NCERT Solution

Here are simplified, step-by-step explanations for the NCERT Solutions for Class 12 Science Biology Chapter 6, “Molecular Basis Of Inheritance”. These solutions are widely favored by Class 12 Science students for Biology as they aid in swiftly completing homework and preparing for exams. All questions and answers from the NCERT Book of Class 12 Science Biology Chapter 6 are available here at no cost, providing invaluable assistance to students. Page No 125: Question 1: Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine. ANSWER: Nitrogenous bases present in the list are adenine, thymine, uracil, and cytosine. Nucleosides present in the list are cytidine and guanosine. Page No 125: Question 2: If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA. ANSWER: According to Chargaff’s rule, the DNA molecule should have an equal ratio of pyrimidine (cytosine and thymine) and purine (adenine and guanine). It means that the number of adenine molecules is equal to thymine molecules and the number of guanine molecules is equal to cytosine molecules. % A = % T and % G = % C If dsDNA has 20% of cytosine, then according to the law, it would have 20% of guanine. Thus, percentage of G + C content = 40% The remaining 60% represents both A + T molecule. Since adenine and guanine are always present in equal numbers, the percentage of adenine molecule is 30%. Page No 125: Question 3: If the sequence of one strand of DNA is written as follows: 5‘-ATGCATGCATGCATGCATGCATGCATGC-3‘ Write down the sequence of complementary strand in 5‘→3‘ direction ANSWER: The DNA strands are complementary to each other with respect to base sequence. Hence, if the sequence of one strand of DNA is 5‘– ATGCATGCATGCATGCATGCATGCATGC − 3’ Then, the sequence of complementary strand in direction will be 3‘– TACGTACGTACGTACGTACGTACGTACG − 5’ Therefore, the sequence of nucleotides on DNA polypeptide in direction is 5‘– GCATGCATGCATGCATGCATGCATGCAT− 3’ Page No 125: 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. ANSWER: If the coding strand in a transcription unit is 5’− ATGCATGCATGCATGCATGCATGCATGC-3’ Then, the template strand in 3’ to 5’ direction would be 3’ − TACGTACGTACGTACGTACGTACGTACG-5’ It is known that the sequence of mRNA is same as the coding strand of DNA. However, in RNA, thymine is replaced by uracil. Hence, the sequence of mRNA will be 5’ − AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3’ Page No 125: Question 5: Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain. ANSWER: Watson and Crick observed that the two strands of DNA are anti-parallel and complementary to each other with respect to their base sequences. This type of arrangement in DNA molecule led to the hypothesis that DNA replication is semi-conservative. It means that the double stranded DNA molecule separates and then, each of the separated strand acts as a template for the synthesis of a new complementary strand. As a result, each DNA molecule would have one parental strand and a newly synthesized daughter strand. Since only one parental strand is conserved in each daughter molecule, it is known as semi-conservative mode of replication. Page No 125: 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. ANSWER: There are two different types of nucleic acid polymerases. (1) DNA-dependent DNA polymerases (2) DNA-dependent RNA polymerases The DNA-dependent DNA polymerases use a DNA template for synthesizing a new strand of DNA, whereas DNA-dependent RNA polymerases use a DNA template strand for synthesizing RNA. Page No 125: Question 7: How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material? ANSWER: Hershey and Chase worked with bacteriophage and E.coli to prove that DNA is the genetic material. They used different radioactive isotopes to label DNA and protein coat of the bacteriophage. They grew some bacteriophages on a medium containing radioactive phosphorus (32P) to identify DNA and some on a medium containing radioactive sulphur (35S) to identify protein. Then, these radioactive labelled phages were allowed to infect E.coli bacteria. After infecting, the protein coat of the bacteriophage was separated from the bacterial cell by blending and then subjected to the process of centrifugation. Since the protein coat was lighter, it was found in the supernatant while the infected bacteria got settled at the bottom of the centrifuge tube. Hence, it was proved that DNA is the genetic material as it was transferred from virus to bacteria. Page No 125: Question 8: Differentiate between the followings: (a) Repetitive DNA and Satellite DNA (b) mRNA and tRNA (c) Template strand and Coding strand ANSWER: (a) Repetitive DNA and satellite DNA Repetitive DNA Satellite DNA 1. Repetitive DNA are DNA sequences that contain small segments, which are repeated many times. Satellite DNA are DNA sequences that contain highly repetitive DNA. (b) mRNA and tRNA mRNA tRNA 1. mRNA or messenger RNA acts as a template for the process of transcription. tRNA or transfer RNA acts as an adaptor molecule that carries a specific amino acid to mRNA for the synthesis of polypeptide. 2. It is a linear molecule. It has clover leaf shape. (c) Template strand and coding strand Template strand Coding strand 1. Template strand of DNA acts as a template for the synthesis of mRNA during transcription. Coding strand is a sequence of DNA that has the same base sequence as that of mRNA (except thymine that is replaced by uracil in DNA). 2. It runs from 3’ to 5’. It runs from 5’to 3’. Page No 125: Question 9: List two essential roles of ribosome during translation. ANSWER: The important functions of ribosome during translation are as follows. (a) Ribosome acts as the site where protein synthesis takes place from individual amino acids. It is made up of two subunits. The smaller subunit comes in contact …

NCERT Solutions for Class 12 Science Biology Chapter 6 – Molecular Basis Of Inheritance Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Biology Chapter 5 on Principles of Inheritance and Variation, featuring clear, step-by-step explanations. Widely favored by class 12 Science students, these Biology solutions serve as valuable aids for efficiently completing homework assignments and preparing for exams. Free access to all questions and answers from the NCERT Book of class 12 Science Biology Chapter 5 is provided here, facilitating easy and cost-free assistance. Page No 93: Question 1: Mention the advantages of selecting pea plant for experiment by Mendel. ANSWER: Mendel selected pea plants to carry out his study on the inheritance of characters from parents to offspring. He selected a pea plant because of the following features. (a) Peas have many visible contrasting characters such as tall/dwarf plants, round/wrinkled seeds, green/yellow pod, purple/white flowers, etc. (b) Peas have bisexual flowers and therefore undergo self pollination easily. Thus, pea plants produce offsprings with same traits generation after generation. (c) In pea plants, cross pollination can be easily achieved by emasculation in which the stamen of the flower is removed without affecting the pistil. (d) Pea plants have a short life span and produce many seeds in one generation. Page No 93: Question 2: Differentiate between the following − (a) Dominance and Recessive (b) Homozygous and Heterozygous (c) Monohybrid and Dihybrid. ANSWER: (a) Dominance and Recessive Dominance Recessive 1. A dominant factor or allele expresses itself in the presence or absence of a recessive trait. A recessive trait is able to express itself only in the absence of a dominant trait. 2. For example, tall plant, round seed, violet flower, etc. are dominant characters in a pea plant. For example, dwarf plant, wrinkled seed, white flower, etc. are recessive traits in a pea plant. (b) Homozygous and Heterozygous Homozygous Heterozygous 1. It contains two similar alleles for a particular trait. It contains two different alleles for a particular trait. 