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Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 10 Cell Cycle and Cell Division Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division

PSEB 11th Class Biology Guide Cell Cycle and Cell Division Textbook Questions and Answers

Question 1.
What is the average cell cycle span for a mammalian cell ?
Answer:
The average cell cycle span for a mammalian cell is 24 hours.

Question 2.
Distinguish cytokinesis from karyokinesis?
Answer:
Cytokinesis is the division of cytoplasm, whereas karyokinesis is the division of nucleus of the cell.

Question 3.
Describe the events taking place during interphase.
Answer:
The interphase, though called the resting phase, is the time during which the cell is preparing for division by undergoing cell growth and DNA replication. The interphase is divided into three further phases
(i) G₁ – phase (Gap 1)
(ii) S – phase (Synthesis)
(iii) G₂ – phase (Gap 2)
G₁ – phase corresponds to the interval between mitosis and initiation of DNA replication. During Ga-phase the cell is metabolically active and continuously grows but does not replicate its DNA. S or synthesis phase marks the period during, which, DNA synthesis or replication takes place. During this time, the amount of DNA per cell doubles.

During the G₂ – phase, proteins are synthesised in preparation for mitosis, while cell growth continues.
Cells in this stage remain metabolically active but no longer proliferate unless called on to do so depending on the requirement of the organism.

Question 4.
What is G₀ (quiescent phase) of cell cycle?
Answer:
G₀ is the quiescent stage of the cell cycle. It is also known as inactive stage of the cell cycle. Cells in this stage remain metabolically remain active, but no longer proliferate unless called on to do so depending on the requirement of the organisms.

Question 5.
Why is mitosis called equational division?
Answer:
Mitosis is the kind of cell division in which daughter cells have the same number and kind of chromosomes as that of parent cell. Mitosis is therefore, also known as equational division.

Question 6.
Name the stage of cell cycle at which one of the following events occur:
(i) Chromosomes are moved to spindle equator.
(ii) Centromere splits and chromatids separate.
(iii) Pairing between homologous chromosomes takes place.
(iv) Crossing over between homologous chromosomes takes place.
Answer:
(i) Prophase,
(ii) Anaphase,
(iii) Zygotene stage of meiosis-I,
(iv) Pachytene stage.

Question 7.
Describe the following:
(i) Synapsis
(ii) Bivalent
(iii) Chiasmata
Draw a diagram to illustrate your answer.
Answer:
(i) Synapsis: During meiosis-I, the process of pairing of two homologous chromosomes is known as synapsis. It is so exact that pairing is not merely* between corresponding chromosomes but between corresponding individual units.

(ii) Bivalent: The complex formed by a pair of synapsed homologous chromosomes is called a bivalent or a tetrad. It consists of four chromatids.

(iii) Chiasmata: The chiasmata formation is the indication of completion of crossing over and beginning of separation of chromosomes. The chiasma is formed when the chromosomal parts begin to repel each other except in the region where these are in contact. Thus, chiasmata formation is necessary for the separation of homologous chromosomes.

Question 8.
How does cytokinesis in plant cells differ from that in animal cells?
Answer:
Cytokinesis in an Animal Cell: It occurs by a process called cleavage. A cleavage furrow appears and at the site of the cleavage furrow the cytoplasm has a ring of microfilaments made of actin associated with molecules of the protein myosin. This ring of proteins will contract causing the furrow to deepen, which will then in turn pinch the cell into two separate cells.

Cytokinesis in a Plant Cell: In a plant cell, vesicles containing cell wall material collect at the middle of the parent cell. They will fuse and form a membranous cell plate. This plate will grow outward and it accumulates more cell wall material. Eventually, the plate will fuse with the plasma membrane and the cell plate contents will join the parental cell wall. This results in two different daughter cells.

Question 9.
Find examples where the four daughter cells from meiosis are equal in size and where they are found unequal in size.
Answer:
The four daughter haploid cells may or may not be equal in size at the end of meiosis-II with telophase-II

Question 10.
Distinguish anaphase of mitosis from anaphase-1 of meiosis. Ans. Differences between anaphase of mitosis and anaphase-I of meiosis.
Answer:

Anaphase of Mitosis	Anaphase-I of Meiosis
1. Each chromosome arranged at the metaphase plate is split simultaneously and the two daughter chromatids, begin to migrate towards the two opposite poles.	The spindle fibres contract and pull the centromeres of homologous chromosomes towards the opposite poles.
2. The centromere of each chromosome is towards the pole with arms of chromosome trailing behind.	The centromere is not divided, so half of the chromosomes of parent nucleus go to one pole and the remaining half in the opposite pole.
3. In this stage (a) centromeres split and chromatids separate. (b) chromatids move to opposite poles.	In this stage (a) homologous chromosomes separate. (b) sister chromatids remain associated at their centromere.

Question 11.
List the main differences between mitosis and meiosis.
Answer:

Mitosis	Meiosis
1. The division occurs in somatic cells.	It occurs in reproductive cells.
2. It is a single division.	It is a double division.
3. The daughter cells resemble each other as well as their mother cell.	The daughter cells neither resemble one another nor their mother cell.
4. Replication of chromosomes occurs before every mitotic division.	Replication of chromosomes occurs only once though meiosis is a double division.
5. Mitosis does not introduce variations.	Meiosis introduces variations.
6. Mitosis is required for growth, repair and healing.	Meiosis is involved in sexual reproduction.

Question 12.
What is the significance of meiosis?
Answer:
Meiosis is the mechanism by which conservation of specific chromosome number of each species is achieved across generations in sexually reproducing organisms.
It also increases the genetic variability in the population of organisms from one generation to the next. Variations are very important for the process of evolution.

Question 13.
Discuss with your teacher about:
(i) haploid insects and lower plants where cell-division occurs, and
(ii) some haploid cells in higher plants where cell-division does
not occur.
Answer:
(i) In some insects and lower plants fertilization is immediately followed by zygotic meiosis, which leads to the production of haploid organisms. This type of life cycle is known as haplontic life cycle.

(ii) The phenomenon of polyploidy can be observed in some haploid cells in higher plants in which cell division does not occur. Polyploidy is a state in which cells contain multiple pairs of chromosomes than the basic set. Polyploidy can be artificially induced in plants by applying colichine to cell culture.

Question 14.
Can there be mitosis without DNA replication in S-phase?
Answer:
No, DNA replication is required for doubling the chromosome number.

Question 15.
Can there be DNA replication without cell division?
Answer:
Yes, DNA replication takes place during the synthesis stage of interphase in the cell cycle. It is totally independent of cell division. After G₂ -phase a cell may or may not enter into the M-phase.

