CBSE 12th Biology 2017 Guess Paper By 4ono · 2017-09-19 · CBSE 12th Biology 2017 Guess Paper By...

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CBSE 12th Biology 2017 Guess Paper

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CBSE 12th Biology 2017 Guess Paper By 4ono.com

TIME - 3HR. | QUESTIONS - 26

THE MARKS ARE MENTIONED ON EACH QUESTION SECTION- A

Q.1. Why is bagging of the emasculated flowers essential during hybridization experiments? 1 mark Ans. Bagging of the emasculated flower is essential to prevent the entry of unwanted pollen during hybridization experiments. Q. 2. Why is it not possible for an alien DNA to become part of a chromosome anywhere along its length and replicate normally? 1 mark Ans. This is because in a chromosome there is a specific DNA sequence called the origin of replication which is responsible for initiating replication. The alien DNA has to be linked to this origin of replication for it to multiply in the host organism. Q. 3. What is the major difference you observe in the offspring’s produced by asexual reproduction and in the progeny produced by sexual reproduction? 1 mark Ans. The progeny formed from asexual reproduction is genetically similar to parents and it does not show variation. But in sexual reproduction, individuals produced as a result of meiosis and gametic fusion exhibit genetic variation and difference from either of the two parents as well as among themselves' Q.4. Name the group of organisms and the substrate they act on to produce biogas. 1 marks Ans. Organisms-Methanogens Substrate-Cellulosic material / cow dung / agriculture waste. Q.5. How is snow-blindness caused in humans? 1 mark Ans. The inflammation of cornea caused by a high dose of UV-B is known as snow- blindness.

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SECTION – B Q. 6. Explain the process of RNA interference. 2 marks Ans. RNA interference take place in all eukaryotic organisarn as a method of cellular defense. It involves silencing of a specific mRNA due to a complementary ds RNA molecule that binds to and prevents translation of the mRNA. Q. 7. (a) Name the Protozoan parasite that causes amoebic dysentery in humans. 2 marks (b) Mention two diagnostic symptoms of, the disease. (c) How is this disease transmitted to others? Ans. (a) Entamoeba histoiytica. (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. Q. 8. List the symptoms of Ascariasis. flow does a healthy person acquire this infection? 2 marks Ans. Ascaris, the common roundworm and wucheria, the filarial worm, are some of the helminths which are known to be pathogenic to Man. Ascaris, an intestinal Parasite causes ascariasis. Symptoms: In this disease include internal bleeding, muscular pain, fever, anemia and blockage of the intestinal passage. Mode of transmission of disease: The eggs of the parasite are excreted along with the faeces of infected person which contaminate soil, water, plants etc. A healthy person acquires this infection through contaminated water, vegetable, fruits etc Q. 9. Explain the different steps involved during primary treatment phase of sewage. 2 marks Ans. The primary phase of sewage treatment involves physical filtration removal of particles by and sedimentation. Different steps- Initially, floating debris of sewage water is removed by sequential filtration. Then, grit (soil + small pebbles) is removed by sedimentation. The remaining solid particles, which have settled down, form the sludge while the suprenatant forms the effluent. Effluent is then taken for secondary treatment.




Q. 10. In angiosperms, zygote is diploid while primary endosperm cell is triploid. Explain. 2 marks Ans. The zygote is formed by the fusion of haploid male and female gametes, therefore it is diploid. Whereas, primary endosperm cell is formed by the fusion of one haploid male gamete with two haploid polar nuclei, therefore it is triploid. SECTION-C Q.11. Study the figure given below and answer the questions that follow: 3 marks

(a) Name the stage of human embryo the figure represents. Ans. (a) Blastocyst (b) Identify 'a' in the figure and mention its function. Ans. 'a' is trophoblast. The trophoblast layer gets attached to endometrium and later forms extra embryonic membrane namely chronic villi. (c) Mention the fate of the inner cell mass after implantation in the uterus. Ans. The inner cell mass differentiates into the embryo. (d) Where are the stem cells located in this embryo? Ans. Inner cell mass. Q.12. (a) Why do the symptoms of malaria not appear immediately after the entry of sporozoites into the human body when bitten by female Anopheles? Explain. 3 marks Ans. (a) Malaria parasite completes its asexual cycle in liver cells and then it attacks the red

blood cells (RBCs) resulting in their rupture. The ruptured RBCs release toxic substance called hemozoin that is responsible for the symptoms of malaria like chill and high fever. Thus, no symptoms appear in the infected person between the period the parasite enters the body till RBCs release hemozoin.




