NCERT Solutions Class 12 Biology Free PDF Download

Table of Contents

Chapter 1 Reproduction in Organisms

Question 1.
Why is reproduction essential for organisms?
Solution:
An organism gives rise to young ones by reproduction. The offspring grow, mature, and in turn, produce new offspring. Thus, there is a cycle of birth, growth, and death. Reproduction enables the continuity of the species, generation after generation. So, therefore reproduction is essential.

Question 2.
Which Is a better mode of reproduction: sexual or asexual? Why?

Solution:
Sexual reproduction is a better mode of reproduction than asexual mode because the former contributes to the evolution of the species by introducing variation in a population and occurs much more rapidly. Variation in a population occurs because of the fusion of male and female gametes (sexual reproduction) carrying different sets of chromosomes.

Question 3.
Why is the offspring formed by asexual reproduction referred to as clones?
Solution:
In sexual reproduction, the offspring is morphologically and genetically identical to the parent and to each other. Hence, it is called a clone.

Question 4.
Offspring formed due to sexual reproduction have better chances of survival. Why? Is this statement always true?

Solution:
The offspring that are produced by sexual reproduction are not genetically identical to their parents. They exhibit variations because they receive chromosomes from two different parents. Since they show variation, they are highly adapted to the changing environment. Asexually produced organisms are genetically identical and all organism show similar adaptations. So, during any calamity, there is a possibility that the whole generation would destroy leading to extinction of species. However, this statement is not true always because of some inborn genetic disorder due to which organism have a risk in their survival, e.g., Haemophilia.

Question 5.
How does the progeny formed from asexual reproduction differ from those formed by sexual reproduction?
Solution:
The progenies have similar genetic make-up and are exact copies of their parents in asexual reproduction but the progenies have different genetic make-up and different from each other and dissimilar to the parent in sexual reproduction.

Variation is absent in asexual reproduction but it is a common phenomenon of sexual reproduction. In asexual reproduction, variation may occur due to mutation whereas variation occurs due to mutation, crossing over, and recombination in sexual reproduction.

Question 6.
Distinguish between asexual and sexual reproduction. Why is vegetative reproduction also considered as a type of asexual reproduction?
Solution:
The differences between asexual and sexual reproduction are given in the following table:
NCERT Solutions for Class 12 Biology Chapter 1 Reproduction in Organisms Q6.1
As vegetative reproduction does not involve two parents and any meiosis and syngamy so it is considered as a type of asexual reproduction. This term is used in the case of plants.

Question 7.
What is vegetative propagation? Give two suitable examples. 

Solution:
In plants, the vegetative propagules (runner, rhizome, sucker, etc.) are capable of producing new offsprings by the process called vegetative propagation. As the formation of these vegetative propagules does not involve both the parents, the process involved is asexual.
Examples:

  • Adventitious buds in the notches along the leaf margins of the Myriophyllum grow to form new plants.
  • Potato tuber having buds when grown develops into a new plant.

Question 8.
Define

  1. Juvenile phase
  2. Reproductive phase
  3. Senescent phase

Solution:

  1. The juvenile phase is the period of growth between the birth of an individual till reaches reproductive maturity. In plant, this is also called the vegetative phase.
  2. The reproductive phase starts after the vegetative phase (juvenile phase) and the organisms produce offspring during this phase. A few plants show unusual flowering phenomena. Some of them such as bamboo species, flower only once in their lifetime, usually after 30-100 years. They produce large number of fruits and die. Among animals such as birds living in nature lay eggs seasonally while birds in captivity (as in poultry farm) can be made to lay eggs throughout the year.
  3. Senescent phase or old age begins at the end of reproductive phase. It is last phase of life span during which there is progressive deterioration in the body and loss in the capability of reproduction. Old age ultimately leads to death of the organism. In plants, it is characterized by the yellowing of leaves and leaf fall.

Question 9.
Higher organisms have resorted to sexual reproduction inspite of its complexity. Why?
Solution:
Higher organisms have resorted to sexual reproduction to:

  • Get over the unfavourable condition
  • Restore high gene pool in a population
  • Restore vigour and vitality of the race and Get proper parental care
  • Introduce variation to enable better adaptive capacity.

Question 10.
Explain why meiosis and gametogenesis are always interlinked?

Solution:
Gametogenesis refers to the process of the formation of gametes. Gametes are haploid cells. If the organism is diploid it undergoes meiotic division to produce haploid gamete. So gametogenesis and meiosis are always interlinked.

Question 11.
Identify each part in a flowering plant and write whether it is haploid (n) or diploid (2n).

  1. Ovary…..
  2. Anther….
  3. Egg…..
  4. Pollen….
  5. Male gamete…..
  6. Zygote…..

Solution:

  1. Ovary – Diploid(2n)
  2. Anther – Diploid(2n)
  3. Egg – Haploid(n)
  4. Pollen – Haploid(n)
  5. Male gamete – Haploid(n)
  6. Zygote – Diploid(2n)

Question 12.
Define external fertilization. Mention its disadvantages.
Solution:
In most aquatic organisms such as majority algae, fishes and amphibians, syngamy occurs in the external medium i.e., outside the body of the organism. This type of gametic fusion is called external fertilisation. A major disadvantage is that the offspring are extremely vulnerable to predators threatening their survival.

Question 13.
Differentiate between a zoospore and a zygote.

Solution:
Zoospores are The microscopic, flagellated (motile)
special asexual reproductive structures found in certain members of the kingdom fungi and simple plants like algae whereas zygote is a diploid cell formed by die fusion of male and female gametes. The zygote is usually non flagellated.
Zoospores are the structures that give rise to new organism whereas zygote is formed after fertilization which develops into a mature organism.

Question 14.
Differentiate between gametogenesis from embryogenesis.
Solution:
The differences between gametogenesis and embryogenesis are given in the following table :
NCERT Solutions for Class 12 Biology Chapter 1 Reproduction in Organisms Q14.1

Question 15.
Describe the post-fertilization changes in a flower.
Solution:
The post-fertilization changes that take place in a flower are as follows:

  • The formation of zygote which later develops into an embryo and a primary endosperm cell which develops into an endosperm takes place.
  • While the sepals, petals and stamens are shed, the pistil remains intact.
  • The fertilized ovule develops into seeds.
  • The ovary matures into a fruit that later develops a thick protective wall, called the pericarp.
  • Seeds after dispersal germinate under favourable conditions which later develop into a new plant.

Question 16.
What is a bisexual flower?
Solution:
A bisexual flower possesses both male and female reproductive structures i.e., stamens and carpels are present in the same flower
e.g.. Rose (Rosa dim), Kikar (Acacia nilotica), China rose (Hibiscus rosa sinensis) etc.

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 bears unisexual flowers?
Solution:
Other unisexual plants are com, papaya, cucumber etc.

Question 18.
Why are offspring of oviparous animals at a greater risk as compared to offspring of viviparous animals?
Solution:
Cucurbit plant bears unisexual flowers as these flowers have either the stamen or the pistil. The staminate flowers bear bright, yellow-colored 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.

Chapter 2 Sexual Reproduction,in Flowering Plants

Question 1.
Name the parts of an angiosperm flower in which development of male and female gametophyte take place.
Solution:
Inside the anther, the cells of microsporangia develop as male gamete. Inside the ovary megasporangial cells develop as female gametes.

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.
Solution:
Differences between microsporogenesis and megasporogenesis are as follows :
NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants Q2.1
During microsporogenesis and Megas-megasporogenesis meiotic cell division occurs which results in haploid gametes – the microspores or pollen grains and megaspores.

Question 3.
Arrange the following terms in- the correct developmental sequence : Pollen grain, sporogenous tissue, microspore tetrad, pollen mother cell, male gametes.
Solution:
Sporogenous tissue → Pollen mother cell → microspore tetrad → pollen grain → male gamete.

Question 4.
With a neat, labelled diagram, describe the parts of a typical angiosperm ovule.
NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants Q4.1
Solution:
An angiosperm ovule consists of the following parts:

  • The ovule is attached to placenta by means of a stalk called funicle or funiculus.
  • The point of attachment of funiculus to the body of ovule is called hilum.
  • The main body of ovule is made of parenchymatous tissue called nucellus.
  • Nucellus is covered on its outside by one or two coverings called integuments and hence ovule is rightly called as integument megasporangium.
  • The integuments cover entire nucellus except for a small pore at upper end, which is called the micropyle. Micropyle is formed generally by inner integument or by both integuments.
  • The place of junction of integuments and nucellus is called chalaza.
  • In inverted ovules (most common type), the stalk or funiculus is attached to the main body of ovule for some distance to form a ridge like structure, called- raphe.
  • In the nucellus of ovule, a large oval cell is present at micropylar end, which is known as embryo sac (female gametophyte), which develops from the megaspore.

Question 5.
What is meant by monosporic development of female gametophyte?
Solution:
The female gametophyte or the embryo sac develops, from a single functional megaspore. This is known as the 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 4 megaspores, only one functional megaspore develops into a female gametophyte, while the remaining 3 degenerates.

Question 6.
With a neat diagram explain the 7-celled, 8 nucleate nature of the female gametophyte
Solution:
NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants Q6.1

The female gametophyte (embryo sac) develops from a single functional megaspore. Thus, the megaspore undergoes three successive mitotic divisions to form 8 nucleate embryo sac. The first mitotic division in the megaspore forms 2 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 redivide to form 8 nucleate stages.

As a result there are 4 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 4 nuclei only 3 differentiate into 2 synergids and one egg cell. Together they are known as egg apparatus. Similarly, at the chalazal end 3 out of 4 nuclei differentiates as antipodal cells. The remaining 2 cells (of the micropylar and chalazal end) move towards the centre and are known as the polar nuclei, which are situated in the centre of the embryo sac. 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.
Solution:
Chasmogamous flowers or open flowers in which anther and stigma are exposed for pollination. Cross-pollination cannot occur in cleistogamous flowers. These flowers remain closed thus causing only self-pollination. In cleistogamous flowers, anthers dehisce inside the closed flowers. So the pollen grains come in contact with stigma. Thus there is no chance of cross¬pollination, e.g., Oxalis, Viola.

Question 8.
Mention two strategies evolved to prevent self pollination in flowers.
Solution:
Two strategies evolved to prevent self-pollination are:

  • Pollen release and stigma receptivity are not synchronized.
  • Anthers and stigma are placed at such positions that pollen doesn’t reach stigma.

Question 9.
What is self-incompatibility? Why does self-pollination not lead to seed formation in self-incompatible species?
Solution:
When the pollen grains of an anther do not germinate on the stigma of the same flower, then such a flower is called self-sterile or incompatible and such condition is known as self¬incompatibility or self-sterility.
The transference of pollen grains shed from the anther to the stigma of the pistil is called pollination. This transference initiate the process of seed formation. Self-pollination is the transfer of pollen grain shed from the anther to stigma of pistil in the same flower. But in some flower self¬pollination does not lead to the formation of seed formation because of the presence of same sterile gene on pistil and pollen grain.

Question 10.
What is bagging technique? How is it useful in a plant breeding programme?
Solution:
It is the covering of female plants with butter paper or polythene to avoid their contamination from foreign pollens during the breeding programme.

Question 11.
What is triple fusion? Where and how does it take place? Name the nuclei involved in triple fusion.
Solution:
Inside the embryo sac, one male gamete fuses with egg cells to form a zygote (2n) and this is called syngamy or true act of fertilisation. This result of syngamy, i.e., zygote (2n) ultimately develops into an embryo.

The second male gamete fuses with 2 polar nuclei or secondary nucleus to form triploid primary endosperm nucleus and this is called triple fusion. The result of triple fusion, i.e., primary endosperm nucleus (3n) ultimately develops into a nutritive tissue for developing embryo called endosperm.

The nuclei involved in this triple fusion are the two polar nuclei or secondary nucleus and the second male gamete.

Question 12.
Why do you think the zygote is dormant for sometime in a fertilised ovule?
Solution:
The zygote is dormant in fertilized ovule for some time because, at this time, endosperm needs to develop. As endosperm is the source of nutrition for the developing embryo, nature ensures the formation of enough endosperm tissue before starting the process of embryogenesis.

Question 13.
Differentiate between:

  1. Epicotyl and hypocotyl;
  2. Coleoptile and coleorhiza;
  3. Integument and testa;
  4. Perisperm and pericarp

Solution:

  1. Differences between epicotyl and hypocotyl are as follows :
    NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants Q13.1
  2.  Differences between coleoptile and coleorhiza are as follows :
    NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants Q13.2
  3. Differences between integument and testa are as follows :
    NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants Q13.3
  4. Differences between perisperm and pericarp are as follows :
    NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants Q13.4

Question 14.
Why is apple called a false fruit? Which part (s) of the flower forms the fruit?
Solution:
Apple is called a false fruit because it develops from the thalamus instead of the ovary (the thalamus is the enlarged structure at the base of the flower).

Question 15.
What is meant by emasculation? When and why does a plant breeder employ this technique?
Solution:
Emasculation is the removal of stamens mainly the anthers from the flower buds before their dehiscence. This is mainly done to avoid self-pollination. Emasculation is one of the measures in the artificial hybridization. Plant breeders employed this technique to prevent the pollination within same flower or to pollinate stigmas with pollens of desired variety.

Question 16.
If one can induce parthenocarpy through the application of growth substances, which fruits would you select to induce parthenocarpy and why ?
Solution:
Oranges, lemons, litchis could be potential fruits for inducing the parthenocarpy because a seedless variety of these fruits would be much appreciated by the consumers.

Question 17.
Explain the role of tapetum in the formation of pollen-grain wall.
Solution:
Tapetum is the innermost layer of the microsporangium. The tapetal cells are multinucleated and polyploid. They nourish the developing pollen grains. These cells contain ubisch bodies that help in the ornamentation of the microspores or pollen grains walls. The outer layer of the pollen grain is called exine and is made up of the sporopollenin secreted by the ubisch bodies of the tapetal cells. This compound provides spiny appearance to the exine of the pollen grains.

Question 18.
What is apomixis and what is its importance ?
Solution:
Apomixis is the process of asexual production of seeds, without fertilization.
The plants that grow from these seeds are identical to the mother plant.

Uses:

  • It is a cost-effective method for producing seeds.
  • It has great use for plant breeding when specific traits of a plant have to be preserved.

Chapter 3 Human Reproduction

Question 1.
Fill in the blanks.

  1. Humans reproduce___(asexually/ sexually).
  2. Humans are ___(oviparous/ viviparous/ ovoviviparous).
  3. Fertilisation is___in humans (external/internal).
  4. Male and female gametes are___ (diploid/haploid).
  5. Zygote is___(diploid/haploid).
  6. The process of release of ovum from a mature follicle is called___
  7. Ovulation is induced by a hormone called ___
  8. The fusion of male and female gametes is called___
  9. Fertilisation takes place in ___
  10. Zygote divides to form___which is implanted in uterus.
  11. The structure which provides vascular connection between foetus and uterus is called ___

Solution:

  1. sexually;
  2. viviparous;
  3. internal;
  4. haploid;
  5. diploid;
  6. ovulation;
  7. LH;
  8. fertilisation;
  9. ampullary-isthmic junction of Fallopian tube;
  10. blastocyst;
  11. placenta

Question 2.
Draw a labelled diagram of male reproductive system.
Solution:
NCERT Solutions for Class 12 Biology Chapter 3 Human Reproduction Q2.1

Question 3.
Draw a labelled diagram of female reproductive system.
Solution:
NCERT Solutions for Class 12 Biology Chapter 3 Human Reproduction Q3.1

Question 4.
Write two major functions each of testis and ovary.
Solution:
Testis:

  1. The seminiferous tubules of the testis produce sperm.
  2. The Leydig cells of testis produce hormones such as androsterone and testosterone, together called androgens.

Ovary:

  1. Produce ovum
  2. Produce ovarian hormones such as estradiol, estrone, and estriol collectively called estrogens.

Question 5.
Describe the structure of a seminiferous tubule
Solution:
Seminiferous tubules are located in the testicles, and are the specific location of meiosis, and the subsequent creation of gametes, namely spermatozoa.

