Find simplified step-by-step explanations in the NCERT Solutions for Class 12 Science Biology Chapter 12: “Biotechnology And Its Applications.” These solutions are highly favored by Class 12 Science students for Biology, making them invaluable for completing homework assignments swiftly and preparing for exams. All questions and answers from Chapter 12 of the NCERT Book for Class 12 Science Biology are available here at no cost, aiding students in their studies. Page No 215: Question 1: Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because − (a) bacteria are resistant to the toxin (b) toxin is immature: (c) toxin is inactive: (d) bacteria encloses toxin in a special sac. ANSWER: toxin is inactive: In bacteria, the toxin is present in an inactive form, called prototoxin, which gets converted into active form when it enters the body of an insect. Page No 215: Question 2: What are transgenic bacteria? Illustrate using any one example. ANSWER: Transgenic bacteria contain foreign gene that is intentionally introduced into its genome. They are manipulated to express the desirable gene for the production of various commercially important products. An example of transgenic bacteria is E.coli. In the plasmid of E.coli, the two DNA sequences corresponding to A and B chain of human insulin are inserted, so as to produce the respective human insulin chains. Hence, after the insertion of insulin gene into the bacterium, it becomes transgenic and starts producing chains of human insulin. Later on, these chains are extracted from E.coli and combined to form human insulin. Page No 215: Question 3: Compare and contrast the advantages and disadvantages of production of genetically modified crops. ANSWER: The production of genetically modified (GM) or transgenic plants has several advantages. (i) Most of the GM crops have been developed for pest resistance, which increases the crop productivity and therefore, reduces the reliance on chemical pesticides. (ii) Many varieties of GM food crops have been developed, which have enhanced nutritional quality. For example, golden rice is a transgenic variety in rice, which is rich in vitamin A. (iii) These plants prevent the loss of fertility of soil by increasing the efficiency of mineral usage. (iv) They are highly tolerant to unfavourable abiotic conditions. (v) The use of GM crops decreases the post harvesting loss of crops. However, there are certain controversies regarding the use of genetically modified crops around the world. The use of these crops can affect the native biodiversity in an area. For example, the use of Bt toxin to decrease the amount of pesticide is posing a threat for beneficial insect pollinators such as honey bee. If the gene expressed for Bt toxin gets expressed in the pollen, then the honey bee might be affected. As a result, the process of pollination by honey bees would be affected. Also, genetically modified crops are affecting human health. They supply allergens and certain antibiotic resistance markers in the body. Also, they can cause genetic pollution in the wild relatives of the crop plants. Hence, it is affecting our natural environment. Page No 216: Question 4: What are Cry proteins? Name an organism that produces it. How has man exploited this protein to his benefit? ANSWER: Cry proteins are encoded by cry genes. These proteins are toxins, which are produced by Bacillus thuringiensis bacteria. This bacterium contains these proteins in their inactive from. When the inactive toxin protein is ingested by the insect, it gets activated by the alkaline pH of the gut. This results in the lysis of epithelial cell and eventually the death of the insect. Therefore, man has exploited this protein to develop certain transgenic crops with insect resistance such as Bt cotton, Bt corn, etc. Page No 216: Question 5: What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency. ANSWER: Gene therapy is a technique for correcting a defective gene through gene manipulation. It involves the delivery of a normal gene into the individual to replace the defective gene, for example, the introduction of gene for adenosine deaminase (ADA) in ADA deficient individual. The adenosine deaminase enzyme is important for the normal functioning of the immune system. The individual suffering from this disorder can be cured by transplantation of bone marrow cells. The first step involves the extraction of lymphocyte from the patient’s bone marrow. Then, a functional gene for ADA is introduced into lymphocytes with the help of retrovirus. These treated lymphocytes containing ADA gene are then introduced into the patient’s bone marrow. Thus, the gene gets activated producing functional T- lymphocytes and activating the patient’s immune system. Page No 216: Question 6: Diagrammatically represent the experimental steps in cloning and expressing an human gene (say the gene for growth hormone) into a bacterium like E. coli ? ANSWER: DNA cloning is a method of producing multiple identical copies of specific template DNA. It involves the use of a vector to carry the specific foreign DNA fragment into the host cell. The mechanism of cloning