Question 16.
Analyse the events during every stage of cell cycle and notice how the following two parameters change?
(i) Number of chromosomes (n) per cell
(ii) Amount of DNA content (C) per cell.
Answer:
(i) The number of chromosomes (n) is double in the prophase, metaphase and anaphase stage of an organism. In the telophase stage, during daughter cell formation, the number of chromosomes get reduced to half.

(ii) Amount of DNA content (C) per cell also changes during different phases of cell division. It is double in the prophase, metaphase and anaphase stage of an organism. In the telophase stage, during daughter cell formation, the number of chromosomes get reduced to half than that of their parents. ’

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 2 Sexual Reproduction in Flowering Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

PSEB 12th Class Biology Guide Sexual Reproduction in Flowering Plants Textbook Questions and Answers

Question 1.
Name the parts of an angiosperm flower in which development of male and female gametophyte take place.
Answer:
The male gametophyte or the pollen grain develops inside the pollen chamber of the anther, whereas the female gametophyte (also known as the embryo sac) develops inside the nucellus of the ovule from the functional megaspore.

Question 2.
Differentiate between microsporogenesis and megasporogenesis. Which type of cell division occurs during these events? Name the structures formed at the end of these two events.
Answer:

Microsporogenesis	Megasporogenesis
1. It is the process of the formation of microspore tetrads from a microspore mother cell through meiosis.	It is the process of the formation of the four megaspores from a megaspore mother cell in the region of the nucellus through meiosis.
2. It occurs inside the pollen sac of the anther.	It occurs inside the ovule.

(b) Both events (microsporogenesis and megasporogenesis) involve the process of meiosis or reduction division which results in the formation of haploid gametes from the microspore and megaspore mother cells.

(c) Microsporogenesis results in the formation of haploid microspores from a diploid microspore mother cell. On the other hand, megasporogenesis results in the formation of haploid megaspores from a diploid megaspore mother cell.

Question 3.
Arrange the following terms in the correct developmental sequence:
Pollen grain, sporogenous tissue, microspore tetrad, pollen mother cell, male gametes.
Answer:
The correct developmental sequence is as follows:
Sporogenous tissue, pollen mother cell, microspore tetrad, pollen grain, male gametes.
During the development of microsporangium, each cell of the sporogenous tissue acts as a pollen mother cell and gives rise to a microspore tetrad, containing four haploid microspores by the process of meiosis (microsporogenesis). As the anther matures, these microspores dissociate and develop into pollen grains. The pollen grains mature and give rise to male gametes.

Question 4.
With a neat, labelled diagram, describe the parts of a typical ’ angiosperm ovule.
Answer:
An ovule is a female megasporangium where the formation of megaspores takes place.

The various parts of a typical angiospermic ovule are as follows :
1. Funiculus: It is a stalk-like structure which represents the point of attachment of the ovule to the placenta of the ovary.

2. Hilum: It is the point where the body of the ovule is attached to the funiculus.

3. Integuments: They are the outer layers surrounding the ovule that provide protection to the developing embryo.

4. Micropyle: It is a narrow pore formed by the projection of integuments. It marks the point where the pollen tube enters the ovule at the time of fertilisation.

5. Nucellus: It is a mass of the parenchymatous tissue surrounded by the integuments from the outside. The nucellus provides nutrition to the developing embryo. The embryo sac is located inside the nucellus.

6. Chalaza: It is the based swollen part of the nucellus from where the integuments originate.

Question 5.
What is meant by monosporic development of female gametophyte?
Answer:
The female gametophyte or the embryo sac develops from a single functional megaspore. This is known as monosporic development of the female gametophyte. In most flowering plants, a single megaspore mother cell present at the micropylar pole of the nucellus region of the ovule undergoes meiosis to produce four haploid megaspores. Later, out of these four megaspores, only one functional megaspoxe develops into the female gametophyte, while the remaining three degenerate.

Question 6.
With a neat diagram explain the 7-celled, 8-nucleate nature of the female gametophyte.
Answer:
Structure of the mature embryo sac

The female gametophyte (embryo sac) develops from a single functional megaspore. This megaspore undergoes three successive mitotic divisions to form eight nucleate embryo sacs.
The first mitotic division in the megaspore forms two nuclei. One nucleus moves towards the micropylar end while the other nucleus moves towards the chalazal end. Then, these nuclei divide at their respective ends and re-divide to form eight nucleate stages. As a result, there are four nuclei each at both the ends i.e., at the micropylar and the chalazal end in the embryo sac. At the micropylar end, out of the four nuclei only three differentiate into two synergids and one egg cell. Together they are known as the egg apparatus.

Similarly, at the chalazal end, three out of four nuclei differentiates as antipodal cells. The remaining two cells (of the micropylar and the chalazal end) move towards the centre and are known as the polar nuclei, which are situated in a large central cell. Hence, at maturity, the female gametophyte appears as a 7-celled structure, though it has 8-nucleate.

Question 7.
What are chasmogamous flowers? Can cross-pollination occur in cleistogamous flowers? Give reasons for your answer.
Answer:
There are two types of flowers present in plants namely Oxalis and Viola – chasmogamous and cleistogamous flowers. Chasmogamous flowers have exposed anthers and stigmata similar to the flowers of other species.

Cross-pollination cannot occur in cleistogamous flowers. This is because cleistogamous flowers never open at all. Also, the anther and the stigma lie close to each other in these flowers. Hence, only self-pollination is possible in these flowers.

Question 8.
Mention two strategies evolved to prevent self-pollination in flowers.
Answer:
Self-pollination involves the transfer of pollen from the stamen to the pistil of the same flower. Two strategies that have evolved to prevent self-pollination in flowers are given on next page:
1. Self-incompatibility: In certain plants, the stigma of the flower has the capability to prevent the germination of pollen grains and hence, prevent the growth of the pollen tube. It is a genetic mechanism to prevent self-pollination called self-incompatibility. Incompatibility may be between individuals of the same species or between individuals of different species. Thus, incompatibility prevents breeding.

2. Protandry: In some plants, the gynoecium matures before the androecium or vice-versa. This phenomenon is known as protogyny or protandry respectively. This prevents the pollen from coming in contact with the stigma of the same flower.

Question 9.
What is self-incompatibility? Why does self-pollination not lead to seed formation in self-incompatible species?
Answer:
Self-incompatibility is a genetic mechanism in angiosperms that prevents self-pollination. It develops genetic incompatibility between ‘ individuals of the same species or between individuals of different
species.

The plants which exhibit this phenomenon have the ability to prevent germination of pollen grains and thus, prevent the growth of the pollen tube on the stigma of the flower. This prevents the fusion of the gametes along with the development of the embryo. As a result, no seed formation takes place.