The lining of seminiferous tubules called germinal epithelium contains two types of cells – primary germ ceils which undergo spermatogenesis to form spermatozoa and columnar indifferent cells (derived from coelomic epithelium) which enlarge to form Sertoli cells. Sertoli cells function as nurse cells for providing nourishment to the developing spermatozoa, phagocvtise defective sperm and secrete protein hormone inhibin (which inhibits FSH secretion).

The seminiferous tubules are situated in testicular lobules. Both ends of the tubule are connected to the central region of the testis and form a network of small ductules called the rete testis.

Question 6.
What is spermatogenesis ? Briefly describe the process of spermatogenesis.
Solution:
Spermatogenesis is the process by which male spermatogonia develop into mature male gamete, spermatozoa. It starts at puberty and usually continues uninterrupted until death, although a slight decrease can be discerned in the quantity of produced sperm with increase in age. The process of spermatogenesis includes the formation of spermatogonia from germinal epithelium (primordial germ cell) through mitosis (multiplication phase). Finally they stop undergoing mitosis, grow and become primary spermatocytes (growth phase). Each spermatocyte undergoes meiosis (maturation phase). First maturation division is reductional, and produces two secondary spermatocytes. The latter divides by equational division (second maturation division) to form four haploid spermatids. Spermatids receive nourishment from the Sertoli cells to form sperms. This step is called spermiogenesis.

During this process one spermatogonium produces four sperms having half number of chromosomes.

Question 7.
Name the hormones involved in the regulation of spermatogenesis.
Solution:
The hormones involved in the regulation of spermatogenesis are:

  • Gonadotropin-releasing hormone
  • Luteinizing hormone (LH)
  • Follicle-stimulating hormone
  • Testosterone.

Question 8.
Define spermiogenesis and spermiation.
Solution:
The transformation of spermatid into sperm is called spermiogenesis. The release of sperm from the seminiferous tubules is called spermiation.

Question 9.
Draw a labelled diagram of sperm
Solution:
NCERT Solutions for Class 12 Biology Chapter 3 Human Reproduction Q9.1

Question 10.
What are the major components of seminal plasma?
Solution:
Secretions of prostate gland, seminal vesicle and Cowper’s gland and , sperms together constitute semen.

Question 11.
What are the major functions of male accessory ducts and glands?
Solution:
The main functions of male accessory ducts and glands are as follows:

1. Functions of accessory ducts:

  • Rete Testis: They transport sperms from seminiferous tubule to Vas efferentia.
  • Vas efferentia: Transports sperms to epdidymis.
  • Epididymis: Sperms are stored here. Maturation of sperms occurs.
  • Vas deference: Transports sperms from the epididymis to the urethra.

2. Functions of glands:

  • Prostate gland: It produces milky secretion which forms a considerable part of the semen. It makes sperm motile.
  • Bulbourethral gland: Its secretion makes the penis lubricated.
  • Seminal vesicle: It secretes mucus and watery alkaline fluid which provide energy to the sperm.

Question 12.
What is oogenesis? Give a brief account of oogenesis.
Solution:
Oogenesis is the production and growth of the ova (egg cell) in the ovary. It starts only after the female has attained puberty. The process is induced by FSH from the anterior pituitary. It leads to the growth of a single Graafian follicle in one of the two ovaries every month. The developing ovary is colonised by primordial germ cells prior to birth which differentiate into oogonia. These enlarge within the follicle under the influence of mitotic division to form primary oocyte containing diploid number of chromosomes. These undergo reductional division (1st meiotic division) to form a secondary oocyte and first polar body. The secondary oocyte proceeds with meiosis II but the division gets arrested until fertilisation occurs. The ‘egg’ is released at secondary oocyte stage under the effect of LH. A second polar body is extruded. The first polar body may also divide to form two polar bodies of equal sizes which do not take part in reproduction and ultimately degenerates. During oogenesis one oogonium produces one ovum and three polar bodies. Polar bodies containing small amount of cytoplasm helps to retain sufficient amount of cytoplasm in the ovum which is essential for the development of early embryo. Formation of polar bodies maintains the half number of chromosomes in the ovum.

Question 13.
Draw a labelled diagram of a section through ovary.
Solution:
NCERT Solutions for Class 12 Biology Chapter 3 Human Reproduction Q13.1

Question 14.
Draw a labelled diagram of a Graafian follicle ?
Solution:
NCERT Solutions for Class 12 Biology Chapter 3 Human Reproduction Q14.1

Question 15.
Give the functions of the following.

  1. Corpus luteum
  2. Endometrium
  3. Acrosome
  4. Sperm tail
  5. Fimbriae

Solution:
The functions of the following:

  1. Corpus luteum secretes a large amount of progesterone which is essential for the maintenance of the endometrium of the uterus.
  2. Endometrium is necessary for the implantation of the fertillised ovum, for contributing towards the making of the placenta and other events of pregnancy.
  3. Acrosome is filled with enzymes that help in dissolving the outer cover of the ovum and entry of sperm nucleus.
  4. Sperm tail facilitates motility of the sperm essential for reaching the ovum to fertilize it.
  5. Fimbriae are fingers-like projections at the mouth of fallopian tubules that help in the collection of the ovum after ovulation.

Question 16.
Identify True/False statements. Correct each false statement to make it true.

  1. Androgens are produced by Sertoli cells.
  2. Spermatozoa get nutrition from Sertoli cells.
  3. Leydig cells are found in ovary.
  4. Leydig cells synthesise androgens.
  5. Oogenesis takes place in corpus luteum.
  6. Menstrual cycle ceases during pregnancy.
  7. Presence or absence of a hymen is not a reliable indicator of virginity Or sexual experience.

Solution:

  1. False: Androgens are produced by interstitial cells or Leydig cells.
  2. True
  3. False: Leydig’s cells are found in the testes (in between the seminiferous tubules).
  4. True
  5. False: Oogenesis takes place in the ovary.
  6. True
  7. True

Question 17.
What is the menstrual cycle? Which hormones regulate the menstrual cycle?
Solution:
The recurring cycle of physiological changes in the uterus, ovaries and other sexual structures that occur from the beginning of one menstrual period through the beginning of the next is called menstrual cycle. The beginning of menstruation is called menarche. Hormones involved in the regulation of menstrual cycle are pituitary or ovarian hormones. These are LH, FSH, estrogen and progesterone.

Question 18.
What is parturition? Which hormones are involved in the induction of parturition?

Solution:

  • The process of delivering of the fully developed fetus or baby at the end of the pregnancy period through vigorous contraction of the uterus is called parturition.
  • Estrogen (amount of estrogen is more than progesterone) and oxytocin are the hormones involved in the induction of parturition.

Question 19.
In our society, women are often blamed for giving birth to daughters. Can you explain why this is not correct?
Solution:
This is not correct that women in our society are often blamed for giving birth to daughters because the sex of the baby is determined by the father, not by the mother. As we know that the chromosome pattern in the human female is XX and that in the male is XY.

Therefore, all the haploid gametes produced by the female (ova) have the sex chromosome X whereas in the male gametes (sperms) the sex chromosome could be either X or Y, hence, 50 per cent of sperms carry the X chromosome while the other 50 per cent carry the Y. After fusion of the male and female gametes the zygote would carry either XX or XY depending on whether the sperm carrying X or Y fertilised the ovum. The zygote carrying XX would develop into a female baby and XY would form a male.

Question 20.
How many eggs are released by a human ovary in a month? How many eggs do you think would have been released if the mother gave birth to identical twins? Would your answer change if the twins born were fraternal?
Solution:
Each ovary develops a number of immature eggs associated with groups of other cells called follicles. Normally, in humans, only one egg is released at one time; occasionally, two or more erupt during the menstrual cycle. The egg erupts from the ovary on the 14th to 16th day of the approximately 28 day menstrual cycle. Identical twins occur when a single egg is fertilised to form one zygote (monozygotic) which then divides into two separate embryos. And if the twins were born fraternal two eggs are released. Fraternal twins (commonly known as “non-identical twins”) usually occur when two fertilised eggs are implanted in the uterine wall at the same time. The two eggs form two zygotes, and these twins are therefore also known as dizygotic as well as “binovular” twins.

Question 21.
How many eggs do you think were released by the ovary of a female dog which gave birth to 6 puppies?
Solution:
One oogonium produces one ovum and three polar bodies. The ovum is the actual female gamete. The polar bodies take no part in reproduction and hence, soon degenerate. In human beings, ovum is released from the ovary in the secondary oocyte stage. So, six ova (eggs), were released by ovary of a female dog which gave birth to 6 puppies.

Chapter 4 Reproductive Health

Question 1.
What do you think is the significance of reproductive health in a society?
Solution:
Significance of reproductive health in society are:

  • Control over the transmission of STDs.
  • Less death due to reproduction-related diseases like-AIDS, cancer of the reproductive tract.
  • Control in a population explosion.
  • Not only the reproductive health of men and women affects the health of the next generation.

Question 2.
Suggest the aspects of reproductive health which need to be given special attention in the present scenario.
Solution:
Providing medical facilities and care to the problems like menstrual irregularities, pregnancy related aspects, delivery, medical termination of pregnancy, STDs, birth control, infertility. Post-natal child maternal management is another important aspect of the reproductive and child health care programme.

Question 3.
Is sex education necessary in schools? Why?
Solution:
Yes, sex education is necessary for schools because:

  • It will provide proper information about reproductive organs, adolescence, safe, hygienic sexual practices, and Sexually Transmitted Diseases (STDs).
  • It will provide the right information to avoid myths and misconceptions about sex-related queries.

Question 4.
Do you think that reproductive health in our country has improved in the past 50 years? If yes, mention some such areas of improvement.
Solution:
The reproductive health in our country has improved in the last 50 years. Some areas of improvement are :

  • Massive child immunization.
  • Increasing use of contraceptives.
  • Better awareness about sex related matters.
  • Increased number of medically assisted deliveries and better post-natal care leading to decreased maternal and infant mortality rates.
  • Increased number of couples with small families.
  • Better detection and cure of STDs and overall increased medical facilities for all sex related problems.

Question 5.
What are the suggested reasons for the population explosion?
Solution:

  • Improved medical facilities
  • Decline in death rate, IMR, MMR
  • Slower decline in birth rate.
  • Longer life span.
  • Lack of 100% family planning and education among the village.

Question 6.
Is the use of contraceptives justified? Give reasons.
Solution:
Yes, the use of contraceptives is justified: To overcome the population growth rate, contraceptive methods are used. It will help in bringing birth rate down & subsequently curb population growth. With the rapid spread of HIV/ AIDS in the country, there is now a growing realization about the need to know about contraception & condoms.

Question 7.
Removal of gonads cannot be considered as a contraceptive option. Why?
Solution:
Removal of gonads not only stops the production of gametes but will also stop the secretions of various important hormones, which are important for bodily functions. This method is irreversible and thus, can not be considered as a contraceptive method.

Question 8.
Amniocentesis for sex determination is banned in our country. Is this ban necessary? Comment.
Solution:
Amniocentesis is a prenatal diagnostic technique to find out the genetic disorders and metabolic disorders of the foetus. Unfortunately, the useful technique of amnio-centesis had been misused to kill the normal female foetuses as it could help detect the sex of foetus also. Hence, this technique is now banned in our country. This ban is necessary as this technique was promoting female foeticide in our country.

Question 9.
Suggest some methods to assist infertile couples to have children.
Solution:
If the couples are enabled birth to the children and corrections are not possible, the couples could be assisted to have children through certain special techniques, commonly known as Assisted Reproductive Technologies (ART). Some methods are given as:

1. In Vitro Fertilization (IVF): In this method, ova from the female and the sperm from the male are collected and induced to form a zygote under simulated conditions in the laboratory. This process is called In Vitro Fertilization (IVF). Some method is given as follows:

  • Zygote Intrafallopian Transfer (ZIFT): The zygote or early embryo with up to 8 blastomeres is transferred into the fallopian tube.
  • Intra-Uterine Transfer (IUT): Embryo with more than 8 blastomeres is transferred into the uterus in females who cannot conceive embryos formed by the fusion of gametes in another female are transferred.
  • Test tube baby: In this method, ova from the donor (female) and sperm from the donor (male) are collected and are induced to form a zygote under simulated conditions in the laboratory. The zygote could then be transferred into the fallopian tube and embryos transferred into the uterus, to complete its further development. The child born from this method is called a test-tube baby.

2. Gamete Intra Fallopian Transfer (GIFT): It is the transfer of an ovum collected from a donor into the fallopian tube 8 another female who cannot produce one, but can provide a suitable environment for fertilization and further development of the embryo.

3. Intra Cytoplasmic Sperm Injection (ICSI) : It is a procedure to form an embryo HI* the laboratory by directly injecting the sperm into an ovum.

4. Artificial Insemination (AI): In this method, the semen collected either from the husband or a healthy donor is artificially introduced into the vegina or into the uterus (Intra Uterine Insemination, IUI). This technique is used in cases where the male is unable to inseminate sperms in the female reproductive tract or due to very low sperm counts in the ejaculation.

5. Host Mothering: In this process, the embryo is transferred from the biological mother to a surrogate mother. The embryo then develops till it is fully developed or partially developed. It is then transferred to the biological mother or into any other. This technique is useful for females in which embryo forms but is not able to develop.

Question 10.
What are the measures one has to take to prevent contracting STDs?
Solution:
Diseases or infections which are transmitted through sexual intercourse are collectively called sexually transmitted diseases (STDs) or reproductive tract infections (RT), e.g., gonorrhea, syphilis, genital herpes, AIDS, etc. The measures that one has to take to prevent from contracting STDs are:

  • Avoid sex with unknown partners/multiple partners.
  • use condoms during coitus.
  • In case of doubt, go to a qualified doctor for early detection and get complete treatment if diagnosed with the disease.

Question 11.
State True/False with an explanation.

  1. Abortions could happen spontaneously too.
  2. Infertility is defined as the inability to produce viable offspring and is always due to abnormalities/defects in the female partner.
  3. Complete lactation could help as a natural method of contraception.
  4. Creating awareness about sex related aspects is an effective method to improve the reproductive health of people.

Solution:

  1. True: One-third of all pregnancies abort spontaneously (called miscarriage) within four weeks of conception and abortion passes unrecognized with menses.
  2. False: Infertility is defined as the inability of the couple to produce viable offspring. It is due to abnormalities/defects in either male or female or both.
  3. True: Complete lactation is a natural method of contraception as during this period ovulation does not occur, but this is limited to a period of 6 months after parturition.
  4. True: Creating awareness in people about sex-related aspects like right information about reproductive organs, accessory organs of reproduction, safe and hygienic sexual practices, birth control methods, care of pregnant women, post-natal care of mother and child, etc., can help in improving the reproductive health of people.

Question 12.
State True/False with an explanation.
(a) Abortions could happen spontaneously too. (True/False)
Answer:
False, Abortion does not happen under normal conditions. It happens accidentally or under the will of Parents.

(b) Infertility is defined as the inability to produce a viable offspring and is always due to abnormalities/defects in the female partner. (True/False)
Answer:
False, Sterility always does not occur due to females sometimes. Males are also responsible for this.

(c) Complete lactation could help as a natural method of contraception. (True/False)
Answer:
True, the Menstrual cycle does not occur after parturition which can act as natural
contraception but this method is functional for a period of six months from parturition.

(d) Creating awareness about sex related aspects is an effective method to improve the reproductive health of dead people. (True/False)
Answer:
True, this creates better reproductive health among people.