2. Genotype for homozygous possess either dominant or recessive, but never both the alleles. For example, RR or rr Genotype for heterozygous possess both dominant and recessive alleles. For example, Rr 3. It produces only one type of gamete. It produces two different kinds of gametes. (c) Monohybrid and Dihybrid Monohybrid Dihybrid 1. Monohybrid involves cross between parents, which differs in only one pair of contrasting characters. Dihybrid involves cross between parents, which differs in two pairs of contrasting characters. 2. For example, the cross between tall and dwarf pea plant is a monohybrid cross. For example, the cross between pea plants having yellow wrinkled seed with those having green round seeds is a dihybrid cross. Page No 93: Question 3: A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced? ANSWER: Locus is a fixed position on a chromosome, which is occupied by a single or more genes. Heterozygous organisms contain different alleles for an allelic pair. Hence, a diploid organism, which is heterozygous at four loci, will have four different contrasting characters at four different loci. For example, if an organism is heterozygous at four loci with four characters, say Aa, Bb, Cc, Dd, then during meiosis, it will segregate to form 8 separate gametes. If the genes are not linked, then the diploid organism will produce 16 different gametes. However, if the genes are linked, the gametes will reduce their number as the genes might be linked and the linked genes will be inherited together during the process of meiosis. Page No 93: Question 4: Explain the Law of Dominance using a monohybrid cross. ANSWER: Mendel’s law of dominance states that a dominant allele expresses itself in a monohybrid cross and suppresses the expression of recessive allele. However, this recessive allele for a character is not lost and remains hidden or masked in the progenies of F1 generation and reappears in the next generation. For example, when pea plants with round seeds (RR) are crossed with plants with wrinkled seeds (rr), all seeds in F1 generation were found to be round (Rr). When these round seeds were self fertilized, both the round and wrinkled seeds appeared in F2 generation in 3: 1 ratio. Hence, in F1 generation, the dominant character (round seeds) appeared and the recessive character (wrinkled seeds) got suppressed, which reappeared in F2 generation. Page No 93: Question 5: Define and design a test − cross? ANSWER: Test cross is a cross between an organism with unknown genotype and a recessive parent. It is used to determine whether the individual is homozygous or heterozygous for a trait. If the progenies produced by a test cross show 50% dominant trait and 50% recessive trait, then the unknown individual is heterozygous for a trait. On the other hand, if the progeny produced shows dominant trait, then the unknown individual is homozygous for a trait. Page No 93: 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. ANSWER: In guinea pigs, heterozygous male with black coat colour (Bb) is crossed with the female having white coat colour (bb). The male will produce two types of gametes, B and b, while the female will produce only one kind of gamete, r. The genotypic and phenotypic ratio in the progenies of F1 generation will be same i.e., 1:1. Page No 93: Question 7: When a cross in made between tall plants with yellow seeds (TtYy) and tall plant with green seed (TtYy), what proportions of phenotype in the offspring could be expected to be (a) Tall and green. (b) Dwarf and green. ANSWER: A cross between tall plant with yellow seeds and tall plant with green seeds will produce (a) three tall and green plants (b) one dwarf and green plant Page No 94: 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? ANSWER: Linkage is defined as the coexistence of two or more genes in the same chromosome. If the genes are situated on …

NCERT Solutions for Class 12 Science Biology Chapter 5 – Principles Of Inheritance And Variation Read More »

Certainly! Here is a revised version: Explore the comprehensive NCERT Solutions for Class 12 Science Biology Chapter 4 on Reproductive Health. These solutions are favored by many students for their clear, step-by-step explanations. Whether you need assistance with homework or effective exam preparation, the Biology Reproductive Health Solutions are a valuable resource. All questions and answers from Chapter 4 of the NCERT Book for Class 12 Science Biology are available here for free, providing students with a convenient and accessible study aid. Page No 66: Question 1: What do you think is the significance of reproductive health in a society? ANSWER: Reproductive health is the total well being in all aspects of reproduction. It includes physical, emotional, behavioural, and social well being. Sexually transmitted diseases such as AIDS, gonorrhoea, etc. are transferred from one individual to another through sexual contact. It can also lead to unwanted pregnancies. Hence, it is necessary to create awareness among people, especially the youth, regarding various reproduction related aspects as the young individuals are the future of the country and they are most susceptible of acquiring sexually transmitted diseases. Creating awareness about the available birth control methods, sexually transmitted diseases and their preventive measures, and gender equality will help in bringing up a socially conscious healthy family. Spreading awareness regarding uncontrolled population growth and social evils among young individuals will help in building up a reproductively healthy society. Page No 66: Question 2: Suggest the aspects of reproductive health which need to be given special attention in the present scenario. ANSWER: Reproductive health is the total well being in all aspects of reproduction. The aspects which have to be given special attention in the present scenarios are (1)Counselling and creating awareness among people, especially the youth, about various aspects of reproductive health, such as sexually transmitted diseases, available contraceptive methods, case of pregnant mothers, adolescence, etc. (2)Providing support and facilities such as medical assistance to people during pregnancy, STDs, abortions, contraceptives, infertility, etc. for building a reproductively healthy society Page No 66: Question 3: Is sex education necessary in schools? Why? ANSWER: Yes, introduction of sex education in schools is necessary. It would provide right information to young individuals at the right time about various aspects of reproductive health such as reproductive organs, puberty, and adolescence related changes, safe sexual practices, sexually transmitted diseases, etc. The young individual or adolescents are more susceptible in acquiring various sexually transmitted diseases. Hence, providing information to them at the right time would help them to lead a reproductively healthy life and also protect them from the myths and misconceptions about various sex related issues. Page No 66: 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. ANSWER: Yes, the reproductive health has tremendously improved in India in the last 50 years. The areas of improvement are as follows. (1) Massive child immunization programme, which has lead to a decrease in the infant mortality rate (2) Maternal and infant mortality rate, which has been decreased drastically due to better post natal care (3) Family planning, which has motivated people to have smaller families (4) Use of contraceptive, which has resulted in a decrease in the rate of sexually transmitted diseases and unwanted pregnancies Page No 66: Question 5: What are the suggested reasons for population explosion? ANSWER: The human population is increasing day by day, leading to population explosion. It is because of the following two major reasons. (a) Decreased death rate (b) Increased birth rate and longevity The death rate has decreased in the past 50 years. The factor leading to decreased death rate and increased birth rate are control of diseases, awareness and spread of education, improvement in medical facilities, ensured food supply in emergency situation, etc. All this has also resulted in an increase in the longevity of an individual. Page No 66: Question 6: Is the