(b) Give the scientific name of material parasite that causes malignant malaria in humans.

Ans. Plasmodium falciparum Q.13. Explain the function of each of the following: 3 marks (a) Coleorhiza (b) Umbilical cord (c) Germ pores Ans. Coleorhiza- Protects the radicle of monocot embryo. Umbilical cord: It develops from two separate fetal origins. The yolk sack and the allantosis both comprise the umbilical cord. It helps in transport of substances to and from the embryo. Germ Pores: The outer wall of pollen grain through which the germ tube or embryo tube makes its exist on Germination. Q. 14. Draw a labelled diagram of the reproductive system in a human female. 3 marks Ans.

Q.15. The diagram above is that of a typical biogas plant. Explain the sequence of events occurring in a biogas plant. Identify a, b and c. 3 marks




Ans. a: Sludge tank

b: Gas holder c: Charge kit The sequence of events occurring in a biogas plant is that the biogas plant is installed basically to get the biogas which is used as fuel for heating cooking and lighting etc. It also yields nitrogen rich slurry that is an excellent source of manure. The biogas plant tank is fed with slurry of dung. A floating cover is placed which over the slurry keeps on rising as the gas is produced in the tank due to the microbial activity of methanogens like Methanobacterium. Methanoghens grow anaerobically on cellulosic material in cow dung to produce large amount of methane, 𝐶𝑂2 and 𝐻2. The biogas plant has an outlet, which is connected to a pipe to supply biogas. The spent slurry is removed through mother outlet and is used as fertilizer.

Q.16. Explain mutualism with the help of any two examples. How is it different from commensalism? 3 marks Ans. 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: (i) Associations between fungi and plants, called mycorrhizae-The plant is benefitted by essential soil nutrients that the fungus absorbs and transfers to the plant through its roots. The fungus in turn derives the benefits of receiving energy yielding carbohydrates from the plant. (ii) Pollination-The flowers of a plant provide sweet, mucilaginous nectar to birds or insects in return for getting help from the birds or insets in spreading them. pollen grains onto other flowers. The plant-pollinator pair often undergoes co-evolution to safeguard against the use of the nectar by other non-useful organisms. Mutualism differs from commensalism in that the latter provides a benefit to just one of the participating species, the benefitted species being called a commensal. Q.17. Women are often blamed for producing female children. Consequently, they are ill-treated and ostracized. How will you address this issue scientifically if you were to conduct an awareness programme to highlight the values involved? 3 marks




Ans. Women are never responsible for determination of the sex of a child. Moreover, it is not appropriate to ill-treat a woman for giving; birth to a girl child, as both males and females are equally important for the balance of nature and continuity of our species. All human-beings have 23 pairs of chromosomes. A human sperm (haploid) has 22 autosomes and one of the two types of sex chromosomes, i.e. either X or Y. On the contrary, human females have 22 autosomes that are exactly same as males and contain two X chromosomes. The sex of an individual is determined by the type of sex chromosome (X or Y) contained by the sperm that fuses with the ovum. If the fertilising sperm has an X chromosome, then the baby would be a female and if a sperm with Y chromosome fuses with the ovum, it will develop into a male child. Thus, males are responsible for determination of the sex of a child. Q. 18. Why is Agrobacterium tumefaciens a good cloning vector ? Explain. 3 marks Ans. The tumor inducing plasmid of Agrobacterium tumefaciens Ti -plasmid has now been modified into a cloning vector which is no more pathogenic to the plants but is still able to use the mechanism to deliver genes of our interest into a variety of plants. Agrobacterium tamifaciens is a soil bacterium which causes diseases in many dicot plants. It is able to deliver a piece of DNA known as T-DNA, to transform the normal cells into tumor cells and direct these tumor cells to produce the chemicals required by the pathogen.

