Your SlideShare is downloading. ×
Studymate Solutions to CBSE Board Examination 2011-2012     Series : SMA/1                                                ...
STUDYmate                                       SECTION-A1.   Mention the unique flowering phenomenon exhibited by Strobil...
STUDYmate                                                  SECTION-B9.   Draw a neat labelled sketch of a replicating fork...
STUDYmate14. (a)     State the difference between meiocyte and gamete with respect to chromosome            number.     (b...
STUDYmate18. (a)    Name the Protozoan parasite that causes amoebic dysentery in humans.     (b)   Mention two diagnostic ...
STUDYmate     Stanley L Miller. Miller designed a glass apparatus comprising a gas flask, a condenser,     and a liquid fl...
STUDYmate     These crystals contain insecticidal protein. These crystals exist as inactive prototoxins     but once an in...
STUDYmateAns. (a)       This representation indicates a normal human because in the respective amino               acid ch...
STUDYmateAns.            71% carbon is found dissolved in oceans. This oceanic reservoir regulates the             amount...
STUDYmate      (b)     Effluent from industries contains large amounts of nutrients. This causes              excessive gr...
STUDYmate         These mitotic divisions are strictly free nuclear, that is, nuclear divisions are          not followed...
STUDYmate                                Round          Wrinkled                                RR             rr       Pa...
STUDYmate       Studymate Solutions to CBSE Board Examination 2011-2012     Series : SMA/1                                ...
STUDYmate     1.     Chronic inflammation of the organs in which they live for many years.     2.     The genital organs a...
STUDYmate     plant and animal life burgeons, and organic remains begin to be deposited on the     lake bottom. Over the c...
STUDYmate                   (I)    National parks                   (II)   Wild life sanctuaries                   (III) B...
STUDYmate       Studymate Solutions to CBSE Board Examination 2011-2012     Series : SMA/1                                ...
STUDYmate                                       SECTION-C19. (a)     Construct a complete transcription unit with promotor...
STUDYmate           In collaboration with R.V. College of Engineering and Banglore city Corporation,           Ahmed Khan ...
STUDYmate            (d)   Prior to the work of Oswald Avery, Colin MacLeod and Maclyn McCarty                  (1933-44),...
STUDYmate                                         OR(a)   Three different allelic forms of ‘I’ gene are                   ...
Upcoming SlideShare
Loading in...5

Bio sol cbse_2011-12


Published on

Published in: Technology, Education
  • Be the first to comment

No Downloads
Total Views
On Slideshare
From Embeds
Number of Embeds
Embeds 0
No embeds

No notes for slide

Transcript of "Bio sol cbse_2011-12"

  1. 1. Studymate Solutions to CBSE Board Examination 2011-2012 Series : SMA/1 Code No. 57/1/1 Candidates must write the Code onRoll No. the title page of the answer-book.  Please check that this question paper contains 6 printed pages.  Code number given on the right hand side of the question paper should be written on the title page of the answer-book by the candidate.  Please check that this question paper contains 30 questions.  Please write down the Serial Number of the questions before attempting it.  15 minutes time has been allotted to read this question paper. The question paper will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the student will read the question paper only and will not write any answer on the answer script during this period. BIOLOGY[Time allowed : 3 hours] [Maximum marks : 70]General Instructions:1. All questions are compulsory.2. This questions paper consists of four Sections A, B, C and D. Section-A contains 8 questions of one mark each, Section-B is of 10 questions of 2 marks each, Section-C is of 9 questions of three marks each and Section-D is of 3 questions of five marks each.3. There is no overall choice. However, an internal choice has been provided in one questions of 2 marks, one questions of 3 marks and one question of 5 marks weightage. A student has to attempt only of the alternatives in such questions.4. Wherever necessary, the diagrams drawn should be neat and properly labelled. -(1)-
