Class 12 Physics Handwritten Notes PDF Chapter 7 – Alternating current – Free PDF Download
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### Download Class 12 Physics Handwritten Notes PDF Chapter 7 Alternating current

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NCERT Solutions for Class 12 Biology free PDF – All Chapters – Free PDF Download

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### CBSE Class 12 Biology Practicals

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NCERT Solutions for Class 12 Chemistry free PDF – All Chapters – Free PDF Download

Hey are you a class 12 Student and looking for ways to Download NCERT Solutions for Class 12 Chemistry free PDF to start your Preparations? If yes then Don’t worry you are at the right place.

If you want to score high in your class 12 Chemistry Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

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## NCERT Solutions for Class 12 Chemistry free PDF Chapterwise Free Download

### All Units of Chemistry class 12

#### Unit I: Solid State

Classification of solids based on different binding forces: molecular, ionic, covalent and metallic solids, amorphous and crystalline solids (elementary idea). Unit cell in two dimensional and three dimensional lattices, calculation of density of unit cell, packing in solids, packing efficiency, voids, number of atoms per unit cell in a cubic unit cell, point defects, electrical and magnetic properties. Band theory of metals, conductors, semiconductors and insulators and n and p type semiconductors.

#### Unit II: Solutions

Types of solutions, expression of concentration of solutions of solids in liquids, solubility of gases in liquids, solid solutions, colligative properties – relative lowering of vapour pressure, Raoult’s law, elevation of boiling point, depression of freezing point, osmotic pressure, determination of molecular masses using colligative properties, abnormal molecular mass, Van’t Hoff factor.

#### Unit III: Electrochemistry

Redox reactions, conductance in electrolytic solutions, specific and molar conductivity, variations of conductivity with concentration, Kohlrausch’s Law, electrolysis and law of electrolysis (elementary idea), dry cell-electrolytic cells and Galvanic cells, lead accumulator, EMF of a cell, standard electrode potential, Nernst equation and its application to chemical cells, Relation between Gibbs energy change and EMF of a cell, fuel cells, corrosion.

#### Unit IV: Chemical Kinetics

Rate of a reaction (Average and instantaneous), factors affecting rate of reaction: concentration, temperature, catalyst; order and molecularity of a reaction, rate law and specific rate constant, integrated rate equations and half-life (only for zero and first order reactions), concept of collision theory (elementary idea, no mathematical treatment). Activation energy, Arrhenious equation.

#### Unit V: Surface Chemistry

Adsorption – physisorption and chemisorption, factors affecting adsorption of gases on solids, catalysis, homogenous and heterogenous activity and selectivity; enzyme catalysis colloidal state distinction between true solutions, colloids and suspension; lyophilic, lyophobic multi-molecular and macromolecular colloids; properties of colloids; Tyndall effect, Brownian movement, electrophoresis, coagulation, emulsion – types of emulsions.

#### Unit VI: General Principles and Processes of Isolation of Elements

Principles and methods of extraction – concentration, oxidation, reduction – electrolytic method and refining; occurrence and principles of extraction of aluminium, copper, zinc and iron

#### Unit VII: Some p -Block Elements

Group -15 Elements:

General introduction, electronic configuration, occurrence, oxidation states, trends in physical and chemical properties; Nitrogen preparation properties and uses; compounds of Nitrogen, preparation and properties of Ammonia and Nitric Acid, Oxides of Nitrogen(Structure only) ; Phosphorus – allotropic forms, compounds of Phosphorus: Preparation and Properties of Phosphine, Halides and Oxoacids (elementary idea only).

Group 16 Elements:

General introduction, electronic configuration, oxidation states, occurrence, trends in physical and chemical properties, dioxygen: Preparation, Properties and uses, classification of Oxides, Ozone, Sulphur -allotropic forms; compounds of Sulphur: Preparation Properties and uses of Sulphur-dioxide, Sulphuric Acid: industrial process of manufacture, properties and uses; Oxoacids of Sulphur (Structures only).

Group 17 Elements:

General introduction, electronic configuration, oxidation states, occurrence, trends in physical and chemical properties; compounds of halogens, Preparation, properties and uses of Chlorine and Hydrochloric acid, interhalogen compounds, Oxoacids of halogens (structures only).

Group 18 Elements:

General introduction, electronic configuration, occurrence, trends in physical and chemical properties, uses.

#### Unit VIII: “d” and “f” Block Elements

General introduction, electronic configuration, occurrence and characteristics of transition metals, general trends in properties of the first row transition metals – metallic character, ionization enthalpy, oxidation states, ionic radii, colour, catalytic property, magnetic properties, interstitial compounds, alloy formation, preparation and properties of K2Cr2O7 and KMnO4.

Lanthanoids – Electronic configuration, oxidation states, chemical reactivity and lanthanoid contraction and its consequences.

Actinoids – Electronic configuration, oxidation states and comparison with lanthanoids.

#### Unit IX: Coordination Compounds

Coordination compounds – Introduction, ligands, coordination number, colour, magnetic properties and shapes, IUPAC nomenclature of mononuclear coordination compounds. Bonding, Werner’s theory, VBT, and CFT; structure and stereoisomerism, importance of coordination compounds (in qualitative inclusion, extraction of metals and biological system).

#### Unit X: Haloalkanes and Haloarenes

Haloalkanes: Nomenclature, nature of C-X bond, physical and chemical properties, mechanism of substitution reactions, optical rotation.

Haloarenes: Nature of C-X bond, substitution reactions (Directive influence of halogen in monosubstituted compounds only).

Uses and environmental effects of – dichloromethane, trichloromethane, tetrachloromethane, iodoform, freons, DDT.

#### Unit XI: Alcohols, Phenols and Ethers

Alcohols: Nomenclature, methods of preparation, physical and chemical properties (of primary alcohols only), identification of primary, secondary and tertiary alcohols, mechanism of dehydration, uses with special reference to methanol and ethanol.

Phenols: Nomenclature, methods of preparation, physical and chemical properties, acidic nature of phenol, electrophillic substitution reactions, uses of phenols.

Ethers: Nomenclature, methods of preparation, physical and chemical properties, uses.

#### Unit XII: Aldehydes, Ketones and Carboxylic Acids

Aldehydes and Ketones: Nomenclature, nature of carbonyl group, methods of preparation, physical and chemical properties, mechanism of nucleophilic addition, reactivity of alpha hydrogen in aldehydes, uses.

Carboxylic Acids: Nomenclature, acidic nature, methods of preparation, physical and chemical properties; uses.

#### Unit XIII: Organic compounds containing Nitrogen

Amines: Nomenclature, classification, structure, methods of preparation, physical and chemical properties, uses, identification of primary, secondary and tertiary amines.

Cyanides and Isocyanides – will be mentioned at relevant places in text.

Diazonium salts: Preparation, chemical reactions and importance in synthetic organic chemistry.

#### Unit XIV: Biomolecules

Carbohydrates – Classification (aldoses and ketoses), monosaccahrides (glucose and fructose), D-L configuration oligosaccharides (sucrose, lactose, maltose), polysaccharides (starch, cellulose, glycogen); Importance of carbohydrates.

Proteins -Elementary idea of – amino acids, peptide bond, polypeptides, proteins, structure of proteins – primary, secondary, tertiary structure and quaternary structures (qualitative idea only), denaturation of proteins; enzymes. Hormones – Elementary idea excluding structure.

Vitamins – Classification and functions.

Nucleic Acids: DNA and RNA.

#### Unit XV: Polymers

Classification – natural and synthetic, methods of polymerization (addition and condensation), copolymerization, some important polymers: natural and synthetic like polythene, nylon polyesters, bakelite, rubber. Biodegradable and non-biodegradable polymers.

#### Unit XVI: Chemistry in Everyday life

Chemicals in medicines – analgesics, tranquilizers antiseptics, disinfectants, antimicrobials, antifertility drugs, antibiotics, antacids, antihistamines.

Chemicals in food – preservatives, artificial sweetening agents, elementary idea of antioxidants. Cleansing agents- soaps and detergents, cleansing action.

If you have any Confusion related to NCERT Solutions for Class 12 Chemistry free PDF then feel free to ask in the comments section down below. To watch Free Learning Videos on Class 12 by Kota’s top Faculties Install the eSaral App

NCERT Solutions for Class 12 Physics free PDF – All Chapters – Free PDF Download
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If you want to score high in your class 12 Physics Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

In this post We have listed NCERT Solutions for Class 12 Physics all Chapters in PDF that are prepared by Kota’s top IITian’s Faculties in a well Structured Format by keeping Simplicity in mind.

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## NCERT Solutions for Class 12 Physics free PDF Chapterwise Free Download

 CHAPTER TOPICS Chapter–1 Electric Charges and FieldsElectric ChargesConservation of charge, Coulomb’s law-force between two point chargesForces between multiple charges superposition principle and continuous charge distribution.Electric fieldElectric field due to a point charge, electric field lines, electric dipole, electric field due to a dipole, torque on a dipole in uniform electric fleld.Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitely long straight wire uniformly charged infinite plane sheet and uniformly charged thin spherical shell (field inside and outside). Chapter–2 Electrostatic Potential and CapacitanceElectric potential, potential difference, electric potential due to a point charge,A dipole and system of chargesEquipotential surfaces, electrical potential energy of a system of two point charges and of electric dipole in an electrostatic field.Conductors and insulators, free charges and bound charges inside a conductor.Dielectrics and electric polarisation, capacitors and capacitanceCombination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the platesEnergy stored in a capacitor. Chapter–3 Current ElectricityElectric current, flow of electric charges in a metallic conductor, drift velocity mobility and their relation with electric current; Ohm’s law electrical resistanceV-I characteristics (linear and non-linear), electrical energy and power, electrical resistivity and conductivityCarbon resistors, colour code for carbon resistorsSeries and parallel combinations of resistorsTemperature dependence of resistance.Internal resistance of a cell, potential difference and emf of a cellCombination of cells in series and in parallelKirchhoff’s laws and simple applicationsWheatstone bridge, metre bridgePotentiometer – principle and its applications to measure potential difference and for comparing EMF of two cells; measurement of internal resistance of a cell Chapter–4 Moving Charges and MagnetismConcept of magnetic field, Oersted’s experimentBiot – Savart law and its application to current carrying circular loopAmpere’s law and its applications to infinitely long straight wireStraight and toroidal solenoids (only qualitative treatment)Force on a moving charge in uniform magnetic and electric fields, CyclotronForce on a current-carrying conductor in a uniform magnetic fieldForce between two parallel current-carrying conductors-definition of ampereTorque experienced by a current loop in uniform magnetic fieldMoving coil galvanometer-its current sensitivity and conversion to ammeter and voltmeter Chapter–5 Magnetism and MatterCurrent loop as a magnetic dipole and its magnetic dipole momentMagnetic dipole moment of a revolving electronMagnetic field intensity due to a magnetic dipole (bar magnet) along its axis and perpendicular to its axisTorque on a magnetic dipole (bar magnet) in a uniform magnetic fieldBar magnet as an equivalent solenoid, magnetic field lines; earth’s magnetic field and magnetic elementsPara-, dia- and ferro – magnetic substances, with examplesElectromagnets and factors affecting their strengths, permanent magnets Chapter–6 Electromagnetic InductionElectromagnetic inductionFaraday’s lawsinduced EMF and currentLenz’s LawEddy currentsSelf and mutual induction Chapter–7 Alternating CurrentAlternating currentsPeak and RMS value of alternating current/voltageReactance and impedanceLC oscillations (qualitative treatment only)LCR series circuit, resonance; power in AC circuits, power factor, wattless currentAC generator and transformer Chapter–8 Electromagnetic WavesBasic idea of displacement currentElectromagnetic waves, their characteristics, their Transverse nature (qualitative ideas only)Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays) including elementary facts about their uses Chapter–9 Ray Optics and Optical InstrumentsRay Optics: Reflection of light, spherical mirrors, mirror formula, refraction of light, total internal reflection and its applications, optical fibers, refraction at spherical surfaces, lenses, thin lens formula, lensmaker’s formula, magnification, power of a lens, combination of thin lenses in contact, refraction of light through a prismScattering of light – blue colour of sky and reddish appearance of the sun at sunrise and sunsetOptical instruments: Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers Chapter-10 Wave OpticsWave front and Huygen’s principle, reflection and refraction of plane wave at a plane surface using wave frontsProof of laws of reflection and refraction using Huygen’s principleInterference, Young’s double slit experiment and expression for fringe widthCoherent sources and sustained interference of lightDiffraction due to a single slit, width of central maximumResolving power of microscope and astronomical telescopePolarisation,Plane polarised lightBrewster’s law, uses of plane polarised light and Polaroids Chapter–11 Dual Nature of Radiation and MatterDual nature of radiationPhotoelectric effectHertz and Lenard’s observationsEinstein’s photoelectric equation-particle nature of lightMatter waves-wave nature of particles, de-Broglie relationDavisson-Germer experiment (experimental details should be omitted; only conclusion should be explained) Chapter–12 AtomsAlpha-particle scattering experimentRutherford’s model of atomBohr model,Energy levels, hydrogen spectrum Chapter–13 NucleiComposition and size of nucleusRadioactivity, alpha, beta and gamma particles/rays and their properties; radioactive decay lawMass-energy relation, mass defect; binding energy per nucleon and its variation with mass number; nuclear fission, nuclear fusion Chapter–14 Semiconductor Electronics: Materials, Devices and Simple CircuitsEnergy bands in conductors, semiconductors and insulators (qualitative ideas only)Semiconductor diode – I-V characteristics in forward and reverse bias, diode as a rectifierSpecial purpose p-n junction diodes: LED, photodiode, solar cell and Zener diode and their characteristics, zener diode as a voltage regulator

### Benefits of NCERT Solutions For Class 12 Physics

If you want to score high in your class 12 Physics Exam then you can use NCERT Class 12 Books and Solutions as your Main study Material. To download Solutions PDF of any Chapter Simply click on the Link of that Particular chapter.

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NCERT Solutions for Class 12 Biology Chapter 16 Environmental Issues PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 16 Environmental Issues PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 16 Environmental Issues PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 16 Environmental Issues in PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 16 “Environmental Issues” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

So, without wasting more time Let’s start.

Question 1: What are the various constituents of domestic sewage? Discuss the effects of sewage discharge on a river.

Solution. Domestic sewage is the waste originating from the kitchen, toilet, laundry, and other sources. It contains impurities such as suspended solid (sand, salt, clay), colloidal material (fecal matter, bacteria, plastic and cloth fiber), dissolved materials (nitrate, phosphate, calcium, sodium, ammonia), and disease-causing microbes. When organic wastes from the sewage enter the water bodies, it serves as a food source for micro-organisms such as algae and bacteria. As a result, the population of these micro-organisms in the water body increases. Here, they utilize most of the dissolved oxygen for their metabolism. This results in an increase in the levels of Biological oxygen demand (BOD) in river water and results in the death of aquatic organisms. Also, the nutrients in the water lead to the growth of planktonic algal, causing algal bloom. This causes deterioration of water quality and fish mortality.

Question 2: List all the wastes that you generate, at home, school or during your Trips to other places, could you very easily reduce? Which would be Difficult or rather impossible to reduce?

Solution. Wastes generated at home include plastic bags, paper napkin, toiletries, kitchen wastes (such as peelings of vegetables and fruits, tea leaves), domestic sewage, glass, etc.

Wastes generated at schools include waste paper, plastics, vegetable and fruit peels, food wrappings, sewage etc.