Question 10.
What is bagging technique? How is it useful in a plant breeding programme?
Answer:
Various artificial hybridisation techniques (under various crop improvement programmes) involve the removal of the anther from bisexual flowers without affecting the female reproductive part (pistil) through the process of emasculation. Then, these emasculated flowers are wrapped in bags to prevent pollination by unwanted pollen grains. This process is called bagging.

This technique is an important part of the plant breeding programme as
it ensures that pollen grains of only desirable plants are used for fertilisation of the stigma to develop the desired plant variety.

Question 11.
What is triple fusion? Where and how does it take place? Name the nuclei involved in triple fusion.
Answer:
Triple fusion is the fusion of the male gamete with two polar nuclei inside the embryo sac of the angiosperm. This process of fusion takes place inside the embryo sac.

When pollen grains fall on the stigma, they germinate and give rise to the pollen tube that passes through the style and enters into the ovule. After this, the pollen tube enters one of synergids and releases two male gametes there.

Out of the two male gametes, one gamete fuses with the nucleus of the egg cell and forms the zygote (syngamy). The other male gamete fuses with the two polar nuclei present in the central cell to form a triploid primary endosperm nucleus. Since this process involves the fusion of three haploid nuclei, it is known as triple fusion. It results in the formation of the endosperm.
One male gamete nucleus and two polar nuclei are involved in this process.

Question 12.
Why do you think the zygote is dormant for sometime in a fertilised ovule?
Answer:
The zygote is formed by the fusion of the male gamete with the nucleus of the egg cell. The zygote remains dormant for some time and waits for the endosperm to form, which develops from the primary endosperm cell resulting from triple fusion. The endosperm provides food for the growing embryo and after the formation of the endosperm, further development of the embryo from the zygote starts.

Question 13.
Differentiate between:
(a) hypocotyl and epicotyl;
(b) coleoptile and coleorrhiza;
(c) integument and testa;
(d) perisperm and pericarp.
Answer:
(a)

Hypocotyl	Epicotyl
1. The portion of the embryonal axis which lies below the cotyledon in a dicot embryo is known as the hypocotyl.	The portion of the embryonal axis which lies above the cotyledon in a dicot embryo is known as the epicotyl.
2. It terminates with the radicle.	It terminates with the plumule.

(b)

Coleoptile	Coleorrhiza
It is a conical protective sheath that encloses the plumule in a monocot seed.	It is an undifferentiated sheath that encloses the radicle and the root cap in a monocot seed.

(c)

Integument	Testa
It is the outermost covering of an ovule. It provides protection to it.	It is the outermost covering of a seed. It provides protection to the young embryo.

(d) Perisperm

Perisperm	Pericarp
It is the residual nucellus which persists. It is present in some seeds such as beet and black pepper.	It is the ripened wall of a fruit, which develops from the wall of an ovary.

Question 14.
Why is apple called a false fruit? Which part(s) of the flower forms the fruit?
Answer:
Fruits derived from the ovary and other accessory floral parts are called false fruits. On the contrary, true fruits are those fruits which develop from the ovary, but do not consist of the thalamus or any other floral part. In an apple, the fleshy receptacle forms the main edible part. Hence, it is a false fruit.

Question 15.
What is meant by emasculation? When and why does a plant breeder employ this technique?
Answer:
Emasculation is the process of removing anthers from bisexual flowers without affecting the female reproductive part (pistil), which is used in various plant hybridisation techniques.

Emasculation is performed by plant breeders in bisexual flowers to obtain the desired variety of a plant by crossing a particular plant with the desired pollen grain. To remove the anthers, the flowers are f covered with a bag before they open. This ensures that the flower is pollinated by pollen grains obtained from desirable varieties only. Later, the mature, viable, and stored pollen grains are dusted on the
bagged stigma by breeders to allow artificial pollination to take place and obtain the desired plant variety.

Question 16.
If one can induce parthenocarpy through the application of growth substances, which fruits would you select to induce parthenocarpy and why?
Answer:
Parthenocarpy is the process of developing fruits without involving the process of fertilisation or seed formation. Therefore, the seedless varieties of economically important fruits such as orange, lemon, water melon etc. are produced using this technique. This technique involves inducing fruit formation by the application of plant growth hormones such as auxins.

Question 17.
Explain the role of tapetum in the formation of pollen-grain wall.
Answer:
Tapetum is the innermost layer of the microsporangium. It provides nourishment to the developing pollen grains. During microsporogenesis, the cells of tapetum produce various enzymes, hormones, amino acids, and other nutritious material required for the development of pollen grains. It also produces the exine layer of the pollen grains, which is composed of the sporopollenin.

Question 18.
What is apomixis and what is its importance?
Answer:
Apomixis is the mechanism of seed production without involving the process of meiosis and syngamy. It plays an important role in hybrid seed production. The method of producing hybrid seeds by cultivation is very expensive for farmers. Also, by sowing hybrid seeds, it is difficult to maintain hybrid characters as characters segregate during meiosis. Apomixis prevents the loss of specific characters in the hybrid. Also, it is a cost-effective method for producing seeds.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 16 Environmental Issues Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 16 Environmental Issues

PSEB 12th Class Biology Guide Environmental Issues Textbook Questions and Answers

Question 1.
What are the various constituents of domestic sewage? Discuss the effects of sewage discharge on a river.
Answer:
Domestic sewage is the waste originating from the kitchen, toilet, laundry, and other sources. It contains impurities such as suspended solid (sand, salt, clay), colloidal, material (faecal matter, bacteria, plastic and cloth fibre), dissolved materials (nitrate, phosphate, calcium, sodium, ammonia), and disease-causing microbes. When organic wastes from the sewage enter the water bodies, it serves as a food source for micro-organisms such as algae and bacteria. As a result, the population of these micro-organisms in the water body increases.

Here, they utilise most of the dissolved oxygen for their metabolism. This results in an increase in the levels of Biological oxygen demand (BOD) in river water and results in the death of aquatic organisms. Also, the nutrients in the water lead to the growth of planktonic algal, causing algal bloom. This causes deterioration of water quality and fish mortality.

Question 2.
List all the wastes that you generate, at home, school or during your trips to other places. Could you very easily reduce the generation of these wastes? Which would be difficult or rather impossible to reduce?
Answer:
Wastes generated at home include plastic bags, paper napkins, toiletries, kitchen wastes (such as peelings of vegetables and fruits, tea leaves), domestic sewage, glass, etc.

Wastes generated at schools include waste paper, plastics, vegetable and fruit peels, food wrappings, sewage etc.
Wastes generated at trips or picnics include plastic, paper, vegetable and fruit peels, disposable cups, plates, spoons etc.