Chapter 5 Principles of Inheritance and Variation

Question 1.
Mention the advantages of selecting pea plant for experiment by Mendel.
Solution:

  1. The plant shows clear-cut contrasting characters.
  2. Hybrids are perfectly fertile.
  3. Genes for the seven contrasting characters are located on seven separate chromosomes.
  4. Easy to cultivate.
  5. The floral structure is suitable for artificial pollination.
  6. Short growth period and life cycle.
  7. Cross-pollination is easy if self-pollination is prevented.
  8. Pure breeding varieties are available

Question 2.
Differentiate between the following:

  1. Dominant and Recessive
  2. Homozygous and Heterozygous
  3. Monohybrid and Dihybrid

Solution:

  1. Differences between dominant and recessive genes are as follows :
    NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation Q2.1
  2. Differences between homozygous and heterozygous are as follows :
    NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation Q2.2
  3. Differences between monohybrid and dihybrid cross are as follows :
    NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation Q2.3

Question 3.
A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
Solution:
A diploid organism heterozygous for 4 loci will have the supported genetic constitution YyRr for two characters. The alleles Y-y and R-r will be present on different 4 loci. Each parent will produce four types of gametes – YR, Yr, yR, yr.

Question 4.
Explain the law of dominance using a monohybrid cross.
Solution:
The Law of dominance states that when a pair of alleles or allelomorphs are brought together in F1 hybrid, then only one of them expresses itself, masking the expression of the other completely. Monohybrid cross was made to study the simultaneous inheritance of a single pair of Mendelian factors. The cross in which only alternate forms of a single character are taken into consideration is called a monohybrid cross. The trait which appeared in the F1 generation was called dominant and the other which did not appear in the F1 population was called recessive.
NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation Q4.1
Thus, when a pair of alleles are brought together in an F1 hybrid, then only one of them expresses itself masking the expression of the other completely. In the above example, in Tt – F1 hybrid (tall) only ‘T’ expresses itself so dominant, and ‘t’ is masked so recessively. Thus, this’ proves and explains the law of dominance.

Question 5.
Define and design a test-cross.
Solution:
The crossing of F1 individuals having dominant phenotype with its homozygous recessive parent is called test cross. The test cross is used to determine whether the individuals exhibiting dominant character are homozygous or heterozygous.
Example: When a tall plant (TT) is crossed with the dwarf plant (tt) in the F1, generation only tall plant (Tt) appears which is then crossed with homozygous recessive (tt) in a test cross.
NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation Q5.1
In the given test cross between tall heterozygous F1 hybrid with dwarf homozygous recessive parent produces tall and dwarf progeny in equal proportion indicating that F : hybrids are heterozygous.

Question 6.
Using a Punnett square, work out the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.
Solution:
When a heterozygous male tall plant (Tt) is crossed with the homozygous dominant female tall plant (TT), we get two types of gametes in males: half with T and a half with t, and in females, we get only one type of gametes i.e., T.
NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation Q6.1
From the Punnett square it is seen that all the progeny in the F generation are tall (Tt), 50% homozygous tall (TT), and 50% heterozygous tall (Tt).

Question 7.
When a cross is made between a tall plant with yellow seeds (TtYy) and a tall plant with the green seed (Ttyy), what proportions of phenotype in the offspring could be expected to be

  1. tall and green
  2. dwarf and green

Solution:
NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation Q7.1
Phenotypes of the offsprings –
Tall Yellow : 3
Tall Green : 3
Dwarf Green: 1
Dwarf Yellow: 1
(a) Proportion of tall and green is 3/8.
(b) Proportion of dwarf and green is 1/8.

Question 8.
Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F1 generation for a dihybrid cross?
Solution:
Two heterozygous parents (i.e. GgLl and GgLl) are crossed and the two loci are linked then the cross will be
NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation Q8.1
This means, if ‘G’ represent grey body (dorhinant), ‘g’ black body (recessive), ‘L’-long (dominant) and ‘I’-dwarf (recessive) then the distribution of phenotypic features in F1 generation will be 3 : 1 i.e. 3/4 will show the dominant feature, grey and long, either in homozygous (GGLL) or in heterozygous (GgLl) condition and 1/4 will show the recessive feature, black and dwarf (ggll).

Question 9.
Briefly mention the contribution of T.H. Morgan in genetics.
Solution:
TH Morgan is a Geneticist who got Nobel Prize.

  • He found fruit fly (Drosophila Melanogaster) to be an experimental material as it was easy to rear and multiply.
  • The established presence of genes over the chromosomes.
  • Principle of linkage and crossing over.
  • Discovered sex linkage and crossing over.
  • He observed mutations.
  • The developed technique of chromosome mapping,
  • Wrote the book “The theory of Gene”.

Question 10.
What is pedigree analysis? Suggest how such an analysis, can be useful.
Solution:
A record of inheritance of certain genetic traits for two or more generations presented in the form of a diagram of family tree is called pedigree. Pedigree analysis is study of pedigree for the transmission of particular trait and finding the possibility of absence or presence of that trait in homozygous or heterozygous state in a particular individual. Pedigree analysis is useful for the following:

  • It is useful for the genetic counsellors to advice intending couples about the possibility of having children with genetic defects like haemophilia, colour blindness, alkaptonuria, phenylketonuria, thalassemia, sickle cell anaemia (recessive traits), brachydactyly and syndactyly (dominant traits).
  • Pedigree analysis indicates that Mendel’s principles are also applicable to human genetics with some modifications found out later like quantitative inheritance, sex linked characters and other linkages.
  • It can indicate the origin of a trait in the ancestors, e.g., haemophilia appeared in Queen Victoria and spread in royal families of Europe through marriages.
  • It helps to know the possibility of a recessive allele to create a disorder in the progeny like thalassemia, muscular dystrophy, haemophilia.
  • It can indicate about the harm that a marriage between close relatives, may cause.
  • It helps to identify whether a particular genetic disease is due to a recessive gene or a dominant gene.
  • In certain cases it may help to identify the genotypes of offspring yet to be born.

Question 11.
How is sex determined in human beings?
Solution:
In humans, there are 23 pairs of chromosomes. 22 pairs of these chromosomes do not take part in sex determination called autosomes. The 23rd pair determines the sex of an individual called allosome or sex chromosome. If it is XX then female, if XY then male. The presence of Y1 makes a person male. Human females produce only 1 type of gamete 22 + X. In males, it could be 22 + X or 22+ Y.
NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation Q11.1

Question 12.
A child has blood group O. If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.
Solution:
If the father has blood group A i.e., IAIA (homozygous) and mother has blood group B i.e., IBIB (homozygous) then all the offsprings will have blood group AB (IAIB) and not blood group O.
NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation Q12.1
NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation Q12.2
Thus the genotypes of the parents of child with blood group O will be IAi and IBi There is the possibility of 3 other types of blood groups of offsprings besides O blood group offspring. They are IAi (blood group A). IBi (blood group B) and IAIB (blood group AB).

Question 13.
Explain the following terms with an example:

  1. Codominance
  2. Incomplete dominance

Solution:
Codominance (1 : 2 : 1) — It is the phenomenon of two alleles (different forms of a Mendelian factor present on the same gene locus on homologous chromosomes) lacking dominant- recessive relationship and are able to express themselves independently when present together.

Example – AB blood group: Alleles for blood group A(IA) and blood group B(IB) are codominant so that when they come together in an individual, they produce blood group AB. It is characterized by the presence of both antigen A (from IA) and antigen B (from IB) over the surface of erythrocytes.
NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation Q13.1

Incomplete dominance (1 : 2 : 1) – It is the phenomenon where none of the two contrasting alleles being dominant so that expression in the hybrid is intermediate between the expressions of the two alleles in the homozygous state. Fphenotypic ratio is 1 : 2 : 1, similar to genotypic ratio. Example-In Mirabilis jalapa (Four o’clock) and Antirrhinum majus (Snapdragon or dog flower), there are two types of flower colour generation are of three types- red, pink and white flowered in the ratio of 1 : 2 : 1. The pink colour apparently appears either due to the mixing of red and white colours (incomplete dominance).
NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation Q13.2

Question 14.
What is point mutation? Give one example.
Solution:
Point mutation is a gene mutation that arises due to a change in a single base pair of DNA.
Example: Sickle-cell anaemia.
Substitution of a single nitrogen base at the sixth codon of the β- globin chain of haemoglobin molecule causes the change in the shape of the R.B.C. from biconcave disc to the elongated shaped, structure which results in sickle cell anaemia.

Question 15.
Who had proposed the chromosomal theory of inheritance?
Solution:
Sutton and Boveri proposed the chromosomal theory of inheritance. The theory believes that chromosomes are vehicles of hereditary information that possess mendelian factors or genes and it is the chromosomes which segregate and assort independently during transmission from one generation to the next.

Question 16.
Mention any two autosomal genetic disorders with their symptoms
Solution:
Cystic fibrosis is an autosomal recessive disorder of infants, children, and young adults that is due to a recessive autosomal allele present on chromosome 7. It is common in Caucasian Northern Europeans and White North Americans. The disease gets its name from the fibrous cysts that appear in the pancreas. In 70% of cases, it is due to the deletion of three bases. It produces a defective glycoprotein. The defective glycoprotein causes the formation of thick mucus in the skin, lungs, pancreas, liver, and other secretory organs. Accumulation of thick mucus in the lungs results in obstruction of airways. Because of it, the disease was also called mucoviscoides, Mucus deposition in the pancreas blocks secretion of pancreatic juice. There is a maldigestion of food with high-fat content in the stool. The liver may undergo cirrhosis and there is impaired production of bile. Vasa deferentia of males undergo atrophy.

Huntington’s disease or Huntington’s chorea is a dominantly autosomal inherited disorder in which muscle and mental deterioration occur. There is gradual loss of motor control resulting in uncontrollable shaking and dance-like movements (chorea). The brain shrinks between 20-30% in size followed by slurring of speech, loss of memory, and hallucinations. Life expectancy averages 15 years from the onset of symptoms. This disorder does not occur till the age of 25 to 55. The defective gene is dominant autosomal, located on chromosome 4. This defective gene has 42 -100 repeats of CAG instead of 10-34 repeats in the normal gene. The frequency of this disorder is 1 in 10000 to 1 in 20000.

Chapter 6 Molecular Basis of Inheritance

Question 1.
Group the following as nitrogenous bases and nucleosides:
Adenine, cytidine, thymine, guanosine, uracil, and cytosine.

Solution:
Adenine, Guanosine, Thymine, Uracil, and Cytosine are nitrogenous bases. (Adenine and Guanosine → Purine, Thymine, Uracil and Cytosine → Pyrimidine) Cytidine is a nucleoside.

Question 2.
If a double-stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA.
Solution:
According to Chargaff’s rule, in a double-stranded DNA, the total number of cytosine molecules will be equal to the number of guanine molecules and the number of adenine molecules will be equal to the number of thymine molecules. Therefore, if a double-stranded DNA has 20 percent of cytosine then the guanine will also be 20 per cent. The remaining 60% will consist of adenine and thymine in equal amount. Thus adenine will be 30%.

Question 3.
If the sequence of one strand of DNA is written as follows:
5′-ATGCATGCATGCATGCATGCA
TGCATGC-3′
Write down the sequence of complementary strand in 5′ -> 3′ direction.
Solution:
5′-GCATGCATGCATGCATGCAT G C ATG CAT-3′.

Question 4.
If the sequence of the coding strand in a transcription unit is written as follows: 5′-ATGCATGCATGCATGCATGCATGCATGC-3′ Write down the sequence of mRNA.
Solution:
If the sequence of coding strand is :
5′ – ATGCATGCATGCATGCATGCATGCATGC – 3′
Then template strand is :
3′ – TACGTACGTACGTACGTACGTACGTACG – 5′
The mRNA will be formed on the template strand in 5′ —> 3’ direction. Thus mRNA sequence will be:
5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′
Thymine in DNA is substituted by uracil in RNA.

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise a semi-conservative mode of DNA replication? Explain.
Solution:
The two strands of DNA show complementary base pairing. This property of DNA led Watson and Crick to suggest a semi-conservative mechanism of DNA replication in which one strand of a parent is conserved while the other complementary strand formed is new.

Question 6.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases
Solution:
DNA dependent DNA polymerases and DNA dependent RNA polymerases.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic
material?
Solution:
They raised 2 types of bacteriophages

  • On radioactive phosphorous (32P)
  • On radioactive sulphur (35S).

35S gets into protein and 32P into DNA When both bacteriophages infected bacteria differently and by shaking them, the viral protein coat was separated

After raising these bacteria it was found that those infected with 32P bacteriophage → radioactivity were found. But with 35S → no radioactivity was found.

Question 8.
Differentiate between the following:

  1. Repetitive DNA and Satellite DNA
  2. Template strand and Coding strand
  3. mRNA and tRNA

Solution:

  1. Differences between repetitive DNA and satellite DNA are as follows:
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.1
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.2
  2. Differences between template strand and coding strand are as follows:
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.3
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.4
  3. Differences between mRNA and tRNA are as follows:
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.5
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.6
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.7
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.8

Question 9.
List two essential roles of ribosome during translation
Solution:
Two essential roles of the ribosome during translation are:

  1. One of the RNA acts as a peptidyl transferase ribozyme for the formation of peptide bonds.
  2. The ribosome provides sites for attachment of mRNA and charged tRNA for polypeptide synthesis.

Question 10.
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
Solution:
The lac operon is regulated by the amount of lactose in the medium where the bacteria are grown. When the amount of lactose is exhausted in the medium, the lac operon shuts down.

Question 11.
Explain (in one or two lines) the function of the followings:

  1. Promoter
  2. tRNA
  3. Exons

Solution:

  1. Promoter: It is located at the 5′ end of the transcription unit and provides site for attachment of transcription factors (TATA Box) and RNA polymerase.
  2. tRNA: It takes part in the transfer of activated amino acids from cellular pool to ribosome so that they can take part in protein formation.
  3. Exons: In eukaryotes, DNA is mosaic of exons and introns. Exons are coding sequences of DNA which are both transcribed and translated.

Question 12.
Why is the Human Genome Project called a mega project?
Solution:
The human genome was a megaproject that aimed to sequence every base in the human genome. The estimated cost of the project would be a billion (1 billion = 100 crores) US dollars.

Question 13.
What is DNA fingerprinting? Mention its application.
Solution:
DNA fingerprinting is the identification of differences in specific regions of DNA sequences based on DNA polymorphism, repetitive DNA, and satellite DNA.
Application of DNA fingerprinting: Settling, paternity disputes and identity of criminal by different DNA profiles in forensic laboratories.

Question 14.
Briefly describe the following:

  1. Transcription
  2. Polymorphism
  3. Translation
  4. Bioinformatics

Solution:
1. Transcription – It is the process of copying genetic information from the anti-sense or template strand of the DNA into RNA. It is meant for taking the coded information from DNA in nucleus to the site where it is required for protein synthesis. Principle of complementarity is used even in transcription. The exception is that uracil is incorporated instead of thymine opposite adenine of template. The segment of DNA that takes part in transcription is called transcription unit. It has three components

    • a promoter,
    • the structural gene and
    • a terminator.

2. Polymorphism – It is the variation at genetic level, arisen due to mutations. Such variations are unique at particular site of
DNA. They occur approximately once in every 500 nucleotides or about 107 times per genome. These are due to deletions, insertions, and single-base substitutions. These alterations in healthy people, occur in non-coding regions of DNA and do not code for any protein but are heritable. The polymorphism in DNA sequences is the basis of genetic mapping of human genome as well as DNA fingerprinting.

3. Translation – It is the mechanism by which the triplet base sequence of mRNA guides the linking of a specific sequence of amino acids to form a polypeptide chain (protein) on ribosomes in the cell cytoplasm. All the protein that a cell needs are synthesised by the cell within itself.
The raw materials required in protein synthesis are ribosomes, amino acids, mRNA, tRNAs and amino acyl tRNA synthetase. Mechanism of protein synthesis involves following steps:

    • Activation of amino acids
    • Charging or aminoacylation of tRNA
    • Initiation
    • Elongation (Polypeptide chain formation)
    • Termination

The ribosomes move along the mRNA ‘reading’ each codon in turn. Molecules of transfer RNA (tRNA), each bearing a particular amino acid, are brought to their correct positions along the mRNA, molecule base pairing occurs between the bases of the codons and the complementary base triplets of tRNA. In this way, amino acids are assembled in the correct sequence to form the polypeptide chain.

