized to more stable Cr³⁺ ion with half-filled 3d³ or t_2g orbital.

Question.30. (a) Illustrate the following name reactions by giving example:
(i) Cannizzaros reaction
(ii)Clemmensen reduction
(b) An organic compound A contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce ToUen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives positive iodoform test. On vigorous i oxidation it gives ethanoic and propanoic acids. Derive the possible structure of compound A.
OR
(a) How are the following obtained?
(i) Benzoic acid from ethyl benzene
(ii)Benzaldehyde from toluene
(b)Complete each synthesis by giving the missing material reagent or products:

Answer: (a) (i) Cannizzaro reaction: When aldehydes which do not have a- hydeogen atom undergo disproportionation reaction, potassium salt of acid and alcohol are formed.


Since hydrogen atoms are double than carbon atoms, therefore, it is likely to be aldehyde or ketone. It does not reduce Tollen’s reagent therefore, it is not an aldehyde. It is a ketone. It reacts with N_aHSO₃ and gives iodoform test therefore, it is a methyl ketone. On vigorous oxidation it gives ethanoic acid and propanoic acid.
Therefore, it must be pentan-2-one, CH₃COCH₂CH₂CH₃

SET II

Note: Except for the following questions, all the remaining questions have been asked in previous sets.

Question.21. What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?
Answer : Multimolecular colloids : They are aggregates of atoms or molecules each having size less than lnm. e.g. Sulphur sol. Their molecular masses are not very high and they are held together by weak Vander Waals forces.
Micromolecular colloids : They themselves are large mole-cules of colloidal dimensions e.g. Starch sol. They have high molecular masses and Vander Waals forces holding the molecules are comparatively stronger.
Associated colloids: They are colloids which behaves as normal electrolytes at low concentration and as colloid at higher concentration, e.g. soaps, detergents etc. Their molecular masses are generally high and higher in the concentration, greater are the Vander Waals forces concentration, they exhibit colloidal properties.

Question.24. Explain the following observations :

1. Fluorine does not exhibit any positive oxidation state..
2.  The majority of known noble gas compounds are those of xenon.
3. Phosphorus is much more reactive than nitrogen.

Answer :

1.  Fluorine is the most electronegative element, has the maximum reduction potential and has no d-orbitals for octet expansion, therefore it shows only a negative oxidation state of-1.
2. Due to low ionization energy and large atomic radius, Xenon readily forms compounds, particularly with fluorine and oxygen.
3. Phosphorus is more reactive than nitrogen because bond dissociation energy to break N = N triple bond is very large as compared to the energy required to break P-P single bond, which makes nitrogen inert and unteactive.

Question.27. How do antiseptics differ from disinfectants? Give one example of each type.
Answer : Antiseptics are the chemical substance that are applied to living tissues to kill or prevent the growth of micro-organisms. They are safe to be applied to living tissues and are generally applied on wounds, ulcers and diseased skin surfaces e.g. Dettol, furacin, soframycin etc.
Disinfectants, on the other hand, are chemical substances which are applied to non-living objects to kill micro-organisms. They are not safe to be applied to the living tissues and are generally used to kill micro-organisms present in the drains, toilets, floors etc. eg., 1% phenol solution or SO₂ in low concentration.

Question.28. (a) Complete the following equations :

(b) Explain the following observations :
(i) Transition elements are known to form many interstitial compounds.
(ii)With the same d⁴ d-orbital configuration Cr²⁺ ion is reducing while Mn³⁺ion is oxidizing.
(iii)The enthalpies of atomization of transition’ elements are quite high.


(b)(i) Because of their ability to accommodate small non- metallic atoms owing to spaces or voids present between atoms due to their variable valency transition elements form many interstitial compounds.
(ii)Cr²⁺ has the configuration which easily changes to d^ due to stable half-filled t2g orbitals. Therefore Cr²⁺ is reducing agent. While Mn²⁺ has stable half-filled d^ configuration. Hence, Mn³⁺easily changes to Mn²⁺ and acts as an oxidizing agent.
(iii)As transition metals certain a large number of unpaired electrons, they have strong interatomic attractions (metallic) bonds. Hence, they have high enthalpies of atomization.

(b)(i) Transition elements generally have unpaired electrons in their d-orbitals and hence due to d → d transition of electrons, their compounds are coloured.
(ii)Due to incompletely filled d-orbitals, the oxidation states of transition metals vary.
(iii)Due to small energy gaps between, 5f,6dand 7s subshells, actinoids exhibit a greater range of oxidation states.

Question.29.(a) What type of a cell is lead storage battery? Write the anode and cathode reactions and the overall reaction occurring in the lead storage battery while operating. 
(b) A voltaic cell is set up at 25°C with the half cells, Al/Al³⁺ (0.001 M) and Ni/Ni²⁺ (0.50 M). Write the equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.

(a) Express the relation amongst the cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to the conductivity of the solution ?
(b) Calculate the equilibrium constant for the reaction

Answer : (a) A lead storage battery is a secondary cell and it can be recharged. During discharging, the cell reactions at anode and cathode are as follows :

(a) The relationship between cell constant of a cell (G*) resistance of the solution in the cell (R) and conductivity (K) is given by –

SET III

Note : Except for the following questions, all the remaining question have been asked in previous sets.

Question.8. What does the part ‘6,6’ mean lh the name nylon-6,6?
Answer : ‘6,6’ refers to the 6 carbon atoms in monomers molecules of nylon-6,6, i.e. adipic acid and hexamethylene diamine.

Question.19. Calculate the freezing point depression expected for 0.0711 m aqueous solution of Na2S04. If this solution actually freezes at -0.320°C, what would be the value of Van’t Hoff factor?

Question.24. Compare the following complexes with respect to their shape, magnetic behaviour and the hybrid orbitals involved :

Question.27. What are the following substances?.Give an example of each

1. Antacid
2. Non-ionic detergent
3. Antiseptics

Answer :

1. Antacids are drugs that help to overcome the problem of hyperacidity by neutralizing the excess acid and raising the pH to. an appropriate level in stomach eg. Rantidine, sodium bicarbonate etc.
2. Non-ionic detergents do not contain any ion in their constitution as they are esters by high molecular mass alcohol which are obtained by reaction between polyethylene glycol and steric acid. e.g. Polyethylene glycol sterate, lawyrl alcohol ethoxylate.
3. Antiseptics are the drugs applied to living tissues in order to prevent microbial growth, e.g, iodoform for wounds, boric acid for eyes.

Question.28. (b) (i) In general the atomic radii of transition elements decreases with atomic number in a given series.
(ii)The E°Mn²⁺ |M for copper is positive (+0.24 V). It is the only metal in the first transition series showing this type of behavior.

OR
(a) What is meant by lanthanoid contraction? What is it due to and what consequences does it have on the chemistry of elements following lanthanoids in the periodic table?
(b) Explain the following observations:
(i)Cu²⁺ is unstable in aqueous solution
(ii) Although CO²⁺ appears to be stable, it is easily oxidized to CO³⁺ ion in the presence of a strong ligand.
(iii) The E°Mn²⁺|Mn value for manganese is much more than expected from the trend of the other elements in the series.

(iii) The E° value for Mn³⁺|Mn²⁺ is much more positive than for Fe³⁺|Fe²⁺ or Cr³⁺|Cr²⁺.        
Answer : (b)
(i) When moving from left to right across a period, because of increase in effective nuclear charge and weak shielding effect of d-electrons the atomic radii decreases.
(ii)Because of its high enthalpy of atomization and low hydration enthalpy. Hence E°Mn²⁺|M for copper is positive.
(iii)This is because of the much larger third ionization
enthalpy of Mn (where the required charge is d⁵to d⁴ )
OR
(a) Lanthanoid contraction is the steady decrease in the
atomic radii of lanthanoids with increasing atomic number due to mutual imperfect or poor shielding of the electrons in the Af orbital – As we move along the lanthanoid series, the effective nuclear charge increases on addition of electrons and the electrons thus added in f-subshell causes imperfect shielding which is unable to counter balance the effect of the increased nuclear charge. As a result, the contraction in size occurs.
Consequences :
(i) Due to lanthanoid contraction, zirconium (Zr) and Hafnium (Hf) have a comparable size. They are also known as chemical twins due to their similar radii.
(ii) Due to lanthanoid contraction, the chemical properties of lanthanoids are very similar due to which their separation becomes very difficult.
(b) (i) Since Cu+ in aqueous solution undergoes dispropertio- nate i.e.
2Cu⁺ (aq)–> 2 Cu²⁺(aq)+Cu (s)
The E° value for this is favourable.
(ii)CO(III) is stablised because of higher crystal field splitting energy [or strong ligand causes pairing of electron to give more stable CO(III) ion]
(iii)The comparatively high value for Mn shows that Mn²⁺( d⁵) is particularly stable.

Question.29. (a) Corrosion is essentially an electrochemical phenomenon. Explain the reaction occurring during the corrosion of iron kept in an open atmosphere.
(b) One half-cell in a voltalic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration. Its other half-cell consists of a zinc electrode dipping in 1.0 M solution of Zn(NO₃ )₂. A voltage of 1.48 V is measured for this cell.

Use this information to calculate the concentration of silver nitrate solution used.
Answer:
(a) Corrosion is the wearing away of a metal due to gases and water vapour present in the atmosphere. It results generally in the formation of oxides, sulphides or carbonates.
In the atmosphere, pure iron surface behaves as a small electro chemical cell in the presence of Oxygen and water vapour.

CBSE Previous Year Solved  Papers  Class 12 Chemistry Outside Delhi 2009

Time allowed: 3 hours                                                                                        Maximum Marks: 70

General Instructions:

1.  All questions are compulsory. 
2.  Questions number 1 to 5 are very short-answer questions and carry 1 mark each.
3.  Questions number 6 to 10 are short-answer questions and carry 2 marks each.
4.  Questions number 11 to 22 are also short-answer questions and carry 3 marks each.
5.  Questions number 23 is a value based question and carry 4 marks.
6.  Questions number 24 to 26 are long-answer questions and carry 5 marks each. 
7.  Use log tables, if necessary. Use of calculators is not allowed.

SET I

Questin.1. How do metallic and ionic substances differ in conducting
electricity ?
Answer : Metals conduct electricity through free electrons while ionic substances conduct electricity in aqueous form through ions.

Questin.2. What is the ‘coagulation’ process? 
Answer : The process of setting of colloidal particles in a colloid is called coagulation. It is carried out by electrophoresis, persistent dialysis on addition of an electrolyte.

Questin.3. What is meant by the term ‘pyrometallurgy’?
Answer: Pyrometallurgy in the process of extracting metal by converting a metal oxide to metallic form on strong heating with a reducing agent.

Questin.4. Why is red phosphorus less reactive than white phosphorus?
Answer : Due to angular strain weak Yander weals forces are present in P₄ molecules of white phosphorus whereas they are tetrahedrally attached by strong covalent bonds in red phosphorus. Therefore, red phosphorus is less reactive than white phosphorus.

Questin.5. Give the IUPAC name of the following compound:

Answer: Hex-l-en-3-ol

Questin.6. Write the structural formula of 1-phenylpentan-l-one.
Answer :

Questin.7. Arrange the following compounds in an increasing order of the basic strength in their aqueous solutions :

Questin.8. What does ‘6,6’ indicate in the name nylon-6,6?
Answer : 6,6 indicates the number of carbon atoms in the each of two monomers of nylon-6,6 and is desired from the two monomers hexamethylene diamine and adipic acid.

Questin.9. What type of cell is a lead storage battery? Write the anode and the cathode reactions and the overall cell reaction occurring in the use of a lead storage battery?
OR
Two half cell reactions of an electrochemical cell are given below:

Construct the redox equation from the two half cell reactions and predict if this reaction favours formation of reactants or product shown in the equation.
Answer: Lead storage battery is a secondary cell. The cathode, anode and overall cell reaction as follows :

Questin.10. Define:

1.  Elementary step in a reaction
2. Rate of reaction

Answer :

1.  If a reaction takes place in a single step with a single transition state then it is called elementary reaction.
2. The change in concentration of the reactants or products per unit time is called rate of reaction.

Questin.11. Describe the underlying principle of each of the following metal refining methods:

1. Electrolytic refining of metals
2. Vapour phase refining of metals

Answer :

1. In electrolytic refining, impure metal is made to act as an anode and a thin sheet of pure metal as cathode, and a soluble salt solution of the metal is used as electrolyte. On passing electric current, the metal from anode passes into the solution and pure metal is deposited on cathode with impurities going into the solution or setting down as anode mud.
2.  The crude metal is first converted into its volatile compound which is then decomposed to give back pure metal.

Questin.12. Complete the following chemical reaction equations :
(i)  X_eF₂+ H₂O→
(ii) PH₃ + H_gCl_2→

Questin.13. Complete the following chemical reaction equations

Questin.14. Which one in the following pairs undergoes S_Nl substitution reaction faster and why?

Questin.18. Differentiate between molecular structures and behaviours of thermoplastic and thermosetting polymers. Give one example of each type.
Answer :

Questin.15. Complete the following reaction equation :

Questin.19. A first order reaction has a rate constant of 0.0051 min^-1. If we begin with 0.10 M concentration of reactant what concentration of the reactant will be left after 3 hours?

Questin.16. Name the four bases present in DNA. Which one of these is not present in RNA?
Answer: Adenine, thymine, guanine and cytosine are present in DNA. Thymine is not present in RNA.

Questin.17. Name the two fat soluble vitamins, their sources and the diseases caused due to their deficiency in diet.
Answer:Vitamin A and’Vitamin K are fat soluble vitamins.
Source of Vitamin A – Carrot, papaya, Disease (Vitamin A) – Night blindness .
Sources of Vitamin K – Cereals, leafy vegetables, Diseases (Vitamin K) – Haemorrhagic conditions.

Questin.20. Silver crystallizes with face-centred cubic unit cells. Each side of the unit cell has a length of 409 pm. What is the radius of an atom of silver? (Assume that each face atom is touching the four comer atoms.)

Questin.21. A copper-silver cell is set up. The copper ion concentration in it is 0.10 M. The concentration of silver ion is not known. The cell potential measured is 0.422 V. Determine die concentration of the silver ion in the cell.

Questin.22. What happens in die following activities and why?

1.  An electrolyte is added to a hydrated ferric oxide sol in water.
2. A beam of light is passed through a colloidal solution.
3. An electric current is passed through a colloidal solution.

Answer:

1.  Coagulation of sol takes place by absorbing OH” ions.
2. The beam of light is scattered by the colloidal particles and the path of light becomes visible. This phenomenon is known as Tyndall effect.
3. Electrophoresis takes place. The positively charged ions move towards cathode and negatively charged ions move towards anode and by loosing their charges get coagulated.

Questin.23. Giving a suitable example for each, explain the following:

1. Crystal field splitting
2. Linkage isomerism
3. Ambidentate ligand

OR
Compare the following complexes with respect to structural shapes of units, Magnetic behaviour and hybrid orbitals involved in units:

Answer :

1. When a transition metal ion is involved in complex formation, the degenerate orbitals split into two sets, one with lower energy and the other with higher energy. This is called crystal field splitting.
2. Linkage isomerism occurs in compounds containing ambidentate ligands, i.e., when more than one atom may function as a donor and get linked with central metal ion.
3. A ligand that has more than one donor atom but only one donor atom is attached to the neutral ion at a time is called ambidentate ligand eg. SCN^–, NO^–₂
   

Questin.24. Explain the following:
(i) The boiling point of ethanol is higher than that of methoxymethane.
(ii)Phenol is more acidic than methanol.
(iii)o-and p-nitrophenols are more acidic than phenol.
Answer :

1. Due to intermolecular hydrogen bonding in ethanol.
2. Phenol looses H+ ions and forms phenoxide ion. Ethanol looses H⁺ ion and forms ethoxide ion. Phenoxide ion is more stable than-ethoxide ion due to resonance. Therefore, phenol is more acidic than ethanol.
3. Due to electron withdrawing character of NO₂ group,
   electron density in the OH bond of the substituted phenol decreases and hence loss of proton becomes easy, so it is more acidic.

Questin.25. How would you account for the following:

1.  Many of the transition elements and their compounds can act as good catalysts.
2. The metallic radii of the third (5 d) series of transitions elements are virtually the same as those of the corresponding member of the second series.
3. There is a greater range of oxidation states among the actinoids than among the lanthanoids.

Answer :

1. Due to presence of unpaired electrons in their incomplete -orbitals and variable oxidation states.
2. Due to lanthanoid contraction in the second series, the atomic radii of the elements of second and third series become almost same, therefore, show same properties.
3. Due to very small energy gap between 5f 6d and 7s – subshells all the electrons take part in bonding and show variable oxidation states.

Questin.26. Complete the following reaction equations :

Questin.27. Describe the following substances with one suitable example of each type :

1.  Non-ionic detergents
2.  Food preservatives
3.  Disinfectants

Answer :

1. Non-ionic detergents do not contain any ions. They are esters of high molecular mass prepared by reaction of polyethylene glycol with stearic acid e.g. Polyethylene glycol stearate.
2.  Preservatives are chemical substances used to protect food from bacteria, yeasts and moulds and avoid its spoilage, e.g. Sodium benzoate, salt-sugar etc.
3. Disinfectants are chemicals which kill microbes but are applied only to non-living objects like floors and drains eg. Phenol, chlorine.

Questin.28. (a) Define the following terms:

1.  Mole fraction
2. van’t Hoff factor

(b) 100 mg of a protein is dissolved in enough water to make 10.0 mL of the solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of protein?
(R = 0.0821 L atm mol^-1 K^-1 and 760 mm Hg = 1 atm.) OR
(a) What is meant by:
1. Colligative properties
2. Molality of a solution
(b) What concentration of nitrogen should be present in a glass of water at room temperature? Assume a temperature of 25°C, a total pressure of 1 atmosphere and mole fraction of nitrogen in air of 0.78
(KH for nitrogen = 8.42 x 10-7 M/mm Hg)
Answer : (a) (i) Mole fraction is the ratio of number of moles of a component to the total number of moles present in the solution.

(a) (1) Colligative, properties are those properties which depend on the number of particles of solute dissolved in a definite amount of solvent.
(2) Molality in the number of moles of solute dissolved in one

Questin.29. (a) Draw the structure of the following:

1. H₂S₂O₈
2. HClO₄

(b) How would you account for the following?

1. NH₃ is a stronger base than PH₃
2. Sulphur has a greater tendency for catenation than oxygen.
3. F₂ is a stronger oxidizing agent than Cl₂.

OR
(a) Draw the structure of the following:

1.  H₂S₂O₇
2. HClO₃

(b) Explain the following observation :

1.  In the structure of HNO₃, the N-0 bond (121 pm) is shorter than the N-OH bond (140 pm)
2. All the P-Cl bond in PCl₅ are not equivalent

(b) (i) Due to small size and high electronegativity of nitrogen ‘compound to phosphorus, electron density on nitrogen is more therefore, it can easily donate electrons and hence it is a stronger base than PH₃.
(ii)Due to small size of oxygen, lone pair of oxygen repel (O- O) bond more than lone pair on sulphur atoms in S-S bond, hence S-S forms strong bond.
(iii) Due to small size and high electronegativity of electron- electron repulsion between lone pair of electrons are very large, therefore, bond dissociation energy of F₂ is less than Cl₂.

(b) (i) The N-O bond has partial double bond character while the N-OH is a single bond in the structure of HNO₃.
(ii) PCl₅ has a trigonal bipyramidal structure with 3 equivalent equatorial and 2 equivalent axial bonds. Due to higher bond pair-bond pair repulsion in case of axial bonds, they are longer than equatorial bonds. .
(iii) IC1 bond is weaker than I-I bond due to which I-I bond breaks easily to form halogen atoms which can easily bring out the reaction, hence it is more reactive than I₂ :

Questin.30. (a) Write chemical equations to illustrate the following name bearing reaction :

1. Cannizzaro’s reaction
2. Hell-Volhard-Zelinsky reaction

(b) Give chemical test to distinguish between the billowing pairs of compound.

1. Propanal and propanone
2. Acetophenone and Benzophenone
3. Phenol and Benzoic acid

OR
(a) How will you bring about the following conversions :

1. Ethanol to 3-Hydroxybutanal
2. Benzaldehyde to benzophenone

(b) An organic compound A has the molecular formula C₆H₁₆0₂. It gets hydrolysed with dilute sulphuric acid and gives a carboxylic acid B and an alcohol C. Oxidation of C with chromic acid also produced B. C
on dehydration reaction gives but-l-ene. Write all equations for the reactions involved.
Answer :
(a) (i) Aldehydes which donot contain a-hydrogen atom undergo self oxidation and reduction with cone, alkali to give alcohol and carboxylic salt.

SET II

Note : Except these following questions, all the remaining questions have been asked in previous set.

Questin.1. Which point defect of its crystals decreases the density of a solid?
Answer:Schottky defect

Questin.8. What is a primary structural feature necessary for a molecule to make it useful in a condensation polymerization reaction?
Answer : The monomers have a specific functional group representing their functionality and they combine through these functional groups to form condensation polymers.

Questin.19. Iron has a body-centred cubic unit cell with a cell edge of 286.65 pm. The density of iron is 7.87 g cmn^-3. Use this information to calculate Avogadro’s number. (At mass of iron = 56 g mol^-1).

Questin.20. For a decomposition reaction the value of rate constant K at two different temperatures, are given below:

Questin.27. Complete the following reaction equations :

Questin.28. (a) Give chemical tests to distinguish between compounds in the following pairs of substances :

1.  Ethanol and propanal
2. Benzoic acid and ethyl benzoate

(b) An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acids. Derive the structure of the compound.
OR
(a) Arrange the following compounds in an increasing order of their indicated property.