  2. 2. STUDYmate SECTION-A1. Mention the unique flowering phenomenon exhibited by Strobilanthus kunthiana (Neelakuranaji). [1]Ans. It flowers once in 12 years.2. How does smoking tobacoo in human lead to oxygen deficiency in their body? [1]Ans. Smoking increases the carbon monoxide content in blood and reduces the concentration of oxygen bound to haeme group of haemoglobin.3. A garden pea plant (A) Produced inflated yellow pod, and another plant (B) of the same species produced constricted green pods. Identify the dominant traits. [1]Ans. Dominant traits are Green and inflated.4. Why is Eichhornia crassipes nicknamed as “Terror of Bengal”? [1]Ans. Eichhornia crassipes is called terror ‘of Bengal’ because it grows very fast in standing water. It drains oxygen from water thus leading to death of marine life.5. Write the location and function of the sertoli cells in humans. [1]Ans. Sertoli cells are located in the germinal epithelium of seminiferous tubules. They provide nourishment to developing spermotozoa.6. Name the following: (a) The semi-dwarf variety of wheat which is high-yielding and disease-resistant. (b) Any one inter-specific hybrid mammal. [1]Ans. (a) Sonalika / Kalyan sova (b) Mule7. Write the similarity between the wing of a butterfly and the wing of a bat. What do you infer from the above with reference to evolution? [1]Ans. Both butter fly and bat use their wings to fly. They are anatomically different. So it is result of convergent evolution.8. Write what do phytophagous insects feed on.Ans. They feed on plants -(2)-
  3. 3. STUDYmate SECTION-B9. Draw a neat labelled sketch of a replicating fork of DNA. [2]Ans. 5 3 Template DNA (parental strands) continuous Discontinuous 5 synthesis 3 synthesis 3 Newly 5 5 synthesised 3 strands10. Where is sporopollenin present in plants? State its significance with reference to its chemical nature? [2]Ans. Sporopollenin is present in exine layer of wall of pollen grain. It is an organic polymer and is resistant to oxidation and leaching.11. (a) Highlight the role of thymus as a lymphoid organ. (b) Name the cells that are released from the above mentioned gland. Mention how they help in immunity. [2]Ans. (a) In thymus immature lymphocytes differnetiate into antigen – sensitive lymphocytes. After maturation in thymus they migrate to secondary lymphoid organs. (b) The cells released from thymus are called T-lymphocytes. These lymphocytes are responsible for cell mediated immunity which defends the body against virus, fungi and some bacteria which has entered the hosts cells. Helper T cells stimulate B-cells to produce antibodies and killer T-cells migrate to site of infection.12. Explain the work carried out by Cohen and Boyer that contributed immensely in biotechnology. [2]Ans. Stanley Cohen and Herbert Boyer constructed first artificial recombinant DNA. They did this by isolating the antibiotic resistant gene by cutting out a piece of DNA from a plasmid which was responsible for giving antibiotic resistance.13. Why do clown fish and sea anemone pair up? What is this relationship called? [2]Ans. This interaction is called commensalism. Here the clown fish lives in the tentacles of sea anemone. The fish gets protected from predators which stay away from stinging tantacles. However anemone does not get any benefit from clown fish. -(3)-
  4. 4. STUDYmate14. (a) State the difference between meiocyte and gamete with respect to chromosome number. (b) Why is a whiptail lizard referred to as parthenogenetic? [2]Ans. (a) Meiocyte (gamete mother cells) is diploid (2n), where as gamete is haploid (n). (b) Whiptail lizard is said to be parthenogenetic because female gamete undergoes development to form new organisms without fertilization.15. Name the plant source of the drug popularly called “smack”. How does it affect the body of the abuser? [2] OR Why is Rhizobium categorized as a ‘symbiotic bacterium’? How does it act as a biofertiliser? [2]Ans. It is obtained from Papver somniferum. Smack is a stronger analgesic than morphine. It reduces heart beat, blood pressure and increases blood sugar. OR Rhizobium lives in root nodules of leguminous plants. This association is mutually beneficial. Rhizobium gets food and shelter and leguminous plant gets nitrogen in return. Since Rhizobium is capable of fixing atmospheric nitrogen so it acts as biofertiliser.16. (a) State the role of DNA ligase in biotechnology. (b) What happens when Meloidegyne incognitia consumes cells with RNAi gene? [2]Ans. (a) Role of DNA ligase in biotechnology is joining of DNA fragments end to end, having same kind of sticky ends. (b) If Meloidegyne incognitia consumes cells with RNA i gene silencing of specific mRNA occurs due to a complementary ds RNA molecule formation that binds to and prevents translation of mRNA (silencing) and thus causing death of the nematode.17. Some organisms suspend their metabolic activities to survive in unfavourable conditions. Explain with the help of any four examples. [2]Ans. Examples: (a) In bacteria, fungi and lower plants various kinds of thick walled spore are formed which help them to survive unfavourable conditions. (b) In higher plants, seeds and some other vegetative reproductive propagules serve as a means to tide over period of stress, besides helping in seed dispersal. (c) Bears going into hibernation during winter. (d) Snails and fishes go into aestivation to avoid summer related head problems and dessication. OR (e) Many zooplankton are known to enter diapause, a stage of suspended development. -(4)-