use of contraceptives justified? Give reasons. ANSWER: Yes, the use of contraceptives is absolutely justified. The human population is increasing tremendously. Therefore, to regulate the population growth by regulating reproduction has become a necessary demand in the present times. Various contraceptive devices have been devised to reduce unwanted pregnancies, which help in bringing down the increased birth rate and hence, in checking population explosion. Page No 66: Question 7: Removal of gonads cannot be considered as a contraceptive option. Why? ANSWER: Contraceptive devices are used to prevent unwanted pregnancy and to prevent the spreading of STDs. There are many methods, such as natural, barrier, oral, and surgical methods, that prevent unwanted pregnancy. However, the complete removal of gonads cannot be a contraceptive option because it will lead to infertility and unavailability of certain hormones that are required for normal functioning of accessory reproductive parts. Therefore, only those contraceptive methods can be used that prevent the chances of fertilization rather than making the person infertile forever. Page No 66: Question 8: Amniocentesis for sex determination is banned in our country. Is this ban necessary? Comment. ANSWER: Amniocentesis is a pre-natal diagnostic technique that is used to determine the sex and metabolic disorders of the developing foetus in the mother’s uterus through the observation of the chromosomal patterns. This method was developed so as to determine any kind of genetic disorder present in the foetus. However, unfortunately, this technique is being misused to detect the sex of the child before birth and the female foetus is then aborted. Thus, to prevent the increasing female foeticides, it is necessary to ban the usage of amniocentesis technique for determining the sex of a child. Page No 66: Question 9: Suggest some methods to assist infertile couples to have children. ANSWER: Infertility is the inability of a couple to produce a baby even after unprotected intercourse. It might be due to abnormalities present in either male or female, or might be even both the partners. The techniques used to assist infertile couples to have children are as follows. (a) Test tube babies This involves …

NCERT Solutions for Class 12 Science Biology Chapter 4 – Reproductive Health Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Biology Chapter 3 on Human Reproduction, featuring easy-to-follow step-by-step explanations. Widely favored by Class 12 Science students, these Biology Human Reproduction Solutions are invaluable for efficiently completing homework assignments and preparing for exams. Access free solutions to all questions from the NCERT Book of Class 12 Science Biology Chapter 3, designed to assist you in mastering the topic. Page No 55: Question 1: Fill in the blanks: (a) Humans reproduce __________. (asexually/sexually) (b) Humans are__________. (oviparous/viviparous/ovoviviparous) (c) Fertilization is __________ in humans. (external/internal) (d) Male and female gametes are __________. (diploid/haploid) (e) Zygote is __________. (diploid/haploid) (f) The process of release of the ovum from a mature follicle is called__________. (g) Ovulation is induced by a hormone called the __________. (h) The fusion of the male and the female gametes is called __________. (i) Fertilization takes place in the __________. (j) The zygote divides to form __________, which is implanted in uterus. (k) The structure which provides vascular connection between the fetus and uterus is called __________. ANSWER: (a) Humans reproduce. (b) Humans are. (c) Fertilization is  in humans. (d) Male and female gametes are. (e) Zygote is. (f) The process of release of the ovum from a mature follicle is called. (g) Ovulation is induced by a hormone called the. (h) The fusion of the male and the female gametes is called. (i) Fertilization takes place in the. (j) The zygote divides to form, which is implanted in uterus. (k) The structure which provides vascular connection between the fetus and uterus is called . Page No 56: Question 2: Draw a labeled diagram of male reproductive system. ANSWER: Page No 56: Question 3: Draw a labeled diagram of female reproductive system. ANSWER: Page No 56: Question 4: Write two major functions each of testis and ovary. ANSWER: Functions of the Testis: (a) They produce male gametes called spermatozoa by the process of spermatogenesis. (b) The leydig cells of the seminiferous tubules secrete the male sex hormone called testosterone. Testosterone aids the development of secondary sex characteristics in males. Functions of the ovary: (a) They produce female gametes called ova by the process of oogenesis. (b) The growing Graffian follicles secrete the female sex hormone called estrogen. Estrogen aids the development of secondary sex characteristics in females. Page No 56: Question 5: Describe the structure of a seminiferous tubule. ANSWER: The production of sperms in the testes takes place in a highly coiled structure called the seminiferous tubules. These tubules are located in the testicular lobules. Each seminiferous tubule is lined by germinal epithelium. It is lined on its inner side by two types of cells namely spermatogonia and sertoli cells respectively. Spermatogonia are male germ cells which produce primary spermatocytes by meiotic divisions. Primary spermatocytes undergo further meiotic division to form secondary spermatocytes and finally, spermatids. Spermatids later metamorphoses into male gametes called spermatozoa. Sertoli cells are known as nurse cells of the testes as they provide nourishment to the germ cells. There are large polygonal cells known as interstitial cells or leydig cells just adjacent to seminiferous tubules. These cells secrete the male hormone called testosterone. Page No 56: Question 6: What is spermatogenesis? Briefly describe the process of spermatogenesis. ANSWER: Spermatogenesis is the process of the production of sperms from the immature germ cells in males. It takes place in seminiferous tubules present inside the testes. During spermatogenesis, a diploid spermatogonium (male germ cell) increases its size to form a diploid primary spermatocyte. This diploid primary spermatocyte undergoes first meiotic division (meiosis I), which is a reductional division to form two equal haploid secondary spermatocytes. Each secondary spermatocyte then undergoes second meiotic division (meiosis II) to form two equal haploid spermatids. Hence, a diploid spermatogonium produces four haploid spermatids. These spermatids are transformed into spermatozoa (sperm) by the process called spermiogenesis. Page No 56: Question 7: Name the hormones involved in regulation of spermatogenesis. ANSWER: Follicle-stimulating hormones (FSH) and luteinizing hormones (LH) are secreted by gonadotropin releasing hormones from the hypothalamus .These hormones are involved in the regulation of the process of spermatogenesis. FSH acts on sertoli cells, whereas LH acts on leydig cells of the testis and stimulates the process of spermatogenesis. Page No 56: Question 8: Define spermiogenesis and spermiation. ANSWER: Spermiogenesis: It is the process of transforming spermatids into matured spermatozoa or sperms. Spermiation: It is the process when mature spermatozoa are released from the sertoli cells into the lumen of seminiferous tubules. Page No 56: Question 9: Draw a labeled diagram of sperm. ANSWER: Page No 56: Question 10: What are the major components of seminal plasma? ANSWER: Semen (produced in males) is composed of sperms and seminal plasma. The major components of the seminal plasma in the male reproductive system are mucus, spermatozoa, and various secretions of accessory glands. The seminal plasma is rich in fructose, calcium, ascorbic acid, and certain enzymes. It provides nourishment and protection to sperms. Page No 56: Question 11: What are the major functions of male accessory ducts and glands? ANSWER: The male accessory ducts are vasa efferentia, epididymis, vas deferens, and rete testis. They play an important role in the transport and temporary storage of sperms. On the contrary, male accessory glands are seminal vesicles, prostate glands, and bulbourethral glands. These glands secrete fluids that lubricate the reproductive system and sperms. The sperms get dispersed in the fluid which makes their transportation into the female body easier. The fluid is rich in fructose, ascorbic acid, and certain enzymes. They also provide nutrients and activate the sperm. Page No 56: Question 12: What is oogenesis? Give a brief account of oogenesis. ANSWER: Oogenesis is the process of the formation of a mature ovum from the oogonia in females. It takes place in the ovaries. During oogenesis, a diploid oogonium or egg mother cell increases in size and gets transformed into a diploid primary oocyte. This diploid primary oocyte undergoes first meiotic division i.e., meiosis I or reductional division to form two unequal haploid cells. The smaller cell is known as the first polar body, while the larger cell is known as the secondary oocyte. This secondary oocyte …