1. Benzoic acid, 4-nitrobenzoic acid, 3,4-dinitrobenzoic acid, 4-methoxybenzoic acid (acid strength)
2. CH₃CH₂CH(Br)COOH,CH₃CH (Br) CH₂COOH (CH₃)₂ CHCOOH, CH₃CH₂CH2COOH (acid strength)

(b) how would you bring about he following conversions:

1.  Propanone to propene
2. Benzoic acid to benzaldefiyde
3. Bromobenzene to 1-phenylethanol

Answer :(a)

1.  Ethanol does not form silver mirror with Tollen’s reagent but propanal gives silver mirror with Tollen’s reagent due to presence of aldehyde group.
2. Benzoic acid produce brisk effervescene with NaHCC>3 Solution while ethyl benzoate does not
   

Determination of structure : Since the compound doesnot reduce Tollen’s reagent and gives a positive iodoform test. So the given compound is Ketone not aldehyde. Since the given compound on vigrous oxidation gives a mixture of ethanoic acid and propanoic acid. Therefore, the given organic compound is methyl Ketone and its structure would be CH₃COCH₂CH₂CH₃

Questin.29. (a) Draw the structures of the following:
(i) H₃PO₂
(ii)BrF₃
(b) How would you account for the following observations:
(i) Phosphorous has a greater tendency for catenation than nitrogen.
(ii)Bond dissociation energy of fluorine is less than that of chlorine.
(iii)No chemical compound of helium is known.
OR
(a) Draw the structure of the following
(i) N₂O₅
(ii)XeOF₄
(b) Explain the following observations:
(i) The electron gain enthalpy of sulphur atom has a greater negative value than that the oxygen atom.
(ii)Nitrogen does not form pentahalides.
(iii)In aqueous solution HI is a stronger acid than HCl.

(b) (i) This is because p—p Single bond is stronger than n—n Single bond.
(ii) Due to small size of fluorine, lone pairs of electrons. Create more repulsion than chlorine. Bond dissociation energy of fluorine is less than Cl₂.
(iii) Valence shell electrons of He can’t be promoted to higher energy level due to lack of p or d orbitals.

(b) (i) This is due to smaller size of oxygen the electron cloud is distributed over a small region of space making electron density high which repels incoming electron.
(ii)Due to small size and absence of empty d-orbitals, Nitrogen does not form pentahalides.
(iii)The strength of acid depend up on its bond strength. Since bond dissociation energy of HCl is greater than HI. Therefore, HI is stronger than HCl.

CBSE previous Year Solved  Papers  Class 12 Biology Delhi 2013

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5.  Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SET-I

SECTION-A

Question.1. An anther with malfunctioning tapetum often fails to produceviable male gametophytes. Give any one reason.
Answer : The tapetum is responsible for nourishing the pollen grains. Tapetum malfunctioning results in insufficient nvitrition of the pollen grains, and will not be able to produce viable male gametophytes.

Question.2. Why sharing of injection needles between two individuals is not recommended ?
Answer: Sharing of injection needle between two individuals could lead to the transfer of blood borne diseases viz. HIV from infected person to a healthy person.

Question.3. Name the enzyme and state its property that is responsible for continuous and discontinuous replication of the two strands of a DNA molecule.
Answer : DNA dependent DNA polymerase is the enzyme. The property t hat is enzyme catalys polymerization only in one direction, i..e., 5′ —> 3′. As a result on template strand with 3’ —> 5′ the replication is continuous while on the template strand with polarity 5′ —> 3′ it is discontinuous.

Question.4. Identify the examples of convergent evolution from the following:
(i) Flippers of penguins and dolphins
(ii) Eyes of octopus and mammals
(iii) Vertebrate brains
Answer : The examples of convergent evolution are :
(i) Flippers of penguins and dolphins.
(ii) Eyes of octopus and mammals.

Question.5. Write the importance of MOET.
Answer : The technique is used to increase the successful rate of production of hybrids in short duration of time. e.g. In MOET cow is given hormones with FSH like activity. These hormones induce follicular maturation and super-ovulation which produces 6-8 eggs per cycle instead of one egg.

Question.6. Why is the enzyme cellulase needed for isolating genetic material from plant ct’Us and not from the animal cells ?
Answer : Plant cells have cell wall of cellulose but animal cells have no cell wall and do not have cellulose. So the enzyme cellulase is required for isolating genetic material from plant cells only.

Question.7. Name the type of biodiversity represented by the following:
(a) 50,000 different strains of rice in India.
(b) Estuaries and alpine meadows in India.
Answer : (a) Genetic diversity (b) Ecological diversity

Question.8. Write the equation that helps in deriving the net primary productivity of an ecosystem.
Answer : Net primary productivity (NPP) in an ecosystem can be derived using the following equation,
NPP = GPP (Gross primary productivity) – R (Respiratory losses)

SECTION-B

Question.9. Geitonogamous flowering plants are genetically autogamous but functionally cross-pollinated. Justify.
Answer : Geitonogamy is a pollination where the pollen grains are transferred from the anther of a flower to the stigma of another flower on the same plant.
Geitonogamous flowering plants are usually autogamous as the gametes come from the same parent plant, but since the pollen grains are being passed to a different flower which requires a pollinating agent, it is an example of cross pollination.

Question.10. When and where do chorionic villi appear in humans ? State their function.
Answer : Chorionic villi appear as finger like projections that arise from the trophoblast layer of blastocyst when it is undergoing implantation.
Functions of Chorionic villi:

1.  It inter-digitates with projections from uterine tissue to form a structure called the placenta, which is the connecting link between the mother and the foetus.
2.  It facilitates the supply of oxygen and nutrients to the embryo.

Question.11. In a cross between two tall pea plants some of the off springs produced were dwarf. Show with the help of Punett Square how this is possible. 
Answer : In a cross between 2 phenotypically tall plants, some of the progeny may become to be phenotypically dwarf. It defines both parents are heterozygous (Tt). The cross is demonstrated below by a Punnett Square.
Genotype of Parents = Tt.
Genotypic ratios of progeny after F] cross= TT, Tt, Tt, tt. Phenotypic ratios= 1 TT+ 2 Tt: 1 tt or 3 Tall: 1 dwarf .

Question.12. A student on a school trip started sneezing and wheezing soon after reaching the hill station for no explained reasons. But, on return to the plains, the symptoms disappeared. What is such a response called ? How does the body produce it?
Answer : The responses such as sneezing and wheezing for no explained reasons is called Allergy. The substance against which the immune system show such exaggerated response are termed as allergens. Allergy„is produced in a body due to release of chemicals like histamine and serotonin from mast cells. Body produces IgE type of antibody against allergens as a response. Most common example of-such allergens are pollen grains, dust etc.

Question.13. Name two commonly used bioreactors. State the importance of using a bioreactor.
Answer : The two most widely used bioreactors are simple stirred – tank bioreactor and sparged stirred- tank bioreactor. The importance of using bioreactors :

1. It gives huge amount for cultures. So, products are pro-duced in large quantity.
2.  It gives optimal conditions like temperature and pH for growth of desired product.

Question.14. Write the function of adenosine deaminase enzyme. State the cause of ADA deficiency in humans. Mention a possible cure for a ADA deficiency patient.
Answer : The enzyme adenosine deaminase (ADA) is very important for the proper functioning of our immune system. The cause of ADA deficiency in humans is deletion of the gene which codes. for ADA. ADA deficiency can be cured permanendy by gene therapy. Then functional ADA gene is inserted in the cells at early embryonic age for permanent cure.

Question.15. Expand the following and mention one application of each :
(i) P.CR (ii) ELISA
Answer : (i) PCR-Polymerase chain reaction is a technique in molecular biology to amplify a gene or a piece of DNA to obtain its several copies. Each cycle of PCR has three steps :
(a) Denaturation of DNA Strand
(b) Primer annealing
(c) Extension of primers
Use : It is extensively used in the process of gene manipulation,
(ii) ELISA (Enzyme Linked Immunosorbent Assay) is a method in molecular biology which utilizes antigens and . antibodies to find infectious diseases. An infection is found by the presence of antigens like proteins or by the synthesis of antibodies against the infection.
This method is broadly used for finding AIDS.
OR
(a) Mention the difference in the mode of action of exonuclease and endonuclease.
(b) How does restriction endonuclease function ?
Answer : (a) Exonuclease removes the nucleotides from the ends of the DNA chain while endonuclease cuts the DNA at the specific positions within the DNA strand.
(b) Each restriction endonuclease finds its specific pallindromic • nucleotide sequences in the DNA and cut the DNA at these specific sites. It does so by binding to the DNA at these sites and cutting both the strands at specific points in their sugar- phosphate backbones, e.g. E CORI cuts the DNA at following palindromic sequences.

Question.16. Name any two source of e-waste and write two different ways for their disposal.
Answer : Two sources of e-waste are :

1. Irreparable computers.
2.  Electronic items like mobile phones, television sets etc. Two different ways of disposal of e-waste :
   (i) Dumping the e-wastes into landfills.
   (ii) Incinerating e-waste i.e. burning the e-wastes com pletely into ashes.
   These ways of disposal pose threat to the environment by, releasing toxic substances into it. So, recycling, of e-waste in an environment-friendly manner is the only solution for its disposal.

Question.17. Why the pyramid of energy is always upright ? Explain.
Answer : The pyramid of energy shows the total quantity of energy used by each trophic level in a given food chain. An energy pyramid is always upright becau.se the total quantity of energy available for utilization at the upper trophic level less than the energy available at the lower levels. This occurs because according to the 10% law of energy transfer, only 10% of the total energy is transmitted from one trophic level to another and when energy transmits frotn a one trophic level to next trophic level, some amount is always lost as heat at each step.
Eventually it is lost to atmosphere and never goes back to Sun.
Question.18. Explain why very small animals are rarely found in polar region ? .
Answer: Loss of body heat is direcdy proportional to the surface area. So, small animals have larger surface area compare to their volume. They lose body heat rapidly in colder areas. Loss of body heat in colder areas can create challenge to their survival. That’s why very small animals are rarely found in polar areas.

SECTION-C

Question.19. Draw a diagram of the microscopic structure of human sperm. Label the following parts in it and write their Functions : (a) Acrosome (b) Nucleus (c) Middle piece
Answer : Structure of human sperm :

Question.20. With the help of any two suitable examples explain the effect of anthropogenic action or organic evolution
Answer: The anthropogenic activities results in increased rate of organic evolution. For example :

1.  The extreme use of herbicides and pesticides causes selection of resistant variety of pests and insects in a short time period. The change in the environment faciliate resistant pests and insects results in their evolution.
2.  The excessive use of antibiotics causes selection of drug resistant microbes. The micro-organisms which are sensitive , to specific antibiotics died but few variants of micro-organisms which develop resistance against the specific antibiotics are survived. This assist in the evolution of deadly lethal micro-organisms.

Question.21. (a) Why is human ABO blood group gene considered a good example of multiple alleles ?
(b) Work out a cross up to F generation only, between a mother with blood group A (Homozygous) and the father with blood group B (Homozygous). Explain the pattern of inheritance exhibited.
Answer : (a) A gene is represented by two alleles. But, for the blood group in humans there are three alleles namely IA, IB and i governing a same character. Thus, it is an example of multiple allele. IA & IB alleles are dominant over 1°. These alleles decide the blood group of person.

Question.22. Describe the structure of a RNA polynucleotide chain having four different types of nucleotides.
Answer : A RNA nucleotide has three main components a nitrogenous base, a ribose sugar and a phosphate group.

1.  The ribose sugar and the phosphates form the backbone of a polynucleotide, chain with nitrogenous bases linked to sugar moiety and projecting from the backbone.
2.  Two types of nitrogenous bases are present i.e. Purines . (Adenine (A) and Guanine (G)) and Pyrimidines
   (Cytosine (C) and Uracil (U)).
3. A nitrogenous base is linked to the ribose sugar through N-glycosidic linkages to form a nucleoside (like adenosine, guanosine or cytidine and uridine).
4.  A phosphate group is linked to 5’-OH of a nucleoside through phosphoester linkage to form a corresponding nucleotide.
5.  Every nucleotide residue has an additional -OH group present at 2’ -position in the ribose.
6.  Many nucleotides are linked through 3 -5′ phospho- diester linkages to each other to form the polynucleotide chain.
7.  The end of the chain which has a free phosphate moiety at 5′- end of ribose sugar is referred to as 5 – end and the other end of the chain having a free 3 – OH group at the ribose sugar is referred to as 3′- end of the polynucleotide chain.
   

Question.23. Differentiate between inbreeding and out-breeding in catde. State one advantage and one disadvantage for each one of them.
Answer : Difference between inbreeding and out-breeding

Inbreeding:
Advantage : Pure breed of progeny.
Disadvantage : It reduces in the fertility and productivity of an organism because of constant inbreeding. This is also called as Inbreeding depression.
Outbreeding:
Advantage : Progeny has proffered features of both the parents.
Disadvantage : The hybrid animal produced is not every time fertile.

Question.24. (a) Why are the fruit juices bought from market clearer as compared to those made at home ?
(b) Name the bioactive molecules produced by Trichoderma polysporum and Monascus purpureas.
Answer: (a) The fruit juices available in the market (the bottled ones) are made clear by treating them with the enzymes – pectinases and proteases. So, they are clearer as compare to those made at home.
(b) Trichoderma polysporum is used to produce immunosuppressive agent cyclosporin A.
Monascus purpureus is used to produce blood-cholesterol lowering agent called statin.

Question.25. (a) Why are transgenic animals so called ?
(b) Explain the role of transgenic animals in (i) Vaccine safety and (ii). Biological products with the help of an example each.
Answer : Transgenic animals are called so as these animals 1 acquire foreign gene that is intentionally inserted into the genome. The foreign gene is inserted in the genome of the organisms by recombinant DNA technology.
Role of transgenic animal in vaccine safety : Now a days transgenic mice are being used in testing the safety of vaccines before they are available to humans use.
Example : Transgenic mice are being used to test the safety of the polio vaccine. If the transgenic mice found doing well and trustworthy, they could substitute the use of monkeys to test the safety of batches of the vaccine.
Role of transgenic animal in production of biological products : Transgenic cow, Rosie is used for the production of human protein-enriched milk, which contained human a-lactalbumin and it was having increased nutrition more suitable for human babies.

Question.26. How have human activities caused desertification ? Explain.
Answer : Following human activities contribute to desertification :

1. Deforestation : To construct road, buildings, new city etc. the trees are being cut down which results in desertification.
2.  Improper farming practices : If same crop is growing again and again over a longer peroid of time makes the soil deprived resulting in the loss of fertility of soil.
3.  Excessive ploughing of field.
4. Soil erosion : Soil erosion by deforestation due to indu- serialization and construction of houses.
5.  Mining actions and leaching of minerals further oblite-rate soil quality results’ in infertile soil.

OR
How does algal bloom destroy the quality of a fresh water body ? Explain.
Answer : An algal bloom is the result of excessive growth of planktons forms in a highly nutrient rich water body. When the planktonic species grow repetitively on the surface of water body, at results in a layer that finally covers the whole surface of the water body. They prevent sunlight to reach to the bottom and make unavailable to submerged aquatic species which is having role in delievering essential nutrients to other aquatic life forms.
Some algal species discharge molecules that are toxic and ‘ lethal for other aquatic organisms. Because of high respiratory rate the biological oxygen demand. (BOD) of the water body increases which causes death of a number of aquatic organisms. Their remains after their death again contribute to the deterioration of water quality.

Question.27. Explain mutualism with the help of any two examples. How is it different from commensalism ?
Answer : Mutualism is a kind of population interaction in which both the participating species derive a benefit from each other’s presence. Examples of mutualism are given below:

1.  Associations between fungi and plants, known as mycorrhizae : The plant gets benefits by soil nutrients that the fungus absorbs and transmits to the plant by its roots. The fungus then derives the benefit of getting energy yielding carbohydrates from plant.
2. Pollination : The flowers provide sweet, mucilaginous nectar to birds or insects on the contrary for getting benefit from the bird or insect in passing their pollen grains onto other flowers. The plant – pollinator’ pair often goes co-evolution to safeguard against the use of the nectar by other non-important organisms.
   Mutualism is different from commensalism in that the latter gives benefit to just one of the participating species, the benefitted species being called a commensal.

SECTION-D

Question.28. (a) Draw a diagrammatic sectional view of a mature anatropous ovule and label the following parts in it:
(1) that develops into seed coat.
(2) that develops into an embryo after fertilization.
(3) that develops into an endosperm in an albuminous seed.
(4) through which the pollen tube gain entry into the embryo sac.
(5) that attaches the ovule to the placenta.
(b) Describe the characteristic features of wind pollinated flowers.
Answer: (a)

1. The part that develops into seed coat- Integument
2.  The part that develops into an embryo after fertilization Embryo sac/ovule
3. The part that develops into an endosperm in an albuminous seed – Nucellus
4.  The part through which the pollen tube gains entry into the embryo sac – Micropyle .
5.  The pollens are dry and unwettable.

OR
(a) Draw a diagrammatic sectional view of the female reproductive system of human and label the parts :
(1) Where the secondary oocytes develop ?
(2) Which helps in collection of ovum after ovulation ?
(3) Where fertilization occurs
(4) Where implantation of embryo occurs ?
(5) Explain the role of pituitary and the ovarian hormones in menstrual cycle in human females.
Answer:(a) The female Reproductive System Uterine fundus

1.  The Ovary
2.  Fimbriae
3.  Fallopian tubes
4. Uterus
   

(B) The role of pituitary and the ovarian hormones in menstrual cycle in human females :
The Menstrual cycle lasts for about 4 -5 days. If fertilization does not occur, the corpus luteum degenerate, and the inner lining of uterus and fallopian tubes which contain soft tissue and blood vesseles sheds off, resulting in menstrual flow. In this phase, the primary follicles grow into the GrafAan follicles results in the regeneration of the endometrium.
Ovarian and pituitary hormones are responsible for following changes:

1.  The release of gonadotropins (LH and FSH) increases . which facilitate follicular growth and the growing follicles produce oestrogen.
2. In the middle of the menstruation cycle (14th day) the LH and FSH are at their peak, assist rupture of the Graffian follicles to discharge ovum. This phase is known as ovulatory phase.
3.  The remains of the Graffian follicles get transformed into the corpus luteum, which secretes progesterone for the maintenance of the endometrium.

Question.29. Describe the asexual and sexual phase of life cycle of Plasmodium that causes malaria in humans.
Answer: Life Cycle of Plasmodium :

1. Plasmodium requires two hosts to complete its life cycle.
2. When female Anopheles mosquito bites a healthy human being, it inject Plasmodium, which lives in its Anopheles mosquito as sporozoite the infectious form.
3. The parasites Plasmodium multiply (asexual reproduction) after transportation to the liver cells and finally rupture the liver cells. Sporozoites parasite are then released in to blood stream.
4.  Bursting of RBCs is accompanied by release of a toxic substance called haemozoin (associated with fever and chills).
5.  In the RBCs, only sporozoites change into gametocytes (sexual stage).
6.  When the diseased person is bitten by a female Anop¬heles mosquito, gametocytes are introduced into the mosquito.
7. Gametocytes fertilise and develop inside the intestine of mosquito to form sporozoites.
8. Sporozoites are stored in the salivary glands of mosquito and are released into the healthy person who is bitten by this mosquito.
   

OR
(a) What is plant breeding ? List the two steps the classical . plant breeding involves.
(b) How has the mutation breeding helped in improving crop varieties ? Give one example where this technique has helped.
(c) How has the breeding programme helped in improving the public nutritional health ? State two example in support of your answer.
Answer : (a) Plant breeding is a method that involves in the crossing of two plants to create the progeny with certain traits in their genes and transfer on to the future generation to create specific plant types, which are more suitable for cultivation, give better yields, and are disease resistant.
Two steps involved in Classical plant breeding are :

1.  Crossing of superior pure lines and
2. Selection of plants with desired characteristics.

(b)Mutation breeding : In this, genetic variations are made, which then creates traits, not found in the parental type. It has benefited in making disease resistant plants by giving resistance against bacterial, fungal and viral diseases.
(c) Breeding helped in improvising the public nutritional health by growing crops that are good in nutrients. This is called bio-fortification of crops. Benefits of bio-fortification are to help :

1. Protein content and quality
2.  Oil content and quality
3. Vitamin content
4.  Micronutrient and mineral content Two examples are :
   (i) Maize hybrids grows 2000 and have 2 times the quantity of lysine and tryptophan compared to other maize hybrids.
   (ii) Atlas 66 (a wheat variety having higher protein content).

Question.30. A child suffering from Thalassemia is born to a normal couple. But the mother is being blamed by the family for delivering a sick baby.
(a) What is Thalassemia ?
(b) How would you counsel the family not to blame the mother for delivering a child suffering from this disease ? Explain.
(c) List the values your counselling can propagate in the families.
Answer : (a) Thalassemia is a autosomal recessive blood disorder. Thalassemia is illustrated by severe anaemia due to production of imperfect haemoglobin chains. Thalassemia is caused by mutations in the genes coding either the alpha, beta or delta chains constituting haemoglobin lead to the synthesis of incorrectly folded haemoglobin that is unable to transport , .oxygen proficiently.
(b) Thalassemia is an autosomal recessive disease, which means the mutation is carried on one of the autosomes, so the carrier can be any one of the two parents. It has an equal chance of coming from the mother or the father, so to just blame the mother for the child’s defect is unfair.
(c) The values of councelling can propagate in the families are :

1.  Provide healthy diet to the child
2. Accepting their child with all his/her positives and negatives
3. None of the parents is responsible for giving birth to a sick baby.
4.  The defect in the gene is caused by a random change in the genes of the child.
5.  Encouraging the child to follow his/her treatment regularly * and lead a happy and normal life.

SET- II

SECTION-A

Question.5. Identify the examples of homologous structures from the following:
(i) Vertebrate hearts
(ii) Thoms in Bougainvillea and tendrils of Cucurbita.
(iii) Food storage organs in sweet potato and potato.
Answer: (ii) Thorns in Bougainvillea and tendrils of Cucurbita

SECTION-B

Question.9. Describe the gene therapy procedure for an ADA-deficient patient.
Answer : Gene therapy of ADA deficiency :

1.  The term gene therapy describes as genetic modification of cells. The material transferred into patient’s cell through genes, gene segment or oligonucleotides.
2.  Lymphocytes isolated from patients blood are cultured in vitro. Functional ADA are then introduced into the cultured lymphocytes.
3.  These lymphocytes are returned back to the patient’s body.
4. The gene must be delivered inside the target cells and
   work properly without causing adverse effects to cure the disease. .
5.  The introduction of isolated gene from bone marrow cell producing ADA into cells at early embryonic stages used can be, a permanent cure for this disorder.