OR

Explain the importance of (a) ori, (b) 𝒂𝒎𝒑𝑹 and (c) rop in the E coli vector shown below:

Ans. a = Ori = origin of replication b = 𝑎𝑚𝑝𝑅 = Ampicillin antibiotic resistance genes. c = rop = It codes for the Proteins involved in replication of the plasmid.

Importance (a) Ori - Ori is a sequence from where replication starts and any pieces of DNA when linked to this sequence can be made to replicate within the host cells. It is also responsible for controlling the copy number of the linked DNA.




(b) 𝑨𝒎𝒑𝑹 = The ligation of alien DNA is carried out at a restriction site present in

any antibiotic resistance gene. (c) It codes for the proteins involved in the replication of the plasmid.

Q.19. Study the population growth curves in the graph given below and answer the questions which follow: 3 marks (i) Identify the growth curves ‘a’ and 'b'.

(ii) Which one of them is considered a more realistic one and why?

(iii) If 𝒅𝑵

𝒅𝒕= 𝒓𝑵 (

𝑲−𝑵

𝑲) is the equation of the logistic growth curve, what does K

stand for? (iv) What is symbolised by N? Ans. (i) (a) = dN/dt = rN ⇒ Population exponential growth curve.

When resources are not limiting the. Growth, plot is exponential.

(b) = 𝑑𝑁/𝑑𝑡 = 𝑟𝑁 (𝐾−𝑁

𝐾)

When resources are limiting the growth, plot is logistic. (ii) Verhulst - Pearl Logistic Growth is more realistic for most animal population because the resources for growth are finite and become Limiting sooner or later. (iii) K is the carring capacity. (iv) N is population density at time t.




Q.20.

(a) Write your observation on the variations seen in the Darwin's finches shown above. Ans. From the original seed-eating features. many other forms with altered beaks arose, enabling them to become insectivorous and vegetarian finches. (b) How did Darwin explain the existence of, different varieties of finches on Galapagos Islands? 3 marks Ans. Darwin explained it as the process of evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography (habitats), called adaptive radiation. Q. 21. 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 is possible. 3 marks Ans.

In this case, the progeny would be 50% tall and 50% dwarf. Q. 22. (a) A NDA 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. 3 marks




Ans. (a) Cytosine and thymine are pyrimidines, According to Chargaff's rule of purine to pyrimidines is equal Number of adenine (A) will be equal to the number of thymine (T)

𝐴 = 𝑇 = 240 ⇒ 𝐴 + 𝑇 = 240 + 240

= 480. Number of cytocine (c) will be equal to the number of guanine (G). G = C ⇒ G + C = total number of nucleotides- Nucleotides containing A and T nitrogenous base

⇒ 1000 − (A + T) ⇒ 1000 − 480 = 520

𝐺 + 𝐶 = 520

𝐺 = 5202⁄ = 260

𝐺 = 260 & 𝐶 = 260 Number of Pyrimidines in the segment processes = C + T = 260 + 240 = 500 (b) Diarammatic sketch of a portion of DNA segment.