Wastes generated at trips or picnics include plastic, paper, vegetable and fruit peels, disposable cups, plates, spoons etc.

Yes, wastes can be easily reduced by the judicious use of the above materials. Wastage of paper can be minimized by writing on both sides of the paper and by using recycled paper. Plastic and glass waste can also be reduced by recycling and re-using. Also, substituting plastics bags with biodegradable jute bags can reduce wastes generated at home, school, or during trips. Domestic sewage can be reduced by optimizing the use of water while bathing, cooking, and other household activities.

Non- biodegradable wastes such as plastic, metal, broken glass, etc are difficult to decompose because micro-organisms do not have the ability to decompose them.

Question 3: Discuss the causes and effects of global warming. What measures need to be taken to control global warming?

Solution. Global warming is defined as an increase in the average temperature of the Earth’s surface.

Causes of global warming: Global warming occurs as a result of the increased concentration of greenhouse gases in the atmosphere. Greenhouse gases include carbon dioxide, methane, and water vapour. These gases trap solar radiations released back by the Earth. This helps in keeping our planet warm and thus, helps in human survival. However, an increase in the amount of greenhouse gases can lead to an excessive increase in the Earth’s temperature, leading to global warming. Global warming is a result of industrialization, burning of fossil fuels, and deforestation.

Effects of global warming: Global warming is defined as an increase in the average temperature of the Earth’s surface. It has been observed that in the past three decades, the average temperature of the Earth has increased by $0.6^{\circ} \mathrm{C}$. As a result, the natural water cycle has been disturbed resulting in changes in the pattern of rainfall. It also changes the amount of rain water. Also, it results in the melting of Polar ice caps and mountain glaciers, which has caused a rise in the sea level, leading to the inundation of coastal regions

Control measures for preventing global warming:

(i) Reducing the use of fossil fuels

(ii) Use of bio-fuels

(iii) Improving energy efficiency

(iv) Use of renewable source of energy such as CNG etc.

(v) Reforestation. (vii) Recycling of materials

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

Solution.

Question 5: Write critical notes on the following:

(a) Eutrophication

(b) Biological magnification

(c) Groundwater depletion and ways for its replenishment

Solution. (a) Eutrophication:- It is the natural ageing process of a lake caused due to nutrient enrichment. It is brought down by the runoff of nutrients such as animal wastes, fertilizers, and sewage from land which leads to an increased fertility of the lake. As a result, it causes a tremendous increase in the primary productivity of the ecosystem. This leads to an increased growth of algae, resulting into algal blooms. Later, the decomposition of these algae depletes the supply of oxygen, leading to the death of other aquatic animal life.

(b) Biological magnification: – To protect the crops from the several diseases and pests, a large number of pesticides are used. These pesticides reach the soil and are absorbed by plants with water and minerals from the soil. Due to rain, these chemicals can also enter water sources and into the body of aquatic plants and animals. As a result, chemicals enter the food chain. Since these chemicals cannot be decomposed, they keep on accumulating at each trophic level. The maximum concentration is accumulated at the top carnivore’s level. This increase in the concentration of pollutants or harmful chemicals with an increase in the trophic level is called biological magnification. For example, high DDT concentrations were found in a pond. The producers (phytoplankton) were found to have $0.04$ ppm concentration of DDT. Since many types of phytoplankton were eaten by zooplankton (consumers), the concentration of DDT in the bodies of zooplankton was found to be $0.23$ ppm. Small fish that feed on zooplankton accumulate more DDT in their body. Thus, large fish (top carnivore) that feed on several small fish have the highest concentration of DDT.

(c) Ground water depletion and ways for its replenishment: – The level of ground water has decreased in the recent years. The source of water supply is rapidly diminishing each year because of an increase in the population and water pollution. To meet the demand of water, water is withdrawn from water bodies such as ponds, rivers etc. As a result, the source of ground water is depleting. This is because the amount of groundwater being drawn for human use is more than the amount replaced by rainfall. Lack of vegetation cover also results in very small amounts of water seeping through the ground. An increase in water pollution is another factor that has reduced the availability of ground water.

Measures for replenishing ground water:-

(i) Preventing over-exploitation of ground water

(ii) Optimizing water use and reducing water demand

(iii) Rain water harvesting

(iv) Preventing deforestation and plantation of more trees

Question 6: Why ozone hole forms over Antarctica? How will enhanced ultraviolet Radiations affect us?

Solution. The ozone hole is more prominent over the region of Antarctica. It is formed due to an increased concentration of chlorine in the atmosphere.

Chlorine is mainly released from chlorofluorocarbons (CFC’s) widely used as refrigerants. The CFC’s magnate from the troposphere to the stratosphere, where they release chlorine atoms by the action of UV rays on them. The release of Chlorine atoms causes the conversion of ozone into molecular oxygen. One atom of chlorine can destroy around 10,000 molecules of ozone and causes ozone depletion.

The formation of the ozone hole will result in an increased concentration of UV – B radiations on the Earth’s surface. UV -B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV $-B$ cause corneal cataract in human beings.

Question 7: Discuss the role of women and communities in protection and conservation of forests.

Solution. Women and communities have played a major role in environmental conservation movements.

(i) Case study of the bishnoi community: TheBishnoi community in Rajasthan strictly believes in the concept of living peacefully with nature. In 1731 , the king of Jodhpur ordered his ministers to arrange wood for the construction of his new palace. For this purpose, the minister and the workers went to bishnoi village. There, a Bishnoi woman called Amrita Devi along with her daughter and hundreds of other Bishnois showed the courage to step forward and stop them from cutting trees. They embraced the trees and lost their lives at the hands of soldiers of the king. This resistance by the people of the village forced the king to give up the idea of cutting trees.

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

Question 8: What measures, as an individual, you would take to reduce environmental pollution?

Solution. The following initiatives can be taken to prevent environmental pollution:

Measures for preventing Air pollution:

(i) Planting more trees

(ii) Use of clean and renewable energy sources such as CNG and bio-fuels

(iii) Reducing the use of fossil fuels

(iv) Use of catalytic converters in automobiles

Measures for preventing water pollution:-

(i) Optimizing the use of water

(ii) Using kitchen waste water in gardening and other household purposes

Measures for controlling Noise pollution:-

(i) Avoid burning crackers on Diwali

(i) Plantation of more trees

Measures for decreasing solid waste generation:-

(i) Segregation of waste

(ii) Recycling and reuse of plastic and paper

(iii) Composting of biodegradable kitchen waste

(iv) Reducing the use of plastics

Question 9: Discuss briefly the following:

(b) Defunct ships and e-wastes

(c) Municipal solid wastes

Solution. (a) Radioactive wastes: – Radioactive wastes are generated during the process of generating nuclear energy from radioactive materials. Nuclear waste is rich in radioactive materials that generate large quantities of ionizing radiations such as gamma rays. These rays cause mutation in organisms, which often results in skin cancer. At high dosage, these rays can be lethal.

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

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

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

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

Question 10: What initiatives were taken for reducing vehicular air pollution in Delhi?

Has air quality improved in Delhi?

Solution. Delhi has been categorized as the fourth most polluted city of the world in a list of 41 cities. Burning of fossil fuels has added to the pollution of air in Delhi.

Various steps have been taken to improve the quality of air in Delhi.

(a) Introduction of CNG (Compressed Natural Gas):By the orderof the supreme court of India, CNG-powered vehicles were introduced at the end of year 2006 to reduce the levels of pollution in Delhi. CNG is a clean fuel that produces very little unburnt particles.

(b) Phasing out of old vehicles

(d) Use of low-sulphur petrol and diesel

(e) Use of catalytic converters

(f) Application of stringent pollution-level norms for vehicles

(g) Implementation of Bharat stage I, which is equivalent to euro II norms in vehicles of major Indian cities.

The introduction of CNG-powered vehicles has improved Delhi’s air quality, which has lead to a substantial fall in the level of $\mathrm{CO}_{2}$ and $\mathrm{SO}_{2}$. However, the problem of suspended particulate matter (SPM) and respiratory suspended particulate matter (RSPM) still persists.

Question 11: Discuss briefly the following:

(a) Greenhouse gases

(b) Catalytic converter

(c) Ultraviolet B

Solution. (a) Greenhouse gases: – Thegreenhouse effect refers to an overall increase in the average temperature of the Earth due to the presence of greenhouse gases. Greenhouse gases mainly consist of carbon dioxide, methane, and water vapour. When solar radiations reach the Earth, some of these radiations are absorbed. These absorbed radiations are released back into the atmosphere. These radiations are trapped by the greenhouse gases present in the atmosphere.. This helps in keeping our planet warm and thus, helps in human survival. However, an increase in the amount of greenhouse gases can lead to an excessive increase in the Earth’s temperature, thereby causing global warming.

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

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

Class 12 Chemistry Notes PDF Free.

Class 12 Biology Book Chapterwise PDF Free.

Class 12 Biology Exemplar Chapterwise PDF Free.

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NCERT Solutions for Class 12 Biology Chapter 15 Biodiversity and Conservation PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 15 Biodiversity and Conservation PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 15 Biodiversity and Conservation PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 15 Biodiversity and Conservation in PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 15 “Biodiversity and Conservation” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

So, without wasting more time Let’s start.

Question 1: Name the three important components of biodiversity.

Solution. Biodiversity is the variety of living forms present in various ecosystems. It includes variability among life forms from all sources including land, air, and water. Three important components of biodiversity are:

(a) Genetic diversity

(b) Species diversity

(c) Ecosystem diversity

Question 2: How do ecologists estimate the total number of species present in the world?

Solution. The diversity of living organisms present on the Earth is very vast. According to an estimate by researchers, it is about seven millions.

The total number of species present in the world is calculated by ecologists by statistical comparison between a species richness of a well studied group of insects of temperate and tropical regions. Then, these ratios are extrapolated with other groups of plants and animals to calculate the total species richness present on the Earth.

Question 3: Give three hypotheses for explaining why tropics show greatest levels of species richness.

Solution. There are three different hypotheses proposed by scientists for explaining species richness in the tropics.

(1) Tropical latitudes receive more solar energy than temperate regions, which leads to high productivity and high species diversity.

(2) Tropical regions have less seasonal variations and have a more or less constant environment. This promotes the niche specialization and thus, high species richness.

(3) Temperate regions were subjected to glaciations during the ice age, while tropical regions remained undisturbed which led to an increase in the species diversity in this region.

Question 4: What is the significance of the slope of regression in a species − area relationship?

Solution. The slope of regression (z) has a great significance in order to find a species-area relationship. It has been found that in smaller areas (where the species-area relationship is analyzed), the value of slopes of regression is similar regardless of the taxonomic group or the region. However, when a similar analysis is done in larger areas, then the slope of regression is much steeper.

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

Solution. Biodiversity is the variety of living forms present in various ecosystems. It includes variability among life forms from all sources including land, air, and water. Biodiversity around the world is declining at a very fast pace. The following are the major causes for the loss of biodiversity around the world.

(i) Habitat loss and fragmentation: Habitats of various organisms are altered or destroyed by uncontrolled and unsustainable human activities such as deforestation, slash and burn agriculture, mining, and urbanization. This results in the breaking up of the habitat into small pieces, which effects the movement of migratory animals and also, decreases the genetic exchange between populations leading to a declination of species.

(ii) Over-exploitation: Due to over-hunting and over-exploitation of various plants and animals by humans, many species have become endangered or extinct (such as the tiger and the passenger pigeon).

(iii) Alien species Invasions: Accidental or intentional introduction of non-native species into a habitat has also led to the declination or extinction of indigenous species. For example, the Nile perch introduced in Lake Victoria in Kenya led to the extinction of more than two hundred species of native fish in the lake.

(iv) Co¬−extinction: In a native habitat, one species is connected to the other in an intricate network. The extinction of one species causes the extinction of other species, which is associated with it in an obligatory way. For example, the extinction of the host will cause the extinction of its parasites.

Question 6: How is biodiversity important for ecosystem functioning?

Solution. An ecosystem with high species diversity is much more stable than an ecosystem with low species diversity. Also, high biodiversity makes the ecosystem more stable in productivity and more resistant towards disturbances such as alien species invasions and floods.

If an ecosystem is rich in biodiversity, then the ecological balance would not get affected. As we all know, various trophic levels are connected through food chains. If any one organism or all organisms of any one trophic level is killed, then it will disrupt the entire food chain. For example, in a food chain, if all plants are killed, then all deer’s will die due to the lack of food. If all deer’s are dead, soon the tigers will also die. Therefore, it can be concluded that if an ecosystem is rich in species, then there will be other food alternatives at each trophic level which would not allow any organism to die due to the absence of their food resource.

Hence, biodiversity plays an important role in maintaining the health and ecological balance of an ecosystem.

Question 7: What are sacred groves? What is their role in conservation? Solution. Sacred groves are tracts of forest which are regenerated around places of worship. Sacred groves are found in Rajasthan, Western Ghats of Karnataka,and Maharashtra, Meghalaya, and Madhya Pradesh. Sacred groves help in the protection of many rare, threatened, and endemic species of plants and animals found in an area. The process of deforestation is strictly prohibited in this region by tribals. Hence, the sacred grove biodiversity is a rich area.

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

Solution. The biotic components of an ecosystem include the living organisms such as plants and animals. Plants play a very important role in controlling floods and soil erosion. The roots of plants hold the soil particles together, thereby preventing the top layer of the soil to get eroded by wind or running water. The roots also make the soil porous, thereby allowing ground water infiltration and preventing floods. Hence, plants are able to prevent soil erosion and natural calamities such as floods and droughts. They also increase the fertility of soil and biodiversity.

Question 9: The species diversity of plants (22 per cent) is much less than that of animals ( 72 per cent). What could be the explanations to how animals achieved greater diversification?

Solution. More than 70 percent of species recorded on the Earth are animals and only 22 percent species are plants. There is quiet a large difference in their percentage. This is because animals have adapted themselves to ensure their survival in changing environments in comparison to plants. For example, insects and other animals have developed a complex nervous system to control and coordinate their body structure. Also, repeated body segments with paired appendages and external cuticles have made insects versatile and have given them the ability to survive in various habitats as compared to other life forms.

Question 10: Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?

Solution. Yes, there are various kinds of parasites and disease-causing microbes that we deliberately want to eradicate from the Earth. Since these micro-organisms are harmful to human beings, scientists are working hard to fight against them. Scientists have been able to eliminate small pox virus from the world through the use of vaccinations. This shows that humans deliberately want to make these species extinct. Several other eradication programmes such as polio and Hepatitis B vaccinations are aimed to eliminate these disease-causing microbes.

Class 12 Chemistry Notes PDF.

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NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem in PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 14 “Ecosystem” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

So, without wasting more time Let’s start.

Question 1: Fill in the blanks.

(a) Plants are called as___________ because they fix carbon dioxide.

(b) In an ecosystem dominated by trees, the pyramid (of numbers) is ___________ type.

(c) In aquatic ecosystems, the limiting factor for the productivity is ______________. (d) Common detritivores in our ecosystem are_________.

(e) The major reservoir of carbon on earth is_________.

Solution. (a) Plants are called as autotrophs because they fix carbon dioxide.

(b) In an ecosystem dominated by trees, the pyramid (of numbers) is of inverted type.

(c) In aquatic ecosystems, the limiting factor for productivity is light.

(d) Common detritivores in our ecosystem are earthworms.

(e) A major reservoir of carbon on Earth is oceans.