  5. 5. STUDYmate18. (a) Name the Protozoan parasite that causes amoebic dysentery in humans. (b) Mention two diagnostic symptoms of the disease. (c) How is this disease transmitted to other? [2]Ans. (a) Entamoeba histolytica. (b) Symptoms of this disease include constipation, abdominal pain, cramps, stools with excessive mucous and blood clots. (c) Houseflies act as mechanical carriers and serve to transmit the parasite from faeces of infected person to food and thus contaminating them. SECTION-C19. It is established that RNA is the first genetic material. Explain giving three reasons. [3] OR (a) Name the enzyme responsible for the transcription of tRNA and the amino acid the initiator tRNA gets linked with. (b) Explain the role of initiator tRNA in initiation of protein synthesis. [3]Ans. RNA was first genetic material because (a) Many evidences suggest that essential life processes such as metabolism, translations, splicing, etc. evolved around RNA. (b) RNA used to act as genetic material as well as catalyst. Many important biochemical reactions in living systems are catalysed by RNA. (c) RNA being catalyst is highly unstable. OR (a) RNA polymerase in procaryotes and RNA polymerase III in Eukaryotes is responsible for transcription of tRNA. Initiator tRNA gets linked with methionine in eukaryotes and formylated methionine in prokaryotes. (b) Initiator tRNA combines with methionine in presence of amino acyl-tRNA synthetas enzymes resulting in formation of charged tRNA. Now this initiator tRNA combines with two subunits of ribosome and mRNA forming translation initiation complex. First mRNA attaches to small subunit of ribosome and charged initiator tRNA. The initiator tRNA joins the initiation codon AUG and signals the start of translation. Now the large subunit of ribosome combines with small subunit. Initiator tRNA lies at the P site of the ribosome.20. State the theory of Biogenesis. How does Miller’s experiment support this theory? [3]Ans. Oparin and Haldane proposed that the first form of life could have come from pre- existing non-living organic molecules (e.g. RNA, protein etc.). According to them origin of live is abiogenesis first but biogenesis there after. Miller’s Experiment: That simple organic compounds could be formed in nature in the manner explained above was experimentally demonstrated in 1953 by -(5)-
  6. 6. STUDYmate Stanley L Miller. Miller designed a glass apparatus comprising a gas flask, a condenser, and a liquid flask interconnected with tubes and fitted with sources of energy. The apparatus simulated the conditions on the primitive earth, including a “reducing atmosphere” and an “ocean”. He circulated in this apparatus a mixture of methane (CH4), ammonia (NH3) and hydrogen (H2) and water vapour (H2O) at 800°C. These gases were believed to prevail in the ancient atmosphere. He provided energy for the interaction of the gases present in the mixture in the form of electric. The electric sparks simulated lightning. Then the gases were condensed in a narrow tube and passed through a liquid flask. Here, energy was provided as heat with an electric heater. He kept the experiment working continuously for 18 days. A mixture of small organic molecules was formed in the gas flask (atmosphere) and was carried by condensation (rain) to the liquid flask (ocean). He found many simple organic compounds which included amino acids, such as glycine, alanine and aspartic acid; adenine and simple sugars such as ribose.21. Name the two different categories of microbes naturally occurring in sewage water. Explain their role in cleaning sewage water into usable water. [3]Ans. Different microbes, occuring in sewage water are aerobic and anaerobic bacteria, Protozoans, and filamentous fungi. The primary effluent is passed into large aeration tanks where it is constainlly agitated. This allows abundant growth of aerobic microbes (bacteria and filamentous fungi) into floes which is a mesh like structure. The growth of these microbes reduces BOD of effluents. Once the BOD is reduced significantly, then the effluent is passed into settling tanks where the bacterial floes are allowed to sediment. This sediment is called activated sludge. A small part of activated sludge is again introduced into large tanks called anaerobic sludge digesters. Here anaerobic bacteria digest the bacterial and fungi in the sludge. This digestion produces methane, H2S and CO2 gas. These gases form biogas. The effluent from secondary treatment is them released into natural water bodies.22. Write the function of each one of the following: (a) (Oviducal) Fimbriae (b) Coleoptile (c) Oxytocin [3]Ans. (a) (Oviducal) Fimbriae: It receives the ova which gets released on ovulation. (b) Coleoptile: It is the protective covering of plumule present inside the monocot seed. (c) Oxytocin: (i) It causes contraction of uterus at the time of child birth. (ii) It also helps in production of milk during lactation period.23. Name the genes responsible for making Bt cotton plants resistant to bollworm attack. How do such plants attain resistance against bollworm attacks? Explain. [3]Ans. The gene responsible for making Bt cotton plant resistant to bollworms is cryIAC and cryIIAb these genes form protein crystals during a particular phase of their growth. -(6)-