SECTION-D Q.23. (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. 4 marks Ans. (a) Fertilization of ovum takes place in the ampullary-isthmic junction of the fallopian tubes. Egg viable for only 24-48 hours after ovulation. Human fertilization is the union of a human egg and sperm, usually occurring in the ampulla of the uterine tube: the result of this union is the production of a Zygote, or fertilized egg, initiating prenatal development. The process of fertilization involves a sperm fusing with an ovum. The most common sequence beings with ejaculation during copulation, followed by ovulation, and finishes with fertilization. Various exceptions to this sequence are possible, including artificial insemination, In vitro fertilization, external ejaculation without, copulation, or copulation shortly after ovulation. Upon encountering the secondary oocyte, the acrosome of the sperm produces enzymes which allow it to burrow through the outer jelly coat of the egg. the sperm plasma than fuses with the egg’s plasma membrane, the sperm head disconnects from its flagellum and the egg travels down the Fallopian tube to reach the uterus. In Vitro Fertilization (IVF) is process by which egg cells are fertilized by sperm outside the womb, in vitro. (b) In Vitro Fertilization (IVF) Process by which egg cells are fertilized by sperm (usually 100,000 sperm/ml) outside the womb, in vitro. IVF is a major treatment in infertility when other methods of assisted reproductivc technology have failed. he process involves hormonally controlling the ovulatory process, removing ova (eggs) from the woman's ovaries and letting sperm fertilise them in a fluid medium. The fertilised egg (zygote) with upto 8 blastomeres is transferred into the fallopion tube (ZIFT: Zygote Intra fallopian Transfer) and embryos with more than 8 blastomeres into the uterus [IUT: Intra Uterine Transfer to complete its futher development.

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.




Ans. (a) Apomixes is a special mechanism of production of seeds by flowering plants without fertilization. The different ways apomictic seeds can develop are: (i) Diploid egg cell is formed without reduction divison which develops into the embryo without fertilization. Eg: Grasses. (ii) Nucellar cells surrounding the embryo sac divide and protrude into the embryo sac and develop into Embryos. Eg: Citrus and mango (b) (i) Apomixes provides an opportunity to the formers to keep using the same hybrid seeds year after year and not buy hybrid seeds every year. This is because, if the hybrid seeds are not made apomictic then they lose hybrid character due to segregation in the next generation. (ii) Apromixis makes use of hybrid seeds more economical. (c)




SECTION – E Q.24. (a) Draw d schematic labelled diagram of a fertilized embryo sac of an Angiosperm. 5 marks Ans. (a)

(b) Describe the stages in embryo development in a dicot plant. Ans.

In a dicot plant one male gamete fuses with egg cell and completes syngamy. This results in the formation of a diploid cell, the Zygote. The Zygote gives rise to the pre-embryo and later globular, heart shaped and mature embryo. A dicot embryo consists of embryonal axis and two cotyledons. The portion of embryonal axis above the cotyledons is the epicotyl and it terminates with the plumule or stem tip. The portion below the level of cotyledons is the hypocotyl that ends in the radicle or root tip. The root tip is covered with a root cap.




OR

(a) Draw a labelled diagram of a sectional view of human seminiferous tubule. (b) Differentiate between gametogenesis in human males and females on the

basis of (i) time of initiation of the process. (ii) products formed at the end of the process.

Ans.(a)




(b)

Spermatogenesis Oogenesis

1. it takes place in the seminiferous tubules of the testes by the repeated division of spermatozoa in male.

2. Spermatogenesis starts at the age of

puberty due to significant increases in the secretion of gonadotropin releasing

hormone. These are of two types

luteinizing hormone follicle stimulating

hormone.

3. It occurs in four stages spermatogenesis → primary spermatocytes → 1𝑠𝑡 meiotic division -+ secondary spermatocytes → 2𝑛𝑑meiotic division spermatids → differentiation → spermatozoa. Spermatogenesis involves the maturation of spermatids in to sperms. Primary spermatocytes give rise to four spermatozoa.

1. It take place in the Graafian follicles of the

ovary by the repeated division of oogonia

in the female.

2. Oogenesis is initiated during the embryonic

development stage. When a couple of

million gamete mother cells are formed

within foetal ovary.

3. It occurs in three stages Oogonia → Mitosis Primary oocyte → 1st meiotic division →secondary oocyte→ ovum. Primary oocyte gives rise to one ovum only and three Polar bodies. Ovum thus, formed are non-motile and generally full of food reserves.




Q.25. (a) Explain the experiment performed be Griffith on Streptococcus pneumonia. What did he conclude from this experiment? 5 marks (b) Name the three scientists who followed up Griffith's experiments. (c) What did they conclude and how? Ans. (a) Frederick Griffith, in a series of a experiments with streptococcus pneumoniae witnessed a miraculous transformation in the bacteria. During the course of his experiment, a living organism had changed in physical form.

