WBBSE Class 10 Life Science And Environment Model Question Paper 2023 Set 3

Model Question Paper Life Science And Environment Set 3

Answer to all questions is compulsory

1. Write the answer in complete sentence by choosing the correct answer for each question with the respective serial number.

Question 1. Select the correct pair-

  1. Cerebrum-Maintenance of the balance of the body
  2. Hypothalamus-Control of intelligence and emotion
  3. Cerebellum-Control of body temperature
  4. Medulla Oblongata-Control of heartbeat and swallowing of food

Answer: The correct Pair is- Medulla Oblongata—Control of heartbeat and swallowing of food.

Question 2. Identify which of the following statements is not true regarding Insulin-

  1. Helps in the absorption of glucose from the blood into most of the somatic cells
  2. Converts glucose into glycogen within liver and muscle cells
  3. Helps in the conversion of fat and protein into glucose
  4. Inhibits the conversion of protein and fat into glucose

Answer: The statement which is not true regarding Insulin is—Helps in the conversion of fat and protein into glucose.

Question 3. Match the words of Column-A with that of Column-B and select which of the following options is correct-

WBBSE Model Question Paper 2020 Life Science And Environment Set 3 Match The Columns.

  1. 1-1, 2-2,3-3
  2. 1-2, 2-3, 3-1
  3. 1-3, 2-1, 3-2
  4. 1-2, 2-1, 3-3

Answer: 3. 1-3, 2-1, 3-2

Question 4 Determine from the answers given below in which phases of karyokinesis during mitotic cell division following two incidents happen- 1. Daughter chromosomes tend to move apart from each other towards their own poles Nuclear membrane and nucleolus disappear

  1. 1. Prophase 2. Anaphase
  2. 1. Anaphase 2. Prophase
  3. 1. Telophase 2. Metaphase
  4. 1. Metaphase 2. Telophase

Answer: 2. During mitotic cell division—

  1. Daughter chromosomes tend to move apart from each other towards their own poles— Anaphase and
  2. Nuclear membrane and nucleolus disappear—Prophase

Question 5. Select which of the following is the feature of cross-pollination-

  1. Occurs within the same flower of the same plant
  2. Agents are not required
  3. Lesser chance of new characters being transmitted
  4. More wastage of pollen grains

Answer: The feature of cross-pollination is— More wastage of pollen grains.

Question 6. Determine the number of DNA molecules that get coiled to form each chromosome in a newly formed daughter cell produced by mitotic cell division in the human body-

  1. 46
  2. 1
  3. 23
  4. Numerous

Answer: 2. The number of DNA molecules that get coiled to form each chromosome in a newly formed daughter cell produced by mitotic cell division in human body is 1.

Question 7. Identify the genotype of guineapig having black colour and rough hair-

  1. BbRr, BBRr
  2. bbRR, bbRr
  3. BBrr, Bbrr
  4. bbrr, bbRr

Answer: 1. The genotype guineapig having black colour and rough hair are—BbRr, BBRr.

Question 8. Decide which of the following two were selected by Mendel as recessive traits-

  1. Colour of the flower-purple, the position of flower-axial
  2. Length of stem-dwarf, a form of ripe seed-wrinkled
  3. Form of the ripe seed-rund, colour of seed- yellow
  4. Position of flower-axial, length of stem- tall

Answer: 2. Length of stem—dwarf and Form of ripe seed—wrinkled were two recessive traits selected by Mendel.

Question 9. Asses from the following, the probable genotype of parents having a haemophilic son and normal daughter-

  1. H || h, hWBBSE Model Question Paper 2020 Life Science And Environment Set 3
  2. H || H, HWBBSE Model Question Paper 2020 Life Science And Environment Set 3
  3. H || H, hWBBSE Model Question Paper 2020 Life Science And Environment Set 3
  4. H || h, HWBBSE Model Question Paper 2020 Life Science And Environment Set 3

Answer: 3. The probable genotype of parents having a haemophilic son and normal daughter are- H || H, hWBBSE Model Question Paper 2020 Life Science And Environment Set 3

Question 10. Parthenium is an exotic species in our country. Other indigenous species cannot survive in such places where it grows. This establishes one of the postulates of Darwin’s theory. Identify the postulate-

  1. Intrasecific struggle
  2. Interspecific struggle
  3. Struggle with environment
  4. Origin of new species

Answer: 2. The correct postulate is Interspecific struggle.

Question 11. Miller and Urey, in their experiment, were able to synthesize some preliminary constituents necessary for the creation of life. Identify the ones which were amino acids among them-

  1. Lactic acid, Acetic acid
  2. Urea, Adenine
  3. Glycine, Alanine
  4. Formic acid, Acetic acid

Answer: The amino acids were—Glycine, and Alanine.

Question 12. Decide for which of the following purposes bees demonstrate waggle dance-

  1. Search for reproductive mates
  2. Inform other worker bees about the direction and the distance of the source of food from the bee hive
  3. Selecting a place for the construction of the new bee hive
  4. Avoid attack by probable enemy

Answer: 2. Bees demonstrate waggle dance to-^ Inform other worker bees about the direction and the distance of the source of food from the bee hive.

Question 13. Identify which of the following is the correct information related to biosphere reserve-

  1. Conservation is promoted to local people and other biotic communities along with the conservation of the ecosystem
  2. National Park and Sanctuary are not included within the biosphere reserve
  3. The presence and participation of local people in the conservation of the ecosystem are not permissible
  4. Its size is usually smaller than a Sanctuary

Answer: 1. The correct information related to biosphere reserve is- conservation is promoted to local people and other biotic communities along with the conservation of the ecosystem.

Question 14. Decide which of the following pair is not correct-

  1. Poaching-Increase the endangeredness of gorilla
  2. Exotic species-Lantana, Tilapia
  3. Determination of hotspot-Number of endemic species and endangered species
  4. Greenhouse gas-Eutrophication

Answer: The wrong pair is—A greenhouse gas— Eutrophication

Question 15. Decide which of the following Project Tiger is located within our state-

  1. Bandipur
  2. Simlipal
  3. Sunderbans
  4. Kanha

Answer: Sunderbans Project Tiger is located within our state.

 

Model Question Paper Life Science And Environment Set 3 Group B

2. Answer any 21 questions out of the 26 questions given below instructed.

Fill in the blanks with proper words in the following sentences (any five):

Question 1. Acharya Jagadish Chandra Bose proved the property of __________ by sending electrical impulses in Mimosa and Desmodium plants.
Answer: Sensitivity

Question 2. If gametes in humans were produced by mitosis instead of meiosis then the number of autosomes in a somatic cell of an offspring would have been __________.
Answer: 88

Question 3. A disease in the human population caused by a recessive gene located in the ‘X’ Chromosome is __________.
Answer: Haemophilia

Question 4. The hoof of modern horse is the transformation of the digit number __________ of their ancestors.
Answer: 3

Question 5. At the __________ phase of the nitrogen cycle, ammonia is converted into nitrite and nitrate by the action of some bacteria.
Answer: Nitrification

Question 6. To produce the bottled cold drinks widely sold in the market, a lot of __________ water is wasted.
Answer: 2.6 Fresh

Decide whether the following statements are true or false (any five):

Question 7. Tropic movement is the movement of growth in plants.
Answer: True

Question 8. Crossing over takes place during mitotic cell division.
Answer: False

Question 9. Mendel used the term gene while describing his experiments related with heredity.
Answer: False

Question 10. The leaf of the Cactus is modified into the spine for the reduction of the rate of transpiration.
Answer: True

Question 11. Rhododendron is an endangered plant species conserved in the Eastern Himalaya hotspot.
Answer: True

Question 12. Choroid helps in the accommodation of the eye by changing the curvature and shape of the lens.
Answer: False

Match the words in Column-A with those which are most appropriate in Column-B and re-write the correct pair mentioning the serial no. of both Columns (any five):

WBBSE Model Question Paper 2020 Life Science And Environment Set 3 Match The Columns

Answer: 13. D, 14. G, 15. A, 16. E, 17. B, 18. C

Answer in a single word or in a single sentence (any six):

Question 19. Choose the odd one and write it: Cerebrum, Hypothalamus, Pons, Thalamus
Answer: Pons

Question 20. Where the Schwann cells are located?
Answer: Schwann cells are found in close contact with axons in the peripheral nerves.

Question 21. A pair of related terms is given below. On the basis of the relationship in the first pair write the suitable word in the gap of the second pair. Mitosis: Radicle :: Meiosis: __________.
Answer: Spore mother cell

Question 22. Which law did Mendel conclude from his dihybrid cross experiment?
Answer: Law of Independent Assortment

Question 23. Give an example of a variation found among healthy persons which is transmitted through generations.
Answer: Rolling tongue and non-rolling tongue

Question 24. How do Chimpanzees break open the hard shells for eating the nuts?
Answer: Chimpanzees put the nut on the flat surface of a hard stone and use another stone or tough piece of wooden branch as a hammer to break the shell of nut.

Question 25. Among the following four terms, one includes the other three. Find it out and write it: SPM, Air Pollution, Greenhouse gas, Lung disease.
Answer: Air pollution

Question 26. Name the practice which jointly the local peoples and forest department maintain for the reclamation of a forest.
Answer: Joint Forest Management (JFM)

 

Model Question Paper Life Science And Environment Set 3 Group C

3. Answer any 12 questions in 2-3 sentences out of 17 questions given below.

Question 1. Distinguish between the functions of hormones and nervous system on the following parameters:

  1. Nature of function
  2. Pace of function
  3. A time span of function
  4. Fate

Answer:

Differences between the endocrine system and nervous system

WBBSE Solutions For Class 10 Life Science And Environment Chapter 1 Response And Physical Co-Ordination In Animals Nervous System Endocrine and nervous system

 

Question 2. ‘A person can see distant objects distinctly but is unable to visualize near objects in a perfect way’-Predict what would be the probable cause and suggest the corrective measure for such a problem.

Answer:

Hypermetropia:

Hypermetropia is a visual impairment in which a person can see any distant object but is unable to find the nearer objects clearly.

1. Characteristic feature:

In this disease, the diameter of the eyeball becomes anteroposteriorly compressed or the cornea becomes too much flattened. As a result, the image is formed beyond the retinal surface. So the patient finds a blurred image of the nearer objects.

2. Remedy:

To rectify hypermetropia, ophthalmologists recommend (+ve power) a convex glass lens. This glass lens creates an image on the retinal surface and makes the vision clear.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 1 Eye As A Sense Organ In Human Hypermetropia and its correction

 

Cataract:

Usually due to ageing, deficiency of certain proteins in the eye lens makes it opaque. This condition is called cataracts.

1. Characteristic feature:

The opacity of the lens increases gradually to make the vision blurred.

2. Remedy:

To solve the problem of cataracts, an eye surgeon can replace the defective lens with a synthetic plastic lens by surgical procedure.

Question 3. Analyse the role of synthetic plant hormones in increasing production and solving the problem of weeds in agriculture.

Answer:

Role of synthetic hormones:

Synthetic hormones are successfully used in agriculture and horticulture. The roles of these synthetic hormones are mentioned below.

1. Developing new plants from stem cuttings:

Cuttings are used for artificial vegetative propagation of different plants like roses, Hibiscus, marigold, Chrysanthemum, ‘etc. After cutting the twigs from a mother plant, a solution of synthetic auxin or auxin powder is applied at the cut end.

Then, these cuttings are planted in moistened soil. By the action of this hormone, adventitious roots grow from the cut end and the cutting grows as an individual daughter plant.

2. Preventing shedding of immature fruits:

Sometimes, immature fruits shed off from the plant if these plants are sprayed with synthetic auxin solution for a few times during ear y developmental phase of the fruits, the rate of immature shedding declines sharply.

Horticulturists spray auxin solution on mango, litchi, grapes, banana, and several other fruit plants to prevent immature shedding of fruits. Synthetic gibberellin and synthetic cytokinin are also effective.

3. Destroying weeds:

Weeds growing in crop fields share water and nutrients with agricultural crops. This affects the quality of production. Scientists have revealed that the application of certain phytohormones destroys dicotyledonous herbs and shrubs.

Application of a synthetic auxin named 2, 4-D effectively kills dicotyledonous weeds from monocot crop (paddy, wheat, etc.) fields.

4. Production of parthenocarpic fruits:

Fruits, produced from the ovary without fertilization do not contain seeds and become larger. These are called parthenocarpic fruits. A treatment of auxin solution before the maturation of flowers triggers the development of the ovary.

As a result, seedless fruits are produced before fertilisation. Synthetic auxin is successfully applied on the plants of guava, grapes, banana, watermelon, etc. to produce seedless fruits.

Synthetic gibberellin is comparatively more effective on tomato plants. These two synthetic phytohormones are used to produce parthenocarpic fruits.

Question 4. LH and ICSH control the secretion of hormones of the reproductive glands of the human body’-judge the validity of the statement.

Answer:

In the female body, Lutenising Hormone (LH) stimulate ovule secretion and the formation of corpus luteum from the Graafian follicle. It also helps in the secretion of progesterone hormone from the corpus luteum.

In the male body, Interstitial Cell Stimulating Hormone (ICSH) stimulate the interstitial cells of Leydig of testes to secrete testosterone.

Question 5. How can you distinguish between the mitosis of plant cells and that of animal cells on the basis of the formation of spindle fibre and the process of cytokinesis?

Answer:

Distinguish between the mitosis of plant cells and that of animal cells are-

WBBSE Solutions For Class 10 Life Science And Environment Chapter 2 Mitotic And Meiotic Differences between mitosis in plant cell and animal cell

 

Question 6. Establish the relationship between the formation of malignant tumours in the human body with the loss of control in the cell cycle.

Answer:

Checkpoints prevent uncontrolled cellular growth and thereby cancer. It mainly checks the genetic as well as physical integrity, if they found any genetic defect, they immediately arrest the cell from going to cellular division.

All cells will be checked by three stages. But if the genes which are responsible for the synthesis of checkpoints are mutated then they lost their control over cell division and the cells are going to divide in an uncontrolled way.

This uncontrolled cellular division gives rise to a cellular lump, called a tumour.

Question 7. ‘The adventitious leaf bud plays a significant role in natural vegetative propagation of plant’-Evaluate the validity of the statement with a proper example.

Vegetative propagation:

Asexual reproduction, in which any portion of the vegetative body of certain plants separates out from the mother’s body and finally grows into a new individual by mitotic cell division, is known as vegetative propagation.

Processes of natural vegetative propagation:

Vegetative propagation occurs by the growth of different parts of plants naturally, which are mentioned below.

1. By leaves:

Certain plants like Bryophyllum, Bigonia, etc. develop buds along the edges of their leaves. These are called leaf buds. These buds grow adventitious roots from their base.

When such a leaf detaches from the plant body and comes in contact with the soil, each of the buds grows into an individual daughter plant.

2. By roots:

Roots of sweet potatoes, pointed gourd, etc. grow adventitious buds. These are called root buds. When detached from the root, these buds grow into new daughter plants.

3. By stem:

Vegetative propagation by stem occurs by two different types of modified stems-

1. Underground modified stems:

The tuber of potato, bulb of onion, the rhizome of ginger, turmeric, etc. are underground stems, modified for storage of food. These organs have buds, which may grow into daughter plants.

2. Sub-aerial modified stems:

Nodes of sub-aerial modified stems of Mentha, Marsilea, Oxalis, Centella, water hyacinth, Chrysanthemum, etc. plants grow adventitious roots. When detached from the mother plant, the rooted branches grow as a daughter plant.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 2 Reproduction natural vegetative propagation

 

Question 8. Show with the help of a cross, who is more important among parents in determining the sex of their offspring.

Answer :

Differentiate between autosome and allosome.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Sex Determination Of Human Differences between autosome and allosome (1)

 

Question 9. ‘Different genotypes produce the same phenotype’-Justify the statement in the form a table by taking an example from the result of dihybrid cross of Pea plant.

Answer :

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Different genotypes produce the same phenotype

 

Dominant phenotypes are expressed in both homozygous and heterozygous forms. So a dominant phenotype may have two genotypes one is homozygous and one is heterozygous yellow seeded plants have a single phenotype with yellow seed colour but its genotype is either YY or Yy.

Question 10. Give your opinion about probable suggestions which can be given to a pair of contenders before marriage in order to prevent the spread of a genetic disease from the society already known to you.

Answer:

The pair of contenders should be suggested to test whether they are carriers or have any particular disease. If both are a carrier for an autosomal disease or one is a carrier and the other have the disease then the marriage should be cancelled.

If they both are normal or if one is normal and the other is a carrier then they can marry.

Question 11. A good number of Tilapia fishes are released in a pond having only different indigenous fish species grown naturally. Think and write which types of the struggle for existence Tilapia fishes have to face in order to survive.

Answer:

The Tilapia fishes have to face three types of struggle for existence in the pond where they are released.

These are

Intra-specific struggle:

It is the struggle among all the newly released tilapia fishes for food and habitat

Inter-specific struggle:

It is the struggle between Tilapia and other naturally grown indigenous fishes for food and habitat.

Struggle with the environment:

The Tilapia fishes struggle with the condition of the pond where they are newly released. Water content, Temperature of water, amount of dissolved oxygen, poisonous substances present in that pond water, etc. are the obstacles that they struggle, to overcome.

Question 12. Prepare a list of roles air sacs of pigeons play to fly in the air.

Answer:

Importance of air sacs in pigeons:

Pigeon is a primary volant animal. It has nine non-vascular and non-muscular air sacs, emerging from the bronchioles, typically helping in flight.

The importance of air sacs in pigeons are-

  1. During the flight, the pigeon needs extra energy, the production of which requires additional oxygen. Oxygen concentration near the ground level is higher than that in the high sky. Before the flight, pigeons fill the air sacs with this oxygen-rich air. When in the high sky, these air sacs supply oxygen-rich air to the lungs.
  2. The air-filled sacs decrease the specific gravity of the body of the pigeon, which is also beneficial for flight.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Air sacs of pigeon

Importance of air bladder in rohu fish:

  1. The air bladder in rohu fish helps it to move up and down in the water. The air bladder changes the buoyancy of the body and thus, assists it to go at different depths of water.
  2. The red gland in the anterior chamber of the air bladder fills gas in it to reduce the specific gravity of the body and thus helps the fish to move upward in water. On the other hand, the rete mirabile of the posterior chamber absorbs the gas to increase the specific gravity of the body of fish and helps the fish to go deeper in the water.

Question 13.

  1. Structure and function
  2. Indicating the nature of evolution

Based on the above two features establish the concept of the analogous organ with the help of a proper example.

Answer:

Analogous organs are those organs which are different in structure and origin but perform similar functions.

Analogous organs indicate convergent evolution. For example, the wings of birds, wings of insects and patagium of bats perform the same function, i.e., they help the organisms to fly but they have different structures.

Wings of birds are modifications of forelimbs, wings of insects are an outgrowth of insects’ exoskeleton and the patagium of bat is actually a fold of skin between forelimbs and hind limbs.

All these structures point towards the fact that in the same environment, different structures may evolve to perform the same function. This further points toward convergent evolution.

Question 14. Relate the following phenomena with the trend of disturbance in the Nitrogen Cycle resulting from different human activities:

  1. Global Warming
  2. Acidification of soil and water of river and lake.

Answer:

Global warming:

Around 40% increase of N20 (nitrous oxide) in the environment is due to human activities. N20 is released in the environment as a result of the combustion of fossil fuel and the overuse of nitrogen-rich fertilizers.

This N2O is a greenhouse gas that absorbs infrared radiation or heat of the sunlight which reflects on the earth and facilitates the greenhouse effect and global warming.

Acidification of soil and water of rivers and lakes:

Oxides of nitrogen get dissolved in rainwater and form nitric acid, which is a major component of acid rain. Acid rain destroys aquatic plants and animals and disrupts the ecological equilibrium of rivers lakes, ponds, etc.

Apart from this, the destruction of forest resources and harm to architecture and monuments take place due to acid rain.

Question 15. Hilsa, Bee, Penguin, Rauwolfia- Assess which are the causes of the endangeredness of the above-mentioned organisms.

Answer:

Hilsa:

Facing threat due to overexploitation of hisa population.

Bee:

Loss of habitat, and excessive use of cell phones that project electromagnetic waves, damage the ability of bees to return to their colony. By that way, it destroys the navigation system of bees.

Penguin:

Penguin faces threats from geological events like a volcanic eruption, pollution, climatic changes and severe weather.

Rauwolfia:

This type of medicinal plant faces threats due to overexploitation.

Question 16. Discuss any two roles of the People’s Biodiversity Register (PBR) in conserving local biodiversity.

Answer:

Joint Forest Management or JFM:

The proper management of biodiversity by the joint action of local people and the forest department of the state government which is approved by the Indian government, is known as Joint Forest Management or JFM.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Role Of JFM And PBR In Biodiversity Conservation JFM or joint Forest national park

 

Role of JFM in the conservation of biodiversity:

In the conservation of biodiversity, the role of JFM is as follows

  1. The committee, formed by the local residents of any forest area, maintains coordination with the state forest department regarding the conservation of the biodiversity of that forest.
  2. The JFM committee members keep a vigil on the forest area along with their daily activities regarding forest fire, encroaching of the forest area, poaching, felling of trees etc. and report it immediately to the forest department in case of any adversity.

People’s Biodiversity Register or PBR:

It is a ready reference prepared by the central forest department in accordance with the Biodiversity Law 2002. This document contains detailed information on the availability of local biodiversity, their habits and habitats, other behaviours, their ecological importance etc.

Role of PBR in the conservation of biodiversity:

In the conservation of biodiversity, the role of PBR is as follows-

  1. PBR provides correct information about the biodiversity of the local areas so that, the rare and endangered animals and plants can be conserved properly.
  2. It makes people aware of the importance of the conservation of biodiversity and involves them directly with the conservation project.
  3. PBR makes the people aware of the sustainable utilisation of forest resources and helps in the economic development of the area.
  4. PBR provides correct geographic information about conserved areas.
  5. Identify endangered species and collect information about the conservation of these species.
  6. Gather knowledge about local species and their cultural and social value.
  7. Gives an idea about the obtained product from species and their value along with their collection and transport procedure.

Question 17. Tabulate any of the four activities which are prohibited in a sanctuary following the provisions of Wildlife Act.

Answer:

The four activities, that are prohibited in the sanctuary are-

  1. Human intervention in the sanctuary is strictly prohibited.
  2. Any human activity which disturbs the wild animals is prohibited.
  3. Poaching, hunting, and fishing is strictly prohibited.
  4. Felling of trees is completely prohibited.

 

Model Question Paper Life Science And Environment Set 3 Group D

Answer 6 questions or their alternatives given below.

Question 1. Draw a neat diagram of the vertical section of the eyeball of the human eye and label the following parts:

  1. Cornea,
  2. Lens,
  3. Vitreous humour,
  4. Retina.

Answer:

Human eyeball:

It js a spherical, fluid-filled portion of the eye. It has two types of structural components—

1. The eye coats:

The structure and function of each component of the eye coats are as follows—

  1. The eye has three coats—fibrous outer coat, vascularised medial coat and neural inner coat.
  2. The fibrous outer coat has two portions, its posterior 5/6th opaque portion is called the sclera and the anterior 1 /6th transparent portion is known as the cornea, the sclera is rigid that keeps the shape of the eye intact. Cornea is a refractive medium that acts as a front window of the eye.
  3. The vascularised medial coat is known as choroid. It is black in colour due to the accumulation of melanin pigment. The cornea is rich in blood capillaries.
  4. The anterior extension of the choroid forms the ciliary body and iris. The ciliary body is a muscular extension of the choroid that holds the lens in position. It remains attached with the suspensory ligament. The pigmented muscular extension of the choroid that is present behind the cornea is called the iris. It has a round aperture at the centre to allow light to enter the eye. This aperture is known as the pupil. Contraction and expansion of iris muscles change the diameter of the pupil to regulate the exposure on the retina.
  5. The neural inner coat known as the retina is composed of two types of photosensitive cells—rod cells and cone cells. The rod cells are sensitive to dim light and the cone cells sense bright light.
  6. fust opposite to the pupil, the retina has a tiny depressed spot, called the macula lutea of yellow spot. This spot contains cone cells only and is capable of creating the sharpest image in bright light. The region of the retina from which the optic nerve emerges is devoid of any photosensitive cells. Therefore, this spot is incapable of sensing light, hence it is called the blind spot.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 1 Eye As A Sense Organ In Human Structure of human eye

 

2. Refractive medium:

The structure and function of each component of the refractive medium is as follows—

  1. The refractive media of eye are—cornea, aqueous humour, lens and vitreous humour.
  2. Cornea is a transparent convex layer, present in front the eye as a window glass.
  3. The thin anterior chamber of eye between the cornea and the lens is filled with a watery fluid called aqueous humour.
  4. Behind the iris, a transparent circular biconvex lens is present, that is held in position by suspensory ligaments and ciliary muscles. The lens creates the image on the retina.
  5. The posterior chamber of the eyeball behind the lens is filled with a viscous matrix, called vitreous humour. Both aqueous and vitreous humour maintains the pressure inside the eyeball and supply nutrition to different parts of the eye.

Or,
Draw a neat diagram of the metaphase of mitotic cell division in an animal cell and label the following parts:

  1. Polar region,
  2. Spindle fibre,
  3. Chromatid,
  4. Centromere.

Answer:

Anaphase of mitosis in plant and animal cells:

The third phase of karyokinesis, in which the daughter chromosomes move from the equatorial plate towards the two opposite poles of the dividing cell, is called anaphase. The events which occur during this phase, in both plant and animal cells, are discussed below.

1. Anaphase in plant cells:

The events occurring during anaphase, in plant cells, are as follows-

  1. The Centromere of each chromosome splits longitudinally, as a result, two sister chromatids separate from each other with their own share of centromere and emerge as two daughter chromosomes.
  2. The Centromere of each daughter chromosome remains attached to the chromosomal spindle fibre.
  3. Inter-polar or Chromosomal continuous spindle fibres connect the two poles of the -spindle fibre spindle.
  4. Gradually, the chromosomal fibres contract. -Chromatid to pull half of the daughter chromosomes towards one separation Inter-zonal pole and the remaining half to the opposite pole. This is a spindle fibre called anaphasic movement.
  5. During anaphasic, movement, metacentric, sub-metacentric, acrocentric and telocentric chromosomes appear like the English letters ‘V’ ‘L’, ‘J’ and ‘I’ respectively, based on the position of centromere on the chromosome.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 2 Mitotic And Meiotic Cell Division Anaphase in plant cell

 

2. Anaphase in animal cells:

The above-mentioned events also occur during anaphase in animal cells. Here, inter-zonal spindle fibres cluster in between separating chromosomes to form columnar stem bodies, which help in anaphasic movement.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 2 Mitotic And Meiotic Cell Division Anaphase in animal cell

 

Question 2. Distinguish between mitotic and meiotic cell divisions in animals on the basis of the following features:-

  1. Site of occurrence
  2. Nature of Division of Chromosomes
  3. A number of cells produced.

Answer:

Meiosis is called as reduction division:

In meiosis, the chromosome number of a diploid mother cell reduces to its half to give rise to a haploid number of chromosome-bearing daughter cells. Thus, meiosis is called reduction division.

Differences between mitosis and meiosis:

WBBSE Solutions For Class 10 Life Science And Environment Chapter 2 Mitotic And Meiotic Differences between mitosis and meiosis

 

Or,
Explain the importance of the following parts of an eukaryotic chromosome:

  1. Centromere
  2. Telomere

Analyse the role of cell divisions in controlling growth, reproduction and repair in an organism.

Answer:

The physical structure of eukaryotic chromosomes:

The morphological features of a chromosome appear ‘most distinctly under a microscope during the metaphase stage of cell division. From this study, we can find 5 parts of a chromosome. These parts are described below.

1. Chromatids:

In a metaphase chromosome, two identical and longitudinal strands are seen. These are chromatids. Two chromatids of the same chromosome are called sister chromatids, which remain attached to a constricted region or centromere.

Each chromatid carries one or a few very fine filaments along its length. These are called chromonemata (singular—chromonema). Each chromonema is composed of a longitudinally arranged coiled DNA.

Along each chromonema, everal spherical linearly arranged bead-like structures are seen, which are called chronometers.

2. Primary constriction and centromere:

Each chromosome has a distinct constricted region at which the sister chromatids remain attached to each other. This is known as primary constriction.

At e primary constriction, a round plate-like and dense heterochromatin structure is seen, which is called the centromere. The centromere has a few adhesive points, called kinetochores, which attach to the spindle fibres during metaphase.

The DNA present in the .centromere is genetically inactive in nature.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 2 Continuity Of Life Chromosome Physiacal Structure of a chromosome

 

3. Secondary constriction:

Other than primary constriction, there are one or a few constricted regions in the chromosomes. These are called secondary constrictions. Generally, the nucleolus is seen affixed to the secondary constriction.

During the telophase of cell division, this region helps to reorganise the nucleolus. Therefore, secondary constriction is also known as nucleolar organiser region or NOR.

4. Satellite:

In a few chromosomes, a bulb-shaped terminal portion is seen beyond the secondary constriction. This is called a satellite or SAT body. The chromosomes with SAT body are called SAT chromosomes.

5. Telomere:

The terminal portions of a chromosome are called telomeres. These are genetically inactive regions of a chromosome. During interphase, telomeres help in DNA replication.

It also prevents the joining of a chromosome with another and controls the ageing and death of a cell.

Significance of cell cycle:

The significance of cell cycle is mentioned below-

1. Controlling cell division:

Certain points of the cell cycle control cell division. These are known as checkpoints. In case of any functional disruption at any of those points, the cell division process becomes uncontrolled, which may lead to tumour formation.

Tumours are of two types-benign tumour and malignant tumour. Benign tumours are harmless but malignant tumour cells invade other tissues through blood or lymph and form tumours there.

This phenomenon is called metastasis, which is a characteristic feature of cancer cells.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 2 Cell Division And Cell Cycle Normal and nucontrolled cell division

 

2. Normal growth and wound healing:

Cell division helps an organism to grow in size. It also assists in wound healing.

Importance of cell division:

The importance of cell cycle are given below.

1. Growth:

The number of cells of an organism increases due to cell division. The daughter cells produced by this process also grow in size. Therefore, the growth of any organism depends directly upon cell division.

2. Reproduction:

Amitosis, mitosis and meiosis help in different types of reproduction processes. Simple unicellular organisms like Amoeba, reproduce by the amitosis process. Mitosis helps in asexual and vegetative reproduction in animals and plants.

By meiotic division, gametes and spores are formed. Therefore, meiosis helps in sexual and asexual reproduction.

3. Wound healing:

Mitosis helps in the repair of wounds and the regeneration of organs in plants and animals.

4. Transfer of genetic characters:

By cell division, the characters of the mother cell are transferred to the daughter cells. In a broader perspective, the newer characters in daughter cells help in adaptation and evolution.

Question 3. Tabulate three pairs of dominant-recessive traits of pea plants as selected by Mendel. State the first law of Mendel as deducted from the experiment of the Monohybrid cross.

Answer :

Three pairs of dominant-recessive traits of pea plant as selected by Mendel

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Recessive traits of pea plant as selected by Mendal

Or,
A colour-blind female married a normal male. Judge the probability of colour blindness among their children in the first filial generation. Show with the help of a cross how the first law of Mendel deviates in the case of the Four-O’ clock plant in F2 generation.

Answer:

Inheritance of colour blindness:

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Some Inheritance of colour blindness

 

Checker board of F1 generation:

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Some Genetic Diseases Checker bord of F1 generation

In the F, generation all the female born will be carriers and all the male born will be colour-blind.

Deviation from Mendel’s first law:

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Some Genetic Diseases Deviation from Mendels first law

 

Checkerboard of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Some Genetic Diseases Checker bord of F2 generation

Phenotypic ratio- Red: Pink: White = 1:2:1
Genotypic ratio-RR: RW: WW = 1:2:1
In the F1, generation the red and white colour is not segregated in a 3:1 ratio rather the phenotypic ratio is 1:2: 1. The phenotypic and genotypic ratios both are the same i.e. 1: 2: 1: Thus in this case Mendel’s 1st law is clearly deviated.

Question 4. With the help of arrow signs, show the major evolutionary events as occurred gradually after the origin of life.

Answer:

The major evolutionary events:

WBBSE solutions For Class 10 Life Science And Environment Evolution Concepts Of Evolution The major evolutionary events

 

Or,
Discuss the following three events as mentioned in the evolutionary theory of Darwin:

  1. Prodigality of production
  2. Origin of variation
  3. Natural selection

‘Adaptation is the change of shape, physiological functions and behaviour of organism’-Justify the statement with the help of any two examples.

Answer:

Darwinian concept on organic evolution:

In 1859, Charles Darwin explained the modern concept of organic evolution in his famous book ‘On. the Origin of Species by Means of Natural Selection, which has become famous as Darwinism or as the theory of natural selection. The basic thoughts of Darwinism are mentioned below.

1. Prodigality of production:

Darwin opined that all organisms increase their population in a geometric rate. For example, a pond snail lays 6 million eggs in a year. A female salmon fish releases 30 million eggs in a reproductive season.

2. Limited food and shelter on earth:

The space on the earth and supply of food do not increase in proportion to the rate of increase in population. Therefore, a scarcity of food and shelter occurs in the living world.

3. Struggle for existence:

Continuous increase in population size and scarcity of food and shelter results in to conflicts among the organisms for survival. Darwin referred to it as the struggle for existence. This struggle is of three types-

  1. Intraspecific struggle-The conflict among the members of a same species population for food, shelter and mating partners.
  2. Inter-specific struggle-The- conflict among the members of different species for resources and shelter.
  3. Environmental struggle- This struggle is for surviving different natural calamities, like drought, flood, earthquake, etc.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 4 Evolution Theories Of Organic Evolution Struggle for existence

 

4. Variation:

Darwin indicated that no two individuals of the same species are identical. There must be some differences between them, which he stated as variations.

5. Survival of the fittest:

Darwin mentioned two different types of variations-favourable and unfavourable. Favourable variations help an organism to adapt to changing environment but unfavourable variations cannot help in their adaptation.

The individuals with unfavourable variations face a defeat in the struggle for existence and become extinct in the course of time. Whereas, the organisms with favourable variations win that struggle and survive on Earth.

Scientist Herbert Spencer denoted this event as the ‘survival of the fittest.

6. Natural Selection:

According to Darwin, organisms with favourable variations are selected by nature for survival because they are fittest to face their surroundings. Darwin explained this event as ‘natural selection.

The naturally selected forms thrive successfully and propagate very fast to increase their population.

7. Origin of new species:

Accumulation of many favourable variations in any group of organisms for generations makes their descendants widely different from their ancestors. Finally, these completely changed descendants emerge as new species.

The canine teeth of carnivorous animals are large and sharp to tear flesh whereas the same are relatively small and blunt in herbivores. This change in the shape of the teeth helps the organisms to consume particular types of food.

Kangaroo rats survive without drinking water for a long time, camels possess some physiological adaptations that allow them to survive a long time without drinking water. These are examples of physiological changes which help the animals to adapt in the desert environment.

Camouflaging is a behavioural adaptation of certain organisms to hide in any particular environment. So it can be called that is the change in shape, physiological function and behaviour of organisms.

Question 5. Evaluate the effects of the following pollutants on the environment and human health:

  1. Non-biodegradable insecticides
  2. Pollen grain
  3. Chemical fertilizers containing phosphate and nitrate
  4. Wastes containing pathogens originated from health centres
  5. Chlorofluorocarbon

Answer:

Effects of Pollutants on the environment and human health

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Topic 2 Environmental Pollution Effects of pollutants on environment and human health

Or,
Speculate the probable causes of each of the following phenomena:

  1. Insomnia, High blood pressure, Partial or Complete deafness
  2. Decrease in the concentration of dissolved oxygen in the water and floating of dead fishes
  3. Decrease in the number of Crocodiles’ Inflammation in the respiratory tract of the lung
  4. Decrease in the number of pollinating insects

Answer:

Probable causes of different phenomena

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Topic 2 Environmental Pollution Probable cause of different phenomena

 

Question 6. ‘The ever-increasing population in the different cities of India is creating the crisis of ground water’-Support the statement with reasons on the basis of your experiences. Construct a concept map to show how the increase in human habitat is influencing the eco-system of Sunderbans.

Answer:

Reasons for the crisis of groundwater:

  1. The underground water level is decreasing due to the excessive use of fresh water. Daily use of water by humans is the reason behind the excess need of water.
  2. A water crisis has been developed because of wastage of water example continuous running of taps at households and municipal or rural water connections of a locality.
  3. In urban areas construction works are taking place, filling ponds, due to which water can not be stored underground. This is also a reason of freshwater scarcity.

Thus based on above mentioned points it can be clearly understood that the ever-increasing population in the different cities of India is creating the crisis of groundwater.

Effects of deforestation:

Four effects of deforestation are-

  1. Soil erosion and desertification,
  2. Global warming,
  3. Loss of biodiversity,
  4. Changes in climatic conditions and changes in the pattern of rainfall.

Problems of Sundarban:

In recent days Sundarban is facing many environmental problems. These are mentioned below.

1. Loss of mangrove forest due to urbanization:

Population pressure has initiated urbanization. Due to this, many areas along the periphery of Sundarban are deforested.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Importance Of Biodiversity Destruction of mangrove foreset in Sundaraban

 

2. Agricultural problem:

Due to soil erosion and siltation, the depth of rivers becomes shallow. During high tide saline water spills over, and floods the agricultural fields to make the soil salty and infertile.

3. Scarcity of fresh water:

The estuarine water is salty. Frequent flood makes inland water bodies like ponds and wells saline. Therefore, sweet water is becoming scarce in the Sundarban area.

4. Destruction of natural habitat:

Deforestation is reducing the natural habitats of large to small animals. As a result, the biodiversity of Sundarban is gradually getting destroyed.

5. Pollution:

The most provocative problem for Sundarban is increasing pollution. All rivers are disposing of gallons of pollutants in the water. Several pollutants are percolating in the soil of the Sundarban area making the soil infertile and toxic.

The oil and grease, spilling from fishing and tourist vessels are polluting the aquatic environment of this area. This pollution is affecting the propagation of fish, prawns, and other aquatic animals.

6. Imbalance in the prey-predator ratio:

Due to overpopulation and pollution, the number of many animal species has declined. This has resulted int o imbalance in prey-predator ratio. The scarcity of natural food compels tigers to enter into human localities in search of food.

7. Rise in water level:

Due to global warming, the glaciers of the polar region are melting. This results into an increase in sea water level. As a result, the delta of Sundarban is going under seawater.

Or,
Summarise which conservation measures have been adopted to increase the population of an endangered mammal exclusively found in the swampy grasslands under the foothills of the Eastern Himalayas. Discuss the role of biodiversity in maintaining the balance of the ecosystem of a river.

Answer:

Causes of decrease in the population of rhino in India

The causes which resulted in the decrease of the population of rhinos are mentioned below.

1. Squeezing of natural habitat:

Due to the increasing human population, the forest areas are being encroached on for human residential purposes. As a result, the normal habitats of rhinos are being decreased. This is one of the most important causes of the decrease in the population of rhinos.

2. Poaching:

The value of rhino horn is very high in the East Asian markets. People, there have a baseless belief that the rhino horn has immense medicinal importance. For this purpose, the poachers kill rhinos indiscriminately and cut their horns.

3. Low birth rate:

The rhinos have a very low birth rate. The gestation period is very long (479 days). That is why the population size does not grow quickly.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Role Of JFM And PBR In Biodiversity Conservation Gorumara national park

 

Measures are taken to protect rhinos

About 85% of the single-horned rhino (Rhinoceros unicornis) are found in Assam. Due to certain human actions, the number of the rhino population was decreasing very fast.

In view of this fact, WWF (World Wildlife Fund for Nature) and the forest department of Assam have jointly taken an initiative named IRV 2020 (Indian Rhino Vision 2020). The objective of this project is to increase the population of rhinos to 3000 by the year 2020.

The success of the project has already been established. The number of rhinos in Assam reached 2544 in 2013. According to the census of 2013, the population of rhinos is exclusively in the Kaziranga National Park reached 2319 in 1855.

Regular surveillance, restricted ecotourism and strong implementation of laws have made the project successful. Rhino conservation has also become an integral part of several other national parks. Gorumara National Park is one of them.

The role of biodiversity in maintaining a balance of the eco-system of a river are

  1. Biodiversity decreases the inter-specific and intra-specific competition of aquatic organisms for food and habitat.
  2. Depletion of certain aquatic organisms due to water pollution hampers the river ecosystem. This can be restored by biodiversity.
  3. Extensive biodiversity provides certain options to nature like killifish, which can withstand with extreme environmental change (water pollution).

WBBSE Class 10 Life Science And Environment Model Question Paper 2023 Set 2

Model Question Paper Life Science And Environment Set 2 Group A

Answer to all questions is compulsory

1. Write the answer in complete sentences by choosing the correct answer for each question with the respective serial number.

Question 1. Select which of the following statement is correctly related to tropic movement-

  1. It is controlled by the intensity of the stimulus
  2. The overall change of place takes place by a plant or part of the plant
  3. This movement is observed in the algae called Volvox
  4. It is an induced movement of curvature controlled by the direction of the stimulus

Answer: 4. The correct statement related to tropic movement is—it is an induced movement of curvature controlled by the direction of the stimulus.

Question 2. Read the following sentences and identify the sentence which is not correct-

  1. AFSH, LH and Prolactin are different types of GTH
  2. Adrenaline decreases cardiac output
  3. Insulin facilitates the absorption of glucose into the cell through the cell membrane
  4. Progesterone helps in the formation of the placenta in the female body

Answer: 2. The statement which is not correct is— Adrenaline decreases cardiac output.

Question 3. Determine the number of Cranial nerves in human body-

  1. 10 Pairs
  2. 12 Pairs
  3. 31 Pairs
  4. 21 Pairs

Answer: 3. The number of cranial nerves in human body is 12 pairs.

Question 4. Identify the correct feature of Amitosis cell division-

  1. Occurs in germ mother cell of sexually reproducing organism
  2. It is called indirect cell division
  3. Chromosome and Spindle fibre are formed
  4. Chromosome and Spindle fibre are not formed

Answer: 4. The correct feature of amitosis cell division is—chromosome and spindle fibre are not formed.

Question 5. Determine which of the following statement is correct regarding the significance of meiosis-

  1. It results in overall growth and development of the body
  2. It results in the growth of reproductive organs and embryo of the organism
  3. It produces haploid gamete containing hereditary variations
  4. It regenerates the organ in the body of some animals if it is lost

Answer: 3. The correct statement regarding the significance of meiosis is—it produces a haploid gamete containing hereditary variations.

Question 6. Select the correct pair and write it-

  1. Multiple fission-Hydra
  2. Fragmentation-Spirogyra
  3. Regeneration-Fern
  4. Budding-Planaria

Answer: 2. The correct pair is—Fragmentation— Spirogyro.

Question 7. Identify which of the following is a dominant trait-

  1. Length of the stem-Dwarf
  2. The shape of the seed-Wrinkled
  3. Colour of the cotyledon-Yellow
  4. Colour of the flower-White

Answer: 3. The dominant trait is—the colour of the cotyledon—Yellow.

Question 8. Assess how many types of gametes are produced from the Pea plant having the genotype RRYY-

  1. One type
  2. Two types
  3. Four types
  4. Three types.

Answer: 1. The pea plant having the genotype RRyy will produce one type of gametes.

Question 9. Select which of the following two genotypes are responsible for the expression of the phenotype wrinkled yellow in Pea plants-

  1. ARRYY and rryy
  2. RRyy and Rryy
  3. BRRYY and RrYy
  4. rrYY and rrYy

Answer: 4. rrYY and rrYy are two genotypes responsible for the expression of the phenotype wrinkled yellow in the Pea plant.

Question 10. Identify which of the following indicates inter-specific struggle-

  1. The struggle among the members of guppy fish for feeding on mosquito larvae
  2. The struggle between snakes and owls for capturing rats for eating
  3. Struggle within a group of deer for feeding on grass at the same location
  4. The struggle between tigers in a jungle for preying on deer

Answer: 2. Struggle between snakes and owls for capturing rats for eating indicates inter¬specific struggle.

Question 11. Determine the term related to the theory of Lamarck-

  1. Struggle for existence
  2. Origin of variation
  3. Inheritance of acquired characters
  4. Natural selection

Answer: 3. The term related to the theory of Lamarck is—Inheritance of acquired characters.