4. Bioinformatics – Bioinformatics is the combination of biology, information technology and computer science. Basically, bioinformatics is a recently developed science which uses information technology to understand biological phenomenon. It broadly involves the computational tools and methods used to manage, analyse and manipulate volumes of biological data.

Chapter 7 Evolution

Question 1.
Explain antibiotic resistance observed in bacteria in light of Darwinian selection theory.
Solution:
Penicillin when discovered was used as an antibiotic against all bacteria. Soon many of these became resistant. This is because alleles of resistance which are already present in bacteria are of no importance in absence of antibiotics. Adjustment to change in environment due to genetic variation is adaptation.

Question 2.
Find out from newspapers and popular science articles any new fossil discoveries or controversies about evolution.
Solution:
Chimps are more evolved than humans (The Times of India):
Chimpanzees are more evolved than humans, a study suggests. There is no doubt that humans are the more advanced species. But a comparison of 14,000 human and chimpanzee genes shows the forces of natural selection have and the greatest impact on our ape cousins.

The researchers’ discovery challenges the common assumption that our large brains and high intelligence were the gifts of natural selection. Humans and chimps followed different evolutionary paths from a common ape ancestor about 5 million years ago. Both underwent changes as the fittest survived to pass their genes on to future generations. But the US study shows that humans possess a ‘substantially smaller’ number of positively-selected genes than chimps.

Question 3.
Attempt giving a clear definition of the term species.
Solution:
A species generally includes a similar organism. Members of this group can show interbreeding. A similar group of genes are found in the members of the same species and this group has the capacity to produce new species. Every species has some cause of isolation which interrupted the interbreeding with the nearest reactional species which is referred to as reproductively isolated.

Question 4.
Try to trace the various components of human evolution (hint: brain size and function, skeletal structure, dietary preference, etc.)
Solution:
Human evolution shows the following trends:
A. Brain size: It increased gradually along with evolution. The brain capacity of Australopithecus africanus – 500 cc, Homo habilis – 700 cc, Homo eredus – 800 – 1300 cc, Homo sapiens sapiens – 1450 cc.

B. Skeletal structure:

  • Dryopithecus was ape-like, without brow ridges, had semierect posture, and prognathous face (having a projecting jaw).
  • Ramapithecus had jaws and teeth like humans (small canines and large molars), prognathous face, and walked on legs
  • Australopithecus africanus had erect posture, human-like teeth, was without chin, with brow ridges, and had a prognathous face.
  • Homo habilis walked nearly erect, had human-like teeth, with brow ridges face was slightly prognathous.
  • Homo erectus had an erect posture, prognathous face, with projecting brow ridges, small canines, and large molar teeth and had a small chin.
  • Homo sapiens had four curves in the vertebral column, orthognathous face (without projecting jaw), forehead broad, chin well developed, walked on the sole.

C. Dietary preference: Dryopithecus and Ram-apithecus were herbivores, Australopithecus africanus and Homo habilis were carnivores, Homo erectus and Homo sapiens were omnivores.

Question 5.
Find out through the internet and popular science articles whether animals other than man have self-consciousness.
Solution:
There are many animals other than humans, which have self-consciousness. An example of an animal being self-conscious is dolphins. They are highly intelligent. They have a sense of self and, they also recognize others among themselves and others. They communicate with each other by whistles, tail-slapping, and other body movements, not dolphins, there are certain other animals such as Crow, Parrot, chimpanzees, Gorilla, Orangutan, etc., which exhibit self-consciousness.

Question 6.
List 5-6 modern-day animals and using the internet resources link it to a corresponding ancient fossil. Name both.
Solution:
The list of few modern-day animals and their corresponding ancient fossils is as follows:
NCERT Solutions for Class 12 Biology Chapter 7 Evolution Q6.1
NCERT Solutions for Class 12 Biology Chapter 7 Evolution Q6.2
NCERT Solutions for Class 12 Biology Chapter 7 Evolution Q6.3

Question 7.
Describe one example of adaptive radiation.
Solution:
Adaptive radiation – Formation of different species from a common ancestor with new species adapting to different geological niches.
Example: Darwin’s finches are Galapagos island have wolves from mainland finches. They underwent changes in the shape, size of beaks, food habits, feathers.

Question 8.
Can we call human evolution adaptive radiation?
Solution:
No, we can not be called human evolution as adaptive evolution.

Question 9.
Using various resources such as your school library or the Internet and discussions with your teacher, trace the evolutionary stages of any one animal say horse.
Solution:
The evolutionary stages of the modern horse are listed in the table given below:
NCERT Solutions for Class 12 Biology Chapter 7 Evolution Q9.1

Chapter 8 Human Health and Diseases

Question 1.
What are the various public health measures, which you would suggest as a safeguard against infectious diseases?
Solution:
Prevention and control of infectious diseases

I. For water-borne diseases like typhoid, amoebiasis, etc.
Practice personal and public hygienic measures.

a. Personal hygienic measures

  • Keeping the body clean
  • Consumption of clean drinking water
  • Eating fresh food

b. Public hygienic measures

  • Proper disposal of waste and excreta
  • Periodic cleaning and disinfection of water reservoirs, pool, tank etc.

II. For air-borne diseases like common cold, pneumonia

  • Avoid close contact with infected persons.
  • Avoid the use of belongings of the infected persons.

III. For vector-borne diseases like malaria

  • Control and eliminate the vectors and their breeding places
  • Introducing larvivorous fishes like Gambusia in ponds that feed on the larvae of the mosquito
  • Avoid stagnation of water around the residential area.
  • Spraying of insecticides in ditches, drainage areas, etc.
  • Protection from a mosquito bite. Use mosquito nets in the doors and windows to prevent the entry of mosquitoes. It is very important in the light of recently widespread diseases like dengue fever, chikungunya etc.

The use of vaccines and immunization programmes has enabled us to eradicate smallpox. Diseases like polio, diphtheria, tetanus etc. have been controlled to an extent by the use of vaccines. Nowadays biotechnology is focussing on the preparation of newer and safer vaccines. A large number of antibiotics are available to treat many infectious diseases.

Question 2.
In which way has the study of biology helped us to control infectious diseases?
Solution:
Study of biology has helped us to know about causes of diseases, carriers of diseases (vectors), effects of diseases on different body functions and above all, means to control diseases. Our immune system plays a major role in preventing diseases.

Question 3.
How does the transmission of each of the following diseases take place ?

  1. Amoebiasis
  2. Malaria
  3. Ascariasis
  4. Pneumonia

Solution:

  1. Through contaminated food and water.
  2. Through Anopheles mosquito.
  3. Through contaminated food and water.
  4. By inhaling the droplets or aerosols released by infected persons.

Question 4.
What measure would you take to prevent water-borne diseases?
Solution:
Water-borne diseases can be prevented by drinking clean water. Water should be free from contamination, suspended and dissolved substances. If water is contaminated it should be boiled and filtered before drinking. Periodic cleaning and disinfection of water reservoirs, pools, and tanks should be done.

Question 5.
Discuss with your teacher what does ‘a suitable gene’ means, in the context of DNA vaccines.
Solution:
‘A suitable gene’ means the gene which is able to produce antigenic polypeptides of the pathogen in bacteria and yeast. Using recombinant DNA technology, it is possible to produce vaccines in large scale for immunisation. Hepatitis B vaccine is produced using this technology.

Question 6.
Name the primary and secondary lymphoid organs.
Solution:
Primary lymphoid organs are bone marrow and thymus. Secondary lymphoid organs are the spleen, lymph nodes, tonsils, Peyer’s patches of the small intestine, and mucosa-associated lymphoid tissues (MALT).

Question 7.
The following are some well-known abbreviations, which have been used in this chapter. Expand each one to its full form.

  1. MALT
  2. CMI
  3. AIDS
  4. NACO
  5. HIV

Solution:

  1. MALT – Mucosal-associated lymphoid tissue.
  2. CMI – Cell-Mediated Immunity
  3. AIDS – Acquired Immuno Deficiency Syndrome
  4. NACO – National AIDS Control Organisation
  5. HIV – Human immunodeficiency virus.

Question 8.
Differentiate the following and give examples of each.

  1. Innate and acquired immunity,
  2. Active and passive immunity

Solution:

  1. : Differences between innate and acquired immunity are as follows:
    NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Diseases Q1.1
  2. Differences between active and passive immunity are as follows:
    NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Diseases Q1.2

Question 9.
Draw a well-labeled diagram of an antibody molecule.
Solution:
NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Diseases Q9.1

Question 10.
What are the various routes by which transmission of the human immunodeficiency virus takes place?
Solution:
Various routes of entry of ADDS virus are:

  • Sexual contact with the infected person.
  • Through placenta (from infected mother to foetus).
  • Transfusion of infected blood or blood products.
  • Sharing infected needles by drug abusers.

Question 11.
What is the mechanism by which the AIDS virus causes a deficiency of the immune system of the infected person?
Solution:
After getting into the body, the virus enters into the macrophages and converts its RNA genome into DNA with the help of a reverse transcriptase enzyme. The viral DNA takes and directs the infected cells to produce more virus particles i.e., the infected macrophages act like an HIV factory. Simultaneously, the HIV attack the T- lymphocytes and replicate and produce more viruses. Then they are released into the blood and attack other T-lymphocytes.

This will lead to a decrease in the number of T-lymphocytes and the patient begins to show the symptoms such as fever, diarrhea, weight loss etc. Subsequently, his immune system weakens and becomes more prone to infections of bacteria (like Mycobacterium), viruses, fungi and even parasites like Toxoplasma. Finally, he is unable to protect himself.

Question 12.
How is a cancerous cell different from a normal cell?
Solution:
Cancerous cell and normal cell are different in the following aspects:
NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Diseases Q12.1
NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Diseases Q12.2

Question 13.
Explain what is meant by metastasis.
Solution:
The rapid growth of cancerous tumour causes overcrowding and disruption of normal cells. It extends to neighbouring tissues. In the last stage, bits of tumour tissue break off and are carried by the circulating blood or lymphs to other parts of the body, where they invade new tissues and start new tumors called secondary tumors. This property is called metastasis. It is fated due to increasing interference with the body’s life processes.

Question 14.
List the harmful effects caused by alcohol/drug abuse.
Solution:
Harmful effects caused by alcohol/drug abuse are as follows:

  • Among youth there is drop in academic performance, lack of interest in personal hygiene, isolation, depression, fatigue, aggressive and rebellious behavior, deteriorating relationships with family and friends, loss of interest in hobbies, change in sleeping and eating habits, fluctuations in weight, appetite, etc.
  • Excessive dose of drugs leads to coma and death due to respiratory failure, heart failure or cerebral haemorrhage.
  • Abusers become mental and cause financial distress to their entire family and friends.
  • They may acquire serious infections like AIDS and hepatitis by taking drugs intravenously.
  • Intake of alcohol/drugs damages nervous system, liver (cirrhosis) and kidney.
  • Drug abuse adversely affects foetus in case of pregnancy, leading to Foetal Alcohol Syndrome (FAS).
  • Continuous use of narcotics and stimulants cause impotency and chromosomal aberrations.
  • Heavy drinking can cause an acute alcoholic myopathy characterised by painful and swollen muscles and high levels of serum creatine phosphokinase (CK). Chronic alcoholic men may show testicular atrophy with shrinkage of the seminiferous tubules and loss of sperm cells.
  • Heavy drinking causes acute and chronic pancreatitis.
  • Alcohol increases RBC size causing a mild anemia.
  • Legal problems occur, such as arrest by police for obtaining and keeping drugs unlawfully.

Question 15.
Do you think that friends can influence one to take alcohol/drugs? If yes, how can one protect himself/herself from such an influence?
Solution:
Yes. This can be avoided by

  • Choosing a good peer group.
  • Discussing ways and means to counteract the presence if any with family elders and teacher/counselors
  • Telling the program of an outing to family.
  • Keeping contact with family while outside the home.

Question 16.
Why is it that once a person starts taking alcohol or drugs, it is difficult to get rid of this habit? Discuss it with your teacher.
Solution:
Once a person starts taking alcohol or drugs, it is difficult to get rid of this habit because he becomes addicted to it. Addiction is a psychological attachment to certain effects such as euphoria and a temporary feeling of well-being. These drive people to consume drugs/alcohol even when these are not needed, or even when their use becomes self-destructive. With repeated use, the tolerance level of the receptors present in the body increases, which consequently leads to a higher dose of drugs/alcohol and addiction.

Thus, the addictive potential of drugs and alcohol pull the user into a vicious circle leading to their regular use from which he/she may not be able to get out.

Question 17.
In your view what motivates youngsters to take alcohol or drugs and how can this be avoided?
Solution:
There are many factors that motivate youngsters to take alcohol or drug. These include:

  • Choosing a good peer group.
  • Discussing ways and means to counteract the presence if any with family elders and teacher/counselors
  • Telling the programme of an outing to family.
  • Keeping contact with family while outside the home.

This can be avoided by the following measures:

  1. Education and counseling: Educating and counseling people to face problems and stresses, and to accept disappointments and failures as a part of life.
  2. Seeking help from parents and peers: Help from parents and peers should be sought immediately so that they can guide appropriately. Help may even be sought from close and trusted friends.
  3. Looking for danger signs: Alert parents and teachers to look for and identify the danger signs. Even friends, if they find someone using drugs or alcohol, should not hesitate to bring this to the notice of parents or teachers in the best interests of the person concerned.
  4. Seeking professional and medical help: Lots of help is available in the form of highly qualified psychologists, psychiatrists, and de-addiction and rehabilitation programmes to help individuals who have unfortunately got in the quagmire of drug/alcohol abuse.
  5. Cross-checking before prescribing and selling drugs: The physicians should prescribe the habituating drugs only to genuine persons and only for the essential duration. Pharmacists should not sell these drugs without the physician’s prescription.
  6. Discipline: Good nurturance with consistent discipline but without suffocating strictness reduces the risk of addictions.
  7. Communication: The child must be able to communicate with the parents seeking clarification of all doubts and discussing problems that arise in studies or develop in the class, with friends, siblings and others.
  8. Appreciation: For even the smallest achievement, good behavior and other activities, the child should be appreciated.
  9. Independent working: Giving responsibility to the child for small tasks and allowing him/her to perform independently. However, guidance should be provided where required.
  10. Avoid undue pressure: Every child has a specific personality with certain preferences and choices. They should be taken care of and respected. No child should be asked to perform beyond threshold limits whether in studies, sports or extracurricular activities.

Chapter 9 Strategies for Enhancement in Food Production

Question 1.
Explain in brief the role of animal husbandry in human welfare.
Solution:

  • Animal husbandry evolves new techniques and technologies for the management of livestock like buffaloes, cows, pigs, horses, cattle, sheep, camels, goats, etc., that are useful to humans.
  • These methods can also be applied to rearing animals like bees, silkworms, prawns, crabs, fishes birds, pigs, cattle, sheep, and camels for their products like milk, eggs, meat, wool, silk, honey, etc.

Role of animal husbandry in human welfare is discussed as follows:

  1. Milk is an important product of farm animals that are consumed as such, in the form of curd, cheese, butter, ice cream, etc. Milk is the only source of animal protein for vegetarians and is a complete food. Most of the milk is obtained from cows and buffalo. Other milk-yielding animals are goat, sheep, camel, and yak.
  2. Egg, like milk, is also a complete food. Chicken and duck are the two major sources of the egg.
  3. Meat is a protein-rich diet that is obtained from all types of livestock, e.g., goat, sheep, pig, cattle, chicken, fish, etc.
  4. Honey is a sweet syrup obtained from the hives of the honey bee. Honey is used in sweetening various preparations.
  5. Fibers like wool and silk are two high-quality fibres which we get from animals. Wool is the hair of sheep, some goats, and rabbits. Silk is a product of silkworms.
  6. The skins of many animals are converted into hides and leather.
  7. Drought animals are trained to carry men and materials besides other functions, e.g., buffalo, bullock, horse, camel, ass, elephant, reindeer, yak.
  8. The rearing of animals provides employment to many persons.
  9. Animal byproducts like horns, feathers, bone, dung, and droppings are all used in developing useful products.