NCERT Solutions for Class 12 Science Biology Chapter 3 – Human Reproduction Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Biology Chapter 2 on Sexual Reproduction in Flowering Plants. These solutions offer clear, step-by-step explanations, making them highly favored among Biology students in Class 12. Ideal for completing homework assignments swiftly and preparing for exams, the Sexual Reproduction in Flowering Plants Solutions cover all questions and answers from the NCERT Book. Access these valuable resources for free to enhance your understanding and excel in your studies. Page No 40: Question 1: Name the parts of an angiosperm flower in which development of male and female gametophyte take place. ANSWER: The male gametophyte or the pollen grain develops inside the pollen chamber of the anther, whereas the female gametophyte (also known as the embryo sac) develops inside the nucellus of the ovule from the functional megaspore. Page No 40: Question 2: Differentiate between microsporogenesis and megasporogenesis. Which type of cell division occurs during these events? Name the structures formed at the end of these two events. ANSWER: (a) Microsporogenesis Megasporogenesis 1. It is the process of the formation of microspore tetrads from a microspore mother cell through meiosis. It is the process of the formation of the four megaspores from a megaspore mother cell in the region of the nucellus through meiosis 2. It occurs inside the pollen sac of the anther. It occurs inside the ovule. (b) Both events (microsporogenesis and megasporogenesis) involve the process of meiosis or reduction division which results in the formation of haploid gametes from the microspore and megaspore mother cells. (c) Microsporogenesis results in the formation of haploid microspores from a diploid microspore mother cell. On the other hand, megasporogenesis results in the formation of haploid megaspores from a diploid megaspore mother cell. Page No 40: Question 3: Arrange the following terms in the correct developmental sequence: Pollen grain, sporogenous tissue, microspore tetrad, pollen mother cell, male gametes ANSWER: The correct development sequence is as follows: Sporogenous tissue – pollen mother cell – microspore tetrad – Pollen grain – male gamete During the development of microsporangium, each cell of the sporogenous tissue acts as a pollen mother cell and gives rise to a microspore tetrad, containing four haploid microspores by the process of meiosis (microsporogenesis). As the anther matures, these microspores dissociate and develop into pollen grains. The pollen grains mature and give rise to male gametes. Page No 40: Question 4: With a neat, labelled diagram, describe the parts of a typical angiosperm ovule. ANSWER: An ovule is a female megasporangium where the formation of megaspores takes place. The various parts of an ovule are – (1) Funiculus – It is a stalk-like structure which represents the point of attachment of the ovule to the placenta of the ovary. (2) Hilum – It is the point where the body of the ovule is attached to the funiculus. (3) Integuments –They are the outer layers surrounding the ovule that provide protection to the developing embryo. (4) Micropyle – It is a narrow pore formed by the projection of integuments. It marks the point where the pollen tube enters the ovule at the time of fertilization. (5) Nucellus – It is a mass of the parenchymatous tissue surrounded by the integuments from the outside. The nucellus provides nutrition to the developing embryo. The embryo sac is located inside the nucellus. (6) Chalazal – It is the based swollen part of the nucellus from where the integuments originate. Page No 40: Question 5: What is meant by monosporic development of female gametophyte? ANSWER: The female gametophyte or the embryo sac develops from a single functional megaspore. This is known as monosporic development of the female gametophyte. In most flowering plants, a single megaspore mother cell present at the micropylar pole of the nucellus region of the ovule undergoes meiosis to produce four haploid megaspores. Later, out of these four megaspores, only one functional megaspore develops into the female gametophyte, while the remaining three degenerate. Page No 40: Question 6: With a neat diagram explain the 7-celled, 8-nucleate nature of the female gametophyte. ANSWER: The female gametophyte (embryo sac) develops from a single functional megaspore. This megaspore undergoes three successive mitotic divisions to form eight nucleate embryo sacs. The first mitotic division in the megaspore forms two nuclei. One nucleus moves towards the micropylar end while the other nucleus moves towards the chalazal end. Then, these nuclei divide at their respective ends and re-divide to form eight nucleate stages. As a result, there are four nuclei each at both the ends i.e., at the micropylar and the chalazal end in the embryo sac. At the micropylar end, out of the four nuclei only three differentiate into two synergids and one egg cell. Together they are known as the egg apparatus. Similarly, at the chalazal end, three out of four nuclei differentiates as antipodal cells. The remaining two cells (of the micropylar and the chalazal end) move towards the centre and are known as the polar nuclei, which are situated in a large central cell. Hence, at maturity, the female gametophyte appears as a 7-celled structure, though it has 8 nucleate. Page No 41: Question 7: What are chasmogamous flowers? Can cross-pollination occur in cleistogamous flowers? Give reasons for your answer. ANSWER: There are two types of flowers present in plants namely Oxalis and Viola − chasmogamous and cleistogamous flowers. Chasmogamous flowers have exposed anthers and stigmata similar to the flowers of other species. Cross-pollination cannot occur in cleistogamous flowers. This is because cleistogamous flowers never open at all. Also, the anther and the stigma lie close to each other in these flowers. Hence, only self-pollination is possible in these flowers. Page No 41: Question 8: Mention two strategies evolved to prevent self-pollination in flowers. ANSWER: Self-pollination involves the transfer of pollen from the stamen to the pistil of the same flower. Two strategies that have evolved to prevent self-pollination in flowers are as follows: (1) In certain plants, the stigma of the flower hasthecapability to prevent the germination of pollen grains and hence, prevent the growth of the pollen tube.It is …