and transfer of gene for growth hormone into E.coli is represented below. Page No 216: Question 7: Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of rDNA technology and chemistry of oil? ANSWER: Recombinant DNA technology (rDNA) is a technique used for manipulating the genetic material of an organism to obtain the desired result. For example, this technology is used for removing oil from seeds. The constituents of oil are glycerol and fatty acids. Using rDNA, one can obtain oilless seeds by preventing the synthesis of either glycerol or fatty acids. This is done by removing the specific gene responsible for the synthesis. Page No 216: Question 8: Find out from internet what is golden rice. ANSWER: Golden rice is a genetically modified variety of rice, Oryza sativa,which has been developed as a fortified food for areas where there is a shortage of dietary vitamin A. It contains a precursor of pro-vitamin A, called beta-carotene, which has been introduced into the rice through genetic engineering. The rice plant naturally produces beta-carotene pigment in its leaves. However, it is absent in the …

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Here are simplified, step-by-step NCERT Solutions for Class 12 Science Biology Chapter 11: Biotechnology – Principles And Processes. These solutions are highly favored by Biology students in Class 12 for their ease of understanding and clarity. They serve as valuable aids for swiftly completing homework assignments and effectively preparing for exams. Crafted by proficient experts, these NCERT Solutions ensure a 100% accuracy rate, providing students with reliable guidance throughout their studies. Page No 205: Question 1: Can you list 10 recombinant proteins which are used in medical practice? Find out where they are used as therapeutics (use the internet). Answer: Recombinant proteins are obtained from the recombinant DNA technology. This technology involves the transfer of specific genes from an organism into another organism using vectors and restriction enzymes as molecular tools. Ten recombinant proteins used in medical practice are − Recombinant protein Therapeutic use 1. Insulin Treatment for type I diabetes mellitus 2. Interferon-α Used for chronic hepatitis C 3. Interferon -β Used against herpes and viral enteritis 4. Coagulation factor VIII Treatment of haemophilia A 5. Coagulation factor IX Treatment of haemophilia B 6. DNAase I Treatment of cystic fibrosis 7. Anti-thrombin III Prevention of blood clot 8. Interferon B For treatment of multiple sclerosis 9. Human recombinant growth hormone For promoting growth in an individual 10. Tissue plasminogen activator Treatment of acute myocardial infarction Page No 205: Question 2: Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate DNA on which it acts, the site at which it cuts DNA and the product it produces. Answer: The name of the restriction enzyme is Bam H 1. Page No 205: Question 3: From what you have learnt, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know? Answer: Enzymes are smaller in size than DNA molecules. This is because DNA contains genetic information for the development and functioning of all living organisms. It contains instructions for the synthesis of proteins and DNA molecules. On the other hand, enzymes are proteins which are synthesised from a small stretch of DNA known as ‘genes’, which are involved in the production of the polypeptide chain. Page No 205: Question 4: What would be the molar concentration of human DNA in a human cell? Consult your teacher. Answer: The molar concentration of human DNA in a human diploid cell is as follows: ⇒ Total number of chromosomes × 6.023 × 1023 ⇒ 46 × 6.023 × 10­­­23 ⇒ 2.77 ×1018 Moles Hence, the molar concentration of DNA in each diploid cell in humans is 2.77 × 10 23 moles. Page No 205: Question 5: Do eukaryotic cells have restriction endonucleases? Justify your answer. Answer: No, eukaryotic cells do not have restriction endonucleases. This is because the DNA of eukaryotes is highly methylated by a modification enzyme, called methylase. Methylation protects the DNA from the activity of restriction enzymes .These enzymes are present in prokaryotic cells where they help prevent the invasion of DNA by virus. Page No 205: Question 6: Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks? Answer: The shake flask method is used for a small-scale production of biotechnological products in a laboratory. On the other hand, stirred tank bioreactors are used for a large-scale production of biotechnology products. Stirred tank bioreactors have several advantages over shake flasks: (1) Small volumes of culture can be taken out from the reactor for sampling or testing. (2) It has a foam breaker for regulating the foam. (3) It has a control system that regulates the temperature and pH. Page No 205: Question 7: Collect 5 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules. Answer: The palindromic sequence is a certain sequence of the DNA that reads the same whether read from 5’ → 3’ direction or from 3’→ 5’ direction. They are the site for the action of restriction enzymes. Most restriction