Yes, wastes can be easily reduced by the judicious use of the above materials. Wastage of paper can be minimised by writing on both sides of the paper and by using recycled paper. Plastic and glass waste can also be reduced by recycling and re-using.

Also, substituting plastic bags with biodegradable jute bags can reduce wastes generated at home, school, or during trips. Domestic sewage can be reduced by optimising the use of water while bathing, cooking, and other household activities. Non-biodegradable wastes such as plastic, metal, broken glass, etc. are difficult to decompose because micro-organisms do riot have the ability to decompose them.

Question 3.
Discuss the causes and effects of global warming. What measures need to be taken to control global warming?
Answer:
Global warming is defined as an increase in the average temperature of the Earth’s surface. Causes of Global Warming: Global warming occurs as a result of the increased concentration of greenhouse gases in the atmosphere. Greenhouse gases include carbon dioxide, methane, and water vapour. These gases trap solar radiations released back by the Earth. This helps in keeping our planet warm and thus, helps in human survival. However, an increase in the amount of greenhouse gases can lead to an excessive increase in the Earth’s temperature, leading to global warming. Global warming is a result of industrialisation, burning of fossil fuels, and deforestation.

Effects of Global Warming: It has been observed that in the past three decades, the average temperature of the Earth has increased by 0.6°C. As a result, the natural water cycle has been disturbed resulting in changes in the pattern of rainfall. It also changes the amount of rainwater. Also, it results in the melting of Polar ice caps and mountain glaciers, which has caused a rise in the sea level, leading to the inundation of coastal regions.

Control Measures for Preventing Global Warming:

• Reducing the use of fossil fuels
• Use of bio-fuels
• Improving energy efficiency
• Use of renewable source of energy such as CNG etc.
• Reforestation.
• Recycling of materials

Question 4.
Match the items given in column A and B:

Column A	Column B
(a) Catalytic converter	(i) Particulate matter
(b) Electrostatic precipitator	(ii) Carbon monoxide and nitrogen oxides
(c) Earmuffs	(iii) High noise level
(d) Landfills	(iv) Solid wastes

Answer:

Column A	Column B
(a) Catalytic converter	(ii) Carbon monoxide and nitrogen oxides
(b) Electrostatic precipitator	(i) Particulate matter
(c) Earmuffs	(iii) High noise level
(d) Landfills	(iv) Solid wastes

Question 5.
Write critical notes on the following:
(a) Eutrophication
(b) Biological magnification
(c) Groundwater depletion and ways for its replenishment
Answer:
(a) Eutrophication:
It is the natural ageing process of a lake caused due to nutrient enrichment. It is brought down by the runoff of nutrients such as animal wastes, fertilisers, and sewage from land which leads to an increased fertility of the lake. As a result, it causes a tremendous increase in the primary productivity of the ecosystem. This leads to an increased growth of algae, resulting, into, algal blooms. Later, the decomposition of these algae depletes the supply of oxygen, leading to the death of other aquatic animal life.

(b) Biological Magnification: Unknowingly some harmful chemicals enter our bodies through the food chain. We use several pesticides and other chemicals to protect our crops from diseases and pests. These chemicals are either washed down into the soil or into the water bodies. From the soil, these are absorbed by the plants along with water and minerals, and from the water bodies, these are taken up by aquatic plants and animals. This is one of the ways in which they enter the food chain. As these chemicals are not degradable, these get accumulated progressively at each trophic level. As human beings occupy the topmost level in any food chain, the maximum concentration of these chemicals get accumulated in our bodies. This phenomenon is known as biological magnification.

(c) Ground Water Depletion and Ways for its Replenishment: The level of groundwater has decreased in the recent years. The source of water supply is rapidly diminishing each year because of an increase in the population and water pollution. To meet the demand of water, water is withdrawn from water bodies such as ponds, rivers etc. As a result, the source of groundwater is depleting.

This is because the amount of groundwater being drawn for human use is more than the amount replaced by rainfall. Lack of vegetation cover also results in very small amounts of water seeping through the ground. An increase in water pollution is another factor that has reduced the availability of groundwater.

Measures for Replenishing Ground Water:

• Preventing over-exploitation of groundwater
• Optimising water use and reducing water demand
• Rainwater harvesting
• Preventing deforestation and plantation of more trees.

Question 6.
Why does ozone hole form over Antarctica? How will enhanced ultraviolet radiation affect us?
Answer:
The ozone hole is more prominent over the region of Antarctica. It is formed due to an increased concentration of chlorine in the atmosphere. Chlorine is mainly released from chlorofluorocarbons (CFC’s) widely used as refrigerants. The CFC’s magnate from the troposphere to the stratosphere, where they release chlorine atoms by the action of UV rays on them.

The release of Chlorine atoms causes the conversion of ozone into molecular oxygen. One atom of chlorine can destroy around 10,000 molecules of ozone and causes ozone depletion. The formation of the ozone hole will result in an increased concentration of UV – B radiations on the Earth’s surface. UV -B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV -B cause corneal cataracts in human beings.

Question 7.
Discuss the role of women and communities in protection and conservation of forests.
Answer:
Women and communities have played a major role in environmental conservation movements.
(i) Case Study of the Bishnoi Community: The Bishnoi community in Rajasthan strictly believes in the concept of living peacefully with nature. In 1731, the king of Jodhpur ordered his ministers to arrange wood for the construction of his new palace. For this purpose, the minister and 1116 workers went to Bishnoi village.

There, a Bishnoi woman called Amrita Devi along with her daughter and hundreds of other Bishnois showed the courage to step forward and stop them from cutting trees. They embraced the trees and lost their lives at the hands of soldiers of the king. This resistance by the people of the village forced the king to give up the idea of cutting trees.

(ii) Chipko Movement: The Chipko movement was started in 1974 in the Garhwal region of the Himalayas. In this movement, the women from the village stopped the contractors from cutting forest trees by embracing them.

Question 8.
What measures, as an individual, would you take to reduce environmental pollution?
Answer:
The following initiatives can be taken to prevent environmental pollution:
Measures for Preventing Air Pollution

• Planting more trees
• Use of clean and renewable energy sources such as CNG and bio-fuels
• Reducing the use of fossil fuels
• Use of catalytic converters in automobiles

Measures for Preventing Water Pollution:

• Optimising the use of water
• Using kitchen wastewater in gardening and other household purposes

Measures for Controlling Noise Pollution:

• Avoid burning crackers on Diwali
• Plantation of more trees

Measures for Decreasing Solid Waste Generation:

• Segregation of waste
• Recycling and reuse of plastic and paper
• Composting of biodegradable kitchen waste
• Reducing the use of plastics.