Question 2: Which one of the following has the largest population in a food chain?

(a) Producers

(b) Primary consumers

(c) Secondary consumers (d) Decomposers

Solution. (d) Decomposers

Decomposers include micro-organisms such as bacteria and fungi. They form the largest population in a food chain and obtain nutrients by breaking down the remains of dead plants and animals.

Question 3: The second trophic level in a lake is-

(a) Phytoplankton

(b) Zooplankton

(c) Benthos

(d) Fishes

Solution. (b) Zooplankton

Zooplankton are primary consumers in aquatic food chains that feed upon phytoplankton. Therefore, they are present at the second trophic level in a lake.

Question 4: Secondary producers are

(a) Herbivores

(b) Producers

(c) Carnivores

(d) None of the above

Solution. (d) None of the above

Plants are the only producers. Thus, they are called primary producers. There are no other producers in a food chain.

Question 5: What is the percentage of photosynthetically active radiation (PAR), in the incident solar radiation.

(a) $100 \%$

(b) $50 \%$

(c) $1-5 \%$

(d) $2-10 \%$

Solution. (b) $50 \%$

Out of total incident solar radiation, about fifty percent of it forms photosynthetically active radiation or PAR.

Question 7: Describe the components of an ecosystem.

Solution. An ecosystem is defined as an interacting unit that includes both the biological community as well as the non-living components of an area. The living and the non-living components of an ecosystem interact amongst themselves and function as a unit, which gets evident during the processes of nutrient cycling, energy flow, decomposition, and productivity. There are many ecosystems such as ponds, forests, grasslands, etc.

The two components of an ecosystem are:

(a) Biotic component: It is the living component of an ecosystem that includes biotic factors such as producers, consumers, decomposers, etc. Producers include plants and algae. They contain chlorophyll pigment, which helps them carry out the process of photosynthesis in the presence of light. Thus, they are also called converters or transducers. Consumers or heterotrophs are organisms that are directly (primary consumers) or indirectly (secondary and tertiary consumers) dependent on producers for their food.

Decomposers include micro-organisms such as bacteria and fungi. They form the largest population in a food chain and obtain nutrients by breaking down the remains of dead plants and animals.

(b) Abiotic component: They are the non-living component of an ecosystem such as light, temperature, water, soil, air, inorganic nutrients, etc.

Question 8: Define ecological pyramids and describe with examples, pyramids of number and biomass.

Solution. An ecological pyramid is a graphical representation of various ecological parameters such as the number of individuals present at each trophic level, the amount of energy, or the biomass present at each trophic level. Ecological pyramids represent producers at the base, while the apex represents the top level consumers present in the ecosystem. There are three types of pyramids:

(a) Pyramid of numbers

(b) Pyramid of energy

(c) Pyramid of biomass

Pyramid of numbers:It is a graphical representation of the number of individuals present at each trophic level in a food chain of an ecosystem. The pyramid of numbers can be upright or inverted depending on the number of producers. For example, in a grassland ecosystem, the pyramid of numbers is upright. In this type of a food chain, the number of producers (plants) is followed by the number of herbivores (mice), which in turn is followed by the number of secondary consumers (snakes) and tertiary carnivores (eagles). Hence, the number of individuals at the producer level will be the maximum, while the number of individuals present at top carnivores will be least.

On the other hand, in a parasitic food chain, the pyramid of numbers is inverted. In this type of a food chain, a single tree (producer) provides food to several fruit eating birds, which in turn support several insect species.

Pyramid of biomass

A pyramid of biomass is a graphical representation of the total amount of living matter present at each trophic level of an ecosystem. It can be upright or inverted. It is upright in grasslands and forest ecosystems as the amount of biomass present at the producer level is higher than at the top carnivore level. The pyramid of biomass is inverted in a pond ecosystem as the biomass of fishes far exceeds the biomass of zooplankton (upon which they feed).

Question 9: What is primary productivity? Give brief description of factors that affect primary productivity.

Solution. It is defined as the amount of organic matter or biomass produced by producers per unit area over a period of time.

Primary productivity of an ecosystem depends on the variety of environmental factors such as light, temperature, water, precipitation, etc. It also depends on the availability of nutrients and the availability of plants to carry out photosynthesis.

Question 10: Define decomposition and describe the processes and products of decomposition.

Solution. Decomposition is the process that involves the breakdown of complex organic matter or biomass from the body of dead plants and animals with the help of decomposers into inorganic raw materials such as carbon dioxide, water, and other nutrients. The various processes involved in decomposition are as follows:

(1) Fragmentation: It is the first step in the process of decomposition. It involves the breakdown of detritus into smaller pieces by the action of detritivores such as earthworms.

(2) Leaching: It is a process where the water soluble nutrients go down into the soil layers and get locked as unavailable salts.

(3) Catabolism: It is a process in which bacteria and fungi degrade detritus through various enzymes into smaller pieces.

(4) Humification: The next step is humification which leads to the formation of a dark-coloured colloidal substance called humus, which acts as reservoir of nutrients for plants.

(5) Mineralization: The humus is further degraded by the action of microbes, which finally leads to the release of inorganic nutrients into the soil. This process of releasing inorganic nutrients from the humus is known as mineralization.

Decomposition produces a dark coloured, nutrient-rich substance called humus. Humus finally degrades and releases inorganic raw materials such as $\mathrm{CO}_{2}$, water, and other nutrient in the soil.

Question 11: Give an account of energy flow in an ecosystem.

Solution. Energy enters an ecosystem from the Sun. Solar radiations pass through the atmosphere and are absorbed by the Earth’s surface. These radiations help plants in carrying out the process of photosynthesis. Also, they help maintain the Earth’s temperature for the survival of living organisms. Some solar radiations are reflected by the Earth’s surface. Only 2-10 percent of solar energy is captured by green plants (producers) during photosynthesis to be converted into food. The rate at which the biomass is produced by plants during photosynthesis is termed as ‘gross primary productivity’. When these green plants are consumed by herbivores, only $10 \%$ of the stored energy from producers is transferred to herbivores. The remaining $90 \%$ of this energy is used by plants for various processes such as respiration, growth, and reproduction. Similarly, only $10 \%$ of the energy of herbivores is transferred to carnivores. This is known as ten percent law of energy flow.

Question 12: Write important features of a sedimentary cycle in an ecosystem.

Solution. Sedimentary cycles have their reservoirs in the Earth’s crust or rocks. Nutrient elements are found in the sediments of the Earth. Elements such as sulphur, phosphorus, potassium, and calcium have sedimentary cycles.

Sedimentary cycles are very slow. They take a long time to complete their circulation and are considered as less perfect cycles. This is because during recycling, nutrient elements may get locked in the reservoir pool, thereby taking a very long time to come out and continue circulation. Thus, it usually goes out of circulation for a long time.

Question 13: Outline salient features of carbon cycling in an ecosystem

Solution. The carbon cycle is an important gaseous cycle which has its reservoir pool in the atmosphere. All living organisms contain carbon as a major body constituent. Carbon is a fundamental element found in all living forms. All biomolecules such as carbohydrates, lipids, and proteins required for life processes are made of carbon. Carbon is incorporated into living forms through a fundamental process called ‘photosynthesis’. Photosynthesis uses sunlight and atmospheric carbon dioxide to produce a carbon compound called ‘glucose’. This glucose molecule is utilized by other living organisms. Thus, atmospheric carbon is incorporated in living forms. Now, it is necessary to recycle this absorbed carbon dioxide back into the atmosphere to complete the cycle. There are various processes by which carbon is recycled back into the atmosphere in the form of carbon dioxide gas. The process of respiration breaks down glucose molecules to produce carbon dioxide gas. The process of decomposition also releases carbon dioxide from dead bodies of plants and animals into the atmosphere. Combustion of fuels, industrialization, deforestation, volcanic eruptions, and forest fires act as other major sources of carbon dioxide.

If you have any Confusion related to NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem PDF then feel free to ask in the comments section down below.

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NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations in PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 13 “Organisms and Populations” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

So, without wasting more time Let’s start.

Question 1: How is diapause different from hibernation?

Solution. Diapause is a stage of suspended development to cope with unfavourable conditions. Many species of Zooplankton and insects exhibit diapause to tide over adverse climatic conditions during their development.

Hibernation or winter sleep is a resting stage where in animals escape winters (cold) by hiding themselves in their shelters. They escape the winter season by entering a state of inactivity by slowing their metabolism. The phenomenon of hibernation is exhibited by bats, squirrels, and other rodents.

Question 2: If a marine fish is placed in a fresh water aquarium, will the fish be able to survive? Why or why not?

Solution. If a marine fish is placed in a fresh water aquarium, then its chances of survival will diminish. This is because their bodies are adapted to high salt concentrations of the marine environment. In fresh water conditions, they are unable to regulate the water entering their body (through osmosis). Water enters their body due to the hypotonic environment outside. This results in the swelling up of the body, eventually leading to the death of the marine fish.

Question 3: Define phenotypic adaptation. Give one example.

Solution. Phenotypic adaptation involves changes in the body of an organism in response to genetic mutation or certain environmental changes. These responsive adjustments occur in an organism in order to cope with environmental conditions present in their natural habitats. For example, desert plants have thick cuticles and sunken stomata on the surface of their leaves to prevent transpiration. Similarly, elephants have long ears that act as thermoregulators.

Question 4: Most living organisms cannot survive at temperature above $45^{\circ} \mathrm{C}^{\circ}$. How are some microbes able to live in habitats with temperatures exceeding $100^{\circ} \mathrm{C} ?$

Solution. Archaebacteria (Thermophiles) are ancient forms of bacteria found in hot water springs and deep sea hydrothermal vents. They are able to survive in high temperatures (which far exceed $\left.100^{\circ} \mathrm{C}\right)$ because their bodies have adapted to such environmental conditions. These organisms contain specialized thermo-resistant enzymes, which carry out metabolic functions that do not get destroyed at such high temperatures.

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

Solution. A population can be defined as a group of individuals of the same species residing in a particular geographical area at a particular time and functioning as a unit. For example, all human beings living at a particular place at a particular time constitute the population of humans.

The main attributes or characteristics of a population residing in a given area are:-

(a) Birth rate (Natality): It is the ratio of live births in an area to the population of an area. It is expressed as the number of individuals added to the population with respect to the members of the population.

(b) Death rate (Mortality): It is the ratio of deaths in an area to the population of an area. It is expressed as the loss of individuals with respect to the members of the population.

(c) Sex ratio: It is the number of males or females per thousand individuals.

(d) Age Distribution: It is the percentage of individuals of different ages in a given population. At any given time, the population is composed of individuals that are present in various age groups. The age distribution pattern is commonly represented through age pyramids.

(e) Population density: It is defined as the number of individuals of a population present per unit area at a given time.

Question 6: If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase (r) of the population?

Solution. A population grows exponentially if sufficient amounts of food resources are available to the individual. Its exponential growth can be calculated by the following integral form of the exponential growth equation:

$N_{t}=N_{0} e^{r t}$

Where,

$N_{t}=$ Population density after time $t$

$N_{O}=$ Population density at time zero

$r=$ Intrinsic rate of natural increase

$e=$ Base of natural logarithms $(2.71828)$

From the above equation, we can calculate the intrinsic rate of increase (r) of a population.

Now, as per the question,

Present population density $=x$

Then,

Population density after two years $=2 x$

$t=3$ years

Substituting these values in the formula, we get:

$\Rightarrow 2 x=x e^{3 r}$

$\Rightarrow 2=e^{3 r}$

Applying log on both sides:

$\Rightarrow \log 2=3 r \log e$

$\Rightarrow \frac{\log 2}{3 \log e}=r$

$\Rightarrow \frac{\log 2}{3 \times 0.434}=r$

$\Rightarrow \frac{0.301}{3 \times 0.434}=r$

$\Rightarrow \frac{0.301}{1.302}=r$

$\Rightarrow 0.2311=r$

Hence, the intrinsic rate of increase for the above illustrated population is 0.2311.

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

Solution. Several plants have evolved various mechanisms both morphological and chemical to protect themselves against herbivory.

(1) Morphological defence mechanisms:

(a) Cactus leaves (Opuntia) are modified into sharp spines (thorns) to deter herbivores from feeding on them.

(b) Sharp thorns along with leaves are present in Acacia to deter herbivores.

(c) In some plants, the margins of their leaves are spiny or have sharp edges that prevent herbivores from feeding on them.

(2) Chemical defence mechanisms:

(a) All parts of Calotropis weeds contain toxic cardiac glycosides, which can prove to be fatal if ingested by herbivores.

(b) Chemical substances such as nicotine, caffeine, quinine, and opium are produced in plants as a part of self-defense.

Question 8: An orchid plant is growing on the branch of mango tree. How do you describe this interaction between the orchid and the mango tree?

Solution. An orchid growing on the branch of a mango tree is an epiphyte. Epiphytes are plants growing on other plants which however, do not derive nutrition from them. Therefore, the relationship between a mango tree and an orchid is an example of commensalisms, where one species gets benefited while the other remains unaffected. In the above interaction, the orchid is benefited as it gets support while the mango tree remains unaffected.

Question 9: What is the ecological principle behind the biological control method of managing with pest insects?

Solution. The basis of various biological control methods is on the concept of predation. Predation is a biological interaction between the predator and the prey, whereby the predator feeds on the prey. Hence, the predators regulate the population of preys in a habitat, thereby helping in the management of pest insects.

Question 10: Distinguish between the following:

(a) Hibernation and Aestivation

(b) Ectotherms and Endotherms

Solution. (a) Hibernation and Aestivation

(b) Ectotherms and Endotherms

Question 11: Write a short note on

(a) Adaptations of desert plants and animals

(b) Adaptations of plants to water scarcity

(d) Importance of light to plants

(e) Effect of temperature or water scarcity and the adaptations of animals.

Solution. (a) Adaptations of desert plants and animals:

Plants found in deserts are well adapted to cope with harsh desert conditions such as water scarcity and scorching heat. Plants have an extensive root system to tap underground water. They bear thick cuticles and sunken stomata on the surface of their leaves to reduce transpiration. In Opuntia, the leaves are entirely modified into spines and photosynthesis is carried out by green stems. Desert plants have special pathways to synthesize food, called CAM $\left(\mathrm{C}_{4}\right.$ pathway). It enables the stomata to remain closed during the day to reduce the loss of Water through transpiration.

Animals found in deserts such as desert kangaroo rats, lizards, snakes, etc. are well adapted to their habitat. The kangaroo rat found in the deserts of Arizona never drinks water in its life. It has the ability to concentrate its urine to conserve water. Desert lizards and snakes bask in the sun during early morning and burrow themselves in the sand during afternoons to escape the heat of the day. These adaptations occur in desert animals to prevent the loss of water.

(b) Adaptations of plants to water scarcity

Plants found in deserts are well adapted to cope with water scarcity and scorching heat of the desert. Plants have an extensive root system to tap underground water. They bear thick cuticles and sunken stomata on the surface of their leaves to reduce transpiration. In Opuntia, the leaves are modified into spines and the process of photosynthesis is carried out by green stems. Desert plants have special pathways to synthesize food, called CAM $\left(\mathrm{C}_{4}\right.$ pathway). It enables their stomata to remain closed during the day to reduce water loss by transpiration.