Question 12. Identify the animal who communicates information with other members of its own group through specific dancing patterns regarding the sources of food-

  1. Chimpanzee
  2. Peacock
  3. Cockroach
  4. Honeybee

Answer: 4. Honey bee communicates information with other members of its own group through specific dancing patterns regarding the sources of food.

Question 13. Decide which of the following if remains for a long time in the environment, the probability of its biomagnification increases-

  1. Newspaper
  2. Faecal matters of animals
  3. Rotten leaves
  4. Chlorinated insecticide

Answer: 4. Chlorinated insecticide if remains for a long time in the environment, the probability of its biomagnification increases.

Question 14. Gorumara, Corbett, Kulik, and Nandadevi choose the correct sequence from the following which is correct for the above forests in sequence-

  1. Biosphere Reserve, Sanctuary, National Park, National Park
  2. National Park, National Park, Biosphere Reserve, Sanctuary
  3. National Park, National Park, Sanctuary, Biosphere Reserve
  4. Sanctuary, Biosphere Reserve, National Park, Sanctuary

Answer: 3. The correct sequence for Gorumara, Corbett, Kutik, and Nandadevi forests is— National Park, National Park, Sanctuary, Biosphere Reserve.

Question 15. If the amount of pollen grains, fungal spores and dust particles increase suddenly in the air, decide which of the following problems may increase-

  1. Tuberculosis
  2. Malaria
  3. Asthma
  4. Dengue

Answer: 2. If the amount of pollen grains, fungal spores and dust particles increase suddenly in the air, problem of asthma will increase.

 

Model Question Paper Life Science And Environment Set 2 Group B

2. Answer any 21 questions out of 26 questions given below as instructed. Fill in the blanks with proper words in the following sentences (any five):

Question 1. By the influence of _______ was the aggregate of some large colloidal molecules during the early phase of the origin of life. hairs on the skin get erected an emergency situation.
Answer: Adrenaline

Question 2. A chromosome is the condensed coiled structure of _______ molecule.
Answer: DNA

Question 3. The sexual reproduction that occurs among two genotypically different organisms belonging to the same species, is called _______.
Answer: Hybridization

Question 4. _______ was the aggregate of some large colloidal molecules during the early phase of the origin of life.
Answer: Coacervate

Question 5. _______is derived from the root of the Sarpagandha plant which is used as medicine for lowering high blood pressure
Answer: Reserpine

Question 6. The ex-situ conservation by keeping sample seeds in liquid nitrogen at a temperature of -196°C is called _______.
Answer: Cryopreservation

Decide whether the following statements are true or false (any five):

Question 7. Dilute urine is profusely eliminated by a person affected with Diabetes Insipidus.
Answer: True

Question 8. The Carpel of flowering plants collects pollen grains with the help of a hairy and sticky style.
Answer: False

Question 9.A pair of sex-determining sex chromosomes are present in the ovum of humans.
Answer: False

Question 10. According to Darwin organisms reproduce at geometric ratios.
Answer: True

Question 11. One-horned rhinoceros is conserved at Manas National Park of West Bengal.
Answer: True

Question 12. Acetylcholine and adrenaline are neurotransmitters.
Answer: False

Match the words in Column-A with those which are most appropriate in Column-B and re-write the correct pair mentioning the serial no of both Columns (any five):

WBBSE Model Question Paper 2019 Life Science And Environment Set 2 Match The Columns

Answer: 13. C, 14. F, 15. A, 16. B, 17. D, 18. E,

Answer in a single word or in a single sentence (any six):

Question 19. Choose the odd one and write it: Glossopharyngeal, Oculomotor, Trigeminal, Oxytocin
Answer: Oxytocin

Question 20. What is the function of the fluid present in the intermediate chamber between the lens and retina of the eyeball?
Answer: Vitreous humor is the fluid present in the intermediate chamber between the lens and retina of the eyeball i.e. vitreous chamber, which helps to form images by refracting the light rays and also maintains the shape of the eyeball.

Question 21. A pair of related terms is given below. On the basis of the relationship in the first pair write the suitable word in the gap of the second pair. Grafting: Mango:: _________ : China Rose
Answer: Stem cutting

Question 22. What is the cause of expression of haemophilia disease only at homozygous conditions?
Answer: Haemophilia is an X-linked recessive disease so it is expressed in females in the homozygous condition (XhXh) and in males in the hemizygous condition (XhY). The female in heterozygous condition (XhXh+) is a carrier for the disease and phenotypically normal, so the disease is not expressed in her body.

Question 23. Write the genotypic ratio obtained at 2nd filial generation in the monohybrid experiment done by Mendel.
Answer: Pure tall: hybrid tall: dwarf =1:2: 1.

Question 24. Which type of evolution is indicated by homologous organs?
Answer: Divergent evolution

Question 25. Among the following four terms, one includes the other three. Find it out and write it: Pesticides used in Agriculture, Typhoid, Water Pollution, liquid wastes from factories.
Answer: Water pollution

Question26. In which type of in-situ conservation practice, the diversity, culture and lifestyle of humans is preserved along with microorganisms, wild animals and wild plants?
Answer: Biosphere Reserve

 

Model Question Paper Life Science And Environment Set 2 Group C

3. Answer any 12 questions in 2-3 sentences out of 17 questions given below.

Question 1. Illustrate with the help of a word diagram the nervous pathway of the way you will open the door by hearing the door bell.

Answer:

When someone rings the bell at the door, we listen to the bell with our ears and open die door. In this case, the receptors present in our ear receive the stimulus and send it to the central nervous system (CNS).

CNS processes that stimulus and sends motor signals to our body parts (muscles of hands and legs) and as a result, we open the door.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 1 Response And Physical Co-Ordination In Animals Nervous Nervous pathway

 

Question 2. State the locations of Meninges and CSF.

Answer:

Location:

Meninges is located just outside the brain and spinal cord as a membranous outer covering.

Functions:

The functions of meninges are as follows

  1. Meninges protects the central nervous system from injury and infection.
  2. Capillaries of the meninges supply blood to the skull and cerebral cortex.

The functions of CSF or cerebrospinal fluid are as follows

  1. CSF supplies nutrients and oxygen to the neural cells of the central nervous system.
  2. It also acts as a cushion to protect the brain and spinal cord against external shocks and pressure.

Question 3. What influences gibberellin hormone exerts on the seeds and internode of plants? Explain.

Answer:

Types of gibberellins:

The chemical name of gibberellin is gibberellic acid (GA). There are about 40 different types of gibberellins isolated from different plants. Some of the common types are GA3, GA7, GAp, etc.

Role of gibberellin in plants:

Gibberellins play different roles in the plant bodies, which are mentioned below.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 1 Response And Chemical Co Ordination In Plants Hormones Role of gibberellin in internodal growth

 

1. Breaking dormancy of seed and bud:

For a certain period, the sign of life is not visible in seeds. This phase is called dormancy of seed. During the dormant stage, the seed contains less gibberellin.

Just before germination, the gibberellin level elevates in it. This phytohormone increases the activity of enzymes in the seed and helps in breaking its dormancy and facilitates germination.

2. Promoting the growth of internodes:

Gibberellin increases the rate of cell division in the intercalary meristem. Due to this, the internodes of the stem increase in length, and thus, the overall height of the plant increases.

3. Increasing the size of the fruit:

Gibberellin quickens the growth of fruits and thus, increases their size. A synthetic form of gibberellin also helps in the formation of seedless or parthenocarpic fruits. Seedless tomatoes, apples, pears, grapes, etc. can be produced by gibberellin treatment.

4. Increasing the size of leaves and flowers:

Gibberellin enhances cell division to increase the size of leaves and flowers.

Question 4. Write the names of the different refractive mediums of eyeballs in proper sequence.

Answer:

The refractive mediums of the eyeball from outside to inside are—cornea → aqueous humour→lens→vitreous humour.

Question 5. During the meiosis cell division, a reduction in the number of chromosomes and exchange of segments between chromatids take place- Analyse what are the significance of these two phenomena.

Answer:

During meiotic division, the number of chromosomes gets reduced from 2n to n for gamete formation. Because during fertilization two cells get fused with each other and form Zygote (2n).

So if these cells have contain 2n chromosomes then the zygote will possess 4n number of chromosome, that is not impossible for the existence of an organism. For that reason, the gamete should contain only half the number of chromosomes (n).

This reduction in chromosome number is only possible because of meiotic division. Another important phenomenon is an exchange of segments between two chromatids.

This exchange can play an important role in genetic recombination, that will give rise to variation in the new progeny. This variation is most important for the newly formed progeny because it helps them tp with the changed environment and to survive. this has an ultimate effect on evolution.

Question 6. Distinguish between asexual and sexual reproduction on the basis of the following features:

  1. Number of Parents
  2. Nature of offspring

Answer:

Principle of micropropagation:

Certain plant cells have a unique ability to grow into a full plant body by means of cell division. This unique ability of plant cells is known as totipotency. Micropropagation is a tissue culture technique, which depends upon totipotency.

The word ‘micro’ means ‘very small. In micropropagation, very small portions are collected from vegetative parts of a plant and are cultured in a proper culture medium inside the laboratory to produce saplings.

Differences between asexual and sexual reproduction

WBBSE Solutions For Class 10 Life Science And Environment Chapter 2 Reproduction Differences between Asexual and Sexual reproduction

Question 7. Identify the stages of Karyokinesis of mitotic cell division depending on the following features:

  1. Alignment of chromosomes along the equator of the cell and their arrangement in a definite pattern
  2. The disintegration of spindle fibres
  3. The disappearance of nuclear membrane and nucleolus
  4. Movement of daughter chromosomes to the opposite poles of the cell

Answer:

  1. Alignment of chromosomes along the equator of the cell and their arrangement in a definite pattern-Metaphase
  2. The disintegration of spindle fibre-Telophase
  3. The disappearance of nuclear membrane and nucleolus-Prophase
  4. Movement of daughter chromosomes to the opposite poles of cell-Anaphase

Question 8. To establish the law of segregation in the case of the Pea plant, you are asked to select characters. Write two such characters having opposite traits.

Answer :

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Characters having oppisite traits

 

Question 9. A daughter is born to a woman carrier for the colour-blind disease who married a colour-blind man. What would be the probability of expression of colour blindness in that girl child? Analyse your answer.

Answer:

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Some Genetic Diseases Carrier colour blind woman and man

 

Checker board

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Some Genetic Diseases Checker board.

 

Colour blind girl =  ½ x 100 = 50%

There will be a 50% chance of the daughter being colour-blind. This is because the X chromosome that she will obtain from her father will definitely bear an allele for colour blindness.

However, the X chromosome that she will obtain from her mother may either bear the allele for colour blindness or that for normal vision. So, she has a 50% chance of being colour-blind.

Question 10. Write the name of two genetic diseases which are expressed in human populations.
Answer:

Two genetic diseases expressed in the human population are-

  1. Thalassemia and
  2. Haemophilia.

Question 11. Tigers are extinct but there are many deer in the jungle. A few tigers are brought from another sanctuary and released in such a jungle. Imagine and write the names of different kinds of struggles for existence those tigers have to be involved in.

Answer:

The tigers, brought from another sanctuary, have to face mainly two types of struggle for existence.

These are-

Inter-specific struggle:

They have to get involved in an inter-specific struggle with deer for food.

The struggle against nature:

Firstly, the tigers will face difficulty in adjusting with the new environment and secondly, as the number of deer is high in that jungle, the jungle will be devoid of bushes which generally provide hiding space to tigers during hunting sessions.

Question 12. Describe any two adaptations in the Sundari Plant for tolerance of salt.
Answer:

The mechanisms used by the Sundari plant to excrete the excess salt from its body are-

  1. Presence of a salt gland in the leaf to excrete excess salt along with water,
  2. Deposition of excess salt in specialized cells of bark and root; later, by means of bark exfoliation and other specific mechanisms, the salt are eliminated from the plant body,
  3. Deposition of salt in mature leaves, in larger amounts, so that it can be eliminated easily by leaf exfoliation,
  4. Deposition of excess salt in the cell sap of vacuoles and eliminate it when needed.

Question 13. Explain how do the chimpanzees hunt termites for food.
Answer:

Importance of behaviour in evolution:

Organisms show neurally controlled reactions in response to any visual, auditory or olfactory signals or any stimulus, like light, sound, smell or any action by other organisms. These responses are called behaviour.

The favourable behaviours which help an organism to survive and propagate on earth, are selected by nature. The effect of these behaviours is so far-reaching that it may help in the course of evolution.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Eating of leaves by Chimpanzee

Adaptive behaviours of chimpanzees:

  1. Chimpanzees break branches of trees and then eat up the leaves to prepare a stick. With the stick, they make a hole in a termite mound. Then, they insert the thin stick into the mound to tease the termite colony inside it. As soon as the termites grab the probing stick, chimpanzees pull it out and lick them up. Few termites, which come out through the hole are eaten up by other chimpanzees. This particular behavioural adaptation of chimpanzees is known as termite fishing.
  2. When a chimpanzee is attacked by parasites, it starts eating medicinal plants. Scientists revealed that chimpanzees can recognise and consume 19 different medicinal plants including Aspilia rudis, when attacked by parasites.
  3. To crack hard nutshell, wild chimpanzees use a piece of rock as an anvil and a strong branch of the tree as a hammer. They keep hard-shelled nuts on the rock and then hammer it with the woody branch to crack it. Finally, they remove the shell and eat the nut inside.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Termite fishing by Chimpanzee

Question 14. List the harmful effects of the following pollutants:

  1. Greenhouse gas
  2. SPM
  3. Chemical fertilizers used in agriculture
  4. Non-biodegradable chemical insecticide

Answer:

The harmful effects of some pollutants are as follows-

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Topic 2 Environmental Pollution Harmful effects of some pollutants

Question 15. Give one example of exotic species causing damage of local biodiversities in each of terrestrial and aquatic ecosystem.
Answer:

In aquatic ecosystems, the exotic species that damage the local biodiversity is-water hyacinth. In terrestrial ecosystems, the exotic species that damage the local biodiversity is American. cockroach.

Question 16. Write the in-situ conservation measures taken for one endangered reptile species and name one site of such conservation in West Bengal.
Answer:

One endangered reptile is a crocodile. Measures taken for the in-situ conservation of crocodiles are-

  1. Protect the remaining population of crocodilians in their natural habitat by creating sanctuaries.
  2. Promote captive breeding.
  3. Prevent poaching of crocodiles and as well as their eggs.

The name of such conservation in West Bengal is the Bhagabatpur crocodile project, Sundarban.

Question 17. “Poaching is one of the fundamental causes of endangeredness of animal diversity in India”-Assess four causes for which animals are poached in India.
Answer:

Four causes of poaching are-

  1. Some animals are mainly killed for their skins and bones, for example, elephants are killed for ivory, and tigers are killed for their bones and skins.
  2. Some animals like deer, pigs, bulls are poached for meet the needs of meat.
  3. The body parts of some animals are also used for medicinal purposes and for home. decor.
  4. Illegally traded items fetch lucrative prices, that encourage poachers to look out for more money.

 

Model Question Paper Life Science And Environment Set 2 Group D

4. Answer 6 questions or their alternatives given below.

Question 1. Draw a neat diagram of the neurone and label the following parts:

  1. Dendron
  2. Node of Ranvier
  3. Myelin sheath
  4. Schwann cell

Answer:

Structure of a typical neurone:

A typical neurone has two structural components—

  1. Cell body or cyton and
  2. Projections.

1. Cell body or cyton:

The structural features of the cell body of a neurone is as follows—

  1. The largest, round or star-shaped portion of the neurone is called the cell body.
  2. The cell body contains thick protoplasm, called neuroplasm, surrounded by a cell membrane.
  3. This cell bears a distinct nucleus within it
  4. Neuroplasm contains different cell organelles like mitochondria, Golgi bodies, etc. Other than these, numerous typical nucleoprotein granules, called Nissl granules are present in the neoplasm.
  5. Fine thread-like neurofibrils remain scattered within the neoplasm.
  6. The centrosome is present in an inactive state, therefore, neurones cannot undergo mitosis.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 1 Response And Physical Co-Ordination In Animals Nervous Structure of neurone

 

2. Projections:

The projections of a neurone are of two types

  1. Axon and
  2. Dendron.

The structural features of these two projections are given below.

Axon:

The structural features of the axon is as follows—

  1. The longer and unbranched protoplasmic projection of a neurone is called an axon.
  2. Axon emerges from a conical portion of the cell body, called the axon hillock.
  3. The semi-liquid cytoplasm present in the axon is called axoplasm, which is externally lined by a cell membrane called axolemma.
  4. Axoplasm does not contain Nissl granules.
  5. In several neurones, the axon has an extra lipid-rich insulating coat outside the axolemma, called myelin sheath.
  6. The myelin sheath has a number of periodic gaps, which are called nodes of Ranvier.
  7. A few tiny nucleated cells remain associated with myelin sheath, these are known as Schwann cells.
  8. The neurones with myelin sheath are called medullated nerve fibres whereas the axons having no myelin sheath are called non-medullated nerve fibres.
  9. Axon terminally gives off a few fine branches, this portion is called the end brush. The end of each branch swells to form a synaptic knob or end button.

Dendron:

The structural features of dendron is as follows—

  1. The shorter and branched protoplasmic projection of a neurone is known as a dendron. Branches of dendrons are known as dendrites.
  2. The base of each dendrone is wide but it gets terminally finer and branched.
  3. Neuroplasm, neurofibrils and Nissl granules are present in the dendron, but it does not have any myelin sheath.

Or,
Draw a neat diagram of anaphase of mitosis of an animal cell and label the following parts:

  1. Polar region
  2. Spindle fibre
  3. Chromatid
  4. Centromere.

Answer:

Anaphase of mitosis in plant and animal cells:

The third phase of karyokinesis, in which the daughter chromosomes move from the equatorial plate towards the two opposite poles of the dividing cell, is called anaphase. The events which occur during this phase, in both plant and animal cells, are discussed below.

1. Anaphase in plant cells:

The events occurring during anaphase, in plant cells, are as follows-

  1. The Centromere of each chromosome splits longitudinally, as a result, two sister chromatids separate from each other with their own share of centromere and emerge as two daughter chromosomes.
  2. The Centromere of each daughter chromosome remains attached to the chromosomal spindle fibre.
  3. Inter-polar or Chromosomal continuous spindle fibres connect the two poles of the -spindle fibre spindle.
  4. Gradually, the chromosomal fibres contract. -Chromatid to pull half of the daughter chromosomes towards one separation Inter-zonal pole and the remaining half to the opposite pole. This is a spindle fibre called anaphasic movement.
  5. During anaphasic, movement, metacentric, sub-metacentric, acrocentric and telocentric chromosomes appear like the English letters ‘V’ ‘L’, ‘J’ and ‘I’ respectively, based on the position of centromere on the chromosome.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 2 Mitotic And Meiotic Cell Division Anaphase in plant cell

 

2. Anaphase in animal cells:

The above-mentioned events also occur during anaphase in animal cells. Here, inter-zonal spindle fibres cluster in between separating chromosomes to form columnar stem bodies, which help in anaphasic movement.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 2 Mitotic And Meiotic Cell Division Anaphase in animal cell

 

Question 2. Why reproduction is important for an organism? “Meiosis is characteristically different from Mitosis”-how would you justify the statement”?

Answer:

Common features of reproduction:

Different phases of reproduction involve mitosis, meiosis or both the processes of cell division. From gametogenesis to the embryonic development of a progeny, cell division is required in every phase.

  1. The molecular basis of reproduction is DNA replication.
  2. An offspring is formed through the union of gametes, germination of spores or division of vegetative cells.

Significance of reproduction:

The significance of reproduction are as follows.

1. Increasing population size:

By means of reproduction, new individuals are born. As a result, the number of individuals in a population increases and the race of a species is maintained.

2. Maintaining the flow of genes:

By reproduction, the genes of a species flow from one generation to the next maintaining the hereditary characteristics.

3. Maintaining ecological balance:

Natural or unnatural death of individuals reduces the population size of a species. By reproduction, this loss is replenished in nature. This process keeps the ratio of different species in an ecosystem constant and thus maintains the ecological balance.

4. Variation:

By sexual reproduction, different genetic variations come into existence. These variations assist any species to adapt in its surroundings and finally help in the evolution of new species.

Meiosis is called as reduction division:

In meiosis, the chromosome number of a diploid mother cell reduces to its half to give rise to a haploid number of chromosome-bearing daughter cells. Thus, meiosis is called reduction division.

Differences between mitosis and meiosis:

WBBSE Solutions For Class 10 Life Science And Environment Chapter 2 Mitotic And Meiotic Differences between mitosis and meiosis

Or,
Describe five processes of asexual reproduction with one suitable example for each.

Answer:

Processes of asexual reproduction:

The following types of asexual reproduction can be seen in organisms-

1. Fission:

In most of unicellular organisms, two or more offspring are produced through mitosis or amitosis. This process is called fission. Some examples are given below.

1. Amoeba:

Binary fission:

In favourable conditions, the nucleus of Amoeba undergoes amitotic division and gives of two nuclei. Its cytoplasm divides longitudinally creating a furrow and forms two offsprings.

Multiple fission:

In unfavourable conditions, the pseudopodia of Amoeba are destroyed and the body gets encapsulated by a cyst wall. Inside the cyst, numerous small spores are formed through multiple fission of the nucleus and cytoplasm.

This type of multiple fission is called sporulation. When the condition becomes favourable, spores come out rupturing the cyst wall and giving rise to new Amoeba offspring.

2. Plasmodium:

In the case of Plasmodium, multiple fission takes place in two stages-Schizont and sporont. In the stomach of female Anopheles mosquito, numerous Plasmodium offsprings are born by the multiple fission called schizogony and sporogony is the schizont and sporont stages respectively.

The Plasmodium offsprings born through schizogony and sporogony are called merozoite and sporozoite respectively.

2. Budding:

Budding is a type of asexual reproduction in which a new organism develops from an outgrowth or bud of the parent body due to cell division at one particular site. Some examples of budding are given below.

Yeast:

Due to the unequal division of the yeast parent cell, a small outgrowth or bud is formed which contains the parent’s nucleus. Later on the bud separates from the parental body and develops into new yeast.

In special cases, the bud of yeast divides many times in the torula stage through the Formulation process and forms mycelium.

Hydra:

In the case of Hydra, the bud is formed outside the parent body (exogenous bud). After the formation of the oral aperture, tentacles, etc. when the offspring matures, it gets detached from the parent body.

3. Fragmentation:

The type of asexual reproduction where the parent body is fragmented into two or more pieces and each fragment turns into new offspring, is called fragmentation.

Example:

The filamentous body of Spirogyra, an alga gets fragmented due to water currents or external damage. Each fragment undergoes mitotic cell division and gives rise to new offspring.

4. Regeneration:

The type of asexual reproduction where a mere body part of a parent organism creates a new offspring, is called regeneration. It is also called morphallaxis.

Example:

If any body part of Planaria, a flatworm gets detached, it forms a new offspring. This process also can be seen in Hydra.

5. Sporulation:

Sporulation is a type of asexual reproduction in which unicellular spores are produced in moss, fern and fungi, they are dispersed and give birth to new offspring. Examples of sporulation are given below.

Fungi:

In fungi, spores with different structures and with or without motility can be seen. These spores germinate in favourable conditions and new fungi are born. Different types of spores are motile zoospores, non-motile aplanospores, thick-walled chlamydospores, oidia, and conidia formed from fragments of filaments, sporangiospores formed in sporangium.

Moss and Fern:

Spores are formed from the sporophytic tissue of the sporophytic plant body of moss. In the sporangium of the sporophytic plant body of fern, spores with similar or different shapes are formed.

Question 3. Show the result of hybridization till F2 generation between a pure black (BB) coarse-haired (RR) guinea pig and a pure white (bb) smooth-haired (rr) guinea pig by a checkerboard. State the conclusion one can reach from this hybridization experiment.

Answer :

Dihybrid cross in Guineapig

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Pure black and white coare haired guinea pig

F1………… Hybrid black and coarse-haired guinea pig.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Hybrid black and coarse haried guiena pig

 

Checker board of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Checker board of F2 generation

 

Phenotype and genotype of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Phenotype and genotype of F2 generation

 

In a cross between pure black and coarse-haired guinea pig and pure white and smooth-haired guinea pig all the F1 individuals will be black and coarse-haired guinea pig because these are dominant traits.

Each of the F1 individuals can produce four types of gametes and their cross produce four types of individuals in a ratio of 9:3:3:1.

Conclusion

The conclusion that can be reached from this hybridisation experiment i.e. —when two organisms with two or more characters with their opposite phenotypes are crossed then the characters not only get separated but remain expressed in all possible combinations independent to each other.

Or,
What symptoms are expressed in a child affected with Thalassaemia? What suggestions are given at the time of genetic counselling in such cases?

Answer:

Symptoms of thalassemia affected child:

  1. Continuous fatigue or weakness,
  2. Shortness of breath,
  3. Yellowing of skin and eyes,
  4. Slow growth and abdominal swelling,
  5. Facial bone deformities and anaemia,
  6. Dark colouration of urine

The suggestions are given at the time of genetic counselling

  1. Not to conduct marriage between carrier male and female.
  2. Not to conduct marriage between thalamic male and normal or carrier female or vice-versa.
  3. Must rest for genetic diseases before marriage or child planning

Question 4. Describe the brief outline of the chemical origin of life mentioning Miller and Urey’s experiment.

Answer:

The experiment of Miller and Urey:

Based on the concept of chemosynthesis of life, proposed by Oparin and Haldane, Stanley Miller and Harold Urey performed an experiment in 1953 in the laboratory of Chicago University.

1. Procedure:

They prepared a highly heat-resistant instrument with two inter-connected chambers of unequal diameters. The larger chamber was fitted with two electrodes and was filled with a mixture of hydrogen, methane and ammonia gases.

The outlet pipe of the larger chamber was fitted with a cooling device and a tap which was used to tap down the contents from it. The smaller chamber was used as a boiler and a connecting pipe was fitted with it to supply vapour to the larger chamber.

Powerful electric sparks were generated in the larger chamber to mimic the lightning-like situation (that prevailed in the primitive atmosphere). The water of the smaller chamber was boiled by heating it externally, thus the newly formed water vapour is pushed into the larger chamber.

The machine was run continuously for one week and the product from the larger chamber was tapped down after condensation.

 

WBBSE solutions For Class 10 Life Science And Environment Evolution Concepts Of Evolution Experimental set up of Miller and Urey

 

2. Observation:

After analysing the product, scientists found a number of amino acids, viz. glycine, alanine, glutamic acid, aspartic acid, etc. in it.

3. Conclusion:

From the result, Miller and Urey concluded that in the primitive, free oxygen-lacking, reducing atmosphere, different inorganic matter reacted with each other to synthesize organic substances.

In this reaction, heat, lightning, ultraviolet ray and other cosmic rays provided the energy that was needed. The organic matter, thus synthesized, acted as the building blocks of cells and finally helped in the origin of life.

Significance of Miller and Urey’s experiment:

Miller and Urey’s experiment supports the abiogenetic theory of the origin of life. According to this theory, life was originated from abiotic or inorganic substances around 3.7 billion years ago.

Or,
Describe two major tenets of Lamarck’s evolutionary theory. How the comparative anatomy of hearts act as a proof in favour of evolutionary theory?

Answer:

Lamarckian concept of evolution:

In 1809, Lamarck proposed the concept of organic evolution in his famous book, titled ‘Philosophie Zoologique. This concept is popularly known as Lamarckism.

Lamarckism is based on the following basic propositions.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 4 Evolution Theories Of Organic Evolution Lengthening of giraffe's neck

 

1. Internal vital force:

An internal vital force or energy in all life forms helps in the development of the body and organs.

2. Development of new organs according to need:

With the change in environment, a need develops among the organisms for morphological and behavioural change. Newly developed organs help the organisms to adapt in the changing environment.

3. Law of use and disuse:

Any organ, which is used more, grows in size and strength, whereas, that which is used less, reduces in size, becomes weak and finally becomes extinct.

Example of use of organ:

Ancestors of giraffes had short necks. To collect the leaves of taller plants, they stretched their necks continuously. This conscious effort made their necks long and eventually, giraffes with longer necks evolved.

Example of disuse of organ:

The ancestors of snakes had limbs. While adapting to the fossorial habitat, their limbs were left unused. Finally, limbs. of snakes were lost during the course of evolution.

4. Inheritance of acquired characters:

Any change in the characteristics of an organism, which an organism achieves by conscious effort, is inherited by its offspring.

5. Formation of new species:

The characters acquired during the lifetime are passed onto the progeny and after a number of generations, new species is produced.

To get leaves from tall trees, ancestors of giraffes stretched their necks. By this process, one might have lengthened it a little, which was inherited directly by its offspring.

This process continued for several thousand generations and finally completely different-looking long-necked giraffes emerged as a new species from their ancestors.

Comparative anatomical evidence in support of organic evolution:

A comparative study on related groups of organisms, clearly reveals the basic structural similarity among few of their organs. It also indicates the evolutionary trend of those organisms.

Here, comparative studies on some organs of different groups of vertebrates are mentioned below.

1. Heart of vertebrates:

The basic structure of the hearts of fishes, amphibians, reptiles, birds and mammals is similar. All of them have an auricle as the blood-receiving chamber and a ventricle as the chamber that expels blood from the heart. However, in the course of evolution, the hearts of different vertebrates encountered some changes.

The heart of fish has one auricle and one ventricle and thus carries only deoxygenated blood along a single circuit. Amphibians have two auricles and one ventricle in their heart. Here, oxygenated and deoxygenated blood get mixed in the ventricle.

Apart from two auricles, the reptilian heart has a partially divided ventricle, in which partial mixing of oxygenated and deoxygenated blood occurs (exception-crocodile has a four-chambered, double-circuit heart).

Hearts of birds and mammals have two auricles and two ventricles. Therefore oxygenated and deoxygenated blood flow through separate circuits.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 4 Evolution Evidence For The Theory Of Evolution Structure of heart of vertibrates

 

Conclusion:

The above-mentioned evidence proves that advanced and complex forms of animals have evolved from primitive and simple forms. The gradual development of the heart and circulatory system of different vertebrates point towards the evolutionary trend as mentioned below.

Fishes → Amphibians → Reptiles → Birds and mammals

2. Forelimbs of different vertebrates:

The basic skeletal structures of the forelimbs of whales, bats, birds, horses and men are all alike. All possess humerus, radius-ulna, carpals, metacarpals and phalanges.

However, these animals use their forelimbs in different ways to adapt in different environments. As whale uses its forelimbs for swimming, birds and bats use them for flying. The horse runs with the help of their forelimbs and man use them for grasping.

Conclusion:

This evidence strongly indicates that all vertebrates have evolved from a single ancestor but diverged from each other to adapt in separate environments.

Question 5. Analyse the role of bacteria at any three steps of nitrogen cycle. “Luxurious and comfortable lifestyle is one of the major reasons of air pollution”-Support the statement with arguments.

Answer:

Nitrogen cycle:

The nitrogen cycle is a bio-geochemical process which involves different phases, which are-nitrogen fixation in soil, entry of nitrogen from soil into living organisms, return of nitrogen from organisms to soil, and liberation of nitrogen in the atmosphere. The phases are explained in the next section.

1. Nitrogen fixation:

By this process, atmospheric nitrogen is trapped in the soil. This occurs by three different processes, which are-

Natural N2 fixation:

During lightning, a huge amount of heat is generated in the atmosphere, when nitrogen and oxygen react gradually to produce nitric oxide (NO) and nitrogen dioxide (NO2).

When rainwater mixes with atmospheric nitrogen dioxide two acids, namely nitrous acid (HNO2) and nitric acid (HNO3) are produced. The reactions are given below-

N2+O2 → 2NO,  2NO+O2 → 2NO2, 2NO2 + H2O → HNO2 + HNO3

These two acids then react with the minerals present in soil and form nitrates (potassium nitrate, calcium nitrate, etc.).

Biological fixation:

Biological nitrogen fixation is done in the following ways-

  1. Soil-dwelling nitrifying bacteria like Azotobacter sp., Clostridium sp., etc. absorb atmospheric nitrogen and trap it in their cells. When these bacteria die, the nitrogenous compounds mix in the soil.
  2. Cyanobacteria like Anabaena sp. and Nostoc sp. absorb nitrogen from the atmosphere and add it to the soil.
  3. Rhizobium sp. present in root nodules of leguminous plants traps atmospheric nitrogen and converts it into nitrogenous compounds. With the death of the host plants, these compounds mix in the soil to increase its nitrogen content.

Industrial fixation:

In fertilizer industries, different nitrogen-rich fertilizers are produced. By the Haber-Bosch method, N2 and H2 are made to react at 300-400°C under 35-100 MPa (Mega Pascal) pressure to produce NH3.

This NH3 and other nitrogen-rich fertilisers when added to soil increases the nitrogen content.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Nitrogen Cycle Industrial fixation

2. Entry of nitrogen from soil into living organisms:

Plants absorb soluble nitrogenous salts from the soil through their roots. This nitrogen takes part in synthesizing plant proteins. Animals get nitrogen either directly or indirectly by consuming plants.

Finally, plant proteins are transformed into animal proteins and remain in the animal bodies.

3. Return of nitrogen from organisms to soil:

Nitrogen returns from the body of organisms to the soil through two different steps. These are-

Ammonification:

The ammonifying bacteria (Bacillus mycoides, Micrococcus sp., etc.) act upon dead animal and plant bodies and the excretory materials to convert the nitrogenous compounds in it into ammonia. This is called ammonification.

Nitrification:

Certain nitrifying bacteria convert ammonia into nitrates through two steps of reaction.

  1. Nitritation-In the first step, Nitrosomonas sp. converts ammonia and ammonium ions into nitrites.
  2. Nitratation-In the next step, soil-dwelling Nitrobacter sp. transforms nitrites into nitrates.

2NH4++3O2 → 2NO2+ 4H+ + 2H2O,   2NO2+O2 → 2NO3

4. Liberation of nitrogen into the atmosphere or denitrification:

In this phase, several bacteria act upon nitrites and nitrates in the soil and liberate free nitrogen into the atmosphere. This biochemical process is called denitrification. Pseudomonas sp., Thiobacillus sp., etc. are a few bacterial species, which help in denitrification.

2NO3  → 2NO2, → NO + N2O → N2

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Nitrogen Cycle Nitrogen in to the atmosphere or denitrification

Explanation to support the statement:

‘Luxurious and comfortable lifestyle is one of the major reasons of air pollution-this statement can be well supported and explained by the facts, like-

  1. CFC used in refrigerators and air conditioners caused global warming, destroys the ozone layer and renders the harmful effects of UV rays on the environment to increase.
  2. Methane gas produced during natural gas extraction causes global warming.
  3. Different oxides of sulphur, nitrogen and smoke from agricultural and industrial field enhances the chances of acid rain and causes different respiratory disorder.

Reason of the increase of methane in the atmosphere:

Methane is produced due to the partial combustion of certain organic matters, putrefaction of dead plants and animals in wetlands and emission from coal mines. These processes increase the percentage of methane in the atmosphere.

Or,
Give two examples of animal biodiversity endangered by the pollution of the river Ganga. Explain any three incidences which are occurring as a result of disruption of the nitrogen cycle.

Answer:

Two examples of animal biodiversity endangered by the pollution of river Ganga are-

  1. Gangetic dolphin and
  2. Gharial.

Human actions affecting the nitrogen cycle:

The human actions affecting the nitrogen cycle are discussed below.

1. Application of nitrogen-rich fertilizers:

Nowadays, farmers apply an excess amount of nitrogen-rich fertilizers in crop fields to increase productivity. Due to this practice, the content of N2O and NO2 is rising alarmingly in the atmosphere.

2. Cultivation of nitrogen-fixing plants:

Leguminous plants like gram, peas, lentils, soya beans, etc. are important sources of plant protein. To fulfil the need of protein, different pulses are being cultivated more and more. These plants efficiently absorb nitrogen from the atmosphere and trap it in the soil.

3. Burning of fossil fuel:

Excessive burning of fossil fuels such as mineral oils, coal, etc. increases nitric oxide (NO), nitrous oxide (N2O), and nitrogen dioxide (NO2) in the air. It increases the supply of nitrogen in the environment.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Nitrogen Cycle Root nodules of pea plant

 

Adverse effects of excess nitrogen in the atmosphere due to human activity:

The adverse effects of excess nitrogen in the atmosphere, due to human activity, are mentioned below.

1. Global warming:

Nitrous oxide (N2O) is a potent greenhouse gas. Accumulation of this gas in air arrests more heat in the atmosphere and causes global warming.

2. Acid rain:

Oxides of nitrogen present in the atmosphere mix with rainwater to produce nitrous and nitric acids. These acids come down on earth as acid rain, which acidifies soil and water bodies to cause serious environmental pollution.

3. Photochemical smog:

Nitric oxide (NO), released as automobile exhaust, reacts with atmospheric hydrocarbons in the presence of sunlight. This reaction produces a polluting gas, named PAN (Peroxy-acetyl nitrate).

It mixes with smoke and creates a very harmful photochemical smog. This reduces visibility through air, and causes inflammation of the respiratory tract and various other respiratory diseases.

Question 6. “Change of climate and global warming is a problem resulting from the increasing human population”-Summarise what impacts it might exert on the environment. “Children and adults both suffer with different respiratory problems during the winter”-Write two names of such problems and state one symptom of each of them.

Answer:

Effect of change of climate and global warming on the environment:

Change of climate and global warming is a problem resulting from the increasing human population. Their effects on the environment are as follows-

  1. Due to the increased level of greenhouse gases in the environment, the number of atmospheric components are being changed. These gases increase global temperature, as a result, polar ice is melting. Polar bears, polar foxes, etc. are becoming endangered as they are losing their habitats.
  2. Due to global warming, sea water level is increasing, as a result, land areas near the sea are getting submerged.
  3. The ecosystem is being destroyed due to global warming, biodiversity is also being decreased. The example-Great barrier reef of Australia is depleting gradually.
  4. Climate change is causing harm to agriculture which in turn causing scarcity of food.
  5. In altered temperatures, harmful germs are getting more active and causing diseases.

Two respiratory problems during winter were found in children and adults

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Environment And Human Population respiratory problems during winter

Or,
You have identified three environmental problems while on research in Sunderbans-

  1. Disbalance in prey-predator number
  2. Destruction of mangrove plants due to urbanization
  3. Submergence of islands due to rising sea level. Analyse what impacts they might exert on biodiversity. Write the location of the Sundaland hotspot and one important species of its biodiversity.

Answer:

Problems of Sundarban:

In recent days Sundarban is facing many environmental problems. These are mentioned below.

1. Loss of mangrove forest due to urbanization:

Population pressure has initiated urbanization. Due to this, many areas along the periphery of Sundarban are deforested.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Importance Of Biodiversity Destruction of mangrove foreset in Sundaraban

 

2. Agricultural problem:

Due to soil erosion and siltation, the depth of rivers becomes shallow. During high tide saline water spills over, and floods the agricultural fields to make the soil salty and infertile.

3. Scarcity of fresh water:

The estuarine water is salty. Frequent flood makes inland water bodies like ponds and wells saline. Therefore, sweet water is becoming scarce in the Sundarban area.

4. Destruction of natural habitat:

Deforestation is reducing the natural habitats of large to small animals. As a result, the biodiversity of Sundarban is gradually getting destroyed.

5. Pollution:

The most provocative problem for Sundarban is increasing pollution. All rivers are disposing of gallons of pollutants in the water. Several pollutants are percolating in the soil of the Sundarban area making the soil infertile and toxic.

The oil and grease, spilling from fishing and tourist vessels are polluting the aquatic environment of this area. This pollution is affecting the propagation of fish, prawns, and other aquatic animals.

6. Imbalance in the prey-predator ratio:

Due to overpopulation and pollution, the number of many animal species has declined. This has resulted in an imbalance in the prey-predator ratio. The scarcity of natural food compels tigers to enter into human localities in search of food.

7. Rise in water level:

Due to global warming, the glaciers of the polar region are melting. This results into an increase in sea water level. As a result, the delta of Sundarban is going under seawater.

One of the most important biodiversity hotspots Sundaland is located in Southeastern Asia, and comprises of the landmass of Malay, Java, Sumatra, Peninsula, Bali, Borneo, Andaman, and Nicobar islands.

The important species of Sundaland is-Sundari plant (plant) or Orangutan (animal).

WBBSE Class 10 Life Science And Environment Model Question Paper 2023 Set 1

Model Question Paper Life Science And Environment Set 1 Group A

Answer to all questions is compulsory

1. Write the answer in complete sentences by choosing the correct answer for each question with the respective serial number.

Question 1. The tentacles on the leaves of an insectivorous plant, called sundew to bend to trap the insect as soon as they come in contact with the insect body. This is-

  1. Seismonasty
  2. Photonasty
  3. Thermonasty
  4. Chemonasty

Answer: 4. The tentacles on the leaves of an insectivorous plant, called sundew bend to trap the insect as soon as they come in contact with the insect body. This is chemonasty.

Question  2. In case of women, the Follicle Stimulating Hormone stimulates the Graafian Follicle of the ovary to secrete the hormone, is-

  1. TSH
  2. Oestrogen
  3. ADH
  4. ACTH

Answer: 2. In the case of women, the Follicle Stimulating Hormone stimulates the Graafian Follicle of the vary to secrete the hormone, is Oestrogen.

Question  3. Which one of the following is the correct reflex arc?

  1. Receptor → Effector Efferent nerve → Nerve Centre → Afferent nerve
  2. Nerve Centre → Receptor → Afferent nerve Effector → Efferent nerve
  3. Efferent nerve →  Receptor → Afferent nerve → Nerve Centre → Effector
  4. Receptor→ Afferent nerve → Nerve Centre → Efferent nerve Effector

Answer: The correct reflex arc is—Receptor → Afferent nerve → Nerve Centre → Efferent nerve Effector.

Question  4. In which stage of mitosis cell division, the nuclear membrane and nucleolus are seen to disappear?

  1. Anaphase
  2. Metaphase
  3. Prophase
  4. Telophase

Answer: 2. In the prophase stage of mitosis cell division, the nuclear membrane and nucleolus are seen to disappear.

Question  5. Which of the following is not a structural component of DNA?

  1. Deoxyribose sugar
  2. Thymine base
  3. Uracil base
  4. Phosphoric acid

Answer: 2. Uracil base is not a structural component of DNA.

Question  6. Which one of the following statements is correct regarding sexual reproduction?

  1. Haploid gamete formation is essential in sexual reproduction
  2. Sexual reproduction depends only on mitosis
  3. In sexual reproduction, offspring could be produced from a single parental organism
  4. The offsprings produced in sexual reproduction are genetically identical with the parental organism

Answer:  1. The correct statement regarding sexual reproduction is haploid gamete formation is essential in sexual reproduction.

Question  7. Which one of the following is the genotypic ratio at F2 generation of Mendel’s monohybrid cross?

  1. 1:2:1
  2. 3:1
  3. 9:3:3:1
  4. 2:1:2

Answer: 1. The genotypic ratio at F2 generation of Mendel’s monohybrid cross is 1:2:1.

Question  8. Which one of the following is not controlled by the autosomal gene of humans?

  1. Roller tongue
  2. Haemophilia
  3. Thalassaemia
  4. Attached ear lobe

Answer:  2. Haemophilia is not controlled by the autosomal gene of humans.

Question  9. What would be the phenotypic ratio in the F2 generation of a monohybrid cross in case of incomplete dominance?

  1. 3:1
  2. 2:1:1
  3. 9:3:3:1
  4. 1:2:1

Answer: The phenotypic ratio in the F2 generation of a monohybrid cross in case of incomplete dominance is 1:2:1.

Question  10. Which one of the following is the intraspecific struggle for the same food?

  1. The struggle between Vulture and Hyaena
  2. The struggle between Eagle and Kite
  3. The struggle among the members of Rohu fish in a pond
  4. Struggle between Egret and Kingfisher

Answer: The intraspecific struggle for the same ‘food is a struggle among the members of Rohu fish in a pond.

Question  11. Which of the following changes did not occur during the course of the evolution of the horse?

  1. Lengthening of limbs
  2. Increase in the length and thickness of all digits in limbs
  3. Increase in the length and thickness of only the third digit in limbs
  4. Increase in size of the whole body

Answer:  Increase in the length and thickness of all digits in limbs did not occur during the course of the evolution of the horse.

Question  12. According to the opinion of the scientists the prebiotic environment of the earth that led to the origin of life was like-

  1. Hot dilute soup in the seawater
  2. Cold dilute soup in the seawater
  3. Cold dilute soup in the river water
  4. Hot dilute soup in the groundwater

Answer: According to the opinion of the scientists the prebiotic environment of the earth that led to the origin of life was like hot dilute soup in the sea water.