Question 9.
Discuss briefly the following:
(a) Radioactive wastes
(b) Defunct ships and e-wastes
(c) Municipal solid Wastes
Answer:
(a) Radioactive Wastes: Radioactive wastes are generated during the process of generating nuclear energy from radioactive materials. Nuclear waste is rich in radioactive materials that generate large quantities of ionising radiations such as gamma rays. These rays cause mutation in organisms, which often results in skin cancer. At high dosage, these rays can be lethal.

Safe disposal of radioactive wastes is a big challenge. It is recommended that nuclear wastes should be stored after pre-treatment in suitable shielded containers, which should then be buried in rocks.

(b) Defunct Ships and E-wastes: Defunct ships are dead ships that are no longer in use. Such ships are broken down for scrap metal in countries such as India and Pakistan. These ships are a source of various toxicants such as asbestos, lead, mercury etc. Thus, they contribute to solid* wastes that are hazardous to health.

E-waste or electronic wastes generally include electronic goods such as computers etc. Such wastes are rich in metals such as copper, iron, silicon, gold etc. These metals are highly toxic and pose serious health hazards. People of developing countries are involved in the recycling process of these metals and therefore, get exposed to toxic substances present in these wastes.

(c) Municipal Solid Wastes: Municipal solid wastes are generated from schools, offices, homes, and stores. It is generally rich in glass, metal, paper waste, food, rubber, leather, and textiles. The open dumps of municipal wastes serve as a breeding ground for flies, mosquitoes, and other disease-causing microbes. Hence, it is necessary to dispose of municipal solid waste properly to prevent the spreading of diseases. Sanitary landfills and incineration are the methods for the safe disposal of solid wastes.

Question 10.
What initiatives were taken for reducing vehicular air pollution in Delhi? Has air quality improved in Delhi?
Answer:
Delhi has been categorised as the fourth most polluted city of the world in a list of 41 cities. Burning of fossil fuels has added to the pollution of air in Delhi.
Various steps have been taken to improve the quality of air in Delhi.

(a) Introduction of CNG (Compressed Natural Gas): By the order of the supreme court of India, CNG-powered vehicles were introduced at the end of year 2006 to reduce the levels of pollution in Delhi. CNG is a clean fuel that produces very little unburnt particles.
(b) Phasing out of old vehicles
(c) Use of unleaded petrol
(d) Use of low-sulphur petrol and diesel
(e) Use of catalytic converters
(f) Application of stringent pollution-level norms for vehicles
(g) Implementation of Bharat stage I, which is equivalent to euro II norms in vehicles of major Indian cities.

The introduction of CNG-powered vehicles has improved Delhi’s air quality, which has lead to a substantial fall in the level of CO₂ and SO₂. However, the problem of suspended particulate matter (SPM) and respiratory suspended particulate matter,(RSPM) still persists.

Question 11.
Discuss briefly the following:
(a) Greenhouse gases
(b) Catalytic converter
(c) Ultraviolet B
Answer:
(a) Greenhouse Gases: The greenhouse effect refers to an overall increase in the average temperature of the Earth due to the presence of greenhouse gases. Greenhouse gases mainly consist of carbon dioxide, methane, and water vapour. When solar radiations reach the Earth, some of these radiations are absorbed.

These absorbed radiations are Released back into the atmosphere. These radiations are trapped by the greenhouse gases present in the atmosphere. This helps in keeping our planet warm and thus, helps in human survival.
However, an increase in the amount of greenhouse gases can lead to an excessive increase in the Earth’s temperature, thereby causing global warming.

(b) Catalytic Converter: Catalytic converters are devices fitted in automobiles to reduce automobile or vehicle pollution. These devices contain expensive metals such as platinum, palladium, and rhodium that act as catalysts.
As the vehicular discharge -passes through the catalytic converter, the unburnt hydrocarbons present in it get converted into carbon dioxide and water. Carbon monoxide and nitric oxide released by catalytic converters are converted into carbon dioxide and nitrogen gas respectively.

(c) Ultraviolet B: Ultraviolet-B is an electromagnetic radiation which has a shorter wavelength than visible light.
It is a harmful radiation that comes from sunlight and penetrates through the ozone hole onto the Earth’s surface.
It induces many health hazards in humans. UV -B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV -B cause corneal cataracts in human beings.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 14 Respiration in Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 14 Respiration in Plants

PSEB 11th Class Biology Guide Respiration in Plants Textbook Questions and Answers

Question 1.
Differentiate between:
(a) Respiration and Combustion
(b) Glycolysis and Krebs’ cycle
(c) Aerobic respiration and Fermentation
Answer:
(a) Differences between Respiration and Combustion

Respiration	Combustion
1. It is the breakdown of complex compounds through oxidation within the cells, leading to release of considerable amount of energy.	Combustion is the complete burning of glucose, which produces CO2 and H2O and yields energy which is given out as heat.
2. It is a controlled biochemical process.	It is an uncontrolled physicochemical process.
3. Many chemical bonds break simultaneously by releasing large amounts of energy.	Chemical bonds break one after another to release energy.
4. Enzymes are involved.	Enzymes are not involved.

(b) Differences between Glycolysis and Krebs’ Cycle

Glycolysis	Krebs’ Cycle
1. It is also called as Embden Meyerhof-Parnas (EMP) pathway.	It is also known as citric acid cycle or tricarboxylic acid cycle.
2. It occurs in the cytosol of prokaryotes as well as eukaryotes.	In eukaryotes, it occurs in matrix of mitochondria, and in prokaryotes, it occurs in cytoplasm.
3. It starts with the oxidation of glucose.	It starts with the oxidation of pyruvic acid.
4. It is an enzyme-controlled 10 steps process by which glucose, fructose or sucrose is reduced to form pyruvic acid with the production of ATP and NADH2.	It involves 8 steps to oxidize two molecules of acetyl CO-A

(c) Differences between Aerobic Respiration and Fermentation

Aerobic Respiration	Fermentation
1. The presence of O2 is required for complete oxidation of organic substances.	Incomplete oxidation of glucose occurs in the absence of oxygen.
2. It releases CO2, water, and a large amount of energy present in the substrate.	Pyruvic acid is converted to CO2 and ethanol and less amount of energy is released.
3. Only two molecules of ATP are produced.	Many molecules of ATP are produced.
4. NADH is oxidized to NAD+ slowly.	This reaction is very vigorous under aerobic respiration.

Question 2.
What are respiratory substrates? Name the most common respiratory substrate.
Answer:
The compounds oxidized during the process of respiration are called respiratory substrates. Carbohydrates, especially glucose, act as respiratory substrates. Fats, proteins, and organic acids also act as respiratory substrates.