Certain organisms are affected by temperature variations. These organisms undergo adaptations such as hibernation, aestivation, migration, etc. to escape environmental stress to suit their natural habitat. These adaptations in the behaviour of an organism are called behavioural adaptations. For example, ectothermal animals and certain endotherms exhibit behavioral adaptations. Ectotherms are cold blooded animals such as fish, amphibians, reptiles, etc. Their temperature varies with their surroundings. For example, the desert lizard basks in the sun during early hours when the temperature is quite low. However, as the temperature begins to rise, the lizard burrows itself inside the sand to escape the scorching sun. Similar burrowing strategies are exhibited by other desert animals. Certain endotherms (warm-blooded animals) such as birds and mammals escape cold and hot weather conditions by hibernating during winters and aestivating during summers. They hide themselves in shelters such as caves, burrows, etc. to protect against temperature variations.

(d) Importance of light to plants

Sunlight acts as the ultimate source of energy for plants. Plants are autotrophic organisms, which need light for carrying out the process of photosynthesis. Light also plays an important role in generating photoperiodic responses occurring in plants. Plants respond to changes in intensity of light during various seasons to meet their photoperiodic requirements for flowering. Light also plays an important role in aquatic habitats for vertical distribution of plants in the sea.

(e) Effects of temperature or water scarcity and the adaptations of animals.

Temperature is the most important ecological factor. Average temperature on the Earth varies from one place to another. These variations in temperature affect the distribution of animals on the Earth. Animals that can tolerate a wide range of temperature are called eurythermals. Those which can tolerate a narrow range of temperature are called stenothermal animals. Animals also undergo adaptations to suit their natural habitats. For example, animals found in colder areas have shorter ears and limbs that prevent the loss of heat from their body. Also, animals found in Polar regions have thick layers of fat below their skin and thick coats of fur to prevent the loss of heat.

Some organisms exhibit various behavioural changes to suit their natural habitat. These adaptations present in the behaviour of an organism to escape environmental stresses are called behavioural adaptations. For example, desert lizards are ectotherms. This means that they do not have a temperature regulatory mechanism to escape temperature variations. These lizards bask in the sun during early hours when the temperature is quite low. As the temperature begins to increase, the lizard burrows itself inside the sand to escape the scorching sun. Similar burrowing strategy is seen in other desert animals.

Water scarcity is another factor that forces animals to undergo certain adaptations to suit their natural habitat. Animals found in deserts such as desert kangaroo rats, lizards, snakes, etc. are well adapted to stay in their habitat. The kangaroo rat found in the deserts of Arizona never drinks water in its life. It has the ability to concentrate its urine to conserve water. Desert lizards and snakes bask in the sun during early morning and burrow in the sand as the temperature rises to escape the heat of the day. Such adaptations can be seen to prevent the loss of water.

Question 12: List the various abiotic environmental factors.

Solution. All non- living components of an ecosystem form abiotic components. It includes factors such as temperature, water, light, and soil.

Question 13: Give an example for:

(a) An endothermic animal

(b) An ectothermic animal

(c) An organism of benthic zone

Solution. (a) Endothermic animal: Birds such as crows, sparrows, pigeons, cranes, etc. and mammals such as bears, cows, rats, rabbits, etc. are endothermic animals.

(b) Ectothermic animal: Fishes such as sharks, amphibians such as frogs, and reptiles such as tortoise, snakes, and lizards are ectothermic animals.

(c) Organism of benthic zone: Decomposing bacteria is an example of an organism found in the benthic zone of a water body.

Question 14: Define population and community.

Solution. A population can be defined as a group of individuals of the same species residing in a particular geographical area at a particular time and functioning as a unit. For example, all human beings living at a particular place at a particular time constitute the population of humans.

Community:

A community is defined as a group of individuals of different species, living within a certain geographical area. Such individuals can be similar or dissimilar, but cannot reproduce with the members of other species.

Question 15: Define the following terms and give one example for each:

(a) Commensalism

(b) Parasitism

(c) Camouflage

(d) Mutualism

(e) Interspecific competition

Solution. (a) Commensalism:Commensalism is an interaction between two species in which one species gets benefited while the other remains unaffected. An orchid growing on the branches of a mango tree and barnacles attached to the body of whales are examples of commensalisms.

(b) Parasitism: It is an interaction between two species in which one species (usually smaller) gets positively affected, while the other species (usually larger) is negatively affected. An example of this is liver fluke. Liver fluke is a parasite that lives inside the liver of the host body and derives nutrition from it. Hence, the parasite is benefited as it derives nutrition from the host, while the host is negatively affected as the parasite reduces the host fitness, making its body weak.

(c) Camouflage:It is a strategy adapted by prey species to escape their predators. Organisms are cryptically coloured so that they can easily mingle in their surroundings and escape their predators. Many species of frogs and insects camouflage in their surroundings and escape their predators.

(d) Mutualism: It is an interaction between two species in which both species involved are benefited. For example, lichens show a mutual symbiotic relationship between fungi and blue green algae, where both are equally benefited from each other.

(e) Interspecific competition: It is an interaction between individuals of different species where both species get negatively affected. For example, the competition between flamingoes and resident fishes in South American lakes for common food resources i.e., zooplankton.

Question 16: With the help of suitable diagram describe the logistic population growth curve.

Solution. The logistic population growth curve is commonly observed in yeast cells that are grown under laboratory conditions. It includes five phases: the lag phase, positive acceleration phase, exponential phase, negative acceleration phase, and stationary phase.

(a) Lag phase: Initially, the population of the yeast cell is very small. This is because of the limited resource present in the habitat.

(b) Positive acceleration phase: During this phase, the yeast cell adapts to the new environment and starts increasing its population. However, at the beginning of this phase, the growth of the cell is very limited.

(c) Exponential phase: During this phase, the population of the yeast cell increases suddenly due to rapid growth. The population grows exponentially due to the availability of sufficient food resources, constant environment, and the absence of any interspecific competition. As a result, the curve rises steeply upwards.

(d) Negative acceleration phase: During this phase, the environmental resistance increases and the growth rate of the population decreases. This occurs due to an increased competition among the yeast cells for food and shelter.

(e) Stationary phase: During this phase, the population becomes stable. The number of cells produced in a population equals the number of cells that die. Also, the population of the species is said to have reached nature’s carrying-capacity in its habitat.

A Verhulst−pearl logistic curve is also known as an S-shaped growth curve.

Question 17: Select the statement which explains best parasitism.

(a) One organism is benefited.

(b) Both the organisms are benefited.

(c) One organism is benefited, other is not affected.

(d) One organism is benefited, other is affected.

Solution. (d) One organism is benefited, other is affected.

Parasitism is an interaction between two species in which one species (parasite) derives benefit while the other species (host) is harmed. For example, ticks and lice (parasites) present on the human body represent this interaction where in the parasites receive benefit (as they derive nourishment by feeding on the blood of humans). On the other hand, these parasites reduce host fitness and cause harm to the human body.

Question 18: List any three important characteristics of a population and explain

Solution. A population can be defined as a group of individuals of the same species, residing in a particular geographical area at a particular time and functioning as a unit. For example, all human beings living at a particular place at a particular time constitute the population of humans.

Three important characteristics of a population are:

(a) Birth rate (Natality): It is the ratio of live births in an area to the population of an area. It is expressed as the number of individuals added to the population with respect to the members of the population.

(b) Death rate (Mortality): It is the ratio of deaths in an area to the population of an area. It is expressed as the loss of individuals with respect to the members of the population.

(c) Age Distribution: It is the percentage of individuals of different ages in a given population. At any given time, a population is composed of individuals that are present in various age groups. The age distribution pattern is commonly represented through age pyramids.

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NCERT Solutions for Class 12 Biology Chapter 12 Biotechnology and its Applications PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 12 Biotechnology and its Applications PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 12 Biotechnology and its Applications PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 12 Biotechnology and its Applications in PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 12 “Biotechnology and its Applications” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 12 Biotechnology and its Applications PDF that you can download to start your preparations anytime.

So, without wasting more time Let’s start.

Question 1: Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because −

(a) bacteria are resistant to the toxin

(b) toxin is immature:

(c) toxin is inactive:

(d) bacteria encloses toxin in a special sac.

Solution. 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.

Question 2: What are transgenic bacteria? Illustrate using any one example.

Solution. 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.

Question 3: Compare and contrast the advantages and disadvantages of production of genetically modified crops.

Solution. 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.

Question 4: What are Cry proteins? Name an organism that produces it. How has man exploited this protein to his benefit?

Solution. Cry proteins are 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 $\mathrm{pH}$ 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 $\mathrm{Bt}$ cotton, Bt corn, etc.

Question 5: What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency.

Solution. 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.

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?

Solution. 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.

Question 7: Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of rDNA technology and chemistry of oil?

Solution. 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.

Question 8: Find out from internet what is golden rice.

Solution. Golden rice is a genetically modified variety of rice, Oryza sativa, which has been developedas 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 endosperm of the seed. This is because beta-carotene pigment helps in the process of photosynthesis while photosynthesis does not occur in endosperm. Since betacarotene is a precursor of pro-vitamin $\mathrm{A}$, it is introduced into the rice variety to fulfill the shortage of dietary vitamin $\mathrm{A}$. It is simple and a less expensive alternative to vitamin supplements. However, this variety of rice has faced a significant opposition from environment activists. Therefore, they are still not available in market for human consumption.

Question 9: Does our blood have proteases and nucleases?

Solution. No, human blood does not include the enzymes, nucleases and proteases. In human beings, blood serum contains different types of protease inhibitors, which protect the blood proteins from being broken down by the action of proteases. The enzyme, nucleases, catalyses the hydrolysis of nucleic acids that is absent in blood.

Question 10: Consult internet and find out how to make orally active protein pharmaceutical. What is the major problem to be encountered?

Solution. Orally active protein pharmaceuticals contain biologically active materials such as peptides or proteins, antibodies, and polymeric beads. It is administrated orally into the body through various formulations. It involves the encapsulation of protein or peptide in liposomes or formulations using penetration enhancers. These proteins or peptides are used for treatment of various diseases and are also used as vaccines. However, the oral administration of these peptides or proteins has some problems related to it. Once these proteins are ingested, the proteases present in the stomach juices denature the protein. As a result, their effect will be nullified. Hence, it is necessary to protect the therapeutic protein from digestive enzymes, if taken orally. This is the reason for the proteins to be injected directly into the target site.

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NCERT Solutions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes in PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 11 “Biotechnology: Principles and Processes” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes PDF that you can download to start your preparations anytime.

So, without wasting more time Let’s start.

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).

Solution. 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 –

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.

Solution. The name of the restriction enzyme is Bam H 1.

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?

Solution. 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.

Question 4: What would be the molar concentration of human DNA in a human cell? Consult your teacher.

Solution. The molar concentration of human DNA in a human diploid cell is as follows:

$\Rightarrow$ Total number of chromosomes $\times 6.023 \times 10^{23}$

$\Rightarrow 46 \times 6.023 \times 10^{23}$

$\Rightarrow 2.77 \times 10^{18}$ Moles

Hence, the molar concentration of DNA in each diploid cell in humans is $2.77 \times 10^{23}$ moles.

Solution. 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.

Question 6: Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?

Solution. 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.

(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.

Question 7: Collect 5 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules.

Solution. The palindromic sequence is a certain sequence of the DNA that reads the same whether read from $5^{\prime} \rightarrow 3$ ‘ direction or from $3 \rightarrow 5$ ‘ direction. They are the site for the action of restriction enzymes. Most restriction enzymes are palindromic sequences.

Five examples of palindromic sequences are:

(1)

(2)

(3)

(4)

(5)

Question 8: Can you recall meiosis and indicate at what stage a recombinant DNA is made?

Solution. 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.

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?

Solution. 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.

Question 10: Describe briefly the followings:

(a) Origin of replication

(b) Bioreactors

(c) Downstream processing

Solution. (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 unidirectional 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, $\mathrm{pH}$, vitamin, oxygen, etc. Bioreactors have an oxygen delivery system, a foam control system, a PH, a temperature control system, and a sampling port to obtain a small volume of culture for sampling.

(c) Downstream processing – Downstream processing is a method of separation and purification of foreign gene products after the completion of the biosynthetic stage. The product is subjected to various processes in order to separate and purify the product. After downstream processing, the product is formulated and is passed through various clinical trials for quality control and other tests.

Question 11: Explain briefly

(a) PCR

(b) Restriction enzymes and DNA

(c) Chitinase

Solution. (a) PCR: – Polymerase chain reaction (PCR) is a technique in molecular biology 10 arripmy a gerie or a piece of DNA to obtain its several copies. It is extensively used in the process of gene manipulation. The process involves in-vitro synthesis of sequences using a primer, a template strand, and a thermostable DNA polymerase enzyme obtained from a bacterium, called Thermus aquaticus. The enzyme utilizes building blocks dNTPs (deoxynucleotides) to extend the primer. In the first step, the double stranded DNA molecules are heated to a high temperature so that the two strands separate into a single stranded DNA molecule. This process is called denaturation. Then, this ssDNA molecule is used as a template strand for the synthesis of a new strand by the DNA polymerase enzyme and this process is called annealing, which results in the duplication of the original DNA molecule. This process is repeated over several cycles to obtain multiple copies of the rDNA fragment.

(b) Restriction enzymes are molecular scissors used in molecular biology for cutting DNA sequences from a specific site. It plays an important role in gene manipulation. The enzymes recognize a specific six-box pair sequence known as the recognition sequence and cut the sequence at a specific site. For example, the recognition site for enzyme ECORI is as follows:

Restriction enzyme are categorized into two types –

(i) Exonuclease – It is a type of restriction enzyme that removes the nucleotide from either 5 ‘ or 3 ends of the DNA molecule.

(ii) Endonuclease − It is a type of restriction enzyme that makes a cut within the DNA at a specific site. This enzyme acts as an important tool in genetic engineering. It is commonly used to make a cut in the sequence to obtain DNA fragments with sticky ends, which are later joined by enzyme DNA ligase.

(c) Chitinase – Chitinase is a class of enzymes used for the degradation of chitin, which forms a major component of the fungal cell wall. Therefore, to isolate the DNA enclosed within the cell membrane of the fungus, enzyme chitinase is used to break the cell for releasing its genetic material.

Question 12: Discuss with your teacher and find out how to distinguish between

(a) Plasmid DNA and chromosomal DNA

(b) RNA and DNA

(c) Exonuclease and Endonuclease

Solution. (a) Plasmid DNA and chromosomal DNA

(b) RNA and DNA

(c) Exonuclease and Endonuclease

If you have any Confusion related to NCERT Solutions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes PDF then feel free to ask in the comments section down below.

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NCERT Solutions for Class 12 Biology Chapter 10 Microbes in Human Welfare PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 10 Microbes in Human Welfare PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 10 Microbes in Human Welfare PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 10 Microbes in Human Welfare in PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 10 “Microbes in Human Welfare” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 10 Microbes in Human Welfare PDF that you can download to start your preparations anytime.

So, without wasting more time Let’s start.

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

Solution. Curd 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.

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

Solution. 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.

Glucose $\frac{\text { without }}{\text { oxygen }}$ alcohol $+$ energy $+\mathrm{CO}_{2}$

(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 $\mathrm{CO}_{2}$ released from the dough gets trapped in the dough, thereby giving it a puffed appearance.

Question 3: In which food would you find lactic acid bacteria? Mention some of their useful applications.

Solution. 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 $\mathrm{B}_{12}$ in curd.

Lactic acid bacteria are also found in our stomach where it keeps a check on the disease-causing micro-organisms.

