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).

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