series bio/sp/1c code no. sp/1-c...sample paper (cbse) series bio/sp/1c code no. sp/1-c sp/1-c...

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Sample Paper (CBSE) Series BIO/SP/1C Code No. SP/1-C SP/1-C ©Educomp Solutions Ltd. 2014- 15 BIOLOGY Time Allowed: 3 hours Maximum Marks: 70 General Instructions: (i) All questions are compulsory. (ii) This question paper consists of four Sections A, B, C , D and E. Section A contains 5 questions of one mark each, Section B is of 5 questions of two marks each, (iii) Section C has 12 questions of three marks each and Section D is of 1 questions of four marks each Section E is of 3 questions of five marks each. (iv) There is no overall choice .However, an internal choice has been provided in one question of 2 marks, one question of 3 marks and all the three questions of 5 marks weightage. A student has to attempt only one of the alternatives in such questions. (v) Wherever necessary, the diagrams drawn should be neat and properly labelled. SECTION A 1. Name the disorder with the following chromosome complement. a. 22 pairs of autosomes + X X Y b. 22 pairs of autosomes + 21st chromosome + XY. 1

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Page 1: Series BIO/SP/1C Code No. SP/1-C...Sample Paper (CBSE) Series BIO/SP/1C Code No. SP/1-C SP/1-C ©Educomp Solutions Ltd. 2014-15 BIOLOGY Time Allowed: 3 hours Maximum Marks: 70 General

Sample Paper (CBSE)

Series BIO/SP/1C Code No. SP/1-C

SP/1-C ©Educomp Solutions Ltd. 2014-15

BIOLOGY

Time Allowed: 3 hours Maximum Marks: 70

General Instructions:

(i) All questions are compulsory.

(ii) This question paper consists of four Sections A, B, C , D and E. Section A

contains 5 questions of one mark each, Section B is of 5 questions of two

marks each,

(iii) Section C has 12 questions of three marks each and Section D is of 1 questions of four marks each Section E is of 3 questions of five marks each.

(iv) There is no overall choice .However, an internal choice has been provided in

one question of 2 marks, one question of 3 marks and all the three questions of

5 marks weightage. A student has to attempt only one of the alternatives in

such questions.

(v) Wherever necessary, the diagrams drawn should be neat and properly

labelled.

SECTION A

1. Name the disorder with the following chromosome complement. a. 22 pairs of autosomes + X X Y b. 22 pairs of autosomes + 21st chromosome + XY.

1

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SP/1-C ©Educomp Solutions Ltd. 2014-15

2. What does ‘R’ represent in the given equation for productivity in an ecosystem ? GPP – R = NPP

3. A boy has been diagnosed with ADA-deficiency. Suggest any

one possible treatment. 4. Identify the two correct statements from the following :

a. Apiculture means apical meristem culture. b. Spinach is iron-enriched. c. Green revolution has resulted in improved pulse-yields. d. Aphids cannot infest rapeseed mustard.

5. Name the type of evolution that has resulted in the

development of structures like wings of butterfly and bird. What are such structures called ?

SECTION B

6. A new breed of sheep was developed in Punjab by crossing two

different breeds of Sheep. Name the two breeds which were crossed and the new breed developed.

7. Why is ‘starter’ added to set the milk into curd? Explain. 8. List four causes of biodiversity loss.

OR

1 1 1 2 2

1

2

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Name two metals used in a catalytic converter. How do they help in keeping the environment clean ?

9. List the different parts of the human oviduct through which

the ovum

10. A potato plant is infected with a virus. Name and explain a method to obtain virus-free potato plants from it.

SECTION C

11. A tRNA is charged with the amino acid methionine.

a. Give the anti-codon of this tRNA. b. Write the Codon for methionine. c. Name the enzyme responsible for binding of amino acid

to tRNA. 12. Hot spots are the regions of exceptionally high biodiversity.

But they have become regions of accidental habitat loss too. Name the three hot spots of our country. Why are they called. Hot spot.?

2 2 2 3

3

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13. In the figure, structure of an antibody molecule is shown.

Observe it and Give the of the following questions. a. Label the parts A, B and C. b. Which cells produce these chemicals? c. State the function of these molecules.

14. Draw a labelled sketch of a typical biogas plant.

OR

a. Name the causative organisms for the following diseases : 1. Elephantiasis 2. Ringworm 3. Amoebiasis

b. How can public hygiene help control such diseases ?

