Here are simplified step-by-step explanations of NCERT Solutions for Class 12 Science Biology Chapter 8 “Human Health And Disease.” These solutions are highly favored among Biology students in Class 12 Science for their ease of understanding and usefulness in completing homework assignments and exam preparation. All questions and answers from Chapter 8 of the NCERT Book for Class 12 Science Biology are available here at no cost, aiding students in their studies. Page No 164: Question 1: What are the various public health measures, which you would suggest as safeguard against infectious diseases? ANSWER: Public health measures are preventive measures which are taken to check the spread of various infectious diseases. These measures should be taken to reduce the contact with infectious agents. Some of these methods are: (1) Maintenance of personal and public hygiene:It is one of the most important methods of preventing infectious diseases. This measure includes maintaining a clean body, consumption of healthy and nutritious food, drinking clean water, etc. Public hygienic includes proper disposal of waste material, excreta, periodic cleaning, and disinfection of water reservoirs. (2) Isolation: To prevent the spread of air-borne diseases such as pneumonia, chicken pox, tuberculosis, etc., it is essential to keep the infected person in isolation to reduce the chances of spreading these diseases. (3) Vaccination: Vaccination is the protection of the body from communicable diseases by administering some agent that mimics the microbe inside the body. It helps in providing passive immunizationto the body. Several vaccines are available against many diseases such as tetanus, polio, measles, mumps, etc. (4) Vector Eradication: Various diseases such as malaria, filariasis, dengue, and chikungunya spread through vectors. Thus, these diseases can be prevented by providing a clean environment and by preventing the breeding of mosquitoes. This can be achieved by not allowing water to stagnate around residential areas. Also, measures like regular cleaning of coolers, use of mosquito nets and insecticides such as malathion in drains, ponds, etc. can be undertaken to ensure a healthy environment. Introducing fish such as Gambusia in ponds also controls the breeding of mosquito larvae in stagnant water. Page No 164: Question 2: In which way has the study of biology helped us to control infectious diseases? ANSWER: Various advancements that have occurred in the field of biology have helped us gain a better understanding to fight against various infectious diseases. Biology has helped us study the life cycle of various parasites, pathogens, and vectors along with the modes of transmission of various diseases and the measures for controlling them. Vaccination programmes against several infectious diseases such as small pox, chicken pox, tuberculosis, etc. have helped eradicate these diseases. Biotechnology has helped in the preparation of newer and safer drugs and vaccines. Antibiotics have also played an important role in treating infectious diseases. Page No 164: Question 3: How does the transmission of each of the following diseases take place? (a) Amoebiasis (b) Malaria (c) Ascariasis (d) Pneumonia ANSWER: Disease Causative organism Mode of transmission a. Amoebiasis Entamoeba histolytica It is a vector-borne disease that spreads by the means of contaminated food and water. The vector involved in the transmission of this disease is the housefly. b. Malaria Plasmodium sp. It is a vector-borne disease that spreads by the biting of the female Anopheles mosquito. c. Ascariasis Ascaris lumbricoides It spreads via contaminated food and water. d. Pneumonia Streptococcus pneumoniae It spreads by the sputum of an infected person. Page No 164: Question 4: What measure would you take to prevent water-borne diseases? ANSWER: Water-borne diseases such as cholera, typhoid, hepatitis B, etc. spread ­­ by drinking contaminated water. These water-borne diseases can be prevented by ensuring proper disposal of sewage, excreta, periodic cleaning. Also, measures such as disinfecting community water reservoirs, boiling drinking water, etc. should be observed. Page No 164: Question 5: Discuss with your teacher what does ‘a suitable gene’ means, in the context of DNA vaccines. ANSWER: A ‘suitable gene’ refers to a specific DNA segment which can be injected into the cells of the host body to produce specific proteins. This protein kills the specific disease-causing organism in the host body and provides immunity. Page No 164: Question 6: Name the primary and secondary lymphoid organs. ANSWER: (a) Primary lymphoid organs include the bone marrow and the thymus. (b) Secondary lymphoid organs are the spleen, lymph nodes, tonsils, Peyer’s patches of small intestine, and appendix. Page No 164: Question 7: The following are some well-known abbreviations, which have been used in this chapter. Expand each one to its full form: (a) MALT (b) CMI (c) AIDS (d) NACO (e) HIV ANSWER: (a) MALT- Mucosa-Associated Lymphoid Tissue (b) CMI- Cell-Mediated Immunity (c) AIDS- Acquired Immuno Deficiency Syndrome (d) NACO- National AIDS Control Organization (e) HIV- Human Immuno Deficiency virus Page No 164: Question 8: Differentiate the following and give examples of each: (a) Innate and acquired immunity (b) Active and passive immunity ANSWER: (a) Innate and acquired immunity Innate immunity Acquired immunity 1. It is a non−pathogen specific type of defense mechanism. 1. It is a pathogen specific type of defense mechanism. 2. It is inherited from parents and protects the individual since birth. 2. It is acquired after the birth of an individual. 3. It operates by providing barriers against the entry of foreign infectious agents. 3. It operates by producing primary and secondary responses, which are mediated by B­−lymphocytes and T-lymphocytes. 4 It does not have a specific memory. 4 It is characterized by an immunological memory. (b) Active and passive immunity Active immunity Passive immunity 1. It is a type of acquired immunity in which the body produces its own antibodies against disease-causing antigens. 1. It is a type of acquired immunity in which readymade antibodies are transferred from one individual to another. 2. It has a long lasting effect. 2. It does not have long lasting effect. 3. It is slow. It takes time in producing antibodies and giving responses. 3. It is fast. It provides immediate relief. 4. Injecting microbes through vaccination inside the body is an example of active immunity. 4. Transfer of antibodies present in the mother’s milk to the infant is an example of …