Question 2.
If your family owned a dairy farm, what measures would you undertake to improve the quality and quantity of milk production?
Solution:
Some of the measures to be followed for proper management of dairy farm are :

  1. Selection of good breeds having high milk yielding potential according to the climatic conditions of the area.
  2. The shed under which the cattle are kept should be well ventilated with an adequate water supply for drinking as well as for washing. Shed should have pucca floor and proper drainage channel.
  3. The feed of the animals should be a balanced diet with right proportions of carbohydrates, fats, proteins, and roughage and it should be given timely in good quantity.
  4. Cleanliness and hygiene comes first for maintaining the livestock’s health and productivity. So, washing cattle and taking precautionary measures while milking are a must.
  5. Inspection, keeping records of the activities and consulting a veterinary doctor for regular checkups of the livestock should be undertaken.

Question 3.
What is meant by the term ‘breed’? What are the objectives of animal breeding?
Solution:
A group of animals which are related by descent to each other and possess similar characteristics like appearance, size, features etc. are said to belong to a breed. The purpose of animal breeding is to produce animals with increased yield, faster growth, improved reproductive rate.

Question 4.
Name the methods employed in animal breeding. According to you which of the methods is best? Why?
Solution:
Animal breeding is producing improved breeds of domesticated animals by improving their genotypes through selective mating. There are two methods of animal breeding, natural breeding which includes inbreeding, out-breeding, cross-breeding, out-crossing, etc., and artificial breeding which involves artificial insemination and multiple ovulation embryo transfer technology (MOET). It involves inseminating the native cows with the semen of superior bulls of exotic or indigenous breeds. Artificial breeding is the best method of breeding because of the following reasons:

  • Semen collected from males may be used immediately or can be frozen and used later.
  • The semen of desired bulls is collected under hygienic conditions, preserved, and sent to all insemination centres throughout the country.
  • Semen collected from one bull can be used to inseminate many cows as fewer sperms are required to achieve conception when semen is deposited artificially. Hence, artificial insemination is very economical.
  • It is healthier as the spread of sexually transmitted diseases can be controlled by this technique.

Question 5.
What is apiculture? How is it important in our lives?
Solution:
Apiculture is the practice of bee-keeping for the production of various products such as honey-bee’s wax, etc. Honey is a highly nutritious food source and is used as an indigenous system of medicines. Other commercial products obtained from honeybees include bee’s wax and bee pollen. Bee’s wax is used for making cosmetics, polishes and is even used in several medicinal preparations. Therefore, to meet the increasing demand of honey, people have started practicing bee-keeping on a large scale. It has become an income-generating activity for farmers since it requires low investment and is labour intensive.

Question 6.
Discuss the role of fishery in the enhancement of food production.
Solution:
Fishery is the rearing, breeding, catching & marketing of fishes and other aquatic animals. Fishes are important food for a large portion of human population. Meat of fishes is a rich source of proteins and other useful substances like polyunsaturated fatty acids (PUFA). The meat of other aquatic animals like prawn, crab is also consumed as food by human beings.

Question 7.
Briefly describe various steps involved in plant breeding.
Solution:
The major steps in breeding a new genetic variety of a crop are as follows:

  1. Collection of variability.
  2. Evaluation and selection of parents.
  3. Cross-hybridization among the selected parents.
  4. Selection and testing of superior recombinants.
  5. Testing, release, and commercialization of new cultivars.

Question 8.
Explain what is meant by biofortification.
Solution:
Biofortification is method of breeding crops with higher levels of vitamins and minerals, or higher proteins and healthier fats in view to improve public health. E.g., iron-fortified rice containing five times more iron than other varieties, wheat variety, Atlas 66 having high protein content, maize varieties having high lysine and tryptophan are produced.

Question 9.
Which part of the plant is best suited for making virus-free plants and why?
Solution:
The terminal bud having apical meristem are the best-suited parts of the plant for making a virus-free plant because they are not infected by a virus.

Question 10.
What is the major advantage of producing plants by micropropagation?
Solution:
Micropropagation is the tissue culture technique used for rapid vegetative multiplication of ornamental plants and fruit trees by using small-sized explants. Because of the minute size of the propagules in the culture, the propagation technique is named micropropagation. This method of tissue culture produces several plants. Each of these plants will be genetically identical to the original plant from which explants were taken. Plants obtained by vegetative propagation of a single plant constitute a somaclonal. The members of a single somaclonal have the same genotype. It is the only process adopted by Indian plant biotechnologists in different industries mainly for the commercial production of ornamental plants like lily, orchids, Eucalyptus, Cinchona, blueberry, etc., and fruit trees like tomato, apple, banana, grapes, potato, Citrus, palm, etc.

Question 11.
Find out what the various components of the medium used for propagation of an explant in vitro are.
Solution:
The major components of the medium for in-vitro propagation are:

  • Water
  • Agar-agar
  • Sucrose
  • Inorganic salts
  • Vitamins
  • Amino acids
  • Growth hormones like Auxin, Cytokinins.

Question 12.
Name any five hybrid varieties of crop plants which have been developed in India.
Solution:
Some of the hybrid varieties of plants in India are:

  • Pusa Gaurav
  • Pusa Sem 2
  • Pusa Sem 3
  • Pusa Sawani
  • Pusa A-4

Chapter 10 Microbes in Human Welfare

Question 1.
Bacteria cannot be seen with the naked eyes, but these can be seen with the help of a microscope. If you have to carry a sample from your home to your biology laboratory to demonstrate the presence of microbes under a microscope, which sample would you carry and why?
Solution:
Curd. It is simple to carry and easily demonstrate the presence of Lactobacillus bacteria.

Question 2.
Give examples to prove that microbes release gases during metabolism.
Solution:

  • Making of dough for bread, dosa, and idli with the help of fermenting microbes. Heat expels the gases and makes the food spongy.
  • Production of biogas

Question 3.
In which food would you find lactic acid bacteria? Mention some of their useful applications.
Solution:
Curd. LAB (Lactic Acid Bacteria) produce acids that coagulate and partially digest the milk protein. LAB also play very beneficial role in the stomach to check disease-causing microbes.

Question 4.
Name some traditional Indian foods made of wheat, rice, and Bengal gram (or their products) which involve use of microbes.
Solution:
Idli, Dhokla, Dosa. Several food items such as dosa, idli, jalebi and bread are prepared by fermentation process in which one or more kinds of microbes are used.

Question 5.
In which way have microbes played a major role in controlling diseases caused by harmful bacteria?
Solution:
Microbes are very useful to combat disease-causing harmful bacteria. A number of antibiotics have been isolated from microorganisms. An antibiotic is a substance which in low concentration inhibits the growth and metabolic activity of pathogenic organisms without harming the host. Penicillin was the first antibiotic to be discovered by Alexander Fleming from fungus Penicillium notation. Antibiotics are obtained from lichens, fungi, eubacteria, and actinomycetes. Some common antibiotics and their sources are as follows :

  1. Polymyxin – Bacillus polymyxa
  2. Chloramphenicol – Streptomyces venezuelae
  3. Neomycin – Streptomyces fradiae
  4. Tetracycline (Terramycin) – Streptomyces rimosus
  5. Cephalosporin – Cephalosporium acremonium

Question 6.
Name any two species of fungus, which are used in the production of the antibiotics
Solution:
Penicillium notatum provides antibiotic penicillin and antibiotic fumagillin is obtained from Aspergillus fumigatus.

Question 7.
What is sewage? In which way can sewage be harmful to us?
Solution:
The wastewater containing large amounts of organic matter and microbes is called sewage. The sewage contains many harmful pathogens. It will pollute the natural water bodies like rivers and streams if it is released into them. This will in turn cause the spreading of many communicable diseases which are transmitted through contaminated food and water.

Question 8.
What is the key difference between primary and secondary sewage treatment?
Solution:
Primary (1°) treatment is a physical process that involves for removal of particulate as settleable particles. Secondary (2°) treatment is purely a biological treatment involving microbial oxidation.

Question 9.
Do you think microbes can also be used as a source of energy? If yes, how?
Solution:

Biogas is a mixture of gas containing methane, hydrogen sulphide and carbon dioxide. Methane is the predominant gas. It is produced by microbial activity and is used as a source of energy. The types of gases produced depend upon the types of organic materials they (microbes) use. The bacteria grow anaerobically on cellulose, produce a large amount of meth­ane along with hydrogen and carbon dioxide. These bacteria are collectively called methanogens.

One of the common methanogens is Methanobacterium. Methanogens are commonly found in sewage. They are also found in the rumen (a chamber of the compound stomach) of cattle where a large amount of cellulosic food is present. These bacteria help in the breakdown of cellu­lose and thus play a vital role in digestion (symbiotic digestion). Thus, the excreta (dung) of cattle, commonly called gobar can be used for the produc­tion of biogas and hence called gobar gas.

Question 10.
Microbes can be used to decrease the use of chemical fertilizers and pesticides. Explain how this can be accomplished.
Solution:
Due to its hazardous nature and anti-environmental effect the use of chemical fertilisers and chemical insecticides are very illegitimate. The development of biofertilisers and bioinsecticides has enabled us to reduce the use of chemical fertilisers and chemical insecticides. Microbes are very important biological agents as biofertilisers and biopesticides.
Microbes as biofertilisers

  1. Free-living nitrogen-fixing bacteria – Azotobactcr, Clostridium.
  2. Free-living nitrogen-fixing cyanobacteria – Anabaena, Nostoc.
  3. Symbiotic nitrogen-fixing bacteria – Rhizobium
  4. Symbiotic nitrogen-fixing cyanobacteria – Azolla – Anabaena
  5. Mycorrhiza – symbiotic association between fungi and roots of a higher plant.

Microbes as bio-pesticides:
Biopesticides are that biological agents that are used to control weeds, insects, and pathogens. The microorganisms used as biopesticides are viruses, bacteria, protozoa, fungi, and mites. Some of the biopesticides are being used at a commercial scale. Most important example is the soil bacterium Bacillus thuringiensis (Bt). Spores of this bacterium produce the insecticidal Cry protein. This bacterium was the first biopesticide to be used on a commercial scale in the world. Through the use of genetic engineering, the scientists have introduced the B. thuringiensis toxin gene into plants. Such plants are resistant to attack by insect pests.

Question 11.
Three water samples namely river water, untreated sewage water and secondary effluent discharged from a sewage treatment plant were subjected to BOD test. The samples were labelled A, B and C; but the laboratory attendant did not note which was which. The BOD values of the three samples A, B and C were recorded as 20mg/L, 8mg/L and 400mg/L, respectively. Which sample of the water is most polluted? Can you assign the correct label to each assuming the river water is relatively clean?
Solution:
If the BOD level of A, B and C are 20 mg/L, 8 mg/L and 400 mg/L, the /most polluted one is sample ‘C’ i.e., untreated sewage water. Sample B is the least polluted and it is river water. Sample A is secondary effluent discharged from a treatment plant.

Question 12.
Find out the name of the microbes from which Cyclosporin A (an immunosuppressive drug) and Statins (blood cholesterol-lowering agents) are obtained.
Solution:
Cyclosporin A – Trichoderma polysporum. Statin – Monascus purpureus.

Question 13.
Find out the role of microbes in the following:

  1. Single-cell protein (SCP)
  2. Soil

Solution:
1. Single-cell protein (SCP) is the protein extracted from cultivated microbial biomass. Edible mushrooms, yeast cells, and blue-green algae provide high protein resources with some medicinal effects. SCP is available in the forms of fresh cells, dry cells, tablets, or extracts.

2. The use of biological methods for controlling plant diseases and pests is called biocontrol. This method has been replaced by the indiscriminate use of chemicals used in the chemical control of pests. The chemicals used for killing pests are toxic and extremely harmful to men and domestic animals. They also pollute the environment and our crop plants.

Question 14.
Arrange the following in the decreasing order (most important first) of their importance, for the welfare of human society. Give reasons for your answer. Biogas, citric acid, penicillin and curd
Solution:
The correct order from viewpoint of human welfare should be as follows: Penicillin > biogas > citric acid > curd Penicillin is an antibiotic, which is used to combat pathogenic microorganisms. Today, we cannot imagine a world without antibiotics because antibiotics have greatly improved our capacity to treat deadly diseases such as plague, whooping cough (kali khansi), diphtheria (gal ghotu) and leprosy (kusht rog), which used to kill millions of people all over the globe. It should be given first priority.

As we are going to face a great crisis of fossil fuels in near future, biogas can be the legitimate and brilliant alternative to fossil fuels. It can be used as fuel for heating, cooking, lighting, power for irrigation and other purposes. It is considered an eco-friendly and pollution-free energy source.
Citric acid is produced through the fermentation carried out by Aspergillus niger on many carbohydrates. Citric acid is used in medicines, dyeing, mirror silvering, manufacture of ink, flavouring and preservation of food and candies.

Curd is prepared by fermentation of milk. Fermentation agents are lactic acid bacteria. Curd is more nutritious than milk as it contains a number of vitamins and organic acids.

Question 15.
How do biofertilisers enrich the fertility of the soil?
Solution:
The use of biological methods for controlling plant diseases and pests is called biocontrol. This method has been replaced by the indiscriminate use of chemicals used in the chemical control of pests. The chemicals used for killing pests are toxic and extremely harmful to men and domestic animals. They also pollute the environment and our crop plants.

Chapter 11 Biotechnology: Principles and Processes

Question 1.
Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate DNA on which it acts, the site at which it cuts DNA and the product it produces.
Solution:
Restriction enzyme – Eco RI
Source – Escherichia coli RY13
NCERT Solutions for Class 12 Biology Chapter 11 Biotechnology Principles and Processes Q1.1

Question 2.
Can you list 10 recombinant proteins which are used in medical practice? Find out where they are used as therapeutics.
Solution:
The proteins produced through rDNA technology and being used in the medical practice include rh-Growth Hormone, r-Human insulin, Erythropoietin, Follicle stimulating hormone, Interferon, Insulin-like growth factor, Tissue Plasminogen Activator, factor VIII, DNase, the Envelope protein of hepatitis B virus.

Question 3.
From what you have learned, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?
Solution:
Enzymes are bigger than DNA as they are proteins and proteins are macromolecules made of amino acids which are bigger than nucleotides. This can also be proved by gel electrophoresis, where denatured protein would not move but denatured DNA will move to a distance. Protein synthesis is regulated by small portions of DNA, called genes.

Question 4.
What would be the molar concentration of human DNA in a human cell? Consult your teacher.
Solution:
The average molecular weight of a nucleotide in human DNA is 130.86. The molecular weight of human DNA will therefore be 6 x 109 nucleotides (based on the human genome project) x 130-86 = 784-56 x 109 gm/mol. The molar concentration of DNA can be calculated accordingly.
The molarity can be calculated as
Molar Concentration = No.ofmoleculesMolecularWeight

Question 5.
Do eukaryotic cells have restriction endonucleases? Justify your answer.
Solution:
Restriction enzymes also called ‘the molecular scissors’ are used to break DNA molecules. These enzymes are present in many bacteria where they function as a part of their defence mechanism called a restriction-modification system. The molecular basis of this system was explained first by W. Arber in 1965. The restriction-modification system consists of two components;

A restriction enzyme which identifies the introduced foreign DNA and cuts into pieces and is called restriction endonuclease,

The second component is a modification enzyme in which methylation is done. Once a base in a DNA sequence is modified by the addition of a methyl group, the restriction enzymes fail to recognise and could not cut that DNA. This is how a bacterium modifies and therefore, protects its own chromosomal DNA from cleavage by these restriction enzymes. Eukaryotic cells do not have restriction endonucleases (or restriction-modification system). The DNA molecules of eukaryotes are heavily methylated by a modification enzyme, called methylase. Eukaryotes exhibit some different mechanisms to counteract viral attacks.