Question.14. (a) How does cleistogamy ensure autogamy ?
(b) State one advantage and one disadvantage of cleistogamy to the plant.
Answer : (a) Cleistogamous flowers do not open at all and pollen from other plants cahnot fall on the stigma of these flowers. In this situation cross pollination cannot be possible and only autogamy occurs. Therefore, clesistogamy ensures autogamy.
(b) Cleistogamy has this advantage that the plant can propagate itself under unfavorable conditions and disadvantage is that, no chances of cross pollination, self pollination occurs therefore chances of variation and evolution of genetically superior progeny is reduced.

Question.17. A young boy when brought a pet dog home started to complain of watery eyes and running nose. The symptoms disappeared when the boy was kept away from the pet.
(a) Name of type of antibody and the chemicals responsible for such a response in the boy.
(b) Mention the name of any one drug that could be given to the boy for immediate relief from such a response.
Answer : (a) The IgE type of antibody and chemicals like histamine and serotonin released from mast cells are responsible for this type of response.
(b) (i) Antihistamine
(ii) Adrenalin and
(iii) Steroids can be given to the boy for immediate relief from such a response.

Question.18. (a) Explain how to find whether an E coli bacterium has transformed or not when a recombinant DNA bearing ampicillin resistant gene is transferred into it.
(b) What does the ampicillin resistant gene act as in the above case ?
Answer : (a) When E. coli with a recombinant DNA will be grown on culture media having ampicillin the transformed cells will survive as they carry recombinant DNA with ampicillin resistant gene; while the non transformed cells will die as they are ampicillin sensitive.
(b) Antibiotic resistant genes such as ampR (ampicillin resistant), acts as selectable markers.

SECTION-C

Question.22. (a) Explain how to overcome inbreeding depression in cattle.
(b) List three advantages of inbreeding in catde.
(c) Name an improved breed of catde.
Answer : (a) Inbreeding depression in cattle can be made overcome by mating selected animals with unrelated superior animals of the same breed. It is called out-breeding.
(b) Advantages of inbreeding in catde :

1.  Maintaining the accumulation of superior quality of breed of organisms.
2. Evolve a pure breed of progeny.
3. Elimination of deleterious alleles as they are not passed to future generations.

(c) Jersey cow is an improved breed of cattle.

Question.27. (a) Draw a diagram of the structure of a human ovum surrounded by corona radiate. Label the following parts : (i) Ovum (ii) Plasma Membrane (iii) Zona Pellucida (b) State the function of Zona Pellucida.
Answer : (a) Diagram of a mature ovum

(b) Zona pellucida layer of the ovum-prevent entry of more than one sperm and inhibit polyspermy in humans. It also facilitate changes in the membrane of the ovum to block the entry of additional sperms, thereby makes sure the entry of only one sperm inside the ovum.

SECTION-D

Question.30. (a) Draw a labeled schematic diagram of the transverse
section of a mature anther of an angiosperm plant.
(b) Describe the characteristic feature of an insect pollinated flower.
Answer: (a) Diagram of transverse section of a mature anther of an angiospermic plant

(b) Characteristic features of insect pollinated flower:

1.  Contains nectar to attract the,animal pollinators
2. Large, colourful and fragrant flowers
3.  Ovule may be one or more
4.  Sticky pollen grains

OR
(a) Describe the events of spermatogenesis with the help of a schematic representation.
(b) Write two differences between spermatogenesis and oogenesis.
Answer : (a) Events of spermatogenesis with the help of schematic diagram

1. Spermatogenesis starts at the age of puberty.
2.  Spermatogonia are present on the inside wall of semniferous tubules multiply by mitotic divisions and increase in number.
3. Each spermatogonium is diploid and contain 46 chromosomes.
4. Some of spermatogonia called primary spermatocytes
   undergo meiosis I to produce 2 equal haploid cells called – secondary spermatocytes with 23 chromosomes each.
5.  Secondary spermatocyte undergo second meiotic division to produce 4 equal haploid spermatids.
6.  Spermatid later transform into spermatozoa.

1. 

SET- III

SECTION-A

Question.6. “Sweet potato tubers and potato tubers are the result of convergent evolution.” Justify the statement.
Answer : Convergent evolution is the process where some diverse organisms individually develop the traits which are similar in function, i.e, sweet potato tubers and potato tubers – are not anatomically similar structures even though they perform similar function, e.g. storage of food and vegetative reproduction but both are different in origin i.e., sweet potato is an adventitious root and potato tuber is an underground stem.

SECTION-B

Question.13. Explain the steps that ensure cross pollination in an autogamous flower.
Answer: Cross pollination in an autogamous flower represents by following steps :

1.  Pollen grains and stigma receptivity should not be synchronized.
2. Anther and stigma are placed at different positions so that pollen cannot come in contact with stigma of the same flower.
3. Self incompatibility also prevents inbreeding.

Question.16. A student on a picnic to a park on a windy day started sneezing and having difficulty in breathing on reaching the park. The teacher enquired whether the student was allergic to something.
(a) What is an allergy ?
(b) Write the two unique characteristics of the system involved in the response observed in the student ?
Answer: (a) Allergy is a hypersensitive reaction of foreign substances by the immune system. The body system of defense against these substances particularly pathogens (the agents of infection) caused allergy. The inflammation of the tissues inside the nose causes allergy after allergens are inhaled.
(b) The two unique characteristics of the system involved in the response observed in the students are :

1.  The immune system recognizes foreign antigens and responds to them.
2.  It has a memory so it remembers antigens.

Question.17. Why and how bacteria can be made ‘competent’?
Answer : The bacterial cells are made competent by treating them with a specific concentration of divalent cations like calcium or magnesium e.g., CaCl2 or MgCl2 This makes the cell wall permeable and bacterial cell takes up the plasmid DNA.

Question.18. (a) Name the deficiency for which first clinical gene therapy was given.
(b) Mention the cause of and one cure for this deficiency.
Answer : (a) ADA (Adenosine deaminase) deficiency.
(b) Cause : The gene coding for enzyme ADA gets deleted leading to deficiency of ADA and problems in immune system. Cure : It can be treated by isolating the gene for ADA from bone marrow cells at embryonic stage.

SECTION-C

Question.23 Draw the following diagrams related to human reproduction and label them.
(a) The zygote after the first cleavage division
(b) Morula stage
(c) Blastocyst stage (sectional view)
Answer : (a) The zygote after the first cleavage division.

Question.29. Draw a diagram of a mature embryo sac of an angiosperm and label the following parts in it.
(i) Filiform apparatus (ii) Synergids
(iii) Central cell (iv) Egg cell
(v) Polar nuclei (vi) Antopodals
(b) Write the fate of egg cell and polar nuclei after fertilization.
Answer: (a)

(b) Fate of Egg cell and polar nuclei: Egg cell fuses with male gametes to form zygote. Zygote finally give rise to Embryo while the polar nuclei fuses with the other male gamete to produce a triploid Primary Endosperm Nucleus (PEN). PEN develops into endosperm. Since two kinds of fusion—syngamy and triple fusion—take place, the process is known as double
fertilisation, and is a characteristic of flowering plants.

CBSE previous Year Solved  Papers  Class 12 Biology Outside Delhi 2015

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5.  Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SET I

SECTION-A

Question.1. How many chromosomes do drones of honeybee possess ?
Name the type of cell division involved in the production of sperms by them.
Answer : Drones of honey bees are haploid and possess 16 chromosomes.
Mitosis is the cell division that is involved in the production of sperms by drones.

Question.2. What is a cistron ?
Answer : Cistron is a section of DNA or RNA molecule that codes for a specific protein or polypeptide.

Question.3. Retroviruses have no DNA However, the DNA of the infected host cell does possess viral DNA. How is it possible ?
Answer : After an attack on the host cell, retro virus enters the micro phages (in case of HIV) where the virus undergoes Teminism or Reverse Transcription to use its RNA genome to form viral DNA in presence of reverse transcriptase enzyme. This viral DNA then incorporates into the host cell DNA, directing the infected cells to prpduce more copies of viruses. In this way, the infected cell possesses viral DNA.

Question.4. Why do children cured by enzyme-replacement therapy for adenosine deaminase deficiency need periodic treatment?
Answer : Children cured by enzyme-replacement therapy for adenosine deaminase (ADA) deficiency need periodic treatment because it is not a completely curative method. In this method, the lymphocytes are retrieved from the blood of the patient. These lymphocytes are then grown in a culture medium. The lymphocytes are then incorporated with functional ADA, i.e. cDNA using a retro viral vector. Since the-lymphocytes have a definite life cycle, there is a need for periodic transfusion of genetically engineered lymphocytes to the patient.

Question.5. List two advantages of the use of unleaded petrol in automobiles as fuel.
Answer : Following are the two advantages of using unleaded petrol as fuel in automobiles :

1. Pollution caused by unleaded petrol is less as it does not release lead compounds from exhaust fumes into the atmosphere.
2. It also helps in preventing health issues like anaemia, loss of appetite, damage to erythrocytes & damage to control Nervous system.

SECTION – B

Question.6. Why do moss plants produce very large number of male gametes ? Give one reason. What are these gametes called?
Answer : Mosses are bryophytes and they need water for . fertilization. They lay their flagellated male gametes that swim across the water to reach the female gamete. In this process, a large number of male gametes are destroyed or lost. Thus, very large number of male gametes are produced by moss plants so that even if some of the gametes get destroyed, the remaining can fertilise the female gamete. The male gametes are called antherozoids.

Question.7. (a) Select the homologous structures from the combinations given below:
(i) Forelimbs of whales and bats
(ii) Tuber of potato and sweet potato
(iii) Eyes of octopus and mammals
(iv) Thorns of Bougainvillea and tendrils of Cucurbita
(b) State the kind of evolution they represent.
Answer:
(a) From the above options which forms homologus structures they are:
(i) Forelimbs of whales and bats are similar in structure but perform different functions of swimming and flying, respectively.
(iv) Thorns of Bougainvillea and tendrils of Cucurbita are both modifications of a stem arising from axillary bud but perform different functions like protection and climbing, respectively.
(b) The evolution represented by homologous organs or structures is divergent evolution as their origin is common but have diverged (became dissimilar) with evolution.

Question.8. (a) Why are the plants raised through micropropagation termed as somaclones ?
(b) Mention two advantages of this technique.
Answer:
(a) The plants obtained by micropropagation are called somaclones because they are genetically identical to the original plant from which they are produced.
(b) The advantages of micropropagation are as follows :
(i) It helps in the propagation of a large number of plants in a short span of time.
(ii) It helps in the production of plants which are disease and pest resistant plants.

Question.9. Explain the different steps involved during primary treatment phase of sewage.
Answer : The primary phase of sewage treatment involves physical removal of particles through filtration andsedimentation Initially, floating debris is removed by sequential Alteration.
Then the grit are removed by sedimentation. All solids that settle form the Primary sludge, and the supernatant forms the effluent. The effluent from the primary setting tank is taken for secondary treatment.

Question.10. What is mutualism ? Mention any two examples where the organisms involved are commercially exploited in agriculture.
OR
List any four techniques where the principle of ex-situ conservation of biodiversity has been employed.
Answer : Mutualism is a relationship between two organisms of different species in which both organisms are benefited.Examples of the organisms involved that are commercially exploited in agriculture are as follows :

1. Commercial exploitation of Rhizobium in agriculture : Continuous growth of crops leads to the nutrient deficiency in soil. Farmers, then, grow leguminous crops containing Rhizobium in its root nodules to replenish the lost nutrients (especially nitrogen) in the soil.
2. Commercial exploitation of Mycorrhiza in agriculture.
   Mycorrhiza is an association of the soil fungus with the roots of the higher plants. Farmers use Mycorrhiza commercially in agriculture as it improves the soil quality and reduces soil erosion by improving plant rooting capacity. Fungal symbiont absorbs and store N2(nitrogen), calcium, phosphorus, and potassium in the fungal mantel, and overall increase in plant growth and development.

OR
Ex-site conservation is the preservation of biological diversity outside their natural habitats. This involves conservation of genetic resources, as well as wild and cultivated or species, and draws on a diverse body of techniques and facilities. Some of these include :

1. Gene banks, e.g. seed banks, sperm and ova banks, field banks.
2.  In vitro plant tissue and microbial culture collections.
3. Captive breeding of animals and artificial propagation of plants, with possible reintroduction into the wild, and
4.  Collecting living organisms for zoos, aquaria, and botanical gardens for research and public awareness.

SECTION – C

Question.11. State what is apomixis. Comment on its significance. How can if be commercially used ?
Answer : Apomixis is the mechanism of formation of new individuals without involving the process of meiosis and syngamy. In this process, gamete- formation is absent. Apomixis is observed in certain plants like grasses, citrus plants, conifers like pine etc. It occurs through agamospermy, parthenogenesis appogamy.Significance of Apomixis : Apomixis provides a better alternative for raising hybrid varieties from hybrid seeds. Hybrid seeds have to be produced every year as these cannot be collected from hybrid plants. The hybrid characters segregate during meiosis. If apomixis is introduced in the hybrid seeds there will be no need of producing seeds every year.
Commercial applications of apomixis are :

1.  By apomixis, hybrid varieties of seeds can be produced, which will provide higher and better yield.
2.  It prevents the loss of specific characteristics in the hybrid plants.
3. This method is a cost-effective method of producing seeds.

Question.12. During a monohybrid cross involving a tall pea plant with a dwarf pea plant, the offspring populations were tall and dwarf in equal ratio. Work out a cross to show how it it possible.
Answer : The given condition is possible only when the tall pea plant is heterozygous and dwarf (small) pea plant is homozygous. This cross can be represented as follows :
Alleles can be similar as in the case of homozygotes (TT) and (tt) can be dissimilar as in the case of the heterozygote it. Since the TT plant is heterozygous for gene controlling one character, it is a monohybrid cross.

Question.13. Explain the significance of satellite DNA in DNA fingerprinting technique.
Answer: Satellite DNA are short sequences of DNA repeated again and again to form long sequences and generally non-coding DNA. It is very significant in DNA fingerprinting for the following reasons:

1.  It is specific for every species
2. It is a highly polymorphic DNA, an example of which is Variable Number of Tandem Repeats (VNTR). This means number of repeats vary in different individuals and these differences can be used for identification purpose.
3. VNTR sequences are inheritable and an individual inherits it from both parents. This can be used for paternity testing.
4. Regions of crossing over can also be easily marked, making identification easier.

Question.14. What does the following equation represent ? Explain f + 2pq + q2 = 1
Answer : The Hardy-Weinberg equation is a mathematical equation that can be used to calculate the genetic variation of a population at equilibrium. The Hardy-Weinberg equation is expressed asp + q = 1 where p is the frequency of the “A” allele and q is the frequency of the “a” allele in the population. In the equation, p2 represents the frequency of the homozygous genotype AA, q2 represents the frequency of the homozygous genotype cut, and 2pq represents the. Creoency of the. heterozygous Aa. Hence, p1 + 2pq + q2 = 1, which is the expression of (p + q)2.
Any deviation in the values of frequency of alleles from the usual value, indicate that there is an evolutionary change in the population.

Question.15. A heavily bleeding and bruised road accident victim was brought to a nursing home. The doctor immediately gave him an injection to protect him against a deadly disease.
(a) Write what did the doctor inject into the patients body.
(b) How do you think this injection would protect the patient against the disease ?
(c) Name the disease against which this injection was given and the kind of immunity it provides.
Answer:
(a) In the patients body, the doctor has injected antiserum containing preformed antibodies against the causative organism or toxin produced by it.
(b) The solution injected by the doctor had antibodies; hence, the injection provide humoral immunity to patient and would protect the patient against the disease.
(c) The disease against which this injection was given is tetanus caused by Clostridium tetani, which usually exists in environment as spores and may gain access to the body through wound.
The kind of immunity that the injection containing
antiserum provides is passive immunity as performed antibodies are used because fast action is required in this emergency case.

Question.16. Enumerate any six essentials of good, effective Dairy Farm Management Practices.
Answer : Six important ways of good and effective dairy farm management practices are as follows :

1.  Choosing the breeds that have high yielding potential is essential.
2. Proper accomodation (including the aspects of hygiene) and adequate water are essential for the care of catde.
3.  Periodic visit by a veterinary doctor should be compulsory for the good health of the cattles.
4.  The fodder given to the cattle should be adequate quantity and good quality (including grains and protein concentrates).
5.  The feeding of cattle should be carried out in a scientific
   manner and become mechanised which reduces chance of direct contact with the handler.
6.  Regular inspection of dairy farms should be done by appointed officials to ensure that all the instructions are being strictly followed.

Question.17. State the medicinal value and the bioactive molecules produced by Streptococcus, Monascus and Trichoderma.
OR
What are methanogens ? How do they help to generate biogas ?
Answer:
Streptococcus : It produces Streptokinase, also called Tissue plasminogen activator (TPA) is used as clot buster to dissolve blood clots in patients who have undergone heart attacks.
Monascus : This fungal species produce statins that help in lowering blood cholesterol, statins, act as inhibitors of enzyme HMG CoA reductase of liver which form mevalonate, required . for cholesterol synthesis.
Trichoderma : It produces cyclosporin A which act as immunosuppressive agent by inhibiting the activation of T-cell response to transplanted organs.
Methanogens are the bacteria found in catde dung (gobar) and in anaerobic sludge during sewage treatment. They grow anaerobically on cellulosic material and produce a large amount of methane (main constituent of bio gas) along with C02 and H . Thus, methanogens are used in biogas production.

Question.18. Rearrange the following in the current sequences to accomplish an important biotechnological reaction :
(a) In vitro synthesis of copies of DNA of interest
(b) Chemically synthesized oligonucleotides
(c) Enzyme DNA-polymerase
(d) Complementary region of DNA
(e) Genomic DNA template
(f) Nucleotides provided
(g) Primers
(h) Thermostable DNA-polymerase (from Thermus aquaticus)
(i) Denaturation of ds-DNA.
Answer : The steps given are the steps involved in PCR (Polymerase chain reaction) correct sequence of the steps are :
(i) Denaturation of ds-DNA
(d) Complementary region of DNA
(c) Enzyme DNA polymerase
(b) Chemically synthesized oligonucleotides
(g) Primers
(h) Thermostable DNA-polymerase (from Thermus acquticus)
(f) Nucleotides Provided
(a) In vitro synthesis of region of DNA of interest

Question.19. Describe any three potential applications of genetically modified plants.
Answer: Three potential applications of genetically modified (GM) plants are as follows :

1.  Nutrient Enrichment
   GM plants provide essential nutrients to people through the consumption of main staple crop.
   Example: Golden rice is a variety of rice that is genetically engineered by inserting genes to produce beta carotene, which is a precursor of Vitamin A. Thus, consumption of this crop helps in the prevention of Vitamin A deficiency diseases.
2. Insect/Pest Resistance
   G.M. Crops Such as BT Cotton are insect|pest resistant Crops. It is a genetically modified variety of cotton that produces a protein called Bt toxin. The Bt toxin act as insecticide & kills cotton balloworms and Larvae of Lepidopterans & Dipterans.
3.  Abiotic Stress Resistance
   GM crops are tolerant to stress conditions such as high and low temperature, salinity and drought.

Question.20. How did an American Company, Eli Lilly use the knowledge of r-DNA technology to produce an insulin ?
Answer : Insulin Consists of 51 amino acids forming two short polypeptide chains chain A having 21 amino acids and chain B with 30 amino acids. The two chains are linked by disulfide bond. In animals including humans, insulin occurs as proinsulin is made of chain A, chain B and chain C (30 amino acids). As the insulin matures, chain C is removed.
The American company, Eli Lily used the following technique to produce synthetic human insulin or humulin. The steps followed were:

1.  The genetic engineering of insulin begins with identification & separation of DNA sequences coding for chain A and chain B.
2. These sequences were then introduced into plasmid (pBR322) of E.coli bacterium. It is referred as factory used in genetic engineering of insulin.
3. In E.coli, p-galactosidase controls the transcription of these genes, therefore, insulin gene needs to be tied to this enzyme. The protein formed by E-coli consists partly of P-galactosidase joined to either A or B chain of insulin.
4.  These are then extracted from P-galactosidase fragment & purified.
5. The two chains are mixed & reconnected in a reaction that forms disulfide bridges resulting in pure synthetic human insulin (humulin).

Question.21. How do snails, seeds, bears, zooplanktons, fiingi and bacteria adapt to conditions unfavourable for their survival ?
Answer: Snails adapt to unfavourable conditions by producing epiphragm during hibernation that covers the opening of its shell and thus prevent desiccation.
Seeds adapt to unfavourable conditions by getting into the state of dormancy.
Bears adapt to unfavourable conditions by hibernating and reducing their body metabolic activities by 75%. Zooplanktons adapt to unfavourable conditions by entering into diapause (stage of suspended development).
Fungi adapt to unfavourable conditions by reducing their metabolic rate and by forming thick-walled spores.
Bacteria adapt to unfavourable conditions by forming endospores.

Question.22. With the help of a flow chart, show the phenomenon of biomagnification of DDT in an aquatic food chain.
Answer: Biomagnification is the increase in the concentration of pollutants or harmful chemicals within each step of the food chain. The levels of biomagnification will be different at different trophic levels. For example, in a pond of water, DDT was sprayed and the producers were found to have 0.04 ppm concentration of DDT. Since many types of planktons are eaten by some fishes and clams, their body accumulates
0.23 ppm of DDT. Sea gull that feeds on clams accumulates- more DDT as one sea gull eats many clams. Hawk, the top carnivore, has the highest concentration of DDT.

SECTION-D

Question.23. Your school has been selected by the Department of Education to organize and host an interschool seminar on “Reproductive Health – Problems and Practices”. However, many parents are reluctant to permit their wards to attend it. Their argument is that the topic is “too embarrassing”.
Put forth four arguments with appropriate reasons and explanation to justify the topic to be very essential and timely.
Answer: Reproductive health is the total well-being in all aspects of reproduction. It includes physical, emotional, behavioural and social well-being of an individual. Therefore, there is an urgent need to educate and discuss topics related to the reproductive health.
Following are the topics., about reproductive health that should be discussed with the students :

1.  Sexually transmitted diseases, such as AIDS and Gonorrhoea, are transferred from one individual to another through sexual contact. Therefore, making the students aware about these diseases will help to jprevent their spread.
2. Lack of knowledge about the reproductive status may lead to unwanted pregnancies. Hence, it is necessary to create awareness among people, especially the youth.
3.  Learning about one’s sexuality at a proper age may help the students to know about the different changes happening in their body; thereby, leading to a better mental and physical state of health.
4.  Counselling and creating awareness about reproductive health also help to curb the problems of infertility, birth control, mortality, etc.