NCERT Solutions for Class 12 Science Biology Chapter 2 – Sexual Reproduction In Flowering Plants Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Biology Chapter 1: “Reproduction In Organisms.” These solutions, accompanied by clear step-by-step explanations, are highly favored by Biology students at the 12th-grade level. Whether you need assistance with homework or effective exam preparation, these solutions for Reproduction In Organisms prove to be a valuable resource. All questions and answers from Chapter 1 of the NCERT Book for class 12 Science Biology are conveniently provided here for free, ensuring a convenient and accessible study aid for students. Page No 17: Question 1: Why is reproduction essential for organisms? ANSWER: Reproduction is a fundamental feature of all living organisms. It is a biological process through which living organisms produce offspring’s similar to them. Reproduction ensures the continuance of various species on the Earth. In the absence of reproduction, the species will not be able to exist for a long time and may soon get extinct. Page No 17: Question 2: Which is a better mode of reproduction sexual or asexual? Why? ANSWER: Sexual reproduction is a better mode of reproduction. It allows the formation of new variants by the combination of the DNA from two different individuals, typically one of each sex. It involves the fusion of the male and the female gamete to produce variants, which are not identical to their parents and to themselves. This variation allows the individual to adapt to constantly changing and challenging environments. Also, it leads to the evolution of better suited organisms which ensures greater survival of a species. On the contrary, asexual reproduction allows very little or no variation at all. As a result, the individuals produced are exact copies of their parents and themselves. Page No 17: Question 3: Why is the offspring formed by asexual reproduction referred to as clone? ANSWER: A clone is a group of morphologically and genetically identical individuals. In the process of asexual reproduction, only one parent is involved and there is no fusion of the male and the female gamete. As a result, the offsprings so produced are morphologically and genetically similar to their parents and are thus, called clones. Page No 17: Question 4: Offspring formed due to sexual reproduction have better chances of survival. Why? Is this statement always true? ANSWER: Sexual reproduction involves the fusion of the male and the female gamete. This fusion allows the formation of new variants by the combination of the DNA from two (usually) different members of the species. The variations allow the individuals to adapt under varied environmental conditions for better chances of survival. However, it is not always necessary that the offspring produced due to sexual reproduction has better chances of survival. Under some circumstances, asexual reproduction is more advantageous for certain organisms. For example, some individuals who do not move from one place to another and are well settled in their environment. Also, asexual reproduction is a fast and a quick mode of reproduction which does not consume much time and energy as compared to sexual reproduction. Page No 17: Question 5: How does the progeny formed from asexual reproduction differ from those formed by sexual reproduction? ANSWER: Progeny formed from asexual reproduction Progeny formed from sexual reproduction 1. Asexual reproduction does not involve the fusion of the male and the female gamete. Organisms undergoing this kind of reproduction produce offspring’s that are morphologically and genetically identical to them. Sexual reproduction involves the fusion of the male and the female gamete of two individuals, typically one of each sex. Organisms undergoing this kind of reproduction produce offspring’s that are not identical to them. 2. Offsprings thus produced do not show variations and are called clones. Offspring’s thus produced show variations from each other and their parents. Page No 17: Question 6: Distinguish between asexual and sexual reproduction. Why is vegetative reproduction also considered as a type of asexual reproduction? ANSWER: Sexual reproduction Asexual reproduction 1 It involves the fusion of the male and female gamete. It does not involves the fusion of the male and the female gamete 2. It requires two (usually) different individuals. It requires only one individual. 3. The individuals produced are not identical to their parents and show variations from each other and also, from their parents. The individuals produced are identical to the parent and are hence, called clones. 4. Most animals reproduce sexually. Both sexual and asexual modes of reproduction are found in plants. Asexual modes of reproduction are common in organisms having simple organizations such as algae and fungi. 5. It is a slow process. It is a fast process. Vegetative propagation is a process in which new plants are obtained without the production of seeds or spores. It involves the propagation of plants through certain vegetative parts such as the rhizome, sucker, tuber, bulb, etc. It does not involve the fusion of the male and the female gamete and requires only one parent. Hence, vegetative reproduction is considered as a type of asexual reproduction. Page No 18: Question 7: What is vegetative propagation? Give two suitable examples. ANSWER: Vegetative propagation is a mode of asexual reproduction in which new plants are obtained from the vegetative parts of plants. It does not involve the production of seeds or spores for the propagation of new plants. Vegetative parts of plants such as runners, rhizomes, suckers, tubers, etc. can be used as propagules for raising new plants. Examples of vegetative reproduction are: 1. Eyes of potato: The surface of a potato has several buds called eyes. Each of these buds when buried in soil develops into a new plant, which is identical to the parent plant. 2. Leaf buds of Bryophyllum: The leaves of Bryophyllum plants bear several adventitious buds on their margins. These leaf buds have the ability to grow and develop into tiny plants when the leaves get detached from the plant and come in contact with moist soil. Page No 18: Question 8: Define (a) Juvenile phase, (b) Reproductive phase, (c) Senescent phase. ANSWER: (a) Juvenile phase: It is the period of growth in …

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Here are the NCERT Solutions for Class 12 Science Chemistry Chapter 7, titled “Chemistry In Everyday Life.” These solutions offer clear and straightforward explanations, making them highly favored by class 12 Science students. They prove to be invaluable for efficiently completing homework assignments and preparing for exams. Free access to all questions and answers from the NCERT Book of Class 12 Science Chemistry Chapter 7 is provided here, ensuring a helpful resource for students. Page No 448: Question 16.1: Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take its doses without consultation with the doctor, Why? ANSWER: Most drugs when taken in doses higher than recommended may cause harmful effects and sometimes, may even lead to death. Hence, a doctor should always be consulted before taking any medicine. Page No 448: Question 16.2: With reference to which classification has the statement, ‘ranitidine is an antacid” been given? ANSWER: The given statement refers to the classification of pharmacological effects of the drug. This is because any drug that is used to counteract the effects of excess acid in the stomach is called an antacid. Page No 450: Question 16.3: Why do we require artificial sweetening agents? ANSWER: A large number of people are suffering from diseases such as diabetes and obesity. These people cannot take normal sugar i.e., sucrose as it is harmful for them. Therefore, artificial sweetening agents that do not add to the calorie intake of a person are required. Saccharin, aspartame, and alitame are a few examples of artificial sweeteners. Page No 453: Question 16.4: Write the chemical equation for preparing sodium soap from glyceryl oleate and glyceryl palmitate. Structural formulae of these compounds are given below. (i)  (ii)  ANSWER: (i) (ii) Page No 453: Question 16.5: Following type of nom-ionic detergents are present in liquid detergents, emulsifying agents and wetting agents. Label the hydrophilic and hydrophobic parts in the molecule. Identify the functional group (s) present in the molecule. ANSWER: Functional groups present in the molecule are: (i) Ether, and (ii) primary alcoholic group Page No 454: Question 16.1: Why do we need to classify drugs in different ways? ANSWER: The