  7. 7. STUDYmate These crystals contain insecticidal protein. These crystals exist as inactive prototoxins but once an insect ingest the inactive toxin it is converted into active form of toxin inside alkaline conditions of gut of insect. This solubilizes the crystals. The active toxin binds to the surface of midgnt epithelial cells and create pores that cause cell swelling and lysis and causes death of the insect.24. Study a part of the life cycle of malarial parasite given below. Answer the questions that follows: [3] (a) Mention the roles of ‘A’ in the life cycle of the malarial parasite. (b) Name the event ‘C’ and the organ where this event occurs. (c) Identify the organ ‘B’ and name the cells being released from it.Ans. (a) A is female Anopheles mosquito that act as vector for Plasmodium. (b) C = fertilization that takes place in mosquito intestine. (c) B = salivary gland cells which are released sporozoites.25. Given below is the representation of amino acid composition of the relevant translated portion of -chain of haemoglobin, related to the shape of human red. [3] Gene ..... CTC .... ..... GAG .... mRNA ... GAG... Val His Leu Thr Pro Glu Glu 1 2 3 4 5 6 7 HbA Peptide (a) Is this representation indicating a normal human or a sufferer from certain related genetic disease? Give reason in support of your answer. (b) What difference would be noticed in the phenotype of the normal and the sufferer related to this gene? (c) Who are likely to suffer more from the defect related to the gene represented the males, the females or both males and females equally? And why? -(7)-
  8. 8. STUDYmateAns. (a) This representation indicates a normal human because in the respective amino acid chain, Glutamic acid is present at the 6th position. (b) In the sufferer who exhibit sickle cell trait, defect is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the 6th position of Beta globin chain of the haemoglobin. (c) Both the males and females suffer equally because sickle cell anaemia is not a sex linked disease. It is an autosomal disease and sickle shaped RBC will cause equal deficiency of oxygen in both males and females.26. By the end of 2002 the public transport of Delhi switched over to a new fuel. Name the fuel. Why is this fuel considered better? Explain. [3]Ans. Delhi Government shifted to CNG (Compressed Natural Gas). CNG is better because it burns more efficiently unlike petrol or diesel, in automobiles and very little of it is left unburnt. Also CNG is cheaper fuel them petrol and diesel.27. Draw a schematic sketch of pBR322 plasmid and label the following in it: (a) Any two restriction sites. (b) Ori and rop genes (c) An antibiotic resistant gene.Ans. Cla I Eco R I Hind III Bam HI Pvu I Pst I Sal I Ori Rop Origin of Pvu II replication Diagrammatic representation of E.coli cloning vector pBR322 showing restriction sites R R (Hind III, Eco RI, Bam HI, Sal I, Pvu II, Pst I, Cla I), Ori V and (amp and tet ). Rop codes for the proteins involved in the replication of the plasmid. SECTION-D28. Explain the carbon cycle with the help of a simplified model. [5] OR Explain how does: [5] (a) a primary succession start on a bare rock and reach a climax community? (b) the algal bloom eventually choke the water body in an industrial area? -(8)-
  9. 9. STUDYmateAns.  71% carbon is found dissolved in oceans. This oceanic reservoir regulates the amount of CO2 in the atmosphere.  Fossil fuel also represent a reservois of carbon.  Carbon cyclic occurs through atmosphere, ocean and through dead and living organisms.  A considerable amount of CO2 is fixed by process of photosynthesis.  Carbon from plants moves thruogh food chains.  Detritus food chain releases it to soil.  Respiration and combustion of fossil fuels, burning of forest firewood and organic debris releases CO2 in atmosphere.  Decomposers pass substantial CO2 to pool by their processing of water materials and dead organic matter of land or ocean.  Some amount of fixed carbon is lost to sediment and removed from circulation.  Human activities have significantly influenced carbon cycle. Deforestation and massive burning of fossil fuel has significantly increased the rate of release of CO2 into atmosphere. OR (a) Such a succession is called Xerarch succession. In this succession, (i) Pioneer specie is lichens which secrete acids to dissolve rock causing weathering and soil formation. (ii) Next seral stage will be bryophytes which can hold in the small amount of soil. (iii) Bryophytes are then succeeded by grasses. (iv) Grasses eventually will pave way for bigger trees which will form stable climax community. This remains stable as long as the environment remains unchanged. -(9)-