Question  13. Which one of the following microbes takes part in nitrification?

  1. Nitrosomonas
  2. Azotobacter
  3. Pseudomonas
  4. Thiobacillus

Answer: Nitrosomonas takes part in ‘ nitrification.

Question  14. An example of Ex-situ conservation is-

  1. Sunderban Tiger Reserve
  2. Corbett National Park
  3. Nilgiri Biosphere Reserve
  4. Cryopreservation

Answer: An example of Ex-situ conservation is cryopreservation

Question  15. Which of the following effect is the result of water pollution?

  1. Global warming
  2. Eutrophication
  3. Deafness
  4. Bronchitis

Answer: Eutrophication is the result of water pollution

 

Model Question Paper Life Science And Environment Set 1 Group B

2. Answer any 21 questions out of 26 questions given below as instructed. Fill in the blanks with proper words in the following sentences (any five):

Question 1. Due to a deficiency of iodine, the synthesis of ________________ hormone is hampered.
Answer: Thyroxine /T4/T3

Question 2. During embryonic development and growth in vertebrates _______________ cell division takes place.
Answer: Mitosis

Question 3. The different _______________ of Pea Plant may show the same phenotype.
Answer: Genotype

Question 4. Homologous organs evolve due to _______________ evolution.
Answer: Divergent

Question 5. The name of one exotic species, that affects the local biodiversity, is _______________.
Answer: Water hyacinth/Parthenium/Tilapia fish.

Question 6. The fine droplets of smoke, ashes, dust, pollen grains etc., suspended in air collectively called _______________, which cause various lung diseases.
SPM or Suspended Particulate Matter.

Decide Whether The Following Statements Are True Or False Any Five:

Question 7. Hypothalamus helps to maintain body balance in humans.
Answer:  False

Question 8. In DNA, Adenine is linked with Guanine by 2.22 In the case of guineapig whether the phenotype hydrogen bonds.
Answer: False

Question 9. In his monohybrid cross experiment, Mendel obtained 75% pure tall Pea Plants in the first filial generation.
Answer: False

Question 10. In Lamarck’s opinion, organisms can acquire any particular inheritable physical trait according to their need.
Answer: True

Question 11. Acid rain is caused by SO2 and NO2 gases formed due to air pollution.
Answer: True

Question 12. Auxin helps the seed to germinate by breaking off dormancy.
Answer: False

Match the words in column-A with those which are most appropriate in column B and re-write the correct pair mentioning the serial no. of both columns (any five):

WBBSE Model Question Paper 2018 Life Science And Environment Set 1 Match the columns

Answers: 13. F, 14. C, 15. E, 16. A, 17. B, 18. G

Answer In A Single Word Or In A Single Sentence Any Six:

Question 19. Choose the odd one and write it: Dwarfism, Goiter, Thalassaemia, Diabetes Mellitus.
Answer: Thalassaemia

Question 20. Mention one function of the lens of the human eyeball.
Answer: The lens creates an image on the retina by the refraction of incident rays of light entering into the eye.

Question 21. A pair of related terms is given below. On the basis of the relationship in the first pair write the suitable word in the gap of the second pair. Mitosis: Radicle::____________: Spore Mother cell.
Answer: Meiosis

Question 22. In the case of guineapig whether the phenotype of the two genotypes bbRR and bbRr is same?
Answer: Yes, the phenotype of the two genotypes bbRR and bbRr will be the same in the case of guineapig.

Question 23. Write the phenotypic ratio obtained in the F2 generation of Mendel’s Dihybrid cross experiment.
Answer: 9 : 3 : 3: 1

Question 24. Mention one adaptation of the Sundari plant for salt tolerance.
Answer: Salt glands present on the leaves of the Sundari plant help in removing excess salt from the plant body.

Question 25. Among the following four terms one includes the other three. Find out that term and write it: Overall knowledge regarding local biological resources, PBR, Traditional belief regarding local biological resources, Sustainable use of local biological resources.
Answer: PBR

Question 26. Name one endangered animal conserved at Singalila National Park.
Answer: Red panda/Clouded leopard

 

Model Question Paper Life Science And Environment Set 1 Group C

3. Answer any 12 questions in 2-3 sentences out of 17 questions given below.

Question 1. Reflex actions occur when the following incidents happen in the everyday life of humans. Write the importance of these two reflex actions:

  1. When food particles enter into the trachea
  2. When foreign particles enter inside the nasal cavity

Answer:

  1. When food particles enter into the trachea, coughing occurs. Through coughing, the food particles are expelled out of the trachea along with pulmonary air.
  2. When foreign particles enter inside the nasal cavity, sneezing occurs. Sneezing helps to expel the foreign particle from th6 nasal cavity.

Question  2. For correction of which defects in the vision of the eye, spectacles of convex and concave lenses are used?
Answer:

Spectacles of convex lenses are used in hyperopia and spectacles of concave lenses are in myopia.

Question  3. List the names of the hormones related to the following functions:

  1. Regulation of sugar level in blood
  2. Stimulation for the secretion of hormones from the Thyroid gland
  3. Causing the growth of the Corpus Luteum and stimulating the secretion of Progesterone hormone in the female body
  4. Increasing blood pressure due to anxiety

Answer:

  1. Insulin and glucagon.
  2. TSH or Thyroid Stimulating Hormone.
  3. LH or Luteinizing Hormone.
  4. Adrenaline or Epinephrine.

Question  4. A resting animal attempts to move. Mention four probable reasons behind it.

Answer:

Driving forces of locomotion;

The driving forces of locomotion are mentioned below.

1. Search of food:

Unlike plants, animals are not capable of synthesising their own food. Therefore animals have to roam around in search of food resources.

2. Escaping predation:

Among animals, prey-predator interaction continues all the time. Here the aggressor is called the predator and the victims are called the prey.

To avoid predation’ prey have to flee from the predator. Therefore, locomotion is essential for prey to escape predation.

3. Migration:

Due to environmental changes, various species move from one place to another this is called migration. To avoid natural calamities, epidemics and other adversities, animals migrate to favourable places. This is a driving force of locomotion.

For example: during rain ants upward in search of a dry habitat.

4. Searching for favourable habitat:

Animals always look out for places with adequate food, water, ambient temperature, light, etc. This is also a purpose of animal locomotion.

5. Searching for a mating partner and favourable mating place:

Animals and a few plants perform locomotion in search of favourable places for reproduction. Animals roam around in searc 1 o suitable mating partners for successful reproduction.

Question 5. Distinguish between autosome and sex chromosome of humans based on the following features:

  1. Nature
  2. Number

Answer:

Differences between autosome and allosome are—

WBBSE Solutions For Class 10 Life Science And Environment Chapter 2 Continuity Of Life Chromosome Differences between Autosome and Allosome

Question  6. What are the processes of asexual reproduction in Spirogyra and Planaria?

Answer:

Spirogyra mediates their reproduction by fragmentation, whereas planaria make it by regeneration.

Question  7. Mention two changes related to vision and bones at late adulthood or the senescence phase of human development.

Answer:

During adulthood or the senescence phase of human development, some noticeable changes are-

  1. Vision: Due to the weakening of eyesight, symptoms of presbyopia can be seen.
  2. Bones: Due to the decaying of bones and joints, symptoms of osteoarthritis and osteoporosis can be noticed.

Question  8. With the help of a cross show how the sex of offspring of humans is determined.
Answer:

Differentiate between autosome and allosome.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Sex Determination Of Human Differences between autosome and allosome (1)

Question  9. One day students read an article in the newspaper on Thalassemia and were very scared to know the fate of a thalassemic patient. Write what kind of measures they can take to eradicate this disease from the population.

Answer:

The measure they can take to eradicate this disease from the population are-

  1. Genetic counselling before marriage and not to conduct marriage between carrier male and female or affected and normal individual.
  2. If genetic counselling is not done before marriage then it must be done before conceiving a child and not to conceive in sensitive cases.

Question  10. What are all probable gametes to be produced from the hybrid plant YyRr produced at F1 generation in a dihybrid cross experiment?

Answer :

The probable gametes to be produced from the hybrid plant YyRr at F1, generation in a dihybrid cross experiment are four types which are YR, Yr, YR, yr.

Question  11. What are the two adaptive significances of air-sacs of a pigeon?

Answer:

The roles of air sacs in volant adaptation of it needs more supply of oxygen to generate extra pigeons are

  1. Pigeon is a volant animals; hence energy during its flight. The air sacs, attached to its lungs, supply extra oxygen and therefore this process enhances the ability of the lungs.
  2. These air sacs hold air to reduce the specific gravity of the body during flight.

Question  12. Write the names of one each vestigial organ present in the vertebral column and alimentary canal in the human body.

Answer:

In the human body, the vestigial organ present in the vertebral column is the coccyx and that present in the alimentary canal is the appendix.

Question 13. Write in chronological sequence the names of four main fossil ancestors during the course of the evolution of the horse.

Answer:

Four important features modified in the evolution of horse are-

  1. Gradual increase in height,
  2. Decrease in the number of digits in both forelimbs and hindlimbs and formation of the hoof,
  3. Increase in the size of crown of molar and pre-molar teeth,
  4. Increase in volume of the cerebral hemisphere.

Question 14. “Nitrogen cycle is affected due to human activities”-state two phenomena to justify it.

Answer:

Human actions affecting nitrogen cycle:

The human actions affecting the nitrogen cycle are discussed below.

1. Application of nitrogen-rich fertilisers:

Nowadays, farmers apply excess amounts of nitrogen-rich fertilisers in crop fields to increase productivity. Due to this practice, the content of N2O and NO2 is rising alarmingly in the atmosphere.

2. Cultivation of nitrogen-fixing plants:

Leguminous plants like gram, peas, lentils, soya beans, etc. are important sources of plant protein. To fulfil the need of protein, different pulses are being cultivated more and more. These plants efficiently absorb nitrogen from the atmosphere and trap it in the soil.

3. Burning of fossil fuel:

Excessive burning of fossil fuels such as mineral oils, coal, etc. increases nitric oxide (NO), nitrous oxide (NO) and nitrogen dioxide (NO2) in air. It increases the supply of nitrogen in the environment.

Question  15. What would be the environmental consequences as a result of recurrent destruction of wetlands and depletion of a number of agricultural lands?

Answer:

Wetlands are full of biodiversity. These help in removing pollutants from water and also control the local Due environment. overpopulation, civilisation and industrialisation these wetlands are getting destroyed.

As a result, water pollution is increasing and local bio-diversity is decreasing. Due to urbanisation, the quantity of urban land is being depleted.

As a result, worldwide food Due to urbanisation, the quantity of urban land is production is decreasing compared to the increasing human population.

Question 16. What are the harmful impacts sound pollution can exert on the ears and heart in the human body?

Answer:

Adverse effects of noise pollution on man:

The adverse effects of noise pollution on humans are mentioned below.

1. Effect on the ear:

Hearing a high-frequency sound for a prolonged period causes the sound-sensory cells of the organ of Corti inactive. In this case, the person may gradually become deaf. Drivers of metro rail face a similar problem.

A sudden sound of more than 150 dB may rupture the eardrums and may damage the cochlea to cause permanent deafness. Survivors of bomb blasts often face similar consequences.

2. Effect on heart:

  1. Hearing high-intensity noise causes excitement which in turn increases heartbeat.
  2. Continuous hearing a noise above 90 dB increases systolic pressure.
  3. Prolonged listening to a sound above 60 dB may cause myocardial infarction to the listener.
  4. Hearing high intensity sound for some time increases the chance of heart attack.

3. Other effects:

  1. Noise pollution leads to mental depression.
  2. High-intensity sound causes a lack of concentration.
  3. It may lead to insomnia.
  4. Noise curtails physical ability and enhances fatigue.
  5. A person may become ill-tempered with the effect of noise pollution.

Adverse effects of noise pollution on animals

The adverse effects of noise pollution on animals are mentioned below.

1. Deafness:

A high-intensity sound may cause deafness in animals.

2. Hampered reproduction:

The noise changes the reproductive behaviour of many animals. The population of birds and several other animals has shrunk significantly due to noise pollution.

3. Effect on foetal growth:

It is experimentally proved that high-intensity sound restricts foetal development in rats. Speculate the probable causes of each of the following phenomena

Question  17. Mention the roles of biodiversity in the following two aspects in the progress of human civilization:

  1. Production of foods
  2. Manufacturing of medicines

Answer:

Production of food:

We all depend on nature for food. Nature can give us a wide range of various organisms that meet the need of food. Different food items provide different essential nutrients like-Vitamin, and minerals along with energy.

Other than that a wide range of animal protein are also acquired from poultry and fishery. All these animals and plants are the gift of the biodiversity of nature.

Manufacturing of medicine:

In the early decade. We mainly depended on plants for medicine. A variety of plant species are provided by biodiversity, mainly used for medicinal purposes. Not only plants, many fungi and body parts of animals are also used as medicine.

For example, we get quinine from the cinchona plant; a vital drug for malaria reserpine from the roots of rauwolfia; maintain blood pressure level to normal, daturine from datura prevents hyperventilation etc.

Along with that, holy basil, and margosa plants are also exert their medical value. Fungi like Penicillium notatum is essential for producing penicillin, an antibiotic that fights against a wide range of bacteria. So biodiversity helps to progress overall human welfare and civilization.

 

Model Question Paper Life Science And Environment Set 1 Group D

4. Write the 6 questions or their alternatives given below.

Question  1. Draw a diagram of a reflex arc and label the following parts:

  1. Receptor
    Sensory nerve
    Nerve centre
    Motor nerve
    Or,
    Draw the morphology of an ideal eukaryotic chromosome and label the following parts:

Chromatid
Centromere
Nucleolar Organizer
Telomere

Answer:

Functional activities of a simple reflex arc:

If a thorn pricks our sole, we withdraw the foot at once. This is an example of a reflex action. The process of this reflex action occurs in the following way—

  1. As the thorn pricks, the receptors in the skin below our soul sense the stimulus.
  2. The receptors at once, generate a sensory impulse and sensory or afferent neurone (afferent nerve) carries it to the spinal cord (nerve centre).
  3. Immediately, the sensory impulse is modified into a motor impulse and sensory or afferent neurone (afferent nerve) carries it to the spinal cord (nerve centre).
  4. Immediately, the sensory impulse is modified into a motor impulse and it is conveyed to the motor or efferent neurone (efferent nerve) by an inter-neurone within the spinal cord.
  5. The efferent neurone instantly carries the motor impulse to the leg muscles (effector).
  6. The muscles of the leg act at once to lift the foot promptly to avert a deeper prick.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 1 Human Nervous System Simple reflex arc

or

Answer:

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 2 Continuity Of Life Chromosome Physiacal Structure of a chromosome

 

Question  2. What types of chemical components are synthesized in different stages of Interphase in a cell cycle? What might happen if normal control at different points of a cell cycle is lost?
Or,
Depict the alternation of generation in a fern with the help of a flowchart. “In case of Prophase and Telophase changes of opposite nature happens”-Write two such changes.

Answer:

Different phases of the cell cycle:

The systematic cycle of events, which continuously occurs from the end of a cell division to the end of the next cell division, is called the cell cycle. A cell cycle has two main phases-

1. Interphase:

The longer phase of a cell cycle in which a new daughter cell grows by synthesis of DNA, proteins, RNA etc. and gets all necessary cellular components to become a fully matured cell, ready for division, is called interphase.

This phase is divided into three sub-phases-G1 or growth-1 phase, S or synthesis phase and G2 or growth-2 phase.

G1, or growth-1 phase: It is the primary phase of interphase, where rRNA, mRNA, tRNA, ribosomes and the materials necessary for DNA synthesis are produced. Doubling of various cell organelles occurs in this phase.

S or synthesis phase: This is the intermediate phase between G1, and G2 phases. During this phase, the synthesis or replication of DNA and the synthesis of histone proteins occur. Here, the DNA content of a chromosome becomes double.

G2 or growth-2 phase: It is the last phase of the interphase. In this phase, the proteins, necessary for upcoming cell division, are synthesised.

2. Mitotic phase:

It is the shortest phase of a cell cycle. This phase is divided into two sub-phases-

Karyokinesis, where the nucleus of a cell divides through prophase, metaphase, anaphase and telophase.

Cytokinesis, in which cytoplasm divides to give rise to two daughter cells.

A human needs 20-24 hours to complete a cell cycle, of which 19-23 hours goes for interphase only and 1 hour for the mitotic phase.

Result of the uncontrolled cell cycle:

If for certain reasons, the checkpoints of cell cycle lost its function, then any diseased or torn-out cells can easily proceed towards cell division in an uncontrolled way and thereby give rise to cancer.

or

Answer:

Alternation of generations:

The cyclic alternation between haploid gametophytic and diploid sporophytic generations in the life cycle of any sexually reproducing organism is called the alternation of generation.

Example:

In plants like moss, fern etc. and protists like Paramoecium, Monocystis etc. distinct alternation of generation is seen.

Importance of mitosis in the alternation of generations:

During the alternation of generation, the cyclic rotation continues between haploid (n) and diploid (2n) generations. Diploid spore mother cells divide by meiosis to give rise to haploid spores.

These spores germinate to grow as haploid (n) gametophytes. The gametophytes carry separate gametangia to produce haploid male and female gametes. These gametes fertilise to form diploid zygotes to initiate the sporophytic generation. That is why meiosis is important for the alternation of generations.

Question 3. Colour of seed and shape of seed of a pea plant-taking these two characters Mendel performed a dihybrid cross. Write the genotypes of pea plants having yellow and round seeds produced in the F2 generation of this experiment. State the law of Independent Assortment of Mendel.
Or,
State the opposite traits for each of the characters of flowers of the pea plant chosen by Mendel for his experiment. “For developing the scientific ideas on heredity the experiments performed by Mendel on peas, and plants are epoch-making.”-Mention three reasons behind his success in performing these experiments.

Answer :

The genotypes of yellow and round seeds of F2 generation are—

  1. YYRR
  2. YYRr
  3. YyRR
  4. YyRr.

Law of independent assortment

Mendel derived this law from the results of his dihybrid cross-experiment. It states that—each pair of alleles segregates independently of other pair of alleles during gamete formation.

In other words, the allele a gamete receives for one gene does not influence the allele received for another gene.

or

Answer :

Opposite traits for the characters of flowers of pea plant chosen by Mendel

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Chacters of flowers of pea plant chosen by Mendel

 

Question  4. How the special feature of RBC in camels is related to their ability to withstand the extreme loss of water? Give examples of the ways Chimpanzees solve their problems intelligently for procuring food and preventing disease.
Or,
Explain the process of Natural Selection as proposed by Darwin with the help of a suitable example. What are the roles of the swim bladder in the aquatic adaptation of rohu fish?

Answer:

The physiological adaptations of camel:

Camel is one of the best-adapted animals in desert environments. The adaptive features of a camel with respect to the conservation of water and modification of RBC are mentioned below.

1. Adaptive features helping in the conservation of water:

  1. The hump on the back of the camel reserves fat, which is oxidised to produce water and energy during their long walk through the desert.
  2. Camels discharge highly concentrated urine to conserve water.
  3. A camel can drink up to 36 litres of water at a time, which is stored in the water reservoir cells of its stomach.
  4. The mucous membrane of the nasal passage absorbs moisture from the air that the camel exhales.
  5. Body temperature of the camel changes in accordance with the external temperature, which reduces perspiratory water loss, even at a high temperature of 49° Celsius.

2. Adaptive modifications of RBC:

The RBCs of camel are very small and oval in shape. This modification facilitates the movement of these blood cells through finer blood capillaries even at very low water content in plasma.

When plasma contains adequate water, the RBCs of camels can expand up to about 240 times of its normal size, by absorbing water by endosmosis. These cells can withstand a high osmotic pressure and do not rupture even after the intake of a huge quantity of water at a time.

or

Answer:

Natural selection:

Natural selection means the selection of a suitable organism in an environment by nature itself. According to Darwin, organisms acquire either favourable or unfavourable variations during their struggle for existence.

Those with favourable variations are selected by the nature. They can adapt perfectly with its environment and come out victorious in the struggle for life. They thrive successfully and propagate quickly.

On the other hand, the organisms, which have unfavourable variation, do not get the favour of natural selection and cannot adapt themselves in their environment. The population of these forms reduces gradually and finally, becomes extinct.

For example, in an ecosystem, some giraffes have long necks and others have short necks. If a situation arises, where there is scarcity of low-lying shrubs, the giraffes with short necks would not get enough food. After a few generations, all the giraffes would have long necks.

Drawbacks of Darwinism:

  1. Darwin mentioned the role of variation in evolution but could not explain the sources of it.
  2. Darwin emphasised on small variations but in reality, small variations play no role in the origin of species.
  3. Darwin supported the concept of survival of the fittest, however, he could not explain the cause of the emergence of the fittest.
  4. According to Darwin, excessive development of any organ facilitates evolution but it is seen that over-specialisation often pushes a species towards extinction.
  5. Darwin did not differentiate between somatic and reproductive variations.
  6. During natural calamities, even the fittest organisms also die, where natural selection does not operate.

Question  5. ‘A lot of indiscriminate activities of humans pollute environment-Support the statement by giving three examples in favour of it. Give examples of a national park and a biosphere reserve situated in West Bengal.
Or,
Show the steps of the Nitrogen Cycle with the help of a flowchart.

Answer:

Three examples in favour of the statement:

A lot of indiscriminate activities of human beings pollute the environment in large scale. These are-

  1. Combustion of fossil fuels causes an excessive increase in the level of greenhouse gases which in turn result global warming.
  2. Excessive use of chemical fertilisers and pesticides causes soil pollution.
  3. Industrial sewage, oil spill, etc. create water pollution. These examples justify the statement correctly.

Different types of pollution:

Based on the nature of the environmental resources, pollution can be classified into four types. These are-

  1. Air pollution,
  2. Water pollution,
  3. Soil pollution and
  4. Noise pollution.

or,

Answer:

Nitrogen cycle:

The nitrogen cycle is a bio-geochemical process which involves different phases, which are-nitrogen fixation in soil, entry of nitrogen from soil into living organisms, return of nitrogen from organisms to soil, and liberation of nitrogen in the atmosphere. The phases are explained in the next section.

1. Nitrogen fixation:

By this process, atmospheric nitrogen is trapped in the soil. This occurs by three different processes, which are-

Natural N2 fixation:

During lightning, a huge amount of heat is generated in the atmosphere, when nitrogen and oxygen react gradually to produce nitric oxide (NO) and nitrogen dioxide (NO2).

When rainwater mixes with atmospheric nitrogen dioxide two acids, namely nitrous acid (HNO2) and nitric acid (HNO3) are produced. The reactions are given below-

N2+O2 → 2NO,  2NO+O2 → 2NO2, 2NO2 + H2O → HNO2 + HNO3

These two acids then react with the minerals present in soil and form nitrates (potassium nitrate, calcium nitrate, etc.).

Biological fixation:

Biological nitrogen fixation is done in the following ways-

  1. Soil-dwelling nitrifying bacteria like Azotobacter sp., Clostridium sp., etc. absorb atmospheric nitrogen and trap it in their cells. When these bacteria die, the nitrogenous compounds mix in the soil.
  2. Cyanobacteria like Anabaena sp. and Nostoc sp. absorb nitrogen from the atmosphere and add it to the soil.
  3. Rhizobium sp. present in root nodules of leguminous plants traps atmospheric nitrogen and converts it into nitrogenous compounds. With the death of the host plants, these compounds mix in the soil to increase its nitrogen content.

Industrial fixation:

In fertilizer industries, different nitrogen-rich fertilizers are produced. By the Haber-Bosch method, N2 and H2 are made to react at 300-400°C under 35-100 MPa (Mega Pascal) pressure to produce NH3.

This NH3 and other nitrogen-rich fertilisers when added to soil increases the nitrogen content.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Nitrogen Cycle Industrial fixation

2. Entry of nitrogen from soil into living organisms:

Plants absorb soluble nitrogenous salts from the soil through their roots. This nitrogen takes part in synthesizing plant proteins. Animals get nitrogen either directly or indirectly by consuming plants.

Finally, plant proteins are transformed into animal proteins and remain in the animal bodies.

3. Return of nitrogen from organisms to soil:

Nitrogen returns from the body of organisms to the soil through two different steps. These are-

Ammonification:

The ammonifying bacteria (Bacillus mycoides, Micrococcus sp., etc.) act upon dead animal and plant bodies and the excretory materials to convert the nitrogenous compounds in it into ammonia. This is called ammonification.

Nitrification:

Certain nitrifying bacteria convert ammonia into nitrates through two steps of reaction.

  1. Nitritation-In the first step, Nitrosomonas sp. converts ammonia and ammonium ions into nitrites.
  2. Nitratation-In the next step, soil-dwelling Nitrobacter sp. transforms nitrites into nitrates.

2NH4++3O2 → 2NO2+ 4H+ + 2H2O,   2NO2+O2 → 2NO3

4. Liberation of nitrogen into the atmosphere or denitrification:

In this phase, several bacteria act upon nitrites and nitrates in the soil and liberate free nitrogen into the atmosphere. This biochemical process is called denitrification. Pseudomonas sp., Thiobacillus sp., etc. are a few bacterial species, which help in denitrification.

2NO3  → 2NO2, → NO + N2O → N2

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Nitrogen Cycle Nitrogen in to the atmosphere or denitrification

 

Question  6. What environmental factors may cause cancer in humans? If the water level in the sea increases enormously due to global warming, what problems the humans and the biodiversity of Sunderban may face?
Or,
Mention the in-situ measures adopted for the conservation of lions in Gir National Park of India. What environmental problems may arise in a locality due to the ever-increasing human population?

Answer:

Problems of Sundarban:

In recent days Sundarban is facing many environmental problems. These are mentioned below.

1. Loss of mangrove forest due to urbanisation:

Population pressure has initiated urbanisation. Due to this, many areas along the periphery of Sundarban are deforested.

2. Agricultural problem:

Due to soil erosion and siltation, the depth of rivers becomes shallow. During high tide saline water spills over, floods the agricultural fields to make the soil salty and infertile.

3. Scarcity of fresh water:

The estuarine water is salty. Frequent flood makes inland water bodies like ponds and wells saline. Therefore, sweet water is becoming scarce in the Sundarban area.

4. Destruction of natural habitat:

Deforestation is reducing the natural habitats of large to small animals. As a result, the biodiversity of Sundarban is gradually getting destroyed.

5. Pollution:

The most provocative problem for Sundarban is increasing pollution. All rivers are disposing of gallons of pollutants in the water. Several pollutants are percolating in the soil of the Sundarban area making the soil infertile and toxic.

The oil and grease, spilling from fishing and tourist vessels are polluting the aquatic environment of this area. This pollution is affecting the propagation of fish, prawns and other aquatic animals.

6. Imbalance in the prey-predator ratio:

Due to overpopulation and pollution, the number of many animal species has declined. This has resulted in to an imbalance in the prey-predator ratio. The scarcity of natural food compels tigers to enter into human localities in search of food.

7. Rise in water level:

Due to global warming, the glaciers of the polar region are melting. This results in to an increase in sea water level. As a result, the delta of Sundarban is going under seawater.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Biodiversity Conservation Of Some Endangered Species Of India

Chapter 5 Biodiversity Conservation Of Some Endangered Species Of India Summary

  • Many organisations take part in the conservation of biodiversity. But one of the organisations which play a major and most important role in biodiversity conservation in India is JFM.
  • The full form of JFM is joint forest management. It is regulated by Govt. of India. It was first implemented successfully in the Arabari forest of Medinipur for the conservation of the sal tree. Local people take part in this programme.
  • Another organisation other than JFM is PBR. The full form of PBR is Peoples’s biodiversity register. According to the biodiversity act 2002, a data bank is prepared, where the comprehensive information on the availability and knowledge of local biological resources and their way of usage are registered is called PBR.
  • Various measures are taken for the conservation of certain animals like lions, tigers, crocodiles, red pandas etc. in India. In Sundarban biosphere reserve tiger is conserved, in Manas national park rhino, in Gir national park lion, in Bhitarkanika, crocodiles are and in Barsi red panda isonserved.

Biodiversity Conservation Of Some Endangered Species Of India

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Chapter 5 Biodiversity Conservation Of Some Endangered Species Of India LAQs

Question 1. Briefly describe the concept of JFM and PBR and the role in the conservation of biodiversity.
Part Question, Discuss any two roles of the People’s Biodiversity Register (PBR) in Conserving local biodiversity
Part Question, What are the roles of forest Management in the conservation of biodiversity?

Answer:

Joint Forest Management or JFM:

The proper management of biodiversity by the joint action of local people and the forest department of the state government which is approved by the Indian government, is known as Joint Forest Management or JFM.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Role Of JFM And PBR In Biodiversity Conservation JFM or joint Forest national park

Role of JFM in the conservation of biodiversity:

In conservation of biodiversity, the role of JFM is as follows

  1. The committee, formed by the local residents of any forest area, maintains coordination with the state forest department regarding the conservation of the biodiversity of that forest.
  2. The JFM committee members keep a vigil on the forest area along with their daily activities regarding forest fire, encroaching of the forest area, poaching, felling of trees etc. and report it immediately to the forest department in case of any adversity.

People’s Biodiversity Register or PBR:

It is a ready reference prepared by the central forest department in accordance with the Biodiversity Law 2002. This document contains detailed information on the availability of local biodiversity, their habits and habitats, other behaviours, their ecological importance etc.

Life Science Class 10 Wbbse

Role of PBR in the conservation of biodiversity:

In conservation of biodiversity, the role of PBR is as follows-

  1. PBR provides correct information about the biodiversity of the local areas so that, the rare and endangered animals and plants can be conserved properly.
  2. It makes people aware of the importance of the conservation of biodiversity and involves them directly with the conservation project.
  3. PBR makes the people aware of the sustainable utilisation of forest resources and helps in the economic development of the area.
  4. PBR provides correct geographic information about conserved areas.
  5. Identify endangered species and collect information about the conservation of these species.
  6. Gather knowledge about local species and their cultural and social value.
  7. Gives an idea about the obtained product from species and their value along with their collection and transport procedure.

Question 2. Mention the causes of the decrease in the tiger population in India. Briefly describe the measures taken to protect tigers.

Answer:

Causes of decrease in tiger population:

The causes which resulted in the decrease of the population of tigers are mentioned below.

1. Poaching:

The value of tiger skin is as high as 12500 $ in the international markets. The bones, teeth, and claws also attract poachers to kill tigers indiscriminately.

2. Squeezing of natural habitat:

With the increase in human population, forest areas are being encroached for building residential places. As a result, the normal habitats of tigers are being squeezed. This is one of the most important causes of the decrease in the tiger population.

3. Problem in reproduction:

Tigers prefer dense forests for breeding. With continuous human intervention, their natural breeding habitat is being disturbed. Moreover, due to the absence of partner, the number of tigers is decreasing.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Role Of JFM And PBR In Biodiversity Conservation Tiger conservation

Measures are taken to protect tigers

In 1969, a general meeting of IUCN was held in New Delhi. Based on the resolution of that meeting, the killing of tigers was banned since 1970. After two years, the Government of India framed and implemented the Wildlife Conservation Act (1972).

In 1973, a committee, named National Tiger Conservation Authority (NTCA) was formed by the Government of India under the recommendation of IUCN and WWF. It was the first step exclusively for the conservation of tigers in India.

Primarily, the conservation project was started in nine sanctuaries in 1973. Now there are 42 different projects running in this country. Out of them Jim Corbett National Park in Uttarakhand is an important tiger conservation centre in India.

Life Science Class 10 Wbbse

Question 3. Mention the causes of the decrease in the population of rhinos in India. Briefly describe the measures taken to protect rhinos.
Part Question, Summarize which conservation measures have been adopted to increase the population of an endangered mammal exclusively found in the swampy grasslands under the foothills of the Eastern Himalayas.

Answer:

Causes of decrease in the population of rhino in India

The causes which resulted in the decrease of the population of rhinos are mentioned below.

1. Squeezing of natural habitat:

Due to the increasing human population, the forest areas are being encroached on for human residential purposes. As a result, the normal habitats of rhinos are being decreased. This is one of the most important causes of the decrease in the population of rhinos.

2. Poaching:

The value of rhino horn is very high in the East Asian markets. People, there have a baseless belief that the rhino horn has immense medicinal importance. For this purpose, the poachers kill rhinos indiscriminately and cut their horns.

3. Low birth rate:

The rhinos have very low birth rate. The gestation period is very long (479 days). That is why the population size does not grow quickly.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Role Of JFM And PBR In Biodiversity Conservation Gorumara national park

Measures are taken to protect rhinos

About 85% of the single-horned rhino (Rhinoceros unicornis) are found in Assam. Due to certain human actions, the number of the rhino population was decreasing very fast.

In view of this fact, WWF (World Wildlife Fund for Nature) and the forest department of Assam have jointly taken an initiative named IRV 2020 (Indian Rhino Vision 2020). The objective of this project is to increase the population of rhinos to 3000 by the year 2020.

The success of the project has already been established. The number of rhinos in Assam reached 2544 in 2013. According to the census of 2013, the population of rhinos is exclusively in the Kaziranga National Park reached 2319 from 1855.

Regular surveillance, restricted ecotourism and strong implementation of laws have made the project successful. Rhino conservation has also become an integral part of several other national parks. Gorumara National Park is one of them.

Life Science Class 10 Wbbse

Question 4. Mention the causes of the decrease in the lion population in India. Briefly describe the measures taken to protect lions.

Answer:

Causes of decrease in the lion population in India:

The causes which resulted in the decrease of the population of lions are mentioned below.

1. Squeezing of natural habitat:

Due to the increasing human population, the forest areas are being. encroached for human residential purposes and cultivation. As a result, the normal habitats of lions are being squeezed. Railway tracks and roadways through the forest result in the accidental death of the lions.

2. Poaching:

The value of lion skin is very high in the international markets. The bones are used in Ayurvedic medicines. Some people like to keep the teeth and nails of lions in their collection as a sign of power. Due to this, the poachers kill lions indiscriminately.

Measures are taken to protect lions:

Centuries back Asiatic lions were available in many West Asian countries, Due to indiscriminate hunting, the population is reduced to a few. King of Junagarh was the first person to take a move towards the conservation of lions.

He banned the hunting of lions in 1910. The ban was continued even after independence. In 1960 and 1970, the Gir forest was declared as a sanctuary and national park respectively.

To protect Asiatic lions, the state government of Gujarat has taken an initiative named Gir Lion Project. Since 1974, the population of lions started increasing. According to the census 2015, the number of lions in Gir National Park is 523.

In the year 2015 gradually the central ministry of Environment formed a high-level committee named ‘Asiatic Lion Re-introduction Project’ for the conservation of Gir lions.

Life Science Class 10 Wbbse

Question 5. Mention the causes of the decrease in the crocodile population in India. Briefly describe the measures taken to protect crocodiles.

Answer:

Causes of decrease in crocodile population in India:

The causes which resulted in the decrease of the population of crocodiles are mentioned below.

1. Poaching:

The demand for crocodile skin is very high in the international market. For this purpose, poachers kill crocodiles indiscriminately.

2. Loss of egg:

Crocodile eggs are often eaten up by jackals. Many people also consume these eggs. Therefore, the birth rate of crocodiles goes down.

3. Killing by a fisherman:

Crocodiles, often trapped in the fishermen’s net are killed. This also decreases the number of crocodiles.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Role Of JFM And PBR In Biodiversity Conservation Crocodile conservation

 

Measures taken to protect crocodiles:

To conserve crocodiles, Crocodile Conservation and Research Project was taken up by the government in 1974. In the next year, another project named Crocodile Breeding and Management Project or Project Crocodile was started.

This project was spread in 11 crocodile sanctuaries and 16 crocodile rehabilitation centres. To train up competent personnel, the Crocodile Breeding and Management Training Centre is established in Hyderabad in 1980.

Bhitarkanika Sanctuary is a renowned crocodile conservation centre in Odisha.

Question 6. Mention the causes of the decrease in the red panda population. Briefly describe the measures taken to protect red pandas.se in red pandas

Answer:

Causes of decrease in red panda population:

The causes which resulted in the decrease of the population of red pand are mentioned below.

1. Loss of habitat:

Regular human interventions in forests for collection of wood, and indiscriminate felling of bamboo trees reduce the size of their natural habitat. They usually feed on the leaves of bamboo. Due to the felling of bamboo trees red pandas are dying of starvation.

2. Poaching:

The soft, reddish brown fur and striated bushy tails of red pandas have always been in very high demand. That is why, poachers indiscriminately kill this beautiful animal.

3. Killing by stray dogs:

The red panda is a slow-moving animal. They are often attacked by stray dogs and are killed.

Measures are taken to protect red pandas:

The red panda is an endangered animal. To keep this animal in the race of life, certain conservational measures have been taken. In 2005, the state governments of Sikkim and Arunachal Pradesh have taken a joint project with the recommendation of the WWF.

Red panda conservation projects are running in Varsey Rhododendron Sanctuary and Kangchenjunga Biosphere Reserve of Sikkim. For this purpose, the collection of bamboo and other wood from these forests are banned completely.

About 200 local residents are trained to prepare fuel briquettes with agricultural wastes and mud, to supplement wood and bamboo.

In Arunachal Pradesh, the habitats of the red panda are taken under the vigil of the Community Conserved Area (CCA), where local people are made involved in the conservation of this beautiful endangered animal.

Question 7. Mention the scientific names, conservation projects and the locations for conservation of the Indian tiger, single-horned rhino, Asiatic lion, gharial, crocodile and red panda.

Answer:

Few endangered animals in India:

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Role Of JFM And PBR In Biodiversity Conservation Few endangered animals in india

Chapter 5 Biodiversity Conservation Of Some Endangered Species Of India SAQs

Question 1. Mention three tiger conservation centres in India.
Answer:

Three tiger conservation centres in India are –

  1. Sundarban in West Bengal,
  2. Corbett tiger reserve in Uttarakhand and
  3. Kanha National Park in Madhya Pradesh.

Question 2. How and where was JFM established?
Answer:

JFM:

In 1971, to prevent indiscriminate felling of trees and grazing in the forest of Arabari in West Midnapore, the forest officer of that time, Dr. Ajit Kumar Banerjee, formed a forest protection committee with the local people.

By this process, the first JFM came into being. The members. planted several saplings of Sal tree in the Arabari area to expand the forest and by 1988, they grew Sal trees of worth of 12.6 million rupees.

Question 3. Name few committees working under JFM.
Answer:

Few committees working under JFM:

There are several committees working under JFM. A few committees under JFM are Forest Protection Committee (FPC), Eco Development Committee (EDC), Village Forest Committee (VFC) etc.

Wbbse Class 10 Life Science Solutions

Question 4. Name a few animals that are protected in Kaziranga National Park.
Answer:

Few animals that are protected in Kaziranga National Park:

Numerous animals are protected in Kaziranga National Park. The animals protected in Kaziranga National Park are one-horned rhinos, sambar, wild hog, hog deer, Indian muntjac etc.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Role Of JFM And PBR In Biodiversity Conservation Kaziranga national park

 

Question 5. What are world heritage sites?
Answer:

World heritage sites:

World Heritage Convention (WHC), a wing of UNESCO, looks out for any natural place or any man-made structure with outstanding natural, geographical or historical importance. To provide special protection to these places, WHC designate them as a world heritage sites.

Example- Sundarban Biosphere Reserve.

Question 6. What is CITES?
Answer:

CITES:

According to the proposal of IUCN (1963) in Washington, a convention named CITES (Convention on International Trade in Endangered Species of Wild Flora and Fauna) was formed in 1975.

Since its inception, this body had banned the illegal trading of endangered or rare animal and plant products and body parts on a global basis. By this action, the indiscriminate hunting and selling of endangered species have been reduced significantly.

Question 7. Why is Sundarban treated as a world heritage site?
Answer:

Sundarban treated as a world heritage site:

The mangrove forest of Sundarban is the largest of its kind. It shelters several endangered fauna and flora including the famous Royal Bengal tiger.

The root system of the dense mangrove trees prevents soil erosion and protects the area from storm. Sundarban is very rich in forest resources. Therefore, Sundarban is designated as a world heritage site.

Wbbse Class 10 Life Science Solutions

Question 8. What is meant by out-of-danger species?
Answer:

Out-of-danger species:

By the process of conservation, the number or population of certain species, which had once become endangered, have increased significantly. These species are treated as out-of-danger species.

Example-Spotted deer.

Question 9. What are extinct species?
Answer:

Extinct species:

The species of plants or animals that can no longer be seen living anywhere in the wild or in captivity, are called extinct species.

Examples- Dodo, passenger penguin, Carolina parakeet etc.

Question 10. What is Project Tiger
Answer:

Project Tiger:

Project Tiger is a project for the conservation of tigers in India sorted by the Indian govt. in 1973. This is maintained by the national tiger project is a conservation authority.

Question 11. What is meant by endangered species?
Answer:

Endangered species:

Due to human intervention, pollution, lack of food and squeezing of native habitats, the rate of reproduction of certain animals and plants is getting reduced to such an extent that, they are on the verge of extinction. These species are called endangered species.

Examples- Nilgiri langur, snow leopard etc.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Role Of JFM And PBR In Biodiversity Conservation Nilgiri langur endangered species

Question 12. Hilsa, Bee, Penguin, and Rauwolfia-Asscess which are the causes of the endangeredness of the above-mentioned organisms.
Answer:

Hilsa:

Facing threat due to overexploitation of hilsa population.

Bee:

Loss of habitat, and excessive use of cell phones that projects electromagnetic waves, damage the ability of bees to return to their colony. By that way, it destroys the navigation system of bees.

Penguin:

Penguin faces threats from geological events like volcanic eruption, pollution, climatic changes and severe weather.

Rauwolfia:

This type of medicinal plant facing threats due to overexploitation.

Question 13. Write down two steps to increase of the number of one-horned rhinos in India. Or, Propose two conservation measures that could be taken to increase the number of Indian one-horned rhinos.
Answer:

Two steps, that are taken to increase the number of one-horned rhinos in India are-

  1. Increase the habitat and food source of one-horned rhino
  2. Arrangement of the breeding programme or increased the number of rhinos by captive breeding.

Question 14. What is a red data book?
Answer:

Red data book:

For the conservation of rare and endangered animals and plants, IUCN publish a book comprising detailed information regarding the rare and endangered organisms all over the earth.

This book is known as the red data book. In this book, the organisms are classified under nine different classes according to their availability status like extinct, endangered, vulnerable etc.

Question 15. What is captive breeding?
Answer:

Captive breeding:

Breeding of animals in a controlled environment and in a scientific way in a sit with consrevation facilities is called captive breeding In the case of rhinos, and red panda this process is applied.

Question 16. What are the steps taken in Gir national park to increase the number of lions?
Or,
Mention the in-situ measures adopted for the conservation of lions in Gir National Park of India.
Answer:

Steps for the conservation of lions in Gir national park to increase the number of lions are-

  1. The natural habitat of Gir national park needs to be protected.
  2. Wild animals and plants should be conserved in Gir national park so that the food. supply and shelter for the lion could be maintained.
  3. Captive breeding are taken as a potential tool to increase the population of lion.
  4. The electric fence should be avoided.

Question 17. What are the main buds to protect biodiversity?
Answer:

Main acts to protect biodiversity are-

  1. The wildlife protection act (1972)
  2. The forest conservation act (1980)
  3. The environment protection act (1986)
  4. Biological diversity act (2002).

Question 18. What are the information documented in PBR?
Or,
Write the major topics of biodiversity documented in the Biodiversity Register (PBR).
Answer:

Major topics documented in PBR are-

  1. PBR provides correct information about the biodiversity of the local areas so that, the rare and endangered animals and plants can be conserved properly.
  2. It makes people aware of the importance of the conservation of biodiversity and involves them directly with the conservation project.
  3. PBR makes the people aware of the sustainable utilisation of forest resources and helps in the economic development of the area.
  4. PBR provides correct geographic information about conserved areas.

Wbbse Class 10 Life Science Solutions

Question 19. Write the in-situ conservation measures taken for one endangered reptile species and name one site of such conservation in West Bengal.
Answer:

One endangered reptile is a crocodile. Measures taken for the in-situ conservation of crocodiles are-

  1. Protect the remaining population of crocodilians in their natural habitat by creating sanctuaries.
  2. Promote captive breeding.
  3. Prevent poaching of crocodiles and as well as their eggs.

The name of such conservation in West Bengal is the Bhagabatpur crocodile project, Sundarban.