3

3

3

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15. Following a severe accident, many charred-disfigured bodies are recovered from the site making the identification of the dead very difficult. Name and explain the technique that would help the authorities to establish the identity of the dead to be able to hand over the dead to their respective relatives.

16. a. What is a bioreactor ? How does it work ?

b. Name two commonly used bioreactors.

17. Explain polygenic inheritance with the help of a suitable example. 18. Name the host plant and its part that Meloidogyne incognita

infects. Explain the role of Agrobacterium in the production of ds-RNA in the host plant. 19. With the help of a schematic diagram, explain the location and the role of the following in a transcription unit : Promoter, Structural gene, Terminator. 20. Morgan carried out several dihybrid crosses in Drosophila and found F2-ratios deviated very significantly from the expected Mendelian ratio. Explain his findings with the help of an example. 21. Describe endosperm development in angiosperm.

3 3 3 3 3 3

3

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22. A few residents in your locality, for business gains, have established small-scale industrial / commercial activities such as pathological labs and fabric dyeing centres without obtaining ‘No objection certificates’ from municipal authorities. Would you support these activities ? Give any three reasons in support of your .

SECTION D 23. Pooja’s grandmother blames her mother for giving birth to three daughters and wants another marriage of her son.

a. Is it right to blame a female for sex determination? b. Justify your with the help of appropriate llustration.

SECTION E

24. a. Draw a simplified model of phosphorus cycling in a

terrestrial ecosystem. b. Write the importance of such cycles in ecosystems.

OR a. Explain the narrowly utilitarian, broadly utilitarian and

ethical arguments in favour of conservation of biodiversity. b. How is designation of certain areas as hotspots a step

towards biodiversity conservation ? Name any two hotspots in India.

3 4 5 5

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25. a. Work out a cross between a tall pea plant bearing violet

flowers (heterozygous for both) with a dwarf pea plant having white flowers. Write the genotypes and phenotypes of the progeny along with their ratios.

b. Name such a cross and state its importance.

OR Explain the process of translation. 26. Schematically represent and explain the events of

spermatogenesis in humans.

OR Angiosperm flowers may be monoecious, cleistogamous or show self-incompatibility. Describe the characteristic features of each one of them and state which one of these flowers promotes inbreeding and outbreeding respectively.

5 5 5 5

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SECTION A

1. a. Klinefelter.s Syndrome

b. Down.s syndrome

2. Respiration

3. Gene Therapy 4. (ii) , (iii) 5. Analogous structures as a result of convergent evolution

SECTION B

6. By crossing Bikaneri ewes and Marino rams, the new breed

Hisardale was developed.

7. The inoculum or starter contain millions of LAB which at suitable temperatures multiply and convert milk to curd. It also improves its nutritional quality by increasing vitamin B12.

8. a. Habitat loss and fragmentation b. Over-exploitation: c. Alien species invasions:

ANSWERS

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d. Co-extinctions:

OR

The two metals used in a catalytic converter are a. Platinum-palladium b. Rhodium

They help in keeping the environment clean by reducing emission of poisonous gases by

1. Converting unburnt hydrocarbons into carbon dioxide and water.

2. Changing Carbon monoxide and nitric oxide to carbon dioxide and nitrogen gas, respectively.

9. a. Funnel-shaped infundibulum.

b. Finger-like projections called fimbriae on the edges of the infundibulum.

c. Wider part of the oviduct called ampulla. d. The last part of the oviduct is the isthmus

10. A potato plant is infected with a virus. Name and explain a method to obtain virus-free potato plants from it. a. Remove the apical and axillary meristem that is free of virus. b. Grow the meristem in vitro to obtain virus-free plants.

SECTION C

11. a. UAC

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b. AUG c. Amino-acyltRNA synthetase. 12. Westerm Ghats and Sri lanka; Indo-Burma; Himalaya called

¡¥biodiversity hot spots¡¦ as they show a. High level of species richness b. High degree of endemism

13. a. A-Antigen binding site B-Light chain b. B-lymphocytes. c. Heavy Chain d. Antibodies provide acquired immune response.

14.

OR a. 1. Wuchereria bancrofti and Wuchereria. malayi

2. Microsporum, Trichophyton and Epidermophyton 3. Entamoeba histolytica

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b. Elephantiasis can be prevented by disinfection of water reservoirs and prevention of mosquito breeding. Ringworm include personal hygiene like keeping the body

clean . Amoebiasis can be prevented by consumption of clean drinking water, food, vegetables, fruits,

15. The technique of DNA fingerprinting would help the

authorities to establish the identity of the dead to be able to hand over the dead to their respective relatives.