classification of drugs and the reasons for classification are as follows: (i) On the basis of pharmacological effect: This classification provides doctors the whole range of drugs available for the treatment of a particular type of problem. Hence, such a classification is very useful to doctors. (ii) On the basis of drug action: This classification is based on the action of a drug on a particular biochemical process. Thus, this classification is important. (iii) On the basis of chemical structure: This classification provides the range of drugs sharing common structural features and often having similar pharmacological activity. (iv) On the basis of molecular targets: This classification provides medicinal chemists the drugs having the same mechanism of action on targets. Hence, it is the most useful to medicinal chemists. Page No 454: Question 16.2: Explain the term target molecules or drug targets as used in medicinal chemistry. ANSWER: In medicinal chemistry, drug targets refer to the key molecules involved in certain metabolic pathways that result in specific diseases. Carbohydrates, proteins, lipids, and nucleic acids are examples of drug targets. Drugs are chemical agents designed to inhibit these target molecules by binding with the active sites of the key molecules. Page No 454: Question 16.3: Name the macromolecules that are chosen as drug targets. ANSWER: The macromolecules chosen as drug targets are carbohydrates, lipids, proteins, and nucleic acids. Page No 454: Question 16.4: Why should not medicines be taken without consulting doctors? ANSWER: A medicine can bind to more than one receptor site. Thus, a medicine may be toxic for some receptor sites. Further, in most cases, medicines cause harmful effects when taken in higher doses than recommended. As a result, medicines may be poisonous in such cases. Hence, medicines should not be taken without consulting doctors. Page No 454: Question 16.5: Define the term chemotherapy. ANSWER: The use of chemicals for therapeutic effect is called chemotherapy. For example: the use of chemicals in the diagnosis, prevention, and treatment of diseases. Page No 454: Question 16.6: Which forces are involved in holding the drugs to the active site of enzymes? ANSWER: Either of the following forces can be involved in holding drugs to the active sites of enzymes. (i) Ionic bonding (ii) Hydrogen bonding (iii) Dipole − dipole interaction (iv) van der Waals force Page No 454: Question 16.7: While antacids and antiallergic drugs interfere with the function of histamines, why do these not interfere with the function of each other? ANSWER: Specific drugs affect particular receptors. Antacids and anti-allergic drugs work on different receptors. This is the reason why antacids and anti-allergic drugs do not interfere with each other’s functions, but interfere with the functions of histamines. Page No 454: Question 16.8: Low level of noradrenaline is the cause of depression. What types of drugs are needed to cure this problem? Name two drugs. ANSWER: Anti-depressant drugs are needed to counteract the effect of depression. These drugs inhibit enzymes catalysing the degradation of the neurotransmitter, noradrenaline. As a result, the important neurotransmitter is slowly metabolised and then it can activate its receptor for longer periods of time. Two anti-depressant drugs are: (i) Iproniazid (ii) Phenelzine Page No 454: Question 16.9: What is meant by the term ‘broad spectrum antibiotics’? Explain. ANSWER: Antibiotics that are effective against a wide range of gram-positive and gram-negative bacteria are known as broad spectrum antibiotics. Chloramphenicol is a broad spectrum antibiotic. It can be used for the treatment of typhoid, dysentery, acute fever, pneumonia, meningitis, and certain forms of urinary infections. Two other broad spectrum antibiotics are vancomycin and ofloxacin. Ampicillin and amoxicillin −synthetically modified from penicillin − are also broad spectrum antibiotics. Page No 454: Question 16.10: How do antiseptics differ from disinfectants? Give one example of each. ANSWER: Antiseptics and disinfectants are effective against micro-organisms. However, antiseptics are applied to the living tissues such as wounds, cuts, ulcers, and diseased skin surfaces, …

NCERT Solutions for Class 12 Science Chemistry Chapter 7 – Chemistry In Everyday Life Read More »

Here are step-by-step explanations for NCERT Solutions for Class 12 Science Chemistry Chapter 6 on Polymers. These solutions are highly favored by Chemistry students for completing homework and exam preparation. All questions and answers from the NCERT Book of Class 12 Science Chemistry Chapter 6, Polymers, are available here for free use. Page No 428: Question 15.1: What are polymers? ANSWER: Polymers are high molecular mass macromolecules, which consist of repeating structural units derived from monomers. Polymers have a high molecular mass (103 − 107u). In a polymer, various monomer units are joined by strong covalent bonds. These polymers can be natural as well as synthetic. Polythene, rubber, and nylon 6, 6 are examples of polymers. Page No 428: Question 15.2: How are polymers classified on the basis of structure? ANSWER: Polymers are classified on the basis of structure as follows: 1. Linear polymers: These polymers are formed of long straight chains. They can be depicted as: For e.g., high density polythene (HDP), polyvinyl chloride, etc. 2. Branched chain polymers: These polymers are basically linear chain polymers with some branches. These polymers are represented as: For e.g., low density polythene (LDP), amylopectin, etc. 3. Cross-linked or Network polymers: These polymers have many cross-linking bonds that give rise to a network-like structure. These polymers contain bi-functional and tri-functional monomers and strong covalent bonds between various linear polymer chains. Examples of such polymers include bakelite and melmac. Page No 428: Question 15.3: Write the names of monomers of the following polymers: ANSWER: (i) Hexamethylenediamine and adipic acid  (ii) (iii) Tetrafluoroethene  Page No 428: Question 15.4: Classify the following as addition and condensation polymers: Terylene, Bakelite, Polyvinyl chloride, Polythene. ANSWER: Addition polymers: Polyvinyl chloride, polythene Condensation polymers: Terylene, bakelite Page No 435: Question 15.5: Explain the difference between Buna-N and Buna-S. ANSWER: Buna − N is a copolymer of 1, 3−butadiene and acrylonitrile. Buna − S is a copolymer of 1, 3−butadiene and styrene. Page No 435: Question 15.6: Arrange the following polymers in increasing order of their intermolecular forces. (i) Nylon 6, 6, Buna-S, Polythene. (ii) Nylon 6, Neoprene, Polyvinyl chloride. ANSWER: Different types of polymers have different intermolecular forces of attraction. Elastomers or rubbers have the weakest while fibres have the strongest intermolecular forces of attraction. Plastics have intermediate intermolecular forces of attraction. Hence, the increasing order of the intermolecular forces of the given polymers is as follows: (i) Buna − S < polythene < Nylon 6, 6 (ii) Neoprene < polyvinyl chloride < Nylon 6 Page No 437: Question 15.1: Explain the terms polymer and monomer. ANSWER: Polymers are high molecular mass macromolecules composed of repeating structural units derived from monomers. Polymers have a high molecular mass (103 − 107u). In a polymer, various monomer units are joined by strong covalent bonds. Polymers can be natural as well as synthetic. Polythene, rubber, and nylon 6, 6 are examples of polymers. Monomers are simple, reactive molecules that combine with each other in large numbers through covalent bonds to give rise to polymers. For example, ethene, propene, styrene, vinyl chloride. Page No 437: Question 15.2: What are natural and synthetic polymers? Give two examples of each type. ANSWER: Natural polymers are polymers that are found in nature. They are formed by plants and animals. Examples include protein, cellulose, starch, etc. Synthetic polymers are polymers made by human beings. Examples include plastic (polythene), synthetic