  10. 10. STUDYmate (b) Effluent from industries contains large amounts of nutrients. This causes excessive growth of free-floating algae causing algal bloom. Algae starts consuming oxygen for its own consumption. This will decrease BOD of water body. This reduced BOD eventually causes death of all aquatic life thus choking the water body.29. [5] (i) Identify the figure that illustrates ovulation and mention the stage of oogenesis it represents. (ii) Name the ovarian hormone and the pituitary hormone that have caused the above mention event. (iii) Explain the changes that occur in the uterus simultaneously in anticipation. (iv) Write the difference between ‘c’ and ‘h’. (v) Draw a labelled sketch of the structure of a human ovum prior to fertilization. OR How does the megaspore mother cell develop into 7-celled, 8 nucleate embryo sac in an angiosperm? Draw a labelled diagram of mature embryo sac. [5]Ans. (i) (a) f = stage of ovulation (b) Meiosis I of oogenesis is completed. (ii) Ovarian hormone = estrogen Pituitary hormone = Luetenizing hormone (iii) Endometrium lining gets thickened and vascularized. (iv) c = Developing graffian follicle h = Degenerating corpus luteum (v) Human ovum prior to fertilization OR  The nucleus of the functional megaspore divides mitotically to form two nuclei which move to the opposite poles, forming the 2-nucleate embryo sac. Two more sequential mitotic nuclear divisions result in the formation of the 4-nucleate and later the 8-nucleate stages of the embryo sac. -(10)-
  11. 11. STUDYmate  These mitotic divisions are strictly free nuclear, that is, nuclear divisions are not followed immediately by cell wall formation. After the 8-nucleate stage, cell walls are laid down leading to the organisation of the typical female gametophyte or embryo sac.  Six of the eight nuclei are surrounded by cell walls and organised into cells; the remaining two nuclei, called polar nuclei are situated below the egg apparatus in the large central cell.  There is a characteristic distribution of the cells within the embryo sac. Three cells are grouped together at the micropylar end and constitute the egg apparatus. The egg apparatus, in turn, consists of two synergids and one egg cell.  The synergids have special cellular thickenings at the micropylar tip called filiform apparatus, which play an important role in guiding the pollen tubes into the synergid.  Three cells are at the chalazal end and are called the antipodals. The large central cell, as mentioned earlier, has two polar nuclie. Thus, a typical angiosperm embryo sac, at maturity, though 8-nucleate is 7-celled.30. What is the inheritance pattern observed in the size of starch grains and seed shape of Pisum sativum? Workout the monohybrid cross showing the above traits. How does this pattern of inheritance deviate from that of Mendelian law of dominance? [5] OR State the aim and describe Messelson and Stahl’s experiment. [5]Ans. A round seed has well developed starch grains whereas wrinkled seeds don’t have starch grains. So, a cross between two will give to an intermediate situations. Suppose R is allele for round shape and r is allele for wrinkled. -(11)-
  12. 12. STUDYmate Round Wrinkled RR rr Parents R r gametes Rr F1 (intermediate) R r Gametes R r Round intermediate wrinkled R RR Rr 1 : 2 : 1 r Rr rr This is an example of incomplete dominance. Whereas in Mendelian inheritance we have complete dominance. OR Matthew Meselson and Franklin Stahl performed the following experiment in 1958: (i) They grew E. coli in a medium containing 15NH4Cl (15N is the heavy isotope of nitrogen) as the only nitrogen source for many generations. The result was that 15 N was incorporated into newly synthesised DNA (as well as other nitrogen containing compounds). This heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient (Please note that 15N is not a radioactive isotope, and it can be separated from 14N only based on densities). (ii) Then they transferred the cells into a medium with normal 14NH4Cl and took samples at various definite time intervals as the cells multiplied, and extracted the DNA that remained as double-stranded helices. The various samples were separated independently on CsCl gradients to measure the densities of DNA. (iii) Thus, the DNA that was extracted from the culture one generation after the transfer from 15N to 14N medium [that is after 20 minutes; E. coli divides in 20 minutes] had a hybrid or intermediate density. DNA extracted from the culture after another generation [that is after 40 minutes, II generation] was composed of equal amounts of this hybrid DNA and of ‘light’ DNA. ×·×·×·×·× -(12)-
  13. 13. STUDYmate Studymate Solutions to CBSE Board Examination 2011-2012 Series : SMA/1 Code No. 57/1/2 UNCOMMON QUESTIONS ONLY SECTION-A1. Cucurbits and papaya plants bear staminate and pistillate flowers. Mention the categories they are put under separately on the basis of the type of flowers they bear.Ans. Cucurbits are monoecious and papaya is Dioecious.4. What is the interaction called between cuscuta and shoe flower blush?Ans. Parasitism5. When do the oogenesis and the spermatogenesis initiate in human females and males respectively?Ans. Oogenesis is initiated during the embryonic development, spermatogenesis begins at puberty.7. State the significance of the study of fossils in evolution.Ans. Study of fossils indicates the geological period in which various life forms were arisen. The calculation of geological period can be done via radioactive dating. SECTION-B13. Draw a schematic diagram of a part of double stranded dinucleotide DNA chain having all the four nitrogenous bases and showing the correct polarity.Ans.14. Name the parasite that causes filariasis in human. Mention its two diagnostic symptoms. How is this disease transmitted to others?Ans. Wuchereria bancrofti and Wuchereria malayi causes filariasis in humans. The two diagnostic symptoms are -(13)-