 

Chapter 5 Biodiversity Conservation Of Some Endangered Species Of India MCQs

Question 1. Rhinoceros is the state animal of—

  1. Tripura
  2. Assam
  3. Uttarakhand
  4. Arunachal Pradesh

Answer: 2. Assam

Question 2. The Tiger project was implemented in India in—

  1. 1967
  2. 1973
  3. 1983
  4. 1987

Answer: 2. 1973

Question 3. The organisation that publishes the list of endangered species is—

  1. BBC
  2. UNO
  3. WWF
  4. IUCN

Answer: 4. IUCN

Question 4. The organisation that publishes the red data book is—

  1. VVYVF
  2. IUCN
  3. IMF
  4. FAO

Answer: 2. IUCN

Question 5. In India, lions are conserved in—

  1. Gir forest
  2. Jim Corbett National Park
  3. Kanha National Park
  4. Jaldapara National Park

Answer: 1. Gir forest

Question 6. In West Bengal, Singalila National Park is situated in the district—

  1. Cooch Behar
  2. South 24 Parganas
  3. Darjeeling
  4. Malda

Answer: 3. Darjeeling

Question 7. In West Bengal, rhinoceros is conserved in—

  1. Snndarban
  2. Jaldapara
  3. Betliuadahari
  4. Forest of Purulia

Answer: 2. Jaldapara

Question 8. The red data book comprises information about—

  1. Animal species of die the Red Sea
  2. Tigers of Sundarban
  3. Endangered species
  4. Revolutionary activities

Answer: 3. Endangered species

Question 9. The natural habitat of the Indian lion is—

  1. Siinlipal Forest of Odisha
  2. Betla forest of Bihar
  3. Gir forest of Gujarat
  4. Kaziranga National Park of Assam

Answer: 3. Gir forest of Gujarat

Question 10. The animal which is conserved in Padmaja Naidu Himalayan Zoological Park is—

  1. Snow leopard
  2. Red panda
  3. Himalayan wolf
  4. All of these

Answer: 4. All of these

Question 11. The full form of JFM is—

  1. Joint Feature Management
  2. Judicial Forest Management
  3. Jungle Fire Management
  4. Joint Forest Management

Answer: 4. Joint Forest Management

Question 12. The district in West Bengal where the concept of JFM was first implemented, is—

  1. West Midnap ore
  2. Howrah
  3. Kolkata
  4. Nadia

Answer: 1. West Midnap ore

Question 13. The year in which the concept of JFM was first implemented is—

  1. 1976
  2. 1985
  3. 1963
  4. 1971

Answer: 4. 1971

Question 14. The number of Asiatic lions in Gir National Park (as per census 2015) is—

  1. Less dian 50
  2. More than 500
  3. More than 1000
  4. Less than 500

Answer: 2. More than 500

Question 15. The objective of IRV is the conservation of

  1. Royal Bengal tiger
  2. Indian rhino
  3. Asiatic lion
  4. Gangetic crocodile

Answer: 2. Indian rhino

Question 16. NTGA works for the welfare and protection of—

  1. Royal Bengal tigers
  2. Indian rhinos
  3. Asiatic lions
  4. Gangetic crocodiles

Answer: 1. Royal Bengal tigers

Question 17. Decide which of the following project tiger is located within our state.

  1. Bandipur
  2. Simlipal
  3. Sundarbans
  4. Kanha

Answer: 3. Sundarbans

Question 18. Kaziranga National Park is famous for the conservation of —

  1. Deer
  2. Elephant
  3. Rhinoceros unicornis
  4. Lion

Answer: 3. Rhinoceros unicornis

Question 19. The biological diversity act in India was introduced in—

  1. 1992
  2. 1996
  3. 2000
  4. 2002

Answer: 4. 2002

Question 20. Singalila National Park is situated in which of the following district—

  1. Cooch Behar
  2. Darjeeling
  3. Malda
  4. South 24 Parganas

Answer: 2. Darjeeling

Question 21. Wildlife protection act was introduced in —

  1. 1962
  2. 1972
  3. 19821992

Answer: 1. 1972

 

Chapter 5 Biodiversity Conservation Of Some Endangered Species Of India VSAQs

Question 1. Write the full form of PBR.
Answer: Peoples’ Biodiversity Register

Question 2. Write down the full form of CITES.
Answer: Convention of International Trade in Endangered Species of Wild Fauna and Flora.

Question 3. Write the full form of IUCN.
Answer: International Union for Conservation of Nature and Natural Resources

Question 4. Write the scientific name of the Indian tiger.
Answer: Panthera Tigris

Question 5. Write the scientific name of the Indian rhino.
Answer: Rhinoceros unicornis

Question 6. Write the scientific name of the Asiatic lion.
Answer: Panthera leo persica

Question 7. Write the scientific name of gharial.
Answer: Gavialis gangeticus

Question 8. Write the scientific name of brackish or saltwater crocodile.
Answer: Crocodylus porosus

Question 9. Write the scientific name of red panda.
Answer: Ailurus fulgens

Question 10. In which part of Odisha there is a crocodile project?
Answer: A crocodile project is situated in Bhitarkanika of Odisha.

Question 11. What is the full form of NTCA?
Answer: National Tiger Conservation Authority

Question 12. Where did the first JFM start?
Answer: The first JFM was started in the Arabari forest of West Midnapore.

Question 13. Who was the first person to conduct JFM?
Answer: Dr Ajit Kumar Banerjee, the Divisional Forest Officer (DFO) of Midnapore, was the first person to conduct JFM.

Question 14. By which law is PBR regulated?
Answer: Biological Diversity Act, 2002

Question 15. Which committee was formed by ‘PBR?
Answer: PBR formed Biodiversity Management Committee (BMC).

Question 16. Write down the names of a few committees that work under JFM.
Answer: Forest protection committee (FPC), Ecodevelopment Committee (EDC), Village forest committee (VFC).

Question 17. Which committee form PBR?
Answer: Biodiversity management committee or BMC

Question 18. Write down the name of the Indian tiger conservation project.
Answer: Project tiger

Question 19. Write down the name of the Indian forest conserving tiger.
Answer: Sundarban

Question 20. Write down the name of a project of Indian rhino conservation.
Answer: Indian rhino vision 2020

Question 21. Name one area of India for rhino conservation.
Answer: Kaziranga national park

Question 22. Give one example of a crocodile project.
Answer: Bhagabatpur crocodile project, West Bengal

Question 23. Write down the name of one captive breeding site of India.
Answer: Nandankanan of Orissa

Question 24. Name one zoological park of West Bengal under Project Red Panda.
Answer: Padmaja Naidu Himalayan zoological park, Darjeeling

Question 25. What is the full form of VFC?
Answer: Village Forest committee

Question 26. Write down the names of some endangered animals of India.
Answer: Indian elephant, Royal Bengal tiger, Red panda, Asian lion

Question 27. What is the full form of IRV 2020
Answer: Indian rhino vision 2020

Question 28. Name one endangered animal conserved at singalila national park.
Answer: Red panda or Clouded leopard.

Question 29. Name the practice which jointly the local peoples and forest department maintain for the reclamation of the forest.
Answer: JFM or Joint Forest Management.

 

Chapter 5 Biodiversity Conservation Of Some Endangered Species Of India Fill In The Blanks

Question 1. In India, about _________ % of rhinoceros inhabit Assam.
Answer: 85

Question 2. The Wildlife Protection Act was implemented in _________.
Answer: 1972

Question 3. FPC is a committee under_________.
Answer: JFM

Question 4. A maximum number of one-horned rhinos inhabit in the state of _________.
Answer: Assam

Question 5. Asiatic lions are conserved in _________ forest of Gujarat.
Answer: Gir

Question 6. _________ types of crocodiles are found in India.
Answer: Three

Question 7. The _________ Wildlife Sanctuary in Odisha has a crocodile breeding centre.
Answer: Bhitarkanika

Question 8. The other name of the red panda is _________
Answer: Firefox

Question 9. The _________ Biosphere Reserve is the present conservation centre for red pandas.
Answer: Kanchenjunga

Question 10. The first JFM of West Bengal was formed in the year _________.
Answer: 1971

Question 11. The highest number of rhinos is found in _________ state of India.
Answer: Assam

Question 12. _________ is conserved under IRV 2020 project.
Answer: Rhino

Question 13. Lion is found in _________ forest of India.
Answer: Gir

Question 14. The red panda is the state animal of _________ state.
Answer: Sikkim

Question 15. _________ sanctuary of Odisha is a crocodile conservation centre.
Answer: Bhitarkanika

Question 16. The first JFM in West Bengal was introduced in _________ forest.
Answer: Arabari

 

Chapter 5 Biodiversity Conservation Of Some Endangered Species Of India True Or False

Question 1. According to the census of 2014, the number of tigers in India is 1412.
Answer: False

Question 2. The only conservation centre for Asiatic lions is situated at Gir, Gujarat.
Answer: True

Question 3. The concept of JFM was initiated in the year 1975.
Answer: False

Question 4. JFM was first introduced in Sundarban of West Bengal.
Answer: False

Question 5. PBR is maintained by local law.
Answer: False

Question 6. Lion is conserved in the Gir forest of India.
Answer: True

Question 7. Indian lion is an endangered animal.
Answer: False

Question 8. One-horned rhinoceros is conserved at Manas National Park of West Bengal.
Answer: False

 

Chapter 5 Biodiversity Conservation Of Some Endangered Species Of India Match The Columns

1.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Role Of JFM And PBR In Biodiversity Conservation Match The Columns 1

Answers:  1-B; 2-D; 3-F; 4-A; 5-C; 6-E

2.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Role Of JFM And PBR In Biodiversity Conservation Match The Columns 2

Answers: 1-F; 2-E; 3-A; 4-C; 5-G; 6-B

 

Chapter 5 Biodiversity Conservation Of Some Endangered Species Of India Find The Odd One Out

Question 1. BMC, PBR, NBA, JFM
Answer: JFM

Question 2. Tikarpada, Bandhabgarh, Sundarban, Kanha.
Answer: Tikarpada

Question 3. Ranthambore National Park, Manas National Park, kaziranga National Park, Pobitora Wildlife Sanctuary
Answer: Ranthambore national park

 

Chapter 5 Biodiversity Conservation Of Some Endangered Species Of India Fill In The Blanks By Looking At The First Pair

Question 1. Biodiversity data: PBR:: Biodiversity management: _______________
Answer: Red panda

Question 2. Bhitarkanika Sanctuary: Crocodile:: Singalila National Park: _______________
Answer: JFM

Question 3. Barsey Rhododendron Sanctuary: Red Panda Gir National Park: _______________
Answer: Lion

Question 4. Rhinoceros unicornis: Assam:: Asian Lion: _______________
Answer: Gujrat

Question 5. Manas National Park: Rhino:: Sundarban: _______________
Answer: Royal Bengal tiger

Question 6. Red Panda: Singalila:: Rhinoceros unicorns:: _______________
Answer: Kaziranga

 

Chapter 5 Biodiversity Conservation Of Some Endangered Species Of India Among The Four Concepts Given Three Of Them Belong To One Find That

Question 1. FPC, EDC, JFM, VFC
Answer: JFM

Question 2. PBR, JFM, Biodiversity conservation, Project tiger
Answer: Biodiversity conservation

Question 3. Lion Conservation, Endangered animal conservation, Red Panda project, Indian rhino vision
Answer: Endangered animal conservation

Question 4. Dangamal, Bhitarkanika, Odisha, Ramtirtha
Answer: Odisha

Question 5. Overall knowledge regarding local biological resources, PBR, Traditional beliefs regarding local biological resources, and sustainable use of local biological resources.
Answer: PBR

 

Chapter 5 Biodiversity Conservation Of Some Endangered Species Of India Advanced Questions And Answers

Question 1. Mention a few lung and accessory respiratory organ-related pathogenic diseases caused by air pollution and state the names of the pathogens and two symptoms of each disease.
Answer:

Lung and accessory respiratory organ-related pathogenic diseases, the pathogens and symptoms

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Role Of JFM And PBR In Biodiversity Conservation Pathogenic diseases , the pathogens and symptoms

 

Question 2. What is meant by recycling?
Answer:

The process, by which waste matters are modified and is reused to produce other usable materials is called recycling. By means of recycling of waste matters, environmental pollution can be kept under control.

Question 3. What is meant by natural air pollutants?
Answer:

There are certain air-polluting substances, which are produced naturally in the environment. These are called natural air pollutants.

Examples- Pollen grains, smoke and dust from volcanoes, methane gas from wetlands, etc.

Question 4. What is meant by man-made air pollutants?
Answer:

Certain human actions produce different materials, which often cause air pollution. These materials are known as man-made or anthropogenic air pollutants.

Example-Smoke emitted from factories, automobiles, etc.

Question 5. What are the types of ecosystem diversity?
Answer: Ecosystem diversity is of three types-

  1. Alpha diversity
  2. Beta diversity and
  3. Gamma diversity

Question 6. What is alpha diversity?
Answer:

Alpha diversity is the mean species diversity in sites or habitats at a local scale. Alpha diversity thus refers to the diversity within a particular ecosystem.

Question 7. What is beta diversity?
Answer:

Beta diversity is the ratio between regional and local species diversity. Beta diversity measures the change in the diversity of species from one environment to another.

Question 8. What is gamma diversity?
Answer:

Gamma diversity is the total species diversity in a landscape. Gamma diversity is a measure of the overall diversity for the different ecosystems within a region.

It reflects the diversity of species occupying individual habitats combined with a measure of the heterogeneity between the habitats.

Question 9. Which is the first national park of the world?
Answer: Yellowstone national park

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Importance Of Biodiversity

Chapter 5 Importance Of Biodiversity Summary

  • Variability of genes, species, and ecosystems in land, water, and air can provide a variety of organisms. This variation is called biodiversity. India is one of the 12 mega-diversity countries in the period.
  • Biodiversity is very important for various reasons. It is related to food production, and medicine formation from a variety of organisms. In fact, it is essential
  • in maintaining the ecological balance, helping to increase rainfall, and controlling climatic changes.
  • Biodiversity has wide economic importance. It supplies raw materials for furniture, and paper and also provides resin, wax, silk, fur, pearl, etc. In fact, it also has a great influence in painting, architecture, and music, to entertain the people.
  • Regions that are rich in biodiversity, like the Himalayas Western Ghats, the Indo-Burma region, and the Sundaland (including Nicobar group of islands) are called hotspots.

Life Science Class 10 Wbbse

  • There are various reasons behind the depletion of biodiversity, like-change in land patterns, extensive use of land for industry, house, and agriculture, hunting, global warming, climatic change, invasion of foreign species, natural calamities, etc.
  • In recent decades, Sundarban has faced lots of problems like-destruction of mangrove plants, agricultural lands gets salty due to floods, scarcity of drinking water, scarcity of habitat, water pollution, and above all increase in sea level submerge the whole area. All these calamities can destroy the ecological balance of Sundarban.

Importance Of Biodiversity

Read and Learn More WBBSE Solutions for Class 10 Life Science And Environment

Chapter 5 Importance Of Biodiversity Long Answer Type Questions

Question 1. Briefly explain the significance of biodiversity.

Answer:

Significance of biodiversity:

The significance of biodiversity are given below.

1. Ecological importance:

In an ecosystem, all organisms are ecologically interdependent on each other. The extinction of an organism will increase or decrease the number of other organisms and will create ecological imbalance. Therefore, a rich biodiversity results in a stable ecosystem.

2. Environmental value:

A rich biodiversity is the indicator of a pollution-free environment. By controlling the environmental pollution the conservation of the biosphere can be maintained.

3. Economic importance:

We are dependent on plants and animals for food, clothing, medicines, and shelter. We get various types of food from different varieties of plants. Other than the food, we get wood, paper, fibers, gums, resin rubbers, and various medicinally important alkaloids from plants.

We extract honey, wax, silk, wool, leather, milk, meat, and egg from animals. Thus, biodiversity is essential for the survival of human beings.

Life Science Class 10 Wbbse

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Importance Of Biodiversity Economically important materials from plants and animals

 

4. Weather control:

Biodiversity is essential to prevent environmental pollution and to control the weather. This process maintains the stability of the biosphere. The tropical rain forests absorb CO, and thus reduce the greenhouse effect and global temperature.

A forest increases moisture in the atmosphere by transpiration and thereby increases the chance of cloud formation and rainfall. Many plants absorb toxic elements and help to detoxify the soil. Several plants and microbes increase the nitrogen content of the soil and thus increase soil fertility.

5. Aesthetic beauty:

Living organisms add natural beauty and liveliness on the earth.

6. Moral value:

Every organism on the earth has the right to live. This concept was accepted by United Nations in the memorandum of the earth summit in Nairobi, 1982.

Life Science Class 10 Wbbse

Question 2. Mention the types of biodiversity. State the names and locations of the biodiversity hotspots in India.

Answer:

Types of biodiversity:

Biodiversity can be classified into three types. These are mentioned below.

1. Genetic diversity:

Several chromosomal and genetic differences occur within a population, which create many structural and behavioral variations within the members of the same species. This is called genetic diversity.

2. Species diversity:

Within a biotic community of a particular area, several species interact with each other. The structural and behavioral variation of the members of these species and the difference in the relative abundance of all the species in a biotic community is known as species diversity.

3. Ecological diversity:

The variety among the organisms, belonging to different interacting ecosystems within a vast geographical area is known as ecological diversity.

Indian biodiversity hotspots:

The names and locations of the biodiversity hotspots in India are mentioned below.

1. Eastern Himalayas:

The North-Eastern states like Sikkim, North Bengal, Assam, and Arunachal Pradesh are included under this hotspot. Due to the abundance of ancient flowering plants in these areas, the Eastern-Himalayas hotspot is known as the ‘cradle of speciation.

2. Indo-Burma:

Other North-Eastern states like Manipur, Tripura, Mizoram, Meghalaya, Nagaland, and south Assam belong to the Indo-Burma hotspot region.

3. Western Ghats and Sri Lanka:

The hilly regions of Western Ghat in Gujarat, Maharashtra, Karnataka, and Kerala belong to this hotspot region.

4. Sundaland:

It comes under the Asia-Pacific region. The Nicobar group of islands belongs to this hotspot region.

Life Science Class 10 Wbbse

Question 3. Explain the causes of the depletion of biodiversity.
Or,
Assess the causes of depletion of biodiversity with proper examples.

Answer:

Causes of depletion of biodiversity:

The causes of the depletion of biodiversity are explained below.

1. Change in the utilization of land:

Due to increased population pressure, the forest areas are cleared and converted into residential or industrial areas. The laying of roadways and railway tracks through forest land is squeezing the natural habitat of wild animals and plants.

Indian elephant, mountain gorilla, orangutan, red panda, etc. species are facing extinction due to habitat loss.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Importance Of Biodiversity Hunting

 

2 Climatic change and global warming:

Indiscriminate felling of trees increases CO2 levels in the atmosphere. Excess CO2 causes the greenhouse effect and increases the atmospheric temperature. Many animals cannot adapt in this changed environment and become extinct.

Due to global warming, the glaciers of the polar region are melting. This results in to an increase in sea level. As a result, the delta of Sundarban is going under seawater.

3. Pollution:

With the modernization of life and industrialization, environmental pollution has increased greatly. Many species have become extinct or surviving in endangered states due to pollution.

Excessive use of pesticides has an immense effect on the depletion of biodiversity. Industrial pollution is a big threat for coral reefs.

4. Hunting and poaching:

Until the middle of the last century, wild animals were hunted for fun. The big cats became endangered due to indiscriminate hunting. Poachers kill animals for tusks, horns, skins, etc. Some tribes celebrate hunting competitions.

These practices cut down the population of different animals. Several species of jaguar and ocelot have been jeopardized by the demand for their fur.

5. Excessive use:

Ayurvedic medicine producers collect huge quantities of medicinal herbs or plant parts from forests. As a result, the number of several medicinally important plants have reduced significantly.

For Example, Coptis teeth-an endangered plant is used to regulate blood pressure.

6. Natural calamity:

Forest fire is the deadliest of the natural calamities, which reduce biodiversity very fast. Volcanic eruptions, floods, tsunamis, earthquakes, etc. are a few other natural calamities, which have reduced biodiversity at different times.

For example-Cave beer are extinct due to this.

7. Invasion of other species:

Invasion of foreign species like water hyacinth, tilapia fish, hybrid catfish, etc. greatly hamper the ecological structure of an area. The American cockroach Periplaneta americana harms the local cockroach of India, i.e., Blatta orientalis.

Question 4. Explain the problems of Sundarban. Similar question, You have identified three environmental problems while on research in Sundarbans-

  1. Disbalance in predator number,
  2. Destruction of mangrove plants due to urbanization,
  3. Submergence of islands due to rising sea level. Analyze what impact they might exert on biodiversity.

Part question, If the water level in the sea increases enormously due to global warming, what problems the humans and biodiversity of Sunderban may face?
Part question, Construct a concept map to show how the increase in human habitat is influencing the ecosystem of Sundarban.

Answer:

Problems of Sundarban:

In recent days Sundarban is facing many environmental problems. These are mentioned below.

1. Loss of mangrove forest due to urbanization:

Population pressure has initiated urbanization. Due to this, many areas along the periphery of Sundarban are deforested.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Importance Of Biodiversity Destruction of mangrove foreset in Sundaraban

 

2. Agricultural problem:

Due to soil erosion and siltation, the depth of rivers becomes shallow. During high tide saline water spills over, and floods the agricultural fields to make the soil salty and infertile.

3. Scarcity of fresh water:

The estuarine water is salty. Frequent flood makes inland water bodies like ponds and wells saline. Therefore, sweet water is becoming scarce in the Sundarban area.

4. Destruction of natural habitat:

Deforestation is reducing the natural habitats of large to small animals. As a result, the biodiversity of Sundarban is gradually getting destroyed.

5. Pollution:

The most provocative problem for Sundarban is increasing pollution. All rivers are disposing of gallons of pollutants in the water. Several pollutants are percolating in the soil of the Sundarban area making the soil infertile and toxic.

The oil and grease, spilling from fishing and tourist vessels are polluting the aquatic environment of this area. This pollution is affecting the propagation of fish, prawns, and other aquatic animals.

6. Imbalance in the prey-predator ratio:

Due to overpopulation and pollution, the number of many animal species has declined. This has resulted int o imbalance in prey-predator ratio. The scarcity of natural food compels tigers to enter into human localities in search of food.

7. Rise in water level:

Due to global warming, the glaciers of the polar region are melting. This results into an increase in sea water level. As a result, the delta of Sundarban is going under seawater.

Question 5. Mention the roles of biodiversity in the following two aspects in the progress of human civilization-

  1. Production of food,
  2. Manufacturing of Medicine.

Answer:

Production of food:

We all are depends on nature for food. Nature can give us a wide range of various organisms that meet the need of food. Different food items provide different essential nutrients like-Vitamin, minerals along with energy.

Other than that a wide range of animal protein are also acquired from poultry and fishery. All these animals and plants are the gift of the biodiversity of nature.

Manufacturing of medicine:

In the early decade. We mainly depended on plants for medicine. A variety of plant species are provided by biodiversity, mainly used for medicinal purposes. Not only plants, but many fungi and body parts of animals are also used as medicine.

For example, we get quinine from the cinchona plant; a vital drug for malaria reserpine from the roots of rauwolfia; maintain blood pressure level to normal, daturine from datura prevents hyperventilation, etc.

Along with that, holy basil, and margosa plants also exert their medical value. Fungi like Penicillium notatum is essential for producing penicillin, an antibiotic that fights against a wide range of bacteria. So biodiversity helps to progress overall human welfare and civilization.

 

Chapter 5 Importance Of Biodiversity Short Answer Type Questions

Question 1. What is meant by species?
Answer:

Species:

Species is a group of closely related, characteristically similar and naturally interbreeding organisms, which are capable of producing fertile offsprings to create population. For example, there are seven species of honey bee in our planet.

Question 2. What is biodiversity?
Answer:

Biodiversity:

The intraspecific, interspecific or ecological variation seen among the different living organisms, inhabiting any terrestrial or aquatic environment, is called biodiversity. It is a measure of the variety of organisms present in different ecosystems.

Example-Coral reef biodiversity.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Importance Of Biodiversity Coral reef biodiversity

 

Question 3. What are endemic species?
Answer:

Endemic species:

The animal or plant species, which remain naturally confined to a particular geographical area, are called endemic species.

For example- The Asiatic lion is endemic to Gir forest, muga silk moth is endemic to Assam.

Question 4. How do you classify biodiversity on the basis of their interrelations?
Answer:

On the basis of interrelation, biodiversity is classified into 3 different types-

  1. Genetic diversity,
  2. Species diversity
  3. Ecological diversity.

Question 5. Name any two organizations, which work for developing awareness on the importance of biodiversity.
Answer:

Several international organizations work for developing awareness on the importance of biodiversity.

Two of these are-

  1. WWF (World Wide Fund for Nature) and
  2. IUCN (International Union for Conservation of Nature and Natural Resources).

Question 6. What is genetic diversity?
Answer:

Genetic diversity:

Several chromosomal and genetic differences occur within a population. These create many structural and behavioral variations within the members of the same species. This is called genetic diversity.

Question 7. What is species diversity?
Answer:

Species diversity:

Within a biotic community of a particular area, several species interact with each other. The structural and behavioral variations of these organisms and the differences in the relative abundance of all these species in a biotic community are known as species diversity.

Question 8. What is ecological diversity?
Answer:

Ecological diversity:

Different interacting ecosystems within a vast geographic area possess a wide variety of species. Such diversity among the organisms in a very large area is known as ecological diversity.

Question 9. What is meant by biodiversity hotspot?
Answer:

Biodiversity hotspot:

There are some regions on the earth which comprise several endangered and rare endemics Western Ghats and Eastern Himalayas are two species, are known as biodiversity hotspots. important biodiversity hotspots in India.

Question 10. How many types can we classify the ecological diversity and what are they?
Answer:

Ecological diversity can be classified into three types.

They are-

  1. Alpha diversity,
  2. Beta diversity and
  3. Gamma diversity.

Question 11. Give two examples of animal biodiversity endangered by the pollution of the river Ganga.
Answer:

Two examples of animal biodiversity endangered by the pollution of river Ganga are-

  1. Gangetic dolphin and
  2. Gharial.

Question 12. Which parts are included in Eastern Himalayas and Sundaland biodiversity hotspots?
Answer:

The forests of Sikkim, North Bengal, and Arunachal Pradesh are included in the Eastern Himalayas biodiversity hotspot. Sumatra, Borneo, and Nicobar Islands are included in the Sundaland biodiversity hotspot.

Question 13. Why is the Eastern Himalayas called a biodiversity hotspot?
Answer:

In Eastern Himalayas, there are several endemic animal and plant species. Here, about 5800 species of plants and 2000 species of animals are found, all of which are endemic. Therefore, Eastern Himalayas is treated as a biodiversity hotspot.

Question 14. Name the biodiversity hotspots in India.
Answer:

In India, there are 4 biodiversity hotspots.

These include

  1. Eastern Himalayas,
  2. Western Ghats and Sri Lanka,
  3. Indo-Burma and
  4. Sundaland.

 

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Importance Of Biodiversity Indian biodiversity hotspots

 

Question 15. Why is Western Ghats and Sri Lanka treated as biodiversity hotspots?
Answer:

In the Western Ghats and Sri Lanka, about 5000 different vascular plant species are found, out of which 1700 are endemic. Moreover, around 985 animal species are found here, out of which 360 are endemic species.

Due to this richness of endemic species, the Western Ghats and Sri Lanka are treated as biodiversity hotspots.

Question 16. Write the location of the Sundaland hotspot and one important species of its biodiversity.
Answer:

One of the most important biodiversity hotspots Sundaland is located in Southeastern Asia, and comprises of the landmass of Malay, Java, Sumatra, Peninsula, Bali, Borneo, Andaman, and Nicobar islands.

The important species of Sundaland is-Sundari plant (plant) or Orangutan (animal).

Question 17. Which Indian states come under the Western Ghats and Sri Lanka hotspots?
Answer:

The Indian states those come under the Western Ghats and Sri Lanka hotspots are Gujarat, Maharashtra, Karnataka, and Kerala.

Question 18. Which are the two conditions imposed by the ‘conservation international’ to declare an area as a biodiversity hotspot?
Or,
Which conditions do you consider to declare one area out of two as a biodiversity hotspot?
Answer:

The two conditions imposed by the ‘conservation international’ to declare an area as a biodiversity hotspot are as follows-

  1. The area must have at least 1500 endemic vascular plant species.
  2. About 70% of the biodiversity of that area should have already been extinct.

Question 19. Discuss the role of biodiversity in maintaining the balance of the ecosystem of a river.
Answer:

The role of biodiversity in maintaining balance of the eco-system of a river are

  1. Biodiversity decreases the inter-specific and intra-specific competition of aquatic organisms for food and habitat.
  2. Depletion of certain aquatic organisms due to water pollution hampers the river ecosystem. This can be restored by biodiversity.
  3. Extensive biodiversity provides certain options to nature like killifish, which can withstand with extreme environmental change (water pollution).

Question 20. Which states come under the Indo-Burma hotspot?
Answer:

A number of states come under the Indo-Burma hotspot. These are Manipur, Mizoram, Tripura, Meghalaya, Arunachal Pradesh, and southern Assam.

Question 21. Mention the location of Sundarban.
Answer:

The location of Sundarban:

Sundarban is located at the southern part of South 24 Parganas of West Bengal and the southern portion of the Khulna district of Bangladesh.

Its geographical position is 21°32′ to 21°55′ northern latitude and 88°42′ to 89°04 eastern longitude. The total surface area of Sundarban is 9630 sq km.

Question 22. Which types of plants are called halophytes or mangrove plants?
Answer:

The plants, which grow in the saline, silty and muddy soil along the bank of estuaries and sea shore are called halophytes or mangrove plants.

Example-Sundari (Heritiera), Garan (Ceriops), etc.

Question 23. Mention the importance of mangrove forests in Sundarban.
Answer:

The importance of mangrove forest in Sundarban are as follows-

  1. The root system of mangrove trees holds the soil to prevent erosion of soil and breaking of the coastline.
  2. This forest protects the coastal areas against storms and supplies wood, honey, etc. to the residents of this area.

Question 24. Name a few endangered plants and animals of Sundarban.
Answer:

The endangered plants of Sundarban are Sundari (Heritiera), Geon (Excoecaria), etc. Among animal species, the endangered ones are the Royal Bengal tiger, fishing cat, etc.

Question 25. What are the reasons of the destruction of the mangrove forest of Sundarban?
Answer:

The reasons of the destruction of mangrove forests are-

  1. Cutting of mangrove trees for place of human habitat.
  2. Natural calamities like cyclones destroy mangrove forests.
  3. Pollution causes the destruction of mangrove plants.
  4. Mangrove plants are cut for prawn culture.

Question 26. Write down a few processes of protecting biodiversity.”
Answer:

A few processes of protecting biodiversity are-

  1. Identifying the endangered species and necessary steps to be taken for their conservation.
  2. Plantation and prevention of local plant destruction.
  3. Increasing awareness among people and sustainable use of forest products.

Question 27. What is a biodiversity treaty?
Answer:

Biodiversity treaty:

In the Earth summit of 1992 in Rio de Janeiro, Brazil, top level representatives from 150 countries unanimously signed a treaty for the conservation of biological diversities (CBD). This is known as biodiversity treaty.

Question 28. Mention the important features of the biodiversity treaty.
Answer:

The important features of the biodiversity treaty are as follows-

  1. Conservation of biodiversity.
  2. Sustainable utilization of the products collected from biodiversity.
  3. Arrangement of impartial utilization of genetic resources from biodiversity.

Question 29. Give four examples of damage of biodiversity as a result of global warming.
Answer:

Due to global warming, four examples of damage of biodiversity are

  1. Destruction of Australian coral reefs leads to loss of marine biodiversity.
  2. In the present decade, the overall temperature of the earth rises about 1-2°C, that in turn melts the ice in polar regions. Because of these; animals of polar regions like penguins, polar bears, and seal are losing their habitat and thereby their number decreases rapidly.
  3. Extinction of certain species of amphibia, fish, snails, shells, and migratory birds are the result of global warming.
  4. Destruction of mangrove forests is also a result of global warming, due to the rise in sea level.

Question 30. Cites two examples of foreign species causing harm to local species.
Answer:

Two examples of foreign species causing harm to local species are-

  1. The American cockroach Periplaneta americana harms the local cockroach of India Blatta orientalis.
  2. The plant water hyacinth coming from Brazil inhibits other aquatic plants.

Question 31. Give one example of exotic species causing damage of local biodiversities in each of terrestrial and aquatic ecosystem.
Answer:

In aquatic ecosystems, the exotic species that damage the local biodiversity is-water hyacinth. In terrestrial ecosystems, the exotic species that damage the local biodiversity is American. cockroach.

Question 32. ‘Poaching is one of the fundamental causes of endangeredness of animal diversity in India’-Assess four causes for which animals are poached in India.
Answer:

Four causes of poaching are-

  1. Some animals are mainly killed for their skins and bones, for example, elephants are killed for ivory, and tigers are killed for their bones and skins.
  2. Some animals like deer, pigs, bull are poached for meet the needs of meat.
  3. The body parts of some animals are also used for medicinal purposes and for home. decor.
  4. Illegally traded items fetch lucrative prices, that encourage poachers to look out for more money.

Question 33. Pollution is the reason of the depletion of biodiversity Justify.
Answer:

Pollution is the reason of the depletion of biodiversity:

  1. Pollution increases the average temperature causing global warming. Due to global warming glaciers melt causing problems for polar animals like polar beer, penguins,s, etc.
  2. Increasing temperature increases the water level of the sea causing the destruction of the mangrove forest of Sundarban.
  3. Pollution causing the destruction of coral reefs.
  4. Pollution harms the health of many animals.

Question 34. Give two examples of biodiversity depletion due to an increase in human population.

Answer:

  1. Deforestation-Plants are cut for habitat and other facilities of human like in Indonesia forests are cut for increasing oil production.
  2. Excessive use-Many plants and animals are used excessively for human welfare and entertainment, depletes biodiversity. – Passenger pigeons become extinct due to this.

 

Chapter 5 Importance Of Biodiversity MCQs

Question 1. The term ‘biodiversity’ was coined by—

  1. Zimmermann
  2. W G Rosen
  3. Charles Darwin
  4. Lamarck

Answer: 2. W G Rosen

Question 2. The diversity of various communities of organisms inhabiting a vast geographical area is known as—

  1. Genetic diversity
  2. α diversity
  3. β diversity
  4. γ diversity

Answer: 4. γ diversity

Question 3. The total number of biodiversity hotspots on Earth is—

  1. 14
  2. 24
  3. 34
  4. 44

Answer: 3. 34

Question 4. The concept of ‘hotspot’ was given by—

  1. David
  2. Simpson
  3. Mayer
  4. Norman Myers

Answer: 4. Norman Myers

Wbbse Class 10 Life Science Solutions

Question 5. The ecological area with very rich biodiversity is called—

  1. Blue spot
  2. Green spot
  3. Cold spot
  4. Hotspot

Answer: 4. Hotspot

Question 6. Out of 34 biodiversity hotspots worldwide, in India there are—

  1. 2
  2. 4
  3. 8
  4. 16

Answer: 2. 4

Question 7. The main cause of the extinction of biodiversity from equatorial regions is—

  1. Deforestation
  2. Erosion of soil
  3. Earthquake
  4. Super cyclone

Answer: 1. Deforestation

Question 8. The plant species present in India is—

  1. 10% of the global plants
  2. 11% of the global plants
  3. 12% of the global plants
  4. 13% of the global plants

Answer: 2. 11% of the global plants

Question 9. A biodiversity hotspot in India is—

  1. Nilgiri hills
  2. Aravalli hills
  3. Western Ghats
  4. Eastern Ghats

Answer: 3. Western Ghats

Question 10. The mangoes available in Malda have different tastes, flavors, fibrous content, colors and sugar content. This is due to—

  1. Hybridization
  2. Species diversity
  3. Induced mutation
  4. Genetic diversity

Answer: 4. Genetic diversity

Wbbse Class 10 Life Science Solutions

Question 11. The medicinal plant used to reduce high blood pressure is—

  1. Sarpagandha (Rauvolfia)
  2. Cinchona
  3. Guava (Psidium)
  4. Kalmegh (Andrographis)

Answer: 1. Sarpagandha (Rauvolfia)

Question 12. In a hotspot, the minimum number of endemic vascular plants that should be present is—

  1. 1500
  2. 1600
  3. 1700
  4. 1800

Answer: 1. 1500

Question 13. In Sundaland, the number of vascular plants present is about—

  1. 25000
  2. 30000
  3. 35000
  4. 40000

Answer: 1. 25000

Question 14. The Indian biodiversity hotspot, .which includes Sikkim, Assam, Arunachal Pradesh, and North Bengal is—

  1. Indo-Burma region
  2. Eastern Hiihalayan region
  3. Western Ghats region
  4. Sundaland

Answer: 2. Eastern Himalayan region

Question 15. Water hyacinth was introduced in India from

  1. Australia
  2. South America
  3. Japan
  4. Nepal

Answer: 2. South America

Question 16. The percentage of endemic plant species in India is about—

  1. 11%
  2. 22%
  3. 33%
  4. 44%

Answer: 2. 22%

Wbbse Class 10 Life Science Solutions

Question 17. The main cause of the declining number of tigers in Sundarban is—

  1. Expansion of residential area
  2. Poaching
  3. Forest management
  4. All of these

Answer: 3. Forest management

Question 18. ‘International Day for biological diversity is celebrated on—

  1. 5th June
  2. 22nd March
  3. 29th November
  4. 22nd May

Answer: 3. 29th November

Question 19. The species that is not abundant in nature but plays a very important role in maintaining ecological balance is called—

  1. Keystone species
  2. Endangered species
  3. Rare species
  4. Vulnerable species

Answer: 4. Vulnerable species

Question 20. Species diversity occurs due to—

  1. Genetic variation
  2. Ecological biodiversity
  3. Animal Diversity
  4. Plant diversity

Answer: 4. Plant diversity

Question 21. Mangrove forest is seen in—

  1. Dry desert area
  2. Hilly area
  3. Plateau
  4. Salty wetland

Answer: 1. Dry desert area

Question 22. The forest area in India is about—

  1. 19 %
  2. 91 %
  3. 52 %
  4. 23 %

Answer: 1. 19 %

Question 23. In India, the highest number of reptiles are found in—

  1. Himalayas
  2. Western Ghats
  3. Deccan plateau
  4. Sundarban

Answer: 4. Sundarban

Question 24. The exogenous plants that have created great ecological problems are—

  1. Parthenium and water hyacinth
  2. Bryophyllum and Begonia
  3. Berry and banana
  4. Shame plant and Calotropis

Answer: 4. Shame plant and Calotropis

Question 25. Two endangered animals of India are—

  1. Indian tiger and musk deer
  2. Indian ass and wild boar
  3. Cat and mongoose
  4. Wild cat and wild dog

Answer: 2. Indian ass and wild boar

Question 26. Endangered plants of India are—

  1. Sundari, pitcher plant, and sandal
  2. Mango, berry, and jackfruit
  3. Paddy, wheat, and millet
  4. Silk cotton, arjun and tamarind

Answer: 1. Sundari, pitcher plant, and sandal

Question 27. Synonym of ecological diversity is—

  1. Genetic diversity
  2. Species diversity
  3. Habitat diversity
  4. Animal diversity

Answer: 1. Genetic diversity

Question 28. The largest mangrove forest is situated in—

  1. Jalpaiguri
  2. Midnap ore
  3. Sundarban
  4. Purulia

Answer: 1. Jalpaiguri

Question 29. Which of the following is not a hotspot?

  1. Indo Burma
  2. Western Ghat
  3. Sundarban
  4. Eastern Himalaya

Answer: 3. Sundarban

Question 30. Sundarban is —

  1. National park
  2. Hotspot
  3. Biosphere reserve
  4. Sanctuary

Answer: 2. Hotspot

Question 31. The keystone species of Sundarban is—

  1. Mangrove
  2. Royal Bengal tiger
  3. Lion
  4. Fish

Answer: 2. Royal Bengal tiger

Question 32. Which of the following plant destroy the biodiversity of an ecosystem —

  1. Mango
  2. Pea
  3. Eucalyptus
  4. Arjun plant

Answer: 3. Eucalyptus

Question 33. The average temperature of the earth increased in the last century due to global warming is—

  1. 4°c
  2. 0.5°C
  3. 0.6°C
  4. 0.7°C

Answer: 3. 0.6°C

Question 34. The largest mangrove forest in the world is located in—

  1. Jalpaiguri
  2. Mednipur
  3. Sundarbans
  4. Purulia

Answer: 3. Sundarbon

Wbbse Class 10 Life Science Solutions

Question 35. Which of the following plant is used to treat high blood pressure—

  1. Rauwolfia
  2. Cinchona
  3. Guava
  4. Ocimum

Answer: 1. Rauwolfia

Question 36. Decide which of the following pair is not correct.

  1. Poaching—increase the endangeredness of gorilla
  2. Exotic Species—Lantana, Tilapia
  3. Determination of hotspot—Number of endemic species and endangered species
  4. Greenhouse gas—Eutrophication

Answer: 4. Greenhouse gas—Eutrophication

Question 37. One endangered species of Eastern Himalayas biodiversity hotspot is—

  1. Lion-tailed macaque
  2. Orangutan
  3. Red panda
  4. Nilgiri thar

Answer: 3. Red panda

Question 38. The most endangered plant in the world is

  1. Fungi
  2. Algae
  3. Moss
  4. Angiosperm

Answer: 4. Angiosperm

Question 39. The main reason of Rauwolfia to be endangered is—

  1. Global warming and climate change
  2. Invasion of foreign species
  3. Excessive use
  4. Pollution

Answer: 3. Excessive use

Question 40. Indian landmass is—

  1. 2% of the global land mass
  2. 4% of the global land mass
  3. 6% of the global land mass
  4. 8% of the global land mass

Answer: 1. 2% of the global land mass

Chapter 5 Importance Of Biodiversity Answer In A Single Word Or Sentence

Question 1. How do you denote the combination of ecosystem and biodiversity?
Answer: The combination of ecosystem and biodiversity is called ecological diversity.

Question 2. What is the basis of a healthy and active ecosystem?
Answer: A rich biodiversity is the basis of a healthy and active ecosystem.

Question 3. What is the species diversity in two different habitats within the same geographical area known as?
Answer: Beta (β) diversity

Question 4. How do you denote the genetic variations among the members of the same species?
Answer: Genetic diversity

Question 5. How do you treat the abundance of relatively different species in a specific area?
Answer: Species diversity

Question 6. Which type of diversity is called turnover diversity?
Answer: Beta (β) diversity.

Question 7. Which type of diversity indicates the abundance of different species in a specific area?
Answer: Alpha (α) diversity

Question 8. How do you denote the diversity of various species inhabiting a vast geographic area?
Answer: Gamma (γ) diversity

Question 9. Name a mangrove plant.
Answer: Sundari (Heritiera littoralis)

Question 10. What is flora?
Answer: Flora is the collection of all species of plants inhabiting an area or a geological time.

Question 11. What is vegetation?
Answer: Vegetation is the generalized idea on the distribution, number, nature, etc. of all plants inhabiting a specific area.

Question 12. How do you denote the places, which are rich in rare and endemic biodiversity?
Answer: The places, which are rich in rare and endemic biodiversity, are called hotspots.

Question 13. In which biodiversity hotspot in India is leaf deer seen?
Answer: Eastern Himalayas

Question 14. Write the full form of BMC.
Answer: Biodiversity Management Committee

Question 15. What is the total geographical area of India?
Answer: 32,87,263 km²

Question 16. According to FSI-2011, what percentage of Indian landmass is covered by forest?
Answer: According to FSI-2011, 21.05% of the Indian landmass is covered by forest.

Question 17. In which portion of the earth does the richest biodiversity occur?
Answer: The richest biodiversity occurs in the tropical rainforest region of the earth.

Question 18. Name a keystone species in West Bengal.
Answer: Royal Bengal tiger

Question 19. What percentage of species of the world is confined to the megadiversity countries?
Answer: 70%

Question 20. Which day is celebrated as world environment day?
Answer: 5th June

Question 21. Which day is Celebrated as world Conservation Day?
Answer: 3rd December

Question 22. When and where the Earth summit took place?
Answer: 1992 in Rio de Janeiro.

