Prasanna

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 12 Environmental Issues Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 12 Environmental Issues

12th Bio Zoology Guide Environmental Issues Text Book Back Questions and Answers

Question 1.
Right to Clean Water is a fundamental right, under the Indian Constitution
(a) Article 12
(b) Article 21
(c) Article 31
(d) Article 41
Answer:
(c) Article 31

Question 2.
With which of the following, the Agenda 21’ of Rio Summit, 1992 is related to?
(a) Sustainable development
(b) Combating the consequences of population
(c) Mitigation norms of Green House Gases (GHG_s) emission
(d) Technology transfer mechanism to developing countries for ‘clean energy’ production
Answer:
(a) Sustainable development

Question 3.
Which among the following awards instituted by the Government of India for individuals or communities from rural areas that have shown extraordinary courage and dedication in protecting Wildlife?
(a) Indira Gandhi Paryavaran Puraskar
(b) Medini Puraskar Yojana
(c) Amrita Devi Bishnoi Award
(d) Pitambar Pant National Award
Answer:
(c) Amrita Devi Bishnoi Award

Question 4.
The ‘thickness’ of Stratospheric Ozone layer is measured in/on:
(a) Sieverts units
(b) Dobson units
(c) Melson units
(d) Beaufort Scale
Answer:
(b) Dobson units

Question 5.
Which among the following is the most abundant Green-House-Gas (GHG) in the Earth’s atmosphere?
(a) Carbon dioxide
(b) Water Vapour
(c) Sulphur Dioxide
(d) Tropospheric Ozone
Answer:
(a) Carbon dioxide

Question 6.
As per 2017 statistics, the highest per capita emitter of Carbon dioxide in the world is
(a) USA
(b) China
(c) Qatar
(d) Saudi Arabia
Answer:
(b) China

Question 7.
The use of microorganism metabolism to remove pollutants such as oil spills in the water bodies is known as ……………..
(a) Biomagnification
(b) Bioremediation
(c) Biomethanation
(d) Bioreduction
Answer:
(b) Bioremediation

Question 8.
The Ozone Day is observed every year on September 16 as on this day in 1987 the …………………. was signed for launching efforts to arrest the depletion of the fragile ozone layer in the stratosphere that prevents the harmful ultra-violet rays of the Sun from reaching the Earth. Fill the correct word in blank.
(a) Montreal Protocol
(b) Geneva Protocol
(c) Kyoto Protocol
(d) Nagoya Protocol
Answer:
(a) Montreal Protocol

Question 9.
Which among the following always decreases in a Food chain across tropic levels?
(a) Number
(b) Accumulated chemicals
(c) Energy
(d) Force
Answer:
(c) Energy

Question 10.
In the E-waste generated by the Mobile Phones, which among the following metal is most abundant?
(a) Copper
(b) Silver
(c) Palladium
(d) Gold
Answer:
(a) Copper

Question 11.
The Hydrochlorofluorocarbons (HCFC_s) are the compounds which have the following molecules:
(a) Hydrogen
(b) Carbon
(c) Chlorine
(d) Fluorine
Answer:
(c) Chlorine

Question 12.
SMOG is derived from:
(a) Smoke
(b) Fog
(c) Both A and B
(d) Only A
Answer:
(c) Both A and B

Question 13.
Excess of fluoride in drinking water causes:
(a) Lung disease
(b) Intestinal infection
(c) Fluorosis
(d) None of the above
Answers:
(c) Fluorosis

Question 14.
Expand (i) CFC (ii) AQI (iii) PAN
Answer:
(i) CFC: Chlorofluorocarbons
(ii) AQI: Air Quality Index
(iii) PAN: Peroxyacetyl nitrate

Question 15.
What is SMOG and how it is harmful for us?
Answer:
Smog is a type of air pollution caused by tiny particles in the air. The word comes from a mixture of the words smoke and fog. Today, smog generally refers to photochemical smog, which is created when sunlight reacts with nitrogen oxides and volatile organic compounds found in fossil fuel emissions from automobiles, factories, and power plants. These reactions create ground-level ozone and particulate matter, reducing visibility. Smog can make breathing more difficult, especially for people with asthma. Smog also affects plants and animals. It damages crops as well as causes health problems in pets, farm animals and human beings. Smog has also been known to cause corrosive damage to buildings and vehicles.