Question 6.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors to have over shake flasks?
Solution:
Shake flask is used for growing microbes and mixing the desired materials on a small scale in the laboratory. However, the large-scale production of a desired biotechnological product requires large stirred tank bioreactors.
Besides better aeration and mixing properties, bioreactors have the following advantages:

  • It has an oxygen delivery system.
  • It has a foam control, temperature, and pH control system.
  • Small volumes of culture can be withdrawn periodically.

Question 7.
Collect 5 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules.
Solution:
palindromes in DNA are base pair sequences that are the same when read forward (left to right) or backward (right to left) from a central axis of symmetry. For example, the following sequences read the same on the two strands in the 5′ → 3′ direction as well as 3′ → 5′ direction.

  1. 5′ – G – G – A – T – C – C – 3′
    3′ – C – C – T – A – G – G – 5′
  2. 5′ – G – A – A – T – T – C – 3′
    5′ – C – T – T – A – A – G – 5′
  3. 5′ – A – A – G – C – T – T – 3′
    3′ – T – T – C – G – A – A – 5′
  4. 5′ – G – T – C – G – A – C – 3′
    3′ – C – A – G – C – T – G – 5′
  5. 5′ – A – C – T – A – G – T – 3′
    3′ – T – G – A – T – C – A – 5′

Question 8.
Can you recall meiosis and indicate at what stage recombinant DNA is made?
Solution:
A recombinant DNA is made in the pachytene stage of prophase I by crossing over during meiosis cell division. Recombination nodules are visible in a synaptonemal complex in the pachytene sub-stage. Crossing over occurs in this time between chromatids than recombinant DNA is formed.

Question 9.
Can you think and answer how a reporter enzyme can be used to monitor the transformation of host cells by foreign DNA in addition to a selectable marker?
Solution:
Transformation is a process through which a piece of DNA is introduced into a host bacterial cell. Normally, the genes encoding resistance to antibiotics such as ampicillin, tetracycline, etc., are considered useful selectable markers to differentiate between transformed and non-transformed bacterial cell. In addition to these selectable markers, an alternative selectable marker has been developed to differentiate transformed and non-transformed bacterial cell on the basis of their ability to produce colour in the presence of a chromogenic substance.

A recombinant DNA is inserted in the coding sequence of an enzyme (5-galactosidase (reporter enzyme). If the plasmid in the bacterium does not have an insert, the presence of a chromogenic substance gives blue coloured colonies, presence of insert results into insertional inactivation of (3-galactosidase and, therefore, the colonies do not produce any colour, these colonies are marked as transformed colonies.

Question 10.
Describe briefly the following :
(a) Origin of replication
(b) Bioreactors
(c) Downstream processing
Solution:

(a) Origin of Replication (Ori):- It is a DNA sequence which is specialised to initiate replication Bacterial chromosomes and plasmids possess a single origin of replication Eukaryote chromosomes to have a number of origin of replication. Replication proceeds bidirectionally from the site of origin of replication. The sequence also possesses nearby replication control which determines the number of copies it would form. Therefore the selected plasmid should have an origin of replication that supports a high copy number.

(b) Bio-Reactor:
Bio reactor used in biotechnology is generally 100-1000 litre cylindrical metal. container with a curved base to facilitate mixing of contents. The culture medium containing all nutrients, salts vitamins, hormones etc. is added along with the inoculum of transformed cells with recombinant DNA. A stirrer helps in mixing and optimum availability of nutrients to culture cells. The supply of oxygen is maintained if the cells function better under aerobic conditions. Foam is kept under control. Gadgets are attached for knowing the temperature and pH of the contents. Corrections are made when required. There is a sampling port where a small volume of culture can be withdrawn to know the growth of cells and concentration of the extractable product.

(c) Downstream processing:
It is the recovery of product from fully grown genetically modified cells, its purification, and preservation. It is carried out after the sampling report indicates the completion of the biosynthetic phase and the presence of the optimum product in the cells. After leaving a part of the cellular mass of inoculum, the rest is crushed and chemically treated to separate the product. The separated product is purified and then formulated with suitable preservatives. Clinical traits are carried out to know its use and any immediate or long-term adverse effect. Every batch of the product has to pass through strict quality control testing. Of course, the procedure and vigour of downstream processing and quality control vary from product to product.

Question 11.
Explain briefly
(a) PCR
(b) Restriction enzymes and DNA
(c) Chitinase
Solution:
(a) Polymerase chain reaction (PCR) is a technique of synthesizing multiple copies of the desired gene (DNA) in vitro. This technique was developed by Kary Mullis in 1985. It is based on the principle that a DNA molecule, when subjected to high temperature, splits into two strands due to denaturation. These single-stranded molecules are then converted to double-stranded molecules by synthesising new strands in presence of enzyme DNA polymerase. Thus, multiple copies of the original DNA sequence can be generated by repeating the process several times. The basic requirements of PCR are, DNA template, two nucleotide primers usually 20 nucleotides long, and enzyme DNA polymerase which is stable at high temperature (usually Taq polymerase):
The working mechanism of PCR is as follows:

First of all, the target DNA (DNA segment to be amplified) is heated to high temperature (94 to 96° C). Heating results in the separation of two strands of DNA. Each of the two strands of the target DNA now acts as a template for the synthesis of a new DNA strand. This step is called denaturation.

Denaturation is followed by annealing. During this step, two oligonucleotide primers hybridise to each single-stranded template DNA in presence of excess synthetic oligonucleotides. Annealing is carried out at lower temperatures (40° – 60°C).

The third and final step is an extension. During this step, the enzyme DNA polymerase synthesizes the DNA segment between the primers. Usually, Taq DNA polymerase, isolated from a thermophilic bacterium Thermus aquatics, is used in most cases. The two primers extend towards each other in order to copy the DNA segment lying between the two primers. This step requires the presence of deoxynucleoside triphosphates (dNTPs) and Mg2+ and occurs at 72°C.
The above-mentioned three steps complete the first cycle of PCR. The second cycle begins with denaturation of the extension product of the first cycle and after completing the extension step, two cycles are completed. If these cycles are repeated many times, the DNA segment can be amplified approximately a billion times, i.e., one billion copies of desired DNA segment are made.

(b) Restriction enzymes are used to break DNA molecules. They belong to a larger class of enzymes called nucleases. Restriction enzymes are of three types – exonucleases, endonucleases, and restriction endonucleases.

Exonucleases: They remove nucleotides from the terminal ends (either 5′ or 3′) of DNA in one strand of the duplex.

Endonucleases: They make cuts at specific positions within the DNA. These enzymes do not cleave the ends and involve only one strand of the DNA duplex.

Restriction endonucleases: These were found by Arber in 1963 in bacteria. They act as “molecular scissors” or chemical scalpels. They recognise the base sequence at palindrome sites in the DNA duplex and cut its strands. Three main types of restriction endonucleases are type I, type II, and Type III. Out of the three types, only type II restriction enzymes are used in recombinant DNA technology because they can be used in vitro to recognise and cut within specific DNA sequences typically consisting of 4 to 8 nucleotides.

(c) Chitinase is a lysing enzyme that dissolves the fungal cell wall. It results in the release of DNA along with several other macromolecules.

Question 12.
Discuss with your teacher and find out how to distinguish between:
(a) Plasmid DNA and chromosomal DNA
(b) Exonuclease and endonuclease
(c) RNA and DNA
Solution:
(a) Plasmid DNA is naked double-stranded DNA that forms a circle with no free ends. It is associated with few proteins. It is smaller than the host chromosome and can be easily separated.

Chromosomal DNA is a double-stranded linear DNA molecule associated with large proteins. This DNA exists in relaxed and supercoiled forms and provides a template for replication and transcription. It has free ends.

(b) Differences between exonucleases and endonucleases are as follows :
NCERT Solutions for Class 12 Biology Chapter 11 Biotechnology Principles and Processes Q12.1

(c) Differences between DNA and RNA are given in the following table:
NCERT Solutions for Class 12 Biology Chapter 11 Biotechnology Principles and Processes Q12.2
NCERT Solutions for Class 12 Biology Chapter 11 Biotechnology Principles and Processes Q12.3

Chapter 12 Biotechnology and Its Applications

Question 1.
Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because
(a) bacteria are resistant to the toxin
(b) toxin is immature
(c) toxin is inactive
(d) bacteria encloses toxin in a special sac.
Solution:
(c) Toxin is inactive: In bacteria, the toxin is present in an inactive form called prototoxin. This gets converted into the active form when it enters the salivary gland of insects having an alkaline medium.

Question 2.
What are transgenic bacteria? Illustrate using any one example.
Solution:
Transgenic bacteria are one that carries a transgene or a foreign gene of interest introduced using recombinant DNA technology. e. g., bacteria carrying the genes for human insulin.

In 1983, Eli Lilly an American company prepared two DNA sequences corresponding to A and B, chains of human insulin and introduced them in plasmids of E. coli to produce insulin chains. Chains A and B were produced separately, extracted and combined by creating disulfide bonds to form human insulin.

Question 3.
Compare and contrast the advantages and disadvantages of the production of genetically modified crops.
Solution:
Advantages of genetically modified crops or transgenic crops are as follows :

  • They are resistant to pests, herbicides and diseases.
  • They help to reduce post-harvest losses.
  • They enhance the nutritional value of food, e.g., a transgenic variety of rice (golden rice) is rich in vitamin A content.
  • Some transgenic plants, e.g., poplar trees are used to clean up heavy metal pollution from contaminated soil.
  • They are efficient in mineral usage and thus prevent early exhaustion of fertility of the soil.

Transgenic crops have several disadvantages also which are mentioned below:

  • Bt toxins expressed in pollen grains of transgenic crops are harmful for useful varieties of insects, e.g., honey bees and butterflies.
  • The foods produced by transgenic crops might cause toxicity and might result in allergies.
  • The bacteria present in human alimentary canal can become resistant to concerned antibiotic by taking up antibiotic resistance gene present in genetically modified food and become difficult to manage.

Question 4.
What are Cry proteins? Name an organism that produces it. How has man exploited this protein to his benefit?
Solution:
Cry proteins are a group of toxic protein which are highly poisonous to deficient types of insects. It is produced by a soil bacterium Bacillus thuringiensis. The genes controlling their formation are called cry genes eg:- Cry I Ab, Cry I Ac, Cry II Ab, The bacterium produces a protein in the crystal form of protoxin. Two cry genes have been incorporated in cotton (Bt cotton) while one has been introduced in corn (Bt corn) As a result Bt Cotton was disease resistant to bollworm and Bt corn was resistant to corn borer.

Question 5.
What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency.
Solution:
Gene therapy is the technique of genetic engineering used to replace a faulty gene with a normal, healthy functional gene. The first clinical gene therapy was given in 1990 to a 4 years old girl with adenosine deaminase deficiency (ADA deficiency). This enzyme is very important for the immune system to function. Severe combined immunodeficiency (SCID) is caused due to a defect in the gene for the enzyme adenosine deaminase. SCID patient lacks functional T-lymphocytes and, therefore, fails to fight the infecting pathogens.
NCERT Solutions for Class 12 Biology Chapter 12 Biotechnology and Its Applications Q5.1
To perform gene therapy, lymphocytes are extracted from the patient’s bone marrow and a normal functional copy of human gene coding for ADA is introduced into these lymphocytes with the help of a retroviral vector. The cells so treated are reintroduced into the patient’s bone marrow. The lymphocytes produced by these cells contain functional ADA genes which reactivate the victim’s immune system. But, as these lymphocytes do not divide and are short-lived, so periodic infusion of genetically engineered lymphocytes is required. This problem can be overcome if stem cells are modified at an early embryonic stage.

Question 6.
Diagrammatically represent the experimental steps in cloning and expressing a human gene (say the gene for growth hormone) into a bacterium like E.coli?
Solution:
The given diagram represents the experimental steps in cloning and expressing a human gene for growth hormone into a bacterium E. coli.
NCERT Solutions for Class 12 Biology Chapter 12 Biotechnology and Its Applications Q6.1

Question 7.
Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of rDNA technology and the chemistry of oil?
Solution:
rDNA technology is a technique of genetic engineering that involves combining DNA from two different sources to produce recombined or recombinant DNA (rDNA). Oils are composed of glycerol and fatty acids. Thus, to produce oil-free seeds genes coding for glycerol or fatty acids should be identified and nucleotide sequences complementary to the sequence of these genes should be inserted adjacent to these genes in the early cells of the endosperm. During transcription, these complementary sequences will produce anti-sense RNAs to the RNAs produced by glycerol or fatty acids gene and will silence these genes. As a result, oil-free seeds will be produced.

Since glycerol is a common component of all the oils whereas various fatty acids combine with glycerol to form oils, thus it will be easier if we silence the gene for glycerol synthesis.

Question 8.
Find out from the internet what is golden rice.
Solution:
Golden rice is a GM rice with increased vitamin A content.

Question 9.
Does our blood have proteases and nucleases?
Solution:
Proteases occur naturally in all organisms. These enzymes are involved in a multitude of physiological reactions from simple digestion of food proteins to highly-regulated cascades (e.g., the blood-clotting cascade, the complement system, apoptotic pathways, and the invertebrate prophenoloxidase activating cascade). Proteases present in blood serum (thrombin, plasmin, Hageman factor, etc.) play important role in blood clotting, as well as in lysis of the clots, and the action of the immune system. Other proteases are present in leukocytes (elastase, cathepsin G) and play several different roles in metabolic control. Nucleases, such as deoxyribonucleases and ribonucleases are found in the blood which helps in the degradation of exogenous deoxyribonucleic acid and ribonucleic acid circulating in the blood.

Question 10.
Consult the internet and find out how to make orally active protein pharmaceuticals. What is the major problem to be encountered?
Solution:
The problem is stomach enzymes and acids. Once you orally ingest a protein, the proteases in your stomach juices (trypsin, chymotrypsin, pepsin) will cleave the holy-hell out of your therapeutic protein and the acids will denature whatever’s left beyond all recognition. This is why proteins like insulin have to be injected.

Chapter 13 Organisms and Populations

Question 1.
How is diapause different from hibernation ?
Solution:
Diapause is different from hibernation. The table below shows the differences between them :
NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations Q1.1

Question 2.
If a marine fish is placed in a freshwater aquarium/will the fish be able to survive? Why or why not?
Solution:
If a marine fish is placed in a freshwater aquarium, then its chances of survival will diminish. This is because their bodies are adapted to high salt concentrations in the marine environment. In freshwater conditions, they are unable to regulate the water entering the body (through osmosis). Water enters their body due to the hypotonic environment outside. This results in the swelling up of the body, eventually leading to the death of the marine fish.

Question 3.
Define phenotypic adaptation. Give one example.
Solution:
Phenotypic adaptation involves non-genetic changes in individuals such as physiological modifications like acclimatization or behavioural changes.

Question 4.
Most living organisms cannot survive at temperatures above 45°C. How are some microbes able to live in habitats with temperatures exceeding 100°C?
Solution:
organisms survive at a temperature range of 0° to 40°C or less. However, there are some notable exceptions. Certain microorganisms live in hot springs and deep-sea hydrothermal vents where temperature far exceeds 100°C. They survive at the high temperature due to the occurrence of branched-chain lipids in their cell membrane that reduces the fluidity of cell membranes and the occurrence of the minimum amount of free water in their cells that provides resistance to high temperature

Question 5.
List the attributes that populations but not individuals possess.
Solution:

  1. Natality
  2. Mortality
  3. Growth forms
  4. Population density
  5. Population dispersion
  6. Population age distribution

Question 6.
If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase (r) of the population?
Solution:
The intrinsic rate of increase(r), can be calculated by the following exponential growth equation:
NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations Q6.1

Question 7.
Name important defence mechanisms in plants against herbivory.
Solution:

  1. Modification of leaves into thorns.
  2. Development of spiny margins on leaves.
  3. Development of sharp silicated edges on leaves.