Question 23. What is the full form of CBD?
Answer: Convention of biological diversity

Wbbse Class 10 Life Science Solutions

Question 24. How many hotspots are there in India?
Answer: Four

Question 25. Write down the names of two hotspots in India.
Answer: Eastern Himalaya and Western Ghat

Question 26. Write down the name of a plant-of Eastern Himalaya.
Answer: Pitcher plant (Nepenthes khasiana)

Question 27. Which day is celebrated as world biodiversity day?
Answer: 22 May

Question 28. Write down the name of a mammal of the Eastern Himalaya.
Answer: Red panda (Ailurus fulgens)

Question 29. Which forest of India is declared as world heritage site by UNESCO?
Answer: Sundarban

Question 30. Sundarban is famous for the conservation of which animal?
Answer: Royal Bengal tiger

Question 31. What percentage of forest area is present in India according to the 2011 report of FSI?
Answer: 21.05%

Question 32. Write the name of the cause of the latest concern regarding the environment at Sundarbans.
Answer: Loss of mangroves and scarcity of drinking water.

Chapter 5 Importance Of Biodiversity Fill In The Blanks

Question 1. Among all living organisms on the earth, about _________ % is present only in 17 countries.
Answer: 70

Question 2. The source of quinine is _________ plant.
Answer: Cinchona

Question 3. _________ is used as a medicine to treat malaria.
Answer: Quinine

Question 4. A beneficial fungus is _________.
Answer: Penicillium

Question 5. Reserpine, the medicine for high blood pressure is extracted from the roots of _________ plant.
Answer: Sarpagandha

Question 6. _________ is a medicinal plant, the extract of which can be used as an immunostimulant.
Answer: Kalmegh

Question 7. The green plants are the _________ of the ecosystem.
Answer: Producers

Question 8. Wood from _________ plant is generally used for making furniture.
Answer: Teak

Question 9. The pulp prepared for the paper industry contains _________ fibers of the plant.
Answer: Cellulose

Question 10. _________ plant is used for paper production.
Answer: Bamboo

Question 11. The_________ secreted from babool, Shirish, etc. plants is used to prepare glue.
Answer: Gum

Question 12. The resin of pine is used to prepare _________ oil.
Answer: Turpentine

Question 13. _________ is a type of resin, used to add flavour in food.
Answer: Asafoetida

Question 14. Bee wax is used in preparing _________.
Answer: Cosmetics

Question 15. Bombyx mori produces _________ silk.
Answer: Mulberry

Question 16. Angora wool is collected from _________ fur.
Answer: Rabbit

Question 17. The biodiversity hotspot in which the Nicobar Islands are included is _________.
Answer: Sundaland

Question 18. The Eastern Himalayas hotspot is called the cradle of _________.
Answer: Speciation

Question 19. In the Eastern Himalayas hotspot, about _________ endemic flowering plant species are found.
Answer: 2000

Question 20. In the Indo-Burma hotspot, about _________ species of mammals are found.
Answer: 430

Question 21. The hotspot of south India is _________.
Answer: Western ghat

Question 22. _________ is a keystone species.
Answer: Royal Bengal tiger

Question 23. _________ ray is entering into the earth due to ozone layer depletion.
Answer: UV ray

Question 24. The term hot stop was comed by_________.
Answer: Norman Myers

Question 25. The name of one exotic species, that affects the local biodiversity, is _________.
Answer: Parthenium

Question 26. _________ is derived from the root of the propaganda plant which is needed as medicine form lowering high blood pressure.
Answer: Reserpine

Chapter 5 Importance Of Biodiversity State True Or False

Question 1. Caffeine is a structural component of wax.
Answer: True

Question 2. Turpentine oil is used in varnish preparation.
Answer: True

Question 3. Resin is a petroleum byproduct.
Answer: False

Question 4. Sal trees are used in paper manufacturing industries.
Answer: False

Question 5. The term ‘biodiversity’ was coined by W G Rosen.
Answer: True

Question 6. A hotspot must have a minimum of 1000 endemic vascular plants.
Answer: False

Question 7. Sundaland is a biodiversity hotspot in India.
Answer: True

Question 8. Sundari trees are the endemic species of the Eastern Himalayas hotspot region.
Answer: False

Question 9. Sundarban is situated at the estuaries of the Ganges, Padma, Brahmaputra, and Meghna.
Answer: True

Wbbse Class 10 Life Science Solutions

Question 10. The soil of the Sundarban area is saline.
Answer: True

Question 11. Medico is a Megadiversity country
Answer: True

Question 12. The use of N2, fertilizer makes the soil acidic
Answer: True

Question 13. The immersion of the island is an environmental problem of Sundarban.
Answer: True

Question 14. The Amazon forest is called the lungs of the earth.
Answer: True

Question 15. Rhododendron is an endangered plant species conserved in the Eastern Himalaya hotspot.
Answer: True

Chapter 5 Importance Of Biodiversity Match The Columns

1.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Importance Of Biodiversity Match The columns 1

Answer: 1-B; 2-C; 3—A; 4-E; 5-D; 6-G

2.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Importance Of Biodiversity Match The columns 2

Answer:  1-B; 2-0; 3-A; 4-C; 5-F; 6-G

 

3.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Importance Of Biodiversity Match The columns 3

Answer: 1-B; 2-D; 3-A; 4-C; 5-E; 6-F

Wbbse Class 10 Life Science Solutions

Chapter 5 Importance Of Biodiversity Find The Odd One Out

Question 1. Coconut plant, Sonneratia plant, Ceriops plant, Heritiera plant.
Answer: Coconut plant

Question 2. Eastern Himalaya, Indo-Burma, Japan, Sundaland
Answer: Japan

Question 3. Manipur, Mizoram, Indo-Burma, Sundaland
Answer: Sundaland

Question 4. Forestation, Deforestation, Hunting, Invasion of foreign species
Answer: Forestation

Question 5. American Cockroach, Lantana camara, Foreign species, Endemic species
Answer: Endemic species

Chapter 5 Importance Of Biodiversity Fill In The Blanks By Looking At The First Pair

Question 1. Ecology Earnst Haeckel :: Biodiversity: _______________
Answer: Walter Rosen

Question 2. Ship: Marino wool::Pinctada: _______________
Answer: Pearl

Question 3. Cinchona: Malaria:: Deak: _______________
Answer: Furniture

Question 4. Megadiversity country: Brazil:: Biodiversity hotspot: _______________
Answer: Eastern Himalaya

Question 5. Water hyacinth: Foreign species :: Muga moth: _______________
Answer: Endemic species

Question 6. Climate control: Importance of biodiversity:: Change of land use: _______________
Answer: Depletion of biodiversity

Question 7. Invasion of foreign species: Destruction of local species: Industrialisation: _______________
Answer: Habitat destruction

Chapter 5 Importance Of Biodiversity Among The Four Concepts Given Three Of Them Belong To One Find That

Question 1. Biodiversity, Genetic diversity, Species diversity, Ecosystem diversity
Answer: Biodiversity

Question 2. India, China, Megadiversity countries, Brazil
Answer: Megadiversity Country

Question 3. Medicine production, Importance of biodiversity, Climate Control, Economic development
Answer: Importance of biodiversity

Question 4. Deforestation, Poaching, Pollution, Reason of depletion of biodiversity
Answer: Reason of depletion of biodiversity

Question 5. Change of land use, Climate change, Destruction of biodiversity, Invasion of foreign species
Answer: Destruction of Biodiversity

Question 6. Habitat destruction, Problems of Sundarbans, Immersion of island, Destruction of Mangrove forest
Answer: Problems of Sundarban

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Biodiversity Conservation

Chapter 5 Biodiversity Conservation Summary

  • Biodiversity conservation is important to maintain sustainable life. This conservation is of two types. In-situ and Ex-situ conservation.
  • Conservation of organisms in their natural habitat is called in-situ conservation. The national park, sanctuary, and biosphere reserve are examples of in-situ conservation.
  • Conservation of endangered species other than their natural habitat is called ex-situ conservation, zoo, botanical garden etc. are examples of ex-situ conservation.
  • Cryopreservation is a special type of ex-situ conservation. By using liquid nitrogen at -196°C temperature, sperm and egg from animals and pollen, seed, etc. from plants are preserved by this process.

Biodiversity Conservation

Read and Learn More WBBSE Solutions for Class 10 Life Science And Environment

Chapter 5 Biodiversity Conservation Long Answer Type Questions

Question 1. What is a biosphere reserve? Mention the structure of a biosphere reserve.

Answer:

Biosphere Reserve:

To minimise the conflict between development and conservation, UNESCO has introduced a new concept. According to this concept, a large landmass or coastal area is demarcated into different zones based on their functional importance.

Here, the conservation of endangered species, ecotourism, educational excursion and human inhabitancy can be carried out, side by side without causing any harm to others. This type of area is called a biosphere reserve.

Example-Sundarban in West Bengal and Nilgiri hills in Tamilnadu are two biosphere reserves.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Biodiversity Conservation Different zones of a biosphere reserve

Structure of a biosphere reserve:

  1. A biosphere reserve has a central core zone for the conservation of wildlife. In this zone, any type of human intervention is completely prohibited. Here, the animals thrive and reproduce freely.
  2. Surrounding, the core area, there is a buffer zone, where ecotourism or educational excursion and research works may be conducted.
  3. The third zone is the transition zone situated beyond the buffer zone, where human inhabitancy, storage of logs and other harvested matters and forest security activities are allowed.

Wbbse Class 10 Life Science Solutions

Question 2. Mention the importance of the conservation of biodiversity.

Answer:

Importance of conservation of biodiversity:

The importance of the conservation of biodiversity is mentioned below.

1. Storage of food:

Biodiversity is the storehouse of food for all the living organisms on the earth. To keep the store intact we must conserve the biodiversity.

2. Source of energy:

Green plants absorb CO2 from the environment and H2O to synthesise carbohydrates as food. In this reaction, solar energy is trapped in the carbohydrate as chemical energy.

Through the food chain, this energy is distributed among all the members of an ecosystem. Therefore, the conservation of biodiversity is essential for the survival of the living world.

3. Store of the gene pool:

Biodiversity is the store of the gene pool. This genetic diversity is essential for the adaptation and evolution of new species.

4. Source of medicine:

Thousands of plants are there with immense medicinal importance. Scientists have found innumerable plants species, which are directly or indirectly involved in the production of medicines.

5. Source of raw material for industries:

We get various raw materials for different industries from plants and animals. Therefore, biodiversity is important for industries also.

6. Ecological importance:

In an ecosystem, different living organisms remain interconnected through the food web. The organisms include producers, consumers and decomposers. If the existence of any of these levels is in danger the whole ecosystem will break.

Therefore, conservation of biodiversity is important to maintain ecological balance.

Question 3. Explain the methods of conservation of biodiversity.

Answer:

Methods of conservation of biodiversity:

Conservation of biodiversity is done in two ways-

  1. In-situ conservation and
  2. Ex-situ conservation.

These are explained below.

1. In-situ conservation:

Protecting animals and plants within their natural habitat to facilitate their natural propagation by reducing undesirable human intervention and indiscriminate hunting, poaching and chopping of trees is known as in-situ conservation.

This type of conservation is practised in national parks, sanctuaries, protected forests and biosphere reserves. The characteristic features of these sites are mentioned below.

1. National park:

The largest of the forest areas, those that remain under the direct control of the national government where poaching, hunting, felling of trees or fishing is completely prohibited to ensure safe living and propagation of some endangered and rare organisms are known as national parks.

Ecotourism is allowed up to a certain limit with due permission from the department of Forestry. According to the national report 2013, 102 forests are demarcated as national parks.

Example-Gorumara National Park (West Bengal), Kanha National Park (Madhya Pradesh).

2. Sanctuary:

A vast forest area, under the control of state government, in which human intervention for poaching, hunting, felling of trees or fishing is completely prohibited to ensure safe living and propagation of some endangered and rare organisms, is known as a sanctuary.

Controlled human intervention in a sanctuary may be allowed for the purpose of the protective survey, census and research works only. According to the national report 2013, there are 515 sanctuaries in India.

Example-Chapramari Wildlife Sanctuary (West Bengal), Palamau Wildlife Sanctuary (Jharkhand).

3. Reserve forest:

According to the Indian Forest Act 1917, there are few forests under the management of the forest department under the state government in which entry of the general public is completely restricted, cutting of trees, hunting, grazing of cattle and scientific researchers are allowed to enter with due permission from the state forest department are known as reserve forest.

Example-Sholayar Reserve Forest of Kerala, Hanamsagar Reserve Forest of Karnataka.

4. Biosphere reserve:

To minimise the conflict between development and conservation, UNESCO introduced a new concept, by which a large landmass or coastal area is demarcated into different zones based on their functional importance and in which conservation of endangered species, ecotourism, educational excursion and human inhabitancy is carried out side by side without causing any harm to others. This type of area is called a biosphere reserve.

Example-Sundarban in West Bengal and Nilgiri hills in Tamilnadu are two biosphere reserves.

2. Ex-situ conservation:

To provide extra protection and close monitoring, some animals and plants are protected in artificially built areas away from their natural habitat. This type of conservation is known as ex-situ conservation.

This type of conservation is done in zoo gardens, botanical gardens, cryopreservation centres etc.

1. Zoo:

Various terrestrial and aquatic animals are protected in zoos under the close surveillance of zoologists and veterinary specialists.

Example-Alipore Zoological Garden.

2. Botanical gardens:

Several rare, endangered and vulnerable plant species are reared in botanical gardens under the care of botanists.

Example-Shibpur Botanical Garden.

3. Cryopreservation centres:

Cryopreservation is a laboratory technique by which tissues, embryos, body parts etc. of plants and animals are preserved in liquefied nitrogen at a temperature of -196°C.

Example-Botanical Survey of India, Shibpur.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Biodiversity Conservation Conservation of Biodiversity

Chapter 5 Biodiversity Conservation Short Answer Type Questions

Wbbse Class 10 Life Science Solutions

Question 1. Mention two objectives of conservation.
Answer:

Two objectives of conservation are as follows-

  1. To render the rare and endangered species of plants and animals a better environment to reproduce at a rapid rate and better survival on the earth.
  2. To ensure of controlled and sustainable utilisation of natural resources.

Question 2. What do you mean by conservation of biodiversity?
Answer:

Conservation of biodiversity:

Conservation of biodiversity is the protection and scientific management of the biosphere by human beings. This process provides maximum benefit for the present generation and the recent ecosystem.

It also means that the biosphere develops its potential to meet the needs of future generations and the future ecosystem as well.

Question 3. What is in-situ conservation?
Answer:

In-situ conservation:

Protecting animals and plants within their natural habitat to facilitate their natural propagation is called in-situ conservation. In-situ conservation reduces undesirable human intervention and indiscriminate hunting, poaching of animals and chopping of trees.

Example-Conservation of tigers and rhinos in national parks or biosphere reserves.

Question 4. Mention the advantages of in-situ conservation.
Answer:

The advantages of in-situ conservation are as follows-

  1. Conservation in natural habitats is less expensive.
  2. In this process, the success rate of reproduction is much higher than in captivity.

Question 5. Mention the disadvantages of in-situ conservation.
Answer:

The disadvantages of in-situ conservation are as follows-

  1. It needs a large forest area or water body.
  2. The diseased animals cannot be sorted out for treatment, therefore, certain diseases like anthrax may become epidemics.

Question 6. Mention the important sites for in-situ conservation.
Answer:

The important sites for in-situ conservation:

There are a few natural sites, designated for in-situ conservation. These are-national parks, biosphere reserves, sanctuaries and protected forests.

Wbbse Class 10 Life Science Solutions

Question 7. What are national parks?
Answer:

National parks:

The largest of the forest areas remain under the direct control of the government of India, where poaching, hunting, felling of trees or fishing is completely prohibited to ensure safe living and propagation of some endangered and rare organisms, but ecotourism is allowed up to a certain limit with due permission from the department of forestry, are known as a national park.

Example-Gorumara National Park, (West Bengal), Kanha National Park (Madhya Pradesh).

Question 8. Give two examples of ex-situ conservation.
Answer:

The two examples of ex-situ conservation are

  1. Zoo (Alipur Zoo) and
  2. Botanical garden (Acharya Jagadish Chandra Bose Indian Botanic Garden).

Question 9. Give examples of a national park and a biosphere reserve situated in West Bengal.
Answer:

An example of a national park, situated in West Bengal is jaldapara national park and a biosphere reserve is Sundarbans biosphere reserve.

Question 10. Mention the characteristic features of a national park.
Answer:

The characteristic features of a national park are as follows-

  1. The national park is a vast area, controlled by the central government for the conservation of rare and endangered animals and plants.
  2. In a national park, hunting, fishing and felling of trees are completely prohibited.

Question 11. Name some animals, which are protected in Corbett National Park in Uttarakhand.
Answer:

A number of rare and endangered animals are protected in Corbett National Park Uttarakhand. These include tiger, elephant, gaur, Indian grey mongoose, pangolin etc.

Question 12. Mention the objectives of establishing a national park.
Answer:

There are few objectives of establishing national parks. These are-

  1. Protection of rare and endangered animals.
  2. Providing adequate food and breeding ground for the animals.
  3. Maintaining a perfectly balanced natural ecosystem.

Wbbse Class 10 Life Science Solutions

Question 13. What is a sanctuary?
Answer:

Sanctuary:

A vast forest area, under the control of state government, in which human intervention, poaching, hunting, felling of trees or fishing is completely prohibited to ensure safe living and propagation of some endangered and rare organisms, is known as a sanctuary.

Controlled human intervention in a sanctuary may be allowed for the purpose of the protective survey, census and research works only.

Example- Chapramari (West Bengal), Palamau (Jharkhand).

Question 14. Mention two characteristic features of a sanctuary.
Answer:

Two characteristic features of a sanctuary are as follows-

  1. The management of a sanctuary remains under the state government.
  2. Human intervention in a sanctuary is completely prohibited. However, entrance is allowed for scientists only after prior permission from the concerned authority.

Question 15. Mention two importance of sanctuary.
Answer:

Two importance of the sanctuary are-

  1. Protection of wildlife against hunters and poachers.
  2. Conservation of endangered and rare species in their natural habitat.

Question 16. Tabulate any of the four activities which are prohibited in a sanctuary following the provisions of the Wildlife Act.
Answer:

The four activities, that are prohibited in the sanctuary are-

  1. Human intervention in the sanctuary is strictly prohibited.
  2. Any human activity which disturbs the wild animals is prohibited.
  3. Poaching, hunting, and fishing is strictly prohibited.
  4. Felling of trees is completely prohibited.

Question 17. What are reserve forests?
Answer:

Reserve forests:

According to the Indian Forest Act, of 1917, there are several forests under the management of the state government in which entry of the public is completely restricted, only scientists are allowed to enter with due permission from the state forest department. Those forests are known as reserve forests.

Example-Sholayar Reserve Forest of Kerala, Hanamsagar Reserve Forest of Karnataka.

Question 18. Mention two significances of biosphere reserve.
Answer:

Two significance of biosphere reserve are as follows-

  1. The species diversity and genetic diversity are preserved in the biosphere reserve.
  2. In a biosphere reserve, the qualitative development of human and nature interaction can be achieved.

Question 19. Mention the characteristic features of the reserve forest.
Answer:

The characteristic features of reserve forest are as follows-

  1. The management of reserve forests is completely a state government affair.
  2. In this forest, the felling of trees, and hunting of animals is not allowed without legal permission.

Life Science Class 10 Wbbse

Question 20. What is ex-situ conservation?
Answer:

Ex-situ conservation:

To provide extra protection and close monitoring, some animals and plants are protected in artificially built areas away from their natural habitat. This is known as ex-situ conservation.

Example- Conservation of plants in botanical gardens or zoos.

Question 21. Mention the advantages of ex-situ conservation.
Answer:

The advantages of ex-situ conservation are as follows-

  1. By this mode of conservation, the organisms are monitored closely, therefore, they can be medically attended as soon as any disease affects them.
  2. Due to lack of competition, the organisms get adequate food and the percentage of neonatal death can be reduced greatly.
  3. Cryopreserved germ cells or spores may be used to produce these organisms in future if their count goes down very fast.

Question 22. What are the disadvantages of ex-situ tages of ex-site conservation?
Answer:

The disadvantages of ex-situ conservation are as follows-

  1. Many animals and plants do not reproduce in captivity. Several plants become weak in the absence of ambient temperature, humidity and soil composition.
  2. Ex-situ conservation is expensive and it needs a good infrastructure.

Question 23. Mention the common places of ex-situ conservation.
Answer:

Ex-situ conservation is done in many places. These are

  1. Gene bank,
  2. Zoo,
  3. Aquarium,
  4. Arboretum, and
  5. Botanical garden etc.

Question 24. Mention two roles of a zoological garden as a place of ex-situ conservation.
Answer:

Role of a zoological garden as a place of ex-situ conservation are as follows-

  1. In a zoo, terrestrial and aquatic animals from different places are kept in suitable artificial habitats to increase their population size
  2. In this place, possible infrastructural and veterinary support are provided to the animals to help them to reproduce in captivity. If population size increases, few of them are set free in their natural habitat.

Question 25. Give two examples of zoological gardens used for ex-situ conservation.
Answer:

Two examples of zoological gardens used for ex-situ conservation are as follows-

  1. Alipur Zoological Garden in West Bengal acts as an ex-situ conservation centre for the Royal Bengal tiger, Asiatic lion, elephant, and crocodiles.
  2. The Indira Gandhi Zoological Park, at Visakhapatnam, Andhra Pradesh, protects sloth bears, wild dogs, puma, and jaguars as ex-situ conservation.

Life Science Class 10 Wbbse

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Biodiversity Conservation Ex-situ conservation centre

 

Question 26. Mention two characteristic features of the botanical garden as a site for ex-situ conservation.
Answer:

The characteristic features of a botanical garden as a site for ex-situ conservation are as follows-

  1. Various rare and endangered plants and medicinal herbs are nurtured in botanical gardens for ex-situ conservation.
  2. The rare and endangered plants are allowed to propagate under scientific surveillance to increase their number.

Question 27. What is meant by cryopreservation?
Answer:

Cryopreservation:

Cryopreservation is a laboratory technique by which any tissue, embryo, and body parts of animals. and plants are preserved in liquefied nitrogen at a temperature of -196°C.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Biodiversity Conservation Cryopreservation

Question 28. Mention the characteristic features of cryopreservation. Give an example of a cryopreservation centre.
Answer:

Features of cryopreservation:

The characteristic features of cryopreservation are as follows-

  1. In this type of preservation, liquefied nitrogen is used to maintain at a temperature of -196°C.
  2. In this process, pollen grains, seeds, gametes etc. are preserved.

Example of a cryopreservation centre:

An example of a cryopreservation centre is the Botanical Survey of India, Shibpur, Howrah.

Question 29. Give examples of two botanical gardens in India used for ex-situ conservation.
Answer:

There are several botanical gardens in India Two botanical gardens in India used for ex-situ conservation are-

  1. Acharya Jagadish Chandra Bose Indian botanic garden at Shibpur, Howrah and
  2. Jawaharlal Nehru Tropical botanical garden and research institute, Trivandrum, Kerala.

Question 30. What is an arboretum?
Answer:

Arboretum:

Arboretum is a method of ex-situ conservation. Here, rare and endangered plants are studied, reared and propagated under scientific surveillance.

Life Science Class 10 Wbbse

Question 31. What is meant by sacred forest?
Answer:

Sacred forest:

Out of religious belief, native tribes of some regions protect a section of a forest along with its animals and plants and treat this block of the forest as a sacred place. This section of forest is called a sacred forest.

Example-In the Thar desert of Rajasthan, the people of the Bishnoi tribes maintain a sacred forest for generations.

Question 32. What are sacred lakes?
Answer:

Sacred lakes:

Out of religious belief, tribes of a few regions of India protect certain water bodies along with all their inhabitants. These water bodies are called sacred lakes.

Example-Khecheopalri Lake of Sikkim is one of such sacred lakes.

Question 33. Distinguish between In-situ conservation and Ex-situ conservation.
Answer:

Distinguishing features between In-situ and Ex-situ conservation are-

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Biodiversity Conservation In-situ and ex- situ conservation

Chapter 5 Biodiversity Conservation MCQs

Question 1. An example of ex-situ conservation is—

  1. National park
  2. Reserve forest
  3. Seed bank
  4. Sanctuary

Answer: 3. Seed bank

Question 2. An example of conservation in the natural environment is—

  1. National park
  2. Zoo
  3. Arboretum
  4. Seed bank

Answer: 1. National park

Question 3. The management of national parks in India is controlled by—

  1. Central government
  2. State government
  3. Local institutions
  4. United Nations

Answer: 1. Central government

Life Science Class 10 Wbbse

Question 4. The central region of a biosphere reserve is known as—

  1. Reserve area
  2. Core area
  3. Buffer area
  4. Transition area

Answer: 2. Core area

Question 5. The division of IUCN, under which wildlife sanctuary is included, is—

  1. I
  2. II
  3. III
  4. IV

Answer: 2. II

Question 6. Simipal and Manas are

  1. Biosphere reserves
  2. Sanctuaries
  3. National parks
  4. Biodiversity hotspots

Answer: 1. Biosphere reserves

Question 7. Which one of the following is not included under in-situ conservation?

  1. Zoo
  2. National park
  3. Biosphere reserve
  4. Sanctuary

Answer: 1. Zoo

Question 8. Kaziranga National Park is especially famous for the conservation of—

  1. Deer
  2. Elephant
  3. One-horned Rhinoceros
  4. Lion

Answer: 3. One-horned Rhinoceros

Question 9. The number of mega-biodiversity nations is—

  1. 10
  2. 13
  3. 15
  4. 17

Answer: 4. 17

Question 10. Cashmere wool is prepared from the fur of—

  1. Sheep
  2. Goat
  3. Rabbit
  4. Cat

Answer: 2. Goat

Question 11. The first national park in India is—

  1. Bandipur National Park
  2. Corbett National Park
  3. Kanha National Park
  4. Periyar National Park

Answer: 2. Corbett National Park

Life Science Class 10 Wbbse

Question 12. The first week of which month is celebrated as wildlife week?

  1. January
  2. Apri 1
  3. July
  4. October

Answer: 4. October

Question 13. Which of the following is the most endangered class?

  1. Pisces
  2. Reptilia
  3. Aves
  4. Mammalia

Answer: 4. Mammalia

Question 14. The most endangered plants in the world are—

  1. Fungi
  2. Algae
  3. Flowering plants
  4. Mosses

Answer: 3. Flowering plants

Question 15. Sperms, ova, seeds, etc. are stored for a prolonged period in liquefied nitrogen at – 196°C. This type of preservation process is known as—

  1. Bioconservation
  2. Cryopreservation
  3. Ex-situ conservation
  4. ln-situ conservation

Answer: 2. Cryopreservation

Question 16. A modern concept of conservation prescribed ‘ in the MAB programme of UNESCO is—

  1. National park
  2. Reserve forest
  3. Sanctuary
  4. Biosphere reserve

Answer: 4. Biosphere reserve

Life Science Class 10 Wbbse

Question 17. The region seen In a biosphere reserve

  1. Core area
  2. Buffer area
  3. Transition
  4. All of these

Answer: 4. All of these

Question 18. The full form of MAB is—

  1. Man and Biology programme
  2. Man and Biosphere programme
  3. Mammals and Biosphere programme
  4. Mammals and Biology programme

Answer: 2. Man and Biosphere programme

Question 19. Conservation of an animal in a national park or wildlife sanctuary is treated as—

  1. National park or sanctuary conservation
  2. Ex-situ conservation
  3. In-situ conservation
  4. Keystone conservation

Answer: 3. In-situ conservation

Question 20. The reserved area in West Bengal, where red panda is available, is—

  1. Kanha National Park
  2. Sajnekhali Wildlife Sanctuary
  3. Singalila National Park
  4. Jaldapara National Park

Answer: 3. Singalila National Park

Question 21. Poaching of the red panda is done due to its—-

  1. Teeth
  2. Bones
  3. Meat
  4. Furry skin

Answer: 4. Furry skin

Life Science Class 10 Wbbse

Question 22. The full form of WWF is—

  1. Worldwide Fund for Nature
  2. World wide Forum
  3. World wide Forest
  4. World wide Friction

Answer: 1. Worldwide Fund for Nature

Question 23. An extinct bird of India is—

  1. Hornbill
  2. Cockatoo
  3. Parrot
  4. Pink-headed duck

Answer: 4. Pink-headed duck

Question 24. Poaching of rhinoceros is done for its—

  1. Meat
  2. Teeth
  3. Claw
  4. Horn

Answer: 4. Horn

Question 25. The state animal of Sikkim is—

  1. Fishing cat
  2. Red panda
  3. Takin
  4. Bear

Answer: 2. Red panda

Question 26. Which of the following component is used in cryopreservation—

  1. Liquid nitrogen
  2. Liquid hydrogen
  3. Dry ice
  4. Liquid oxygen

Answer: 1. Liquid nitrogen

Question 27. Gene bank is an example of—

  1. Ex-situ Conservation
  2. In-situ conservation
  3. Hot spot
  4. None of the above

Answer: 1. Ex-situ Conservation

Question 28. Gorumara national park is located in—

  1. Madhyapradesh
  2. Uttar Pradesh
  3. Kerala
  4. West Bengal

Answer: 4. West Bengal

Question 29. Which of the following pair is not right—

  1. Gir-Gujrat
  2. Kanha Uttarpradesh
  3. Corbett-Uttarakhand
  4. Kaziranga-Assam

Answer: 2. Kanha Uttarpradesh

Question 30. Simlipal, Sundarban and Manas are—

  1. Biosphere reserve
  2. Sanctuary
  3. National Park
  4. Cold spot

Answer: 1. Biosphere reserve

Question 31. Nilgiri biosphere reserve is located in—

  1. Karnataka
  2. Sikkim
  3. Orissa
  4. Rajasthan

Answer: 1. Karnataka

Question 32. According to the census of 2015, the number of Royal Bengal tigers present in Indian Sundarban is—

  1. 900
  2. 2000
  3. 86
  4. 500

Answer: 3. 86

Question 33. An example of ex-situ conservation is —.

  1. Sundarban Tiger reserve
  2. Corbett National Park
  3. Nilgiri Biosphere reserve
  4. Cryopreservation

Answer: 4. Cryopreservation

Question 34. Gorumara, Corbett, Kulik, Nandadevi-Choose the correct sequence from the following which is correct for the above forests sequence—

  1. Biosphere Reserve, Sanctuary, National Park, National Park
  2. National Park, National Park, Biosphere Reserve, Sanctuary
  3. National park, National Park, Sanctuary, Biosphere Reserve
  4. Sanctuary, Biosphere Reserve, National Park, Sanctuary

Answer: 3. National park, National park, Sanctuary, Biosphere reserve

Question 35. Identify which of the following is the correct information related to biosphere reserve—

  1. Conservation is promoted to local people and other biotic communities along with the conservation of the ecosystem
  2. National Park and sanctuary are not included within the biosphere reserve
  3. The presence and participation of local people for the conservation of the ecosystem are not permissible
  4. Its size is usually smaller than a sanctuary

Answer: 1. Conservation is promoted to local people and other biotic communities along with the conservation of the ecosystem

 

Chapter 5 Biodiversity Conservation Answer In A Single Word Or Sentence

Question 1. Which is the first biosphere reserve in India?
Answer: Nilgiri Biosphere Reserve

Question 2. At present, how many sanctuaries are there in India?
Answer: 515 sanctuaries

Question 3. In which year was Jaldapara Wildlife Sanctuary considered as national park?
Answer: Jaldapara Wildlife Sanctuary was considered as national park in the year 2012.

Question 4. Name two international organisations, which work for the conservation of biodiversity.
Answer: IUCN and WWF are two international organisations which.

Question 5. Write down the full form of NEERI.
Answer: National Environmental Engineering Research Institute.

Question 6. How many categories are there in the red list of IUCN?
Answer: Nine categories are there in the red list of IUCN.

Question 7. What is the full form of UNDP?
Answer: The full form of UNDP is United Nations Development Programme.

Question 8. What is the full form of FAO?
Answer: Food and Agriculture Organisation.

Question 9. What is the full form of NBPGR?
Answer: National Bureau of Plant Genetic Resources.

Question 10. According to the list of IUCN, under which category are the Indian national parks included?
Answer: According to the list of IUCN, Indian national parks are included under category II.

Question 11. What is the main function of IUCN?
Answer: Conservation of biodiversity all over the globe.

Question 12. According to the list of IUCN, under which category are the reserve forests included?
Answer: The reserve forests are included under category IV or VI.

Question 13. In which year the Forest Conservation Act was implemented in India?
Answer: The Forest Conservation Act was implemented in India in 1980.

Question 14. Where is the Valley of Flowers National Park located?
Answer: The Valley of Flowers National Park is located at Uttarakhand.

Question 15. In which country was the Dodo bird found before its extinction?
Answer: Dodo bird was found in Mauritius before its extinction.

Question 16. Name a reserve forest in West Bengal.
Answer: Chapramari

Question 17. Name a sanctuary located in the Birbhum district of West Bengal.
Answer: Ballavpur Wildlife Sanctuary

Question 18. Name the sanctuary located in North 24 Parganas of West Bengal.
Answer: Bibhutibhushan Wildlife Sanctuary

Question 19. In which state of India is Palamau Wildlife Sanctuary situated?
Answer: Jharkhand

Question 20. What is cryoinjury?
Answer: Sometimes the cellular structure is damaged during ice formation in cryopreservation, this is called cryoinjury.

Question 21. In which temperature cryopreservation is done.
Answer: -196°c

Question 22. Write down the name of a botanical garden.
Answer: Acharya Jagadish Chandra Bose Indian botanical garden or botanic garden.

Question 23. Where does Gir national park is located?
Answer: Gujrat

Question 24. Name one bird sanctuary of West Bengal.
Answer: Chintamani kar bird sanctuary [Narendrapur]

Question 25. Write down the name of a zoo of West Bengal.
Answer: Zoological Garden, Alipore

Question 26. How many biosphere reserves in India come under category V in the IUCN schedule?
Answer: In India, 18 Biosphere reserves come under category V in the IUCN schedule.

Question 27. In which type of in-situ conservation practice, the diversity, culture and lifestyle of human is preserved anlog with microorganisms, wild animals and wild plants?
Answer: Biosphere reserve

Question 28. Which is the largest national park in India?
Answer: The Desert National Park of Jaisalmer in Rajasthan is the largest (3162 km2) national park in India.

Question 29. Which is the smallest national park in India?
Answer: South Button Island National Park in Andaman is the smallest (5 km2) national park in India.

Question 30. Where is the first marine national park in India situated?
Answer: India’s first marine national park is situated at the Gulf of Kachchh in Jamnagar, Gujarat.

Question 31. How many marine national parks are there in India?
Answer: There are five marine national parks in India.

Question 32. On which day is World Environment Day celebrated?
Answer: 5th June

Question 33. Which day is celebrated every year as World Conservation Day?
Answer: 3 December

 

Chapter 5 Biodiversity Conservation Fill In The Blanks

Question 1. Due to the rise in sea level,_________ forest area is declining gradually.
Answer: Mangrove

Question 2. In 1997, UNESCO declared _________ as a World Heritage site.
Answer: Sundarban

Question 3. Conservation on any organism in a biosphere reserve is an example of _________ conservation.
Answer: In-situ

Question 4. Jaldapara National Park is situated in the state of _________.
Answer: West Bengal

Question 5. According to Wildlife Protection Act, _________ any state government can declare an area as a reserve forest.
Answer: 1927

Question 6. A zoological garden is an _________ conservation centre.
Answer: Ex-situ

Question 7. Acharya Jagadish Chandra Bose Indian Botanic Garden is situated in district _________.
Answer: Howrah

Question 8. In the year _________, the first national park was established in India.
Answer: Brazil

Question 9. The largest national park of India is _________.
Answer: Jim Corbett.

Question 10. The biosphere reserve is a _________ conservation.
Answer: 10. In-situ

Question 11. Lloyd’s botanical garden is situated in _________ district of west Bengal.
Answer: Darjeeling

Question 12. The full form of MAB is _________.
Answer: Men and Biosphere programme

Question 13. The bharatpur bird sanctuary is situated in _________ state.
Answer: Rajasthan

Question 14. A national park of West Bengal is _________.
Answer: Jaldapara

Question 15. The ex-situ conservation by keeping sample seeds in liquid nitrogen at a temperature of -196°C is called _________.
Answer: Cry op reservation

Question 16. _________ is a biosphere reserve located in West Bengal.
Answer: Sundarban

 

Chapter 5 Biodiversity Conservation State True Or False

Question 1. Kangaroo is an endangered species of India.
Answer: False

Question 2. The temperature for cryopreservation is -196°C.
Answer: True

Question 3. The central portion of a biosphere reserve is called the buffer region.
Answer: False

Question 4. Acharya Jagadish Chandra Bose Indian Botanic Garden is situated at Shibpur, Howrah.
Answer: True

Question 5. The Sundarban area of West Bengal is a biosphere reserve.
Answer: True

Question 6. Brazil is the country with the richest biodiversity in the world.
Answer: True

Question 7. Chapamari is a national park of West Bengal.
Answer: False

Question 8. The national park is a type of In-situ conservation
Answer: True

Question 9. In Manas national park of Assam Royal Bengal tiger is conserved.
Answer: False

Question 10. One-horned rhinoceros is conserved at Manas National Park of West Bengal.
Answer: False

 

Chapter 5 Biodiversity Conservation Match The Columns

1.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Biodiversity Conservation Match the columns 1

Answer: 1-D; 2-F; 3-E; 4-A; 5-C; 6-G

2.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Biodiversity Conservation Match the columns 2

Answer: 1-D; 2-E; 3-A; 4-C; 5-B; 6-F

 

Chapter 5 Biodiversity Conservation Find The Odd One Out

Question 1. Maharashtra, Kerala, Andhra Pradesh, Karnataka
Answer: Andhra Pradesh

Question 2. The national park, Zoological garden, Biosphere reserve, Sanctuary
Answer: Zoological garden

Question 3. Zoo, Botanical garden, Cryopreservation, Biosphere reserve
Answer: Biosphere

Question 4. The core region, Buffer region, Transition region, Botanical garden
Answer: Botanical garden

Question 5. Vitrification, Solid, Liquid nitrogen, Zoological garden
Answer: Zoo garden

Question 6. Nilgiri, Nokrek, Manas, Chapramari
Answer: Chapramari

Life Science Class 10 Wbbse

Chapter 5 Biodiversity Conservation Fill In The Blanks By Looking At The First Pair

Question 1. Sundarban: Biosphere reserve:: Jaldapara: _________
Answer: National Park

Question 2. Kaziranga, Assam: National park :: Manas, Assam: _________
Answer: Wildlife sanctuary

Question 3. Zoo: Ex-situ conservation:: Sanctuary: _________
Answer: In-situ conservation

Question 4. Corbett National Park: In-situ conservation:: Indian Botanic Garden: _________
Answer: Ex-situ conservation

Question 5. Plant: Botanical garden:: Seed and gamete _________
Answer: Cryopreservation

Question 6. Kanha: Madhya Pradesh:: Corbett: _________
Answer: Uttarakhand

 

Chapter 5 Biodiversity Conservation Among The Four Concepts Given Three Of Them Belong To One Find That

Question 1. The national park, In-situ conservation, Wildlife sanctuary, Reserve forest.
Answer: In-situ conservation

Question 2. Botanical garden, Cryopreservation, Zoological garden, Ex-situ conservation.
Answer: Ex-situ Conservation

Question 3. Dachigam forest, Kanha forest, National park, Gorumara forest.
Answer: National park

Question 4. Ovum preservation, Sperm preservation, Ovarian tissue preservation, Cryopreservation.
Answer: Cryopreservation.

Question 5. Parmadan forest, Wildlife sanctuary, Bethuadahari forest, Senchal forest.
Answer: Wildlife sanctuary.

Question 6. Core region, Buffer region, Transition region Biosphere reserve
Answer: Biosphere reserve

Question 7. Vitrification, -196°C Germplasm conservation, Cryopreservation.
Answer: Cryopreservation

Question 8. Embryo, Meristematic tissue, Seed, Cryopreservation.
Answer: Cryopreservation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Nitrogen Cycle

Chapter 5 Nitrogen Cycle Summary

  • Our environment is composed of both living and non-living things. Nitrogen is an important element for organisms. 78.09% of the air is nitrogen. Nitrogen, like other elements, also moves through biotic and abiotic components of the ecosystem, which is known as the nitrogen cycle or N2 cycle.
  • The main stages of the nitrogen cycle are-biological and non-biological fixation of nitrogen from the atmosphere, entry of nitrogen into the organism from the soil, and re-entry of nitrogen into the soil through ammonification, nitrification, and denitrification or release of N2 into the air.
  • The conversion of atmospheric nitrogen into a nitrogenous compound is called nitrogen fixation. The atmospheric nitrogen is fixed by several prokaryotes, i.e. some soil bacteria (Azotobacter and Clostridium), some cyanobacteria (Nostoc, Anabaena), and some symbiotic bacteria (Rhizobium).
  • Also, atmospheric nitrogen is converted into nitrous oxide by the electrical energy of lightning, which is brought down by rain and is used by plants.

Nitrogen Cycle Summary

Life Science Class 10 Wbbse

  • Ammonifying bacteria like Bacillus mycoides convert excretory substances of organisms and dead organic matter into ammonia. Nitrifying bacteria like Nitrosomonas convert ammonia, into nitrate and nitrite called nitrification.
  • Finally, nitrate and nitrite is converted into N2 and released into the atmosphere by denitrifying bacteria like Pseudomonas, and Thiobacillus. Human activities, such as the use of fertilizers greatly increase the amount of nitrogen that is cycling between the living world and soil, water, and atmosphere.
  • This results in increased global concentrations of N2O, which causes global warming, and substantial acidification of water bodies and soil.

Read and Learn More WBBSE Solutions for Class 10 Life Science And Environment

Chapter 5 Nitrogen Cycle Long Answer Type Questions

Question 1. Describe the nitrogen cycle with the help of a flowchart. Similar question, Show the steps of the nitrogen cycle with the help of a flow chart. Part question, Analyse the role of bacteria at any three steps of the nitrogen cycle.

Answer:

Nitrogen cycle:

The nitrogen cycle is a bio-geochemical process which involves different phases, which are-nitrogen fixation in soil, entry of nitrogen from soil into living organisms, return of nitrogen from organisms to soil, and liberation of nitrogen in the atmosphere. The phases are explained in the next section.

1. Nitrogen fixation:

By this process, atmospheric nitrogen is trapped in the soil. This occurs by three different processes, which are-

Natural N2 fixation:

During lightning, a huge amount of heat is generated in the atmosphere, when nitrogen and oxygen react gradually to produce nitric oxide (NO) and nitrogen dioxide (NO2).

Life Science Class 10 Wbbse

When rainwater mixes with atmospheric nitrogen dioxide two acids, namely nitrous acid (HNO2) and nitric acid (HNO3) are produced. The reactions are given below-

N2+O2 → 2NO,  2NO+O2 → 2NO2, 2NO2 + H2O → HNO2 + HNO3

These two acids then react with the minerals present in soil and form nitrates (potassium nitrate, calcium nitrate, etc.).

Biological fixation:

Biological nitrogen fixation is done in the following ways-

  1. Soil-dwelling nitrifying bacteria like Azotobacter sp., Clostridium sp., etc. absorb atmospheric nitrogen and trap it in their cells. When these bacteria die, the nitrogenous compounds mix in the soil.
  2. Cyanobacteria like Anabaena sp. and Nostoc sp. absorb nitrogen from the atmosphere and add it to the soil.
  3. Rhizobium sp. present in root nodules of leguminous plants traps atmospheric nitrogen and converts it into nitrogenous compounds. With the death of the host plants, these compounds mix in the soil to increase its nitrogen content.

Industrial fixation:

In fertilizer industries, different nitrogen-rich fertilizers are produced. By the Haber-Bosch method, N2 and H2 are made to react at 300-400°C under 35-100 MPa (Mega Pascal) pressure to produce NH3.

This NH3 and other nitrogen-rich fertilisers when added to soil increases the nitrogen content.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Nitrogen Cycle Industrial fixation

2. Entry of nitrogen from soil into living organisms:

Plants absorb soluble nitrogenous salts from the soil through their roots. This nitrogen takes part in synthesizing plant proteins. Animals get nitrogen either directly or indirectly by consuming plants.

Finally, plant proteins are transformed into animal proteins and remain in the animal bodies.

3. Return of nitrogen from organisms to soil:

Nitrogen returns from the body of organisms to the soil through two different steps. These are-

Life Science Class 10 Wbbse

Ammonification:

The ammonifying bacteria (Bacillus mycoides, Micrococcus sp., etc.) act upon dead animal and plant bodies and the excretory materials to convert the nitrogenous compounds in it into ammonia. This is called ammonification.