Steps/procedure in DNA fingerprinting .

. Extraction of DNA - using high speed refrigerated centrifuge. . Amplification - many copies are made using PCR

. Restriction Digestion - using restriction enzymes DNA is cut into fragments.

. Separation of DNA fragments - using electrophoresis-agarose polymer gel.

. Southern Blotting : Separated DNA sequences are transferred onto nitrocellulose or nylon membrane.

. Hybridisation : The nylon memberane exposed to radio active probes.

. Autoradiography : The dark bands develop at the probe site.

16. a. A bioreactor is a vessel in which raw materials are biologically converted into specific products, individual enzymes, etc.,using microbial plant, animal or human cells.

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It provides the optimal conditions for achieving the desired product by providing optimum growth conditions (temperature, pH, substrate, salts, vitamins, oxygen).

It has an agitator system, an oxygen delivery system, a foam control system, a temperature control system, pH

control system and sampling ports so that small volumes of the culture can be withdrawn periodically.

b. 1. Simple stirred-tank bioreactor; 2. Sparged stirred-tank bioreactor

17. In polygenic inheritance, a cross between two pure breeding parents produces an intermediate trait in F1.

In F2 generation, apart from the two parental types, there are several intermediates. Gradiations, show a bell shaped curve. F1 hybrid form 8 kinds of gamete in each sex giving 64 combination in F2 having 7 genotype and phenotype. Polygenic inheritance of skin tone 3 loci : each has two possible alleles : Aa, Bb, Cc, each capital allele adds one unit of darkness, each lower case allele adds nothing. When parents with intermediate tone are crossed Offspring produced can have tone darker or lighter than either parent

18. Meloidegyne incognitia infects the roots of tobacco plants

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Nematode specific genes were introduced into the host plant using Agrobacterium as a vector producing both sense and antisense RNA in the host cells. These two RNAs formed an inactive double stranded RNA (dsRNA) that silences mRNA

by the process called RNA interference (RNAi). This result in protection of the transgenic plant from the parasite which could not survive in the transgenic host.

19.

20. A dihybrid heterozygous round, yellow seeded garden pea (Pisum

sativum) was crossed with a double recessive plant. a. What type of cross is this? b. Work out the genotype and phenotype of the progeny. c. What principle of Mendel is illustrated through the result of

this cross? Answer : a. It is a dihybrid test cross b. Parent RrYy (Round Yellow) rryy (Wrinkled green) Gametes RY , Ry , rY , ry X ry Gametes RY Ry rY ry F1 progeny ry RrYy Rryy rrYy rryy Round, Round and Wrinkled Wrinkled, Yellow Green Yellow Green

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Phenotypic ratio : 1 : 1 : 1 : 1 Genytopic ratio : 1 : 1 : 1 : 1 c. It illustrates the Principle of independent assortment.

21. The nucleus of the functional megaspore (n) undergoes three

successive mitotic cell division which results the formation of eight nucleate stage of embryo sac (free nuclera division) q The cell wall formation starts at eight nuclear stages. Three cells are grouped together at micropylar end to form the egg apparatus (2synergids + 1 egg cell). q Three cells are grouped at chalazal end, called antipodal cells. q The remaining 2 nuclei are called polar nuclei move to the centre of embryo sac, called central cell. Thus typical angiospermic embryo sac at maturity is 8 nucleated and 7 celled.

22. Fabric dyeing centres with no objection certificates’ from municipal authorities are likely to cause: a. Noise pollution that may cause sleeplessness, increased heart

beating, altered breathing pattern, thus considerably stressing humans.

b. Water pollution by releasing toxic substances in industrial waste waters that can undergo biomagnification in the aquatic food chain.

c. Pathological labs with no objection certificates’ from municipal authorities may generate hazardous wastes that contain disinfectants and other harmful chemicals, and also pathogenic micro-organisms.

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SECTION D

23. a. It is not right to blame a female for sex determination. b. 1. In males two types of gametes are produced in which 50

per cent of the total sperm produced carry the X- chromosome and the rest 50 per cent has Y-chromosome.

2. Females produce only one type of ovum with an X- chromosome.

3. There is an equal probability of fertilisation of the ovum with the sperm carrying either X or Y chromosome.

In case the ovum fertilises with a sperm carrying X- chromosome the zygote develops into a female (XX) and the fertilization of ovum with Y-chromosome carrying sperm results into a male offspring.