SECTION – E

Question.24. (a) Plan an experiment and prepare a flow chart of the steps that you would follow to ensure that the seeds are formed only from the desired sets of pollen grains. Name the type of experiment that you carried out.
(b) Write the importance of such experiments.
OR
Describe the roles of pituitary and ovarian hormones during the menstrual cycle in a human female.
Answer : The experiment or technique that is carried out to obtain seeds from a desired set of pollen grains is referred as Artificial Hybridization. In this technique, only desired pollen grains are used in pollination and the stigma of the desired plant is kept protected from contamination from the unwanted pollen. This involves two processes :
(a) Emasculation and (b) Bagging.
(a) Emasculation : In this, the anthers are removed by fine forceps in their bud condition from the bisexual flowers of plants selected as female parent Such flowers are called emasculated flowers.
(b) Bagging : This involves covering the stigma of both emasculated and non-emasculated flowers with butter paper of polythene in their bud condition to prevent contamination from unwanted pollen.
When the stigmas of emasculated flowers mature, the bags are removed for a while. The stigmas are dusted with pollen grains fo desired male plants by means of a brush. The flowers are rebagged till the fruits develop. This is how desired seeds are obtained.

(b) Artificial hybridization is important for the following reasons :
It ensures that the crops produced have the desired characteristics.
It helps to improve the crop yield.
It helps to yield commercially superior varieties.
OR
The cycle of events, starting from one menstruation till the next is called the menstrual cycle. Following are the changes that  are about by ovarian and pituitary hormones in the phases of menstrual cycle.

1.  Menstrual phase : It lasts for 3-5 .days. The unfertilised ovum alongwith ruptured uterine epithelium, some blood & mucus is discharged through the vaginal orifice & this is called Menstruation. The decrease in the level of progesterone & estrogen in the event of failure of fertilization stimulates the hypothalmus & anterior pituitary to release GnRH and FSH, to start the phase of next menstrual cycle.
2. Follicular/proliferative phase : It exhibits following changes:
   (a) Under the stimulation of GnRH of hypothalamus there is increased secretion of FSH from the anterior pituitary. The FSH stimulates the ovary and bring about the change| maturation of 1° follicle (6-12 in number) into a Graffian follicle. The follicular cells of Grafian follicle secrete estrogen.
   (b) This estrogen produced, stimulates the growth, maintenance & normal functioning of secondary sex organs.
   (c) The estrogen also aids in repairing the endometrial lining for 2 days & later this lining becomes more vascular in 7-9 days.
   (d) The end of phase marks the inhibition of FSH secretion & stimulation of LH secretion.
3.  Ovulatory phase : Process of release of mature egg cell is called as ovulation. Ovuiationis controlled by increased level of LH. LH starts the change of empty Graffian follicle into corpus luteum and secretion of progesterone from corpus luteum which is responsible for maintenance of endometrium. Ovulatory phase is characterised by high levels of estrogen & progesterone.
4. Luteal/Secretory phase : It secretes progesterone by suppressing the secretion of FSH & inhibits the maturation of follicle & ovulation. The uterine endometirum gets ready for implantation. In absence of GnRH, levels of LH & FSH falls, thus decreasing progesterone levels & prepare to start the menstrual phase.

Question.26. (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.
OR
“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.
Answer : (a) Following are the attributes that a population
has but an individual organism does not have :
Birth rate : per capita Births.
Death rate : per capita deaths.
Sex ratio : Ratio of number of males to females in a population.
(b) Population density means numbers of individuals present per unit area. Population density can be measured by determining the population size.
Different ways to measure population density are :

1.  Direct method : It involves the counting of organisms in the given area, e.g., in order to determine the number of bacteria growing in a Petri dish, their colonies are counted.
2. Quadrat method : It is the method that involves the use of square of particular dimensions to measure the number of organisms, e.g., the number of Parthenium plants in a given area can be measured using the quadrat method.
3.  Indirect method : In this method, there is no need to count the organism individually, e.g., the number of fishes caught per trap gives the measure of their total density in a given water body.

OR
Pyramid of Energy
The graphical representation of the amount of energy trapped at different trophic levels of food chain in an ecosystem is called pyramid of energy. It represents the total amount of energy consumed by each trophic level. The amount of energy transferred through food to successive higher levels decreases with each level. This means that a greater amount of energy is available at the producer level than at the primary consumer level (herbivores). This is because some amount of energy is lost at each trophic level. The energy that is lost is almost 80-90% (by the second law of thermodynamics) of the energy available at the lower trophic level. Thus, this pyramid is always upright or straight. The law states that only 10% of chemical energy is retained at each trophic level and this law is known as the 10 percent law. This 10% of energy is only available to the next trophic level.

Pyramid of Biomass
Biomass is the renewable organic (living) material or the energy contained inside plants and animals. It is measured as both fresh weight and dry weight. It is measured in gram/ (meter)2 or calories/(meter)2. A graphical representation that shows the amount of biomass present at each tropic level in a unit area at any point of time is called pyramid of biomass. The biomass pyramids are of two types : upright and inverted. Pyramids of biomass in a terrestrial ecosystem is an upright or straight pyramid. In this type of pyramid, the combined weight of producers is larger than the combined weight of consumers.

SET-II

SECTION – A

Question.3. State the cause of adenosine deaminase enzyme deficiency.
Answer : Adenosine deaminase (ADA) enzyme deficiency is caused by the mutation in the ADA gene. This gene is present on chromosome number 20 and provides instruction for producing the enzyme adenosine deaminase. ADA deficiency is inherited as an autosomal recessive trait.

SECTION – B

Question.8. Explain the process of secondary treatment given to the primary effluent up to the point it shows significant change in the level of biological oxygen demand (BOD) in it.
Answer: Secondary treatment refers to those treatment process that use biological process to convert dissolved suspended and colloidal organic wastes to more stable solids that can either be removed by settling or discharged to the environment without causing harm. Supernatant from the primary treatment is passed into large aeration tanks during secondary treatment. In these tanks, the effluent is agitated mechanically and air is pumped into it. This causes vigorous growth of the bacteria that lead to the formation of floes, containing bacteria and fungal filaments in a mesh-like structure. While growing, these microbes consume the major part of organic matter in the effluent; it decreases the biological oxygen demand (BOD), the BOD of sewage or waste water is reduced significantly, the effluent if then passed into a setding tank where the bacterial ‘floes’ are allowed to sediment, called activated sludge.

Question.10. In mosses sexual reproduction cannot complete without water. A moss plant is unable to complete its life-cycle in a dry environment. State two reasons.
Answer : Mosses cannot complete their life cycle in a dry environment because of the following reasons :
1. Water acts as a medium for flagellated sperm to reach the egg and undergo fertilization.
2. Mosses cannot absorb water because their roots are rudimentary, thus, for their survival, they need to grow in moist environment.

SECTION – C

Question.14. Two independent monohybrid crosses were carried out , involving a tall pea plant with a dwarf pea plant. In the first . cross, the offspring population had equal number of tall and dwarf plants, whereas in the second cross it was different. Work out the crosses, and explain giving reasons for the difference in the off sping populations.
Answer : First Cross (Case I)
In the first cross between the tall plant and the dwarf plant, equal number of tall and dwarf plants are produced. This is because the tall plant is heterozygous dominant (Tt) and dwarf plant is homozygous recessive (tt). The progenies will be 50 % tall and 50 % dwarf.
Second Cross (Case II)
In the second cross between the tall plant and the dwarf plant, all the offsprings will be tall. This is because the tall plant is  homozygous dominant (TT) and dwarf plant is homozygous recessive (tt).
The two monohybrid crosses can be represented as follows :

Question.19. Explain co-evolution with reference to parasites and their hosts. Mention any four special adaptive features evolved in parasites for their parasitic mode of life.
Answer : Co-evolution is a term used to describe the mutual changes in two or more species, usually one following the other, that affect their interactions. In terms of the relation of host and parasite, it can be explained as follows :
A parasite is an organism that derives nourishment and protection from other living organism known as host, but in doing so, it also harms the host. The host evolves over a long period of time to protect itself from parasite, while parasite evolves so that it can find another way to derive nutrition from the host and hence, the cycle continues.
Four special adaptive features evolved in parasites for their parasitic mode of life are as follows :

1.  Parasites have suckers as organs for attachment such that help them to firmly attach to the host body and help in deriving nutrition from them. For example suckers and hooks in Taenia solium.
2. Parasites are covered by protective body covering, that is tegument (in case of Taenia solium) and cuticle (in case of Ascaris lumbricoides) to protect them from harmful effects of digestive enzymes of the host.
3.  All parasitic organisms usually lack locomotory structures, they do not require to move in search of food.
4.  High reproductive capacity of parasitic organisms ensures the continuation of parasitic race.

Question.21. With the help of a flow-chart exhibit the events of eutrophication.
Answer : Enrichment of water body by excessive nutrients is known as eutrophication.

SECTION – E

Question.26. (a) Why are colour blindness and thalassemia categorised as Mendelian disorders ? Write the symptoms of these diseases seen in people suffering from them.
(b) About 8% of human male population suffers from colour blindness whereas only about 0.4% of human female population suffers from this disease. Write an explanation to show how it is possible.
OR
Explain the process of transcription in prokaryotes. How is the process different in eukaryotes ? 
Answer : (a) Colour blindness and thalassaemia are categoried as Mendelian disorders because they are caused by mutation in a single gene. Their mode of inheritance follows the principles of Mendelian genetics. Mendelian disorders can be:
(i) autosomal dominant (muscular dystrophy)
(ii) autosomal recessive (thalassaemia)
(iii) sex linked (colour blindness)
Symptoms of Thalassaemia
(a) growth problems – not putting on weight or growing in height.
(b) anaemia – red blood cell deficiency, leading to tiredness, weakness and shortness of breath.
(c) jaundice – yellowing of the skin and whites of the eyes.
(d) swollen abdomen (tummy) – this is caused by an enlarged liver of spleen.
Symptoms of Colour Blindness
(a) Difficulty distinguishing between colors
(b) Inability to see shades or tones of the same color
(c) Rapid eye movement (in rare cases)
The events of eutrophication are as follows :
(b) Colour blindness is a X-linked recessive disorder. Males have higher chances of getting affected in comparison to females because males have only one X with Y chromosome while females have two X chromosomes (XX). Thus, for a female to get affected by colour blindness, she has to have the mutant gene on both the X chromosomes while males can be affected if they carry it on the single X chromosome.

From the given table, it can be concluded that females have very less probability of getting this disease as compared to males. Females will be colour-blind only when either both parents are affected or male is affected and female is carrier, while males can be colour-blind even if female is carrier and male is normal.
OR
The transfer of genetic information from DNA to mRNA is known as Transcription. It is done by formation of RNA over the template of DNA. It creates single stranded RNA which has a coded information similar to the sense or coding strand of DNA (with exception of U in place ofT).
The segment of DNA that takes part in transcription is called Transcription Unit. It has 3 components—a promoter, structural

1. Promoter region : It is the proximal area of transcription unit, which provides sites for attachment of transcription factors (CT factor) & RNA polymerase. It is present upstream at 5’ end of coding & 3’ end of template strand. It has AT rich area called TATA box or pribnow Box.
2.  Structural gene: It is the area of template strand involved in transcription. It is polycistronic in prokaryotes.
3.  Terminator region : It is the distal end of transcription unit. It is present downstream at 5’ end of template strand. It provides site for attachement of termination factor (rho-factor) for release of RNA polymerase.
   The main enzyme that takes part in transcription is called RNA polymerase. RNA polymerase has a sigma factor for recogniyng the start signal of promoter region. The remaining part of RNA polymerase is called Core enzyme.

Mechanism of Transcription :

1.  Activation of Ribonucleotides : 4 types of
   ribonucleotides take part in synthesis of RNA over DNA. These are AMP, GMP, UMP & CMP. These are found in nucleoplasm. These are converted to active triphosphates by energy, phosphate & phosphorylase enzyme.
2.  DNA Template : The DNA strand which function as a template for RNA synthesis is the Template or Antisense strand. It has 3’ —» 5’ polarity & transcription proceeds in 5’ —» 3’ direction. The separation of template strand does not require specific enzymes as in DNA replication. Ill transcription, only the RNA polymerase travels along the template strand.
3. Initiation : RNA polymerase enzyme binds to the promoter region. The CT (Sigma) factor recognises the recognises the promoter site and initiates transcription, by using triphosphates.
4. Elongation : (a) The activated ribonucleotides come to lie opposite to the template strand, complementarily & starts pairing. This releases phosphate radical & energy.
   (b) The core enzyme with the help of energy & Mg2* ions, bonds the adjacent nucleotides to form RNA chain. The RNA remians attached to the enzymes in a small region. As the enzyme moves along the DNA, the RNA chain alongates till terminator region is reached.
5.  Termination : (a) When enzyme reaches, terminator region, the RNA polymerase, by means of Rho (p) factor & NUS G. This results in termination of transcription.
   (b) p factor has ATPase activity. After RNA separation, the sense & antisense strands of DNA re-establish Fl-bonds.

Difference in Eukaryitic Transcription :

1.  In Eukaryotes, the promoter site is recognised by presence of specific nucleotide sequence called TATA box (Hogness Box) on template strand.
2.  In Eukaryotes, RNA polymerase is of 3 types-I, II & III.
3. Monocistronic structural genes are present which have uninterrupted coding sequences, i.e. the information on gene is split. The coding sequences are called exons & the non-coding sequences occuring as interruption in exons are
   called Introns. All introns have GU at 5’ end and AG at 3’ end.
   
   

CBSE previous Year Solved  Papers  Class 12 Biology Outside Delhi 2014

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5.  Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SET-I

SECTION-A

Question.1. Name the part of the flower which the tassels of the corn-cob represent.
Answer : Female reproductive parts-are: stigma and style.

Question.2. Mention any two contrasting traits with respect to seeds in pea plant that were studied by Mendel.
Answer: The two contrasting traits with respect to seeds in pea plant that were studied by Mendel are :
(i) Seed shape: round and wrinkled.
(ii) Seed colour: yellow and green.

Question.3. Why is secondary immune response more intense than the primary immune response in human?
Answer: The primary immune response to antigen occurs on the first occasion and generate memory B and T cells with a high specificity for the inducing antigen. The secondary response, mediated by B cells with the help of T cells, quickly produces high-affinity and antigen-specific antibodies against pathogens.

Question.4. Why is it not possible for an alien DNA to become part of a chromosome anywhere along its length and replicate normally?
Answer : Alien DNA requires specific sequence called recognition sites (sites where restriction enzymes cut DNA) to ligate itself with host chromosome. ‘ Recognition sites sequence should be close to the origin of replication (ori sequence where DNA replication starts). This site is necessary for the binding of DNA polymerase to start replication. As this site is not be present in alien DNA molecules, so an alien piece of DNA cannot replicate normally by attaching to any DNA.

Question.5. State the role of C peptide in human insulin.
Answer: Human insulin is produced as a pro-hormone. The 31 amino acid C-peptide of proinsulin is important for the biosynthesis of insulin. It helps in maintaining the level of active insulin.

Question.6. Name the enzymes that are used for the isolation of DNA from bacterial and fungal cells for recombination DNA technology.
Answer: Enzymes used to isolate DNA from bacteria: Lysozyme Enzymes used to isolate DNA from fungi: Chitinase

Question.7. State Gause’s Competitive Exclusion Principle.
Answer : The principle that when two species compete for the same resources within an environment, one of them will eventually outcompete and displace the other. The displaced species may become locally extinct, by either migration or death.

Question.8. Name the type of association that the genus Glomus exhibits with higher plants.
Answer : Mycorrhizal is the type of association.

SECTION – B

Question.9. Why are the human testes located outside the abdominal cavity ? Name the pouch in which they are present.
Answer : Testes lie in the scrotum outside the abdominal cavity. It is so as it keeps the temperature l-3degree C(approx. 35°) less than that of the body (i.e about 37°C). A lesser temperature is required for the sustenance of sperm which is provided by the scrotum.

Question.10. In Snapdragon, a cross between true-breeding red flowered (RR) plants and true-breeding white flowered (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 this phenomenon known as ?
Answer:

(a) The appearance of pink flower is not known as blending because it is due to partial influence of allele for white colour over the allele for red colour. On self-crossing the Fj plants, the F2 progeny are three types of plants-red flowered, pink flowered & white flowered in the ratio of 1 : 2 : 1. The occurrence of red & white flowered plants in F2 generation, indicate that the two alleles (red & white flower colour) and not blended but partially expressed as pink flower plants.
(b) The above example is known as incomplete dominance.

Question.11. With die help of one example, explain the phenomena of co-dominance and multiple allelism in human population.
Answer: A condition in which two different alleles for a genetic trait in a heterozygote are fully expressed thereby resulting in offspring with a phenotype that is neither dominant nor recessive.
When three or more alternative forms of a particular gene existing in a population, it is called multiple allelism.
Example : A typical example showing co-dominance is the ABO blood group system. For instance, a person having IA allele and IB allele will have a blood type AB because both the IA and IB alleles are co-dominant with each other.
ABO blood group is controlled by I gene. The gene I has 3 different alleles IA, IB and i. IA and IB produce two different types of sugars on the plasma membfane of red blood cells. The gene I does not produce any sugars. IA and IB are completely dominant over i. When IA and’lB’ are present together, they express their own type of sugars.

Question.12. Write the scientific name of the fruit-fly. Why did Morgan
prefer to work with fhiit-flies for’his experiments ? State any three reasons.
OR
Linkage and crossing-over of genes are alternative of each other. Justify with the help of an example.
Answer : Scientific name of fruit-fly: Drosophila melanogaster. Morgan used fruitfly for his experiment because :
(i) The fruit-fly could be grown on a simple synthetic medium inside the laboratory
(ii) The life cycle of a fruit-fly is about only two weeks.
(iii) A single mating could produce a large number of progeny offsprings.
Answer : (i) There is some linkage between all genes located on the same chromosome. The linkage strength depends on the percentage of the distance between the two, But linkage can be easily broken by crossing over.
(ii) When genes located on the same chromosome, then . there is possibility of two situations, either a crossing
over between the two genes or no crossing between two genes.
(iii) Crossing over always occurs if genes are located very far from each other – 50% recombinants, 50% parental. Example :Morgan hybridized yellow-bodied, white-eyed females to brown-bodied, red eyed males and intercrossed their F progeny. He found that the genes for white and yellow were very lightly linked and showed only 13% recombinant while white and miniature wing showed 37.2% recombination.

Question.13. List of symptoms of Ascariasis. How does a healthy person acquire this infection?
Answer: Symptoms of Ascariasis include : Worms in stool, coughing up worms, loss of appetite, fever. Severe symptoms of Ascariasis include : Vomiting, shortness of breath, swelling of the abdomen, severe stomach pain, and intestinal blockage.
Mode of Transmission:
(1) It is transmitted by improper disposal of human stool containing the eggs of Ascaris.
(2) Healthy persons may get infection from contaminated water, vegetables, fruits, other food articles & fomites.

Question.14. Explain the significant role of the genus Nucleopolyhedrovirus in an ecological sensitive area.
Answer: The nucleopolyhedrovirus, a sub group of Bacu- loviruses is a virus. It affects insects, predominantly moths and butterflies and used as a biological control agent. It has been used as a pesticide for crops infested by insects specially arthropods. Though this virus is species specific, making it effective under certain circumstances and there no negative effect on plants, mammals, birds, fish or other is non-target insects.

Question.15. How does a restriction nucleases function ? Explain.
Answer : Restriction nucleases are of two different types- endonucleases cut at a specific position inside DNA strand. Exonuclease remove nucleotides from the end of a DNA. strand. Restriction endonucleases recognize short, usually palindromic (meaning the base sequence reads the same backwards and forwards), sequences of 4—8 bp and, in the presence of Mg2+, cleave the DNA within or in close proximity to the recognition sequence. For example, EcoRI digestion produces “sticky” ends.

Whereas, Smal restriction enzyme cleavage produces “blunt” ends:

Question.16. How have transgenic animals proved to be beneficial in:
(a) Production of biological products (b) Chemical safety testing.
Answer : (a) Production of biological products :
The transgenic farm mammal was produced, a sheep called ‘Rosie cow’, had a human gene that expressed high levels of the human protein alpha-1-antitrypsin. The protein, which missing in humans, can lead to a rare form of emphysema.
(b) Chemical safety testing Transgenic animals, toxicity-sensitive transgenic animals have been produced for chemical safety testing. Transgenic animals can also be used to test the identity and purity of human proteins used as drugs. A transgenic » animal that makes a human protein (e g human insulin) will recognise this substance as its own and will therefore not produce an immune response against it.

Question.17. Describe the mutual relationship between fig tree and wasp and comment on the phenomenon that operates in their relationship.
Answer : Mutual relationship : Fig tree and wasp shows mutualism between them. The interaction in which both the interacting species get benefit of each other called mutualism. Fig flower is pollinated only by wasp and not by any other species. Female wasp lays eggs inside the developing fruit and also uses the developing seeds within the fruit for nourishing its larvae. Co-evolution exists between their close specific tight relationship.

Question.18. Construct an age pyramid which reflects an expanding
growth status of human population.
Answer:

Expanding pyramids of human population :

A population at any given time is composed of different age groups. These three groups include :

1.  Pre-reproductive
2. Reproductive
3. Post-reproductive
4.  If the age distribution (percent individuals of a given age or age group) is plotted for the population, the resulting structure is called an age pyramid.
5.  In human – beings, the age pyramids show the age distribution of male and female in a combined diagram.
6. In expanding pyramid, individuals in reproductive age group are more in number so the pyramid is expanding.

