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CBSE ANNUAL PAPER - 2000

BIOLOGY

(SET-I)

Time Allowed : 3 Hours

Max.Marks : 70

General Instructions :
(i) All questions are compulsory.

(ii) Marks for each question are indicated against it.

(iii) Q. Nos. 1-11 are very short answer type carrying one mark each only. Answer them in one word one sentence.

(iv) Q. Nos. 12-23 are short answer questions carrying 2 marks each. It is appropriate to answer them in about 30-50 words.

(v) Q. Nos. 24-28 are short answer questions carrying 3 marks each. It is appropriate to answer them in about 30-50 words,.

(vi) Q. Nos. 29-32 are long answer questions carrying 5 marks each. It is appropriate to answer them in about 40-100 words.

(viii) Do not unnecessarily make the answer lengthier than desired.

Section - A

Q. Nos. 1-8 are of very short answer type carrying 1 mark each. Answer them in one word/ one sentence having 1 - 20 words each.

Q.1.

What is meant by trisomic condition ?

Ans.

It is a condition in which three chromosomes of a particular type exist together.

Q.2.

State the presence or absence of sinus venous in a fish and a mammal.

Ans. Fish - sinus venous present. Mammal - sinus venous absent.
Q.3.

Which type of cartilage is present between vertebrae to allow limited movement ?

Ans. Hyaline cartilage
Q.4.

Name any two breeds of wild rock pigeon that have been developed through artificial selection.

Ans. Jacobin, Pouter.
Q.5. Which chromosomes are called autosomes ?
Ans. Somatic chromosomes are called autosomes.
Q.6. When is tumour referred to as malignant ?
Ans.

Tumour is called malignant when it spreads to distant places.

Q.7.

What are the two factors that contribute for the disassociation of oxyhaemoglobin in the arterial blood to release molecular oxygen in an active tissue ?

Ans. (i) Lack of ATP,(ii) Lack of Oxygen.
Q.8.

Which one of the following is fatal unless oral rehydration therapy is administered: (i) Whooping cough, (ii) Cholera, (iii) Tetanus

Ans. Cholera

Section - B

Q. 9-18 are of short answer type carrying 2 marks each. Answer them in approximately 20-30 words each.

Q.9. What are the two functions of DNA polymerase ?
Ans. The two functions of DNA polymerase are -

(i) To synthesize new DNA strand in 5 - 3 direction.

(ii) To do proof reading and DNA repair.

Q.10.

Stanley Miller and Harold C. Urey performed an experiment by recreating in the laboratory the probable conditions of the atmosphere of the primitive earth.

(i) What was the purpose of the experiment ?

(ii) In what forms was the energy supplied for the chemical reactions to occur ?

(iii) For how long the experiment was run continuously ?

Ans. (i) To show that life originated from inanimate.

(ii) Electrodes.

(iii) A week.

Q.11.

How does the mustard inflorescence differ from the banana inflorescence in arrangement ? Give the technical term for each.

Ans. Mustard Inflorescence Banana Inflorescence
(i) It is called Raceme. (i) It is called spadix.

(ii)Flowers are borned
in acropetal succession
and are pedicallate.

(ii)Flowers one borne
in acropetal succession
but are sessile.

Q.12.

Name the basic nitrogenous catabolite of proteins produced in birds. In what form is it is eliminated from their body ? What is the advantage of this type of excretion ?

Ans.

The basic nitrogenous catabolite of proteins produced in birds is uric acid. It is eliminated from their body in the form of solid. This type of excretion is useful to conserve water.

Q.13.

Describe the theory of inheritance of acquired character. Who disproved it experimentally ?

Ans.

Jean Baptist de Lamarck gave the theory of inheritance of acquired characters. It has three main features -

1. The effect of environment : The environment changes the organism for its better adaptation.

2. Use and disuse of body parts : When the organism adapts itself for a new environment the organs that are more in use are changed and this change is seen in the next generations.

3. Inheritance of acquired characters : When these changes are regularly inherited to the next generations the organism represents a new species.

Weismann disproved this theory. He showed that even after cutting the tail of the rats for several generations, no rat was born without a tail. This means that disuse of tail did not lead to loss of the tail in the offsprings.