Nitrification:

Certain nitrifying bacteria convert ammonia into nitrates through two steps of reaction.

  1. Nitritation-In the first step, Nitrosomonas sp. converts ammonia and ammonium ions into nitrites.
  2. Nitratation-In the next step, soil-dwelling Nitrobacter sp. transforms nitrites into nitrates.

2NH4++3O2 → 2NO2+ 4H+ + 2H2O,   2NO2+O2 → 2NO3

4. Liberation of nitrogen into the atmosphere or denitrification:

In this phase, several bacteria act upon nitrites and nitrates in the soil and liberate free nitrogen into the atmosphere. This biochemical process is called denitrification. Pseudomonas sp., Thiobacillus sp., etc. are a few bacterial species, which help in denitrification.

2NO3  → 2NO2, → NO + N2O → N2

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Nitrogen Cycle Nitrogen in to the atmosphere or denitrification

Question 2. ‘Nitrogen cycle is affected due to human activities’-State two phenomena to justify it. Explain the adverse effects of excess nitrogen in the atmosphere, liberated due to human activity. Similar question, Explain any three incidences which are occurring as a result of the disruption of the nitrogen cycle.

Answer:

Human actions affecting the nitrogen cycle:

The human actions affecting the nitrogen cycle are discussed below.

1. Application of nitrogen-rich fertilizers:

Nowadays, farmers apply an excess amount of nitrogen-rich fertilizers in crop fields to increase productivity. Due to this practice, the content of N2O and NO2 is rising alarmingly in the atmosphere.

2. Cultivation of nitrogen-fixing plants:

Leguminous plants like gram, peas, lentils, soya beans, etc. are important sources of plant protein. To fulfill the need of protein, different pulses are being cultivated more and more. These plants efficiently absorb nitrogen from the atmosphere and trap it in the soil.

3. Burning of fossil fuel:

Excessive burning of fossil fuels such as mineral oils, coal, etc. increases nitric oxide (NO), nitrous oxide (N2O), and nitrogen dioxide (NO2) in the air. It increases the supply of nitrogen in the environment.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Nitrogen Cycle Root nodules of pea plant

Adverse effects of excess nitrogen in the atmosphere due to human activity:

The adverse effects of excess nitrogen in the atmosphere, due to human activity, are mentioned below.

1. Global warming:

Nitrous oxide (N2O) is a potent greenhouse gas. Accumulation of this gas in air arrests more heat in the atmosphere and causes global warming.

2. Acid rain:

Oxides of nitrogen present in the atmosphere mix with rainwater to produce nitrous and nitric acids. These acids come down on earth as acid rain, which acidifies soil and water bodies to cause serious environmental pollution.

3. Photochemical smog:

Nitric oxide (NO), released as automobile exhaust, reacts with atmospheric hydrocarbons in the presence of sunlight. This reaction produces a polluting gas, named PAN (Peroxy-acetyl nitrate).

It mixes with smoke and creates a very harmful photochemical smog. This reduces visibility through air, and causes inflammation of the respiratory tract and various other respiratory diseases.

Question 3. Explain the microbial fixation of atmospheric nitrogen in the nitrogen cycle.

Answer:

Microbial fixation of atmospheric nitrogen in the nitrogen cycle:

Certain microorganisms are capable of absorbing atmospheric nitrogen and trapping it in soil. This mechanism of trapping of atmospheric nitrogen with the help of microorganisms is called biological nitrogen fixation. Two types of microorganisms take part in this process, which are described here.

Life Science Class 10 Wbbse

1. Nitrogen fixation by free-living microbes:

Few free-living soil-dwelling bacteria like Clostridium sp., Azotobacter sp., and few cyanobacteria like Anabaena sp. (a symbiont with Azolla as well), Nostoc sp., etc. absorb atmospheric nitrogen and convert it into nitrogenous compounds within their cell.

As the microbes die these nitrogenous compounds mix in the soil to increase its nitrogen content.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Nitrogen Cycle Azotobacter

2. Nitrogen fixation by symbiotic microbes:

Rhizobium sp. is a soil-dwelling bacterium, which enters into the root tissues of leguminous plants and stays there as a symbiont. As the bacterium enters in to the host tissue, the root cells start to divide quickly to form nodules.

Rhizobium, inhabiting in these root nodules, absorbs atmospheric nitrogen and converts it into various nitrogenous compounds. A portion of these nitrogenous compounds is supplied to the host plant and the rest is stored in the microbial cell body.

When these plants die and are decomposed, these nitrogenous compounds get mixed in soil, to increase the fertility of the soil.’

Chapter 5 Nitrogen Cycle Short Answer Type Questions

Question 1. What is meant by the bio-geochemical cycle?
Answer:

Bio-geochemical cycle:

The continuous cycling of different elements (C, H, N, O, P, S, Ca, etc.) among biotic and abiotic components of the environment is called a bio- geochemical cycle. The nitrogen cycle is an important biogeochemical cycle in the environment.

Question 2. What will happen if the bio-geochemical cycle is disrupted?
Answer:

The bio-geochemical cycle is disrupted:

If the bio-geochemical cycle is disrupted, the balance of the important elements in the environment would have been disturbed. This may lead to natural disasters.

Question 3. Why deficiency of basic elements does not occur in the environment?
Answer:

Deficiency of basic elements never occurs in the environment, because living organisms never uptake any element permanently from the environment. The organisms release such elements as compounds through excretion.

Alternatively, these are liberated in the environment by the decomposition of the dead and.. decaying organic matter.

Question 4. What is the nitrogen cycle?
Answer:

Nitrogen cycle:

The nitrogen cycle is a bio-geochemical cycle, which describes how nitrogen circulates in the atmosphere, the soil, and living organisms. So, by the help of this biogeochemical cycle, the content of nitrogen in atmosphère and in living organisms remains unaltered.

Question 5. Mention the importance of nitrogen in living organisms.
Answer:

The importance of nitrogen in living organisms:

The main structural component of a living body is protein. Nitrogen, on the other hand, is the main structural element of protein. That is why, nitrogen is important for all living organisms.

Question 6. What is meant by nitrogen fixation?
Answer:

Nitrogen fixation:

Atmospheric nitrogen is trapped in to the soil in the form of different nitrogenous compounds by a number of physicochemical or biochemical processes. These processes are collectively known as nitrogen fixation.

Life Science Class 10 Wbbse

Question 7. What is meant by biological nitrogen fixation?
Answer:

Biological nitrogen fixation:

Certain blue-green algae or cyanobacteria and a few bacteria absorb nitrogen directly from the atmosphere and reduce it to ammonia with the help of an enzyme, nitrogenase.

This ammonia then assimilates in the protoplasm of these microbial organisms. This biochemical trapping of atmospheric nitrogen is called biological nitrogen fixation.

Question 8. What is formed by the reaction between atmospheric nitrogen and oxygen during lightning?
Answer:

During lightning, a large amount of heat is generated in the sky, when nitrogen and oxygen react gradually to produce nitric oxide (NO) and nitrogen dioxide (NO2). The reactions are given below.

N2+O2 → 2NO,  2NO+O2 →2NO2

Question 9. What happens when rainwater mixes with atmospheric nitrogen dioxide?
Answer:

When rainwater mixes with atmospheric nitrogen dioxide two acids, namely nitrous acid (HNO,) and nitric acid (HNO3) are formed. The reaction is mentioned below.

2NO2 + H2O→ HNO2 + HNO3

Question 10. What is ammonification?
Answer:

Ammonification:

The biochemical process, by which dead animal and plant bodies or nitrogenous waste matters are acted upon by certain bacteria to produce ammonia, is known as ammonification. Bacillus mycoides, Micrococcus sp., etc. are a few examples of ammonifying bacteria.

Question 11. What is nitrification?
Answer:

Nitrification:

The biochemical process, by which soil ammonia is first converted into nitrite and finally into nitrate, is known as nitrification. Nitrosomonas, Nitrobacter, etc. are a few examples of nitrifying bacteria that help in this process.

Question 12. What is denitrification?
Answer:

Denitrification:

The biochemical process, by which nitrate or any other nitrogenous compounds break to liberate free nitrogen into the atmosphere, is known as denitrification. Certain bacteria like Thiobacillus, Pseudomonas, etc. are denitrifying in nature.

Question 13. Mention the role of bacteria in the nitrogen cycle.
Answer:

The role of bacteria in the nitrogen cycle are as follows-

  1. Certain bacteria trap atmospheric nitrogen and reduce it to form ammonia.
  2. A few others, then convert this ammonia into nitrite.
  3. Certain other bacteria, then transform nitrite into nitrate.
  4. A few other bacteria break nitrates to liberate elementary nitrogen into the atmosphere.

Question 14. Mention the role of bacteria in nitrification.
Answer:

The role of bacteria in nitrification:

Bacteria like Bacillus mycoides, Micrococcus sp., etc. reduce nitrogen from decomposing organic matter to produce ammonia. Nitrosomonas sp. then converts the ammonia into nitrites.

Nitrobacter sp. acts upon these nitrites and oxidizes them to stable nitrates. This is how bacteria help in nitrification.

Life Science Class 10 Wbbse

Question 15. Mention the importance of soil-living microbes.
Answer:

The importance of soil-living microbes:

Soil-living microbes help in the decomposition of dead organisms and other organic matter. By this process, several simple organic compounds and elements mix in the soil to prepare hummus. This humus gradually increases nutrients in the soil and makes it more fertile.

Question 16. Mention the role of bacteria in increasing soil fertility.
Answer:

The role of bacteria in increasing soil fertility:

Free-living soil bacteria like Azotobacter, Clostridium, etc., and symbiotic bacteria like Rhizobium absorb atmospheric nitrogen and trap it in the form of nitrogenous compounds.

These compounds increase the nitrogen content in the soil, thus, increasing soil fertility.

Question 17. Write the names of two microorganisms participating in the process of biological nitrogen fixation.
Answer:

The two microorganisms participating in the process of biological nitrogen fixation are- Azotobacter (free-living) and Rhizobium (symbiotic).

Question 18. What is hummus?
Answer:

Hummus:

After the death of animals and plants, the bodies are decomposed and various organic matters mix in the soil. This brownish, sticky organic matter, which is generally acidic in nature, is known as humus.

Question 19. What is leghaemoglobin?
Answer:

Leghaemoglobin:

Leghaemoglobin is a pinkish-red pigment present in the nitrogen-absorbing root nodules of leguminous plants. The structure of this pigment and its capacity of absorbing oxygen is similar to that of hemoglobin.

This pigment is present in the space between the symbiotic bacterial cell and the surrounding host cells.

Question 20. Why do the leaves of insectivorous plants have different modifications?
Answer:

Insectivorous plants grow in such soil, where nitrogen content is significantly low. To compensate for the short supply of nitrogen, these plants trap insects and digest them to absorb protein from their body.

To trap insect prey efficiently, the leaves of insectivorous plants show various types of modification. Example-Pitcher plant, sundew, etc.

Question 21. What are decomposers?
Answer:

Decomposers:

Decomposers are microorganisms, which putrefy or decompose the dead organisms and excretory materials to liberate simple elementary components from complex organic matter.

Example-Different bacteria and fungi.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Nitrogen Cycle Decomposers fungi

Question 22. What role do the decomposers play in the environment?
Answer:

Decomposers putrefy dead organisms and excretory materials and convert complex organic matters into simple matters. This process maintains the reserve of elements in the environment, thus, maintaining the balance among different elementary nutrients.

Other than this, decomposers keep the environment clean by disposing of the decaying matter.

Question 23. Relate the following phenomena with the trend of disturbance in the nitrogen cycle resulting from different human activities-

  1. Global warming,
  2. Acidification of soil and water of rivers and lakes.

Answer:

Global warming:

Around 40% increase of N20 (nitrous oxide) in the environment is due to human activities. N20 is released in the environment as a result of the combustion of fossil fuel and the overuse of nitrogen-rich fertilizers.

This N2O is a greenhouse gas that absorbs infrared radiation or heat of the sunlight which reflects on the earth and facilitates the greenhouse effect and global warming.

Wbbse Class 10 Life Science Solutions

Acidification of soil and water of rivers and lakes:

Oxides of nitrogen get dissolved in rainwater and form nitric acid, which is a major component of acid rain. Acid rain destroys aquatic plants and animals and disrupts the ecological equilibrium of rivers lakes, ponds, etc.

Apart from this, the destruction of forest resources, harm to architecture and monuments take place due to acid rain.

Question 24. How does acidification affect the environment?
Answer:

Air-polluting gases like nitric oxide (NO), other nitrogenous oxides, sulfur oxides, etc. mix. with rainwater and come down as acid rain.

This acidic rainwater mixes with soil and different water bodies to acidify them which harm the how acidification affects the environment. various components of the environment. This is how acidification affects the environment.

Question 25. Mention the significance of the nitrogen cycle.
Answer:

The significance of the nitrogen cycle are as follows-

  1. The nitrogen cycle helps to maintain the nitrogen balance in the environment.
  2. The amount of nitrogen necessary for the synthesis of protein is supplied to the living bodies from the nitrogen cycle. By this process, the nitrogen cycle helps in sustaining the race of life.

Chapter 5 Nitrogen Cycle Multiple Choice Question And Answers

Question 1. The most abundant gas in the atmosphere is—

  1. Oxygen
  2. Hydrogen
  3. Carbon dioxide
  4. Nitrogen

Answer: 4. Nitrogen

Question 2. Which of the following is not involved with the nitrogen cycle?

  1. Nitrification
  2. Eutrophication
  3. Ammonification
  4. Denitrification

Answer: 2. Eutrophication

Question 3. The plant, which traps and consumes insects to fulfill the need for nitrogen, is—

  1. Sarpagandha [Rauvolfia)
  2. Nayantara (Catharanthus)
  3. Sunflower
  4. Pitcher plant

Answer: 4. Pitcher plant

Question 4. Plants can absorb nitrogen from the soil in the form of—

  1. NO2 only
  2. N2O
  3. NO
  4. NO2 and NO3

Answer: 4. NO2 and NO3

Question 5. A nitrogen-fixing bacterium is—

  1. Nitrosomonas
  2. Azotobacter
  3. Nitrobacter
  4. Mycobacterium

Answer: 2. Azotobacter

Question 6. The microorganism inhabiting the root nodules of a sweet pea plant is—

  1. Nostoc
  2. Anabaena
  3. Chlamydomonas
  4. Rhizobium

Answer: 4. Rhizobium

Question 7. Ammonification is the formation of—

  1. Ammonia from nitrates by decomposers
  2. Ammonia from nitrogen
  3. Ammonia from amino adds
  4. Ammonia from nitrates by nitrogen fixers

Answer: 3. Ammonia from amino adds

Question 8. Which of the following is not a nitrogen-fixing bacterium?

  1. Rhizobium
  2. Mycobacterium
  3. Nostoc
  4. Azotobacter

Answer: 2. Mycobacterium

Question 9. The majority of nitrogen fixation occurs by—

  1. Biological nitrogen-fixing organisms
  2. The natural process by lightning
  3. Volcanic eruptions
  4. Haber-Bosch process

Answer: 1. Biological nitrogen-fixing organisms

Question 10. Which of the following components is increased in soil due to the cultivation of leguminous plants in it?

  1. Oxygen
  2. Carbon dioxide
  3. Sulfur dioxide
  4. Nitrogen

Answer: 4. Nitrogen

Question 11. Application of nitrogen in soil—

  1. Reduces fertility
  2. Increases fertility
  3. Has no effect on fertility
  4. Increases pH

Answer: 2. Increases fertility

Question 12. The percentage of nitrogen in the atmosphere is about—

  1. 0.08
  2. 20.60
  3. 1.40
  4. 78.09

Answer: 4. 78.09

Question 13. The component of air, which neither plants nor animals can absorb directly, is—

  1. Oxygen
  2. Carbon dioxide
  3. Nitrogen
  4. Both 1 and 2

Answer: 3. Nitrogen

Question 14. The enzyme by which nitrogen-fixing bacteria anaerobically convert atmospheric nitrogen into ammonia is—

  1. Nitrogenase
  2. Amylase
  3. Pectinase
  4. Galactase

Answer: 1. Nitrogenase

Question 15. The conversion of ammonia into nitrate is called—

  1. Ammonification
  2. Denitrification
  3. Calcification
  4. Nitrification

Answer: 4. Nitrification

Question 16. The blue-green algae capable of absorbing nitrogen from the atmosphere is—

  1. Chlamydomonas
  2. Clostridium
  3. Anabaena
  4. Rhodochorton

Answer: 3. Anabaena

Question 17. A nitrogen-releasing bacterium is—

  1. Thiobacillus
  2. Rhizobium
  3. Nostoc
  4. Diatom

Answer: 1. Thiobacillus

Question 18. The bacteria, which can trap at”losp^ nitrogen in the soil in the form of nitrogenous compounds, are called—

  1. Symbiotic bacteria
  2. Ammonifying bacteria
  3. Nitrogen-fixing bacteria
  4. Saprophytic bacteria

Answer: 3. Nitrogen-fixing bacteria

Question 19. Bacillus mycoides is a type of—

  1. Symbiotic bacterium
  2. Ammonifying bacterium
  3. Denitrifying bacterium
  4. Nitrifying bacterium

Answer: 2. Ammonifying bacterium

Question 20. Which of the following is correct regarding the nitrogen cycle?

  1. N2 cycle is a bio-geochemical cycle
  2. The major reservoir of N2 is the atmosphere
  3. N2 is the most abundant nutrient for plants
  4. All of these

Answer: 4. All of these

Question 21. A soil-dwelling free-living nitrogen-fixing bacterium is—

  1. Clostridium
  2. Nostoc
  3. Anabaena
  4. Diatom

Answer: 1. Clostridium

Question 22. Nitrate in the soil is transformed into ammonia and then into free nitrogen by bacterial action. This process is called—

  1. Nitrification
  2. Ammonification
  3. Denitrification
  4. Symbiosis

Answer: 3. Denitrification

Wbbse Class 10 Life Science Solutions

Question 23. Which of the following nitrogenous compounds is increased in the atmosphere due to human activity?

  1. N2O
  2. NO
  3. NH3
  4. All of these

Answer: 4. All of these

Question 24. A nitrogenous greenhouse gas is—

  1. NH3
  2. N2O
  3. HNO3
  4. All of these

Answer: 2. N2O

Question 25. The nitrogenous compounds responsible for ozone depletion are—

  1. NH4OH and HNO3
  2. NO and NO2
  3. HNO2 and HNO3
  4. NH3 and HNO2

Answer: 2. NO and NO2

Question 26. The nitrogenous compound responsible for acid rain is—

  1. NO2
  2. NO
  3. HNO3
  4. All of these

Answer: 4. All of these

Question 27. Which of the following phase of the nitrogen cycle Pseudomonas is associated with?

  1. Nitrogen fixation
  2. Nitrification
  3. Denitrification
  4. Amnonification

Answer: 3. Denitrification

Question 28. Which of the following microbes take part in nitrification?

  1. Nitrosomonas
  2. Azotobacter
  3. Pseudomonas
  4. Thiobacillus

Answer: 1. Nitrosomonas

Chapter 5 Nitrogen Cycle Answer In A Single Word Or Sentence

Question 1. Name the process of cyclic rotation of various elements and inanimate matters in the biosphere.
Answer: Bio-geochemical cycle

Question 2. Mention the role of the biogeochemical cycle.
Answer: Bio-geochemical cycle maintains the balance among different elements of the environment.

Question 3. Name a gas that is essential for all plants but they cannot absorb it directly from the atmosphere.
Answer: Nitrogen

Question 4. Name a denitrifying bacterium.
Answer: Thiobacillus denitrificans

Question 5. Mention the types of microorganisms which help in the biological fixation of nitrogen in the soil.
Answer: Biological nitrogen fixation occurs in soil by free-living microorganisms and symbiotic microorganisms.

Question 6. Name the pinkish-red pigment present in the root nodules of leguminous plants.
Answer: Leghaemoglobin

Question 7. Name a nitrogen-fixing free-living bacterium.
Answer: Clostridium

Question 8. Name a nitrogen-fixing symbiotic bacterium.
Answer: Rhizobium leguminosarum

Question 9. Name a nitrifying bacterium.
Answer: Nitrosomonas

Question 10. In which form does nitrogen stay in soil?
Answer: Nitrogen stays in the soil in the form of nitrate and nitrite salts.

Question 11. Give examples of two cyanobacteria, which are capable of fixing nitrogen from the atmosphere.
Answer: Anabaena and Nostoc

Question 12. Which role is played by fungi in the environment?
Answer: Fungi play the role of decomposers in the environment.

Question 13. Name two plants which feed on insects to obtain nitrogen.
Answer: Pitcher plant and sundew plant

Question 14. Name the process where nitrates are converted into free nitrogen.
Answer: Denitrification

Question 15. Name the two steps of nitrification.
Answer: Nutrition and nitration

Wbbse Class 10 Life Science Solutions

Question 16. By which process of nitrification, nitrite is produced from ammonia?
Answer: Nutrition

Question 17. Which type of bacteria converts nitrates into free nitrogen?
Answer: Denitrifying bacteria

Question 18. Name the process of ammonia formation by the decomposition of dead and decaying organic matter.
Answer: Ammonification

Question 19. Name two non-leguminous nitrogen-fixing plants.
Answer: Pine and Azolla

Question 20. Which enzyme helps to catalyze the reactions for N2 fixation?
Answer: Nitrogenase

Question 21. Write down the names of two nitrogen-containing chemical fertilisers.
Answer: Urea [CO(NH2)2] and ammonium nitrate (NH4NO3)

Question 22. Name the process by which industrial nitrogen fixation occurs.
Answer: Haber-Bosch process

Question 23. Give an example of two natural problems created due to the abnormality of N2 cycle.
Answer: Two natural problems created due to the abnormality of N2 cycle are acid rain and global warming (due to increased concentration of greenhouse gas.)

Question 24. NH3 \(\stackrel{\mathbf{A}}{\longrightarrow}\) NO2 \(\stackrel{\mathbf{B}}{\longrightarrow}\) NO3 write down the names of two bacteria suitable in positions A and B.
Answer: A-Nitrosomonas, B-Nitrobacter

Question 25. Give an example of one human activity that hampers the N2 cycle.
Answer: The use of nitrogen-rich fertilizers hamper the N2 cycle.

Chapter 5 Nitrogen Cycle Fill In The Blanks

Question 1. The process of trapping atmospheric nitrogen in the soil is known as _________.
Answer: Nitrogen fixation

Question 2. _________ develops root nodules in the root system of leguminous plants.
Answer: Rhizobium

Question 3. Anabaena is an example of _________.
Answer: Cyanobacteria

Question 4. During lightning,_________ is produced in the atmosphere by the combination of nitrogen and oxygen.
Answer: No

Question 5. Nitrosomonas is a type of _________ bacterium.
Answer: Nitrifying

Question 6. _________ is a bacterium which helps in the conversion of nitrite into nitrate.
Answer: Nitrobacter

Question 7. Another name of ammonification is _________.
Answer: Mineralisation

Question 8. Pea, gram, and other pulses are known as _________ plants.
Answer: Leguminous

Question 9. Rhizobium is a _________ nitrogen-fixing bacterium.
Answer: Symbiotic

Question 10. NO, of atmosphere mixes with rainwater to produce _________ and nitric acid.
Answer: Nitrous

Question 11. Azotobacter is a _________ nitrogen-fixing bacterium.
Answer: Free-living

Question 12. At the _________ phase of the nitrogen cycle, ammonia is converted into nitrite and nitrate by the action of some bacteria.
Answer: Nitrification

Chapter 5 Nitrogen Cycle State True Or False

Question 1. Nitrosomonas is a denitrifying bacterium.
Answer: False

Question 2. The Haber-Bosch reaction is a natural process of nitrogen fixation.
Answer: False

Question 3. Clostridium is a free-living nitrogen-fixing bacterium.
Answer: True

Question 4. Ammonia dissolves in water to produce nitrous acid and nitric acid.
Answer: False

Question 5. The bacterial species helping in the ammonification process is Bacillus mycoides.
Answer: True

Question 6. Nostoc is capable of absorbing atmospheric nitrogen.
Answer: True

Question 7. Azotobacter releases free nitrogen from nitrates, trapped in soil.
Answer: False

Question 8. The content of nitrogen in the atmosphere is 28.09%.
Answer: False

Question 9. Anabaena is an example of a symbiotic microorganism.
Answer: True

Wbbse Class 10 Life Science Solutions

Question 10. The other name of ammonification is mineralization.
Answer: True

Question 11. Carbon dioxide is a nitrogenous greenhouse gas.
Answer: False

Question 12. Bacteria act as decomposers but fungi do not.
Answer: False

 

Chapter 5 Nitrogen Cycle Match The Columns

1.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Nitrogen Cycle Match The Columns 1

Answer: 1-B; 2-C; 3-D; 4-A; 5-E; 6-G

2.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 5 Nitrogen Cycle Match The Columns

Answer: 1-E; 2-D; 3-B; 4-A; 5-C; 6-F

 

Chapter 5 Nitrogen Cycle Find The Odd One Out

Question 1. Azotobacter, Rhizobium, Clostridium, Anabaena
Answer: Anabaena

Question 2. Nitric acid, Nitric oxide, Ammonia, Nitrogen dioxide
Answer: Ammonia

Question 3. Nitrification, Denitrification, Industrial N2 fixation, Nitrite formation
Answer: Industrial N2 fixation

Question 4. Nitrosomonas, Nitrococcus, Clostridium, Nitrobacter
Answer: Clostridium

Chapter 5 Nitrogen Cycle Fill In The Blanks By Looking At The First Pair

Question 1. Clostridium Free-living bacterium Rhizobium:_________
Answer: Symbiotic bacterium

Question 2. Sweet pea plant: Rhizobium :: Azolla:_________
Answer: Anabaena

Question 3. Nitrogen fixation: Azotobacter:: Nitrification: _________
Answer: Nitrobacter

Question 4. Ammonifying bacteria: Bacillus mycoides:: _________: Nitrosomonas
Answer: Nitrifying bacteria

Question 5. Nitric oxide: Photochemical smog:: Nitrous oxide: _________
Answer: The greenhouse effect

Question 6. Nitrobacter: Nitrification:: Bacillus mycoides: _________
Answer: Ammonification

Chapter 5 Nitrogen Cycle Among The Four Concepts Given Three Of Them Belong To One Find That

Question 1. The action of Bacillus ramosus, Breakdown of amino acids, Action of Bacillus mycoides, Ammonification
Answer: Ammonification

Question 2. Nitritation and nitratation, Action of Nitrobacter, Nitrification, Action of Nitrosomonas
Answer: Nitrification

Question 3. Nitrosomonas, Nitrococcus, Nitrobactor, Nitrifying bacteria
Answer: Nitrifying bacteria

Question 4. Biological N2 fixation, Rhizobium, Azolla, Azotobacter
Answer: Biological N2 fixation

Question 5. Clostridium, Nostoc, Rhizobium, N2 fixing microbes
Answer: N2 fixing microbes

Question 6. Acid rain, Greenhouse gas formation, PAN formation, Abnormality of N2, cycle
Answer: Abnormality of N2, cycle

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation

Chapter 4 Adaptation Summary

  • The reaction of an organism in response to environmental stimuli is called its behaviour. Behaviour helps in the adaptation and evolution of an organism.
  • Adaptation is the structural or functional changes of an organism which help the organism to survive in a new environment or their constantly changing current environment.
  • Morphological adaptations can be observed in the followings-
    1. The leaves of some desert plants turn into spines, even the number of leaves also decreases. These are to reduce the rate of transpiration in water deficit conditions.
    2. The air bladder is the buoyancy organ in fish which helps the fish to maintain its depth while swimming.
    3. Air sacs in birds help them by increasing oxygen exchange capacity and lessen the specific gravity of birds thus enabling them to fly higher.

Adaptation

Wbbse Class 10 Life Science Solutions

  • Physiological adaptation is observed in the Sundari plant, camel, etc. Removing excess salt by bark exfoliation and through salt glands present in leaves, conducting oxygen intake via pneumatophore, etc. are some physiological processes which enable the Sundari plant to survive in coastal saline soil.
  • RBC of the camel is small, oval and can expand up to 240% after water absorption. Besides, water production by oxidizing the fat stored in the hump, releasing concentrated urine, etc. help camel to survive in the extreme condition of a desert region.
  • Chimpanzees and honey bees show behavioural adaptation. Chimpanzees prepare sticks from twigs and use these as fishing tools to extract termites from their mound and eat them. They use rocks as a hammer to break the shells of nuts.
  • When they get infected by parasites they use herbs as medicines to treat themselves. Honey bees perform typical patterns of dance, like waggle dance, and circular dance, to indicate the direction and distance of food source from the hive.

Read and Learn More WBBSE Solutions for Class 10 Life Science And Environment

Chapter 4 Adaptation Long Answer Type Questions

Question 1. What is the relation between adaptation and evolution? Mention the functions of the spine in the cactus to combat the arid environment.

Answer:

Relation between adaptation and evolution:

Adaptation is the evolutionary process which enables an organism to survive in its habitat.

1. Adaptation-a step to evolution:

To survive in a changing environment, organisms change their morphological, behavioural and physiological features accordingly. Those, which can adjust themselves successfully in a changing environment, get the favour of natural selection to survive and propagate.

The favourable adaptive features are inherited by successive generations, which help in the origin of a new species or assist in evolution. Practically, adaptation is a step towards evolution.

Wbbse Class 10 Life Science Solutions

2. Evolution-a greater notion:

Evolution is a very long and complex process. It is a continuous, ongoing process, since the origin of life on Earth. On the other hand, adaptation occurs when any organism faces any problem related to its survival due to any change in the environment.

Therefore, evolution is a greater notion in comparison to adaptation.

Functions of the spine in cactus:

The main functions of the spine in cactus are as follows-

  1. The spines are modifications of leaves which are transformed to prevent transpiration.
  2. Spines also act as defence organs against herbivores.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Spines of Cactus

Question 2. What is the importance of air sacs in pigeons? Write down the role of the air bladder in rohu fish in maintaining body balance in the water. Part question, Prepare a list of roles air sacks of pigeons to play to fly in the air.

Answer:

Importance of air sacs in pigeons:

Pigeon is a primary volant animal. It has nine non-vascular and non-muscular air sacs, emerging from the bronchioles, typically helping in flight.

The importance of air sacs in pigeons are-

  1. During the flight, the pigeon needs extra energy, the production of which requires additional oxygen. Oxygen concentration near the ground level is higher than that in the high sky. Before the flight, pigeons fill the air sacs with this oxygen-rich air. When in the high sky, these air sacs supply oxygen-rich air to the lungs.
  2. The air-filled sacs decrease the specific gravity of the body of the pigeon, which is also beneficial for flight.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Air sacs of pigeon

Importance of air bladder in rohu fish:

  1. The air bladder in rohu fish helps it to move up and down in the water. The air bladder changes the buoyancy of the body and thus, assists it to go at different depths of water.
  2. The red gland in the anterior chamber of the air bladder fills gas in it to reduce the specific gravity of the body and thus helps the fish to move upward in water. On the other hand, the rete mirabile of the posterior chamber absorbs the gas to increase the specific gravity of the body of fish and helps the fish to go deeper in the water.

Wbbse Class 10 Life Science Solutions

Question 3. Briefly describe the adaptive features of the Sundari tree.

Answer:

Adaptation of Sundari tree:

Sundari trees grow in saline muddy soil at coastal regions or river delta. This soil is rich in mineral salts. The oxygen content of this saline soil is also very low. Halophytes, like Sundari trees, have the following adaptations for surviving in this soil.

1. Adaptive changes in roots:

  1. To get adequate oxygen from the soil for the root system, several lateral branches of roots grow against gravity and emerge vertically above the soil. These aerial roots have numerous pores on the surface to absorb atmospheric oxygen. This type of root is called a pneumatophore.
  2. Sundari trees have plank-like lateral projections, called root buttresses, extending from the root base to support the tree from all sides to stand on soft soil. This type of root in Sundari trees helps the plants to stand erect on muddy soil at the coastal region.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Pneumatophores

2. Adaptive changes in stem:

  1. The surface of the stem is coated with a thick and waxy cuticle layer. This coating prevents transpiratory water loss through the stem surface.
  2. For better transportation of water and food, their vascular system is highly developed.

3. Adaptive changes in leaves:

  1. The number of stomata is reduced and the existing stomata are sunken to reduce the rate of transpiration.
  2. Leaf lamina is covered with a waxy substance to prevent transpiration.

Question 4. Mention the physiological adaptations of a camel. Part question, How the special feature of RBC in camels is related to their ability to withstand the extreme loss of water?

Wbbse Class 10 Life Science Solutions

Answer:

The physiological adaptations of camel:

Camel is one of the best-adapted animals in desert environments. The adaptive features of a camel with respect to the conservation of water and modification of RBC are mentioned below.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Camel

1. Adaptive features helping in the conservation of water:

  1. The hump on the back of the camel reserves fat, which is oxidised to produce water and energy during their long walk through the desert.
  2. Camels discharge highly concentrated urine to conserve water.
  3. A camel can drink up to 36 litres of water at a time, which is stored in the water reservoir cells of its stomach.
  4. The mucous membrane of the nasal passage absorbs moisture from the air that the camel exhales.
  5. Body temperature of the camel changes in accordance with the external temperature, which reduces perspiration water loss, even at a high temperature of 49° Celsius.

2. Adaptive modifications of RBC:

  1. The RBCs of camel are very small and oval in shape. This modification facilitates the movement of these blood cells through finer blood capillaries even at very low water content in plasma.
  2. When plasma contains adequate water, the RBCs of camels can expand up to about 240 times of its normal size, by absorbing water by endosmosis. These cells can withstand high osmotic pressure and do not rupture even after the intake of a huge quantity of water at a time.

Question 5. Why is behaviour important for evolution? Mention three different behaviours of chimpanzees, which help in their adaptation.
Part question, Give an example of ways chimpanzees solve their problems intelligently for procuring food and preventing diseases.
Part question, Explain how do the chimpanzee hunt termites for food. Part question, How do chimpanzees break open the hard shells for eating the nuts?

Answer:

Importance of behaviour in evolution:

Organisms show neurally controlled reactions in response to any visual, auditory or olfactory signals or any stimulus, like light, sound, smell or any action by other organisms. These responses are called behaviour.

The favourable behaviours which help an organism to survive and propagate on earth, are selected by nature. The effect of these behaviours is so far-reaching that it may help in the course of evolution.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Eating of leaves by Chimpanzee

Adaptive behaviours of chimpanzees:

  1. Chimpanzees break branches of trees and then eat up the leaves to prepare a stick. With the stick, they make a hole in a termite mound. Then, they insert the thin stick into the mound to tease the termite colony inside it. As soon as the termites grab the probing stick, chimpanzees pull it out and lick them up. Few termites, which come out through the hole are eaten up by other chimpanzees. This particular behavioural adaptation of chimpanzees is known as termite fishing.
  2. When a chimpanzee is attacked by parasites, it starts eating medicinal plants. Scientists revealed that chimpanzees can recognise and consume 19 different medicinal plants including Aspilia rudis, when attacked by parasites.
  3. To crack hard nutshell, wild chimpanzees use a piece of rock as an anvil and a strong branch of the tree as a hammer. They keep hard-shelled nuts on the rock and then hammer it with the woody branch to crack it. Finally, they remove the shell and eat the nut inside.

Wbbse Class 10 Life Science Solutions

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Termite fishing by Chimpanzee

Question 6. Describe with a diagram how honeybees pass messages to other members of the hive about the source of food.
Or,
How a worker bee of a beehive communicates with other worker bees, regarding the source and location of food?

Answer:

Indication of honeybees about the source of food:

In a honeycomb, there are two types of worker bees, of which, scouts go in search of food and foragers fetch nectar. When any scout bee finds a source of food, it returns to its hive and performs a typical pattern of dance in front of the foragers to pass the message.

There are two patterns of bee dances, waggle dance and circular dance, which are discussed below.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Waggle dance of honeybee

1. Waggle dance:

If the source of food is at a distance of more than 75 metres, scout bees perform a typical dancing flight, known as the waggle dance.

In this dance, they fly in alternate left and right directions to create an ‘8’ shaped course and at the straight middle portion of the course, jerk the abdomen. An upward flight indicates that the source of food is along the direction of the sun.

When the source of food is at the opposite direction to the sun, they perform the dancing flight in a downward direction. If they get the source at an angle of 30° from the hive, they perform the waggle dance at the same angle as the axis of the hive.

2. Circular dance:

When they find the food at a distance within 50-75 metres, the scouts perform the circular dance. As soon as the message is passed to the foragers, they leave the hive in a swarm and fly towards the source of food.

Chapter 4 Adaptation Short Answer Type Questions

Question 1. What is adaptation?
Answer:

Adaptation:

Adaptation is the morphological, physiological or behavioural adjustment of an organism to survive in a changing environment. Survival of any organism in a habitat depends upon its perfect adaptation to that environment.

Question 2. What is the objective of adaptation?
Answer:

The objective of adaptation:

The main objective of adaptation of an organism is to survive in a changing environment. The adaptation may be a morphological, physio- logical or behavioural adjustment with its surrounding environment.

Question 3. Adaptation is the change of shape, physiological functions and behaviour of an organism’-Justify the statement with the help of any two examples.
Answer:

Adaptation is the change of shape, physiological functions and behaviour of an organism:

The canine teeth of carnivorous animals are large and sharp to tear flesh whereas the same are relatively small and blunt in herbivores. This change in the shape of the teeth helps the organisms to consume particular types of food.

Kangaroo rats survive without drinking water for a long time, camels possess some physiological adaptations that allow them to survive a long time without drinking water. These are examples of physiological changes which help the animals to adapt in the desert environment.

Camouflaging is a behavioural adaptation of certain organisms to hide in any particular environment. So it can be called that is the change in shape, physiological function and behaviour of organisms.

Question 4. What is acclimatisation?
Answer:

Acclimatisation:

The physiological changes occurring in the organisms in accordance with the changes in its environment are collectively called acclimatisation.

For example, people living at high altitudes, have higher RBC count in blood and greater total lung capacity to acclimatise with low oxygen density in the atmosphere.

Question 5. What is meant by behaviour?
Answer:

Behaviour:

Behaviour simply refers to the expression of any organism in an environment. It can be defined as the neurally controlled reaction, shown by an organism in response to any environmental signal or any specific stimulus or any action by other organisms.

Question 6. How is behaviour related to the environment?
Answer:

Behaviour is related to the environment in the with the environment by its own behaviour.

following ways-

  1. Every organism coordinates
  2. The behaviour of an organism changes in accordance with the change in the environment.

Question 7. What is ethology?
Answer:

Ethology:

The branch of science that deals with the study of the behaviour of different animals, from small insects to large mammals, is called ethology. By studying ethology, one can have an idea on the nature of the expression of an animal in response to any external or internal stimulus.

Question 8. Distinguish between adaptation and evolution.
Answer:

The distinguishing features between adaptation and evolution are-

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Distinguishing features between Adapation and evolution

Question 9. What is the relation between behaviour and evolution?
Answer:

The relation between behaviour and evolution:

The behaviour of any organism helps it to survive and propagate in nature. As the environment changes, the organisms also change their behaviour to adapt in it.

If these changes are selected by nature and propagate to the next progeny for several generations, new species evolve that possess this changed behaviour.

Question 10. Define morphological adaptation and give an example.
Answer:

Morphological adaptation:

The permanent and beneficial physical changes, occurring either internally or externally in the body of an organism to help it survive in a changing environment, are called morphological adaptation.

Example-Modification of leaves of cactus into spines is an example of morphological adaptation.

Question 11. What is meant by xerophytic adaptation?
Answer:

Xerophytic adaptation:

The morphological and physiological changes, which occur in plants to befit them in arid environments, are collectively known as xerophytic adaptation. This adaptation is seen in desert plants, like cacti, babul, Aloe vera etc.

Question 12. Which plants are called xerophytes?
Answer:

The plants, which are adapted to survive in physically dry soil and in the atmosphere with very low moisture content, are known as xerophytes.

Wbbse Class 10 Life Science Solutions

Example- Cactus, Asparagus, babul, etc.

Question 13. Which adaptive features are seen in cacti to reduce water loss due to transpiration?
Answer:

To reduce water loss by transpiration, the following adaptations are seen in cacti-

  1. The size and number of leaves reduce. Sunken stomata are present in leaves.
  2. In some cacti, leaves are modified into spines to reduce the rate of transpiration.
  3. The phylloclade is covered with cuticle to seal stomata thus preventing water loss.

Question 14. Which morphological structures modify into spines of cactus and thorns of wood apples?
Answer:

Spines of cactus are the modification of leaves. On the other hand, the thorns of wood apples are a modified form of axillary buds.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Throne of wood apple

Question 15. Who are the primary aquatic animals?
Answer:

The primary aquatic animals:

The aquatic animals, which have evolved from aquatic ancestors, are called primary aquatic animals. Example-Generally all fishes are primarily aquatic animals.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Primary aquatic animal fish

Question 16. Who are called secondary aquatic animals?
Answer:

Secondary aquatic animals:

Certain animals which had terrestrial ancestors but have adapted themselves to aquatic life for certain reasons, are called secondary aquatic animals.

Example-Whale, sea cow, etc.

Question 17. Mention two aquatic adaptations of fishes.
Answer:

Two aquatic adaptations of fishes are as follows-

  1. Fishes have air bladders to move in different depths of water.
  2. To swim in different directions and maintain balance in the water, fishes have different types of fins.

Question 18. What are the roles of the swim bladder in the aquatic adaptation of rohu fish?
Answer:

The roles of the swim bladder in the aquatic adaptation of rohu fish

The swim bladder or air bladder in fish is a gas-filled internal organ and acts as a hydrostatic organ. It can increase or decrease the buoyancy of a fish by respectively increasing or decreasing the volume of gas within it.

Thus, an air bladder helps a fish to go up to different depths of water without floating upward or sinking downwards.

Wbbse Class 10 Life Science Solutions

Question 19. What is rete mirabile?
Answer:

Mirabile:

Rete mirabile is a network of blood capillaries in the posterior chamber of the air bladder of fish. It absorbs gas from the air bladder to increase the specific gravity of the body of a fish and helps it to go deeper in the water.

Question 20. What problems a fish would have faced if it lacked a spindle-shaped body and lateral line sense organ?
Answer:

If a fish lacked a spindle-shaped body, it would have faced more physical resistance while swimming. If there had been no lateral line sense organ, a fish could not have felt pressure, temperature and depth of water.

Question 21. Who are called primary volant animals?
Answer:

Primary volant animals:

The animals, adapted to aerial life, which have descended from flying ancestors, are called primary volant animals.

Example-Pigeon, bat, etc.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation primary volant animal bat

Question 22. Who are called secondary volant animals?
Answer:

Secondary volant animals:

There are certain volant animals, which have evolved from non-flying ancestors. These animals are called secondary volant animals.

Examples- Flying fish, flying squirrels, etc.

Question 23. How is the body shape of a pigeon modified for flying?
Answer:

For a flying bird-like pigeon, an aerodynamic body shape is needed. This body shape helps to overcome air resistance during fast flight. Hence pigeon has become spindle-shaped. to overcome the air-resistance, the body of a

Question 24. Mention the roles of air sacs in the volant adaptation of pigeons.
Or,
What are the two adaptive significances of air sacks of pigeons?
Answer:

The roles of air sacs in the volant adaptation of it needs more supply of oxygen to generate extra pigeons are

  1. Pigeon is a volant animals; hence energy during its flight. The air sacs, attached to its lungs, supply extra oxygen and therefore this process enhances the ability of the lungs.
  2. These air sacs hold air to reduce the specific gravity of its body during flight.

Question 25. What will happen to a pigeon if its air sacs become inactive?
Answer:

If a pigeon has inactive air sacs, it will face the following problems-

  1. Body weight will increase and double respiration will not occur.
  2. Body balance and centre of gravity of body will get disrupted during flight.
  3. It will not be able to collect highly oxygenated air from ground level to use it during high flight.

Question 26. Which organs have disappeared from the pigeon’s body to make it lighter?
Answer:

For flying, an animal always requires a lighter body. So to make the body of a pigeon lighter, certain organs have disappeared. These organs are teeth, gall bladder, rectum, right ovary and fallopian tube in females and penis in males.

Question 27. Define physiological adaptation and give an example.
Answer:

Physiological adaptation:

The permanent and beneficial changes in various physiological activities of an organism to facilitate its survival in a changing environment, is called physiological adaptation.

Example-Development of pneumatophores in the Sundari tree is an example of physiological adaptation.