SECTION-C

Question.19. Make a list of any three outbreeding devices that flowering plants have developed and explain how they help to encourage cross-pollination ?
OR
Why are angiosperm anther called dithecous ? Describe the structure of its microsporangium.
Answer : The three outbreeding devices to encourage, cross-pollination
(i) Protoandry : The pollen grain and stigma of the flower mature at two different times, so that pollen release and stigma receptively are not simultaneous.
(ii) Protogyny : Mechanical barrier on the stigmatic surface of flowers, so that the anther and stigma cannot come in contact with each other of same flower.
(iii) Self incompatibility : The receptive stigma retard the growth of the pollen tube of fallen mature pollen grains of the same flower.
OR
Angiosperm anther is bilobed. Each lobe has two theca (microsporangium) so it is known as dithecous. Structure of microsporangium: The transverse section of a typical microsporangium are circular in outline. The microsporangium surrounded by four separate wall layers: epidermis, endothecium, middle layers and tapetum. The innermost wall layer tapetum provide nourishment to developing pollen grains. Tapetum cells are multi-nuclei ‘ and have dense cytoplasm. The outer three wall layers perform the function of protection and help iif dehiscence of anther to release the pollen. When the anther is young, a group of compactly arranged homogeneous cells called the sporogenous tissue occupies the centre of each microsporangium.

Question.20. If implementation of better techniques and new strategies are required to provide more efficient care and assistance to people, then why is there a statutary ban on amniocentesis ? Write the use of this technique and give reason to justify the ban.
Answer: Amniocentesis is a prenatal technique of diagnosing the genetic & metabolic disorders of the foetus by taking out a small quantity of amniotic fluid. Amniotic fluid contains foetal cells, placental cell, foetal enzymes, proteins & other biochemicals. Foetal cells give information about the sex of the foet us and any abnormality in the chromosomes. It the foet us suffers from in curable genetic & metabolic disorders, then the foetus needs to be aborted through MTP.
But this very useful technique has been misused to know the sex of the developing foetus & destroying the same if the foetus is female. Therefore, the test has been banned except it few centres & this ban is justified.

Question.21. Why is pedigree analysis done in the study of human genetics ? State the conclusions that can be drawn from it.
Answer : Pedigree analysis is the study of family history about the inheritance of a particular trait. It can be used to draw the inheritance of a specific trait, abnormality or disease in humans because control crosses are not possible in case of human being.
Conclusions:

1. Identification of the recessive or dominant nature of a specific trait could be done by pedigree analysis.
2.  The trait is linked to sex chromosome or autosomal can be find out by pedigree’ analysis, for example, haemophilia is a sex linked recessive disease. X-linked recessive trait shows transmission from carrier female to – male progeny.
3.  The pattern of inheritance of Mendelian disorders can be traced in a family by pedigree analysis. For example, most common Mendelian disorders are haemophilia, cystic
   fibrosis, sickle cell anaemia, colour blindness, phenylketonuria, thalassemia, myotonic dystrophy (autosomal dominant trait), etc.

Question.22. Identify ‘a’, ‘b’, ‘c’, ‘d’, ‘e’, and ‘f’ in the table given below:

Answer: a. (i) Palm is broad with characteristics palm crease; short statured with small round head.
(ii) Physical, mental, psychomotor development is retarded,
(b) Both (c) Klinefelter’s syndrome (d) Male
(e) (i) Short stature and underdeveloped feminine character,
(ii) Such females are sterile as ovaries are rudimentary. They also do not have well developed secondary sexual characters.
(f) Female

Question.23. Community service department of your school plans a visit to a slum area near the school with an objective to educate the slum dwellers with respect to health and hygiene.
(a) Why is there a need to organise such visits ?
(b) Write the steps you will highlight, as a member of this department, in your interaction with them to enable them to lead a healthy life.
Answer: (a) The well-being of each human being depends on their environment. In slum areas individuals live in congested and insanitary conditions. In this type of conditions they are more susceptible to suffer from diseased condition. So there is a need to organize such visits to educate them about the importance of health and hygiene.
(b) Steps to enable slum dwellers guide to healthy life

1.  Use of mosquito nets while sleeping, get wire mesh fixed to doors and windows, prevent water logging, regularly change water of water-coolers to avoid mosquito breeding.
2.  Wash hands before eating and after toilet use, maintain the environment clean so that flies do not breed. Disinfect water by chlorine tablets if it is drawn from well or any other source.
3. Clean toilets and use disinfectants regulary.
4. Educate people about the benefit of vaccine which are available at the health centres such as DPT for diphtheria, pertusis (whooping cough) and tetanus, polio vaccine. MMR vaccine for measles, mumps, rubella.

Question.24. The following graph shows the species —area relationship. Answer the following questions as directed.

(a) Name the naturalist who studied the kind of relationship shown in the graph. Write the observations made by him.
(b) Write the situations as discovered by the ecologists when the value of ‘Z’ (slope of the line) lies between
(i) 0.1 and 0.2 (ii) 0.6 and 1.2
What does ‘Z’ stands for ?
(c) When would the slope of the line ‘b’ become steeper ?
Answer : (a) Species area relationship was studied by Alexander Von Humboldt. He made a observation that within a region, species richness increased with increasing explored area but only upto a limit.
(b) (i) Z = 0.1 to 0.2: the stope of regression lines are similar the slope of regression is steepel when are analyse the species area relationship among very large areas like entire countinent.
(ii) Z = 0.6 to 1.2 : for large area for example entire continent.
(c) The slope of the line b become steeper when species area relationship is analyzed in a very large area like the entire continents.

Question.25. Name and describe the technique that helps in separating the DNA fragments formed by the use of restriction endonuclease.
Answer: Agarose gel electrophoresis is used to separate DNA * fragments formed by restriction endonuclease.
Agarose gel electrophoresis : The DNA cleavage by restric¬tion endonucleases which results in DNA fragments. Electrophoresis is a technique used to separate and sometimes purify nucleic acids that differ in size, charge or conformation. As such, it is one of the most widely-used techniques in biochemistry and molecular biology. When DNA molecules are placed in an electric field. DNA molecules are negatively charged due to their phosphate backbone, and migrate toward the anode. The DNA fragments separate (resolve) according to their size through sieving effect’ provided by the agarose gel. Hence, the smaller the fragment size, the further it moves. The separated DNA fragments can be visualized only after staining the DNA with a compound # known as ethidium bromide followed by exposure to UV radiation (pure DNA fragments cannot be seen in the visible light and without staining). The bands are cut from the gel and extracted by using a convenient technique. This step is called elution. The eluted DNA fragments are then purified and used in constructing recombinant DNA by joining them with cloning vectors.

Question.26.State the function of a reservoir in a nutrient cycle. Explain the simplified model of carbon cycle in nature.
Answer : Function of a reservoir : To meet with the deficit which occurs due to imbalance in ttie rate of influx and efflux of nutrients.

Question.27. Since the origin of life on 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.
Answer : (i) The current extinction “sixth extinction” rates are estimated to be 100 to 1000 times faster than in pre human times.
(ii) Human activities in ecosystem are mainly responsible for sixth extinction.
Main reason for this extinction is :
(a) Habitat loss and fragmentation, (b) Over exploitation
(c) Alien species introduction (d) Co extinction
(iii) Afforestation: Creation of sacred groves in which all the trees and wild life are venerated and given total protection.
(iv) By preventing habitat loss : Zoological park, botanical gardens, wildlife sanctuaries can also help to overcove such extinction.
(v) By the use of Diverse species.
(vi) By in-situ conservation & ex-situ conservation.

SECTION-D

Question.28. (a) Where does fertilization occur in humans ? Explain the events that occur during this process.
(b) A couple where both husband and wife are producing
functional gametes, but the wife is still unable to conceive, is seeking medical aid. Describe any one method that you can suggest to this couple to become happy parents.
Answer: (a) In humans fertilisation of male & female gamete occurs in the junction of ampulla and isthmus of fallopain tube. The various events which occur during the fusion of gametes are :
(A) Acrosomal reaction : As sperm comes in contact with the egg surface, it secretes/release enzyme Hyaluronidase which dissolves the corona radiata.
(ii) As sperm reaches Zona pellucida, the acrosome release Acrosin/zona lysin and dissolves zona pellucida.
(iii) Compatibility reaction also stimulates development of an outgrowth by the oocyte called Fertilisation cone. Egg cell has fertilizin protein and sperm has antifertilizing protein.
(B) Sperm Entry : Sperm head comes in contact with the fertilisation cone. Sperm & egg membrane dissolve at this point and components of head (nucleus), neck and middle piece of sperm enter the cytoplasm of egg. Tail of sperm is left out.
(C) Zona Reaction : The zona pellucida stiffens after ,, entry and does not allow any other sperm to enter. The phenomenon called Monospermy.
(D) Activation of oocyte to ovum : Egg is secondary ” oocyte stage undergo meiosis II by removal by MPF & development of APC/APF resulting in mature ovum/ ootid and second polar body.
(E) Karyogamy : It is the find stage of fertilisation. The sperm nucleus fuses with the egg nucleus. Nuclear envelops breakdown forming a spindle & thus form the zygote under laboratory condition. The zygote or early embryo is transferred in the fallopian tube.

(b) Couple able to produce functional gamete but unable to conceive can assist to have children through one of following techniques commonly “called as — Assisted Reproductive Technologies (ART).
In vitro fertilization followed by embryo transfer : Ova from the wife/donor and sperms from the husband/ donor male are collected and fused to form zygote under laboratory condition. The zygote or early embryo is transferred in the fallopian tube.
OR
(a) Explain the different ways apomictic seeds can develop.
Give an example of each.
(b) Mention one advantage of apomictic seeds to farmers.
(c) Draw a labelled mature stage of a dicotyledonous embryo.
Answer : (a) Different ways apomictic seeds development;
(i) The diploid egg cell is formed without reduction division and develops into an embryo without fertilization. Example : Grasses/Asteracease.
(ii) Nuclear cells surrounding the embryo sac start dividing and protrude into the embryo sac and develop into the embryo, for example, citrus and mango. They have more than one embryo in a seed known as polyembryony.
(b) Advantage of apomictic seeds to farmers :
As apomictic seed formation does not involves meiosis and fertilization, they are genetically identical to their parents.
If the hybrid seeds become apomictic they will maintain their traits generation after generation. As does not involves meipsis so lack of segregation of characters & not involves fertilization so no recombination and trait will be maintained for several generations, so the farmers can use these apomictic seeds to raise new crop year after year.
(c)

Question.29. (a) Describe the various steps of Griffiths experiment that led to the conclusion of the ‘Transforming Principle’.
(b) How did the chemical nature of the ‘Transforming Principle’ get established ?
OR
Describe how the lac operon operates, both in the presence and absence of inducer in E.coli.
Answer : (a) Transformation is change of genetic material of an organism by obtaining genes from other organism (dead relative). Fredrick Griffith a British bacteriologist in 1928, carried out experiments on ‘Transforming principle’. He worked with strains of streptococcus pneurnonical. These are two strains of this bacteria :
(i) Virulent or S. Strain, which produces smooth colony and has the capacity to cause the disease (pneurnonical).
(ii) Non-virulent or R-strain, which produces rough colony and does not cause pneumonia.
Griffith’s experiment was carried out as follows:

1. R-type (strain) of live bacteria injected in the mice No disease observed in mice.
   Mice + R-strain (live) ——-> No disease in mile
2.  Live S-strain of bacteria injected in the mice-mice has occurence of pneumonia & dies. .-
   Mice + S-strain (live) ——–> Disease seen & mice dies
3.  Heat-killed s-strain bacteria injected in mice.
   No disease observed. Mice survives
   Mice + s-strain (Heat-killed) No disease Mice Survives
4.  Mice injected with a mixture of heat killed s-strain of bacteria and live R-strain. Mice dies of pneumonia.
   Mice + s-strain (heat killed) ——–> Disease occurs + R-strain (live) Mice dies
   On observing the blood of mice, it showed presence of both R-strain & S-strain live bacteria. Occurrence of live s-strain was possible only th rough a change transformation of R-strain into s-strain through transfer of biochemical substance.

(b) Oswald TAvery, Collin Macleod & Maclyn McCarty in 1944 established the chemical nature of transforming principle.

1. The heat killed s-strain of bacteria and separated their’ components – DNA, proteins & Carbohydrates
2. The DNA component was segregated into two. One with hydrolysing enzyme DNA-ase & the other without it.
3. They then mixed these components of s-strain with live R-strain in Separate culture media.
4.  There was no change in three culture having additions of neat- killed s-stiain carbohydrates, proteins & DNA (with DNAase).
5. But the fourth culture medium having neat-killed s-strain DNA without DNAase showed presence oflive s-strain bacteria.
   It was concluded that the live S-strain bacteria must have been formed from R-strain with the help of DNA of S-strain. Thus, DNA is the genetic material was established.
   

OR
In E.Coli, the breakdown of lactose requires three enzymes. These enzymes are synthesized together in a coordinated manner & the unit is known as lac operon. Since the addition of lactose itself stimulates the production of lactose itself stimulates the production of required enzymes. It is also referred as Inducible system. It gets switched off in normal conditions.
The genes involved in lac operon are as follows :

1.  Strcutural Genes : These genes code for the proteins needed by the cell which include enzymes or other proteins having structural functions. Lac operon has three structural genes :
   (a) Lap z : Gene coding for enzyme b-galactosidase for splitting lactose into glucose & galactose.
   (b) Lacy y : Gene coding for enzyme permease/Galactoside permease which is required for entry of lactose.
   (c) Lac a ; Gene coding for enzyme Transacteylase/ Galactoside acetylase.
   The three structural genes of lac operon produce a single polycistronic mRNA.
2. Operator gene (O): It gives passage to RNA polymerase when the structural genes are to express themselves. Normally, it is covered by a repressor & is in off position.
3.  Promoter gene (p) : This, gene is the recognition centre/ initiation point for RNA polymerase of the operon.
4.  Regulator gene (i) : It is also called inhibitory gene. It produces a repressor protein that binds the operator gene, when the substrate (lactose) is not available, so as to keep it non-functional. It prevents the passage of RNA polymerase from promoter to structural gene.
5.  Repressor (p) : It is a small portion formed by regulator gene which binds to operator gene & blocks the passage of RNA polymerase towards the structural genes. It has two allosteric sites, one for attaching to operator gene & second for binding to the inducer.
6.  Inducer : It is a chemical which attaches to repressor, and changes the shape of operator binding sites so that the repressor rem ains no more attached to the operator. Mechanism of Lac Operon :
   (I) In the absence of induce (lactose) : The lac operon is generally off which is ensured by the formation of repressor by the regulator gene which blocks the operator gene. Thus, there is no transcription & no enzymes are produced. Operon is switched aff.
   (II) On the other hand, when inducer (lactose) is added, the repression protein (produced by gene i) gets bound is removed from the operator. RNA Polymerase is now allowed to act & the transcription of lac genes take place. The operon is now switched ON. All the three genes are transcribed to form a single mRNA strand. It is a polycistronic mRNA. This process continues till the inducer .is consumed. Once inducer finishes, the repressor again blind the operator gene & switches OFF the operon.
   

Question. 30. With advancements in genetics, molecular biology and tissue culture, new traits have been incorporated into crop plants.
Explain the main steps in breeding a new genetic variety of a crop.
OR
(a) State the objective of animal breeding.
(b) List the importance and limitations of inbreeding. How can the limitations be overcome ?
(c) Give an example of a new breed each of cattle and poultry.
Answer : Different steps in breeding a new crop variety.

1. Collection of variability: Genetic variability is essential for breeding program. If genetic variability is not present than new variety can not be develops thus it is pre requisite condition for breeding. The collection of all the different alleles for all genes in a given crop is called germplasm collection.
2.  Evaluation and selection of parent: Different germplasm is evaluated for desired trait and plants having the desired character are selected as parent. The selected plants are multiplied and pure lines obtained, which are used for hybridization.
3.  Cross hybridization among the selected plant : The selected plants are hybridized to combine the character of two different parents.
4.  Selection and testing of superior recombinants : On the basis of presence of desired character in hybrid, superior recombinants are selected. Plants are then self pollinated for several generations.
5. Testing, release and commercialization of new cultivars: These new recombinant are evaluated for their yield and different agro climatic condition (such as quality and disease resistance) for several years along with best available local check variety. If these lines are superior than local check then they are released for commercial cultivation. .

OR
(a) Objective of animal breeding: To increase the yield of animal & improving the desirable qualities of product.
(b) Importance of Inbreeding :

1.  Superior male & superior female of same breed are identified for mating.
2.  To evolve a pure line of animal.
3. Exposes harmful recessive gene that are eliminated by selection.
4.  Also accumulates superior genes & elimination of less desirable gene.
5.  Increases the productivity of inbreed population. Limitation of Inbreeding :
   Continued inbreeding specially closed inbreeding . usually reduces fertility & productivity. This is called as inbreeding depression.

(c) New breed of cattle —> Hisardale, and New breed of poultry —> Leghorn.

SET-II

SECTION-A

Question.1. Why is Gambusia introduced into drains and ponds ?
Answer: Gambusia introduced into drains and ponds because they feed on mosquito larvae.

Question.7. Why are analogous structures a result of conveigent evolution ?
Answer : Analogous structures are not anatomically similar though they perform similar functions so they are a result of convergent evolution.

Question.8. Name the vegetative propagules in the following:
(a) Agave (b) Bryophyllum
Answer : (a) Agave : Bulbils.
(b) Bryophyllum : nodes (leaves).

SECTION-B

Question.11. State the difference between the structural gene in a Transcription Unit of Prokaryotes and Eukaryotes.
Answer : Prokaryote structural genes consist of only exons (functional) while eukaryotes consists of both introns and exons. Introns are removed by the process of splicing before translation.
Prokaryotes are having polycistronic and continuous structural genes while eukaryotes have monocistronic and split.

13. Write the location and function of the following in human testes:
(a) Sertoli Cells (b) Leydig Cells 
Answer : (a) Sertoli Cells The sertoli cells are located within the seminiferous tubules. Their task is the creation of a hemato-testicular barrier and the nourishment of the spermatozoa.
(b) Leydig Cells: Leydig cells, also known as iriterstitial cells of Leydig, are found adjacent to the seminiferous tubules in the testicle. They produce testosterone in the presence of luteinizing hormone (LH).

SECTION – C

Question.21. A woman has certain queries as listed below, before starting with contraceptive pills. Answer them.
(a) What do contraceptive pills contain and how do they acts as contraceptive ?
(b) What schedule should be followed for taking these pills ?
Answer : (a) Contraceptive pills contains progesterone and estrogen combination. This disrupts hormone patterns needed for pregnancy and affects the ovaries and the development of the uterine lining, making pregnancy less likely. They prevent ovulation (the egg leaving the ovary and moving into the fallopian tube). They block the hormones
needed for the egg to be able to be fertilized. They may affect the lining of the uterus and thus alters sperm transport, i which prevents sperm from reaching the egg to fertilize it.
(b) The pills have to be taken daily for a period of 21 days starting from the fifth days of menstrual cycle to the 25 th day. After a gap of 7 days (during which menstruation occurs), it has to be repeated in the same pattern till the female desires to prevent conception.

Question.24. Two types of aquatic organisms in a lake show specific growth pattems as shown below, in a brief period of time. The lake is adjacent to an agricultural land extensively supplied with fertilizers.

Answer the questions based on the fact given above :
(i) Name the organisms depicting the pattern A and B.
(ii) State the reason for the growth pattern seen in A.
(iii) Write the effects of the growth patterns seen above.
Answer : (i) A—>Planktonic Algae (free floating); B—>Fish
(ii) The reason for the growth pattern in ATPresence of large amount of nutrients in fertilizers in water causes excessive growth of planktonic (free – floating) algae known as Algal Bloom which consumes a lot of Oxygen and nutrients. As a result there is a sharp decline in the dissolve oxygen in the lake.
(iii) The increase in BOD (Biochemical Oxygen Demand) due to algal bloom which causes deterioration of the water quality which results in fish mortality. Some bloom forming
algae are extremely toxic to human beings and animals also.

Question.26. Explain, giving three reasons, why tropics show greatest levels of species diversity.
Answer :

1. Tropical latitude have remained relatively undisturbed for million of years so they have greatest level of species diversity.
2.  Tropical environment are less seasonal, relatively more constant and predictable. Such constant environment promotes niche specialization and lead to a greater species diversity.
3. There is more solar energy available in the tropics which contributes to higher productivity, so indirectly contribute to greater diversity.

SECTION – D

Question.28. Describe the Hershey and chase experiment. Write the conclusion drawn by the scientists after their experiment.
Answer : Experiments by Hershey and Chase in the 1950’s using the bacteriophage T2 and E. coli cells demonstrated that DNA is the genetic material of the bacteriophage.

1. Hershey and Chase conducted their experiments on the T2 phage, a virus whose structure had recendy been shown by electron microscopy.
2.  The phage consists of a protein shell containing its genetic material. The phage infects a bacterium by attaching to its outer membrane and injecting its genetic material and leaving its empty shell attached to the bacterium.
3.  In their first set of experiments, Hershey andChase labeled the DNA of phages with radioactive Phosphorus-32 (the element phosphorus is present in DNA but not  present in any of the 20 amino acids from which proteins are made).
4. They allowed the phages to infect E. coli (Escherichia coli), observed that the transfer of P32 labeled phage DNA into the cytoplasm of the bacterium.
5.  In their second set of experiments, they labeled the phages with radioactive Sulfur-35 (Sulfur is present in the amino acids cysteine and methionine, but not in DNA).
6.  Following infection of E. coli they then sheared the viral protein shells off of infected cells using a high – speed blender and separated the cells and viral coats by using a centrifuge.
7.  After separation, the radioactive S35 tracer was observed , id the protein sheik, but not in the infected bacteria,
   supporting the hypothesis that the genetic material which infects the bacteria was DNA and not protein.

Conclusion :

1.  Hershey and chase concluded that DNA,not protein was the genetic material. They determined that a protective protein coat formed around the bacteriophage, but that the internal DNA is conferred its ability to produce progeny inside a bacteria.
2.  They showed that, in growth, protein has no function, while DNA has some function. Only 20% of the P32 (radioactive) remained outside the cell and it was incorporated with DNA in the cell’s genetic material. All of S35(radioactive) in the protein coats remained outside the cell, it was not incorporated into the cell, and protein was not the genetic material.
   

OR
“Work out a typical Mendelian dihybrid cross and state the law that he derived from it.
Mendelian Dihybrid Cross : a cross between two parents that differ by two pairs of alleles (AABBXaabb) and he derived it from law of independent assortment.