Question 3.
Give the schematic representation of glycolysis.
Answer:
Glycolysis

• The scheme of reactions was given by Embden, Meycrhof and Pumas, hence it is also called EMP Pathway; it occurs in the cytoplasm of cells.
• In this process, glucose is partially oxidized/ converted into two molecules of pyruvic acid.
• Glucose is phosphorylated with the help of ATP into Glucose-6-Phosphate, catalyzed by the enzyme hexokinase.
• Glucose-6-Phosphate is converted into its isomer, Fructose-6-Phosphate, catalyzed by isomerase.
• Fructose-6-Phosphate is then phosphorylated with the use of ATP into Fructose 1,6-bisphosphate, catalyzed by
  phosphofructokinase.
• Fructose 1,6-bisphosphate is split into one molecule of 3-pho sphoglyceraldchyd and one molecule of
  dihydroxy-acetone phosphate; these two products are interconvertible.
• 3-phosphoglyceraldehyde is oxidized to 1, 3 bi phosphoglycerate, where NAD is reduced to NADH.
• 1,3-bisphosphoglycerate is split into 3-phosphoglycerate along with the formation of ATP, catalyzed by phosphoglycerate kinase.

• 3-Phosphoglycerate is subsequently converted into 2-phosphoglycerate and then into Phosphoenol Pyruvate (PEP).
• PEP is converted into pyruvate along with the formation of ATP, catalyzed by the enzyme pyruvate kinase.

In this pathway, ATP molecules are formed in two ways :

1. Direct/substrate phosphorylation of ADP to ATP.
2. Oxidation of NADH to NAD ’ and ATP formation through electron transport (oxidative phosphorylation).

Four molecules of ATP are formed directly in the following steps:

• When 1, 3 bi phosphoglyceric acids is converted into 3-phosphoglyceric acid and
• When phosphoenolpyruvate is converted into pyruvic acid.

Two molecules of ATP are consumed, one in each of the following steps:

1. Phosphorylation of glucose to glucose-6- phosphate and
2. Conversion of fructose-6-phosphate into fructose 1, 6 bisphosphates.

When 3-phosphoglyceraldehyde (PGAL) is converted into 1, 3 bi phosphoglycerates, two redox equivalents are removed in the form of two hydrogen atoms from PGAL and transferred to NAD⁺, which is reduced to NADH ⁺ H⁺.
Oxidation of two molecules of NADH to NAD yields six molecules of ATP during electron transport. The end products of glycolysis are two molecules each of pyruvic acid, NADH, and ATP.

Question 4.
What are the main steps in aerobic respiration? Where does it take place?
Answer:
The main steps in aerobic respiration are as follows :

• Glycolytic breakdown of glucose into pyruvic acid.
• Oxidative decarboxylation of pyruvic acid to acetyl Co-A (acetyl coenzyme-A).
• Krebs’ cycle.
• Terminal oxidation and phosphorylation in respiratory chain.
  It occurs inside the mitochondrial matrix.

Question 5.
Give the schematic representation of an overall view of Kreb’s cycle.
Answer:
Tricarboxylic Acid Cycle or Kreb’s Cycle:

• It is known as tricarboxylic acid cycle, since the first product, citric acid (hence citric acid cycle) is a tricarboxylic acid (has three COOH groups).
• The sequence of reactions in citric acid cycle was first followed by Sir Hans Kreb; hence it is also known as Kreb’s cycle.
• During this cycle of reactions, 3 molecules of NAD and one molecule of FAD are reduced to NADH and FADH respectively.
• Respiration in Plants 115:
• One ATP molecule is produced directly from GTP formed by substrate-level phosphorylation.

The reactions of aerobic oxidation of Pyruvic acid can be summed up as :
Pyruvic acid + 4NAD+ + FAD+ + 2H₂O + ADP + Pi → 3CO₂ + 4NADH + 4H⁺ + FADH₂ + ATP

Question 6.
Explain ETS.
Answer:
Electron Transport System (ETS):

• ETS occurs in the electron transport particles (ETP) on the inner surface of the inner membrane of mitochondria.
• NADH formed in glycolysis and citric acid cycle are oxidized by NADH-dehydrogenase (complex I) and the electrons are transferred to ubiquinone, via FMN.
• Ubiquinone also receives reducing equivalents via FADH generated during the oxidation of succinate by succinate dehydrogenase (complex II).
• The reduced ubiquinone called ubiquinol is then oxidized by transfer of electrons to cytochrome c, via cytochrome be bc₁ -complex (complex III).
• Cytochrome c acts as a mobile carrier between complex III and complex IV.
• Complex IV refers to cytochrome c oxidase complex containing cytochromes a and a3 and two copper centres.
• When the electrons are shunted over the carriers via complex I to IV in the electron transport chain, they are coupled to ATP synthetase (complex V) for the formation of ATP from ADP and IP.
• Oxygen functions as the terminal acceptor of electrons and is reduced to water along with the hydrogen atoms.

Question 7.
Distinguish between the following:
(a) Aerobic respiration and anaerobic respiration.
(b) Glycolysis and fermentation.
(c) Glycolysis and citric acid cycle.
Answer:
(a) Differences between Aerobic and Anaerobic Respiration

Aerobic Respiration	Anaerobic Respiration
1. Presence of oxygen is essential.	Occurs without oxygen.
2. Complete oxidation of substrates occurs into CO2 and H2O.	Incomplete degradations of substrates.
3. End products are non-toxic.	End products are toxic when accumulated in large amount.
4. Involves glycolysis, Krebs’ cycle, and terminal oxidation.	Involves only glycolysis followed by incomplete breakdown of pyruvic acid.
5. 36 ATP molecules are produced from each glucose molecule.	Only two molecules of ATP are produced from each glucose molecule.

(b) Differences between Glycolysis and Fermentation:

Glycolysis	Fermentation
1. it occurs in cytoplasm of the cell and in all living things.	it occurs in yeast and muscle cells in animals when oxygen is not sufficient for cellular respiration.
2. Glucose undergoes partial oxidation to form two molecules of pyruvic acid.	The enzyme-like pyruvic acid decarboxylase and alcohol dehydrogenase catalyze these reactions.
3. The end products are CO2, H2O and energy.	The end products are CO2 and ethanol and in animal cells lactic acid is released.

(c) Differences between Glycolysis and Krebs’ Cycle

Glycolysis	Krebs’ Cycle
1. It is also called as Embden Meyerhof-Parnas (EMP) pathway.	It is also known as citric acid cycle or tricarboxylic acid cycle.
2. It occurs in the cytosol of prokaryotes as well as eukaryotes.	In eukaryotes, it occurs in matrix of mitochondria and in prokaryotes, it occurs in cytoplasm.
3. It starts with the oxidation of glucose.	It starts with the oxidation of pyruvic acid.
4. It is an enzyme-controlled 10 steps process by which glucose, fructose or sucrose is reduced to form pyruvic acid with the production of ATP and NADH2.	It involves 8 steps to oxidize two molecules of acetyl CO-A

Question 8.
What are the assumptions made during the calculation of net gain of ATP?
Answer:
The calculations of net gain of ATP can be made only on certain assumptions :

• There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
• The NADH synthesized in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
• None of the intermediates in the pathway are utilised to synthesize any other compound.
• Only glucose is being respired-no other alternative substrates are entering in the pathway at any of the intermediary stages.