  14. 14. STUDYmate 1. Chronic inflammation of the organs in which they live for many years. 2. The genital organs are also often affected resulting in gross deformities. Pathogens are transmitted to a healthy person through the bite by the female mosquito vectors.15. Name the source of streptokinase. How does this bio-reactor molecule function in our body? OR How do mycorrhizae act as biofertilizers? Explain. Name a genus of fungi that forms a mycorrhizal association with plants.Ans. Streptokinase produced by the bacterium streptococcus and modified by genetic engineering is used as a clot buster for removing clots from the blood vessels of patients who have undergone myocardial infection leading to heart attack. OR Mycorrhizae are associations between fungi and the roots of higher plants. The fungi help the plant in the absorption of essential nutrients from the soil while the plant in turn provides the fungi with energy yielding carbohydrates. Boletus is the soil fungus that forms a mycorrhizal association with plants. SECTION-C19. Write the function of each of the following (a) Middle piece in human sperm. (b) Tapetum in anthers. (c) Luteinizing hormone in human males.Ans. (a) The middle piece possesses numerous mitochondria, which produces energy for the movement of tail that facilitate sperm motility essential for fertilisation. (b) Microsporangium are surrounded by 4 wall layers, epidermis, endothecium, middle layer and tapetum. Outer 3 wall layers performs the function of protection and help in dehiscence of anther to release the pollen. The innermost wall layer is the tapetum, it nourishes the developing pollen grains. Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus. (c) Luteinizing hormone acts at the Leydig cells and stimulates the synthesis and secretion of androgens.26. How does an algal bloom cause eutrophication of a water body? Name the weed that can grow in such a eutrophic lake.Ans. Presence of large amounts of nutrrients in waters also causes execessive growth of planktonic algae called an algal bloom which imparrts a distinct colour to the wate rbodies. Algal blooms causes deterioration of the water quality and fish mortality. Eutrophication is the natural aging of a lake by nutrient enrichment of its water. Streams draining into the lake increases nutrients such as nitrogen and phosphorous which encourage the growth of aquatic organisms. As the lake’s fertility increases, -(14)-
  15. 15. STUDYmate plant and animal life burgeons, and organic remains begin to be deposited on the lake bottom. Over the centuries as silt and organic debris pile up, the lake grows shallower and warmer, with warm-water organisms supplanting those that thrive in a cold environment marsh plants take root in the shallows and begin to fill in the original lake basin. Eventually, the lake gives way to large masses of floating plants, finally converting into land. Eichhornia crassipes water hyacinth can grow in entrophic lake.28. (a) Draw a ‘pyramid of numbers’ of a situation where a large population of insects feed upon a very big tree. The insects in turn, are eaten by small birds which in turn are fed upon by big birds. (b) Differentiate giving reasons, between the pyramid of biomass of the above situation and the pyramid of numbers that you have drawn. OR (a) What are the two types of desirable approaches to conserve biodiversity? Explain with examples bringing out the difference between the two types. (b) What is the association between the bumble bee and its favourite orchid ophrys? How would extinction or change of one would affect the other?Ans. (a) Pyramid of Number Large birds Small birds Insects Tree This is spindle shaped pyramid. Pyramid of biomass Large birds Small birds Insects Tree This is irregular shaped pyramid. (b) In case of pyramid of number for first two steps number increases but them it decreases. Whereas in pyramid of biomass there is no clear trend. OR (a) Two types of desirable approaches to conserve biodiversity are: (i) In-situ conservation: In-situ conservation is the most appropriate method to maintain species of wild animals and plants in their natural habitats. This approach includes protection of total ecosystems through a network of protected areas. These are the biogeographical areas where biological diversity along with natural/cultural resources are protected, maintained and managed. The common natural habitats (protected areas) that have been set for in- situ conservation of wild animals and plants include – -(15)-