Question 8.
An orchid plant is growing on the branch of the mango tree. How do you describe this interaction between the orchid and the mango tree?
Solution:
An orchid growing as an epiphyte on a branch of mango tree is an example of commensalism. Commensalism is the relationship between individuals of two species of which one is benefited and the other is almost unaffected, i.e., neither benefited nor harmed. A commensal may get shelter (protection), or ride, or support instead of or in addition to food. Epiphytes are space parasites, they use trees only for attachment and manufacture their own food by photosynthesis. In Vanda, an epiphytic orchid, a special kind of aerial roots (hanging roots) hang freely in the air and absorb moisture with the help of their special absorptive tissue called velamen.

Question 9.
What is the ecological principle behind the biological control method of managing pest insects?
Solution:
Predation is the means of biological control to manage pest insects where predators prey upon pests and regulate their numbers in the habitat.

Question 10.
Distinguish between the following:

  1. Hibernation and Aestivation
  2. Ectotherms and Endotherms

Solution:

  1. Differences between hibernation and aestivation are as follows :
    NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations Q10.1
  2. Differences between ectotherms and endotherms are as follows:
    NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations Q10.2

Question 11.
Write a short note on :
(a) Adaptations of desert plants and animals
(b) Adaptations of plants to water scarcity
(c) Behavioral adaptations in animals
(d) Importance of light to plants
(e) Effect of temperature or water scarcity and the adaptations of animals.
Solution:
a. Desert plants are called xerophytes. They have adaptations for increased water absorption, reduction in transpiration and water storage. Many desert plants have a thick cuticle on their leaf surfaces and have their stomata arranged in deep pits to minimise water loss through transpiration. They also have a special photosynthetic pathway that enables their stomata to remain closed during day time. In desert plants like Opuntia, leaves are reduced to spines. Animals of dry areas may use metabolic water and reduce water loss bypassing nearly solid faeces and urine.

b. Xerophytes have special adaptations to withstand prolonged periods of drought. These are of four types – ephemerals, annuals, succulents and non-succulent perennials.

  • Ephemerals (drought escapers): Plants which live for a brief period and complete their life cycle during the rains.
  • Annuals (drought evaders): Plants which continue to live for a few
    months even after rains in hot dry conditions. They have modifications to reduce transpiration.
  • Succulents (drought resistants): Plants have fleshy organs to store large amounts of water. They have a very thick cuticle, sunken stomata which open during night only.
  • Non-succulent perennials: These are true xerophytes. They have an extensive root system to absorb the maximum amount of water. They possess waxy coatings on leaves, sunken stomata, reduced leaf blades etc. to reduce transpiration.

c. The animals with variable temperatures called poikilotherms are affected by temperature variations. They are also called ectotherms. They show different adaptations like hibernation, aestivation, periodic activity, winter eggs, and migration.

d. Sun is the ultimate source of energy for most of the organisms on this earth. Light is the visible range of the electromagnetic spectrum. Light (400 nm-700nm) is effective in photosynthesis and is called photo-synthetically active radiation or PAR. The intensity of light, duration of light, etc. are also influencing the growth of plants.

e. Animals live in arid regions show two kinds of adaptations

  1. Reducing loss of water from their bodies.
  2. Ability to tolerate arid conditions.

Question 12.
List the various abiotic environmental factors.
Solution:
Abiotic factors are non-living factors and conditions of the environment which influence the survival, function and behaviour of organisms. Various abiotic factors are :

(i) Temperature – Temperature is one of the most important environmental factors. The average temperature varies seasonally. It ranges from subzero level in polar areas and high altitudes to more than 50°C in tropical deserts in summer and exceeds 100°C in thermal springs and deep-sea hydrothermal vents.

(ii) Water – Next to temperature, water is the most important factor which influences the life of organisms. The productivity and distribution of land plants are dependent upon the availability of water. Animals are adapted according to water availability. E.g., aquatic animals are ammonotelic while xerophytic animals excrete dry feces and concentrated urine.

(iii) Light – Plants produce food through photosynthesis for which sunlight is essential to the source of energy. Light intensity, light duration and light quality influences the number of life processes in organisms, such as – photosynthesis, growth, transpiration, germination, pigmentation, movement and photoperiodism.

(iv) Humidity – Humidity refers to the moisture (water vapour) content of the air. It determines the formation of clouds, dew and fog. It affects the land organisms by regulating the loss of water as vapour from their bodies through evaporation, perspiration and transpiration.

(v) Precipitation – Precipitation means rainfall, snow, sleet or dew. Total annual rainfall, seasonal distribution humidity of the air and amount of water retained in the soil are the main criteria that limit the distribution of plants and animals on land.

(vi) Soil – The soil is one of the most important ecological factor called the edaphic factor. It comprises of different layers called horizons. The upper weathered humus containing part of soil sustains terrestrial plant life.

Question 13.
Give an example for:

  1. An endothermic animal
  2. An ectothermic animal
  3. An organism of the benthic zone.

Solution:

  1. Hedgehog
  2. Frog
  3. Sponges

Question 14.
Define population and community.
Solution:
Population: A population is a group of individuals of the same species, which can reproduce among themselves and occupy a particular area in a given time.

Community: It is an assemblage of several populations in a particular area and time and exhibits interaction and interdependence through trophic relationship.

Question 15.
Define the following terms and give one example for each.
(a) Commensalism
(b) Parasitism
(c) Camouflage
(d) Mutualism
(e) Interspecific competition
Solution:

a. Commensalism is an interspecific interaction between individuals of two species where one species is benefitted and the other is not affected.
e. g. Orchid and mango tree.

b. Parasitism is an interspecific interaction between individuals of two species where generally small species is benefitted and the large species are affected, e.g. Malarial parasite and human beings.

c. Camouflage: It is the ability of the animals to blend with the surroundings or background. In this way, animals remain unnoticed for protection or aggression. An example is a stick insect.

d. Mutualism is an interspecific interaction between individuals of two species where both the interacting species are benefitted in an obligatory way. e.g. Pollination in plants by animals.

e. Interspecific competition: It is an interaction between individuals of two species where both the interacting species are affected, e.g. Monarch butterfly and Queen monarch.

Question 16.
With the help of a suitable diagram describe the logistic population growth curve.
Solution:
Logistic population growth curve or S-shaped or sigmoid growth curve is shown by the populations of most organisms. It has the following phases: lag phase, log phase, exponential phase and stationary phase. In lag phase there is little or no increase in population. In log phase increase in population starts and occurs at a slow rate in the beginning. During exponential phase, increase in population becomes rapid and soon attains its full potential rate. This is due to the constant environment, availability of food and other requirements of life in plenty, absence of predation and interspecific competition and no serious intraspecific competition so that the curve rises steeply upward. The growth rate finally slows down as environmental resistance increases.

Finally, the population becomes stable during the stationary phase because now the number of new cells produced almost equals to the number of cells that die. Every population tends to reach a number at which it becomes stabilized with the resources of its environment. A stable population is said to be in equilibrium, or at saturation level. This limit in population is a constant K and is imposed by the carrying capacity of the environment. The sigmoid growth form is represented by the following equation :
NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations Q16.1
r = intrinsic rate of natural increase
N = population density at time t; K = carrying capacity.

Question 17.
Select the statement which best explains parasitism.
(a) One organism is benefited.
(b) Both the organisms are benefited
(c) One organism is benefited, other is not affected
(d) One organism is benefited, other is affected.
Solution:
(d) One organism is benefited, other is affected,

Question 18.
List any three important characteristics of a population and explain.
Solution:
The three important characteristics of a population are:

  1. Birth and death rate
  2. Age structure
  3. Sex ratio

(i) The birth rate (natality) of a population refers to the average number of young ones produced per unit time (usually per year). In the case of humans, it is commonly expressed as the number of births per 1,000 individuals in the population per year. The death rate (mortality) of a population is the average number of individuals that die per unit time (usually per year). In humans, it is commonly expressed as the number of deaths per 1,000 persons in a population per year.

(ii) The age structure of a population is the percentage of individuals of different ages such as young, adult and old. Age structure is shown bv organisms in which individuals of more than one generation coexist. The ratio of various age groups in a population determines the current reproductive status of the population. It also indicates what may be expected in the future. The population is divided into three age groups; pre-reproductive, reproductive and post-reproductive.

(iii) The sex ratio of a population refers to the number of females per thousand male individuals. There were 933 females per 1,000 males in our country in the 2001 census. The number of females in a population is very important as it is often directly related to the number of births. The number of males may be less significant because in many species a single male can mate with several females.

Chapter 14 Ecosystem

Question 1.
Fill in the blanks
(a) Plants are called as ___ because they fix carbon dioxide.
(b) In an ecosystem dominated by trees, the pyramid (of numbers) is ___ type.
(c) In aquatic ecosystem, the limiting factor for the productivity is ___
(d) Common detritivores in our ecosystem are ____
(e) The major reservoir of carbon on earth is ___
Solution:
(a) producers
(b) inverted or spindle
(c) light
(d) saprotrophs
(e) oceans

Question 2.
Which one of the following has the largest population in a food chain?
(a) Producers
(b) Primary consumers
(c) Secondary consumers
(d) Decomposer’s
Solution:
(d) decomposer’s

Question 3.
The second trophic level in a lake is
(a) phytoplankton
(b) zooplankton
(c) benthos
(d) fishes.
Solution:
(b) zooplankton

Question 4.
Secondary producers are
(a) herbivores
(b) producers
(c) carnivores
(d) none of these
Solution:
(a) herbivores

Question 5.
What is the percentage of photosynthetically active radiation (PAR), in the incident solar radiation.
(a) 100%
(b) 50%
(c) 1 – 5%
(d) 2 – 10%
Solution:
(b) 50%

Question 6.
Distinguish between
(a) Grazing food chain and detritus food chain
(b) Upright and inverted pyramid
(c) Litter and detritus
(d) Production and decomposition
(e) Food chain and food web
(f) Primary and secondary productivity
Solution:
(a) Differences between grazing food chain and detritus food chain are as follows
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 6.1

(b) Differences between upright and inverted pyramids are as follows :
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 6.2

(c) Differences between litter and detritus are as follows :
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 6.3

(d) Differences between production and decomposition are as follows :
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 6.4

(e) Differences between food chain and food web are as follows:
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 6.5

(f) Differences between primary productivity and secondary productivity are as follows :
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 6.6

Question 7.
Describe the components of an ecosystem.
Solution:
Ecosystem: The system resulting from the interaction between organisms and their environment is called an ecosystem.
(a) Producers: Organisms, which can synthesize their own food are included under producers, e.g., Volvox, Pandorina, Oedogonium, Saggitaria, Utricularia, Azolla, Trapa, Lemna, Typha, Nymphaea etc. form the producer class of the pond ecosystem.

(b) Consumers:

  • Primary consumer: Animals, which feed on producers are included in this category e.g., Daphnia, Cyclops, Paramoecium, Amoeba, and small fishes.
  • Secondary consumers: Primary consumers also serve as food for water snakes, a few tortoises, few types of fish, etc. hence, these are carnivores.
  • Tertiary consumers: Secondary consumers also serve as food for aquatic birds like kingfishers, cranes, big fish and these together form a top-class carnivorous group and called tertiary consumers.

(c) Decomposers: All producers and consumers die and accumulate on the floor of the pond. Even the waste material and feces of these animals get accumulated on the floor of the pond. Similarly, the floor of the pond is also occupied by decomposers, which include bacteria and fungi. These decomposers decompose complex organic compounds of then- bodies into simpler forms which are finally mixed with the soil of the floor of ponds. These are again absorbed by the roots of producer plants and thus matter is recycled.

Question 8.
Define ecological pyramids and describe with examples, pyramids of number and biomass.
Solution:
An ecological pyramid is a graphic representation of an ecological parameter, as a number of individuals present in various trophic levels of a food chain with producers forming the base and top carnivores the tip. Ecological pyramids were developed by Charles Elton (1927) and are, therefore, also called Eltonian pyramids.

There are three types of ecological pyramids, namely,

  • Pyramid of numbers
  • Pyramid of biomass
  • Pyramid of energy

Pyramid of numbers: It is a graphic representation of the number of individuals per unit area of various trophic levels stepwise with producers at the base and top carnivores at the tip. In a grassland, the producers, which are mainly grasses, are always maximum in number. This number then shows a decrease towards the apex, as the primary consumers (herbivores) like rabbits, mice, etc. are lesser in number than the grasses; the secondary consumers, snakes, and lizards are lesser in number than the rabbits and mice. Finally, the top (tertiary) consumers hawks or other birds, are the least in number. Thus, the pyramid becomes upright.
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 8.1

Pyramid of biomass: The amount of living organic matter (fresh and dry weight) is called biomass. Here, different trophic level of the ecosystem are arranged according to the biomass of the organisms. In grassland and forest, there is generally a gradual decrease in biomass of organisms at successive levels from the producers to the top carnivores. Thus these pyramids are upright. But in pond ecosystem, it is inverted because the biomass gradually increases from the producers to carnivores.
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 8.2

Question 9.
What is primary productivity? Give a brief description of factors that affect primary productivity.
Solution:
The rate of biomass production is called productivity.
It is expressed in terms of g-2yr-1  or(Kcal-m-2) yr-1 to compare the productivity of ecosystems.
It can be divided into Gross Primary Productivity (GPP) and Net Primary Productivity (NFP).

Gross Primary Productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilized by plants in respiration.

Gross primary productivity minus respiration losses (R), is the Net Primary Productivity (NPP). GPP – R=NPP.
Primary productivity depends on:

  • The plant species inhabiting a particular area.
  • The environmental factors.
  • Availability of nutrients.
  • Photosynthetic capacity of plants.

Question 10.
Define decomposition and describe the processes and products of decomposition.
Solution:
Decomposition is the breakdown of dead or wastes organic matter by micro-organisms. Decomposition is both physical and chemical in nature. Processes involved in decomposition are – fragmentation, catabolism & leaching.

  • Fragmentation – The process primarily due to the action of detritus feeding invertebrate (detritivores) causes it to break into smaller particles. The detritus gets pulverized when passing through the digestive tracts of animals. Due to fragmentation, the surface area of detritus particles is greatly increased.
  • Catabolism – Enzyme degradation of detritus into simpler organic substances by bacteria and fungi.
  • Leaching – The process by which nutrients, chemicals, or contaminants are dissolved & carried away by water, or are moved into a lower layer of soil.

Various inorganic and organic substances are obtained by decomposition. Inorganic substances are obtained in the process of mineralization while organic substances are obtained in humification. A dark coloured amorphous substance called humus is formed by decomposition. Humus is highly resistant to microbial action & undergoes extremely slow decomposition. It serves as a reservoir of nutrients.

Question 11.
Give an account of energy flow in an ecosystem.
Solution:
Ecosystems require a constant input of energy as every component of an ecosystem is regularly dissipating energy.

Two laws of thermodynamics govern this flow of energy. According to the first law of thermodynamics, energy can be transferred as well as transformed but is neither created nor destroyed. According to the second law of thermodynamics, every activity involving energy transformation is accompanied by the dissipation of energy. Except for deep hydrothermal ecosystems, the source of energy in all ecosystems is solar energy. 50% of the solar energy incident over the earth is present in PAR (photosynthetically active radiation).

Energy flow in an ecosystem is always unidirectional or one way, i.e., solar radiation → producers → herbivores → carnivores. It cannot pass in the reverse direction. There is a decrease in the content and flow of energy with the rise in trophic level. Only 10% of energy is transferred from one trophic level to the next.
Producer biomass (1000 K cal) → Herbivore biomass (100 K cal) → Carnivore I biomass (10 Kcal) Carnivore II biomass (1 Kcal)
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 11.1

Question 12.
Write important features of a sedimentary cycle in an ecosystem
Solution:
Sedimentary Biogeochemical cycle:- It is the circulation of a biogeochemical between the biotic and abiotic compound of an ecosystem is a nongaseous being lithosphere or sediments of the earth. Sedimentary cycles occur in the case of phosphorus, calcium, magnesium, zinc, copper, etc.