Question 28. Which plants are called halophytes?
Answer:

The plants, growing in the saline soil of the coastal areas or delta regions, are called halophytes. Example-Sundari, garan, Hetal, etc.

Question 29. Mention two adaptive features of the Sundari tree.
Answer:

Two main adaptive features of Sundari tree are-

  1. Presence of specialised root structures known as pneumatophores.
  2. A thick layer of cuticle on the stem surface and leaves with fewer stomata to minimise water loss.

Question 30. How Sundari plant excretes the excess salt of its body? Similar question, Mention one adaptation of the Sundari plant for salt tolerance.
Or,
Describe any two adaptations in the Sundari plant for tolerance of salt.

Answer:

The mechanisms used by Sundari plant to excrete the excess salt from its body are-

  1. Presence of salt gland in the leaf to excrete excess salt along with water,
  2. Deposition of excess salt in specialized cells of bark and root; later, by means of bark exfoliation and other specific mechanisms, the salt are eliminated from the plant body,
  3. Deposition of salt in mature leaves, in larger amounts, so that it can be eliminated easily by leaf exfoliation,
  4. Deposition of excess salt in the cell sap of vacuoles and eliminate it when needed.

Question 31. What are pneumatophores?
Answer:

Pneumatophores:

Halophytes, like Sundari, Garan, etc., grow in highly saline and muddy soil with very low oxygen content. To get adequate oxygen for the root system, some lateral branches of their roots grow against gravity and emerge vertically above the soil.

These aerial roots have numerous pores on the surface to absorb atmospheric oxygen and such roots are known as pneumatophores or breathing roots.

Question 32. How do pneumatophores help Sundari trees to adapt in saline soil?
Answer:

Sundari trees grow in highly saline muddy coastal soil. This type of soil has very low oxygen content and water logging during high tide is very common in this region. Therefore, the root system does not get enough oxygen.

Pneumatophores are aerial roots which are negatively geotropic, thus helping Sundari trees to absorb aerial oxygen for the root system and help these plants to survive in the low oxygen-containing saline soil.

Question 33. What is meant by viviparous germination?
Answer:

Viviparous germination:

The process of seed germination where the seed germinates within the fruit while it is still attached with the plant is known as viviparous germination. Saline soil is not conducive for the germination of seeds.

To overcome this obstacle, viviparous germination Occurs in halophytes, like garan, Rhizophora, etc.

Question 34. What is the advantage of viviparous germination?
Answer:

The advantage of viviparous germination:

In viviparous germination, the seed germinates within the fruit and the radicle comes out of it as a pointed dagger-shaped structure even before its shedding from the tree.

As it grows some inches long, the fruit drops from the branch and the pointed tip of the radicle penetrates the soft soil keeping the developing embryo above the saline soil. Thus, viviparous germination protects the developing embryo from a hostile environment.

Question 35. Why Sundari tree is called a ‘looking glass tree’?
Answer:

Sundari tree is called a ‘looking glass tree’:

The leaves of Sundari trees are covered with a thick cuticle layer and a thin and smooth layer of wax coating. Besides, the lower surface of the leaves have numerous glossy silvery hairs.

When sunlight reflects from these leaf surfaces, it gives a beautiful glassy look, hence Sundari tree is called a ‘looking glass tree’.

Wbbse Class 10 Life Science Solutions

Question 36. What are root buttresses?
Answer:

Root buttresses:

Root buttresses are plank-like lateral projections extending from the root base to support a tree from all sides to stand on soft soil.

This type of root in trees, like Sundari, helps it to stand erect on muddy soil at the delta region.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Root buttress

Question 37. Mention two adaptive features of a camel.
Answer:

Two adaptive features of camel are-

  1. A hump, containing fat, is present on its back.
  2. RBCs are nucleated, small in size and oval in shape.

Question 38. Mention two adaptations of camels in conserving water.
Answer:

To conserve water, camels have the following adaptive features-

  1. Camels eliminate highly concentrated urine.
  2. The mucous membrane of the inner lining of the nasal passage of camels effectively absorbs moisture from the air that the camels exhale.

Question 39. Why erythrocytes of camels do not burst due to a huge intake of water after a prolonged thirst?
Answer:

The erythrocytes of camels are oval in shape. Each of these red blood cells has a unique ability. to expand about 240 times of its normal size. Therefore, these cells can withstand a very high osmotic pressure.

Because of this, they do not burst even after a huge intake of water to quench a prolonged thirst. This is a physiological adaptation found in camels to survive in the xeric habitat.

Question 40. Define behavioural adaptation and give an example.
Answer:

Behavioural adaptation:

The behavioural changes of any organism, which help them to survive and better adapt to the environment, are collectively known as behavioural adaptation.

Example-Termite fishing, from termite mounds with thin sticks, made from twigs, is an example of the behavioural adaptation of chimpanzees.

Question 41. What information on chimpanzees were obtained from the research of Jane Goodall?
Answer:

The research on chimpanzees by Jane Goodall revealed that wild chimpanzees show several human-like behaviours. She found that chimpanzees collect termites from termite mounds using thin sticks.

She also noticed that these wild animals consume medicinal plants, when they are infected by parasites.

Question 42. How do wild chimpanzees break hard nutshells?
Answer:

Wild chimpanzees use a piece of rock as an anvil and a strong branch of the tree as a hammer. They keep hard-shelled nuts on the rock and hammer it with the woody tree branch to crack the hard nutshell.

Question 43. What was Karl Von Frisch famous for?
Answer:

Karl Von Frisch:

German scientist, Karl Von Frisch was famous for his studies of communication among honeybees. He was the first to interpret the signals given by scout honeybees to foragers of the hive by performing typical dancing flights to indicate the distance and the direction of the source of food.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Karl Van Frisch

Question 44. What information do the scout honeybees pass to the forager bees of the hive by performing a typical bee dance?
Answer:

By performing a typical bee dance, scout honeybees inform forager bees about the distance and direction of the source of food. The dancing pattern even indicates the angle of the food source with the sun.

Question 45. What is meant by the waggle dance of honey-bees?
Answer:

The waggle dance of honey-bees:

If scout honeybees find a source of food at a distance of more than 75 metres from the hive, they perform a typical dancing flight in a certain pattern in front of their hive to pass the message to the forager bees.

They fly in alternate left and right directions to make an ‘8’ shaped course and at the middle portion of the course, they jerk the abdomen. This typical dance is known as the waggle dance.

Wbbse Class 10 Life Science Solutions

Chapter 4 Adaptation Multiple Choice Questions And Answers

Question 1. The survival strategy of living organisms is —

  1. Adaptation
  2. Reproduction
  3. Heredity
  4. Mutation

Answer: 1. Adaptation

Question 2. When attacked by parasites, which of the following animals eats the medicinal plant Aspilia rudis to get rid of it?

  1. Fish
  2. Chimp anzee
  3. Camel
  4. Tiger

Answer: 2. Chimp anzee

Question 3. The neurally controlled reaction, shown by an organism in response to any environmental signal is —

  1. Reproduction
  2. Evolution
  3. Migration
  4. Behaviour

Answer: 4. Behaviour

Question 4. Which of the following helps an animal to survive and reproduce?

  1. Migration
  2. Evolution
  3. Mutation
  4. Adaptation

Answer: 4. Adaptation

Question 5. The structural, behavioural and physiological change of an organism in the course of evolution is called —

  1. Mutation
  2. Adjustment
  3. Adaptation
  4. Accommodation

Answer: 3. Adaptation

Question 6. An example of morphological adaptation is —

  1. Production of sweat to cool the body
  2. Air sacks of pigeon
  3. Hibernation in frog
  4. Camel reserve fat in the hump

Answer: 2. Air sacks of pigeon

Question 7. A physiological adaptation Is —

  1. Air sacks of pigeon
  2. Pneumatophore of Sundari tree
  3. Air bladder of rohu fish
  4. Thom of cactus

Answer: 2. Pneumatophore of Sundari tree

Question 8. The process by which an organism can adjust with the changing environment is known as—

  1. Adaptation
  2. Evolution
  3. Mutation
  4. Variation

Answer: 1. Adaptation

Question 9. The stem of cactus is called —

  1. Rhizome
  2. Phylloclade
  3. Phyllode
  4. Underground stem

Answer: 2. Phylloclade

Question 10. The plants growing in the desert are known as —

  1. Xerophytes
  2. Hydrophytes
  3. Epiphytes
  4. Mesophytes

Answer: 1. Xerophytes

Question 11. Pneumatophores or breathing roots are seen in —

  1. Sundari (Heritiera sp.) tree
  2. Cactus
  3. Pea plant
  4. Date palm tree

Answer: 1. Sundari (Heritiera sp.) tree

Question 12. In fish, the organ, that helps it to move in different depths of water is —

  1. Air bladder
  2. Fins
  3. Scales
  4. Tail

Answer: 1. Air bladder

Question 13. The lung activity of pigeons is increased by —

  1. Keel bone
  2. Feather
  3. Air sacs
  4. Tail

Answer: 3. Air sacs

Question 14. Which is not a water-conserving adaptation of camels?

  1. Scanty sweating
  2. Concentrated urination
  3. Consumption of medicinal plants
  4. Water storage cells in the stomach

Answer: 3. Consumption of medicinal plants

Question 15. The animal having a nucleus in RBC is

  1. Camel
  2. Bird
  3. Man
  4. Mice

Answer: 1. Camel

Question 16. Myotome muscles are present in—

  1. Human
  2. Snake
  3. Fish
  4. Pigeon

Answer: 3. Fish

Question 17. Air sacks are present in —

  1. Seals
  2. Pigeons
  3. Dolphins
  4. Whales

Answer: 2. Pigeons

Question 18. ‘Waggle’ means—

  1. Wobble/Shake
  2. Circle
  3. Angle
  4. Swim

Answer: 1. Wobble/Shake

Question 19. Which of the following is not a flight muscle of a pigeon?

  1. Pectoralis major
  2. Pectoralis minor
  3. Myotome
  4. Coracobrachialis

Answer: 3. Myotome

Question 20. Which of the following plants is not a halophyte?

  1. Sundari'(Heritiera sp.) tree
  2. Cactus
  3. Garan tree
  4. Geona tree

Answer: 2. Cactus

Question 21. The organs that, help fishes in aquatic locomotion are—

  1. Lateral line sense organ
  2. Gills
  3. Venous heart
  4. Fins

Answer: 4. Fins

Question 22. The materials, which are difficult for the plants to collect from physiologically dry soil, are—

  1. CO2 and water
  2. Minerals and water
  3. Water and O2
  4. CO2 and minerals

Answer: 3. Water and O2

Question 23. Identify the animal who communicates information with other members of its own . group through specific dancing patterns regarding the sources of food.

  1. Chimpanzee
  2. Cockroach
  3. Peacock
  4. Honey bee

Answer: 4. Honey bee

Question 24. Which of the following takes part in waggle dance?

  1. Queen bee
  2. Worker bees
  3. Drones
  4. King bee

Answer: 2. Worker bees

Question 25. Decide for which of the following purposes bees demonstrate waggle dance—

  1. Search for reproductive mates
  2. Inform other worker bees about the direction and the distance of the source of food from the bee hive
  3. Selecting a place for the construction of the new bee hive
  4. Avoid attack by probable enemy

Answer: 2. Inform other worker bees about the direction and the distance of the source of food from the bee hive

Question 26. To indicate a distant source of nectar, bees perform—

  1. Angular dance
  2. Square dance
  3. Waggle dance
  4. Rotation dance

Answer: 3. Waggle dance

Question 27. The plant, which is adapted to physiologically dry soil, is—

  1. Cactus
  2. Lotus
  3. Sundari
  4. Pine

Answer: 3. Sundari

Question 28. The skin of the animals living in desert areas is—

  1. Very thin
  2. Thin
  3. Moderately thick
  4. Very thick

Answer: 4. Very thick

Question 29. The sole of the foot is padded with thick, soft muscle in—

  1. Camel
  2. Horse
  3. Buffalo
  4. Donkey

Answer: 1. Camel

Question 30. The animal, having eyelids lined by long eyelashes, is—

  1. Dog
  2. Cat
  3. Camel
  4. Fox

Answer: 3. Camel

Question 31. This animal which has a water-conserving system in the body is—

  1. Cow
  2. Horse
  3. Camel
  4. Chimpanzee

Answer: 3. Camel

Question 32. The main hydrostatic organ of fish is—

  1. Fin
  2. Lateral line sense organ
  3. Operculum
  4. Air bladder

Answer: 4. Air bladder

Question 33. Which of the following absorbs the gas of the swim bladder of bony fishes?

  1. Red gland
  2. Anterior chamber
  3. Gastric gland
  4. Rete mirabile

Answer: 4. Rete mirabile

Question 34. The common adaptive feature found in Sundari and cactus to conserve water is—

  1. Reduction of the number of leaves
  2. Modification of leaves into spines
  3. Development of pneumatophore
  4. Development of cuticle on the epidermis

Answer: 4. Development of cuticle on the epidermis

Question 35. A plant having root buttresses is—

  1. Garan
  2. Watermelon
  3. Cactus
  4. Mango

Answer: 1. Garan

Question 36. A plant growing in physically dry soil is

  1. Sundari
  2. Cactus
  3. Mango
  4. Pea plant

Answer: 2. Cactus

Question 37. In which animal, the size of RBC may increase up to 240 times of its normal size?

  1. Human
  2. Rat
  3. Camel
  4. Chimpanzee

Answer: 3. Camel

Question 38. Who discovered the language of honey bees?

  1. Niko Tinbergen
  2. Karl Von Frisch
  3. Jane Goodall
  4. Konrad Lorenz

Answer: 2. Karl Von Frisch

Chapter 4 Adaptation Answer In A Single Word Or Sentence

Question 1. To which group of plants does the cactus belong?
Answer: Xerophytes

Question 2. Why do the leaves of cactus modify into spines?
Answer: To prevent excess water loss by transpiration

Question 3. Mention one morphological adaptation of cactus to prevent transpiration.
Answer: The modification of leaves of cactus into the spine

Question 4. In which plant do you find phylloclade?
Answer: In cactus, like Opuntia

Question 5. The leaves of cactus modify into spines. In that case, which organ helps it in photosynthesis?
Answer: The green, thick and succulent stem of cactus helps in photosynthesis.

Question 6. Which part of the cactus grows elaborately to absorb water?
Answer: Root

Question 7. What do you call the changes that occur in organisms to adjust themselves to the changing environment?
Answer: Adaptations

Question 8. What do you call the adaptation that occurs in organisms to adapt themselves in the environment of their origin?
Answer: Primary adaptation

Question 9. What do you call the adaptation that occurs in organisms to adapt themselves in the environment other than their site of origin?
Answer: Secondary adaptation

Question 10. What do you call the two separate types of adaptations that occur in organisms living in two different environments?
Answer: Dual adaptation

Question 11. Who carried out extensive research on honeybees to find out their language?
Answer: Karl Von Frisch

Question 12. Which behavioural scientist was the first to carry out long research on the behaviour of wild chimpanzees?
Answer: Jane Goodall

Question 13. How many air sacs are associated with the lungs of pigeons?
Answer: Nine

Question 14. Mention one adaptive feature of the Sundari (Heritiera) tree or halophytic plants.
Answer: The presence of pneumatophores

Question 15. Name two plants with pneumatophores.
Answer: Sundari and Garan

Question 16. Name an animal with primary aquatic adaptation.
Answer: Fish

Question 17. Name an animal with secondary aquatic adaptation.
Answer: Whale

Question 18. Which fin of a fish helps to change its direction?
Answer: Tail fin

Question 19. Name an aquatic mammal.
Answer: Whale

Question 20. Name an animal with primary volant adaptation.
Answer: Pigeon

Question 21. Name an animal with secondary volant adaptation.
Answer: Flying squirrel

Question 22. How many paired and unpaired air sacs are present in pigeons?
Answer: A pigeon has 4 paired and 1 unpaired air sac.

Question 23. Which animal can absorb water with its skin?
Answer: A desert lizard, named Moloch Horridus

Question 24. Name a desert mammal.
Answer: Camel

Question 25. Which animals use thin sticks to collect termites from termite mounds?
Answer: Chimpanzees

Question 26. Which chemical plays an important role in controlling the behaviour of honeybees?
Answer: Pheromones

Question 27. Which type of dance do honeybees perform to indicate a source of food at a distance between 50-75 metres?
Answer: Circular dance

Question 28. Which type of dance do honeybees perform to indicate a source of food at a distance of more than 75 metres?
Answer: Waggle dance

Question 29. In which year did Karl Von Frisch receive the Nobel Prize for discovering the language of bee dance?
Answer: 1973

Question 30. How does the course of the waggle dance of honeybees appear?
Answer: The course of the waggle dance of honeybees appears like the English numerical ‘8.

Question 31. Why do honeybees perform waggle dance?
Answer: Honeybees perform a waggle dance to inform other members of the hive about the distance (more than 75 metres) and direction of the source of food, i.e., nectar.

Question 32. Name a plant with root buttresses.
Answer: Sundari (Heritiera fomes)

Question 33. Which portion of the body of a fish possesses the air bladder?
Answer: A fish possesses an air bladder at the upper portion of the abdominal cavity in between the intestine and kidneys.

Question 34. Which animal discharges concentrated urine to conserve body fluid?
Answer: Camel

Question 35. What is the utility of the oval RBCs in camels?
Answer: Oval RBCs of camels can move through fine blood capillaries even with very low water content in plasma.

Question 36. What is pneumatophore?
Answer: The pore on the pneumatophores of halophytes, through which it absorbs oxygen, is called a pneumatophore.

Question 37. To break the nutshells, chimpanzees use hammers and anvils. What do they use as an anvil?
Answer: Rock platform

Question 38. Name a plant with stilt roots.
Answer: Sundari

Question 39. What is a phylloclade?
Answer: Flattened, thick, green, fleshy, photosynthetic stem of the cactus is called a phylloclade.

Question 40. Which portion of the air bladder produces gas?
Answer: Red gland

Chapter 4 Adaptation Fill In The Blanks

Question 1. The leaves of cactus are modified into __________.
Answer: Spines

Question 2. In spite of being a halophyte,__________ trees do not show viviparous germination.
Answer: Sundari

Question 3. The presence of __________ in the eyes of a pigeon provides it with a sharper vision.
Answer: Pectin

Question 4. Sundari is a perennial __________.
Answer: Halophyte

Question 5. Cacti grow in a desert land, therefore, these are called __________ plants.
Answer: Xerophytic

Question 6. The body cells of __________ contain mucilage, hence these plants are capable of conserving excess water.
Answer: Cactus

Question 7. The animals, which inhabit in aquatic environment since their origin are called __________ aquatic animals.
Answer: Primary

Question 8. The animals which possess adaptive features to perform aerial movement are called __________ animals.
Answer: Volant

Question 9. Certain animals can fly for a brief period. These animals are called __________ volant animals.
Answer: Secondary

Question 10. While flying, the __________-shaped body of a pigeon facilitates it to overcome aerial resistance.
Answer: Spindle

Question 11. The ‘V-shaped __________ muscles at two lateral sides of the body help fish in swimming.
Answer: Myotome

Question 12. The air bladder of rohu fish is an example of __________ adaptation.
Answer: Morphological

Question 13. While flying, __________ the of pigeon help to supply extra energy.
Answer: Air sacs

Question 14. Excessive saline soil of the coastal area is called __________ dry soil.
Answer: Physiologically

Life Science Class 10 Wbbse

Question 15. Bee dance is performed only by the __________ bees.
Answer: Worker

Question 16. The __________ of Sundari trees help it in absorbing atmospheric oxygen for respiration.
Answer: Pneumatophores

Question 17. The red blood cells of camel can expand upto __________ times of their normal size.
Answer: 240

Chapter 4 Adaptation State True Or False

Question 1. The air bladder of fish helps it in respiration.
Answer: False

Question 2. The leaf of the cactus is modified into the spine for the reduction of the rate of transpiration.
Answer: True

Question 3. The shape of the course of the ‘waggle dance’ movement of honeybees is like the English numerical ‘8’.
Answer: True

Question 4. The RBCs of camel are small and oval in shape.
Answer: True

Question 5. Rete mirabile is present in the air bladder of fish.
Answer: True

Question 6. Pigeon has 13 air sacs associated with the lungs.
Answer: False

Question 7. If the source of food is 100 meters away, honeybees perform a circular dance.
Answer: False

Question 8. Xerophytic adaptation is seen in pigeons.
Answer: False

Question 9. Pneumatophores are seen in Sundari trees.
Answer: True

Question 10. Flying squirrel is a primary volant animal.
Answer: False

Question 11. Rohu fish is a primary aquatic animal.
Answer: True

Chapter 4 Adaptation Match The Columns

1.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Match The Columns 1

Answer: 1. D, 2. E, 3. B, 4. C, 5. A, 6. G

2.

WBBSE Solutions For Class 10 Life Science and Environment Chapter 4 Adaptation Match The Columns 2

Answer: 1. C, 2. E, 3. B, 4. A, 5. F, 6. D

Chapter 4 Adaptation Find The Odd One Out

Question 1. Phylloclade, Mucilage cells in stem, Storage of excess food, Elaborate root system
Answer: Storage of excess food

Life Science Class 10 Wbbse

Question 2. Pneumatic bones, Air sacs, Air bladder, Absence of urinary bladder
Answer: Air bladder

Question 3. The thick wax coating on the stem, Highly developed root system, Wide and large leaves, and Mucilage cells in the stem
Answer: Wide and large leaves

Question 4. Leaf spine, phylloclade, Well developed root, Pneumatophore
Answer: Pneumatophore

Question 5. Red gland, Air bladder, Rete mirabile, Air sacs
Answer: Air sacs

Question 6. Small and oval RBC, Fat in hump, Presence of air sacs, Xeric adaptation
Answer: The presence of air sacs

Question 7. Round dance, Waggle dance, Use of fishing tool, Scout bee
Answer: Use of the fishing tool

Question 8. Fishing tools, Use of rock as a hammer, Use of herbs, Air bladder
Answer: Air bladder

Chapter 4 Adaptation Fill In The Blanks By Looking At The First Pair

Question 1. Pneumatophore of Sundari tree: Breathing:: Hump of camel: ________________
Answer: Storage of fat

Question 2. Source of food is 50-60 metres away: Circular dance: Source of food is 150 metres away: ________________
Answer: Waggle dance

Question 3. Water sacs in the stomach: To store extra water:: Hump on back: ________________
Answer: To store extra fat

Question 4. Use of long fine sticks: Collection of termites from termite mound:: Use of hardwood and stone: ________________
Answer: Breaking hard nutshell

Question 5. Opuntia: Xerophyte:: Heritiera: ________________
Answer: Halophyte

Question 6. Termite eating dance: Chimpanzee Waggle dance: ________________
Answer: Honey bee

Question 7. Pigeon: Air sack:: Rohu fish: ________________
Answer: Air bladder

Question 8. Salt gland: Salt excretion:: Red gland: ________________
Answer: Gas production

Life Science Class 10 Wbbse

Chapter 4 Adaptation Among The Four Concepts Given Three Of Them Belong To One Find That

Question 1. The elaborate root system, Sparse leaves, Mucilage containing cells in the stem, and Xerophytic adaptation.
Answer: Xerophytic adaptation

Question 2. Pneumatic bones, Volant adaptation, Additional air sacs with lungs, Absence of urinary bladder and rectum.
Answer: Volant adaptation

Question 3. Water sacs in the stomach, Water conservation, Thick hairy skin, Concentrated urine.
Answer: Water conservation

Question 4. Air bladder, Red gland, Rete mirabile, Pneumatic duct
Answer: Air bladder

Question 5. Mangrove plant, Heritiera sp., Rhizophora sp., Pneumataphore
Answer: Mangrove plant

Question 6. Food collection, Animal behaviour, Round dance, Waggle dance
Answer: Animal behaviour

Chapter 4 Adaptation Advanced Questions And Answers

Question 1. What is the abiogenetic theory of life?
Answer: Greek philosopher Aristotle and Spanish physiologist, Van Helmont separately opined that life has originated spontaneously from inorganic substances in the earth. This theory is called the abiogenetic theory or biopoiesis.

Question 2. What is the venous heart? Give example.
Answer: The heart through which only de-oxygenated blood is transported is called the venous heart.

Example-Heart of fish.

Question 3. Can Darwinian selection theory explain antibiotic resistance, a property of bacteria? Support your answer with reasons.
Answer: Yes. the antibiotic resistance property of bacteria can be explained by Darwinian selection theory. According to Darwin, individuals with favourable variations are better adapted than individuals with less favourable variations.

Darwin stated that nature selects the individuals with favourable variation as these individuals are better evolved to survive in the existing environment.

When a bacterial population comes in contact with antibiotics, those which are sensitive to antibiotic dies, whereas a few bacterial colonies that are resistant to an antibiotic, survive.

Hence, the resistant bacteria multiply quickly, thereby increasing their number. Here it is observed that the advantage of an individual over another helps in the struggle for existence.

Question 4. Define adaptive radiation.
Answer: The process of rapid diversification of an organism from an ancestral species into a new form, during the change of environment which provides new resources or alters biotic interaction, or opens a new environmental niche, is termed as adaptive radiation in evolutionary biology.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Sex Determination Of Human

Chapter 3 Sex Determination Of Human Summary

The male contains 44 autosomes and two sex chromosomes—X and Y in humans (44A+XY), so two types of gametes are formed during sperm formation through meiosis, Y containing (22A+Y) and X containing (22A+X).

Females contain 44 autosomes and two X chromosomes. So, the female can produce only one type of female gamete (22A+X) through meiosis.

Males are called heterogametic as they produce two types of gametes and females are called homogametic as they produce only one type of gamete.

Embryo produced by the fusion of the Y chromosome containing sperm and female gamete develops into a male. On the other hand, an embryo produced by the fusion of an X chromosome containing sperm and a female gamete develops into a male.

Sex Determination Of Human

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Chapter 3 Sex Determination Of Human Long Answer Type Questions

Question 1. Describe the method of sex determination in humans.

Answer :

Sex determination in human

Sex determination in humans is controlled in two ways—primary sex determination and secondary sex determination.

1. Primary sex determination:

A normal human somatic cell carries 22 pairs of autosomes and 1 pair of sex chromosomes or allosomes. Autosomes control bodily features and sex chromosomes play a major role in sex determination.

The human female carries 22 pairs of autosomes and a pair of X chromosomes or the sex-determining chromosomes. Their chromosomal distribution can be represented as 44A+XX.

Human male, on the other hand, has one X and a Y chromosome as the sex-determining chromosomes along with 22 pairs of autosomes. In the case of males, the chromosomal set of a somatic cell is denoted as 44A+XY.

Meiotic cell division during gamete formation divides the chromosomes into two equal halves In the case of females, all gametes (ova) have identical chromosomal distribution, i.e., (22A+X) In males, two types of gametes are formed, the X-carrying gymnosperm (22A+X) and the Y-carrying androsperm (22A+Y).

When an androsperm (22A+Y) fertilizes an ovum (22A+X), a zygote is formed, which develops to form a male baby. On the other hand, the union of a gynosperm (22A+X) and an ovum results into the formation of a female baby. Therefore, the father of a child determines his or her sex.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Sex Determination Of Human Sex determination in human

2. Secondary sex determination:

After primary sex determination by sex chromosomes, male an female sex hormones play an important role in the development of respective gonads and the expression of the secondary sex characters.

Wbbse Class X Life Science

Chapter 3 Sex Determination Of Human Short Answer Type Questions

Question 1. Why are human females called homogametic?
Answer :

Human females called homogametic:

The chromosomal distribution of human females is 44A+XX. It means the somatic cell of a human female carries 22 pairs of autosomes and a pair of X chromosomes.

After meiosis, every ovum gets 22 autosomes and an X chromosome (22A+X). With respect to chromosomal type, a normal human female can produce identical gametes and is therefore called homogametic.

Wbbse Class 10 Life Science Solutions

Question 2. Why human males are called heterogametic?
Answer :

Human males are called heterogametic:

The chromosomal distribution of human males is 44A+XY. It means the somatic cell of a human male carries 22 pairs of autosomes and two different sex chromosomes, X and Y. After meiosis of the male germ mother cell, two different types of male gametes or sperms are formed.

Half of the sperms get 22 autosomes and an X chromosome (22A+X) and the remaining half get 22 autosomes and a Y chromosome (22A+Y). With respect to chromosomal type, a normal human male can produce two different types of gametes and is therefore called heterogametic.

Question 3. Mention the role of a mother in the sex determination of a human baby.
Answer :

The role of a mother in the sex determination of a human baby:

The human female is homogametic. All ova carry X chromosomes, so the mother plays no role in the sex determination of a human baby.

Question 4. What are sex-linked genes? Mention the types.
Answer :

Sex-linked genes:

Genes, present on either X or Y chromosomes, are known as sex-linked genes.

Types:

The genes present on the X chromosome are called X-linked and those present on the Y chromosome are called Y-linked genes.

Question 5. Mention the role of a father in the sex determination of a human baby.
Answer :

The role of a father in the sex determination of a human baby:

The human male is heterogametic. He has two types of sperms, endosperm (carrying the Y chromosome) and gymnosperm (carrying X chromosome).

The union of ovum and andosperm produces a male baby but the union of ovum with gymnosperm produces a female baby. Therefore, only the father plays a role in the sex determination of a human baby.

Question 6. What is the TDF gene?
Answer :

TDF gene:

In the case of male of human, there is a gene called testes determining factor in Y chromosome. Its alternative name is SRY and it helps in male child formation.

Wbbse Class X Life Science

Question 7. Why 22A+Y sperm is called androsperm and 22A+X is called gymnosperm?
Answer :

When 22A+Y sperm fuse with female gametes the embryo thereby formed, develop into a male child. So it is called endosperm as ‘andro’ means male.

On the other hand, if a 22A+X sperm fuse with female gametes the embryo thereby formed develop into a female child. So it is called gynosperm as ‘gyno’ means female.

Question 8. With the help of a cross show how the sex of the offspring of humans is determined. Similar Question: In many families, mothers are labeled as responsible for the birth of their daughter’s child. Demonstrate with the help of a cross, that this belief is not justified. Similar Question: Show with the help of a cross, who is more important among parents in determining the sex of their offspring.
Answer :

Differentiate between autosome and allosome.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Sex Determination Of Human Differences between autosome and allosome (1)

Chapter 3 Sex Determination Of Human MCQs

Question 1. What would be the probability of having a boy and a girl child from a marriage between a male and a female?

  1. 2:1
  2. 3:2
  3. 1:1
  4. 1:2

Answer: 3.  1:1

Question 2. Which of the following chromosomal sets is applicable for a normal human male?

  1. 44A + XX
  2. 44A + XY
  3. 44A + XXY
  4. 44A + XYY

Answer: 2.  44A + XY

Wbbse Class 10 Life Science Solutions

Question 3. The chromosomal setup of a normal human female is—

  1. 44A + XX
  2. 44A + XY
  3. 22A + X
  4. 22A + Y

Answer: 1. 44A + XX

Question 4. The number of chromosomes in a normal human somatic cell is—

  1. 44
  2. 23
  3. 22
  4. 46

Answer: 4. 46

Question 5. The number of autosomes in a normal human somatic cell is—

  1. 44
  2. 1
  3. 2
  4. 46

Answer: 1. 44

Wbbse Class X Life Science

Question 6. The number of autosomes in a normal human germ cell is—

  1. 44
  2. 22
  3. 23
  4. 46

Answer: 2.  22

Question 7. The sex chromosome present in the human male gamete is—

  1.  X or Y
  2. Only X
  3. Only Y
  4. X and Y

Answer: 1. X or Y

Question 8. Number of sex chromosomes present in a normal human somatic cell is—

  1. 46
  2. 1
  3. 44
  4. 2

Answer: 4. 2

Wbbse Class 10 Life Science Solutions

Question 9. The number of chromosomes in the germ mother cell of a human is—

  1. 46
  2. 23
  3. 44
  4. 22

Answer: 1. 46

Question 10. Determination of sex by the qualitative character of chromosomes is known as—

  1. Primary sex determination
  2. Secondary sex determination
  3. Both 1 and 2
  4. Tertiary sex determination

Answer: 1. Primary sex determination

Question 11. Which of the following does not denote a human sperm?

  1. 22A+X
  2. 22A+Y
  3. n
  4. 44A+XY

Answer: 4. 44A+XY

Question 12. Which of the following denotes endosperm?

22A+X
22A+Y
22A+Z
None of the above

Answer: 2. 22A+Y

Question 13. Which of the following gametes obtained from males is responsible for a girl child?

  1. 22A+Y
  2. 22A+X
  3. 44A+Y
  4. 44A+X

Answer: 2. 22A+X

Question 14. A man has three girl children, the chance percentage of the fourth child being a girl is—

25%
50%
75%
100%

Answer: 2.  50%

Chapter 3 Sex Determination Of Human Answer In A Single Word Or Sentence

Question 1. A male gamete carries 22A+Y chromosomes. What will be the sex of the progeny obtained after its fertilization?
Answer: If a male gamete carries 22A+Y chromosomes, the progeny will be male after its fertilization.

Question 2. What are holandric genes?
Answer: The genes, that are present on the Y chromosome, especially in mammals, are called holandric genes. The gene responsible for hairy pinna is an example of the holandric gene.

Question 3. Give an example of a Y-linked character.
Answer: Hairy pinna in human males is an example of a Y-linked character.

Question 4. What is sex determination?
Answer: The mechanism of determining the sex of an organism is known as sex determination.

Question 5. What is the name of the Y chromosome carrying sperm?
Answer: The y chromosome carrying sperm is known as an endosperm.

Question 6. What is the name of the X chromosome carrying sperm?
Answer: The x chromosome carrying sperm is known as the gymnosperm.

Wbbse Class X Life Science

Question 7. What will be the sex of a human baby if the X chromosome carrying sperm of the father unites with the ovum of the mother?
Answer: By the union of an X chromosome carrying the sperm of a father and the ovum of the mother, a female baby will be born.

Question 8. Which sperm of the father is responsible for the birth of a girl child?
Answer: An X chromosome carrying the gymnosperm of the father is responsible for the birth of a girl child.

 

Chapter 3 Sex Determination Of Human Fill In The Blanks

Question 1. The other name of the sex chromosome is ______
Answer: Allosome

Question 2. Considering the type of gamete production, ______ males are in nature.
Answer: Heterogametic

Question 3. Considering the type of gamete production, ______ females are in nature.
Answer: Homogametic

Question 4. Y chromosome is also known as ______
Answer: Androsome

Question 5. SRY is located ______ on a chromosome.
Answer: Y

 

Chapter 3 Sex Determination Of Human State True Or False

Question 1. Human sex chromosomes are of two types.
Answer: True

Question 2. Human germ cells carry 22 autosomes.
Answer: True

Question 3. The chromosomal distribution in a human female germ mother cell is 44A+XX.
Answer: True

Question 4. The chromosomal distribution of andros¬perm is 22A+Y.
Answer: True

Wbbse Class 10 Life Science Solutions

Question 5. The chromosomal distribution of gynosperm is 44A+X.
Answer: False

Question 6. Both human males and females are heterogametic in nature.
Answer: False

Question 7. If a human male gamete has 22A+Y chromosomes, after fertilization, it will produce a male offspring.
Answer: True

Question 8. A pair of sex-determining sex chromosomes are present in the ovum of humans.
Answer: False

Wbbse Class X Life Science

Chapter 3 Sex Determination Of Human Match The Columns

1.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Sex Determination Of Human Match the columns

Answer: 1. A, 2. F, 3. B, 4. E, 5. D, 6. C

Chapter 3 Sex Determination Of Human Find The Odd One Out

Question 1. Sex determination, Y chromosome, Sperm,
Answer: Autosome

Question 2. Female, Homogametic, 44A+XX, Heterogametic
Answer: Heterogametic

Question 3. Sperm, Y chromosome, Male, Female
Answer: Female

Question 4. 22A+X, 22A+Y, 44A+XY, 23A+X
Answer: 23A+X

Question 5.44A+XY, 44A+XX, Heterogamete, Male
Answer: 44A+XX

Chapter 3 Sex Determination Of Human Fill In The Blanks By Looking At The First Pair

Question 1. 46 chromosomes: Somatic Cell:: 23 chromosomes: ______
Answer: Gamete

Question 2. Daughter: 44A+XX :: Son: ______
Answer: 44A+XY

Question 3. 22A+Y: Androsperm :: 22A+X: ______
Answer: Gynosperm

Question 4. Human Autosome: 44:: Human chromosome: ______
Answer: 2

Question 5. 44A+XY : Heterogametic male :: 44A+XX : ______
Answer: Homogametic female

Wbbse Class X Life Science

Chapter 3 Sex Determination Of Human Among The Four Concepts Given Three Of Them Belong To One Find That

Question 1. The x chromosome, Y chromosome, Allosome, Sex determination
Answer: Sex determination

Question 2. Egg, 22A+X, Female, Homogametic
Answer: Female

Question 3. Male, Sperm, 22A+Y, Heterogametic sex
Answer: Male

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel’s Laws And Their Deviation

Chapter 3 Mendel’s Laws And Their Deviation Summary

  • Mendel performed experiments on pea plants to study inheritance patterns. He chose seven characters of pea plants like stem height, seed colour and shape, pod colour and shape, flower colour and position.
  • Each character has a dominant and a recessive phenotype. He performed a monohybrid cross with a single character and a dihybrid cross with two characters.
  • Mendel derived his first law i.e. the law of segregation from the monohybrid cross. This law states that characters never get mixed but rather get segregated in the second generation.
  • Mendel’s second law i.e. the law of independent assortment came from the dihybrid cross. This law states that characters not only segregate but rather are expressed in all possible combinations.
  • Deviation from Mendel’s principle is exhibited by certain organisms in terms of certain characters. The inheritance of flower colour of Mirabilis jalapa (4 o’clock plant) is one such example where the F1 hybrid organism shows an intermediate phenotype of two parents is and none of the parental phenotypes is completely dominant over the other.

Mendel's Laws And Their Deviation Summary

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Chapter 3 Mendel’s Laws And Their Deviation Long Answer Type Questions

Question 1. Explain the two conclusions of Mendel’s monohybrid cross. Rough coat (RR) is dominant and smooth coat (rr) is a recessive character for guinea pigs. If two heterozygous rough-coated guinea pigs are crossed, what will be the types of offspring in F1 generation?

Wbbse Class X Life Science
Answer:

Two conclusions of Mendel’s monohybrid cross Analysing the results of the monohybrid cross of garden pea plants, Mendel reached two conclusions. These are popularly known as the law of dominance and the law of segregation.

1. Law of dominance:

In a cross between two parents that are pure for contrasting traits, only one of the two expresses itself in the first filial generation. This is controlled by a dominant factor that helps to express the dominant trait by suppressing the recessive one. Here, the expressed one is the dominant trait and the suppressed one is the recessive trait.

2. Law of segregation:

This law states that during the formation of gametes, two alleles controlling each character, move apart due to the separation of the homologous chromosomes during meiosis. Thus, each gamete receives only one allele of each character on a random basis.

Monohybrid cross between heterozygous rough-coated guinea pigs

Let us mark the rough coat (dominant) as ‘RR’ and the smooth coat (recessive) as ‘rr

Therefore, the genotype of a heterozygous rough-coated guinea pig should be denoted as ‘Rr. The cross between two heterozygous rough-coated Rr guinea pigs is schematically represented here.

From this cross, we get two different phenotypes, 75% or 3/4th part of the progeny are rough-coated and the remaining 25% or 1/4th part are smooth-coated. Therefore, the phenotypic ratio is- Rough coat: Smooth coat = 3:1.

The genotypic ratio of this cross is- Pure rough coat: Hybrid rough coat: Pure smooth coat = 1:2:1.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Monohybrid cross between heterozygous rough coated gulinea pig

Checkerboard of F1 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Checker board of F1 generation

Question 2. In a cross between two pea plants bearing, pure violet [dominant] and pure white [recessive] flowers, what will be the ratio of pure violet and pure white-flowered plants in the F2 generation?
Answer:

The ratio of pure violet and pure white-flowered plants in F2 generation

Let us represent the dominant pure violet flower character as (W) and the recessive pure white flower character as (w).

In a cross between the pure violet (W) and pure white (w) flowered pea plants, all Fj progeny grow violet flowers. This is due to the dominant nature of the violet flower colour. If self¬pollination is allowed in the F1 plants, two different phenotypes will appear in the F2 generation.

Among the F2 progeny, 3/4th parts of plants grow violet flowers and the remaining 1/4th part of plants bear white flowers. Of these 3/4th parts of plants beaming violet flowers, 1/4th part of plants grow pure violet and 2/4th parts of plants grow hybrid violet flowers.

Therefore, among the F2 progeny, 1/4th part of plants (25%) grow pure violet, 2/4th parts of plants (50%) grow hybrid violet and the remaining 1/4th part of plants (25%) grow pure white flowers. The cross is schematically represented here with a checkerboard.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Ratio of pur vioet and pure white flowered plants

Checker board of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Checker board of F2 generation of ratio

The phenotypic and genotypic ratio of pea plant of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Phenotype and genotypic ratio of pea plant of F2 generation

Therefore, the ratio of violet and white flowered pea plants (phenotypic ratio) is 3: 1. The ratio of pure violet, hybrid violet and pure white flowered pea plants of the F2 generation (genotypic ratio) is 1: 2: 1. The ratio of pure violet and pure white flowered plants of the F2 generation is 1: 1.

Question 3. Explain with a checkerboard, how mendel reached the conclusion to his monohybrid cross experiment.
Answer :

The conclusion of Mendel’s monohybrid cross

Garden pea plants have seven different genetic characteristics, each having two distinct opposite traits.

For the monohybrid cross experiment, Mendel selected only one character, that is the height of the garden pea plant. The two opposite traits are denoted as “IT for pure tall and ‘tt’ for the pure dwarf. The steps of the experiment are mentioned here.

1. The first filial generation:

Mendel collected Pure tall plant Pure dwarf plant pollens from a flower of a pure tall pea plant and placed it on the stigma of some emasculated flowers (flowers, artificially. converted into unisexual female flowers by removing the anthers) of dwarf pea plants.

The progeny plants produced from this cross were the first filial or F1 generation, all of which were tall in phenotype.

Wbbse Class X Life Science

2. The second filial generation:

Mendel then allowed self-pollination among the F1 plants and got the F2 progeny with two different phenotypes. 75% of these were tall and 25% were dwarf. Hence the phenotypic ratio is Tall: Dwarf = 3:1., Among the F2 plants, 25% were pure tall, 50% were hybrid tall and 25% were pure dwarf.

Therefore, the genotypic ratio of a monohybrid cross is Pure tall: Hybrid tall: Pure dwarf = 1:2:1.

The cross is schematically represented here with checkerboard.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation The conclusion of Mendels momohybrid cross

Checker board of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Checker board of F2 generation in Mendals monohybrid cross

The phenotypic and genotypic ratio of pea plant of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Phenotypic and genotypic ratio of pea plant of f2 generation 3

Conclusion:

The experimental result shows that during hybridisation the parental factors, which come together in F1 generation, do not get mixed. During gamete formation of F1 generation, the factors segregate from each other and each gamete carries one factor. Thus, by analysing the result of the monohybrid cross, Mendel framed the law of segregation.

Question 4. Explain the monohybrid cross of an animal and draw a conclusion from this experiment.
Answer :

The monohybrid cross of an animal :

Monohybrid cross of an animal and its conclusion Monohybrid cross can be performed on guinea pig. The black coat (BB) of a guinea pig is a dominant trait over the white coat (bb).

In a cross between pure black and white coated guinea pigs, all F1 offsprings are black (Bb). These hybrid black F1 guinea pigs give birth to F2 offspring with different phenotypes, black and white. Out of this 75% is black and 25% is white. Therefore the phenotypic ratio is 3:1. Among F2 offspring, three different genotypes are found.

These are pure black (BB), hybrid black (Bb) and pure white (bb) in the ratio of 1:2:1.

The monohybrid cross of a guinea pig is schematically represented here

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Monohybrid cross of animal and its conclusion

Checker board of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Checker board of F2 generation in Monohybrid cross of an animal

The phenotypic and genotypic ratio of the guinea pig of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Phenotypic and genotypic ratio of guinea pig of F2 generation.

Conclusion:

The result of the monohybrid cross in between black and white guinea pigs maintains absolute similarity with the Mendelian monohybrid cross and supports the law of segregation.