Question 4: Name some traditional Indian foods made of wheat, rice and Bengal gram (or their products) which involve use of microbes.

Solution. (a) Wheat:

2. Rice: Product: Idli, dosa

2. Bengal gram:

Product: Dhokla, Khandvi

Question 5: In which way have microbes played a major role in controlling diseases caused by harmful bacteria?

Solution. 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.

Question 6: Name any two species of fungus, which are used in the production of the antibiotics.

Solution. 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:

Question 7: What is sewage? In which way can sewage be harmful to us?

Solution. 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.

Question 8: What is the key difference between primary and secondary sewage treatment?

Solution.

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

Solution. 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.

Question 10: Microbes can be used to decrease the use of chemical fertilisers and pesticides. Explain how this can be accomplished.

Solution. 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 bacteria, whereas Anabena, Nostoc and Oscillitoria are examples of nitrogen-fixing cyanobacteria. Bio-fertilizers are cost effective and eco-friendly.

Microbes can also act as bio-pesticides to control insect pests in plants. An example of bio-pesticides is Bacillus thuringiensis, which produces a toxin that kills the insect pests. Dried bacterial spores are mixed in water and sprayed in agricultural fields. When larvae of insects feed on crops, these bacterial spores enter the gut of the larvae and release toxins, thereby it. Similarly, Trichoderma are free living fungi. They live in the roots of higher plants and protect them from various pathogens.

Baculoviruses is another bio-pesticide that is used as a biological control agent against insects and other arthropods.

Question 11 :Three water samples namely river water, untreated sewage water and secondary effluent discharged from a sewage treatment plant were subjected to $B O D$ test. The samples were labelled $A, B$ and $C ;$ but the laboratory attendant did not note which was which. The BOD values of the three samples $A, B$ and $C$ were recorded as $20 \mathrm{mg} / \mathrm{L}, 8 \mathrm{mg} / \mathrm{L}$ and $400 \mathrm{mg} / \mathrm{L}$, respectively. Which sample of the water is most polluted? Can you assign the correct label to each assuming the river water is relatively clean?

Solution. Biological oxygen demand $(\mathrm{BOD})$ is the method of determining the amount of oxygen required by micro-organisms to decompose the waste present in the water supply. If the quantity of organic wastes in the water supply is high, then the number of decomposing bacteria present in the water will also be high. As a result, the BOD value will increase

Therefore, it can be concluded that if the water supply is more polluted, then it will have a higher BOD value. Out of the above three samples, sample $\mathbf{C}$ is the most polluted since it has the maximum BOD value of $400 \mathrm{mg} / \mathrm{L} .$ After untreated sewage water, secondary effluent discharge from a sewage treatment plant is most polluted. Thus, sample $A$ is secondary effluent discharge from a sewage treatment plant and has the $\mathrm{BOD}$ value of $20 \mathrm{mg} / \mathrm{L}$, while sample $\mathrm{B}$ is river water and has the $\mathrm{BOD}$ value of $8 \mathrm{mg} / \mathrm{L}$.

Hence, the correct label for each sample is:

Question 12:Find out the name of the microbes from which Cyclosporin A (an immunosuppressive drug) and Statins (blood cholesterol lowering agents) are obtained.

Solution.

Question 13: Find out the role of microbes in the following and discuss it with your teacher.

(a) Single cell protein (SCP)

(b) Soil

Solution. (a) Single cell protein (SCP)

A single cell protein is a protein obtained from certain microbes, which forms an alternate source of proteins in animal feeds. The microbes involved in the preparation of single cell proteins are algae, yeast, or bacteria. These microbes are grown on an industrial scale to obtain the desired protein. For example, Spirulina can be grown on waste materials obtained from molasses, sewage, and animal manures. It serves as a rich supplement of dietary nutrients such as proteins, carbohydrate, fats, minerals, and vitamins. Similarly, micro-organisms such as Methylophilus and methylotrophus have a large rate of biomass production. Their growth can produce a large amount of proteins

(b) Soil

Microbes play an important role in maintaining soil fertility. They help in the formation of nutrient-rich humus by the process of decomposition. Many species of bacteria and cyanobacteria have the ability to fix atmospheric nitrogen into usable form. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirillium and Azotobocter are free living nitrogen-fixing bacteria, whereas Anabena, Nostoc, and Oscillitoria are examples of nitrogen-fixing cyanobacteria.

Question 14: Arrange the following in the decreasing order (most important first) of their importance, for the welfare of human society. Give reasons for your answer.

Biogas, Citric acid, Penicillin and Curd

Solution. The order of arrangement of products according to their decreasing importance is:

Penicillin- Biogas – Citric acid – Curd

Penicillin is the most important product for the welfare of human society. It is an antibiotic, which is used for controlling various bacterial diseases. The second most important product is biogas. It is an eco-friendly source of energy. The next important product is citric acid, which is used as a food preservative. The least important product is curd, a food item obtained by the action of lactobacillus bacteria on milk

Hence, the products in the decreasing order of their importance are as follows:

Penicillin- Biogas – Citric acid – Curd

Question 15: How do biofertilisers enrich the fertility of the soil?

Solution. Bio-fertilizers are living organisms which help in increasing the fertility of soil. It involves the selection of beneficial micro-organisms that help in improving plant growth through the supply of plant nutrients. These are introduced to seeds, roots, or soil to mobilize the availability of nutrients by their biological activity. 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 bacteria, whereas Anabena, Nostoc, and Oscillitoria are examples of nitrogen-fixing cyanobacteria. Bio-fertilizers are cost effective and eco-friendly.

If you have any Confusion related to NCERT Solutions for Class 12 Biology Chapter 10 Microbes in Human Welfare PDF then feel free to ask in the comments section down below.

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NCERT Solutions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food production PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food production PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food production PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food production in PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 9 “Strategies for Enhancement in Food production” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food production PDF that you can download to start your preparations anytime.

So, without wasting more time Let’s start.

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

Solution. 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.

Question 2: If your family owned a dairy farm, what measures would you undertake to improve the quality and quantity of milk production?

Solution. 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.

Question 3: What is meant by the term ‘breed’? What are the objectives of animal breeding?

Solution. 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.

Question 4: Name the methods employed in animal breeding. According to you which one of the methods is best? Why?

Solution. 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.

Question 5: What is apiculture? How is it important in our lives?

Solution. Apiculture is the practice of bee-keeping for the production of various products such as honey, bee’s wax, etc. Honey is a highly nutritious food source and is used as an indigenous system of medicines. 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 beekeeping on a large scale. It has become an income generating activity for farmers since it requires a low investment and is labour intensive.

Question 6: Discuss the role of fishery in enhancement of food production.

Solution. 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 economy. This is because a large part of the Indian population is dependent on fishes as a source of food, which is both cheap and high in animal protein. A Fishery is an employment generating industry especially for people staying in the coastal areas. Both fresh water fishes (such as Catla, Rohu, etc) and marine fishes (such as tuna, mackerel pomfret, etc.) are of high economic value

Question 7: Briefly describe various steps involved in plant breeding.

Solution. Plant breeding is the process in which two genetically dissimilar varieties are purposely crossed to produce a new hybrid variety. As a result, characteristics from both parents can be obtained in the hybrid plant variety. Thus, it involves the production of a new variety with the desired characteristics such as resistance to diseases, climatic adaptability, and better productivity. The various steps involved in plant breeding are as follows:

(a). Collection of genetic variability: Genetic variability from various wild relatives of the cultivated species are collected to maintain the genetic diversity of a species. The entire collection of the diverse alleles of a gene in a crop is called the germplasm collection.

(b). Evaluation of germplasm and selection of parents: The germplasm collected is then evaluated for the desirable genes. The selected plants with the desired genes are then used as parents in plant breeding experiments and are multiplied by the process of hybridization.

(c). Cross-hybridization between selected parents: The next step in plant breeding is to combine the desirable characters present in two different parents to produce hybrids. It is a tedious job as one has to ensure that the pollen grains collected from the male parent reach the stigma of the female parent.

(d). Selection of superior hybrids: The progenies of the hybrids having the desired characteristics are selected through scientific evaluation. The selected progenies are then self-pollinated for several generations to ensure homozygosity.

(e). Testing, release, and commercialization of new cultivars: The selected progenies are evaluated for characters such as yield, resistance to diseases, performance, etc. by growing them in research fields for at least three growing seasons in different parts of the country. After thorough testing and evaluation, the selected varieties are given to the farmers for growing in fields for a large-scale production.

Question 8: Explain what is meant by biofortification.

Solution. Biofortification is a process of breeding crops with higher levels of vitamins, minerals, proteins, and fat content. This method is employed to improve public health. Breeding of crops with improved nutritional quality is undertaken to improve the content of proteins, oil, vitamins, minerals, and micro-nutrients in crops. It is also undertaken to upgrade the quality of oil and proteins. An example of this is a wheat variety known as Atlas 66, which has high protein content in comparison to the existing wheat. In addition, there are several other improved varieties of crop plants such as rice, carrots, spinach etc. which have more nutritious value and more nutrients than the existing varieties

Question 9: Which part of the plant is best suited for making virus-free plants and why?

Solution. Apical and axillary meristems of plants is used for making virus-free plants. In a diseased plant, only this region is not infected by the virus as compared to the rest of the plant region. Hence, the scientists remove axillary and apical meristems of the diseased plant and grow it in vitro to obtain a disease-free and healthy plant.

Virus-free plants of banana, sugarcane, and potato have been obtained using this method by scientists.

Question 10: What is the major advantage of producing plants by micropropagation?

Solution. Micropropagation is a method of producing new plants in a short duration using plant tissue culture.

Some major advantages of micropropagation are as follows:

(a) Micropropagation helps in the propagation of a large number of plants in a short span of time.

(b) The plants produced are identical to the mother plant.

(c) It leads to the production of healthier plantlets, which exhibit better disease-resisting powers.

Question 11: Find out what the various components of the medium used for propagation of explants in vitro are?

Solution. The major components of medium used for propagation of explants in vitro are carbon sources such as sucrose, inorganic salts, vitamins, amino acids, water, agar-agar, and certain growth hormones such as auxins and gibberellins.

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

Solution. The five hybrid varieties of crop plants which have been developed in India are:

Class 12 Chemistry Notes Free PDF.

Class 12 Biology Book Chapterwise Free PDF.

Class 12 Biology Exemplar Chapterwise Free PDF.

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NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Disease PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Disease PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Disease PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Disease in PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 8 “Human Health and Disease” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Disease PDF that you can download to start your preparations anytime.

So, without wasting more time Let’s start.

Question 1. What are the various public health measures, which you would suggest as safeguard against infectious diseases?

Solution. Various public health measure to safeguard infections are:

• Maintainance of public and personal hygiene

• Practising immunisation.

• Creating awareness about the various vaccines.

Question 2. In which way has the study of biology helped us to control infectious diseases?

Solution. Biology has given us a lot of information such as awareness about the preventive measures for a disease. A lot of vaccines have been developed by understanding the method of infection by the micro-organism. People are immunised against several deadly diseases such as smallpox. In case the disease occurs, antibiotics help in keeping the spread of infection in control.

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

(a) Amoebiasis

(b) Malaria

(c) Ascariasis

(d) Pneumonia

Solution. (a) Amoebiasis: Ingestion of quadrinucleate cysts of Entamoeba histolytica found in contaminated food and water results in amoebiasis. The cysts are found in faeces of patients, which enters food and water.

(b) Malaria: It is a vector-borne disease where Plasmodium (a malarial parasite) is carried by female Anopheles mosquito and transmit the disease from patient to a healthy person when a mosquito bites the latter.

(c) Ascariasis: Ingestion of contaminated food and water infected with embryonated eggs of Ascaris worms.

(d) Pneumonia: It is an airborne disease transmitted by the sputum and droplets released when a patient coughs.

Question 4. What measures would you take to prevent water-borne diseases?

Solution. Waterborne diseases can be prevented by following methods:-

• Consumption of pure drinking water.

• Periodic disinfection of waterborne diseases.

• Maintaining personal and public hygiene.

Question 5. Discuss with your teacher what does ‘a suitable gene’ means, in the context of DNA vaccines.

Solution: Suitable gene is the segment of DNA which is inserted into the host. This DNA forms the infection-causing proteins of the micro-organism. The antibodies are generated by our body against these proteins hence giving us immunity for a future encounter with the disease-causing microorganism.

Question 6. Name the primary and secondary lymphoid organs.

Solution. Primary lymphoid organs are:

• Bone marrow

• Thymus

The secondary lymphoid organ includes:

• Spleen

• Tonsils

• Peyer’s patches

• Lymph Nodes

• Vermiform appendix

• Mucosa-associated lymphoid tissue (MALT)

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

(a) MALT

(b) CMI

(c) AIDS

(d) NACO

(e) HIV

Solution. (a) MALT-Mucosa associated lymphoid tissue

(b) CMI – Cell-mediated immunity

(c) AIDS- Acquired immunodeficiency syndrome

(d) NACO-National AIDS control organisation

(e) HIV-Human immunodeficiency virus

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

(a) Innate and acquired immunity

(b) Active and passive immunity

Solution: (a) Innate and acquired immunity

(b) Active and passive immunity

Question 9. Draw a well-labeled diagram of an antibody molecule.

Solution.

Question 10. What are the various routes by which transmission of human immunodeficiency virus takes place?

Solution. The routes by which transmission of HIV takes place is:

• Sexual contact with an infected person

• Use of contaminated needles for drug abuse.

• Transfusion of infected blood.

• Infected mother to fetus.

Question 11. What is the mechanism by which the AIDS virus causes deficiency of the immune system of the infected person?

Solution. • HIV enters the body through macrophage.

• The viral RNA genome forms viral DNA using reverse transcriptase

• Viral DNA enters the host cell to produce viral particles and turns the macrophage into the viral factory.

• After infecting macrophages, the virus infects T helper lymphocytes and produce more virus particles.

• T helper lymphocytes start decreasing in number.

• The decrease in the number of T lymphocytes makes the person susceptible to infections. This is known as acquired immunodeficiency.

Question 12. How is a cancerous cell different from a normal cell?

Solution. A cancerous cell does not have the property of contact inhibition, and hence, they divide uncontrollably causing tumours.

• Cancer cells differentiate, whereas normal cell does not undergo differentiation.

• Cancer cells can undergo metastasis while this property is absent in normal cells.

Question 13. Explain what is meant by metastasis.

Solution. Metastasis is the property of tumour cells to lodge in various parts of the body and cause another set of tumour cells. It is the transfer of malignant tumour from a primary to the secondary region of a body through lymph or blood. The cancerous cells migrate through blood or lymph to other regions of the body after getting released from the malignant tumour, which accumulates to form secondary tumour growth. The migration and settling of malignant tumour breaks are called metastasis.

Question 14. List the harmful effects caused by alcohol/drug abuse.

Solution. Harmful effects of drug abuse are:

• Recklessness, vandalism, abnormal social behaviour

• Mental depreciation and disorders in psychological behaviour.

• Excessive abuse of alcohol leads to coma and death.

• Respiratory failure, heart failure, and cerebral hemisphere.

• Financial distress and mental stress.

• Serious disease such as AIDS, hepatitis B.

• Nervous damage and liver cirrhosis.

• Masculation in females and female characters in males due to anabolic steroids.

• Stunted growth

Question 15. Do you think that friends can influence one to take alcohol/drugs? If yes, how may one protect himself/herself from such an influence?