4. Thus, it is evident that it is the genetic makeup of the sperm that determines the sex of the child.

It is also evident that in each pregnancy there is always 50 per cent probability of either a male or a female child.

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24. a.

b. The importance of such cycles in ecosystems is as follows: 1. Recycling of nutrients through the various components of

an ecosystem indefinitely.

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2. Gaseous (eg.CO2 ,Nitrogen) as well as sedimentary nutrients(sulphur ,Phospherous) are cycled.

3. Important nutrients like sulphur ,Phospherous whose reservoir is located in Earth’s crust is made available to living organisms.

4. It helps to meet with the deficit which occurs due to imbalance in the rate of influx and efflux of nutrients.

OR

a. 1. The narrowly utilitarian arguments for conserving biodiversity are the direct economic benefits from nature like food(cereals, pulses, fruits), firewood, fibre, construction material, industrial products (tannins, lubricants, dyes, resins, perfumes ) and products of medicinal importance. 2. The broadly utilitarian argument is the provision of 20

per cent of the total oxygen in the earth’s atmosphere through photosynthesis by the rich biodiversity of Amazon forest Pollination by bees, bumblebees, birds and bats helps to provide fruits or seeds. Provision of aesthetic pleasures like walking through thick woods, watching spring flowers in full bloom or waking up to a bulbul’s song in the morning. 3. The ethical argument is that we share biodiversity with

millions of plant, animal and microbe species. Every species has an intrinsic value, even if it may not be of current or any economic value to us. We also have a moral

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duty to care for their well-being and pass on our biological legacy in good order to future generations.

b. As ‘biodiversity hotspots’ regions have very high levels of

species richness and high degree of endemism. Hence are a step towards biodiversity conservation.

Two hotspots in India are Western Ghats and Himalayas.

SECTION E

25. a. When a tall pea plant bearing violet flowers(heterozygous for both) with a dwarf pea plant having white flowers. Phenotype: All Tall and violet Genotype ratio is 1 : 1 : 1 : 1. a. Such a cross is called a test cross. The test cross is used to find the genotype of an organism.

OR Translation refers to the process of polymerisation of amino acids

to form a polypeptide. The order and requence of amino acids are defined by the sequence of bases in the mRNA.

First step is - charging of tRNA or aminoacylation of tRNA-here amino acids are activated in the presence of ATP and linked to specific tRNA. Initiation - Ribosome binds to mRNA at the start codon (AUG) that is recognised by the initiator tRNA.

Elongation phase - Here complexes composed of an amino acid linked to tRNA, sequentially bind to the appropriate codon in

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mRNA by forming complementary base pairs with tRNA codon. The ribosomes move from codon to codon along with the mRNA. Amino acids are added one by one, translated into polypeptide sequences. Termination - Release factors binds to the stop codon, terminating translation and releasing the complete polypetide from the ribosome.

26.

a. In testis at puberty spermatogonia produce sperms by the process of spermatogenesis .

b. The spermatogonia multiply by mitotic division and increase in numbers.

c. Each spermatogonium is diploid and contains 46 chromosomes.

d. Some of the spermatogonia called primary spermatocytes periodically undergo meiosis.

e. A primary spermatocyte completes the first meiotic division (reduction division) leading to

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f. formation of two equal haploid cells called secondary spermatocytes having only 23 chromosomes each.

g. The secondary spermatocytes undergo the second meiotic division to produce four equal, haploid

h. spermatids i. The spermatids are transformed into spermatozoa (sperms)

by the process called j. spermiogenesis.

OR

cleistogamous flowers characteristics: a. They do not open at all. b. The anthers and stigma lie close to each other. c. When anthers dehisce in the flower buds, pollen grains come

in contact with the stigma to effect pollination. It is autogamous

d. They produce assured seed-set even in the absence of pollinators.

Monoecious characteristics: a. The flowers are unisexual. b. Both male and female flowers are present on the same plant

(eg. castor and maize ) c. It prevents autogamy but not geitonogamy. d. self-incompatibility : e. It is a genetic mechanism that prevents self-pollen (from the

same flower or other flowers of the same plant) from fertilizing

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the ovules by inhibiting pollen germination or pollen tube growth in the pistil.

cleistogamous flowers promote inbreeding and monoecious, cleistogamous promotes outbreeding.