Question 13.
Outline salient features of carbon cycling in an ecosystem.
Solution:
Carbon constitutes 49 percent of the dry weight of organisms and is next only to water. 71 percent of carbon is found dissolved in oceans. This ocean reservoir regulates the amount of carbon dioxide in the atmosphere. Fossil fuels also represent a reservoir of carbon. Carbon cycling occurs through the atmosphere, ocean, and living and dead organisms. 4 x 1013 kg of carbon is fixed in the biosphere through photosynthesis annually.

A considerable amount of carbon returns to the atmosphere as Co2 through respiratory activities of the producers and consumers. Decomposers also contribute substantially to the CO2 pool by their processing of waste materials and dead organic matter of land or oceans. Some amount of fixed carbon is lost to sediments and removed from circulation. Burning of wood, forest fire and combustion of organic matter, fossil fuels, volcanic activity are additional sources for releasing Co2 into the atmosphere.

Human activities have significantly influenced the carbon cycle. Rapid deforestation and the massive burning of fossil fuels for energy and transport have significantly increased the rate of release of carbon dioxide into the atmosphere.
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 13.1

Chapter 15 Biodiversity and Conservation

Question 1.
Name the three important components of biodiversity
Solution:
The three important components of biodiversity are genetic diversity, species diversity and ecological diversity. These components are the basic building blocks of biodiversity. These are intimately linked and may have common elements.

Question 2.
How do ecologists estimate the total number of species present in the world?
Solution:
The diversity of living organisms present on the earth is very vast. According to an estimate by researchers, it is about seven million. The total number of species present in the world is calculated by ecologists by statistical comparison between species richness of a well-studied group of insects of temperate and tropical regions. Then, these ratios are extrapolated with other groups of plants and animals to calculate the total species richness present on the earth.

Question 3.
Give three hypotheses for explaining why tropics show the greatest levels of species richness.
Solution:
The three hypotheses for higher species richness in tropical areas are:

  1. Prolong evolutionary time – Temperate areas have undergone frequent glaciation in the past. It killed most of the species. No such disturbance occurred in the tropics where species continued to flourish and evolve undisturbed for millions of years.
  2. Favourable environment – There are no unfavourable seasons in the tropics. The continued favourable environment has helped tropical organisms to gain more niche specialisation and increased diversity.
  3. More sunlight – More solar energy is available in the tropics. This promotes higher productivity and increased biodiversity.

Question 4.
What is the significance of the slope of regression in a species-area relationship?
Solution:
The slope of regression/regression co-efficient of species-area relationship indicates that species richness decreases with a decrease in area.

  • The regression coefficient is between 0.1 – 0.2 regardless of taxonomic group or region, eg: Plants in Britain, Birds in California.
  • But in large areas like continents value is eg:- Frugivorous birds, mammals is tropical forests.

Question 5.
What are the major causes of species losses in a geographical region?
Solution:
The major causes of species losses in a geographical area are:

  1. Habitat loss and fragmentation
  2. Overexploitation
  3. Alien species invasion
  4. Co-extinctions
  5. Disturbance and degradation
  6. Pollution
  7. Intensive agriculture and forestry.

Question 6.
How is biodiversity important for ecosystem functioning?
Solution:
An ecosystem with high species diversity is much more stable than an ecosystem with low species diversity. Also, high biodiversity makes the ecosystem more stable in productivity and more resistant to disturbances such as alien species invasions and floods.

If an ecosystem is rich in biodiversity, then the ecological balance would not get affected. Various trophic levels are connected through food chains. If anyone organism or all organisms of any one trophic level is illed, then it will disrupt the entire food chain. For example, in a food chain, if all plants are killed, then all deer will die due to the lack of food.

If all deer are dead, soon the tigers will also die. Therefore, it can be concluded that if an ecosystem is rich in species, then there will be other food alternatives at each trophic level which would not allow any organism to die due to the absence of their food resource. Hence, biodiversity plays an important role in maintaining the health and ecological balance of an ecosystem.

Question 7.
What are sacred groves? What is their role in conservation?
Solution:
Sacred groves are forest patches around places of work. These are held in high esteem by tribal communities/state or central government. Tribals do not allow to cut even a single branch of trees in these sacred groves. Preserved over the course of many generations, sacred groves represent native vegetation in a natural or near-natural state & thus is rich in biodiversity & harbour many rare species of plants & animals. This is the reason why many endemic species flourish in these regions.

Question 8.
Among the ecosystem services are control of floods and soil erosion. How is this achieved by the biotic components of the ecosystem?
Solution:

  • Control of soil erosion: Plant roots hold the soil particles tightly and do not allow the topsoil to be drifted away by winds or moving water. Plants increase the porosity and fertility of the soil.
  • Control of floods: It is carried out by retaining water and preventing runoff rainwater. Litter and humus of plants function as sponges thus, retaining the water which percolates down and gets stored as underground water. Hence, the flood is controlled.

Question 9.
The species diversity of plants (22 percent) is much less than that of animals (72 percent). What could be the explanations for how animals achieved greater diversification?
Solution:
Scientists recorded 22% of plant species diversity including algae, fungi, bryophytes, pteridophytes, gymnosperms, and angiosperms. But they recorded 72% of animal species (including insects, mollusks, fishes, mammals, birds etc.) diversity. Plants have the less adaptive capacity as compared to animals. Animals show locomotory movements and can move from one place to another to suit the environment and also in search of food. On the contrary, plants are fixed. Moreover, animals have well organised body structure with various organs to help adjust to the environment.

Question 10.
Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?
Solution:
Yes, there are various kinds of parasites and disease-causing microbes that we deliberately want to eradicate from the earth. Since these micro-organisms are harmful to human beings, scientists are working hard to fight against them. Scientists have been able to eliminate the smallpox virus from the world through the use of vaccinations. This shows that humans deliberately want to make these species extinct. Several other eradication programmes such as polio and hepatitis B vaccinations are aimed to eliminate these disease-causing microbes.

Chapter 16 Environmental Issues

Question 1.
What are the various constituents of domestic sewage ? Discuss the effects of sewage discharge on a river.
Solution:
The domestic sewage contains every-thing that goes down the drain into the sewer of the house. The various constituents of domestic sewage are suspended solids, colloidal particles, pathogenic contaminants and dissolved materials. Suspended solids are sand and silt. Colloidal particles include clay, faecal matter, fine fibres of paper and cloth. Pathogenic contaminants are eggs of coliforms and enterococci. Dissolved materials includes inorganic nutrients such as nitrates, phosphates, ammonia, sodium and calcium. Effects of sewage discharge on a river :

  • Water becomes unfit for bathing and drinking and also for domestic or industrial use as it becomes colored, turbid with a lot of particulate matter floating on water.
  • The domestic sewage adds nitrates and phosphates into the river. These nitrates and phosphates encourage a thick bloom of blue green algae, which depletes the oxygen content of the water during night. This suffocates the fish and other aquatic life. Consequently river become highly polluted.

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?
Solution:
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 school include waste paper, plastics, vegetable and fruit peels, food wrapping, 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 minimized 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 optimizing 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 microorganisms do not 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?
Solution:

  • Global warming is a rise in the mean temperature of the lower atmosphere and the earth’s surface. Causes – increase in the quantity of radioactively active greenhouse gases CO2, CH4, N2O, CFCs. They allow heat waves to reach the surface and prevent their escape.
  • They are produced by combustion of fossil fuels, biomass [CO2]; burning of nitrogen-rich fuels [N2O]; paddy fields, fermentation in cattle and wetlands [CH4]; refrigerators, aerosols, drying, cleaning [CFCs].
  • Effects: Heating of earth surface [mean temperature is increased] Climatic changes e.g.: El Nino effect.
  • Increased melting of polar ice caps and Himalayan snowcaps. Increased sea levels and coastal areas will submerge.
  • Measures – Decreased use of fossil fuels, improve the efficiency of energy usage, Reduce deforestation, plant trees Control of man-made sources of greenhouse gases like vehicles, aerosol sprays.

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
Solution:
(a) – (ii); (b) – (i); (c) – (iii); (d) – (iv).

Question 5.
Write critical notes on the following :
(a) Eutrophication
(b) Biological magnification
(c) Groundwater depletion and ways for its replenishment
Solution:
(a) Eutrophication: The natural aging process of lakes by nutrient enrichment of their water. In young lake water is cold and clear and supports only little life. With time, streams introduce nutrients into lake which increases lakes’ fertility and encourages aquatic growth. Over centuries silts and organic debris pile up, and lake becomes shallow and warmer. It supports plants and later gets converted into land. Lakes span depends on the climate, size of lake.

(b) Biological magnification: Industrial wastes released into water contain toxic substances, such as arsenic, cadmium, lead, zinc, copper, mercury, and cyanides, besides some salts, acids, and alkalies. All these materials can prove harmful for our health.
NCERT Solutions for Class 12 Biology Chapter 16 Environmental Issues 5.1
They may reach the human body directly with contaminated food or indirectly by way of plants and other animals. The concentration of the toxic materials increases at each trophic level of a food chain. This is called biological magnification. River water may have a very low concentration of DDT, but the carnivorous fish in that river may contain a high concentration of DDT and become unfit for eating by man. Mercury discharged into rivers and lakes is changed by bacteria to the neurotoxic form called methyl mercury. The latter is highly poisonous and may be directly absorbed by fish.

(c) Groundwater depletion and ways for its replenishment: Groundwater depletion is defined as long-term water-level decline caused by sustained groundwater pumping. The volume of ground water in storage is decreasing in many areas of the world in response to pumping. Some of the negative effects of groundwater depletion include increased pumping costs, deterioration of water quality, reduction of water in streams and lakes.
Some ways for water replenishment are:

  • Reduction in consumption: Sprinkler and subsurface irrigation techniques reduce the amount of water used in irrigation.
  • Rain water harvesting: Rain water collected over roofs is allowed to pass into the ground through deep water pipes.

Question 6.
Why does the ozone hole form over Antarctica? How will enhanced ultraviolet radiation affect us?
Solution:
Ozone hole forms over Antarctica where no one lives and no pollution is present but not over Newyork, Bangalore etc., (polluted cities). It is because CFCs and ozone-depleting substances (ODS) released worldwide accumulates in the stratosphere and drifts towards, Antarctica in winters (July – August) when temperatures is -’85° C in Antarctica.

In winters polar ice clouds are formed over Antarctica. It provides a catalytic surface for (CFCs and other ODS to release CL and other free radicals that breakdown ozone layer forming an ozone hole during spring in presence of sunlight. In summer, the ozone hole disappears due to mixing of air worldwide.

Ozone holes allow UV radiations (UVA & UVB) to reach earth’s surface. Which was earlier reflected by the ozone layer. UVB damages DNA, skin cells and causes mutations and skin cancers respectively. UVB even causes corneal damage (Snow Blindness).

Question 7.
Discuss the role of women and communities in protection and conservation of forests.
Solution:
Forest Conservation and Management:
It is time to think deeply and act seriously in order to protect this vital natural resource. Some of the measures of conservation are

  1. Social forestry programme: It was started in 1976 and involves the affor­estation on public and common lands for fuel, fodder, timber for agricul­tural equipment and fruits. These are mainly meant for rural people.
  2. Agroforestry programme: It involves the multiple use of same land for agriculture, forestry and animal husbandary. Taungya System and Jhum are examples.
    • Taungya System: It involves growing agricultural crops between planted trees.
    • Jhum (Slash and burn agriculture): It involves felling and burning of forests, followed by the cultivation of crops for a few years. Later the cultivation is abandoned for the growth of forests. It is a traditional agroforestry system.
  3. Urban forestry programme: It involves afforestation in urban land ar­eas e.g. along the roads, big parks, big compounds etc. with ornamental and fruit trees.
  4. Commercial forestry: It involves planting of fast-growing trees on avail­able land to fulfill commercial demand.
  5. Conservation forestry: It involves protection of degraded forest to allow recoupment of their flora and fauna.

Reforestation: It is the process of restoring a forest that once existed, but was removed at some point of time in the past. Reforestation may occur naturally in a deforested area. The above-said methods speed up the refor­estation programme.

Question 8.
What measures, as an individual, you would take to reduce environmental pollution?
Solution:
To reduce environmental pollution, we should change our habits and lifestyle so as to reduce the use of disposable materials. We should use preferably those items which can easily be recycled and also minimise the use of fossil fuels. We should also take measures to improve the quality of air by using CNG gases wherever possible instead of using diesel or petrol. We should also use the catalytic converter in our vehicles.

Question 9.
Discuss briefly the following:
(a) Radioactive wastes
(b) Defunct ships and e-wastes
(c) Municipal solid wastes
Solution:
a. Radioactive waste materials are released from thermonuclear explosions. Radioactive isotopes, such as radium-226, thorium- 232, potassium-40, uranium-235, carbon-14, etc. are spread all over the world and contaminate air, soil, water, vegetation and animals.

b. Irrepairble electronic goods and computers are called electronic wastes (e-waste).
Ships that are no longer in use or that are to be dismantled are called defunct ships. Asbestos, Polychlorinated Biphenyl (PCB) produced during dismantling defunct ship cause serious health hazards especially cancer.

c. Municipal solid wastes are wastes from homes, offices, stores, schools, hospitals, etc., that are collected and disposed of by the municipality.

Question 10.
What initiatives were taken for reducing vehicular air pollution in Delhi? Has air quality improved in Delhi?
Solution:
Under the direction of Supreme Court of India, the State Government of Delhi took the following measures to improve the quality of air:

  • Switching over the entire fleet of public transport buses from diesel to CNG (Compressed Natural Gas) by the end of 2002.
  • Phasing out of old vehicles.
  • Use of unleaded petrol.
  • Use of low sulphur petrol and diesel.
  • Use of catalytic converters in vehicles.
  • Application of Euro II norms for vehicles.

Because of above mentioned measures adopted by the Government the air quality of Delhi has improved with a substantial fall in S02, CO, Nox level between 1997-2005.

Question 11.
Discuss in brief the following:
(a) Green house gases
(b) Catalytic converter
(c) Ultraviolet B
Solution:
(a) Green house gases: The gases which are transparent to solar radiation but retain and partially reflect back long wave heat radiations are called greenhouse gases. Green house gases are essential for keeping the earth warm and hospitable. They are also called radiatively active gases. They prevent a substantial part of long wave radiations emitted by earth to escape into space. Rather green house gases radiate a part of this energy back to the earth. The phenomenon is called greenhouse flux. Because of greenhouse flux, the mean annual temperature of the earth is 15°C. In its absence, it will fall to – 18°C.

However, recently the concentration of greenhouse gases has started rising to result in an enhanced greenhouse effect that is resulting in increasing the mean global temperature. It is called global warming. A regular assessment of the abundance of greenhouse gases and their impact on the global environment is being made by IPCC (Intergovernmental Panel on Climate Change). The various green house gases are CO2 (warming effect 60%), CH4 (effect 20%) , chlorofluorocarbons or CFCs (14%) andT nitrous oxide (N2O, 6%). Others of minor significance are water vapors and ozone.

(b) Catalytic converter: Catalytic converters are devices that are fitted into automobiles for reducing the emission of gases. These have expensive metals (platinum – palladium, and rhodium) as catalysts. As the exhaust passes through the catalytic converters, unburnt hydrocarbons are converted into CO2 and H2O and carbon monoxide and nitric oxide are changed to CO2 and N2 respectively. Vehicles fitted with catalytic converters should be run on unleaded petrol as leaded petrol would inactivate the catalyst in the converters.

(c) Ultraviolet B – UV-B having 280-320nm wavelength. Their harmful radiations penetrate through the ozone hole to strike the earth. On earth, these can affect human beings and other animals by causing :

  • Skin cancer
  • Blindness and increased incidence of cataract in eyes, and
  • Malfunctioning of the immune system.
  • Higher number of mutations.

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