Question 5. Describe. Mendel’s dihybrid cross experiment and explain the conclusion drawn from it.
Answer :

Dihybrid cross experiment:

For his dihybrid cross experiment, Mendel selected two different characters of garden pea plants, each having two distinct contrasting traits. The two characters are—

  1. Colour of  seed or cotyledon and
  2. The shape of seed. For colour of seed, the contrasting traits are yellow (YY) and green (yy), where yellow one is dominant over green. For the shape of the seed, the opposite traits are round (RR) and wrinkled (rr), where the round is dominant over wrinkled.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Dihybrid cross experiment

1. First filial generation:

Mendel crossed pure round and yellow seed-bearing pea plants (RRYY) with wrinkled and green seed-bearing pea plants (rryy). From this cross, he got the F1 hybrids with yellow cotyledons and round seeds (YyRr).

2. Second filial generation:

Mendel allowed self-pollination among the F1 plants and F2 offspring were produced. These were of four different phenotypes—round yellow, wrinkled yellow, round green, and wrinkled green in the ratio of 9:3:3:1. By analysing the genotypes of these F2 plants, Mendel found nine different combinations in the ratio of 1:2:2:4; 1:2:1:2:1.

Mendel’s dihybrid cross experiment is schematically represented here.

Checkerboard of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Checker board of F2 generation in Dihybrid cross experiment

Phenotype and genotype of F2 generation of Mendelian dihybrid cross

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Phenotype and F2 genotype of generation of Mendelian dihybrid cross

Conclusion:

From the results of this experiment, Mendel came to the following conclusions.

  1. The factors, responsible for any of the cotyledon colours and shapes of the seeds, do not get mixed in the F1 progeny. All four factors were segregated during the gamete formation of the F1 plants.
  2. The gametes, which carry the colour and shape regulating factors of pea seeds in a segregated manner, unite independently in all possible combinations to form the F2 progeny. Based on these conclusions, Mendel framed his second law of heredity or the law of independent assortment.

Question 6. Mention the seven pairs of contrasting characters of garden pea plants as selected by Mendel.
Answer :

Seven pairs of contrasting characters of garden pea plant:

For his experiment on heredity, Mendel selected seven pairs of opposite characters of garden pea plants, which are mentioned below in a table.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Constrasting characters of garden pea plant 6

Question 7. Mention the causes of Mendel’s success in his experiments on heredity with garden pea plants. Which types of gametes unite to form a male baby?
Answer :

Causes of Mendel’s success in his experiments on heredity with garden pea plants

  1. Mendel selected the pure breed of pea plants for a particular character after two years of continuous successive self-breeding.
  2. The seven pairs of factors (genes), responsible for the selected contrasting traits, are present on seven separate sets of homologous chromosomes. This feature helps in the independent assortment of the factors.
  3. Pea plants are easy to breed, the life cycle is short and therefore, Mendel could study several generations in a short time. Pea flowers are bisexual, and naturally self-pollinating, but Mendel successfully manipulated it for artificial cross¬pollination. Fie was also successful in preventing any undesired pollination.
  4. Mendel, though selected seven traits of the pea plant, but worked on one or two at a time. This helped him in the collection and analysis of the data.
  5. The characters, selected by Mendel, were sharply contrasting and showed complete dominance.
  6. Mendel used many samples at a time for breeding to get a large number of progeny, which helped him for statistical analysis of the data and to get the correct ratio between two or more contrasting characters.

Gametes are required to produce a male baby

Union of an ovum with 22 autosomes and an X chromosome (22A+X) and a sperm with 2 autosomes and a Y chromosome (22A+Y) can produce a male baby with a genotype (44A+XY).

Question 8. What is a hybrid organism? What will be the phenotype of the F1 offspring from a cross between a hybrid black and a pure white guinea pig? Explain your answer.
Answer :

Hybrid organism

The progeny, produced from a cross between parents bearing two contrasting traits with respect to a single or more character, is called a hybrid organism.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation cross between a Hybrid black and a pure white guinea pig

Checker board of F1 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Checker board of F1 generation in Hybrid black and pure white guinea

Hybridisation experiment on guinea pig

In a cross between hybrid black (Bb) and pure pHHH white (bb) guinea pigs, .two different b ‘ phenotypes will be seen among the F1 offsprings. 50% of the offspring will have “b ~ black coat and the remaining 50% will be white coated. Therefore, the phenotypic ratio will be Black: White = 1:1. The cross is schematically represented here.

Explanation:

The genotype of the hybrid black guinea pig is ‘Bb’. Therefore, this parent produces two different types of gametes, ‘B’ and ‘b! The other parent, a pure white guinea pig is homozygous recessive having a genotype of ‘bb! This parent produces only ‘b’ type of gametes. Union of ‘B’ and ‘b’ gametes produce hybrid black ‘Bb’ offspring.

On the other hand, the fertilisation between two ‘b’ gametes from two parents gives rise to pure white ‘bb’ offspring. The checker board of the F1 generation of this cross clearly shows that the proportion of hybrid black and pure white offspring is equal.

Question 9. The black coat of the guinea pig is dominant over the white coat. If a cross is performed between two hybrid black guinea pigs, what will be the result in the F1 generation? Which types of gametes unite to form a female baby?
Answer :

Crossing of hybrid black guinea pig

Let the allele for the dominant black coat colour be ‘B’ and the recessive white coat colour be ‘b’. So, the genotype of a hybrid guinea pig can be denoted as ‘Bb’.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation crossing of hybrid black guinea pig

A hybrid black-coated guinea pig may produce two different types of gametes ‘B’ 1 and ‘b’. Therefore, the F1 progeny produced from a cross between two hybrid black guinea pigs will have two different phenotypes. 75% of them will be black and the remaining 25% will be white.

Hence, the phenotypic ratio for Black :

white is 3:1. If the genotypes of F1 progeny are analysed on a checkerboard, we will see three different types—25% of it is pure black, 50% is hybrid black and the remaining 25% is pure white. Therefore, the genotypic ratio for pure black: hybrid black: pure white is 1:2:1.
The cross and the checkerboard is shown here.

checker board of F1, generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Checker board of F1 generation in Hybrid black guinea

Gametes are required to produce a female baby

A Union of an ovum with 22 autosomes and an X chromosome (22A+X) and a sperm with 22 autosomes and an X chromosome (22A+X) can produce a female baby with genotype 44A+XX.

Question 10. Describe a dihybrid cross-experiment on an animal (guinea pig) and analyse the result.
Answer :

Dihybrid cross-experiment on animal

In the dihybrid cross experiment guinea pig, the coat colour and fur character are considered. The two opposite traits for coat colour are black and white. The opposite traits for fur characters are rough and smooth. The black coat (BB) and rough fur (RR) are dominant over the white coat (bb) and smooth fur (rr) respectively.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Dihybrid cross experiment on animal

1. First filial generation:

First filial generation:

A cross is done between a black and rough-furred generation male and a white and smooth-furred female guinea pigs to get the F1 offspring. Eventually, all these hybrid offsprings are with black coats and rough fur (BbRr).

2. Second filial generation:

The F1 offspring are allowed to breed among themselves. They produce four different types of phenotypes. 9 animals of which are black-rough furred, 3 animals are black-smooth furred, 3 animals are white-rough furred and remaining 1 animal is white-smooth furred.

Therefore, the phenotypic ratio of this dihybrid cross is 9:3:3:1.

The F1 hybrids produce four different types of gametes— BR, Br, bR and br. These four different varieties of male and female gametes unite with each other in all possible combinations to produce F2 generation with nine different types of genotypes in the ratio of 1:2:2:4:1:2:1:2:1.

The checkerboard for the above-mentioned cross is represented here.

Checker board of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Checker board of F2 generation in Dihybrid cross experiment on animal

Phenotype and genotype of F2 generation of Mendelian dihybrid cross

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Phenotype and F2 genotype of generation of Mendelian dihybrid cross

Conclusion:

The result obtained from this experiment on the dihybrid cross of guinea pigs is the same as that of the Mendelian dihybrid cross of garden pea plants. This experimental result confirms that the law of independent assortment is also applicable to the heredity of the animals like guinea pigs.

Question 11. Why did Mendel select garden pea plants for his experiments on heredity? Or, Mention the main causes of the selection of pea plants sativum) by Mendel for his experiments on heredity.
Answer :

Causes of Selection of pea plants by Mendel

  1. Pea flower is naturally self-pollinating. The stamens and pistils always remain closed inside the petals, therefore, cross-pollination by insects and air does not occur. However, these flowers can be manipulated for artificial cross-pollination for experimental purposes.
  2. The bisexual pea flowers can be emasculated and transformed into female flowers by simply cutting the stamens and removing the anthers.
  3. Pea plants have distinct . contrasting phenotypic features and the characters remain unchanged for generations.
  4. The life cycle of pea plants is short, hence, a number of generations can be studied within a few years. The dormancy of seed is brief, therefore, a new generation can grow quickly.
  5. The hybrids of pea plants are fertile and these are suitable for both selling and crossing to produce the next generations.
  6. Each pea plant produces several seeds, therefore, analysis of the result can be done more accurately.

Question 12. Describe the experimental works of Mendel on garden pea plants.
Answer :

Mendel’s work on garden pea plant

  1. Mendel first visually selected seven distinct contrasting characters of the pea plant.
  2. To become sure of the homozygosity of the parental generation, he allowed the selfing of selected character-bearing plants for several successive generations.
  3. Before crossing, Mendel gently opened the petal cover to expose the stamens and cut the anthers before they mature. By this process, he emasculated the bisexual flowers to prevent undesired self-pollination.
  4. For crossing, Mendel collected pollens from a selected flower from a parental plant with a fine brush and placed it delicately on the stigma of the emasculated flowers of the other parental plant, bearing the opposite trait. He performed reciprocal crossing between two parental pea plants.
  5. After crossing, he covered the cross-pollinated flower with fine muslin cloth to prevent any undesired pollination.
  6. He collected seeds from the resultant fruits of the cross and sowed them to get the generation.
  7. Mendel meticulously studied the concerned characters of each generation and statistically analysed the result to reach a conclusion.

Through this process, Mendel performed separate experiments on different sets of contrasting traits before framing his famous laws of heredity.

Question 13. What is incomplete dominance? Explain with a suitable example.
Answer :

Incomplete dominance

In a cross between two pure contrasting trait-bearing parents, sometimes none of the traits dominates completely over the other so the phenotypic expressions of both the parents’ characters are subdued partially in the F1 generation. This event is treated as incomplete dominance.

Explanation of incomplete dominance

Among plants, incomplete dominance is clearly noticed in four o’clock (Mirabilis jalapa) plants. In this plant, two contrasting flower colours are red and white. Let us mark the pure red flower as ‘RR’ and pure white as ‘. When these two contrasting flower-bearing plants are crossed, the F1 generation grows neither red nor white flowers, instead, they grow pink flowers.

Here, phenotypic expressions of flower colour of both parents are subdued to some extent. When these F1 plants are self-pollinated, three different phenotypes are produced. These are- 25% red, 50% pink and 25% white. The genotypes of red, pink and white flower-bearing plants are ‘RR, ‘Rr’ and ‘rr. Here, both phenotypic and genotypic ratios are identical, i.e., 1: 2: 1, which is a deviation from the Mendelian monohybrid cross.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Explanation of incomplete dominance

Checker board of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Checker board of F2 generation in incomplete dominance

The phenotypic and genotypic ratio of four o’clock flower of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Phenotypic and genotypic ratio of four 0 clock flower of F2 generation

Question 14. Distinguish between—monohybrid & dihybrid cross, homozygous and heterozygous state.
Answer :

Differences between monohybrid and dihybrid cross

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Differences between monohybrid and dihybrid cross

Differences between homozygous and heterozygous state

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Differences between Homozygous state and Heterozygous state

Question 15. State the differences between pure and hybrid organisms. Mention differences between complete and incomplete dominance.
Answer :

Differences between pure and hybrid organisms

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Differences between pure and hybrid organisms

Differences between complete and incomplete dominance

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Differences between complete and incomplete dominance

Question 16. Colour of seed and shape of seed of a pea plant—taking these two characters Mendel performed a dihybrid cross. Write the genotypes of pea plants having yellow and round seeds produced in the F2 generation of this experiment. State the law of Independent Assortment of Mendel.
Answer :

The genotypes of yellow and round seeds of F2 generation are—

  1. YYRR
  2. YYRr
  3. YyRR
  4. YyRr.

Law of independent assortment

Mendel derived this law from the results of his dihybrid cross-experiment. It states that—each pair of alleles segregates independently of other pair of alleles during gamete formation.

In other words, the allele a gamete receives for one gene does not influence the allele received for another gene.

Question 17. State the opposite traits for each of the characters of flowers of the pea plant chosen by Mendel for his experiment. “For developing the scientific ideas on heredity the experiments performed by Mendel on pea plants are epochmaking.”—Mention three reasons behind his success in performing these experiments.
Answer :

Opposite traits for the characters of flowers of pea plant chosen by Mendel

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Chacters of flowers of pea plant chosen by Mendel

Three reasons behind success of Mendel’s experiment

  1. Mendel concentrated only on one or two characters at a time and kept accurate records. This prevented confusion and puzzle.
  2. The pea plant is an annual and self-pollinated plant. So plants are homozygous. Therefore the pure line of pea plants were easily available.
  3. The genes controlling characters chosen by Mendel are located on different chromosomes. This prevents the linkage and gave accurate results.

Question 18. With the help of a checker board show the types of offspring that might be produced in a cross between a hybrid black guinea pig and a pure white guinea pig. State the law of segregation as proposed by Mendel.
Answer :

Monohybrid cross in Guineapig

The black and white guinea pig will be produced by the cross in 1:1 ratio.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Monohybrid cross in Guineapig

Checkerboard

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Checker board in Monohybrid cross in Guineeapig

Phenotypic ratio – Black: White – 1: 1
Genotypic ratio – Bb: bb – 1: 1

Law of segregation

The law of segregation states that the two alleles for a heritable character segregate (separate from each other) during gamete formation and end up in different gametes.

Question 19. Tabulate three pairs of dominant-recessive traits of pea plant as selected by Mendel. State the first law of Mendel as deduced from the experiment of a monohybrid cross.
Answer :

Three pairs of dominant-recessive traits of pea plant as selected by Mendel

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Recessive traits of pea plant as selected by Mendal

Question 20. Show the result of hybridisation till F2 generation between a pure black (BB) coarse-haired (RR) guinea pig and a pure white (bb) smooth-haired (rr) guinea pig by a checkerboard. State the conclusion one can reach from this hybridisation experiment.
Answer :

Dihybrid cross in Guineapig

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Pure black and white coare haired guinea pig

F1………… Hybrid black and coarse-haired guinea pig.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Hybrid black and coarse haried guiena pig

Checkerboard of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Checker board of F2 generation

Phenotype and genotype of F2 generation

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Phenotype and genotype of F2 generation

In a cross between pure black and coarse-haired guinea pig and pure white and smooth-haired guinea pig all the F1 individuals will be black and coarse-haired guinea pig because these are dominant traits.

Each of the F1 individuals can produce four types of gametes and their cross produce four types of individuals in a ratio of 9:3:3:1.

Conclusion

The conclusion that can be reached from this hybridisation experiment i.e. —when two organisms with two or more characters with their opposite phenotypes are crossed then the characters not only get separated but remain expressed in all possible combinations independent to each other.

Chapter 3 Mendel’s Laws And Their Deviation Short Answer Type Questions

Question 1. Mention the Mendelian laws of heredity.
Answer :

Mendel proposed two laws of heredity, they are-

  1. Law of Segregation and
  2. Law of independent assortment.

Question 2. How did Mendel explain the occurrence of all tall pea plants in the F1, generation of his monohybrid cross experiment?
Answer :

In the monohybrid cross between pure tall and pure dwarf pea plants, all progeny of the F1 generation came out as tall.

From this observation, Mendel concluded that the tall character is a dominant one, which was expressed by the suppression of the recessive dwarf character. From this finding, Mendel framed a law of heredity, known as the law of dominance.

Question 3. Which type of offspring will come out of a Which type of monohybrid cross between a pure black and pure white guinea pig? Why does it happen?
Answer :

In a cross between a pure black and a pure white guinea pig, all F1 offspring will be hybrid black.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation White and black guinea pigs

In guinea pigs, the black coat colour is dominant over the white coat colour. Therefore, the gene for black coat colour will express itself by suppressing the gene for white coat colour.

Question 4. Which law of heredity did Mendel frame from the results of his famous monohybrid cross experiment? State the law.
Answer :

Law of heredity from monohybrid cross:

From the results of the monohybrid cross experiment on garden pea plants, Mendel framed the law of segregation.

Law of segregation:

During hybridisation, the factors of each character are transmitted to the F1 generation from their parents but do not get mixed. During gamete formation, these factors segregate from each other, so that each gamete receives only one factor for each character.

Question 5. What is a test cross?
Answer :

Test cross:

To determine the genotype of an individual exhibiting the dominant phenotype of a trait, it is crossed with an individual that is homozygous recessive for that trait. This cross is known as the test cross.

Example- To test the genotype of the tall F, pea plants in Mendel’s monohybrid cross, these can be crossed with homozygous recessive dwarf (tt) pea plants.

Question 6. What is a back cross?
Answer :

Back cross:

A cross between any of the parents or any individual with a genotype, identical to any of the parents, and a member of the F1, generation or any individual with a similar genotype to that of the F1 progeny is known as the back cross.

Example- A cross between “TT” and “Tt’ and ‘tt’ and ‘T’ are examples of backcrosses with respect to Mendel’s monohybrid cross.

Question 7. Explain: ‘Every test cross is a back cross’.
Answer :

‘Every test cross is a back cross’:

Across between any of the parents ‘TT’ or ‘tt’ and a member of the F, generation “Tt’ is a back cross. In the Mendelian monohybrid cross, one of the parents is homozygous recessive (tt).
When this recessive homozygous (tt) parent is used for back cross with a F1 progeny (Ti), it becomes a test cross.

Therefore, every test cross may be treated as a back cross

Question 8. What is incomplete dominance?
Answer :

Incomplete dominance:

Incomplete dominance is a genetic event where one of the alleles does not dominate completely over the other, but both the alleles express themselves partially when both are present together in the hybrid.

Question 9. What is a checkerboard? Why is it known as the Punnett square?
Answer :

Checkerboard:

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation R C Punnett

 

Checker board:

The tabular representation, of a genetic cross show- ing all genotypes of offspring, produced from every possible combination of gametes, is known as checkerboard.

Punnett Square:

Geneticist Reginald C Punnett first represented a cross in tabular form. Therefore, checker board is also known as the Punnett Square, after his name.

Question 10. Give an example of incomplete dominance in a plant and an animal.
Answer :

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Mirabilis jalapa

Example of incomplete dominance in a plant:

From a cross between red and white flowered four o’clock (Mirabilis jalapa) plants, we get pink flowers in F, generation.

Example of incomplete dominance in an animal:

If the pure black Andalusian hen is crossed with a pure white cock, the F, chicks will be blue in colour.

Question 11. Mention the contrasting traits of any two characters of the garden pea plant.
Answer :

Two characteristics of garden pea plants are height and flower colour. Opposite traits of height are tall and dwarf. Opposite traits of flower colour are violet and white.

Question 12. What will be the phenotypes of the two pea plants, when their genotypes are ‘TT’ and tt? Mention the chromosomal distribution of normal human males and females.
Answer :

Phenotypes of ‘TT’ and ‘tt’:

The phenotype of “TT” will be tall and ‘tt’ will be dwarf.

Chromosomal distribution of humans:

Chromo- somal distribution of normal human males is 44A+XY and that of females is 44A+XX.

Question 13. The genotype of a sweet pea plant bearing round and yellow seeds is ‘RrYy’. Which types of gametes may be produced from this plant?
Answer :

From a pea plant with the genotype ‘RrYy, four different types of gametes may be produced. These are ‘RY ‘Ry, ‘ry’ and ‘ry.

Question 14. In a dihybrid cross experiment, the black and rough-coated male is crossed with a white and smooth-coated female guinea pig. What will be the phenotypes of the F2 progeny?
Answer :

In a cross between a black and rough-coated male with a white and smooth-coated female guinea pig, the F, progeny will be of four types.

These are-

  1. Black and rough-coated,
  2. Black and smooth coated,
  3. White and rough coated and
  4. White and smooth coated.

Question 15. What is a genome?
Answer :

Genome:

All the genes present in the haploid set of chromosomes of an individual organism are collectively known as the genome. The genome is carried through the chromosomes of a gamete from an individual parent to its offspring.

Question 16. Mention some human characters and their respective dominant and recessive traits.
Answer :

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Charatcers and their respective dominant and recessive traits

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation curly and straight hair in human

Question 17. How did Mendel explain the combinations of different dominant and recessive factors in F2 generation of a dihybrid cross?
Answer :

In a dihybrid cross, all different types of gametes, carrying various combinations of dominant and recessive factors, unite randomly in all possible combinations to create the F2, offspring.

During this process, no factor affects the expression of any other. Mendel explained this random distribution of factors as an independent assortment.

Question 18. What are the limitations of law of segregation?
Answer :

The limitations of law of segregation:

  1. When one character is controlled by more than one gene then in that case, segregation may take place in a different ratio other than monohybrid cross.
  2. In case of incomplete dominance segregation takes place but not in a ratio of monohybrid cross.

Question 19. What is checker board?
Answer :

Checker board:

Checker board is a method of expressing different combinations of genotypes through a tabular representation of the genetic cross. It was discovered by Reginald Punnet.

Question 20. Dwarf plants are always pure’-Justify it.
Answer :

Dwarf phenotype is a recessive phenotype. In the presence of a dominant allele, the recessive one is not expressed. Therefore recessive character is always expressed in homozygous. So dwarf plants are always homozygous or pure.

Question 21. State with an example how dominant trait is expressed in the experiment of hybridisation.
Answer :

The expression of the dominant trait can be explained with the example of hybridisation between pea plants bearing white and purple-coloured flowers.

In the case of peas, the purple flower colouration is controlled by a dominant gene (designated here as ‘P’), while the white colouration is controlled by a recessive gene (designated here as ‘p’).

When these (PP and pp) were bred to create the first generation, the offsprings were pea plants bearing purple flowers, with a genotype Pp. So, although both alleles were passed down, the recessive white colour alleles were masked by the dominant purple colour alleles

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation dominant trait is expressed in the experiment of hybridisation

Question 22. What would be the ratio of phenotype and genotype of F2 generation in a monohybrid experiment in case of incomplete dominance?
Answer :

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Explanation of incomplete dominance

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation monohybrid experiment in case of incomplete dominance

F2 generation

Genotypic ratio
– RR: RW : WW
– 1: 2: 1

Question 23. What are all probable gametes to be produced from the hybrid plant YyRr produced at F1 generation in a dihybrid cross experiment?
Answer :

The probable gametes to be produced from the hybrid plant YyRr at F1, generation in a dihybrid cross experiment are four types which are YR, Yr, YR, yr.

Question 24. To establish the law of segregation in case of pea plants, you are asked to select characters. Write two such characters having opposite traits.
Answer :

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Characters having oppisite traits

Question 25. Different genotypes produce the same phenotype-Justify the statement in the form of a table by taking an example from the result of the dihybrid cross of pea plant.
Answer :

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Different genotypes produce the same phenotype

Dominant phenotypes are expressed in both homozygous and heterozygous forms. So a dominant phenotype may have two genotypes one is homozygous and one is heterozygous yellow seeded plants have a single phenotype with yellow seed colour but its genotype is either YY or Yy.

Question 26. State the limitations of the Law of segregation.
Answer :

The limitations of the Law of segregation are-

  1. When a character is controlled by more than one pair of alleles, then though segregation occurs, it may not follow the Law of Segregation.
  2. In the case of incomplete dominance, though segregation occurs, the phenotypic ratio deviates from that of Mendel’s monohybrid cross.

Question 27. State the limitations of the Law of independent assortment.
Answer :

The limitations of the Law of independent assortment are-

  1. Often the genes on the same. chromosomes have a tendency of being transmitted together. This is called linkage. In this case, Law of independent assortment is not followed.
  2. Sometimes, during gametogenesis, homologous chromosomes may pass to the same gamete. This is called non-disjunction. This phenomenon can not be explained by the Law of independent assortment.

Question 28. Differentiate between monohybrid and dihybrid cross.
Answer :

Difference between monohybrid and dihybrid cross:

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation differentiate between monohybrid and dihybrid cross

Chapter 3 Mendel’s Laws And Their Deviation Multiple Choice Questions B Answers [Mcq]

Question 1. The laws of heredity were framed by

  1. Hugo de Vries
  2. Darwin
  3. Lamarck
  4. Mendel

Answer: 4. Mendel

Question 2. Mendelian concept deviates in

  1. Law of segregation
  2. Law of dominance
  3. Law of incomplete dominance
  4. Law of independent assortment

Answer : 3. Law of incomplete dominance

Question 3. A genetic cross between hybrid offspring and one of the homozygous parental types is known as

  1. Back cross
  2. Test cross
  3.  Reciprocal cross
  4.  Final cross

Answer: 1. Back cross

Question 4.In a cross between red (WW) and white (ww) flower-bearing evening primrose plants, all plants of F1 generation bear pink (Ww) flowers. It is an example of

  1. Hybridisation
  2. Dominance
  3. Incomplete dominance
  4. Mutation

Answer : 3. Incomplete dominance

Question 5. A cross is performed between white (recessive) and violet (dominant) flower-bearing garden pea plants. What would be the percentage of white flower-bearing plants F2 generation?

  1. 100%
  2. 25%
  3. 50%
  4. 75%

Answer: 2. 25%

Question 6. What would be the genotypic ratio of the offspring from a test cross of an organism having an AaBb genotype?

  1. 1:1:1:1
  2. 1:2:1
  3. 3:1
  4. 9:3:3:1

Answer: 1. 1:1:1:1

Question 7. The findings of Mendel on heredity and the laws of inheritance are collectively known as

  1. Mendelism
  2. Monohybrid cross
  3. Genetics
  4. Dihybrid cross

Answer: 1. Mendelism

Question 8. The genotype of a pure dwarf pea plant is

  1. TT
  2. Tt
  3. tt
  4. Ttt

Answer : 3. tt

Question 9. How many pairs of contrasting traits of plants did Mendel select for carrying out his experiments on heredity?

  1. One pair
  2. Two pairs
  3. Seven pairs
  4. Nine pairs

Answer : 3. Seven pairs

Question 10. As per Mendel’s experiment on monohybrid cross, a cross is performed between a heterozygous tall (Tt) and a homozygous dwarf (tt) garden pea plant. What would be the percentage of tall pea plants in the first filial (F1,) generation?

  1. 25%
  2. 50%
  3. 75%
  4. 100%

Answer: 4. 100%

Question 11. The phenotypic ratio in F2 generation of Mendel’s monohybrid cross is

  1. 1:2:1
  2. 3:1
  3. 9:3:3:1
  4. 1:1

Answer : 3. 3:1

Question 12. The phenotypic ratio in F2 generation of Mendel’s dihybrid cross is—

  1. 1:2:1
  2.  3:1
  3. 9:3:3:1
  4. 1:1

Answer : 3. 9:3:3:1

Question 13. What would be the percentage of dwarf pea plants in F1 generation resulting from a cross between two hybrid tall pea plants?

  1. 25%
  2. 50%
  3. 75%
  4. 100%

Answer: 1. 25%

Question 14. Dwarf pea plants are always—

  1. Homozygous
  2. Heterozygous
  3. Hybrid tall
  4. Hemizygous

Answer: 1.  Homozygous

Question 15. Name the Mendelian law that is derived from the dihybrid cross.

  1. Law of segregation
  2. Law of incomplete dominance
  3. Law of dominance
  4. Law of independent assortment

Answer: 4.  Law of independent assortment

Question 16. A cross is performed between two hybrid tall pea plants. What would be the percentage of tall and dwarf offspring respectively, in the F1 generation of this cross?

  1. 50% and 25%
  2. 25% and 25%
  3. 75% and 25%
  4. 50% and 50%

Answer : 3.  75% and 25%

Question 17. The pea plants, produced from the seeds resulting from a cross between hybrid tali (Tt) and pure dwarf (tt) pea plants will be—

  1. All tall
  2. All dwarf
  3. 50% tall and 50% dwarf
  4. 75% tall and 25% dwarf

Answer : 3.  50% tall and 50% dwarf

Question 18. The numerical ratio of pure black and pure white offspring obtained in the F2 generation of a monohybrid cross between pure black (BB) and pure white (bb) guinea pig, is—

  1. 1:1
  2. 1:2
  3. 2:1
  4. 3:1

Answer: 1. 1:1

Question 19. The numerical ratio of pure white and hybrid black offsprings obtained in the F2 generation of a monohybrid cross between pure black (BB) and pure white (bb) guinea pig, is—

  1. 1:1
  2. 1:2
  3. 3:1
  4. 2:1

Answer: 2. 1:2

Question 20. The ratio of pure genotypes produced in F2 generation of a monohybrid cross is—

  1. 1:1
  2. 1:2
  3. 1:3
  4. 3:1

Answer: 1. 1:1

Question 21. The percentage of hybrid tall offspring obtained from a cross between two hybrid tall pea plants is—

  1. 25%
  2. 50%
  3. 75%
  4. 100%

Answer: 2. 50%

Question 22. The ABO blood group in humans is an example of— a”

  1. Dominance
  2. Codominance
  3. Polygene
  4. Incomplete dominance

Answer: 2. Codominance

Question 23. Mendel’s Law of segregation is derived from—

  1. Monohybrid cross
  2. Dihybrid cross
  3. Hybridisation
  4. Variation

Answer: 1. Phenotypic

Question 24. The 1:2:1 ratio in the F2 generation of Mendel’s monohybrid cross is—

  1. Phenotypic
  2. Genotypic
  3. Homozygotic
  4. Heterozygotic

Answer: 2. Genotypic

Question 25. Self-pollination is possible in garden pea flowers because it is a—

  1. Male flower
  2. Female flower
  3. Unisexual flower
  4. Bisexual flower

Answer: 4.  Bisexual flower

Question 26. Which of the following indicates the test cross of Mendel’s monohybrid experiment?

  1. Tt x TT
  2. Tt x Tt
  3. Tt x tt
  4. TT x TT

Answer:  3. Tt x tt

Question 27. In which of the following cases, both phenotypic and genotypic ratios are 1:2:1 in F2 generation of a monohybrid cross?

  1. Complete dominance
  2. Incomplete dominance
  3. Super dominance
  4. Mendelian dominance

Answer:  2. Incomplete dominance

Question 28. The allelic state or the genetic constitution of an organism is known as—

  1. Allelomorph
  2. Gene
  3. Genotype
  4. Phenotype

Answer : 3. Genotype

Question 29. Which of the following is a dominant character?

  1. Stem height-Dwarf
  2. Shape of seed-Wrinkled
  3. Cotyledon colour-Yellow
  4. Flower colour-White

Answer : 3. Cotyledon colour-Yellow

Question 30. Which of the following is a recessive character?

  1. Wrinkled seed
  2. Purple flower
  3. Yellow coloured seed
  4. Axillary flower

Answer: 1. Wrinkled seed

Question 31. o’ is a symbol of—

  1. Bisexual
  2. Male
  3. Female
  4. None of the above

Answer: 2. Male

Question 32. Which of the following cross will produce a 1:1 ratio?

  1. TTxtt
  2. TTxTt
  3.  tTtxTt
  4. Ttxtt

Answer: 4. Ttxtt

Question 33. How many genotypes are possible of a tall pea plant?

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 2. 2

Question 34. How many individuals out of 500 will be heterozygous in the cross Tt x Tt?

  1. 100
  2. 250
  3. 350
  4. 500

Answer: 2. 250

Question 35. How many types of genotypes are obtained in Mendel’s monohybrid cross in the F2 generation?

  1. 1
  2. 2
  3. 3
  4. 4

Answer : 3. 3

Question 36. A test cross is done between—

  1. F1x F1
  2. F1 x dominant parent
  3. F1 x recessive parent
  4. None of the above

Answer : 3. F1 x recessive parent

Question 37. The phenotypic ratio of offspring obtained from a cross between a homozygous (bb) and heterozygous (Bb) individual—

  1. 3:1
  2. 2:1
  3. 1:1
  4. 1:2

Answer : 3.  1:1

Question 38. How many types of gametes will be produced from YyRr?

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 4.  4

Question 39. How many types of gametes will be produced from the genotype YYRR?

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 1. 1

Question 40. How many types of genotypes are obtained in Mendel’s dihybrid cross F2 generation?

  1.  4
  2. 6
  3. 8
  4. 9

Answer: 4. 9

Question 41. Which of the following traits in the pea plant is recessive?

  1. Wrinkled seed
  2. Yellow coloured seed
  3. Purple coloured flower
  4. Axial flower

Answer: 1. Wrinkled seed

Question 42. How many types of gametes are formed from pea plants having genotype YyRr?

  1. 1
  2. 4
  3. 2
  4. 3

Answer: 2. 4

Question 43. Which of the following is the genotypic ratio generation of Mendel’s monohybrid cross?

  1. 1:2:1
  2. 3:1
  3. 9:3:3:1
  4. 2:1:2

Answer: 1. 1:2:1

Question 44. What would be the phenotypic ratio in the F2 generation of a monohybrid cross in case of incomplete dominance?

  1. 3:1
  2. 2:1:1
  3. 9:3:3:1
  4. 1:2:1

Answer: 4. 1:2:1

Question 45. Asses how many types of gametes are produced from the pea plant having the genotype RRYY.

  1. One type
  2. Four types
  3. Two types
  4. Three types

Answer: 1. One type

Question 46. Identify which of the following is a dominant trait—

  1. Length of the stem—Dwarf
  2. The shape of the seed—Wrinkled
  3. Colour of the cotyledon—Yellow
  4. Colour of the flower—White

Answer : 3. Colour of the cotyledon—Yellow

Question 47. Select which of the following two genotypes are responsible for the expression of the phenotype wrinkled yellow in pea plants—

  1. RRYY and rryy
  2. RRYy and RrYy
  3. RRyy and Rryy
  4. rrYY and rrYy

Answer: 4. rrYY and rrYy

Question 48. Identify the genotype of a guinea pig having black colour and rough hair—

  1. BbRr, BBRr
  2. BBrr, Bbrr
  3. bbRR, bbRr
  4. bbrr, bbRr

Answer: 1.  BbRr, BBRr

Question 49. Decide which of the following two were selected by Mendel as recessive traits—

  1. Colour of flower—purple, the position of the flower—axial
  2. Length of stem—dwarf, form of ripe seed—wrinkled
  3. Form ol ripe seed—round, colour of seed—yellow
  4. Position of flower—axial, length of stem—tall

Answer: 2. Length of stem—dwarf, form of ripe seed—wrinkled

 

Chapter 3 Mendel’s Laws And Their Deviation VSAQs

Question 1. Write down the scientific name of the garden pea plant.
Answer: Pisum sativum

Question 2. What is the phenotypic ratio in F2 generation of a monohybrid cross?
Answer: The phenotypic ratio in F2 generation of a monohybrid cross is, Dominant: Recessive=31.

Question 3. What is the genotypic ratio in F2 generation of a monohybrid cross?
Answer: The genotypic ratio in F2 generation of a monohybrid cross is—Pure dominant : Hybrid: Pure recessive = 1:2:1.

Question 4. Give an example of a dominant character of guinea pig.
Answer: Black coat colour is a dominant character of guinea pig.

Question 5. In a monohybrid cross, the phenotypic and genotypic ratio of F2 generation are 3:1 and 1:2:1 respectively. What does this observation indicate?
Answer: This observation indicates the complete dominance of the concerned character.

Question 6. What does ‘Tt’ indicate in Mendel’s experiment?
Answer: In Mendel’s experiment, ‘Tt’ indicates the genotype of a hybrid tall pea plant.

Question 7. Which type of pea plant is denoted by ‘TT’ in Mendel’s experiment?
Answer: In Mendel’s experiment, ‘TT’ indicates a pure tall pea plant.

Question 8. Which type of pea plant is denoted by ‘tt’ in Mendel’s experiment?
Answer: In Mendel’s experiment, ‘tt’ indicates a pure dwarf pea plant.

Question 9. Mention a character and its opposite traits found in guinea pig.
Answer: Coat colour is a character of guinea pig, which has two opposite traits—black and white.

Question 10. Which is the dominant colour of flowers of garden pea plants?
Answer: The dominant colour of flowers of garden pea plant is violet.

Question 11. Which ratio indicates the external features of the progeny of a hybrid cross?
Answer: Phenotypic ratio

Question 12. In which journal Mendel’s paper on the heredity of garden pea plants was published?
Answer: Mendel’s paper on the heredity of garden pea plants was published in the Journal of Natural History Society at Brunn.

Question 13. What does the ratio of 3:1 indicate in genetics?
Answer: The ratio of 3:1 indicates the phenotypic ratio of the F2 progeny of the monohybrid cross.

Question 14. What does the ratio ail :2:1 indicate in genetics?
Answer: The ratio of 1:2:1 indicates the genotypic ratio of the F2 progeny of monohybrid cross

Question 15. What does the ratio of ot9:3:3:1 indicate in genetics?
Answer: The ratio of 9:3:3:1 indicates the phenotypic ratio ot the F2 progeny of dihybrid cross.

Question 16. Which Mendelian law is derived from the result ot a monohybrid cross experiment?
Answer: The law of segregation is derived from the result ol Mendel s monohybrid cross experiment.

Question 17. Which Mendelian law is derived from the result of a dihybrid cross experiment?
Answer: The law of independent assortment is derived from the result of Mendel’s dihybrid cross experiment.

Question 18. Who first designed the checker board?
Answer: Reginald C Punnett, a British geneticist, first designed the checker board.

Question 19. What does Mendelian factors mean?
Answer: Mendelian factors are the alleles of genes, which are responsible for expression of various hereditary characters

Question 20. What is mulatto?
Answer: The medium complexioned offspring, bom to pure black and pure white parents, is called mulatto

Question 21. In case of guinea pig whether the phenotype of the two genotypes bbRR and bbRr is same?
Answer: Yes, the phenotype of the two genotypes bbRR and bbRr will be the same in case of guinea pig.

Question 22. Write the phenotypic ratio obtained in the F2 generation of Mendel’s dihybrid cross experiment.
Answer: 9:3:3:1

Question 23. Write the genotypic ratio obtained at 2nd filial generation in the monohybrid experiment done by Mendel.
Answer: 1:2:1

Question 24. Which law did Mendel conclude from his dihybrid cross experiment?
Answer: Law of independent assortment.

 

Chapter 3 Mendel’s Laws And Their Deviation Fill In The Blanks

Question 1. An individual which carries both dominant and recessive alleles is called _______ individual.
Answer: Hybrid

Question 2. Mendel’s first law of heredity is known as the law of ______
Answer : Segregation

Question 3. Mendel’s second law of heredity is known as the law of _______
Answer: Independent assortment

Question 4. In a monohybrid cross between red and white flower-bearing evening primrose plants, the phenotypic ratio of the F2 progeny is ______
Answer: 1:2:1

Question 5. The parental generation of a monohybrid cross is marked as ______ generation.
Answer : P

Question 6. Mendel performed ______ pollination between the flowers of pure tall and pure dwarf pea plants.
Answer: Cross

Question 7. To get F2 generation, Mendel allowed ______ pollination among F1 plants.
Answer: Self

Question 8. The yellow colour of cotyledons of pea plants is a ______character.
Answer: Dominant

Question 9. Hybridisation between two traits of a single character is known as ______ cross.
Answer: Monohybrid

Wbbse Class X Life Science

Question 10. Permanent and heritable change in the number and structure of gene or chromosome is known as _______
Answer: Mutation

Question 11. Mendel selected ______ number of characters of pea plant for his experiment.
Answer: Seven

Question 12. Different ______ of an individual may show the same phenotype.
Answer: Genotypes

Question 13. The white and smooth fur of a guinea pig is a __________ character.
Answer: Recessive

Question 14. The ratio of TT:Tt:tt in monohybrid cross is ______
Answer: 1:2:1

 

Chapter 3 Mendel’s Laws And Their Deviation State True Or False

Question 1. A temporary change in the number and structure of a gene or chromosome is known as mutation.
Answer: False

Question 2. Mendel selected seven pairs of opposite characters of the garden pea plant for his experimental observations.
Answer: True

Wbbse Class X Life Science

Question 3. The scientific name of evening primrose is Mirabilisjalapa.
Answer: False

Question 4. In F2 generation of Mendel’s dihybrid cross, both the phenotypic and genotypic ratio is 9: 3:3:1.
Answer: False

Question 5. In Mendel’s monohybrid cross, all F1 offsprings have pure dominant character.
Answer: False

Question 6. In genetics, cross denotes the union of a male and a female gamete.
Answer: True

Question 7. Round seed is a recessive character.
Answer: False

Question 8. Each character selected by Mendel was controlled by multiple alleles.
Answer: False

Question 9. In pea plant, both selfing and cross-pollination can be done.
Answer: True

Question 10. Mendel obtained 75% tall plant in F1 generation of a monohybrid cross.
Answer: False

Wbbse Class 10 Life Science Solutions

Question 11. Back cross always means test cross.
Answer: False

Question 12. In the Evening primrose plant incomplete dominance is seen.
Answer: True

Question 13. Recessive character is always expressed in homozygous conditions.
Answer: True

Wbbse Class X Life Science

Question 14. The dominant phenotype may be both homozygous or heterozygous.
Answer: True

Question 15. In his monohybrid cross experiment, Mendel obtained 75% pure tall pea plants in the first filial generation.
Answer: False

Question 16. Mendel used the term ‘gene’ while describing his experiments related with heredity.
Answer: False

 

Chapter 3 Mendel’s Laws And Their Deviation Match The Columns

1.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Match the columns 1

Answer: 1. B, 2. C, 3. E, 4. F, 5. A, 6. G

2.

WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel's Laws And Their Deviation Match the columns 2

Answer: 1. B, 2. F, 3. D, 4. E, 5. A, 6. G

 

Chapter 3 Mendel’s Laws And Their Deviation Find The Odd One Out

Question 1. Grefcn colour of the pod, White colour of the flower, Round shape of seed, Axial position of the flower
Answer: The white colour of the flower

Question 2. Crossing over, Meiosis, Genetic recombination, Mitosis
Answer: Mitosis

Question 3. Law of segregation, Law of independent assortment, Law of use and disuse, Law of dominance
Answer: Law of use and disuse

Wbbse Class X Life Science

Question 4. 9:3:3:1, independent assortment, 3:1 BbSs x BbSs
Answer: 3:1

Question 5. TT, Tt, tt, rr
Answer: rr

Wbbse Class 10 Life Science Solutions

Question 6. Round wrinkled, Yellow wrinkled, Green wrinkled, Tall
Answer: Tall

Question 7. Purple flower x white flower Yellow seed x green seed, Round seed x wrinkled seed, Yellow Round Seed x green wrinkled seed
Answer: Yellow Round seed x Green Wrinkled seed

Question 8. Purple flower, White flower, Green pea seed, Axial flower
Answer: White flower

 

Chapter 3 Mendel’s Laws And Their Deviation Fill In The Blanks By Looking At The First Pair

Question 1. Monohybrid cross: Law of segregation:: Dihybrid cross: ______
Answer: Law of independent assortment

Question 2. Purple flower: Dominant character:: White flower:______
Answer: Recessive character

Wbbse Class X Life Science

Question 3. Monohybrid phenotype ratio : 3:1 :: Dhybridphenotypic ratio :______
Answer: 9:3:3:1

Question 4. Mendelian inheritence: Pea plant :: Mendel’s deviation:______
Answer: Mirabilis Jalapa

Question 5. Mendel’s first law: Law of segregation:: Mendel’s second law: ______
Answer: Law of independent assortment

Question 6. Cross of P generation : Cross-pollination :: Cross of F1 generation:______
Answer: Self-pollination

Question 7. Axial flower: Dominant Character:: Terminal flower: ______
Answer: Recessive character

 

Chapter 3 Mendel’s Laws And Their Deviation Among The Four Concepts Given, Three Of Them Belong To One. Find That

Question 1. Monohybrid cross, Dihybrid cross, Hybridisation technique, Crossbreeding
Answer: Hybridisation technique

Question 2. TT x tt, Monohybrid Cross, Tt x Tt, BB x bb
Answer: Monohybrid Cross

Wbbse Class 10 Life Science Solutions

Question 3. Green Round, F2 phenotype, Yellow Wrinkled, Yellow Round
Answer: F2 Phenotype

Question 4. Mirabilis jalapa, Pink flower, Incomplete dominance, 1:2:1
Answer: Incomplete dominance

Question 5. Purple flower, Tall plant, Yellow seed, Dominant Character
Answer: Dominant character