enzymes are palindromic sequences. Five examples of palindromic sequences are: Page No 205: Question 8: Can you recall meiosis and indicate at what stage a recombinant DNA is made? Answer: Meiosis is a process that involves the reduction in the amount of genetic material. It is two types, namely meiosis I and meiosis II. During the pachytene stage of prophase I, crossing over of chromosomes takes place where the exchange of segments between non-sister chromatids of homlogous chromosomes takes place. This results in the formation of recombinant DNA. Page No 205: Question 9: Can you think and answer how a reporter enzyme can be used to monitor transformation of host cells by foreign DNA in addition to a selectable marker? Answer: A reporter gene can be used to monitor the transformation of host cells by foreign DNA. They act as a selectable marker to determine whether the host cell has taken up the foreign DNA or the foreign gene gets expressed in the cell. The researchers place the reporter gene and the foreign gene in the same DNA construct. Then, this combined DNA construct is inserted in the cell. Here, the reporter gene is used as a selectable marker to find out the successful uptake of genes of interest (foreign genes). An example of reporter genes includes lac Z gene, which encodes a green fluorescent protein in a jelly fish. Page No 206: Question 10: Describe briefly the followings: (a) Origin of replication (b) Bioreactors (c) Downstream processing Answer: (a) Origin of replication −Origin of replication is defined as the DNA sequence in a genome from where replication initiates. The initiation of replication can be either uni-directional or bi-directional. A protein complex recognizes the ‘on’ site, unwinds the two strands, and initiates the copying of the DNA. (b) Bioreactors − Bioreactors are large vessels used for the large-scale production of biotechnology products from raw materials. They provide optimal conditions to obtain the desired product by providing the optimum temperature, pH, vitamin, oxygen, etc. Bioreactors have an oxygen delivery system, a foam control system, …

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Here are simplified, step-by-step explanations of NCERT Solutions for Class 12 Science Biology Chapter 10: Microbes In Human Welfare. These solutions for Microbes In Human Welfare are widely favored by Class 12 Science students studying Biology, as they are helpful for completing homework quickly and preparing for exams. All questions and answers from the NCERT Book of Class 12 Science Biology Chapter 10 are provided here for free. Page No 189: Question 1: Bacteria cannot be seen with the naked eyes, but these can be seen with the help of a microscope. If you have to carry a sample from your home to your biology laboratory to demonstrate the presence of microbes under a microscope, which sample would you carry and why? ANSWER: Curd can be used as a sample for the study of microbes. Curd contains numerous lactic acid bacteria (LAB) or Lactobacillus. These bacteria produce acids that coagulate and digest milk proteins. A small drop of curd contains millions of bacteria, which can be easily observed under a microscope. Page No 189: Question 2: Give examples to prove that microbes release gases during metabolism. ANSWER: The examples of bacteria that release gases during metabolism are: (a) Bacteria and fungi carry out the process of fermentation and during this process, they release carbon dioxide. Fermentation is the process of converting a complex organic substance into a simpler substance with the action of bacteria or yeast. Fermentation of sugar produces alcohol with the release of carbon dioxide and very little energy. (b) The dough used for making idli and dosa gives a puffed appearance. This is because of the action of bacteria which releases carbon dioxide. This CO2 released from the dough gets trapped in the dough, thereby giving it a puffed appearance. Page No 189: Question 3: In which food would you find lactic acid bacteria? Mention some of their useful applications. ANSWER: Lactic acid bacteria can be found in curd. It is this bacterium that promotes the formation of milk into curd. The bacterium multiplies and increases its number, which converts the milk into curd. They also increase the content of vitamin B12 in curd. Lactic acid bacteria are also found in our stomach where it keeps a check on the disease-causing micro-organisms. Page No 189: Question 4: Name some traditional Indian foods made of wheat, rice and Bengal gram (or their products) which involve use of microbes. ANSWER: (a) Wheat: Product: Bread, cake, etc. Product: Idli, dosa Product: Dhokla, Khandvi Page No 189: Question 5: In which way have microbes played a major role in controlling diseases caused by harmful bacteria? ANSWER: Several micro-organisms are used for preparing medicines. Antibiotics are medicines produced by certain micro-organisms to kill other disease-causing micro-organisms. These medicines are commonly obtained from bacteria and fungi. They either kill or stop the growth of disease-causing micro-organisms. Streptomycin, tetracycline, and