The phenotypes and general genotypes from this cross can be represented in the following manner :

SET III

SECTION-A

Question.2. Name the stage of cell division where segregation of an independent pair of chromosomes occurs.
Answer: The stage of cell division in which the segregation of an independent pair of chromosomes occurs Anaphase of meiosis I.

Question.3. Write an alternate source of protein for animal and human nutrition.
Answer: Single cell protein is an alternative source of protein for animal and human nutrition.

Question.4 Give an example of a plant which came into India as a contaminant and is a cause of pollen allergy.
Answer : Plant came into India as a contaminant and is a cause of pollen allergy is Parthenium.

SECTION-B

Question.16. Explain the two factors responsible for conferring stability to double helix structure of DNA.
Answer: Factors responsible for conferring stability to double helix structure of DNA:

1.  Presence of Hydrogen bond in between base pair stack * confers stability to DNA.
2. Presence of thymine at the plage of uracil gives more stability to DNA.

Question.18. Write the effect of the high concentration of L.H. on a mature Graafian follicle.
Answer : High levels of Luteinizing Hormone (LH) induces rupture of mature Graafian follicle and causes release of ovum known as ovulation.

SECTION-C

Question.24. (a) Explain adaptive radiation with the help of suitable example.
(b) Cite an example where more than one adaptive radiations have occurred in an isolated geographical area. Name the type of evolution your example depicts and state why it is so named.
Answer : (a) Adaptive radiation or divergent evolution: Different species are evolved in a given geographical area starting from the single point and literally radiating to other habitats in that area.
Ex : In the Australian region, marsupials each different from the other evolved from an ancestral stock, but all within the Australian island continent when more than one adaptive radiation appeared to have occurred in an isolated geographical area, one can call this convergent evolution.
(b) Convergent evolution : Convergent evolution is the process by which unrelated or distandy related organisms evolve similar body forms, coloration, organs, and adaptations. Natural selection can result in evolutionary convergence under several different circumstances. Species can converge in sympatry, as in mimicry complexes among insects, especially butterflies.
Ex : Marsupial fauna of Australia and the placental mammals of the Old World. The two lineages are clades—that is, they each share a common ancestor that belongs to their own group, and are more closely related to one another than to any other clade— but very similar forms evolved in each isolated population.

Question.25. (a) Name any two copper releasing IUDs.
(b) Explain how do they acts as effective contraceptives in human females.
Answer : (a) Two copper releasing IUDs are CuT, Cu7, Multiload 375.
(b) CuT is a method of intrauterine devices (IUTs). These devices are administered by the doctor in the uterus through vagina. The CuT is a copper releasing device which increases phagocytosis of sperms within the uterus and the cu ions released suppress sperm motility and fertilizing capacity of the sperm. So they acts as effective contraceptives in human females.

Question.27. (a) State how the constant internal environment is beneficial to organisms.
(b) Explain any two alternatives by which organisms can overcome stressful external conditions.
Answer : (a) The constant internal environment is beneficial to organisms because there is a continuous interaction between the organisms and the environment. An organism, fully adapted to environmental conditions in which they survive, grows and reproduces. The environment can be defined as the total of all physical and biotic conditions which influence the responses of organism.
(b) The two alternatives by which organisms can overcome stressful external conditions are :
(i) Direct Factors (ii) Indirect Factors
(i) , Direct factors : These factors influence the organisms
directly e.g., light, temperature, humidity and soil nutrients etc.
(ii) , Indirect factors: These factors affect organisms indirectly
by modifying other factors e.g., soil organisms, wind etc.

SECTION-D

Question.28. Explain the process of sewage water treatment before it can be discharged into natural water bodies. Why is this treatment essential ?
OR
Explain the process of replication of a retrovirus after it gains entry into the human body.
Answer : There are agronomic and economic benefits of wastewater used in agriculture. Irrigation with wastewater can increase the available water supply or release better quality supplies for alternative uses.
Sewage treatment generally involves three stages, called primary, secondary and tertiary treatment.

1.  Primary Treatment : In primary treatment, the incoming flow is slowed in large tanks which allow the dirt, gravel, and other heavier components of the waste stream to settle out. Grease, oil, and other floatables are also removed here. Rotating arms simultaneously remove the settled solids from the bottom and the separated floatables from the top. Both pollutants are pumped into large heated holding silos, called digesters.
2.  Secondary Treatment: This treatment removes dissolved and suspended biological matter. Secondary treatment is typically performed by indigenous, water-borne micro-organisms in a managed habitat. It may require a separation process to remove the micro-organisms from the treated water prior to discharge or tertiary treatment.
3.  Tertiary Treatment : It is sometimes disinfected chemically or physically (for example, by lagoons and microfiltration) prior to discharge into a stream, river or it can be used for the irrigation of a green way or park. If it is sufficiently clean, it can also be used for groundwater recharge or agricultural purposes.
   This treatment is essential because the sewage water contains large amount of pathogenic microbes, organic matters.

OR
HIV multiplies in human body first in macrophages during inhibition period & later in Helper T-cells during which symptoms of AID oppear.
(I) Cycle in Macrophages :

1.  After gaining entry into the human body, the HIV passes to all parts through blood & other body fluids.
2.  When it comes in contact with macrophage, the gp 120 spilce of virus binds with CD4 receptor of the macrophage.
3.  A conformational change aids the virus to attach to another co-receptor called CCR5.
4.  This triggers change in cell membrane of the macrophage which then endocytose HIV.
5.  Once inside, it sheds the protective cover is shed. This frees the RNA along with reverse transcriptase in the cytoplasm of macrophage.
6.  It synthnesizes copy of DNA from which a complement DNA is produced.
7. The double stranded DNA attaches to host DNA in the form of provirus. It then directs the host cell machinery to form genomic RNA & mRNA.
8.  mRNA synthesizes viral proteins including reverse transcriptase. Genomic RNA & viral proteins are packet together to form the virus. In this way, several copies of virus are formed.
9.  These viruses then bud out of the macrophages by the process of exocytosis. These then invade new macrophages to further replicate.

The HIV undergoes similar cycle of replication in the Helper T-cells.

1. The virus first attaches to CD4 receptor hy it gp 120. The complex then comes in contact with conceptor GXCR4.
2.  The virus then passes into the cytoplasm ofT-lymphocytes through endocytosis.
3. Inside the cytoplasm of T-cell, the virus coat is shed the naked RNA alonwith copy DNA & then complement DNA, which then gets integrated to host DNA as provirus.
4.  The provirus directs the synthesis of two types of RNA- genomic & mRNA. mRNA forms vital proteins (including reverse transcriptase). Genomic RNA &viral proteins are packed together to form the virion.
5. The virion comes in contactwith the surface oflymphocyte, raptures its cell membrane & come out. These further infect healthy cells.
6.  Thus, number of T-cells decline, compromising the immune system of the body.
   

CBSE previous Year Solved  Papers  Class 12 Biology Delhi 2016

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5.  Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SECTION-A

Question.1. According to de-vries what is saltation ?
Answer : Saltation is single step, large mutation.

Question.2. Excessive nutrients in a fresh water body cause fish mortality. Give two reasons.
Answer: Excessive nutrients result in excessive algal growth which produce toxins in water. Water quality becomes poor as Dissolved Oxygen decreases leading to increase in BOD.

Question.3. Suggest the breeding method most suitable for animals that are below average in milk productivity.
Answer : To improve productivity any one of the following methods can be followed :
Outbreeding/Outcrossing/Cross-breeding/artificial insemination/hybridisation etc.

Question.4. State a difference between a gene and an allele.
Answer : Gene : It contains information that is required to express a particular trait.
Allele : Genes which code for a pair of contrasting traits.

Question.5. Suggest a technique to a researcher who needs to separate fragments of DNA.
Answer : Gel electrophoresis.

SECTION—B

Question.6. Explain the significance of meiocytes in a diploid organism.
Answer : (i) Meiocytes undergo meiosis or gametogenesis to produce haploid gametes.
(ii) They help to restore diploidy through zygote formation or syngamy.

Question.7. Mention the kind of biodiversity of more than a thousand varieties of mangoes in India represent. How is it possible ?
Answer : Varieties of mango show genetic diversity. Single species of mango show high diversity at genetic level over its distributional range. .

Question.8. List the events that reduce the Biological Oxygen Demand (BOD) of a primary effluent during sewage treatment.
Answer : (i) Effluent from primary settling tank is passed into aeration tank agitated mechanically and air is pumped into it.
(ii) This allows vigorous growth of aerobic microbes into floes which consume major part of organic matter in the effluent.

Question.9. Discuss the role the enzyme DNA ligase plays during DNA replication.
Answer : (i) Discontinuous DNA fragments are joined or sealed by DNA ligase.
(ii) DNA ligase adds on nucleotide in the usual 5’ to 3’ direction along the DNA strand.

Question.10. Name the causative organism of the disease amoebiasis. List three symptoms of the disease.
OR
Identify ‘A’, ‘B’, ‘C’and ‘D’ in the given table.

Answer: Amoebiasis is caused by Entamoeba histolytica.
The symptoms of this disease are : constipation, abdominal pain, cramps, stools with excess mucous and blood clots.
OR
A—Wheat
B—Black rot/Curl blight black rot
C—White rust
D—Pusa Komal

SECTION—C

Question.11. Why is breast-feeding recommended during the initial period of an infant’s growth ? Give reasons.
Answer: During initial period of infant’s growth, colostrum is produced. It is rich in nutrients. It is also rich in antibodies (IgA) which provide passive immunity to thfc new born.

Question.12. Give an example of an autosomal recessive trait in humans. Explain its pattern of inheritance with the help of a cross.
Answer : Sickle cell anaemia is an autosomal recessive trait disease than can be transmitted from parents to the off spring when both the partners are carrier for the gene. The disease is controlled by a single pair of allele, HbA and HBS. Out of three possible genotypes only homo2ygous individuals for Hbs (Hbs Hbs) show the diseased phenotype white heterozygous (HbAHbs) individuals are carrier of the disease.

Question.13. Describe the experiment that helped Louis Pasteur to dismiss the theory of spontaneous generation of life.
Answer : Leuis Pasteur took two pre-sterilised flasks with killed yeast. One flask was sealed while the other was kept open to air. Differential growth of life was observed in the flasks-life was found only in the open flask’. It proved that life comes from pre-existing life (theory of biogenesis).

Question.14. Plant breeding technique has helped sugar industry in North India. Explain how.
Answer: Saccharum barberi was originally grown in North India, but had poor sugar content and yield. Sugar cane grown in South India, Saccharum officinarum had thicker stems and higher sugar content but did not grow well in North India. The two species were crossed to get desirable qualities of high yield, thick stems, high sugar and ability to grow in N orth India.

Question.15. Suggest and describe a technique to obtain multiple copies of a gene of interest in vitro.
Answer : PCR : Polymerase Chain Reaction.
Multiple copies of the gene of interest is synthesised in vitro using two sets of primers and enzyme DNA polymerase. The enzyme extends the primers using nucleotides provided and genomic DNA as template. The process of DNA replication is repeated several times for amplification of DNA with the help of thermostable DNA polymerase which remains active during high temperature induced denaturation of double stranded DNA.

Question.16. What is a GMO ? List any five possible advantages of GMO to a farmer.
Answer : Those plants, bacteria, fungi or animals whose genes have been altered by manipulation are called Genetically Modified Organisms (GMOs).
Advantages :

1. Tolerance to abiotic stresses like cold, drought, salt, heat etc.
2.  Reduce reliance on chemical pesticides.
3.  Reduced post harvest losses.
4. Increased efficiency of mineral usage by plants.
5. Enhanced nutritional value.
6. To create tailor made plants.

Question.17. During a school trip to ‘Rohtang Pass’, one of your classmate suddenly developed ‘altitude sickness’. But, she recovered after sometime.
(a) Mention one symptom to diagnose the sickness.
(b) What caused the sickness ?
(c) How could she recover by herself after some time ?
Answer : (a) The symptoms may be nausea, fatigue or heart palpitation.
(b) The sickness was caused due to low atmospheric pressure which prevails at high altitude. The body does not get enough oxygen.
(c) The body compensates low oxygen availability by
increasing RBC production, decreasing the binding affinity of haemoglobin and by increasing breathing rate.

Question.18. How has RNA technique helped to prevent the infestation of roots in tobacco plants by a nematode ?
Answer: Using Agrobacterium vectors, nematode specific genes were introduced into the host plant. This DNA produced both sense and anti sense RNA in the host cells. These two RNAs being complementary to each other formed a double strand (dsRNA) that initiated RNAi and thus silenced the specific mRNA of the nematode. Hence the parasite could not survive in the transgenic host.

Question.19. “In a food-chain, a trophic level represents a functional level, not a species.” Explain.
OR
(a) Name any two places where it is essential to install electrostatic precipitators. Why it is required to do so ?
(b) Mention one limitation of the electrostatic precipitator.
Answer : Position of a species in any trophic level is determined by the function performed by that mode of nutrition of species in a particular food chain. A given species may occupy more than one trophic level in the same ecosystem at a given time. If the function of the mode of nutrition of species changes, its position shall change in the trophic levels. The same species can be at the primary level of consumer in one food chain and at the secondary consumer level in another food chain in the same ecosystem at a given time.
OR
(a) Electrostatic precipitators can be installed in thermal power plants, smelters or other particulate matter releasing industries. They are important for removing particulate matter.
(b) Limitations :

1. Very, very small particulate matter which are less than 2.5 micrometres are not removed.
2.  The velocity of air between the plates must be low enough to allow the dust to fall.
3.  It cannot work without electricity.
   (Any one can be mentioned)

Question.20. Prior to a sports event blood & urine samples of sports persons are collected for drug tests.
(a) Why is there a need to conduct such tests ?
(b) Name the drugs the authorities usually look for.
(c) Write the generic names of two plants from which these drugs are obtained.
Answer : (a) To detect drug abuse or use of banned drugs cannabinoids, narcotic analgesic, diuretics, hormones or drugs used to accelerate performance, increase muscle strength etc.
(b) Cannabinoids/cocaine/coka alkaloid/coke/crack/
hashish/charas/ganja etc. ’
(c) Cannabis/Atropa/Erythoxylem/Datura etc.

Question.21. Describe the experiment that helped demonstrate the semi-conservative mode of DNA replication.
Answer.

Heavy Hybrid Light Hybrid
Meselson and Stahl grew E. coli in a medium containing\( ^{ 15 }{ N{ H }_{ 4 }Cl }\) (\(N^{ 15 }\) is the heavy isotopes of nitrogen) for many generations to get\( N^{ 15 }\) incorporated into DNA. Then the cells were transferred into \( ^{ 15 }{ N{ H }_{ 4 }Cl }\). The extracted DNA was centrifuged in a CsCl density gradients to measure the densities of DNA. The DNA extracted from the culture after one generation (20 minutes) showed intermediate or hybrid density. The DNA extracted after two generations (40 minutes) showed equal amounts of hybrid and of ‘light’ DNA.

Question.22. Given below is a list of six micro-organisms. State their usefulness to humans.
(a) Nucleopolyhedrovirus
(b) Saccharomyces cerevisiae
(c) Monascus purpureus
(d) Trichoderma polysporum
(e) Penicillium notation
(f) Propionibacterium sharmanii
Answer : (a) As bio control agents for Integrated Pest Management.
(b) It is used in bread making/brewing industry or for production of ethanol.
(c) It is a cholesterol lowering agent.
(d) It produces Cyclosporin A which is an immuno-suppressive agent.
(e) It produces antibiotic penicillin.
(f) It produces large holes in swiss cheese by releasing large amount of CO2.

SECTION—D
Question.23. Reproductive and Child Healthcare (RCH)
programmes are currently in operation. One of the major tasks of these programmes is to create awareness amongst people about the wide range of reproduction related aspects. As this is important and essential for building a reproductively healthy society.
(a) “Providing sex education in schools is one of the ways to meet this goal.” Give four points in support of your opinion regarding this statement.
(b) List any two ‘indicators’ that indicate a reproductively healthy society.
Answer : (a) It is a means of providing right information to the young so as to discourage children from believing in myths and misconceptions about sex related aspects.
Knowledge is also imparted about reproductive organs, adolescence and related changes, safe hygienic practices, STD/AIDS, available birth control options, care of pregnant mothers, post-natal care, importance of breast feeding, sex abuse and sex related crimes. .
(b)

1.  Decreasae in IMR (Infant Mortality Rate), MMR (Maternal Mortality Rate).
2.  Increase in number of couples with small families, better detection and cure of STDs.
3.  Total well being in all aspects of reproduction, normal emotional and behavioural interaction among all sex related aspects.

SECTION—E
Question.24. (a) Explain the post-pollination events leading to seed production in angiospersms.
(b) List the different types of pollination depending upon the source of pollen grain.
OR
(a) Briefly explain the events of fertilization and implantation in an adult human female.
(b) Comment on the role of placenta as an endocrine gland.
Answer : (a) As a result of pollen-pistil interaction, germination of pollen tube takes’ place carrying two male gametes. One male gamete fuses with the egg cell (syngamy), while the other fuses with two polar nuclei to form primary endosperm nucleus (PEN). The zygote develops into an embryo while the PEN develops to form endosperm. After double fertilisation, the ovule matures into a seed while the ovary matures into a fruit.
(b) Different types of pollination depending upon the source of pollen grain are :

1.  Autogamy: Transfer of pollen grains from the anther to the stigma of the same flower.
2.  Geitonogamy : Transfer of pollen grain from the anther to the stigma of another flower of the same plant.
3.  Xenogamy : Only types of pollination which brings genetically different types of pollen grains to the stigma.

OR
(a) Fertilization : A sperm comes in contact with the zona pellucida layer ovum and induces changes to block entry of additional sperms. The entry of sperm induces completion of meiosis II leading to the formation of anootid and second polar body. The haploid nucleus of the sperm and that of the ovum fuse to form a dipolid zygote.
Implantation : The trophoblast layer of the blastocyst attaches to the endometrium of the uterus. The uterine
cells divide rapidly and cover the blastocyst which becomes embedded in the endometrium and implantation is completed.
(b) Placenta acts as an endocrine tissue and produces several hormones like :

1.  human chorionic gonadotropin (hCG)
2. human placental lactogen (hPL)
3.  Estrogens, progestogens, etc.

Question.25 . (a) How are the following formed and involved in DNA packaging in a nucleus of a cell ?
(i) Histone octomer
(ii) Nucleosomes
(iii) Chromatin
(b) Differentiate between Euchromatin and Hetero-chromatin.
OR
Explain the role of lactose as an inducer in a facoperon.
Answer : (a) (i) Eight molecules of positively charged basic proteins called histones are organised to form histone octomer.
(ii) Negatively charged DNA is wrapped around positively charged histone octomer to give rise to a nucleosome.
(iii) Nucleosomes constitute repeating unit of a structure:

Lactose is the substrate for the enzyme beta galactosidase and it regulates switching ON and OFF of the operon. In the presence of an inducer such as lactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase access to the promoter and transcription proceeds.

Question.26. (a) Why should we conserve biodiversity ? How can we do it ?
(b) Explain the importance of biodiversity hot-spots and sacred groves.
OR
(a) Represent diagrammatically three kinds of age pyramids for human populations.
(b) How does an age pyramid for human population
at given point of time helps the policy-makers in planning for future.
Answer: (a) (i) Narrowly utilitarian : We derive economic benefits from nature food (cereals/pulses/fruits). We also get firewood, fibre, construction material, industrial products (tannins, lubricants, dyes, resins, perfumes), products of medicinal importance etc.
Broadly utilitarian : We get 20% of the total O2 from the Amazon rain forests. Pollination is also achieved. We derive several aesthetic pleasures from them.
Ethical Argument : Millions of species of plants, animals and microbes share this planet with us. We need to realise that every species has an intrinsic value. We have a moral duty to care for their well being and pass on our biological legacy to future generations.
(ii) In situ conservation in biosphere reserves, national parks, sanctuaries, sacred groves etc.
Ex situe conservation in zoological parks, botanical gardens, safari parks, cryoprese-rvation, seed banks, tissue culture etc.
(b) Three of these hotspots are Western Ghats and Sri lanka, Indo-Burma and Himalaya-cover our country’s exceptionally high biodiversity regions.
Sacred Groves : They are tracts of forests containing wild life which are venerated and given total protection. Such sacred groves are found in Khasi and Jaintia Hills in Meghalaya, Aravalli hills of Rajasthan, Chand and Baster areas of Madhya Pradesh. In Meghalaya, the sacred groves are the last refuges for Meghalaya, for a large number of rare and threatened plants.
OR
(a)
Post-reproductive Reproductive Pre-reproductive Expanding
(b) Age pyramid analysis of a population helps in planning and health, education, transport, infrastructure, finance, food or employment.

SET-II

SECTION—A

Question.3. Give an example of a human disorder that is caused due to a single gene mutation.
Answer : Sickle cell anaemia/Thalassemia/Phenylketonuria

SECTION—B

Question.8. Explain the importance of syngamy and meiosis in a sexual life cycle of an organism.
Answer : Syngamy : It ensures restoration of diploid chromosome number through zygote formation. Variations are an important characteristic of this process.
Meiosis : Gamete formation takes place as a result of meiosis which involves reduction in chromosome number or haploidy. It also leads to variations (due to crossing over).

Question.9. List the events that lead to biogas production from waste water whose BOD has been reduced significantly.
Answer : After significant reduction of BOD, the effluent is passed into a settling tank where the floes are allowed to sediment to form activated sludge. This sludge is pumped into anaerobic sludge digesters where anaerobic bacteria digest the microbes of the sludge to release a mixture of gases such as methane, H2S and CO2. These gases form biogas which can be used as a source of energy.
10. Why the plants that inhabit a desert are not found in a mangrove ? Give reasons.
Answer : Desert plants are not adapted to survive in saline or aquatic conditions prevailing in a mangrove. Plants are conformers. They are also stenothermal. They cannot maintain constant internal environment. The osmotic concentration of their body fluids affect the kinetics of enzymes through basal metabolism.

SECTION—C

Question.12. Differentiate between somaclones and somatic hybrids. Give one example of each.
Answer : Somaclones are produced through micro¬’ propagation or tissue culture. They are genetically identical, e.g., apple, tomato or banana.
Somatic hybrids are produced by fusion of protoplast of two different plants. They are genetically dissimilar e.g., Pomato (hybrid of potato and tomato).