Q.14.

Differentiate between morula and blastula of mammals.

Ans. Morula Blastula
(i) It is 16-32 celled structure. (i) It is 32-64 celled structure.
(ii) It is solid. (ii) It is hollow.
(iii) It doesn't have cavity. (iii) It has cavity blastocoel.
Q.15.

What is the biological significance of Azolla pinnate in agriculture ?

Ans.

The biological significance of Azolla pinnate in agriculture is that it helps in nitrogen fixation and increases the field of rice crop.

Q.16.

Give any two reasons to justify that ginger is a modified stem.

Ans.

The two reasons that justify that ginger is a modified stem are -

(i) It bears nodes and internodes.

(ii) It bears scaly leaves.

Q.17.

Define biogeography. How do Darwin's Finches provide the biogeographical evidence in favour of evolution ?

Ans.

Biogeography is the study of distribution of animals and plants in the world. Darwin's finches were found on Golapagos islands. They differ in size and beak shape as they adapted themselves according to the available food in that island.

Q.18.

Explain the structural and functional significance of fovea in the human eye.

Ans.

The structural and functional significance of fovea in human eye is

(i) Fovea consists of cones only.

(ii) It is used to focus a particular object.

Section - C

Q. Nos. 19-27 are of short answer type carrying 3 marks each. Answer them in approximately in 30-50 words each.

Q.19.

What is meant by R cells and S cells with which Frederick Griffith carried out his experiments on Diplococus pneumoniae ? What did he prove from these experiments.

Ans.

In 1928, an English bacteriologist, Fredrick Griffth conducted experiments with the bacterium that causes pneumonia, Diplococcus Pneumonia. There are two main types of this bacteria, smooth (S) and rough (R) S-S-type cells have a capsule around capsule, are virulent and cause pneumonia. The R-type bacteria do not have the capsule and are avirulent and harmless. The S-type bacteria when injected into mice cause pneumonia and subsequent death of the animal whereas the R-type bacteria when injected into mice and found to be harmless. Griffith found that the S-type bacteria when killed by heat prior to injection into mice did not cause the disease. However, when heat killed S-type bacteria were injected into mice along with avirulent, R-type bacteria, mice suffered from pneumonia and died. Examination of the dead mice revealed the death occurred due to the living S-type bacteria. Griffith concluded from this experiment that the avirulent R-type bacteria had been transformed into S-type, capsule variety. In other words, the dead S-type bacteria had transmitted their virulence to the living R-type avirulent bacteria. This experiment showed the some component of the bacterial cell is responsible for the phenomenon of transformation.

Q.20.

Draw a diagram of T.S. of a part of somniferous tubule of testis of an adult human male and label any six parts in it.

Ans.
Q.21.

How does industrial melanism bring out the action of natural selection ?

Ans.

Biston betularia (Peppered moth) exists in two strains white ( dull grey) and black. It was found on tree trunks in England in abundance before industrial revolution. As the tree - barks were covered by whitish lichens thus white moths were able to hide them from predating birds.

But with industrialisation, barks of trees got covered by black sooth of smoke and converted to blackish colour as a result white moths were easily identified by predator birds. But black moths (just 10 % of total ) could escape themselves slowly and gradually through hundred of years, black coloured moths ( carbonaria) increased to 90% and white moths decreased.

Gradual replacement of coal by oil and electricity and measures to control soot formation, the trees recovered and become lighter in shade hence again population of white moths increased. It showed and proved how nature has played its role in deciding the type of population to exist according to their adaption towards the environment.

Q.22.

How does intestinal juice contribute in the digestion of proteins ? What provides alkaline pH in the small intestine ?

Ans.

1. Entero peptidase - It activates pancreatic trypsinogen to trypsin and trypsin breaks proteins into peptides.

2. Amino peptidases - It separates amino acids from peptides.

3. Di peptidases - They hydrolyse dipetides into amino acids. Alkaline medium is provided by brunner's glands in the intestines.

Q.23.

An RNA strand has a series of condoms out of which three are mentioned below :(i) AUG (ii) UUU and (iii) UAG.

(a) What will these condoms be translated into ?

(b) What are the DNA condons that would have transcribed these RNA condons ?