penicillin are common antibiotics. Penicillium notatum produces chemical penicillin, which checks the growth of staphylococci bacteria in the body. Antibiotics are designed to destroy bacteria by weakening their cell walls. As a result of this weakening, certain immune cells such as the white blood cells enter the bacterial cell and cause cell lysis. Cell lysis is the process of destroying cells such as blood cells and bacteria. Page No 189: Question 6: Name any two species of fungus, which are used in the production of the antibiotics. ANSWER: Antibiotics are medicines that are produced by certain micro-organisms to kill other disease-causing micro-organisms. These medicines are commonly obtained from bacteria and fungi. The species of fungus used in the production of antibiotics are: Antibiotic Fungus source 1. Penicillin Penicillium notatum 2. Cephalosporin Cephalosporium acremonium Page No 189: Question 7: What is sewage? In which way can sewage be harmful to us? ANSWER: Sewage is the municipal waste matter that is carried away in sewers and drains. It includes both liquid and solid wastes, rich in organic matter and microbes. Many of these microbes are pathogenic and can cause several water- borne diseases. Sewage water is a major cause of polluting drinking water. Hence, it is essential that sewage water is properly collected, treated, and disposed. Page No 189: Question 8: What is the key difference between primary and secondary sewage treatment? ANSWER: Primary sewage treatment Secondary sewage treatment 1. It is a mechanical process involving the removal of coarse solid materials. 1. It is a biological process involving the action of microbes. 2. It is inexpensive and relatively less complicated. 2. It is a very expensive and complicated process. Page No 189: Question 9: Do you think microbes can also be used as source of energy? If yes, how? ANSWER: Yes, microbes can be used as a source of energy. Bacteria such as Methane bacterium is used for the generation of gobar gas or biogas. The generation of biogas is an anaerobic process in a biogas plant, which consists of a concrete tank (10−15 feet deep) with sufficient outlets and inlets. The dung is mixed with water to form the slurry and thrown into the tank. The digester of the tank is filled with numerous anaerobic methane-producing bacteria, which produce biogas from the slurry. Biogas can be removed through the pipe which is then used as a source of energy, while the spent slurry is removed from the outlet and is used as a fertilizer. Page No 189: Question 10: Microbes can be used to decrease the use of chemical fertilisers and pesticides. Explain how this can be accomplished. ANSWER: Microbes play an important role in organic farming, which is done without the use of chemical fertilizers and pesticides. Bio-fertilizers are living organisms which help increase the fertility of soil. It involves the selection of beneficial micro-organisms that help in improving plant growth through the supply of plant nutrients. Bio-fertilizers are introduced in seeds, roots, or soil to mobilize the availability of nutrients. Thus, they are extremely beneficial in enriching the soil with organic nutrients. Many species of bacteria and cyanobacteria have the ability to fix free atmospheric nitrogen. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirillium and Azotobocter are free living nitrogen-fixing …

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Here are step-by-step explanations of NCERT Solutions for Class 12 Science Biology Chapter 9, “Strategies For Enhancement In Food Production.” These solutions are highly favored among Biology students of Class 12 for their simplicity and effectiveness in aiding homework completion and exam preparation. You can access all questions and answers from the NCERT Book of Class 12 Science Biology Chapter 9 without any cost. Page No 178: Question 1: Explain in brief the role of animal husbandry in human welfare. ANSWER: Animal husbandry deals with the scientific management of livestock. It includes various aspects such as feeding, breeding, and control diseases to raise the population of animal livestock. Animal husbandry usually includes animals such as cattle, pig, sheep, poultry, and fish which are useful for humans in various ways. These animals are managed for the production of commercially important products such as milk, meat, wool, egg, honey, silk, etc. The increase in human population has increased the demand of these products. Hence, it is necessary to improve the management of livestock scientifically. Page No 178: Question 2: If your family owned a dairy farm, what measures would you undertake to improve the quality and quantity of milk production? ANSWER: Dairy farm management deals with processes which aim at improving the quality and quantity of milk production. Milk production is primarily dependent on choosing improved cattle breeds, provision of proper feed for cattle, maintaining proper shelter facilities, and regular cleaning of cattle. Choosing improved cattle breeds is an important factor of cattle management. Hybrid cattle breeds are produced for