  16. 16. STUDYmate (I) National parks (II) Wild life sanctuaries (III) Biosphere reserves (IV) Several wetlands, mangroves and coral reefs (V) Sacred grooves and lakes. (ii) Ex-situ conservation: Ex-situ conservation includes the following: (I) Sacred plants and home gardens. (II) Seed banks, field gene banks, cryopreservation. Botanical gardens, Arborata, Zoological gardens, Aquaria. All these approaches help to conserve species and population diversity outside the natural habitats. In-situ Ex-situ 1. It is the most appropriate 1. In this approach, threatened method to maintain species of animals and plants are not wild animals and plants in their preserved in their natural natural habitat. habitat and placed in special setting, where they can be protected and special care. 2. No such technique can be used 2. Gametes of threatened species for In-situ conservation. can be preserved in viable and fertile condition for long periods using cryopreservation techniques. 3. In this method organism 3. Reproduction in captivity often alongwith its entire habitat is slows down and may not give preserved. Chances of desired results. fertilization and propagation are higher. 4. Eg. National Park, Wild Life 4. Eg. Seed banks. Sancturies. (b) Association between the bumble bee and its favourite orchid ophrys is mutualism. Orchids show a bewildering diversity of floral patterns many of which have evolved to attract the right pollinator insect (bees and bumblebees) and ensure guaranteed pollination by it. If the female bee’s colour patterns change even slightly for any reason during evolution, pollination success will be reduced unless the orchid flower co-evolves to maintain the resemblance of its petal to the female bee. ×·×·×·×·× -(16)-
  17. 17. STUDYmate Studymate Solutions to CBSE Board Examination 2011-2012 Series : SMA/1 Code No. 57/1/3 UNCOMMON QUESTIONS ONLY SECTION-A1. Mention the difference between spermiogenesis and spermiation. [1]Ans. Spermiogenesis: Transformation of spermatids into spermatozoa. Spermiation: The process of release of spermatozoa from Sertoli cells into the cavity of sperminiferous tubules.3. What is an interaction called when an orchid grows on a mango plant? [1]Ans. Commensalism.4. Write the names of two semi-dwarf and high yiuelding rice varieties developed in India after 1966. [1]Ans. Jaya/Ratna.6. Mention the unique feature with respect to flowering and fruiting in bamboo species. [1]Ans. Bamboo species flower only once in their lifetime.8. State the significance of biochemical similarities amongst diverse organism in evolution. [1]Ans. Biochemical similarities point to the same shared ancestory as structural similarities among diverse organisms. SECTION-B15. Mention the importance of Lactic acid bacteria to humans other than setting milk into curd. [2] OR How do methanagens help in producing biogas? [2]Ans. (a) Lactic acid bacteria play a very beneficial role in checking disease causing microbes. (b) Methanogens such as Methanobacterium act on excreta of cattle and grow anaerobically, producing large amount of methane along with CO 2 and H2. Methane is the main predominant gas of bio gas. -(17)-
  18. 18. STUDYmate SECTION-C19. (a) Construct a complete transcription unit with promotor and terminator on the basis of the hypothetical template strand given below: A T GCA T GCA T A C (b) Write the RNA strand transcribed from the above transcription unit along with its polarity. [3] OR How are the structural genes inactivated in lac operon in E. Coli? Explain. [3]Ans. (a) A T G C A T G C A T A C 3 5 Promoter Terminator 5 3 T A C G T A C G T A T G (b) RNA strand transcribed from above transcription unit. 5 3 UA C GU A CGU A UG (c) In the absence of lactose, no inactivation of repressor occurs and hence the repressor (which is constitutively produced from i gene) binds to the operator region of the operon; thus preventing RNA polymerase from transcribing the operon thus inactivating of the production of structural genes in E.coli occurs.20. Write the function of each of the following: [3] (a) Seminal vesicle (b) Scutellum (c) Acrosome of human spermAns. (a) Seminal vesicle: Constitute seminal plasma which is rich in fructose calcium and enzymes. (b) Scutellum: Is the papery cotyledon of the monocot seed and acts as a passage for movement of nutrients from the endosperm to the developing embryo. (c) Acrosome of human sperm: Contains hydrolytic enzymes that help in penetration of egg during fertilization.25. (a) Why are the colourful polysterene and plastic packagings used for protecting the food, considered an environmental menace? (b) Write about the remedy found for the efficient use of plastic waste by Ahmed Khan of Bangalore. [3]Ans. (a) Colourful polysterene and plastic packaging used for protecting the food are considered to be environmental menace as plastic is non-biodegradable and its recycling process is very costly and includes manual participation thus exposing workers to toxic substances produced during recycling process. (b) The remedy found for efficient use of plastic waste by Ahmed Khan of Bangalore was making of Polyblend, a fine powder of recycled modified plastic. This mixture is mixed with the bitumen, and is used to lay roads. -(18)-