But this kind of assumptions are not really valid in a living system. All pathways work simultaneously and do not take place one after another. Substrates enter the pathways and are withdrawn from it as end when necessary; ATP is utilized as and when needed; enzymatic rates are controlled by multiple means. In overall steps, there is a net gain of 36 ATP molecules during aerobic respiration of one molecule of glucose.

Question 9.
Discuss ‘the respiratory pathway is an amphibolic pathway.’
Answer:
Aniphibolic Pathway: Glucose is the favored substrate for respiration. All carbohydrates are usually first converted into glucose before they are used for respiration. Other substrates can also be respired but then they do not enter the respiratory pathway at the first step.

Since respiration involves breakdown as well as synthesis of substrates, the respiratory process involves both catabolism and anabolism. That is why respiratory pathway is considered to be an amphibolic pathway rather than as a catabolic one.

Question 10.
Define RQ. What is its value for fats?
Answer:
Respiratory quotient (RQ) or respiratory ratio can be defined as the ratio of the volume of CO₂ evolved to the volume of O₂ consumed during respiration. The value of respiratory quotient depeñds on the type of respiratory substrate. Its value is one for carbohydrates. However, it is always less than one for fats as fats consume more oxygen for respiration than carbohydrates.

It can be illustrated through the example of tripalmitin fatty acid, which consumes 145 molecules of O₂ for respiration while 102 molecules of CO₂ are evolved.The RQ value for tripalmitin is 0.7.

Question 11.
What is oxidative phosphorylation?
Answer:
It refers to the formation of ATP in the mitochondria, utilizing the energy obtained by the oxidation of organic molecules.

Question 12.
What is the significance of step-wise release of energy in respiration?
Answer:
During oxidation within a cell, all the energy contained in respiratory substrates is not released free in a single step. It is released in a series of slow step-wise reactions controlled by enzymes and it is trapped as chemical energy in the form of ATP. The significance of this step-wise release of energy is that some energy is used to synthesize ATP. This ATP is stored for the later utilization wherever required. Hence, ATP acts as the energy currency of cell. The energy

stored in ATP can be utilized for following:

• In various energy-requiring processes of organisms.
• The carbon skeleton produced during respiration is used as precursors for the synthesis of other molecules.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 1 Reproduction in Organisms Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 1 Reproduction in Organisms

PSEB 12th Class Biology Guide Reproduction in Organisms Textbook Questions and Answers

Question 1.
Why is reproduction essential for organisms?
Answer:
Reproduction is a fundamental feature of all living organisms. It is a biological process through which living organisms produce offsprings (young ones) similar to them. Reproduction ensures the continuance of various species on the Earth. In the absence of reproduction, the species will not be able to exist for a long time and may soon get extinct.

Question 2.
Which is a better mode of reproduction sexual or asexual? Why?
Answer:
Sexual reproduction is a better mode of reproduction. It allows the formation of new variants by the combination of the DNA from two different individuals, typically one of each sex. It involves the fusion of the male and the female gamete to produce variants, which are not identical to their parents and to themselves. This variation allows the individual to adapt constantly changing and challenging environment. Also, it leads to the evolution of better suited organisms which ensures greater survival of a species. On the contrary, asexual reproduction allows very little or no variation at all. As a result, the individuals produced are exact copies of their parents and themselves.

Question 3.
Why is the offspring formed by asexual reproduction referred to as clone?
Answer:
A clone is a group of morphologically and genetically identical individuals. In the process of asexual reproduction, only one parent is involved and there is no fusion of the male and the female gamete. As a result, the offsprings so produced are morphologically and genetically similar to their parents and are thus, called clones.

Question 4.
Offspring formed due to sexual reproduction have better chances of survival. Why? Is this statement always true?
Answer:
Sexual reproduction involves the fusion of the male and the female gamete. This fusion allows the formation of new variants by the combination of the DNA from two (usually) different members of the species. The variations allow the individuals to adapt under varied environmental conditions for better chances of survival.

However, it is not always necessary that the offspring produced due to sexual reproduction has better chances of survival. Under some circumstances, asexual reproduction is more advantageous for certain organisms. For example, some individuals who do not move from one place to another and are well settled in their environment. Also, asexual reproduction is a fast and’ a quick mode of reproduction which does not consume much time and energy as compared to sexual reproduction.

Question 5.
How does the progeny formed from asexual reproduction differ from those formed by sexual reproduction?
Answer:

Progeny formed from asexual reproduction	Progeny formed from sexual reproduction
1. Asexual reproduction does not involve the fusion of the male and the female gamete. Organisms undergoing this kind ofreproduction produce offsprings that are morphologically and genetically identical to them.	Sexual reproduction involves the fusion of the male and the female gamete of two individuals, typically one of each sex. Organisms undergoing this kind of reproduction produce offsprings that are not identical to them.
2. Offsprings thus produced do not show variations and are called clones.	Offsprings thus produced show variations from each other and their parents.

Question 6.
Distinguish between asexual and sexual reproduction. Why is
vegetative reproduction also considered as a type of asexual reproduction?
Answer:

Sexual reproduction	Asexual reproduction
1. It involves the fusion of the male and the female gamete.	It does not involves the fusion of the male and the female gamete.
2. It requires two (usually) different individuals.	It requires only one individual.
3. The individuals produced are not identical to their parents and show variations from each other and also, from their parents.	The individuals produced are identical to the parent and are hence, called clones.
4. Most animals reproduce sexually. Both sexual and asexual modes of reproduction are found in plants.	Asexual modes of reproduction are common in organisms having simple organisations such as algae and fungi.
5. It is a slow process.	It is a fast process.

Vegetative reproduction is a process in which new plants are obtained without the production of seeds or spores. It involves the propagation of plants through certain vegetative parts such as the rhizome, sucker, tuber, bulb, etc. It does not involve the fusion of the male and the female gamete and requires only one parent. Hence, vegetative reproduction is considered as a type of asexual reproduction.

Question 7.
What is vegetative propagation? Give two suitable examples.
Answer:
Vegetative propagation is a mode of asexual reproduction in which new plants are obtained from the vegetative parts of plants. It does not involve the production of seeds or spores for the propagation of new plants. Vegetative parts of plants such as runners, rhizomes, suckers, tubers, etc. can be used as propagules for raising new plants.