improved productivity. Therefore, it is essential that hybrid cattle breeds should have a combination of various desirable genes such as high milk production and high resistance to diseases. Cattle should also be given healthy and nutritious food consisting of roughage, fibre concentrates, and high levels of proteins and other nutrients. Cattle’s should be housed in proper cattle-houses and should be kept in well ventilated roofs to prevent them from harsh weather conditions such as heat, cold, and rain. Regular baths and proper brushing should be ensured to control diseases. Also, time-to-time check ups by a veterinary doctor for symptoms of various diseases should be undertaken. Page No 178: Question 3: What is meant by the term ‘breed’? What are the objectives of animal breeding? ANSWER: A breed is a special variety of animals within a species. It is similar in most characters such as general appearance, size, configuration, and features with other members of the same species. Jersey and Brown Swiss are examples of foreign breeds of cattle. These two varieties of cattle have the ability to produce abundant quantities of milk. This milk is very nutritious with high protein content. Objectives of animal breeding: (i) To increase the yield of animals. (ii) To improve the desirable qualities of the animal produce. (iii) To produce disease-resistant varieties of animals. Page No 178: Question 4: Name the methods employed in animal breeding. According to you which one of the methods is best? Why? ANSWER: Animal breeding is the method of mating closely related individuals. There are several methods employed in animals breeding, which can be classified into the following categories: (A) Natural methods of breeding include inbreeding and out-breeding. Breeding between animals of the same breed is known as inbreeding, while breeding between animals of different breeds is known as out-breeding. Out-breeding of animals is of three types: (a). Out-crossing: In this type of out-breeding, the mating of animals occurs within the same breed. Thus, they have no common ancestors up to the last 4-5 generations. (b). Cross-breeding: In this type of out-breeding, the mating occurs between different breeds of the same species, thereby producing a hybrid. (c). Interspecific hybridization: In this type of out-breeding, the mating occurs between different species. (B) Artificial methods of breeding include modern techniques of breeding. It involves controlled breeding experiments, which are of two types:- (a). Artificial insemination: It is a process of introducing the semen (collected from the male) into the oviduct or the uterus of the female body by the breeder. This method of breeding helps the breeder overcome certain problems faced in abnormal mating. (b). Multiple ovulation embryo technology (MOET): It is a technique for cattle improvement in which super-ovulation is induced by a hormone injection. Then, fertilization is achieved by artificial insemination and early embryos are collected. Each of these embryos are then transplanted into the surrogate mother for further development of the embryo. The best method to carry out animal breeding is the artificial method of breeding, which includes artificial insemination and MOET technology. These technologies are scientific in nature. They help overcome problems of normal mating and have a high success rate of crossing between mature males and females. Also, it ensures the production of hybrids with the desired qualities. This method is highly economical as a small amount of semen from the male can be used to inseminate several cattle. Page No 178: Question 5: What is apiculture? How is it important in our lives? ANSWER: Apiculture is the practice of bee-keeping for the production of various products such as honey, bee’s wax, etc. Honey is a highly nutritious food source and is used as an indigenous system of medicines. It is useful in the treatment of many disorders such as cold, flu, and dysentery. Other commercial products obtained from honey bees include bee’s wax and bee pollen. Bee’s wax is used for making cosmetics, polishes, and is even used in several medicinal preparations. Therefore, to meet the increasing demand of honey, people have started practicing bee-keeping on a large scale. It has become an income generating activity for farmers since it requires a low investment and is labour intensive. Page No 178: Question 6: Discuss the role of fishery in enhancement of food production. ANSWER: Fishery is an industry which deals with catching, processing, and marketing of fishes and other aquatic animals that have a high economic value. Some commercially important aquatic animals are prawns crabs, oysters, lobsters, and octopus. Fisheries play an important role in the Indian …

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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 …

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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 …

NCERT Solutions for Class 12 Science Biology Chapter 7 – Evolution Read More »

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 »