Question.17. A couple with normal vision bear a colour blind child. Work out a cross to show how it is possible and mention the sex of the affected child.
Answer :

The affected child is male.

Question.19. In certain seasons we sweat profusely while in some other season we shiver. Explain.
Answer : Mammals are able to maintain homeostasis means which ensure constant body temperature.

1.  In summer, the outside temperature is higher than the body temperature. Hence sweating causes cooling by evaporation of sweat.
2. In winter the outside temperature is lower than the body temperature. Hence shivering is an involuntary exercise which produces heat.
3.  Both the above exercises help to regulate our body temperature.

SECTION—E

Question.26. List the criteria a molecule that cart act as genetic material must fulfill. Which one of the criteria are best fulfilled by DNA or by RNA thus making one of them a better genetic material than the other ? Explain.
OR
(a) Differentiate between analogy and homology giving one example each of plant and animal respectively.
(b) How are they considered as an evidence in support of evolution ?
Answer : (i) The genetic material should be able to carry out replication or generate a replica.
(ii) It should be chemically or structurally stable.
(iii) It should provide scope for slow mutation.
(iv) It should be able to express itself as characters.
Out of the two,clearly, DNA is more stable because of the following factors :
• Presence of H and not OH at 2’ position.
• Presence of thiamine instead of uracil.
• It is less reactive.
• It is structurally more stable because of its double stranded structure with hydrogen bonding.
• DNA is slower to mutate than RNA.
• Complementary strands of DNA further resist changes by evolving a process of repair.
OR
(a) Homology : Those structures which have similar origin but perform different functions show homology.
e.g., Forelimbs of mammals, heart of vestebrates, brain of vertebrates etc.
Thorns of bougainvilleas and tendrils of cucurbits. Analogy : Those structure which have a different origin, but perform similar functions show analogy, e.g., Wings of bat and birds, flippers of penguin and dolphin, eye of octopus and mammals etc.
Sweet potato and potato tuber.
(b) Homology shows common ancestry and divergent evolution.
Analogy does not show common ancestry. It shows convergent evolution.

SET-III

SECTION — A

Question.5. Give an example of a codon having dual functional]
Answer: AUG codes for methionine and also act as initiator codon.

SECTION — B

Question.7. Distinguish between the roles of flocks and anaerobic sludge digesters jn sewage treatments.
Answer.

Question.9. Plants that inhabit a rain-forest ‘are not found in a wetland. Explain.
Answer : Plant inhabiting a rain forest are not adapted to survive in aquatic conditions or wetlands. Plants are conformers. They are stenothermal. They cannot maintain a constant internal environment or temperature. The osmotic concentration of their body fluids affects the kinetics of enzymes through basal metabolic activity.

Question.10. Angiosperms bearing unisexual flowers are said to be either monoecious or dioecious. Explain with the help of one example each.
Answer : Monoecious : Plants bear both male and female unisexual flowers on the same plant. e.g., cucurbits, coconut, maize etc.
Dioecious : Plants bear Cither male or female unisexual flowers on different plants, e.g. papaya, date palms etc.

SECTION—C

Question.13. (a) Name any two fowls other than chicken reared in a poultry farm.
(b) Enlist four important components of poultry farm management.
Answer : (a) Ducks, turkey, geese etc.
(b)

1. Selection of disease free and suitable breeds.
2.  Proper and safe farm conditions.
3.  Proper food and water.
4. Maintenance of hygiene and health care.

Question.18. Explain with the help of suitable examples the three different ways by which organisms overcome their stressful conditions lasting for short duration.
Answer. Three different ways are :

1.  Migration: Organisms can move away temporarily from stressful habitat to a more hospitable area and return when stressful period is over. e.g. humans moving from Delhi to Shimla during summer.
2.  Spore Formation : Various kinds of thick walled spores are formed which germinate on availability of suitable environment, e.g., bacteria, fungi etc.
3.  Dormancy: Seeds or Vegetative reproductive structures help to tide over stress by reducing their metabolic activity e.g., seeds or vegetative reproductive structures of higher plants.
4.  Hibernation : It takes place during winter e.g., bear.
5. Aestivation : It takes place during summer to avoid heat and dessication e.g., Snails/fish etc.
6.  Diapause : Under unfavourable conditions, zooplanktons enter a stage of suspended metablic activity e.g., zooplanktons.
   (Note : Mention any three from the above)

Question.22. How would you find genotype of a tall pea plant bearing white flowers ? Explain with the help of a cross. Name the type of cross you would use.
Answer : Test cross should be used.

SECTION—E

Question.24. Answer the following questions based on Hershey and Chases’s experiments 
(a) Name the kind of virus they worked with and why ?
(b) Why did they use two types of culture media to grow viruses in ? Explain.
(c) What was the need for using a blender and later a centrifuge during their experiments ?
(d) State the conclusion drawn by them after the experiments.
OR
(a) How did Darwin explain adaptive radiation ? Give another example exhibiting adaptive radiation.
(b) Name the scientist who influenced Darwin and how ?
Answer : (a) They worked with bacteriophage which infect bacteria because they want to discover whether it was protein or DNA from the viruses that entered the bacteria.
(b) They used two types of culture media in order to make protein of viruses radioactive with the help of 35S in one case, and DNA molecule in virus radioactive by using 32P in the other case. This was done to identify which one of the two had entered into the bacteria during viral infection.
(c) Blender was used to separate viral protein coats that – were still attached to the surface of bacteria.
Centrifuge was used to separate lighter supernatent containing viral protein coats from denser residue containing bacteria.
(d) They concluded that DNA is the genetic material that is passed from virus to bacteria.
OR
(a) Darwin observed that from original seed eating features in finches, altered breaks arose enabling them to become insectivorous and vegetarian finches.
Adaptive Radiation : Is the process of evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography.
e.g., Australian marsupials, placental mammals in Australia.
(b) Thomas Malthus. Population size grows exponentially. However population size remains limited due to limited natural resources leading to competition.

CBSE Class 10 Science Practical Skills – Introduction

Theoretical knowledge supplemented by practical knowledge, helps the student anchor the science concepts firmly into their mind. It is this that makes lab experiments an integral part of school curriculum and laboratories, the primary workshops where theoretical aspects are verified.

General Lab Instructions 

1. Always carry practical record book, laboratory manual, notebook (to record observations), pen, pencil, scale, eraser, sharpener, scissors, forceps, scalpel, dissecting needle, brush, blades, etc. to the laboratory.
2. Maintain discipline and silence in the lab.
3. Keep your seat neat and tidy.
4. Follow the instructions given by the teacher attentively.
5. Perform the experiment carefully and note down the observations.
6. Handle the lab apparatus and glassware with care.
7. Do not manipulate your results in any circumstances.
8. Use newspaper or blotting paper while using stains so as not to spoil the table.
9. Draw the diagrams neatly and label them correctly. Use fine HB pencil while drawing.
10. Do not play, whistle or stoop into your partner’s set-up.

Maintaining Records
Since specified marks are reserved for the presentation of the record file in the Annual Practical Examination, it is important to maintain neatness and clarity while recording instructions and observations.

How to record:

Plotting A Graph  
All experiments should include measurements. In all measurements, data can be presented in tabular and/or graphical form.
Tabular form is presented in the following forms:

1. headed columns,
2. numbered rows, and
3. mixed format.

To make the dependence between two quantities more visible or to visualise easily, results are presented in graphical form.
Graph is the most convenient way to show the dependence between the quantities clearly. It is also a quick way to make an average set of observations. Therefore, a graph is a straight line or a curve showing the variation of two variable quantities (or their powers or functions) of which one varies due to change in the other. It enables us to

• determine the value of a quantity which is not observed during the experiment.
• deduce the mathematical relationships between the two quantities which were earlier not known.
• verify laws like Ohm’s law, Boyle’s law, etc.
• perceive the error in the result at a glance.
• calibrate certain instruments, such as ammeter, voltmeter and to determine their true readings.

To plot a graph between two variables (independent and dependent variables) following seven steps are followed :

1. Select the title
2. Draw two axes with proper origin
3. Label of both the axes
4. Mention proper units of the quantities on each axis,
5. Select and mention the scale used on each axis
6. Plot of points with crosses or dots and circles,
7. Draw of best fit straight line or smooth curve.

In plotting the graph between dependent and independent variables, the dependent variable should be plotted as ordinate on the y-axis e.g., in the verification of Boyle’s law where volume is measured for various values of pressure. Here, pressure is independent variable and volume is the dependent variable.
The scale used should be convenient for calculation work and occupy a wide sweep of the space available. If scale chosen will be too large, it will tend to highlight the errors of observation and hide the relationship between the variables taken along abscissa (x-axis) and ordinate (y-axis). Each point on the graph shows an actual observation but departure of the point from the final curve is a measure of experimental error in that observation.
Whenever we have to get some information, it should be obtained from a straight line graph because it is more accurately drawn and deductions from such graph are more reliable than the curved graphs. If the relationship between two variables is not a linear relationship, try to obtain a linear relationship by plotting powers of one or other or both of the quantities e.g., in a simple pendulum experiment a plot of time period T against length of the pendulum L is a parabolic graph while T² versus L is a straight line.

Determination of the slope of a line:
To calculate the slope, take two points on the line well apart. Read the values on x-axis and y-axis respectively at these two points drawing the abscissa and ordinate clearly.

Determination of the slope of a curve:
To determine the slope of a curve at a point P on the curve, a tangent is drawn at that point. Take a strip of plane mirror and place it edgewise such that its reflecting surface MN is over the point P. See the image of the portion PQ of the curve in the mirror and adjust its position by slightly turning about the point P such that the image of PQ coincides with the portion PR of the curve. This will happen when the mirror is exactly normal to the curve at point P. Mark a straight line MPN along the edge of the mirror. Now draw a line MPN’ perpendicular to MPN at point P with the help of a protractor. This line MPN’ is tangent to the curve at point P. Now, draw a perpendicular N’L from N’ on the x-axis. From the AM’N’L

Similarly, if you take different points on the curve, you will observe that the inclination or slope of different parts of the curve is different.

Handling A Microscope  

1. Clean the lenses and metal parts including stage with tissue paper or muslin cloth.
2. Do not remove objective lenses from nosepiece.
3. Be careful so that objective lens should not touch the slide in any case. While adjusting, do not move the lens downward, it should always be moved from bottom towards upward direction.
4. Keep the microscope covered when not in use.
5. Always hold the microscope with both the hands in an upright position while carrying it.
6. First focus the slide in low power and then only change to high power.
7. Do not use coarse adjustment screw while using high power objective lens.
8. Keep the microscope in upright position, do not move the stage in a slanting position.

Lab Safety 

General Guidelines

1. Act responsibly at all times in the laboratory.
2. Follow all written and verbal instructions carefully. In case you do not understand a direction or any part of the experiment, ask your teacher before proceeding.
3. Never work in the laboratory without the presence of the teacher.
4. While entering a science room, do not touch any equipment, chemical or other material in the laboratory area until you are instructed to do so.
5. Perform only those experiments which are authorized by your teacher.
6. Do not use laboratory glassware as containers for food or beverages.
7. Always work in a well-ventilated area.
8. Keep the work area clean and tidy at all times.
9. Be alert. Notify the teacher immediately of any unsafe conditions you observe.
10. Dispose all chemical waste properly. Never mix chemicals in sink drains. Sinks are to be used only for water. Check with your teacher for disposal of chemicals and solutions.
11. Set up and use the equipments as directed by your teacher.
12. Keep hands away from face, eyes, mouth, and body while using chemicals or any lab equipment. Wash your hands with soap and water after performing all experiments.
13. Experiments must be performed at all times.
14. Know the locations and operating procedures of all safety equipments including: first aid kit(s) and fire extinguisher. Know where the fire alarm and the exits are located.
15. Whenever you use chemical, heat substances or glassware, wear safety goggles.
16. Contact lenses should never be worn in the laboratory.
17. Dress properly during a laboratory activity. Long hair, dangling jewellery and loose or baggy clothing are a hazard in the laboratory.
18. A lab coat or smock should be worn during laboratory experiments.
19. If a chemical splashes in your eye(s) or on your skin, immediately flush with running water for at least 20 minutes.
20. All chemicals in the laboratory are to be considered dangerous. Avoid handling chemicals with fingers. Always use a tweezer. When making an observation, keep at least 1 foot away from the specimen.
21. Do not taste or smell any chemicals.
22. Check the label on all chemical bottles twice before taking any of the contents. Take only as much chemical as you need.
23. Never put back the unused chemicals to their original container.
24. Never remove chemicals or other materials from the laboratory area.
25. Examine glassware before each time you use. Never use chipped, cracked or dirty glassware.
26. Do not immerse hot glassware in cold water. The glassware may shatter.
27. Do not place hot apparatus directly on the laboratory desk. Always use an insulated pad.
28. Heated glassware remains very hot for a long time. They should be set aside in a designated place to cool, and picked up with caution. Use tongs or heat protective gloves if necessary.
29. Never look into a container that is being heated.

Lab Safety Symbols
Be it a middle school science lab or an advanced genetic research laboratory, the lab safety symbols are mandatory. Students should understand them and do the needful, so as to maintain a safe place to work in.

Poison Safety: The universal cross-bone symbol indicates danger and to follow safety tips. The poison symbol is same as the danger sign, which is used for labelling a poisonous substance in a bottle, test tube or package.

Eye Safety: The eye safety symbol is very easy to identify. If you notice a goggle sign outside a lab, it indicates the necessity to wear safety glasses before entering the lab. Putting safety goggles is a part of chemistry lab safety rules.

Fire Safety: As the name suggests, fire safety symbol is put in the lab so that students take care of themselves when they are around flames. This personal safety symbol is a flame without any signs attached to it.

Sharp Object Safety: Using needles, scalpels or other sharp instruments is quite common in biology experiments. The sharp object safety symbol is a hand with a cut in the forefinger. This is to alert the students while using sharp objects.

Biohazard Alert: Biological hazard refers to dangers from living organisms. Appeared in biology and microbiology labs, the biological hazard symbol is to make people aware about the harmful effects of bacteria and alike pathogens used in practical.

Radioactive Safety: Radioactive and radiation symbols are nearly similar to each other. These symbols appear as if some rays are radiated from the centre to the peripheral parts. Often, they are put with bright colours, so that people can see from a distance.

Thermal Safety: Heating and boiling are common steps for science practical. The thermal safety symbol is very simple and resembles a hand glove. It is used to remind students to take precaution while handling a hot object.

Open Flame Alert: Students using open flame should take precaution while conducting flame involved experiments. The symbol is a flame with a diagonal line over it. The objective is to avoid explosion and fire, which may be caused due to negligence in handling open flames.

Chemical Safety: You will find this in the list of chemistry and biold’gy lab safety symbols. The symbol is a bottle with something pouring from it. Chemical safety sign is put up, if the chemicals used in experiments are detrimental to the skin.

First Aid: The first aid symbol attached in laboratories can be as simple as a white coloured plus sign in a green background, a first aid box, a heart sign with a defibrillator or other complex figures which are specific to a particular theme.

Some Common Lab Apparatus

Science Practical SkillsScience LabsMath LabsMath Labs with Activity

CBSE previous Year Solved  Papers  Class 12 English Delhi 2012

Time allowed : 3 hours                                                                                           Maximum Marks: 100
General Instructions :

1. This paper is divided into three sections : A, B and C. All the sections are compulsory.
2. Separate instructions are given with each section and question, wherever necessary. Read these instructions very carefully and follow them faithfully.
3. Do not exceed the prescribed word limit while answering the questions.

SET I

SECTION – A
READING

Question.1. Read the passage given below and answer the questions ‘ that follow:

1. While there is no denying that the world loves a winner, it is important that you recognise the sighs of stress in your behaviour and be healthy enough to enjoy your success. Stress can strike anytime, in a fashion that may leave you unaware of its presence in your life. While a certain amount of pressure is necessary for performance, it is important to be able to recognize your individual limit. For instance, there are some individuals who accept competition in a healthy fashion. There are others who collapse into weeping wrecks before an exam or on comparing mark- sheets and finding that their friend has scored better.
2.  Stress is a body reaction to any demands or changes in its , internal and external environment. Whenever there is a change in the external environment such as temperature, pollutants, humidity and working conditions, it leads to stress. In these days of competition when a person makes up his mind to surpass what has been achieved by others, leading to an imbalance between demands and resources,
   it causes psycho-social stress.-It is a part and parcel of everyday life.
3.  Stress has a different meaning, depending on the stage of life you are in. The loss of a toy or a reprimand from the parents might create a stress shock in a child. An adolescent who fails in examination may feel as if everything has been lost and life has no further meaning. In an adult the loss of his or her companion, job or professional failure may appear as if there is nothing more to be achieved.
4.  Such signs appear in the attitude and behaviour of the individual, as muscle tension in various parts of the body, palpitation and high blood pressure, indigestion and hyper-acidity. Ultimately the result is self-destructive behaviour such as eating and drinking too much, smoking excessively, relying on tranquilisers. There are other signs of stress such as trembling, shaking, nervous blinking, dryness of throat and mouth and difficulty in swallowing.
5. The professional under stress behaves as if he is a perfectionist. It leads to depression, lethargy and weakness. Periodic mood shifts also indicate the stress status of the students, executives and professionals.
6.  In a study sponsored by World Health Organisation and carried out by Harvard School of Public Health, the global burden of diseases and injury indicated that stress diseases
   and accidents are going to be the major killers in 2020.
7.  The heart disease and depression — both stress diseases- are going to rank first and second in 2020. Road traffic accidents are going to be the third largest killers. These accidents are also an indicator of psycho-social stress in a fast-moving society. Other stress diseases like ulcers, hypertension and sleeplessness have assumed epidemic proportions in modern societies.
8.  A person under stress reacts in different ways and the
   common ones are flight, fight and flee depending upon the nature of the stress and capabilities of the person. The three responses can be elegantly chosen to cope with the stress so that stress does not damage the system and become distress.
9.  When a stress crosses the limit, peculiar to an individual, it lowers his performance capacity. Frequent crossings of the limit may result in chronic fatigue in which a person’ feels lethargic, disinterested and is not easily motivated to achieve anything. This may make the person mentally undecided, confused and accident prone as well. Sudden exposure to un-nerving stress may also result in a loss of memory. Diet, massage, food supplements, herbal medicines, hobbies, relaxation techniques and dance movements are excellent stress busters.

(a) (i) What is stress ? What factors lead to stress ?
Answer : Stress is a body reaction to any demands or changes in its internal and external environment.
Whenever there is a change in the external .environment such as temperature, pollutants, humidity and working conditions, it leads to stress and when there is an imbalance between demands and resources, it causes psycho-social stress.
(b) Which words in the above passage mean the same as the following ?
(i) fall down (para 1)
Answer: Collapse
(ii) Rebuke (para 3)
Answer: Reprimand
(iii) Inactive (para 9)
Answer: Lethargic

Question.2. Read the passage given below and answer the questions that follow:
Research has shown that the human mind can process words at the rate of about 500 per minute, whereas a speaker speaks at the rate of about 150 words a minute. The difference between the two at 350 is quite large.
So a speaker must make every effort to retain the attention of the audience and the listener should also be careful not to let his mind wander. Good communication calls for good listening skills. A good speaker must necessarily be a good listener.
Listening starts with hearing but goes beyond. Hearing, in other words is necessary, but is not a sufficient condition for listening. Listening involves hearing with attention. Listening is a process that calls for concentration. While listening, one should also be observant.
In other words, listening has to do with the ears, as well as with the eyes and the mind. Listening is to be understood as the total process that involves hearing with attention, being observant and making interpretations. Good communication is essentially an interactive process. It calls for participation and involvement. It is quite often a dialogue rather than a monologue. It is necessary to be interested and also show or make it abundantly clear that one is interested in knowing what the other person has to say.
Good listening is an art that can be cultivated. It relates to skills that can be developed.
A good listener knows the art of getting much more than what the speaker is trying to convey. He knows how to prompt, persuade but not to cut off or interrupt what the other person has to say. At times the speaker may or may not be coherent, articulate and well-organised in his thoughts and expressions. He may have it in his mind and yet he may fail to marshal the right words while communicating his thought. Nevertheless a good listener puts him at ease, helps him articulate and facilitates him to get • across the message that he wants to convey. For listening to be effective, it is also necessary that barriers to listening are removed. Such barriers can be both physical and psychological. Physical barriers generally relate to hindrances to proper hearing whereas psychological barriers are more fundamental and relate to the interpretation and evaluation of the speaker and the message.
(a) On the basis of your reading of the above passage, make notes in points only, using abbreviations wherever necessary. Supply a suitable title.
1. Notes
(i) Diff. between mind and spkr.
(a) mind press fstr than spkr .
(b) good communicant rq rs good lisng skills
(c) good spkr shud be good lisnr
(ii) Diff. between lisng and hearing.
(a) hearing not sufficient
(b) lisng involves attntn and concentration
(c) lisng requires participation and involvement
(iii) good lisng is an art
(a) good lisnr knows better than spkr
(b) when spkr fails, lisnr helps
(c) barriers for lisng shud be removed
Title : Qualities of a Good Listener
(b) Write a summary of the above passage in about 80 words.

SUMMARY

The processing of words can be done by human mind at a faster rate than the speaker speaks. Therefore speaker must be attentive and listener should also be careful. Good speaker must be a good listener. Listening is hearing with concentration. Good listening is an art and a good listener grasps faster than what the speaker is trying to tell. When speaker fails, listener helps. But for listening, certain physical and psychological barriers should be removed.