When streptococcus preumoniae bacteria are grown on a culture plate, some produce smooth shiny colonies (S) while other produce rough colonies (R). This is because the S strain bacteria have a mucous (polysaccharide) coat, while R strain does not. S strain mice die from pneumonia but R strain mice did not develop pneumonia. S (strain)-Inject into mice → Mice die. R (strain)-Inject into mice → Mice Live. S strain (heat-killed)-Inject into mice →{ice live. S strain (heat-killed)-Inject into mice-Mice die. + R strain (live) He concluded that the R strain bacteria have been transformed by the heat-killed S-strain bacteria. (b) Prior to the work of Oswald Avery, Colin Macleod and Maclyn McCarty the genetic material was thought to be a protein. They worked to determine the Biochemical nature of ‘transforming principle’ in Griffith's experiment. They purified biochemical (proteins, DNA, RNA etc.)from the heat-killed S cell into 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.

OR

Two blood samples A and B picked up from the crime scene were handed over to the forensic department for genetic finger printing. Describe how the technique

of genetic finger printing is carried out. How will it be confirmed whether the sample belonged to the same individual or two different individuals? Ans. DNA Fingerprinting: This technique was discovered by Alec Jeffreys in 1985. The technique involved following steps: (i) Isolation and extraction of DNA from the cell by centrifugation.

(ii) By the help of enzyme restriction endonuclease, DNA molecules are digested. The fragment also contains VNTRs. (iii) The small DNA fragments are separated through gel electrophoresis set-up that

contains agarose polymer gel. (iv) the separated DNA fragments are transferred from electrophoresis plate to synthetic membrane like nitrocellulose or nylon membrane sheet called Southern blotting. (v) The DNA probes are added which target a specific nucleotide sequence which is complementary to them and this process is called hybridization. (vi) The nylon membrane is exposed to an X-ray film and dark orange coloured bands developed at sites where probes have bound to the DNA fragments. This is

known as autoradiography on comparing the DNIA print of blood samples A and B, it can be confirmed that the blood sample picked up from the crime scene belongs to the same individual or to different individuals.




Q. 26. Describe in sequence the events that lead to the development of a 3-celled pollen grain from microscope mother cell in angiosperms. 5 marks Ans. Microsporogenesis: (a) The interior of the microsporangium is filled with sporogenous cells. These sporogenous cells divide meiotically to form microspore tetrads.

(b) So, the cells of sporogenous tissue are called pollen or microspore mother cell (PMC) because they undergo meiosis to produce pollen grains. (c) The process of formation of microspore from a pollen mother cell through meiosis is called microsporogenesis. (e) When the anther matures, and dehydrates, the microspores differentiate into pollen Grains.

Pollen grains (a) They develop from PMC due to meiotic division. (b) They represent the male gametophyte. (c) Pollen grains are generally spherical in outline. (d) They possess two prominent layered walls. (e) The outer layer is called exine and the inner layer is called intine. (f) The exine is the outermost hard layer chemically composed of sporopollenin which

is one of the most resistant organic material. (g) The exine has prominent aperture formed due to absence of sporopollenin through which the pollen tube develops called germ pore. (h) The innermost thin layer chemically composed of cellulose and pectin is called

intine. (i) The newly differentiated pollen grain has central nucleus and dense cytoplasm. (j) Vacuoles develop thereby pushing the nucleus towards the periphery. (k) The protoplast then divides mitotically forming two unequal cell - bigger

vegetative which is rich in food reserve and smaller generative cell with dense cytoplasm and a nucleus. Pollen mother cells undergo meosis to form a microspore tetrad by the process called microsporogenesis. The microspores dissociate from each other and develop into pollen grains. The protoplast (pollen grain) then divides mitotically to form two unequal cells- the bigger vegetative cell and smaller generative cell.



CBSE 12th Biology 2017 Guess Paper By 4ono · 2017-09-19 · CBSE 12th Biology 2017 Guess Paper By...

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