  19. 19. STUDYmate In collaboration with R.V. College of Engineering and Banglore city Corporation, Ahmed Khan proved that blends of Polyblend and Bitumen, when used to lay roads enhanced the bitumen’s water repellant properties and helped to increase road life by a factor of three. SECTION-D30. Name the scientists who proved experimentally that DNA is the genetic material. Describe their experiment. [5] OR (a) List the three different allelic forms of gene ‘I’ in humans. Explain the different phenotypic expressions, controlled by these three forms. (b) A woman with blood group ‘A’ marries a man with blood group ‘O’. Discuss the possibilities of the inheritance of the blood groups in the folllowing starting with ‘yes’ or ‘no’ for each: (i) They produce children with blood group ‘A’ only. (ii) They produce children some with ‘O’ blood group and some with ‘A’ blood group. [5]Ans. The two scientists that proove DNA is the genetic material are: (i) Griffith Transformation Experiment (a) In 1928, Frederick Griffith, in a series of experiments with Streptococcus pneumoniae (bacterium responsible for pneumonia), witnessed a miraculous transformation in the bacteria. (b) When Streptococcus pneumoniae (pneumococcus) bacteria are grown on a culture plate, some produce smooth shiny colonies (S) while others produce rough colonies (R). This is because the S strain bacteria have a mucous (polysaccharide) coat, while R strain does not. Mice infected with the S strain (virulent) die from pneumonia infection but mice infected with the R strain do not develop pneumonia. S strain  Inject into mice  Mice die R strain  Inject into mice  Mice live Griffith was able to kill bacteria by heating them. He observed that heat- killed S strain bacteria injected into mice did not kill them. S strain (heat-killed)  Inject into mice  Mice live When he injected a mixture of heat-killed S and live R bacteria, the mice died. S strain (heat-killed) + R strain (live)  Inject into mice  Mice die Moreover, he recovered living S bacteria from the dead mice. (c) He concluded that the R strain bacteria had somehow been transformed by the heat-killed S strain bacteria. Some ‘transforming principle’, transferred from the heat-killed S strain, had enabled the R strain to synthesise a smooth polysaccharide coat and become virulent. This must be due to the transfer of the genetic material. -(19)-
  20. 20. STUDYmate (d) Prior to the work of Oswald Avery, Colin MacLeod and Maclyn McCarty (1933-44), the genetic material was thought to be a protein. They worked to determine the biochemical nature of ‘transforming principle’ in Griffith’s experiment. (e) They purified biochemicals (proteins, DNA, RNA, etc.) from the heat-killed S cells to see which ones could transform live R cells into S cells. They discovered that DNA alone from S bacteria caused R bacteria to become transformed. (f) They also discovered that protein-digesting enzymes (proteases) and RNA- digesting enzymes (RNases) did not affect transformation, so the transforming substance was not a protein or RNA. Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation. They concluded that DNA is the hereditary material. (ii) Hershey and Chase Experiment (a) DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952). They worked with viruses that infect bacteria called bacteriophages. (b) The bacteriophage attaches to the bacteria and its genetic material then enters the bacterial cell. The bacterial cell treats the viral genetic material as if it was its own and subsequently manufactures more virus particles. Hershey and Chase worked to discover whether it was protein or DNA from the viruses that entered the bacteria. (c) They grew some viruses on a medium that contained radioactive phosphorus and some others on medium that contained radioactive sulfur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. Similarly, viruses grown on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur. (d) Radioactive phages were allowed to attach to E. coli bacteria. Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge. (e) Bacteria which was infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria. Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicates that proteins did not enter the bacteria from the viruses. DNA is therefore the genetic material that is passed from virus to bacteria. -(20)-
  21. 21. STUDYmate OR(a) Three different allelic forms of ‘I’ gene are A (i) i (ii) I B (iii) I Genotype Blood Group ii O A A A I I;I i A B B B I I ;I i B A B I I AB(b) Possibility I A A I I ii (woman) (man) A A I I i IA i IA i A A i I i I i Yes. A Blood group only Possibility II A I i ii (woman) (man) A I i i IA i ii A i I i ii Yes. 50% A Blood Group 50% O Blood Group ×·×·×·×·× -(21)-