Examples of vegetative reproduction are given below:
1. Eyes of potato: The surface of a potato has several buds called eyes. Each of these buds when buried in soil develops into a new plant, which is identical to the parent plant.

2. Leaf buds of Bryophyllum: The leaves of Bryophyllum plants bear several adventitious buds on their margins. These leaf buds have the ability to grow and develop into tiny plants when the leaves get detached from the plant and come in contact with moist soil.

Question 8.
Define:
(a) Juvenile phase,
(b) Reproductive phase,
(c) Senescent phase.
Answer:
(a) Juvenile Phase: All organisms have to reach a certain stage of growth and maturity in their life, before they can be reproduce sexually. This phase of growth is called the juvenile phase or vegetative phase in plants.

(b) Reproductive Phase: When the juvenile phase is over the organisms enter the period of reproductive phase or sexual maturity. It is indicated by showing various morphological and physiological changes such as development of secondary sexual characters in animals and by flowering in plants. This is the actual period of the life span of any organism when it is capable of producing offsprings. This phase is of variable duration in different organisms.

(c) Senescent Phase : This is the final and third stage of growth cycle. It can be considered as the end of reproductive phase. It is accompanied by reduction in functional capacity and increase in cellular break down and metabolic failures. It ultimately leads to death.

Question 9.
Higher organisms have resorted to sexual reproduction in spite of its complexity. Why?
Answer:
Although sexual reproduction involves more time and energy, higher organisms have resorted to sexual reproduction in spite of its complexity. This is because this mode of reproduction helps introduce new variations in progenies through the combination of the DNA from two (usually) different individuals. These variations allow the individual to cope with various environmental conditions and thus, make the organism better suited for the environment. Variations also lead to the evolution of better organisms and therefore, provide better chances of survival. On the other hand, asexual reproduction does not provide genetic differences in the individuals produced.

Question 10.
Explain why meiosis and gametogenesis are always interlinked?
Answer:
Meiosis is a process of reductional division in which the amount of genetic material is reduced. Gametogenesis is the process of the formation of gametes. Gametes produced by organisms are haploids (containing only one set of chromosomes), while the body of an organism is diploid. Therefore, for producing haploid gametes (gametogenesis), the germ cells of an organism undergo meiosis. During the process, the meiocytes of an organism undergo two successive nuclear and cell divisions with a single cycle of DNA replication to form the haploid gametes.

Question 11.
Identify each part in a flowering plant and write whether it is haploid (n) or diploid (2n).
(a) Ovary …………………………
(b) Anther ……………………..
(c) Egg ………………………….
(d) Pollen ……………………..
(e) Male gamete ……………….
(f) Zygote ………………..
Answer:
(a) diploid (2n)
(b) diploid (2n)
(c) haploid (n)
(d) haploid (n)
(e) haploid (n)
(f) diploid (2n)

Question 12.
Define external fertilisation. Mention its disadvantages.
Answer:
External fertilisation is the process in which the fusion of the male and the female gamete takes place outside the female body in an external medium, generally water. Fish, frog, starfish are some organisms that exhibit external fertilisation.

Disadvantages of external fertilisation
In external fertilisation, eggs have less chances of fertilisation. This can lead to the wastage of a large number of eggs produced during the process.
Further, there is an absence of proper parental care to the offspring, which results in a low rate of survival in the progenies.

Question 13.
Differentiate between a zoospore and a zygote.
Answer:

Zoospore	Zygote
1. A zoospore is a motile asexual spore that utilises the flagella for movement.	A zygote is a noh-motile diploid cell formed as a result of fertilisation.
2. It is an asexual reproductive structure.	It is formed as a result of sexual reproduction.
3. Zoospores are formed in simple plants like, algae or fungi.	Zygote is formed in complex organism.

Question 14.
Differentiate between gametogenesis from embryogenesis.
Answer:

Gametogenesis	Embryogenesis
1. It is the process of the formation of haploid male and female gametes from diploid meiocytes (gamete mother cell) through the process of meiosis.	It is the process of the development of the embryo from the repeated mitotic divisions of the diploid zygote.
2. Gametes may be either homogametes or heterogametes.	Animals may be either oviparous or viviparous.

Question 15.
Describe the post-fertilisation changes in a flower.
Answer:
As a result of double fertilisation in flowering plants, zygote (2n) and the primary endosperm nucleus (3n) is produced. The calyx, corolla, stamens and style wither away. The calyx may persist or even show growth in certain cases. The post fertilisation changes which take place are

• Endosperm formation
• Embryo formation
• Seed formation and
• Fruit formation.

The primary endosperm nucleus becomes active and forms a nutritive vegetative tissue. The endosperm at the expense of food present in the nucellus, Endosperm may be completely used up by the developing embryo (non-endospermic seeds e.g., pea) or may persist in the seed (endospermic seed e.g., castor). The zygote, waits for sometime till the formation of endosperm and then develops into embryo, by withdrawing nutrition from the endosperm. Ultimately the ovules are transformed into seeds and the ovary becomes a fruit. The formation of fruit helps in the nourishment and protection to the developing seeds and later helps in seed dispersal. Under favourable conditions the seeds germinate to form new plants.

Question 16.
What is a bisexual flower? Collect five bisexual flowers from your neighbourhood and with the help of your teacher find out their common and scientific names.
Answer:
A flower that contains both the male and female reproductive structure (stamen and pistil) is called a bisexual flower.
Examples of plants bearing bisexual flowers are:

1. Water lily {Nymphaea odorata)
2. Rose (Rosa multiflora)
3. Hibiscus (Hibiscus Rosa-sinensis)
4. Mustard (Brassica nigra)
5. Petunia (Petunia hybrida)

Question 17.
Examine a few flowers of any cucurbit plant and try to identify the staminate and pistillate flowers. Do you know any other plant that hears unisexual flowers?
Answer:
Cucurbit plant bears unisexual flowers as these flowers have either the stamen or the pistil. The staminate flowers bear bright, yellow coloured petals along with stamens that represent the male reproductive structure. On the other hand, the pistillate flowers bear only the pistil that represents the female reproductive structure.
Other examples of plants that bear unisexual flowers are corn, papaya, cucumber, etc.

Question 18.
Why are offspring of oviparous animals at a greater risk as compared to offspring of viviparous animals?
Answer:
Oviparous animals lay eggs outside their body. As a result, the eggs of these animals are under continuous threat from various environmental factors. On the other hand, in viviparous animals, the development of the egg takes place inside the body of the female. Hence, the offspring of an egg-laying or oviparous animal is at greater risk as compared to the offspring of a viviparous animal, which gives birth to its young ones.