Solution. Yes, friends can influence one in taking alcohol/drugs. Following measures can be taken:

(i) Avoid undue peer pressure.

(ii) Get counseling from a counselor.

(iii) Practice good habits.

(iv) Seek medical help.

Question 16. Why is that once a person starts taking alcohol or drugs, it is difficult to get rid of this habit? Discuss it with your teacher.

Solution. Upon taking alcohol or drugs the person feels either energetic or calm. This happens due to the sudden increase in neurotransmitters. When these drugs are withdrawn the person feels uneasiness hence feel like consuming it again, making it difficult to quit. This tendency of the body to manifest a characteristic and unpleasant feeling is referred to as withdrawal syndrome.

Question 17. In your view what motivates youngsters to take to alcohol or drugs and how can this be avoided?

Solution. Youngsters are in a very vulnerable state of mental and psychological development, the reasons the fall in prey of drug and alcohol abuse are:

Curiosity to experiment which later becomes an escape to facing the problem.

There is a perception that teenagers feel very progressive if they consume alcohol or smoke.

Consumption of drug is common in order to cope up with exam pressure.

Television, movies, newspaper promotes the perception that consumption of alcohol is part of a good lifestyle.

Use of the drug is also common in the cases where it is unsupportive and unstable family structure and family pressure.

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NCERT Solutions for Class 12 Biology Chapter 7 Evolution PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 7 Evolution PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 7 Evolution PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 7 Evolution in PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 7 “Evolution” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

So, without wasting more time Let’s start.

Question 1: Explain antibiotic resistance observed in bacteria in light of Darwinian selection theory.

Solution. 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.

Question 2: Find out from newspapers and popular science articles any new fossil discoveries or controversies about evolution.

Solution. 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.

Question 3: Attempt giving a clear definition of the term species

Solution. Species can be defined as a group of organisms, which have the capability to interbreed in order to produce fertile offspring. Question 4: Try to trace the various components of human evolution (hint: brain size and function, skeletal structure, dietary preference, etc.)

Solution. The various components of human evolution are as follows.

(i) Brain capacity

(ii) Posture

(iii) Food / dietary preference and other important features

Question 5: Find out through internet and popular science articles whether animals other than man have self-consciousness.

Solution. There are many animals other than humans, which have self consciousness. An example of an animal being self conscious is dolphins. They are highly intelligent. They have a sense of self and they also recognize others among themselves and others. They communicate with each other by whistles, tail-slapping, and other body movements. Not only dolphins, there are certain other animals such as crow, parrot, chimpanzee, gorilla, orangutan, etc., which exhibit self-consciousness.

Question 6: List 10 modern-day animals and using the internet resources link it to a corresponding ancient fossil. Name both.

Solution. The modern day animals and their ancient fossils are listed in the following table.

Question 7: Practise drawing various animals and plants.

Solution. 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.

Solution. 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.

Solution. 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.

Question 10: Using various resources such as your school library or the internet and discussions with your teacher, trace the evolutionary stages of any one animal say horse.

Solution. The evolution of horse started with Eohippus during Eocene period. It involved the following evolutionary stages.

(i) Gradual increase in body size

(ii) Elongation of head and neck region

(iii) Increase in the length of limbs and feet

(iv) Gradual reduction of lateral digits

(v) Enlargement of third functional toe

(vi) Strengthening of the back

(vii) Development of brain and sensory organs

(viii) Increase in the complexity of teeth for feeding on grass

The evolution of horse is represented as

(i) Eohippus It had a short head and neck. It had four functional toes and a splint of 1 and 5 on each hind limb and a splint of 1 and 3 in each forelimb. The molars were short crowned that were adapted for grinding the plant diet.

(ii) Mesohippus

It was slightly taller than Eohippus. It had three toes in each foot.

(iii) Merychippus

It had the size of approximately 100 cm. Although it still had three toes in each foot, but it could run on one toe. The side toe did not touch the ground. The molars were adapted for chewing the grass.

(iv) Pliohippus

It resembled the modern horse and was around $108 \mathrm{~cm}$ tall. It had a single functional toe with splint of $2^{\text {nd }}$ and $4^{\text {th }}$ in each limb.

(v) Equus

Pliohippus gave rise to Equus or the modern horse with one toe in each foot. They have incisors for cutting grass and molars for grinding food.

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NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basic of Inheritance PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basic of Inheritance PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basic of Inheritance PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basic of Inheritance in PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 6 “Molecular Basic of Inheritance” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basic of Inheritance PDF that you can download to start your preparations anytime.

So, without wasting more time Let’s start.

Question 1. Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil, and Cytosine

Solution: Cytidine and Guanosine are the nucleosides.

Adenine, Cytosine, Thymine and Uracil are the nitrogenous bases.

Question 2. If a double-stranded DNA has 20 percent of cytosine, calculate the percentage of adenine in DNA.

Solution. According to Chargaff’s rule, DNA from any cell of any organisms should have a 1:1 ratio of pyrimidine and purine bases $(\mathrm{A}=\mathrm{T}$ and $\mathrm{G}=\mathrm{C}$ ).

The given percent of cytosine is $20 .$ So, the percent of guanine will also be 20 , according to Chargaff’s rule.

Hence, Percent of $(\mathrm{C}+\mathrm{G})=20+20=40 \%$

So, $100=40+(\mathrm{A}+\mathrm{T})$

$(A+T)=100-40=60 \%$

Hence, $\mathrm{A}$ is $30 \%$ and $\mathrm{T}$ is $30 \%$.

Therefore, the calculated percentage of adenine in DNA would be $30 \%$ if $20 \%$ of cytosine is present in DNA.

Question 3. If the sequence of one strand of DNA is written as follows: 5’ATGCATGCATGCATGCATGCATGC-3′ Write down the sequence of the complementary strand in the $3^{\prime} \rightarrow 5^{\prime}$ direction.

Solution. The $3^{\prime} \rightarrow 5^{\prime}$ direction of the sequence of the complementary strand is:

3′-TACGTACGTACGTACGTACGTACGTACG- $5^{\prime}$

Question 4. If the sequence of the coding strand in a transcription unit is written as follows: $5^{\prime}$ ATGCATGCATGCATGCATGCATGCATGC-3′, write down the sequence of $\mathrm{m}$ RNA.

Solution. During transcription, the template strand ( $3^{\prime} \rightarrow 5$ ‘) of DNA codes for m-RNA and hence the sequence of template strand and m-RNA would be complementary to each other whereas the coding strand $\left(5^{\prime} \rightarrow 3^{\prime}\right)$ of DNA will be same as m-RNA (except that $\mathrm{T}$ is replaced by $\mathrm{U}$ in m-RNA).

Here, the sequence of a coding given is:

5′-ATGCATGCATGCATGCATGCATGCATGC-3′

Hence, the sequence of m-RNA would be:

5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′

Question 5. Which property of DNA double helix led Watson and Crick to hypothesise the semi-conservative model of DNA replication? Explain.

Solution. The nitrogenous bases form a complementary pair between the two polynucleotide chains of DNA was observed by Watson and Crick. Based on the X-ray diffraction data, they proposed that DNA consists of a double helix of two chains with nitrogen bases on the inside and sugar-phosphate on the outside.

Then, they proposed that the two chains are antiparallel with $5^{\prime}-3$ ‘ orientation of the other. The appearance of the two chains is like that of a rope ladder twisted helically with rigid steps twisted into a spiral.

This property of a double helix model of DNA led them to hypothesise the semiconservative model of DNA replication, where the two strands separate and act as a template for the synthesis of a new complementary strand, individually. Consequently, each daughter DNA molecule would have one parental strand and one newly synthesized daughter strand. As only one parental strand is conserved in each daughter molecule, it is called a “semi-conservative mode of replication”.

Question 6. Depending on the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesized from it (DNA or RNA), list the types of nucleic acid polymerases.

Solution. The list of the types of nucleic acid polymerases are as following:

(A) DNA-dependent DNA polymerase: It uses a DNA template to catalyze the polymerization deoxynucleotides for synthesising a new DNA strand.

(B) DNA-dependent RNA polymerase: It uses a DNA template strand for synthesising RNA.

Question 7. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?

Solution. Alfred Hershey and Martha Chase in 1952 performed their experiments on bacteriophages. These are the viruses which infect bacteria. The bacteriophage infects the bacteria by entering the genetic material into the bacteria. The viral genetic material becomes a part of the bacterial cell which produces further viral particles.

Aim of the experiment was to find out whether protein or DNA entered the bacteria.

Virus particles were grown on medium containing radioactive sulfur and radioactive phosphorus as sulfur was the part of protein and phosphorus was the part of DNA.

Procedure:

The steps of the experiments are as following:

Infection: The bacteriophage were made to infect the bacteria.

Blending: The viral coats were removed by agitating in the blender.

Centrifugation: The viral particles were separated from the bacteria by spinning in a centrifuge.

Observation: It was observed that the bacteria contained radioactive phosphorus while the radioactive sulfur was found in the protein coat of the virus.

Conclusion: It was concluded that it was DNA from the virus which entered the bacterial cells. Hence, it is the DNA which forms the genetic material and gets transferred from one generation to another.

The Hershey – Chase experiment

Question 8. Differentiate between the following:

(A) Repetitive DNA and Satellite DNA

(B) mRNA and tRNA

(C) Template strand and Coding strand

Solution. (A)

(B)

(C)

Question 9. List two essential roles for a ribosome during translation.

Solution. The following essential roles for a ribosome during translation are:

(i) During translation, ribosome binds to mRNA and provides a platform for joining of amino acids.

(ii) During the process of translation, the ribosome also acts as a catalyst for the formation of a peptide bond.

Question 10. In the medium where $E$. coli was growing, lactose was added, which induced the lac operon. Then why does lac operon shut down sometime after the addition of lactose to the medium?

Solution. After adding lactose in the medium where E.coli was growing, the betagalactosidase enzyme is synthesised in $E . c o l i$. This enzyme hydrolyses lactose into galactose and glucose. These sugars are used by bacteria use as a source of energy. In the absence of lactose, the regulatory gene produces a repressor protein which binds on the operator gene and makes it inactive. This prevents RNA polymerase from transcribing the operon. Therefore, lac operon shuts down sometime after the addition of lactose in the medium.

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

(A) Promoter

(B) tRNA

(C) Exons

Solution. (A) Promoter: Promoter is present in the 5 ‘ region of the coding strand of the DNA. It is the region of the DNA which contains certain conserved sequences. The sigma unit of RNA polymerase identifies these conserved sequences. Later causing the binding of RNA polymerase.

(B) t-RNA: t-RNA is referred to as the transfer RNA. It has a major role during the process of translation. It reads the genetic code and carries the amino acid which is required by the -RNA for translation.

(C) Exon: Human genome has split gene arrangement it has the region of the hn- RNA which codes for the protein is referred to as exon whereas the region which does not code for anything is known as introns.

Question 12. Why is the Human Genome Project called a mega project?

Solution. The aim of the human genome project was to sequence $3 \times 10^{9} \mathrm{bp}$ of the human genome. The estimated cost of the project was 9 billion US dollar (3 US dollar per base pair). The database required to store the information of DNA sequence from a single cell was estimated to be 3300 books ( estimating 1000 pages per book and per page having 1000 letters). The enormous amount of data expected to be generated required high-speed computational devices for data storage and retrieval, and analysis. Hence, considering the requirement and the magnitude of the project, it was called a mega project.

Question 13. What is DNA fingerprinting? Mention its application. Solution. DNA Fingerprinting is a technique which is used to find out the variations in individuals of a population at the DNA level. It works on the principle of polymorphism in the DNA sequences.

The applications of DNA fingerprinting are as follows:

It used in establishing a connection in the family

It helps in solving disputes regarding paternity.

It is used in forensic science.

It helps in tracing the evolutionary relationship between the organisms.

Question 14. Briefly describe the following:

(A) Transcription

(B) Polymorphism

(C) Translation

(D) Bioinformatics

Solution. (A) Transcription is a process in which mRNA is synthesised from doublestranded DNA. It takes place in the cytoplasm in case of prokaryotes, and it occurs in the nucleus in case of eukaryotes. The first step of transcription is initiation. In this step, the sigma unit of RNA polymerase identifies the Promoter region in the DNA. The RNA polymerase binds to the DNA and carries out the addition of nucleotides to the RNA this process is referred to as elongation. When the messenger RNA is fully synthesised, the final termination is caused by rho protein. This protein has helicase activity and causes the termination of the RNA-DNA hybrid.

(B) Polymorphism refers to change at the genetic level, which arises due to mutations. The allele sequence variation is called DNA polymorphism if more than one variant at a locus occurs in the human population. As mutations in these sequences may not have any immediate effect, the probability of such variation to be observed in a non-coding DNA sequence is higher. It has an important role in evolution and speciation. These mutations keep on accumulating in each generation and form one of the bases of polymorphism.

(C) The translation is a process of polymerisation of amino acids to form a polypeptide. Messenger RNA, transfer RNA, and ribosomal RNA participates in the process of translation. Messenger RNA provides the sequence for amino acids in the form of codons. Transfer RNA brings the amino acids and forms the peptide bond. Ribosomal RNA forms the entire translation machinery and initiate the process of translation. Further, the elongation takes place by the addition of amino acids followed by termination upon completion of polypeptide chain formation.

(D) Bioinformatics is a branch of science which uses computation to extract the information about the biological data. The use of computational device helps in the collection, storage, retrieval and analysis of the data.

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NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation in PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 5 “Principles of Inheritance and Variation” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation PDF that you can download to start your preparations anytime.

So, without wasting more time Let’s start.

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

Solution. 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.

Question 2: Differentiate between the following −

(a) Dominance and Recessive

(b) Homozygous and Heterozygous

(c) Monohybrid and Dihybrid.

Solution. (a) Dominance and Recessive

(b) Homozygous and Heterozygous

(c) Monohybrid and Dihybrid

Question 3: A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced? Solution. 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.

Question 4: Explain the Law of Dominance using a monohybrid cross.

Solution. 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 $F_{1}$ 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 $\mathrm{F}_{1}$ generation were found to be round (Rr). When these round seeds were self fertilized, both the round and wrinkled seeds appeared in $\mathrm{F}_{2}$ generation in $3: 1$ ratio. Hence, in $\mathrm{F}_{1}$ generation, the dominant character (round seeds) appeared and the recessive character (wrinkled seeds) got suppressed, which reappeared in $\mathrm{F}_{2}$ generation.

Question 5: Define and design a test – cross?

Solution. 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.

Question 6: Using a Punnett square, work out the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.

Solution. 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, $\mathrm{B}$ and $\mathrm{b}$, while the female will produce only one kind of gamete, $\mathrm{r}$. The genotypic and phenotypic ratio in the progenies of $F_{1}$ generation will be same i.e., $1: 1$.

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.

Solution. 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

Question 8 Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in $\mathrm{F}_{1}$ generation for a dihybrid cross?

Solution. Linkage is defined as the coexistence of two or more genes in the same chromosome. If the genes are situated on the same chromosome and lie close to each other, then they are inherited together and are said to be linked genes.

For example, a cross between yellow body and white eyes and wild type parent in a Drosophila will produce wild type and yellow white progenies. It is because yellow bodied and white eyed genes are linked. Therefore, they are inherited together in progenies.

Question 9: Briefly mention the contribution of T.H. Morgan in genetics.