Ans. (a) (i) AUG - Methionine

(ii) UUU - pheny alanine

(iii) UAG - Nonsense condon

(b) (i) AUG - TAC

(ii) UUU - AAA

(iii) UAG - ATC

Q.24.

List any four symptoms shown by a Down's syndrome afflicted child. Explain the cause of this disorder.

Ans. For symptoms of Down's Syndrome are -

1. Rounded face

2. Broad forehead

3. Flatened nasal bridge

4. Open mouth

It oooccurs due to trisomy of 21st Chromosome.

Q.25.

Answer the following with reference to the anatomy of dicot stem :

(i) Where exactly are the cambial cells located in the vascular bundles ?

(ii) What is the name given to such a bundle ?

(iii) How are the exylem vessels arranged ?

(iv) What type of cells constitute the pith ?

Ans. (i) In between xylem and phloem cells.

(ii) Collateral and open.

(iii) Endarch conditions.

(iv) Parenchyma.

Q.26.

Explain any three chemical barriers that offer non-specific defence mechanism.

Ans.

Three chemicals barriers that offer non-specific defence mechanism are - (i) Sweat (ii) Sebum (iii) Lysozyme.

Q.27.

Trace the development of microsporocyte in the anther to a mature pollen - grain.

Ans.

Anther consists of four anther sacs. Anther sac is made up of sporogenous tissue. It consists of cytoplasmic cells. The cells divide mitotically and forms microspore mother cells. Each microspore mother cell undergoes meiosis to form four haploid microspore.

The four microspore formed from a microspore mother cell are arranged in a tetrahedral form. Each microspore or pollen grain has two layers, an outer thick exine and an inner thin intine. The pollen grain divides to form a generative cell and a vegetative cell. The generative cell give rise to two male gametes and a vegetative cellforms pollen tube nucleus that direct pollen tube towards micropyle for fertilization.

Section - D

Q. Nos. 28-30 are long answer types carrying 5 marks each. Answer them in approximately 80-120 word each.

Q.28.

In the case of snapdragon ( Antirrhinum majus) a plant with red flowers was crossed with another plant with white flowers. Trace the inheritance of flower colour up to the F2 generation indicating the genotypes and phenotypes at each level. What special feature do you note in the genotypic and phenotypic ratios in F2 generation ?

Ans.
Q.29.

Name the hormone that regulates each of the following and mention the source of it :

(i) Urinary elimination of water

(ii) Storage of glucose as glycogen

(iii) Sodium and potassium ion metabolism.

(iv) basal metabolic rate

(v) Descent of testes into the scrotum

Ans. (i) ADH Posterior pituitary
(ii) Insulin Pancreas
(iii) Minero corticoid Adrenal gland
(iv) T3 and T4 Thyroid
(v) Tetosterone Testes
Q.30.

Describe the C4 pathway in a paddy plant. How is this pathway an adaptive advantage to the plant ?

Ans.

The anatomy of leaves of C4 plants is different from leaves on C3 plants. This type of anotomy is called Kranz anatomy. In the leaves of such Plants, Palisade tissue is absent. There is bundle sheath around the vascular bundles. The chloroplasts in the bundle sheath cells present are large and without or less developed groove whereas in the mesophyll cells the chloroplasts are small but with well developed groove.

CO2 taken from atmosphere is accepted by a 3 carbon compound. Pyruvate in the chloroplasts of mesophyll cells leading to the formation of 4-C Compound oxaloacelate converted to another 4-C Compound the malic acid. It is transported to the chloroplasts of bundle sheath cells. Here malic acid (C4) is converted to Pyruvic acid (C3) with the release of CO2. Thus concentration of CO2 increases in the bundle sheath cells. These cells contain enzymes of Calvin cycle. Because of high concentration of CO2, RuBP Carboxylase participate in Calvin cycle and not in photorespiration . Sugar formed in Calvin cycle is transported into the phloem.

Pyretic acid generated in the bundle sheath cells re-enters mesophyll cells and regenerates phosphoenol pyruvic acid by consuming one ATP.

Since the conversion results in the formation of AMP ( not ADP), needs 12 glucose molecule, wheres C4 Pathway requires 30 ATP.

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