fibres (nylon 6, 6), synthetic rubbers (Buna − S). Page No 437: Question 15.3: Distinguish between the terms homopolymer and copolymer and give an example of each. ANSWER: Homopolymer  Copolymer  The polymers that are formed by the polymerization of a single monomer are known as homopolymers. In other words, the repeating units of homopolymers are derived only from one monomer. For example, polythene is a homopolymer of ethene. The polymers whose repeating units are derived from two types of monomers are known as copolymers. For example, Buna−S is a copolymer of 1, 3-butadiene and styrene. Page No 437: Question 15.4: How do you explain the functionality of a monomer? ANSWER: The functionality of a monomer is the number of binding sites that is/are present in that monomer. For example, the functionality of monomers such as ethene and propene is one and that of 1, 3-butadiene and adipic acid is two. Page No 437: Question 15.5: Define the term polymerisation. ANSWER: Polymerization is the process of forming high molecular mass (103 − 107u) macromolecules, which consist of repeating structural units derived from monomers. In a polymer, various monomer units are joined by strong covalent bonds. Page No 437: Question 15.6: Is , a homopolymer or copolymer? ANSWER:  is a homopolymer because it is obtained from a single monomer unit, NH2−CHR−COOH. Page No 437: Question 15.7: In which classes, the polymers are classified on the basis of molecular forces? ANSWER: On the basis of magnitude of intermolecular forces present in polymers, they are classified into the following groups: (i) Elastomers (ii) Fibres (iii) Thermoplastic polymers (iv) Thermosetting polymers Page No 437: Question 15.8: How can you differentiate between addition and condensation polymerisation? ANSWER: Addition polymerization is the process of repeated addition of monomers, possessing double or triple bonds to form polymers. For example, polythene is formed by addition polymerization of ethene. Condensation polymerization is the process of formation of polymers by repeated condensation reactions between two different bi-functional or tri-functional monomers. A small molecule such as water or hydrochloric acid is eliminated in each condensation. For example, nylon 6, 6 is formed by condensation polymerization of hexamethylenediamine and adipic acid. Page No 437: Question 15.9: Explain the term copolymerisation and give two examples. ANSWER: The process of forming polymers from two or more different monomeric units is called copolymerization. Multiple units of each monomer are present in a copolymer. The process of forming polymer Buna−S from 1, 3-butadiene and styrene is an example of copolymerization Nylon 6, 6 is also a copolymer formed by hexamethylenediamine and adipic acid. Page No 437: Question 15.10: Write the free radical mechanism for the polymerisation of ethene. ANSWER: Polymerization of ethene to polythene consists of heating or exposing to …

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Explore the highly sought-after NCERT Solutions for Class 12 Science Chemistry Chapter 5: Biomolecules. These step-by-step solutions offer clear explanations, making them a favorite among Chemistry students. Whether you’re completing homework assignments or gearing up for exams, the Biomolecules Solutions for Class 12 Science provide a valuable resource. All questions and answers from Chapter 5 of the NCERT Book for Class 12 Science Chemistry are readily available here for free, ensuring a convenient and effective study experience. Page No 412: Question 14.1: Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain. ANSWER: A glucose molecule contains five −OH groups while a sucrose molecule contains eight −OH groups. Thus, glucose and sucrose undergo extensive H-bonding with water. Hence, these are soluble in water. But cyclohexane and benzene do not contain −OH groups. Hence, they cannot undergo H-bonding with water and as a result, are insoluble in water. Page No 412: Question 14.2: What are the expected products of hydrolysis of lactose? ANSWER: Lactose is composed of β-D galactose and β-D glucose. Thus, on hydrolysis, it gives β-D galactose and β-D glucose. Page No 412: Question 14.3: How do you explain the absence of aldehyde group in the pentaacetate of D-glucose? ANSWER: D-glucose reacts with hydroxylamine (NH2OH) to form an oxime because of the presence of aldehydic (−CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH2OH to give an oxime. But pentaacetate of D-glucose does not react with NH2OH. This is because pentaacetate does not form an open chain structure. Page No 417: Question 14.4: The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain. ANSWER: Both acidic (carboxyl) as well as basic (amino) groups are present in the same molecule of amino acids. In aqueous solutions, the carboxyl group can lose a proton and the amino group can accept a proton, thus giving rise to a dipolar ion known as a zwitter ion. Due to this dipolar behaviour, they have strong electrostatic interactions within them and with water. But halo-acids do not exhibit such dipolar behaviour. For this reason, the melting points and the solubility of amino acids in water is higher than those of the corresponding halo-acids. Page No 417: Question 14.5: Where does the water present in the egg go after boiling the egg? ANSWER: When an egg is boiled, the proteins present inside the egg get denatured and coagulate. After boiling the egg, the water present in it is absorbed by the coagulated protein through H-bonding. Page No 422: Question 14.6: Why cannot vitamin C be stored in our body? ANSWER: Vitamin C cannot be stored in our body because it is water soluble. As a result, it is readily excreted in the urine. Page No 422: Question 14.7: What products would be formed when a nucleotide from DNA containing thymine is hydrolysed? ANSWER: When a nucleotide from the DNA containing thymine is hydrolyzed, thymine β-D-2-deoxyribose and phosphoric acid are obtained as products. Page No 422: Question 14.8: When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA? ANSWER: A DNA molecule is double-stranded in which the pairing of bases occurs. Adenine always pairs with thymine, while cytosine always pairs with guanine. Therefore, on hydrolysis of DNA, the quantity of adenine produced is equal to that of thymine and similarly, the quantity of cytosine is equal to that of guanine. But when RNA is hydrolyzed, there is no relationship among the quantities of the different bases obtained. Hence, RNA is single-stranded. Page No 423: Question 14.1: What are monosaccharides? ANSWER: Monosaccharides are carbohydrates that cannot be hydrolysed further to give simpler units of polyhydroxy aldehyde or ketone. Monosaccharides are classified on the bases of number of carbon atoms and the functional group present in them. Monosaccharides containing an aldehyde group are known as aldoses and those containing a keto group are known as ketoses. Monosaccharides are further classified as trioses, tetroses, pentoses, hexoses, and heptoses according to the number of carbon atoms they contain. For example, a ketose containing 3 carbon atoms is called ketotriose and an aldose containing 3 carbon atoms is called aldotriose. Page No 423: Question 14.2: What are reducing sugars? ANSWER: Reducing sugars are carbohydrates that reduce Fehling’s solution and Tollen’s reagent. All monosaccharides and disaccharides, excluding sucrose, are reducing sugars. Page No 423: Question 14.3: Write two main functions of carbohydrates in plants. ANSWER: Two main functions of carbohydrates in plants are: (i) Polysaccharides such as starch serve as storage molecules. (ii) Cellulose, a polysaccharide, is used to build the cell wall. Page No 423: Question 14.4: Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose ANSWER: Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose Disaccharides: Maltose, lactose Page No 423: Question 14.5: What do you understand by the term glycosidic linkage? ANSWER: Glycosidic linkage refers to the linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule. For example, in a sucrose molecule, two monosaccharide units, ∝-glucose and β-fructose, are joined together by a glycosidic linkage. Page No 423: Question 14.6: What is glycogen? How is it different from starch? ANSWER: Glycogen is a carbohydrate (polysaccharide). In animals, carbohydrates are stored as glycogen. Starch is a carbohydrate consisting of two components − amylose (15 − 20%) and amylopectin (80 − 85%). However, glycogen consists of only one component whose structure is similar to amylopectin. Also, glycogen is more branched than amylopectin. Page No 423: Question 14.7: What are the hydrolysis products of (i) sucrose and (ii) lactose? ANSWER: (i) On hydrolysis, sucrose gives one molecule of ∝-D glucose and one molecule of β- D-fructose. (ii) The hydrolysis of lactose gives β-D-galactose and β-D-glucose. Page No 423: Question 14.8: What is the basic structural difference between starch and cellulose? ANSWER: Starch consists of two components …