Solution. Morgan’s work is based on fruit flies (Drosophila melanogaster). He formulated the chromosomal theory of linkage. He defined linkage as the co-existence of two or more genes in the same chromosome and performed dihybrid crosses in Drosophila to show that linked genes are inherited together and are located on X-chromosome. His experiments have also proved that tightly linked genes show very low recombination while loosely linked genes show higher recombination.

Question 10: What is pedigree analysis? Suggest how such an analysis, can be useful.

Solution. Pedigree analysis is a record of occurrence of a trait in several generations of a family. It is based on the fact that certain characteristic features are heritable in a family, for example, eye colour, skin colour, hair form and colour, and other facial characteristics. Along with these features, there are other genetic disorders such as Mendelian disorders that are inherited in a family, generation after generation. Hence, by using pedigree analysis for the study of specific traits or disorders, generation after generation, it is possible to trace the pattern of inheritance. In this analysis, the inheritance of a trait is represented as a tree, called family tree. Genetic counselors use pedigree chart for analysis of various traits and diseases in a family and predict their inheritance patterns. It is useful in preventing hemophilia, sickle cell anemia, and other genetic disorders in the future generations.

Question 11: How is sex determined in human beings?

Solution. Human beings exhibit male heterogamy. In humans, males (XY) produce two different types of gametes, $X$ and $Y$. The human female ( $X X$ ) produces only one type of gametes containing $X$ chromosomes. The sex of the baby is determined by the type of male gamete that fuses with the female gamete. If the fertilizing sperm contains $X$ chromosome, then the baby produced will be a girl and if the fertilizing sperm contains $Y$ chromosome, then the baby produced will be a boy. Hence, it is a matter of chance that determines the sex of a baby. There is an equal probability of the fertilizing sperm being an $X$ or $Y$ chromosome. Thus, it is the genetic make up of the sperm that determines the sex of the baby.

Question 12:A child has blood group O. If the father has blood group A and mother

blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.

Solution. The blood group characteristic in humans is controlled by three set of alleles, namely, $I^{A}, I^{B}$, and i. The alleles, $I^{A}$ and $I^{B}$, are equally dominant whereas allele, i, is recessive to the other alleles. The individuals with genotype, $\mathrm{I}^{\mathrm{A}} \mathrm{I}^{\mathrm{A}}$ and $\mathrm{I}^{\mathrm{A}} \mathrm{i}$, have blood group A whereas the individuals with genotype, $\left.\left.\right|^{B}\right|^{B}$ and $I^{B}$ i, have blood group $B$. The persons with genotype $\left.\left.\right|^{A}\right|^{B}$ have blood group $A B$ while those with blood group $O$ have genotype ii

Hence, if the father has blood group A and mother has blood group B, then the possible genotype of the parents will be

Father Mother

A cross between homozygous parents will produce progeny with AB blood group.

A cross between heterozygous parents will produce progenies with AB blood group $\left(\mathrm{A}^{\mathrm{A}} \mathrm{B}^{\mathrm{B}}\right)$ and $\mathrm{O}$ blood group (ii)

Question 13: Explain the following terms with example

(a) Co-dominance

(b) Incomplete dominance

Solution. (a) Co-dominance

Co-dominance is the phenomenon in which both the alleles of a contrasting character are expressed in heterozygous condition. Both the alleles of a gene are equally dominant. ABO blood group in human beings is an example of co-dominance. The blood group character is controlled by three sets of alleles, namely, $\mathrm{I}^{A}, \mathrm{I}^{\mathrm{B}}$, and $\mathrm{i}$. The alleles, $\mathrm{I}^{\mathrm{A}}$ and $\mathrm{I}^{\mathrm{B}}$, are equally dominant and are said to be codominant as they are expressed in AB blood group. Both these alleles do not interfere with the expression of each other and produce their respective antigens. Hence, $A B$ blood group is an example of co-dominance.

2. Incomplete dominance

Incomplete dominance is a phenomenon in which one allele shows incomplete dominance over the other member of the allelic pair for a character. For example, a monohybrid cross between the plants having red flowers and white flowers in Antirrhinum species will result in all pink flower plants in $\mathrm{F}_{1}$ generation. The progeny obtained in $\mathrm{F}_{1}$ generation does not resemble either of the parents and exhibits intermediate characteristics. This is because the dominant allele, $R$, is partially dominant over the other allele, r. Therefore, the recessive allele, $r$, also gets expressed in the $F_{1}$ generation resulting in the production of intermediate pink flowering progenies with Rr genotype.

Question 14: What is point mutation? Give one example.

Solution. Point mutation is a change in a single base pair of DNA by substitution, deletion, or insertion of a single nitrogenous base. An example of point mutation is sickle cell anaemia. It involves mutation in a single base pair in the beta-globin chain of haemoglobin pigment of the blood. Glutamic acid in short arm of chromosome II gets replaced with valine at the sixth position.

Question 15: Who had proposed the chromosomal theory of inheritance?

Solution. Sutton and Boveri proposed the chromosomal theory of inheritance in 1902. They linked the inheritance of traits to the chromosomes.

Question 16: Mention any two autosomal genetic disorders with their symptoms.

Solution. Two autosomal genetic disorders are as follows.

1. Sickle cell Anaemia

It is an autosomal linked recessive disorder, which is caused by point mutation in the beta-globin chain of haemoglobin pigment of the blood. The disease is characterized by sickle shaped red blood cells, which are formed due to the mutant haemoglobin molecule. The disease is controlled by $\mathrm{Hb}^{\mathrm{A}}$ and $\mathrm{Hb}^{\mathrm{S}}$ allele. The homozygous individuals with genotype, $\mathrm{Hb}^{\mathrm{S}} \mathrm{Hb}^{\mathrm{S}}$, show the symptoms of this disease while the heterozygous individuals with genotype, $\mathrm{Hb}^{\mathrm{A}} \mathrm{Hb}^{\mathrm{S}}$, are not affected. However, they act as carriers of the disease.

Symptoms

Rapid heart rate, breathlessness, delayed growth and puberty, jaundice, weakness, fever, excessive thirst, chest pain, and decreased fertility are the major symptoms of sickle cell anaemia disease. (b) Down’s syndrome

It is an autosomal disorder that is caused by the trisomy of chromosome 21.

Symptoms

The individual is short statured with round head, open mouth, protruding tongue, short neck, slanting eyes, and broad short hands. The individual also shows retarded mental and physical growth.

Class 12 Chemistry Notes PDF.

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NCERT Solutions for Class 12 Biology Chapter 4 Reproductive Health PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 4 Reproductive Health PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 4 Reproductive Health PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 4 Reproductive Health in PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 4 “Reproductive Health” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

So, without wasting more time Let’s start.

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

Solution: Reproductive health refers to the total well-being of mental, physical, social and behavioural aspects of the reproduction. It includes proper functioning of the reproductive organs and normal emotional and behavioural interaction regarding sex-related aspects.

Significance of reproductive health in society includes:

• Awareness about the normal functioning of reproductive organs.

• Use of birth control methods to keep the family size in check.

• Use of contraception to prevent sexually transmitted diseases from gettingtransmitted.

• Eliminating the differences between a male and a female child.

• Emphasising on the benefits of breastfeeding and the importance of postnatal care.

Question 2. Suggest aspects of reproductive health which need to be given special attention in the present scenario.

Solution. Aspects of reproductive health which needs to be given special attention in the present scenario are:

Importance of contraception in preventing sexually transmitted diseases.

By providing counselling to teenagers about the physical changes during puberty.

By giving importance to post-childbirth care of females.

Benefits of breastfeeding to young ones.

Question 3. Is sex education necessary in schools? Why?

Solution. Yes, sex education is necessary for school because of it very essential for a child to receive the right information about the changes in the adolescence, safe sex practices, appropriate use of contraception methods through a proper channel.

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

Solution. There has been a tremendous improvement in the reproductive health of India in the past 50 years. The areas where the development has occurred include:

• The government has introduced programs such as family planning to introduce the concept of a small family.

• Programs regarding mother and child care were introduced.

• Hospitals have been equipped with machines and facilities which were required in childbirth and newborn care.

• Improved version of contraception has been introduced.

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

Solution. The population is increasing at a rapid pace because there has been an improvement in the medical facilities and supply of food, awareness about well-being and hygiene, which has led to an increase in birth rate and a decrease in the death rate.

Question 6. Is the use of contraceptives justified? Give reasons.

Solution. Contraceptives are beneficial in several ways. Following are the reasons:

Mechanical contraceptive devices have a significant role in preventing sexually transmitted disease.

Intrauterine contraceptive devices help a couple in keeping space between two children.

They prevent females from unwanted pregnancies and abortions.

They help the couple to keep the family size in control.

Question 7. Removal of gonads cannot be considered as a contraceptive option. Why?

Solution. Contraception is a method of avoiding fertilisation for a certain period or to control the no. of progenies. It can be temporary or permanent. However, the gonads continue to produce gametes and hormones. Whereas, removal of gonads cannot be considered as a contraceptive.

Gonads refer to ovaries and testes. Ovaries produce female gamete “ovum” and female sex hormones such as “estrogen” and “progesterone”. Testes produce male gamete “sperm” and male hormone “testosterone”. Hence, if gonads are removed the secretion of sex hormones from them will get affected, causing several other health problems.

Question 8. Amniocentesis for sex determination is banned in our country. Is this ban necessary? Comment.

Solution. Amniocentesis is a technique in which amniotic fluid of the developing fetus is taken, and the cells of the fetus are analysed. Use of amniocentesis for fetal sex determination is banned to keep a check on the female foeticide, which would happen after the fetal sex determination. Detection of genetic disorders in the foetus such as Haemophilia, Sickle cell anaemia, Down syndrome is done using this technique.

Question 9. Suggest some methods to assist infertile couples to have children.

Solution. Methods that can help infertile couples to have children are:

• The couple can undergo diagnosis after finding the problem they can undergo treatment in specialised health care units known as infertility clinics.

• The couple can take the help of specialised techniques known as “Assisted reproductive technologies”(ART). These techniques carry out the fertilisation artificially hence known as In-vitro fertilisation. The in-vitro fertilisation techniques are of two types

1. IFT-Zygote intrafallopian transfer: In this method, mature ova are fertilised with the sperms in artificial condition. The zygote that is formed is placed into the fallopian tube for further development.

2. GIFT -Gamete intrafallopian transfer: In this method, the ovum from a donor female is placed into the fallopian tube of a female who cannot produce ovum but can undergo fertilisation and support fetal development.

ICSI-Intra cytoplasmic injections: In this method, the sperm nucleus is directly injected into the ovum.

IUI- Intrauterine insemination: In this method, the sperm is injected into the uterus. This method is performed when males have low sperm count.

Question 10. What are the measures one has to take to prevent from contracting STDs?

Solution. The measures taken by the individual to prevent sexually transmitted disease are:

• Avoid sex with unknown partners or multiple partners.

• Use a barrier method to prevent fluid contact.

• In case any symptoms of infections are suspected one should go to the doctor and get the necessary tests done.

Question 11. State True/False with explanation

(A) Abortions could happen spontaneously too. (True/False)

(B) Infertility is defined as the inability to produce a viable offspring and is always due to abnormalities/defects in the female partner. (True/False)

(C) Complete lactation could help as a natural method of contraception. (True/False)

(D) Creating awareness about sex-related aspects is an effective method to improve the reproductive health of the people. (True/False)

Solution. (A) True. Abortions could happen spontaneously due to stress or imbalance in hormones.

(B) False. Infertility is defined as the inability to produce a viable offspring and is not always due to abnormalities/defects in the female partner as males can be sterile due to abnormalities such as low sperm count and impotence. (C) True. Complete lactation could help as a natural method of contraception because when the mother is breastfeeding the hormone prolactin is involved in the process of lactation; this hormone counteracts with the Luteinizing hormone required for the ovulation process.

(D) True. Creating awareness about sex-related aspects is an effective method to improve the reproductive health of the people. (True/False)

Question 12. Correct the following statements :

(A) Surgical methods of contraception prevent gamete formation.

(B) All sexually transmitted diseases are completely curable.

(C) Oral pills are very popular contraceptives among rural women.

(D) In E. T. techniques, embryos are transferred into the uterus.

Solution: (A) Surgical methods of contraception include tubectomy in females and vasectomy is males In this method, the fallopian tubes, and vas deferens is cut and then ligated in females and males respectively. These methods do nor stop gamete formation as the surgery is not performed in gonads.

(B) The sexually transmitted disease includes Gonorrhoea, syphilis, genital herpes, chlamydiosis, genital warts, trichomoniasis, and HIV.Out of these diseases, Hepatitis B, genital herpes, HIV infection are not completely curable.

(C) Contraception is not much popular among rural women. Oral pills are mainly used by females staying in urban areas.

(D) In embryo transfer technique, the embryo is transferred into the fallopian tube at the eight-cell stage. If the embryo is more than eight cells, it is implanted into the uterus.

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NCERT Solutions for Class 12 Biology Chapter 3 Human Reproduction PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 3 Human Reproduction PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 3 Human Reproduction PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 3 Human Reproduction PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 3 “Human Reproduction” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

So, without wasting more time Let’s start.

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_____________.

(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_____________.

Solution. (a) Humans reproduce sexually.

(b) Humans are viviparous .

(c) Fertilization is internal in humans.

(d) Male and female gametes are haploid.

(e) Zygote is diploid.

(f) The process of release of the ovum from a mature follicle is called ovulation.

(g) Ovulation is induced by a hormone called the luteinizing hormone.

(h) The fusion of the male and the female gametes is called fertilization

(i) Fertilization takes place in the fallopian tube.

(j) The zygote divides to form blastocyst , which is implanted in uterus.

(k) The structure which provides vascular connection between the fetus and uterus is called placenta

Question 2: Draw a labeled diagram of male reproductive system

Solution.

Question 3: Draw a labeled diagram of female reproductive system.

Solution.

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

Solution. 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.

Question 5: Describe the structure of a seminiferous tubule.

Solution. he 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.

Question 6: What is spermatogenesis? Briefly describe the process of spermatogenesis.

Solution. 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.

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

Solution. 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.

Question 8: Define spermiogenesis and spermiation.

Solution. 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.

Question 9: Draw a labeled diagram of sperm.

Solution.

Question 10: What are the major components of seminal plasma?

Solution. 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.

Question 11: What are the major functions of male accessory ducts and glands?

Solution. 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.

Question 12: What is oogenesis? Give a brief account of oogenesis.

Solution. 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 undergoes second meiotic division i.e., meiosis II or equational division and gives rise to a second polar body and an ovum. Hence, in the process of oogenesis, a diploid oogonium produces a single haploid ovum while two or three polar bodies are produced.

Question 13: Draw a labeled diagram of a section through ovary.

Solution.

Question 14: Draw a labeled diagram of a Graafian Follicle?

Solution.

Question 15: Name the functions of the following.

(a) Corpus luteum

(b) Endometrium

(c) Acrosome

(d) Sperm tail

(e) Fimbriae

Solution. (a) Corpus luteum − Corpus luteum is formed from the ruptured Grafiaan follicle. It secretes progesterone hormone during the luteal phase of the menstrual cycle. A high level of progesterone inhibits the secretions of FSH and LH, thereby preventing ovulation. It also allows the endometrium of the uterus to proliferate and to prepare itself for implantation.

(b) Endometrium − It is the innermost lining of the uterus. It is rich in glands and undergoes cyclic changes during various phases of the menstrual cycle to prepare itself for the implantation of the embryo.