SECTION-B
(ADVANCElt WRITING SKILLS)

Question.3.Your school has planned an excursion to Lonavala near Mumbai during the autumn holidays. Write a notice in not more than 50 words for your school notice board, giving detailed information and inviting the names of those who are desirous to join. Sign as Naresh/Namita, Head boy/ Head girl, D.Y. English School, Thane, Mumbai.
OR
C.RR Senior Secondary School, Meerut is looking for a receptionist for the school. Draft an advertisement in not more than 50 words to be published in classified columns of Hindustan limes. You are Romola Vij, principal of the school.
Answer:

Question.4. Your school commerce Association organized a seminar for class XII students of the schools of your zone on the topic, ‘Rising prices create a crisis.’ As coordinator of the programme, write a report in 100-125 words for your school “* magazine, you are piyush/priya of ABC School, Agra.
OR
A new indoor gymnasium has recently been constructed and inaugurated at APJ International School at Goa. As special correspondent of ‘The Hindu,’draft a report in 100-125 words on the gymnasium and the inauguration ceremony
Answer:

RISING PRICES CREATE A CRISIS
A REPORT

14th July,20XX
A seminar for class XII students of the schools of our zone was organized to discuss on the topic, ‘Rising prices create a crisis’. It was attended by some very eminent Economists and politicians. All the speakers expressed their views on the consequences of rising prices. They pointed out that India is a poor country having mostly villages. Majority of the population lives below the poverty line; they are even unable to fulfill their basic requirements. The prices of the essential commodities have increased at an alarming rate. The poor people are unable to afford even a single meal. The purchase of vegetables, fruits, rice etc. have gone beyond the reach of common man. The necessities have taken the form of luxuries and rupee has lost its value. Now it becomes the duty of the government to take action against undue price hike and stop black marketing to control the rising prices.
Priya
Coordinator ABC School, Agra.
OR

INAUGURATION OF
GYMNASIUM
BY: ‘THE HINDU’CORRESPONDENT

14th July,20XX
A new indoor gymnasium has recently been constructed at APJ International School at Goa. It was inaugurated by Mr. Ravi Kant, Principal of the school. It is a proud addition to the institution. The construction has been possible due to the grant approval by the Education Minister. The gymnasium is spread over an area of lOOsq.ft.and is equipped with all the modern equipment and is fully air-conditioned. On the occasion of the inauguration, a speech on the utility of gymnastics was delivered by the gym instructor who has just received the advanced training from abroad. He motivated the students to take the maximum advantage of the gymnasium which are now practiced up to Olympic level. Overall it was a grand ceremony which was attended by the students as well as their parents.

Question.5. You are Pritam/Priti,27,W.E.A Karol Bagh, Delhi. You have decided to shift your residence to faridabad and have decided to discontinue your membership of Brain Trust Library, Karol Bagh. Write a letter to the librarian, requesting him to cancel your membership and refund your security deposit of? 5000 explaining your inability to continue your membership.
OR
You are Anu/Arun,13, W.E.A. Karol Bagh, New Delhi. You feel very strongly about the illtreatment meted out to stray dogs at the hands of callous and indifferent people. Write a letter to the editor of a national daily giving your views on why some people behave in such a manner and how these dogs should be treated.
Answer:
27, WE.A. Karol Bagh,
Delhi.
20th June,2014
Librarian,
Brain Trust Library,
Karol Bagh,
Delhi.
Subject: Cancellation of membership Dear Sir,
! This is to bring to your notice that I took the membership of your library in the year 2012.1 found it very helpful as it contains the books which cover almost all the areas of knowledge, so I encouraged few of my friends also to join it and take the advantage. Even my grandfather became a regular reader of your books. But unfortunately I’d have to’ discontinue my membership as my father has got transferred to another city and I’ll not be able to continue anymore. So I request you to cancel my membership and refund the security deposit of ? 5000 I am enclosing my membership card no.AG30002659.
Thank you
Yours truly Priti
OR
13,W.E.A. Karol Bagh New Delhi.
27th March,20XX The Editor,
The Times of India, ‘
New Delhi.
Subject: Cruetly to Stray Dogs ‘ Dear Sir,
I would like to share my views through the columns of your esteemed newspaper regarding the ill-treatment and neglect to the animals and in general, stray dogs around us. The sight is heart — rending and shocking.
Being a human being, it is our moral duty to look around us and stop such cruel activities. Every individual must be sensible enough to realize that these animals also form the integral part of our eco-system. If one cannot take care of animals then he should not at least tease them. They can inform any of the organizations which can be contacted.
The animals cannot express their suffering or pain but they also feel. They need our attention, care and love. We should spread awareness through campaigns, posters and plays, etc.
Yours truly .
Anu

Question.6. Spurt of violence previously unknown in Indian schools makes it incumbent on the educationists to introduce value education effectively in schools. Write an article in 150-200
words expressing your views on the need of value education. You are Anu/Arun.
OR
Regular practices of yoga can help in maintaining good health and even in the prevention of so many ailments. Write a speech in 150-200 words to be delivered in the morning assembly on the usefulness of yoga.
Answer :
Value Education-Today’s Need
By Anu
Education makes a man perfect. The overall development of personality whether it’s spiritual, physical, social, mental etc., all depends on education only. It inculcates ethics and moral values where there is no place for violence. But in the modern world of competitions, children are deprived of these values. Conflicting values a’nd moral dilemmas have become common in our daily lives. It has confused a child a lot and there is a sense of inadequacy to face the life situations. Media has negative influence on their personalities and to uplift the students morally and spiritually has become the topmost priority of the educational institutions. Till now, the institutions were concerned with the academic progress of the child, there was no place of value education in the curriculum, but the decreasing level of morality has shocked everybody. Illegal means of getting the things done have become very common and are in fashion. So there is need of acute awareness and bring forth the generation which is fruitful in the building of nation. This could be possible only when there is introduction of value education in schools. Although many schools have already introduced this in the curriculum but there is widespread need of it. Only then children can be developed into a stable, well rounded and morally integrated personalities.
OR
Dear Friend,
I feel privileged to get an opportunity to speak on the benefits of yoga. Today’s sedentary lifestyle has given invitation to many diseases where man has limited time to spare for his health. But practicing yoga helps to not get stuck in mental noise. It helps in maintaining balance between mental and social status of human being. With regular practice of yoga, we tend to become more sensitive to the kind of food our body asks. A few minutes of yoga helps in getting relief from stress. It helps detox the body and de-stress the mind. We all want to live in a peaceful atmosphere and among nature. But we hardly realize that peace can be found within us. Yoga is the best way to calm a disturbed mind. On the other hand, yoga postures massage organs and strengthen muscles; breathing techniques and meditation release stress and improve immune system. Yoga and meditation work on keeping the mind happy and peaceful. When we feel drained out of energy, a few minutes spent on yoga provides freshness and energy. It also helps in improving body posture.
In the end, I would like to tell that yoga is a continuous process, so keep practicing !

SECTION-C
(TEXT BOOK)

Question.7. (a) Read the extract given below and answer the questions
that follow:
The stunted, unlucky heir Of twisted bones, reciting a fathers gnarled disease, His lesson, from his desk. At back of the dim class One unnoted, sweet and young. His eyes live in a dream, Of squirrel’s game, in tree room, other than this.
(i) Who is this ‘unlucky heir’ and what has he inherited ?
Answer : The boy who has twisted bones and is sitting in the slum classroom is the unlucky heir. He has inherited the gnarled disease of twisted bones from his father.
(ii) What is the stunted boy reciting ?
Answer : The stunted boy is reciting his father’s gnaried disease and his lessons.
(iii) Who is sitting at the back of the dim class ?
Answer : At the back of the dim class, there is an unnoted, sweet and young dreamer who is dreaming of squirrel’s game, is sitting.
OR
For once on the face of the Earth
Let’s not speak in any language.
Let’s stop for one second.
And not move our arms so much.
(i) Why does the poet want us to keep quite ?
Answer : The poet wants us to keep quiet to retrospect and do self-assessment for mental peace and save human race from doom.
(ii) WTiat does he want us to do for one second ?
Answer : He wants us to remain quiet for one second and stop moving our arms. In fact he wants us to stop any kind of activities.
(iii) What does he mean by ‘not move our arms’?
Answer : By not moving our arms means no activity or movement, so as to harm anybody Symbolically by this he means no war or violence.
(b) Answer any three of the following in 30-40 words each :
(i) Why are the young trees described as sprinting ?
Answer : The young trees are compared to the poet’s old mother who is old and pale whereas a tree is young and full of life. Tree seems to be sprinting while her mother is very aged.
(ii) How is a thing of beauty joy forever?
Answer : With the passage of time beauty increases and it becomes constant. It never passes into nothingness but
multiplies and gives joy forever. It leaves a concrete impact on our minds and soul.
(iii) Why did Aun t Jennifer choose to embroider tigers on the panel?
Answer: Aunt Jennifer wanted to express her feelings through , the embroidery. She had some hidden desire as she wanted to be fearless, proud and chivalrous like tigers. She actually expressed her pain on the panel and showed how she would like her life to be.

Question.8. Answer the following in 30-40 words each :
(a) What changes did the order from Berlin cause in the school ?
Answer : There was an order from Berlin that French would not be taught in school anymore. Instead German would be taught by a new teacher. Thlt was probably their last French lesson. It was a Sunday morning and there was no noise or activity. The French teacher, Mr. Hamel seemed to be calm and he was becoming emotional. He was wearing a special dress and all the villagers were sitting on the back benches of the classroom.
(b) Why was Douglas determined to get over his fear of water?
Answer: Douglas could not enjoy water sports like canoeing, boating and swimming because of fear of water, therefore he wanted to overcome his fear. Although he wanted to get into the waters of cascades but his fear held him back. He felt paralyzed due to fear at the very thought of water, so he decided to overcome it.
(c) How were Shukla and Gandhiji received in Rajendra Prasad’s house ?
Answer : Rajkumar Shukla, a poor peasant of Champaran wanted Gandhiji to come to champaran to help the poor sharecroppers. From Calcutta, on a fixed day, they went to Patna to meet Rajendra Prasad, but he was out of town. Servants knew Rajkumar Shukla as a poor peasant; therefore they thought the same for Gandhiji and took him to be untouchable. They were asked to sit on the ground and did not allow Gandhiji to draw water from the well.

Question.9. Answerthe following 125-150 words :
(a) How are the attitude of the ironmaster and his daughter different ? Support your answer from the text.
OR
(b) Compare and contrast Sophie and Jansie highlighting their temperaments and aspirations.
Answer : The attitude of the ironmaster was totally different from his daughter Edla. It can be proved by several instances from the story. According to ironmaster, the peddler was an old acquaintance. He mistook him so he invited him to his house to spend Christmas. But when the peddler refused, he sent his daughter to persuade the peddler to come to his . house as she had the quality of convincing. She was able to win the confidence of the peddler who ultimately agreed to come to her home.
When the ironmaster realized that the peddler was not the one who was thinking after he got his hair cut and shaving done, he immediately asked him to go out of his house. However his daughter asked peddler to stay with them that dayfor Christmas. She had a feeling of sympathy for the ‘poor hungry’ man. She offered him her father’s suit as Christmas gift. It was her attitude that changed peddler’s mind and left with her the money he had stolen.
OR
Sophie and Jansie were classmates. They belonged to lower middle class. Jansie was practical, whereas Sophie was a dreamer. Jansie knew the realities of life and she advised Sophie also not to indulge in fantasies. Sophie was totally unaware of the realities of life and dreamt of big things. She wanted to have a boutique of her own. She also wanted to become an actress as there was lot of money in that profession. She wanted to be a fashion designer also. Overall it can be said that she wanted to live a grand and luxurious life. But there was least possibility of fulfilling her dreams. Jansie warned her to remain on ground but she remained busy in dreaming. On the other side Jansie knew her limitations. She was not a daydreamer like Sophie. She also knew that both of them were likely to work in a biscuit factory in future. She tried to explain this thing many times to Sophie who was overambitious. That was the difference between them. ‘

Question.10. Answer tke following in 125-150 words:
(a) How did the Tiger king stand in danger losing his kingdom? How was he able to avert the danger ?
Answer : Tiger king was told by the astrologer that he would be killed by a tiger, so he started hunting and killing tiger. If anybody disobeyed him, he would be punished severely. A high ranking British official Wanted to kill the tigers. Maharaja did not give him the permission and told him that he could kill any animal except tigers. The officer wished to, get photograph of himself holding a gun beside tiger’s body. But Maharaja refused that also. As he had not allowed him to fulfill his desire, he was in a danger of losing his kingdom. He then discussed the issue with his Deewan. So he offered a birbe of 50 diamond rings of the value of three lakhs to the wife of the British officer. She became happy and kept all of them. This way he was able to retain his kingdom.

Question.11. Answer the following in 30-40 words each:
(a) Did Hana think the Japanese tortured their prisoners of war? Why?
Answer: Different mouth told different stories about the war, so it was difficult to know for Hana whether they were true. The newspaper reports said that people received Japanese soldiers gladly with cries of joy.
(b) How did the Wizard help Roger Skunk?
Answer : Roger told wizard that nobody played with him due to his foul smell. So he put forth his desire to smell like roses. Holding his magic wand, wizard began to matter some magical words and soon Roger began to smell like roses.
(c) How does Mr. Lamb keep himself busy when it is a bit cool ?
Answer : When it cool, Mr. Lamb used to bring a ladder and a stick to pull the crab apples down for making Jelly. He liked sitting in his garden, hearing the humming sound of bees and enjoyed reading books. He made toffees also with honey.
(d) Who was carter ? What did the Governor ask him to do ?
Answer : Carter was a prison officer. Government asked him to take McLeery with him to search for the absconder. On being asked by the governor that who had seen Evans at prison gate, Stephen told him that it was him.

SET II

Note : Except for the following questions, all the remaining questions have been asked in previous sets.

Question.4. Your school conducted a seminar on ‘How to prevent cruelty towards animals’, in which 40 city CBSE schools took part. As coordinator of the programme, write a report in 100-125 words for the school magazine. You are Vikram/Vidhi of C.P.S. Senior Secondary School, Bangalore.
OR
You witnessed a road accident near Nalbandh Chowaraha at Agra in which a bus and a scooter were involved. Write a report for Amar Ujala in 100-125 words. Sign as Vivek/ Vimla, Special Correspondent.
Answer: How to Prevent Cruetly
Towards Animals
Report By : Vidhi (Coordinator)
21st May, 20XX
C.P.S. Senior Secondary School, Banglaore conducted a seminar on the topic ‘How to prevent cruetly towards animals’. It was attended by 40 city CBSE schools which took part in the discussion. It also saw some eminent personalities like Education Minister along with the chief guest Mrs. Maneka Gandhi, noted environmentalist and politician. The activists laid emphasis on the fact that we should try to stop cruetly towards animals. They too have equal right on earth as we hurpans have. Animal abuse happens to cats, dogs, livestock animals every time and everywhere such as circuses, races, laboratories etc. Laws concerning animal cruelty are designed to prevent needless cruelty to animals. It was also conveyed that if anyone witness someone abusing any animals, one can make complains as it is illegal in every state. Suggestions were given to avoid using animal products like cosmetics, drugs, clothes etc. Thus seminar received a great applaud and was endearing for all. .
OR
A Road Accident
Report By : Vimla (Special Correspondent)
Agra, 24th December
Last week I witnessed a shocking road accident while coming hack home from school in which a bus and scooter were involved. The accident took place near Nalbandh Chowraha at Agra. A folly crowded roadways bus which was coming from the University side, was trying to overtake another bus and in the process hit a scooter. There were two persons on the scooter, a man and a woman; probably husband and wife. Wife died on the spot and the husband was rushed to hospital immediately where sources told that his condition was critical. It was horrible sight to see as the scooter was completely damaged. Most of the passengers of the bus were also injured badly and they were taken to a nearby hospital. Bus driver was about to flee but the crowd on the busy road reacted immediately and prevented him from escaping. The angry mob beat him brutally and handed over to the local police.

SECTION – C

Question.8. Answer the following in 30-40 words each
(b) Who is Mukesh ? WTiat is his dream ?
Answer : Mukesh is the son of a poor bangle maker from Firozabad. He does not want to follow the family’tradition of making bangles. His dream is to become a motor mechanic and a car driver.
(c) Why did the peddler decline the invitation of the ironmaster ?
Answer: The peddler declined the invitation of the ironmaster as he was having thirty stolen kronors. Secondaly ironmaster mistook him for an old regimental comrade and was afraid that after finding the truth, he would be sent to jail.

Question.9. Answer the following in 125-150 words:
Attempt a charactersketch of Sophie as a woman who lives in her dreams.
OR
How did Douglas develop an aversion to water ?
Answer : Sophie is a young girl who always lives in a world of fantasy. She belongs to a lower middle class family. Like other adolescents, she too has dreams. She wants to own a boutique, become an actress, a fashion designer or a manager. She does not know where the money will come from for all her dreams, she lives far from reality. Even her friend Jansie warns her but she is given to her wild thoughts. She ignores them all. No one shows interest in what she thinks, so she confides in her brother Geoff who is an introvert. She also likes Danny Casey, a young football player from Ireland. She always thinks of him. She even thinks that she has met him and he has promised to meet her again. She goes to the place where she expects to meet him but finding him nowhere, she feels disappointed. This way or thoughts keep hovering in her mind.
OR
Douglas was afraid of water since childhood. It happened so when he was four year old, he went to the beach with his father in California. The waves knocked him down and swept over him. He went down into the water and was out of breath.
Again at the YMCA pool, a misadventure took place. He was learning swimming there. A strong boy threw him in the deep water one day when he was sitting at the edge of the pool. Douglas slowly went to the bottom of the pool in the same sitting position. He tried three times to come out of the water but failed. He was now in great panic. He was completely exhausted and his lungs were aching. Ultimately he fainted but was saved. This incident had a negative impact on him and fear of water stayed with him. He could not sleep or eat for many days and remained away from water.

Question.11. Answer the following in 30-40 words each :
(a) Why had Han a to wash the wounded man herself ?
Answer : Hana had to wash the wounded man herself as her domestic maid Yumi refused to do so because the man was an enemy. Not only this, but all the servants were against Dr. Sadao and his wife Hana in this respect, so they all left the house. :
(b) What were the contents of the small brown suitcase that McLeery carried ?
Answer: There were many things in the small brown suitcase that McLeery carried. They included a sealed questions paper envelope, a yellow coloured invigilation form, a paper knife, Bible, authentication card, a semi inflated rubber tube and a copy of ‘’The Church Times. ’

SET III

Note : Except for the following questions, all the remaining questions have been asked in previous sets.

Question.4. Your school celebrated international year of chemistry 36 schools from Delhi participated in various competitions. Chairman CBSE was the chief Guest. As coordinator draft a report in 100-125 words for your school magazine. You are Ram/Ramya of G.B. Senior Secondary School, Delhi.
OR
You witnessed a bomb blast in a Delhi market when you went there for shopping along with your parents for Diwali purchases. Write a report in 100-125 words for your school magazine.
Answer : International Year of Chemistry
A Report
20th Aug. 20XX,
Recently our school celebrated international year of Chemistry. Children of* 36 schools from Delhi took part in different competitions. CBSE chairman was the chief guest who was given a warm welcome by our Principal. Our Principal in his welcome speech inspired the future generation by emphasizing on the need of studying Chemistry and its utility for the modern technology. The honorable chief guest also addressed the children by telling them the Scientific wonders and achievements. He further told Chemistry to be an important subject, and an integral part of Science. Essay writing competition was conducted for the seniors. They , were given the topics half an hour before. Science teachers from different schools laid emphasis on its significance as it has changed the pattern of life. In the end winners were awarded prizes in the form of laptop, smartphone and tables. Ramya (Coordinator)
G.B. Senior Secondary School
OR
Bomb Blast in the Market
A Report
25 th October, 20XX, (Delhi)
Day by day incidents of bomb blast have been increasing. It seems as if they are a part of daily news. Two days ago I went to market with my parents for Diwali shopping. The market was beautifully decorated and huge crowd of people was busy in shopping. Everything appeared to be normal when suddenly we heard a blast from nearby area. Within fraction of minutes everything went haphazard. Before we could understand anything, the shrieking sounds from all the sides started coming. The whole atmosphere changed completely. Nearly four shops were hit badly and fire broke out in a restaurant gulping the other shops in flames. Women, and children were lying on the floor, seriously hurt. Overall I can say that whole situation was appalling and pathetic.

SECTION C

Question.8. Answer the following in 30-40 words each:
(b) Is Sabeb happy working at the tea stall ? Why/Why not?
Answer : Saheb is not at all happy working at the tea stall. When he is asked by the narrator whether he likes his job, he became sad. The canister in his hands seemed to be heavier than the plastic bag that he used to carry as a rag-picker.
(c) Why was the Crofter so friendly and talkative with the .
Peddler ?
Answer : Crofter had always lived alone; there was no one to talk to him. So when, peddler reached his cottage. He felt very happy as he needed a company. He wanted to share his feelings and because, peddler had to jpend the night over there, he listened to him peacefully.

Question.9. Answer the following in 125-150 words:
(a) How did Douglas try to save himself from drowning in the YMCA pool ?
Answer : Douglas was terribly afraid of water due to an incident in which he had nearly died in the swimming pool. He was so afraid that he avoided going near water for many days. But he liked the water activities like canoeing, wade in the waters of cascade, fishing etc. The fear of water seized him to do any activity. It destroyed his joy of all activities. So he thought to overcome his fear and engaged an instructor to learn swimming. The instructor taught him all the tricks of becoming a good swimmer. Once instructor left him alone in the pool when he was sure that Douglas would be able to swim on his own, Douglas swam the length of the pool up and down but he was not satisfied so he went to Lake Wentworth. The old fear returned when he had put his face under water and could see nothing except bottomless water. To overcome his fear, he dived into the lake and swam to the other shore. He shouted with joy as he had triumphed.
OR
(b) Draw a characters sketch of Sophie’s father.
Answer : Sophies father was a hard working man who did not believe in what Sophie tells him. He knew that she had a habit of indulging in wild fantasies. He was an aggressive
person and had authoritative kind of personality. -He was a sports lover and took his children to watch football matches. His face reflected his whole days labor and it seemed grimy and sweated. He was fun loving too and it was obvious when he went to pub to celebrate United Team’s victory. He had a plump face and thick neck. His face and arms were shiny and pink. He showed contempt when Geoff told him that Sophie met Danny Casey and started talking about him and his talent. But when he got to know from Sophie that Danny was going to buy a shop, he showed signs of disgust which scared Sophie.

Question.11. Answer the following in 30-40 words :
(a) What help did Dr. Sadao seek from Hana while operating on the wounded white man ?
Answer : Dr. Sadao was alone to operate the wounded white man so he sought help from his wife Hana who brought towel for him, helped him in turning the white man, giving anesthesia and holding cotton soaked in anesthesia near his nostrils.
(d) What did the detective superintendent inform the Governor about Evans?
Answer : The detective superintendent informed the Governor that McLeery had spotted Evans drive off along Elsfield way. After getting the car number they started chasing but lost track and thought that Evans might have come back to the city.