NCERT Solutions for Class 12 Science Biology Chapter 8 – Human Health And Disease Read More »

Here are simplified, step-by-step explanations of NCERT Solutions for Class 12 Science Biology Chapter 7 on Evolution. These solutions have gained popularity among Class 12 Science students for Biology Evolution. They prove useful for swiftly completing homework assignments and preparing for exams. All questions and answers from the NCERT Book of Class 12 Science Biology Chapter 7 are available here at no cost. Page No 142: Question 1: Explain antibiotic resistance observed in bacteria in light of Darwinian selection theory. ANSWER: Darwinian selection theory states that individuals with favourable variations are better adapted than individuals with less favourable variation. It means that nature selects the individuals with useful variation as these individuals are better evolved to survive in the existing environment. An example of such selection is antibiotic resistance in bacteria. When bacterial population was grown on an agar plate containing antibiotic penicillin, the colonies that were sensitive to penicillin died, whereas one or few bacterial colonies that were resistant to penicillin survived. This is because these bacteria had undergone chance mutation, which resulted in the evolution of a gene that made them resistant to penicillin drug. Hence, the resistant bacteria multiplied quickly as compared to non-resistant (sensitive) bacteria, thereby increasing their number. Hence, the advantage of an individual over other helps in the struggle for existence. Page No 142: Question 2: Find out from newspapers and popular science articles any new fossil discoveries or controversies about evolution. ANSWER: Fossils of dinosaurs have revealed the evolution of reptiles in Jurassic period. As a result of this, evolution of other animals such as birds and mammals has also been discovered. However, two unusual fossils recently unearthed in China have ignited a controversy over the evolution of birds.Confuciusornis is one such genus of primitive birds that were crow sized and lived during the Creataceous period in China. Page No 142: Question 3: Attempt giving a clear definition of the term species ANSWER: Species can be defined as a group of organisms, which have the capability to interbreed in order to produce fertile offspring. Page No 142: Question 4: Try to trace the various components of human evolution (hint: brain size and function, skeletal structure, dietary preference, etc.) ANSWER: The various components of human evolution are as follows. (i) Brain capacity (ii) Posture Name Brain capacity Posture Food Features 1. Dryopithecus africans — Knuckle walker,walked similar to gorillas and chimpanzees (was more ape-like) Soft fruit and leaves Canines large, arms and legs are of equal size 2. Ramapithecus — Semi-erect (more man-like) Seeds, nuts Canines were small while molars were large. 3. Australopithecus africanus 450 cm3 Full erect posture, height (1.05 m) Herbivorous (ate fruits) Hunted with stone weapons, lived at trees, canines and incisors were small 4. Homo habilis 735cm3 Fully erect posture, height (1.5 m) Carnivorous Canines were small. They were first tool makers. 5. Homo erectus 800-1100 cm3 Fully erect posture, height(1.5-1.8 m ) Omnivorous They used stone and bone tools for hunting games. 6. Homo neanderthalnsis 1300-1600 cm3 Fully erect posture, height (1.5-1.66 m) Omnivorous Cave dwellers, used hides to protect their bodies, and buried their dead 7. Homo sapiens fossilis 1650 cm3 Fully erect posture with height (1.8 m) Omnivorous They had strong jaw with teeth close together. They were cave dwellers, made paintings and carvings in the caves. They developed a culture and were called first modern men. 8. Homo sapiens sapiens 1200-1600 cm3 Fully erect posture, height (1.5-1.8 m ) Omnivorous They are the living modern men, with high intelligence. They developed art, culture, language, speech, etc. They cultivated crops and domesticated animals. Page No 142: Question 5: Find out through internet and popular science articles whether animals other than man have self-consciousness. ANSWER: 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 only dolphins, there are certain other animals such as crow, parrot, chimpanzee, gorilla, orangutan, etc., which exhibit self-consciousness. Page No 142: Question 6: List 10 modern-day animals and using the internet resources link it to a corresponding ancient fossil. Name both. ANSWER: The modern day animals and their ancient fossils are listed in the following table. Animal Fossil 1. Man Ramapithecus 2. Horse Eohippus 3. Dog Leptocyon 4. Camel Protylopus 5. Elephant Moerithers 6. Whale Protocetus 7. Fish Arandaspis 8. Tetrapods Icthyostega 9. Bat Archaeonycteris 10. Giraffe Palaeotragus Page No 142: Question 7: Practise drawing various animals and plants. ANSWER: Ask your teachers and parents to suggest the names of plants and animals and practice drawing them. You can also take help from your book to find the names of plants and animals. Page No 142: Question 8: Describe one example of adaptive radiation. ANSWER: Adaptive radiation is an evolutionary process that produces new species from a single, rapidly diversifying lineage. This process occurs due to natural selection. An example of adaptive radiation is Darwin finches, found in Galapagos Island. A large variety of finches is present in Galapagos Island that arose from a single species, which reached this land accidentally. As a result, many new species have evolved, diverged, and adapted to occupy new habitats. These finches have developed different eating habits and different types of beaks to suit their feeding habits. The insectivorous, blood sucking, and other species of finches with varied dietary habits have evolved from a single seed eating finch ancestor. Page No 142: Question 9: Can we call human evolution as adaptive radiation? ANSWER: No, human evolution cannot be called adaptive radiation. This is because adaptive radiation is an evolutionary process that produces new species from a single, rapidly diversifying lineage, which is not the case with human evolution. Human evolution is a gradual process that took place slowly in time. It represents an example of anagenesis. Page No 142: Question 10: Using various …

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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 …

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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 …

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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 …

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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 …

NCERT Solutions for Class 12 Science Biology Chapter 1 – Reproduction In Organisms Read More »

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 »