NCERT Solutions for Class 12 Science Chemistry Chapter 5 – Biomolecules Read More »

Explore the comprehensive NCERT solutions for Class 12 Science Chemistry Chapter 4 on Amines, complete with easy-to-follow step-by-step explanations. Widely favored among Class 12 Science students, these Chemistry Amines Solutions are invaluable for efficiently tackling homework assignments and gearing up for exams. Free of charge, you can access all questions and answers from Chapter 4 of the NCERT Book for Class 12 Science Chemistry, ensuring a convenient resource for your academic needs. Page No 384: Question 13.1: Classify the following amines as primary, secondary or tertiary: (i) (ii) (iii) (C2H5)2CHNH2 (iv) (C2H5)2NH ANSWER: Primary: (i) and (iii) Secondary: (iv) Tertiary: (ii) Page No 384: Question 13.2: (i) Write structures of different isomeric amines corresponding to the molecular formula, C4H11N (ii) Write IUPAC names of all the isomers. (iii) What type of isomerism is exhibited by different pairs of amines? ANSWER: (i), (ii) The structures and their IUPAC names of different isomeric amines corresponding to the molecular formula, C4H11N are given below: (a) CH3-CH2-CH2-CH2-NH2 Butanamine (10) (b)  Butan-2-amine (10) (c)  2-Methylpropanamine (10) (d) 2-Methylpropan-2-amine (10) (e) CH3-CH2-CH2-NH-CH3 N-Methylpropanamine (20) (f) CH3-CH2-NH-CH2-CH3 N-Ethylethanamine (20) (g) N-Methylpropan-2-amine (20) (h) N,N-Dimethylethanamine (3°) (iii) The pairs (a) and (b) and (e) and (g) exhibit position isomerism. The pairs (a) and (c); (a) and (d); (b) and (c); (b) and (d) exhibit chain isomerism. The pairs (e) and (f) and (f) and (g) exhibit metamerism. All primary amines exhibit functional isomerism with secondary and tertiary amines and vice-versa. Page No 387: Question 13.3: How will you convert? (i) Benzene into aniline (ii) Benzene into N, N-dimethylaniline (iii) Cl−(CH2)4−Cl into hexan-1, 6-diamine? ANSWER: (i) (ii) (iii) Page No 396: Question 13.4: Arrange the following in increasing order of their basic strength: (i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH (ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2 (iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2. ANSWER: (i) Considering the inductive effect of alkyl groups, NH3, C2H5NH2, and (C2H5)2NH can be arranged in the increasing order of their basic strengths as: Again, C6H5NH2 has proton acceptability less than NH3. Thus, we have: Due to the −I effect of C6H5 group, the electron density on the N-atom in C6H5CH2NH2 is lower than that on the N-atom in C2H5NH2, but more than that in NH3. Therefore, the given compounds can be arranged in the order of their basic strengths as: (ii) Considering the inductive effect and the steric hindrance of the alkyl groups, C2H5NH2, (C2 H5)2NH2, and their basic strengths as follows: Again, due to the −R effect of C6H5 group, the electron density on the N atom in C6H5 NH2 is lower than that on the N atom in C2H5NH2. Therefore, the basicity of C6H5NH2 is lower than that of C2H5NH2. Hence, the given compounds can be arranged in the increasing order of their basic strengths as follows: (iii) Considering the inductive effect and the steric hindrance of alkyl groups, CH3NH2, (CH3)2NH, and (CH3)3N can be arranged in the increasing order of their basic strengths as: In C6H5NH2, N is directly attached to the benzene ring. Thus, the lone pair of electrons on the N−atom is delocalized over the benzene ring. In C6H5CH2NH2, N is not directly attached to the benzene ring. Thus, its lone pair is not delocalized over the benzene ring. Therefore, the electrons on the N atom are more easily available for protonation in C6H5CH2NH2 than in C6H5NH2 i.e., C6H5CH2­NH2 is more basic than C6H5NH2. Again, due to the −I effect of C6H5 group, the electron density on the N−atom in C6H5CH2NH2 is lower than that on the N−atom in (CH3)3N. Therefore, (CH3)3N is more basic than C6H5CH2NH2. Thus, the given compounds can be arranged in the increasing order of their basic strengths as follows. Page No 396: Question 13.5: Complete the following acid-base reactions and name the products: (i) CH3CH2CH2NH2 + HCl  (ii) (C2H5)3N + HCl  ANSWER: (i)  (ii) (C2H5)3N Triethylamine+ HCl → (C2H5)3N+HCl−TriethylammoniumchlorideC2H53N Triethylamine+ HCl → C2H53N+HCl-Triethylammoniumchloride Page No 396: Question 13.6: Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution. ANSWER: Aniline reacts with methyl iodide to produce N, N-dimethylaniline. With excess methyl iodide, in the presence of Na2CO3solution, N, N-dimethylaniline produces N, N, N−trimethylanilinium carbonate. Page No 396: Question 13.7: Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained. ANSWER: Page No 396: Question 13.8: Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid. ANSWER: The structures of different isomers corresponding to the molecular formula, C3H9N are given below: (a)  Propan-1-amine (10) (b) Propan-2-amine (10) (c) (d) N,N-Dimethylmethanamine (30) 10amines, (a) propan-1-amine, and (b) Propan-2-amine will liberate nitrogen gas on treatment with nitrous acid. Page No 399: Question 13.9: Convert (i) 3-Methylaniline into 3-nitrotoluene. (ii) Aniline into 1,3,5-tribromobenzene. ANSWER: (i) (ii) Page No 400: Question 13.1: Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (i) (CH3)2 CHNH2 (ii) CH3(CH2)2NH2 (iii) CH3NHCH(CH3)2 (iv) (CH3)3CNH2 (v) C6H5NHCH3(vi) (CH3CH2)2NCH3 (vii) m−BrC6H4NH2 ANSWER: (i) 1-Methylethanamine (10 amine) (ii) Propan-1-amine (10 amine) (iii) N−Methyl-2-methylethanamine (20 amine) (iv) 2-Methylpropan-2-amine (10 amine) (v) N−Methylbenzamine or N-methylaniline (20 amine) (vi) N-Ethyl-N-methylethanamine (30 amine) (vii) 3-Bromobenzenamine or 3-bromoaniline (10 amine) Page No 400: Question 13.2: Give one chemical test to distinguish between the following pairs of compounds. (i) Methylamine and dimethylamine (ii) Secondary and tertiary amines (iii) Ethylamine and aniline (iv) Aniline and benzylamine (v) Aniline and N-methylaniline. ANSWER: (i) Methylamine and dimethylamine can be distinguished by the carbylamine test. Carbylamine test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines. Methylamine (being an aliphatic primary amine) gives a positive carbylamine test, but dimethylamine does not. (ii) Secondary and tertiary amines can be distinguished by allowing them to react with Hinsberg’s reagent (benzenesulphonyl chloride, C6H5SO2Cl). Secondary amines react with Hinsberg’s reagent to form a product that is insoluble in an alkali. For example, N, N−diethylamine reacts with Hinsberg’s reagent to form N, N−diethylbenzenesulphonamide, which is insoluble in an alkali. Tertiary amines, however, do not react with Hinsberg’s reagent. (iii) Ethylamine and aniline can be distinguished using the azo-dye test. A dye is obtained when aromatic amines react with HNO2 (NaNO2 + dil.HCl) at 0-5°C, followed by a reaction with the alkaline solution of 2-naphthol. The dye is usually yellow, red, or orange in colour. Aliphatic amines give a brisk effervescence due (to the evolution of N2 gas) under similar conditions. ​ (iv) Aniline …

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