(c) Acrosome − It is a cap-like structure present in the anterior part of the head of the sperm. It contains hyaluronidase enzyme, which hydrolyses the outer membrane of the egg, thereby helping the sperm to penetrate the egg at the time of fertilization.

(d) Sperm tail − It is the longest region of the sperm that facilitates the movement of the sperm inside the female reproductive tract.

(e) Fimbriae − They are finger-like projections at the ovarian end of the fallopian tube. They help in the collection of the ovum (after ovulation), which is facilitated by the beating of the cilia.

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

(a) Androgens are produced by Sertoli cells. (True/False)

(b) Spermatozoa get nutrition from Sertoli cells. (True/False)

(c) Leydig cells are found in ovary. (True/False)

(d) Leydig cells synthesise androgens. (True/False)

(e) Oogenesis takes place in corpus luteum. (True/False)

(f) Menstrual cycle ceases during pregnancy. (True/False)

(g) Presence or absence of hymen is not a reliable indicator of virginity or sexual experience. (True/False)

Solution. (a) Androgens are produced by Sertoli cells. (False)

Androgens are produced by Leydig cells found in seminiferous tubules of the testis.

(b) Spermatozoa get nutrition from Sertoli cells. (True)

(c) Leydig cells are found in ovary. (False)

Leydig cells are found in the seminiferous tubules of the testis.

(d) Leydig cells synthesise androgens. (True)

(e) Oogenesis takes place in corpus luteum. (False)

Oogenesis takes place in the ovary.

(f) Menstrual cycle ceases during pregnancy. (True)

(g) Presence or absence of the hymen is not a reliable indicator of virginity or sexual experience. (True)

Question 17: What is menstrual cycle? Which hormones regulate menstrual cycle?

Solution. The menstrual cycle is a series of cyclic physiologic changes that take place inside the female reproductive tract in primates. The whole cycle takes around 28 days to complete. The end of the cycle is accompanied by the breakdown of uterine endothelium, which gets released in the form of blood and mucous through the vagina. This is known as menses.

The follicle stimulating hormone (FSH), luteinizing hormone (LH), estrogen, and progesterone are the various hormones that regulate the menstrual cycle. The level of FSH and LH secreted from the anterior pituitary gland increases during the follicular phase. FSH secreted under the influence of RH (releasing hormone) from the hypothalamus stimulates the conversion of a primary follicle into a graafian follicle. The level of LH increases gradually leading to the growth of follicle and secretion of estrogen. Estrogen inhibits the secretion of FSH and stimulates the secretion of luteinizing hormone. It also causes the thickening of the uterine endometrium. The increased level of LH causes the rupturing of the graafian follicle and release the ovum into the fallopian tube. The ruptured graafian follicle changes to corpus luteum and starts secreting progesterone hormone during the luteal phase. Progesterone hormone helps in the maintenance and preparation of endometrium for the implantation of the embryo. High levels of progesterone hormone in the blood decrease the secretion of $\mathrm{LH}$ and FSH, therefore inhibiting further ovulation.

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

Solution. Parturition is the process of giving birth to a baby as the development of the foetus gets completed in the mother’s womb. The hormones involved in this process are oxytocin and relaxin. Oxytocin leads to the contraction of smooth muscles of myometrium of the uterus, which directs the full term foetus towards the birth canal. On the other hand, relaxin hormone causes relaxation of the pelvic ligaments and prepares the uterus for child birth.

Question 19: In our society the women are often blamed for giving birth to daughters.

Can you explain why this is not correct?

Solution. All human beings have 23 pairs of chromosomes. Human males have 22 pairs of autosomes and contain one or two types of sex chromosome. They are either $X$ or $Y$. On the contrary, human females have 22 pairs of autosomes and contain only the $X$ sex chromosome. The sex of an individual is determined by the type of the male gamete (X or $Y)$, which fuses with the $X$ chromosome of the female. If the fertilizing sperm is $X$, then the baby will be a girl and if it is $Y$, then the baby will be a boy. Hence, it is incorrect to blame a woman for the gender of the child.

Question 20: How many eggs are released by a human ovary in a month? How many eggs do you think would have been released if the mother gave birth to identical twins? Would your answer change if the twins born were fraternal?

Solution. An ovary releases an egg every month. When two babies are produced in succession, they are called twins. Generally, twins are produced from a single egg by the separation of early blastomeres resulting from the first zygotic cleavage. As a result, the young ones formed will have the same genetic make- up and are thus, called identical twins.

If the twins born are fraternal, then they would have developed from two separate eggs. This happens when two eggs (one from each ovary) are released at the same time and get fertilized by two separate sperms. Hence, the young ones developed will have separate genes and are therefore, called non-identical or fraternal twins.

Question 21: How many eggs do you think were released by the ovary of a female dog

which gave birth to 6 puppies?

Solution. Dogs and rodents are polyovulatory species. In these species, more than one ovum is released from the ovary at the time of ovulation. Hence, six eggs were released by the ovary of a female dog to produce six puppies.

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NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 2 “Sexual Reproduction in Flowering Plants” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

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In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants PDF that you can download to start your preparations anytime.

So, without wasting more time Let’s start.

Question 1: Name the parts of an angiosperm flower in which development of male and female gametophyte take place.

Solution. The male gametophyte or the pollen grain develops inside the pollen chamber of the anther, whereas the female gametophyte (also known as the embryo sac) develops inside the nucellus of the ovule from the functional megaspore.

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

Solution.

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.

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

Question 3: Arrange the following terms in the correct developmental sequence: Pollen grain, sporogenous tissue, microspore tetrad, pollen mother cell, male gametes

Solution. 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.

Question 4: With a neat, labelled diagram, describe the parts of a typical angiosperm ovule.

Solution. An ovule is a female megasporangium where the formation of megaspores takes place.

The various parts of an ovule are –

Funiculus – It is a stalk-like structure which represents the point of attachment of the ovule to the placenta of the ovary.

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

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

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.

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.

Chalazal – It is the based swollen part of the nucellus from where the integuments originate.

Question 5: What is meant by monosporic development of female gametophyte?

Solution. 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.

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

Solution.

The female gametophyte (embryo sac) develops from a single functional megaspore. This megaspore undergoes three successive mitotic divisions to form eight nucleate embryo sacs.

The first mitotic division in the megaspore forms two nuclei. One nucleus moves towards the micropylar end while the other nucleus moves towards the chalazal end. Then, these nuclei divide at their respective ends and re- divide to form eight nucleate stages. As a result, there are four nuclei each at both the ends i.e., at the micropylar and the chalazal end in the embryo sac. At the micropylar end, out of the four nuclei only three differentiate into two synergids and one egg cell. Together they are known as the egg apparatus. Similarly, at the chalazal end, three out of four nuclei differentiates as antipodal cells. The remaining two cells (of the micropylar and the chalazal end) move towards the centre and are known as the polar nuclei, which are situated in a large central cell. Hence, at maturity, the female gametophyte appears as a 7-celled structure, though it has 8 nucleate.

Question 7: What are chasmogamous flowers? Can cross-pollination occur in cleistogamous flowers? Give reasons for your answer.

Solution. There are two types of flowers present in plants namely Oxalis and Violachasmogamous and cleistogamous flowers.

Chasmogamous flowers have exposed anthers and stigmata similar to the flowers of other species. Cross-pollination cannot occur in cleistogamous flowers.

This is because cleistogamous flowers never open at all. Also, the anther and the stigma lie close to each other in these flowers. Hence, only self-pollination is possible in these flowers.

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

Solution. 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:

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 a genetic mechanism to prevent self-pollination called self- incompatibility. Incompatibility may be between individuals of the same species or between individuals of different species. Thus, incompatibility prevents breeding.

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

Question 9: What is self-incompatibility? Why does self-pollination not lead to seed formation in self-incompatible species?

Solution. Self-incompatibility is a genetic mechanism in angiosperms that preventsself-pollination. It develops genetic incompatibility between individuals of the same species or between individuals of different species.

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

Question 10: What is bagging technique? How is it useful in a plant breeding programme?

Solution. Various artificial hybridization techniques (under various crop improvement programmes) involve the removal of the anther from bisexual flowers without affecting the female reproductive part (pistil) through the process of emasculation. Then, these emasculated flowers are wrapped in bags to prevent pollination by unwanted pollen grains.

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

Question 11: What is triple fusion? Where and how does it take place? Name the nuclei involved in triple fusion.

Solution. Triple fusion is the fusion of the male gamete with two polar nuclei inside the embryo sac of the angiosperm.

This process of fusion takes place inside the embryo sac. When pollen grains fall on the stigma, they germinate and give rise to the pollen tube that passes through the style and enters into the ovule. After this, the pollen tube enters one of synergids and releases two male gametes there. Out of the two male gametes, one gamete fuses with the nucleus of the egg cell and forms the zygote (syngamy). The other male gamete fuses with the two polar nuclei present in the central cell to form a triploid primary endosperm nucleus. Since this process involves the fusion of three haploid nuclei, it is known as triple fusion. It results in the formation of the endosperm.

One male gamete nucleus and two polar nuclei are involved in this process.

Question 12: Why do you think the zygote is dormant for sometime in a fertilized ovule?

Solution. The zygote is formed by the fusion of the male gamete with the nucleus of the egg cell. The zygote remains dormant for some time and waits for the endosperm to form, which develops from the primary endosperm cell resulting from triple fusion. The endosperm provides food for the growing embryo and after the formation of the endosperm, further development of the embryo from the zygote starts.

Question 13: Differentiate between:

(a) Hypocotyl and epicotyl;

(b) Coleoptile and coleorrhiza;

(c) Integument and testa;

(d) Perisperm and pericarp.

Solution. (a)

(b)

(c)

(d)

Question 14: Why is apple called a false fruit? Which part(s) of the flower forms the fruit?

Solution. Fruits derived from the ovary and other accessory floral parts are called false fruits. On the contrary, true fruits are those fruits which develop from the ovary, but do not consist of the thalamus or any other floral part. In an apple, the fleshy receptacle forms the main edible part. Hence, it is a false fruit.

Question 15: What is meant by emasculation? When and why does a plant breeder employ this technique?

Solution. Emasculation is the process of removing anthers from bisexual flowers without affecting the female reproductive part (pistil), which is used in various plant hybridization techniques.

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

Question 16: If one can induce parthenocarpy through the application of growth substances, which fruits would you select to induce parthenocarpy and why?

Solution. Parthenocarpy is the process of developing fruits without involving the process of fertilization or seed formation. Therefore, the seedless varieties of economically important fruits such as orange, lemon, water melon etc. are produced using this technique. This technique involves inducing fruit formation by the application of plant growth hormones such as auxins.

Question 17: Explain the role of tapetum in the formation pollen-grain wall.

Solution. Tapetum is the innermost layer of the microsporangium. It provides nourishment to the developing pollen grains. During microsporogenesis, the cells of tapetum produce various enzymes, hormones, amino acids, and other nutritious material required for the development of pollen grains. It also produces the exine layer of the pollen grains, which is composed of the sporopollenin.

Question 18: What is apomixis and what is its importance?

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

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NCERT Solutions for Class 12 Biology Chapter 1 Reproduction in Organism PDF – eSaral

# NCERT Solutions for Class 12 Biology Chapter 1 Reproduction in Organism PDF

Hey, are you a class 12 student and looking for ways to download NCERT Solutions for Class 12 Biology Chapter 1 Reproduction in Organism PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 12 Biology Chapter 1 Reproduction in Organism PDF that are prepared by Kota’s top IITian Faculties by keeping Simplicity in mind.

If you want to learn and understand class 12 Biology Chapter 1 “Reproduction in Organism” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 12 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 12 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

So, without wasting more time Let’s start.

Question 1: Why is reproduction essential for organisms?

Solution. 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.

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

Solution. 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.

Question 3: Why is the offspring formed by asexual reproduction referred to as clone?

Solution. A clone is a group of morphologically and genetically identical individuals.

In the process of asexual reproduction, only one parent is involved and there is no fusion of the male and the female gamete. As a result, the offsprings so produced are morphologically and genetically similar to their parents and are thus, called clones.

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

Solution. Sexual reproduction involves the fusion of the male and the female gamete. This fusion allows the formation of new variants by the combination of the DNA from two (usually) different members of the species. The variations allow the individuals to adapt under varied environmental conditions for better chances of survival.

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

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

Solution.

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

Solution.

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.

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

Solution. egetative 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.

Question 8: Define

(a) Juvenile phase,

(b) Reproductive phase,

(c) Senescent phase.

Solution. (a) Juvenile phase:

It is the period of growth in an individual organism after its birth and before it reaches reproductive maturity.

(b) Reproductive phase:

It is the period when an individual organism reproduces sexually.

(c) Senescent phase:

It is the period when an organism grows old and loses the ability to reproduce.

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

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

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

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

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

(a) Ovary _______________

(b) Anther _____________________

(c) Egg ________________

(d) Pollen _________________

(e) Male gamete __________________

(f ) Zygote ____________________

Solution. (a) Ovary Diploid $(\underline{2 n})$.

(b) Anther Diploid $(2 n)$

(c) Egg Haploid $(\underline{n})$

(d) Pollen Haploid $(\underline{n})$.

(e) Male gamete $\underline{\text { Haploid }}(\underline{n})$

(f) Zygote Diploid $(2 n)$

Question 12: Define external fertilization. Mention its disadvantages.

Solution. External fertilization is the process in which the fusion of the male and the female gamete takes place outside the female body in an external medium, generally water. Fish, frog, starfish are some organisms that exhibit external fertilization.

In external fertilization, eggs have less chances of fertilization. This can lead to the wastage of a large number of eggs produced during the process.

Further, there is an absence of proper parental care to the offspring, which results in a low rate of survival in the progenies.

Question 13: Differentiate between a zoospore and a zygote.

Solution.

Question 14: Differentiate between gametogenesis from embryogenesis.

Solution.

Question 15: Describe the post-fertilization changes in a flower.

Solution. Fertilization is the process of the fusion of the male and the female gamete to form a diploid zygote. After fertilization, the zygote divides several times to form an embryo. The fertilized ovule forms a seed. The seed contains an embryo, enclosed in a protective covering, called the seed coat. As the seed grows further, other floral parts wither and fall off. This leads to the growth of the ovary, which enlarges and ripens to become a fruit with a thick wall called the pericarp.

Question 16: What is a bisexual flower? Collect five bisexual flowers from your neighborhood and with the help of your teacher find out their common and scientific names.

Solution. A flower that contains both the male and female reproductive structure (stamen and pistil) is called a bisexual flower. Examples of plants bearing bisexual flowers are:

(1) Water lily ( Nymphaea odorata)

(2) Rose (Rosa multiflora )

(3) Hibiscus (Hibiscus Rosa-sinensis )

(4) Mustard ( Brassica nigra)

(5) Petunia (Petunia hybrida)

Question 17: Examine a few flowers of any cucurbit plant and try to identify the staminate and pistillate flowers. Do you know any other plant that bears unisexual flowers?

Solution.

Cucurbit plant bears unisexual flowers as these flowers have either the stamen or the pistil. The staminate flowers bear bright, yellow coloured petals along with stamens that represent the male reproductive structure. On the other hand, the pistillate flowers bear only the pistil that represents the female reproductive structure.

Other examples of plants that bear unisexual flowers are corn, papaya, cucumber, etc.

Question 18: Why are offspring of oviparous animals at a greater risk as compared to offspring of viviparous animals?

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