Question 16.
List all the wastes that you generate, at home, school or during your trips to other places. Could you very easily reduce the generation of these wastes? Which would be difficult or rather impossible to reduce?

Yes, we can reduce the generation of daily wastes in our life. For example, using jute/cloth bags instead of plastic bags, using the plastic drum as flower pots, etc.
E-wastes and plastic wastes are generally difficult to biodegrade.

Question 17.
Discuss the causes and effects of global warming. What measures need to be taken to control global warming?
Answer:

1. Phase down or ban the use of CFCs (CFC free refrigerants).
2. Minimizing the use of chemicals such as halons and halocarbons.
3. Creating awareness about ozone-depleting agents.
4. UV rays may penetrate deep into the skin and can lead to premature skin aging and wrinkling of the skin; suppression of the immune system, skin cancer (melanoma), and chronic effects leading to eye damage.
5. DNA damage can result from free radicals and reactive oxygen and. photons can damage the DNA itself.

Question 18.
What would Earth be like without the greenhouse effect?
Answer:
The greenhouse effect is vital for the sustenance of life. Greenhouse gases like CO₂, water vapour etc absorb some of the reflected sun’s radiation and radiate back it to the Earth surface, thus maintaining the Earth’s warm condition. Without this effect, life on Earth would be difficult or rather impossible for existence or become hostile to most living organisms.

Question 19.
Write notes on the following:
a. Eutrophication
b. Algal Bloom
Answer:
a. Eutrophication
Eutrophication refers to the nutrient enrichment in water bodies leading to a lack of oxygen and will end up in the death of aquatic organisms.
b. Algal Bloom
Algal Bloom is excess growth of algae due to abundant excess nutrients imparting distinct color to water.

Question 20.
What effect can fertilizer runoff have on an aquatic ecosystem?
Answer:
When run-off from land containing nutrients’ reaches water bodies like lakes, it results in a dense growth of plant life. This phenomenon is called Eutrophication.

Question 21.
How can we control eutrophication?
Answer:
Eutrophication can be controlled by reducing the use of fertilizers in agricultural lands, checking the runoff from fields, planting vegetations along the stream beds thereby the nutrients will be uptaken by plants.

Question 22.
Why does an ozone hole form over Antarctica?
Answer:
Ozone hole in Antarctica is due to special atmospheric and chemical conditions that existed there and nowhere else in the globe.

Question 23.
Mention the causes of enhanced ultraviolet radiation.
Answer:
UV rays may penetrate deep into the skin and can lead to premature skin aging and wrinkling of skin; suppression of the immune system, skin cancer (melanoma) and chronic effects leading to eye damage. DNA damage can result from free radicals and reactive oxygen and photons can damage the DNA itself.

Question 24.
Discuss the role of women in the protection and conservation of forests.
Answer:
Women and communities have played a crucial role in environmental conservation programmes.

Example 1: Chipko movement was started in 1974 in India. In this women from the village hugged the trees and prevented the felling of trees by contractors.

Example 2: Amrita Devi, from Khejarli village of Jodhpur district, Rajasthan sacrificed her life to maintain Bishoni Dharma. Here the King of Jodhpur ordered his men to get the wood of Khejri trees by cutting them to construct his new palace. On hearing this Amrita Devi and many others had hugged the trees to save them from cutting. But her effort went on vain. Amrita Devi and others were killed by King’s men. This resistance by the people of the village forced the King to give up the idea of cutting trees.

Question 25.
Discuss the role of an individual to reduce environmental pollution.
Answer:

• Decrease waste generation
• Use efficient transportation
• Reduce energy consumption
• Planting trees

Question 26.
How does recycling help to reduce pollution?
Answer:
Recycling prevents the emission of many greenhouse gases which are the major cause of global warming. It also reduces water pollutants and saves energy.

Question 27.
What is the primary purpose of the Kyoto Protocol?
Answer:
The primary purpose of Kyoto Protocol is to reduce the onset of global warming by reducing the concentration of greenhouse gases in the atmosphere.

Question 28.
In what way Peyang conserves the forest?
Answer:
The ‘Forest man of India’, Jadav Payeng who created 1,360 acres of dense and defiant forest was born in Arunasapori (a river island on the Brahmaputra). He had just completed his Class X exams in1979 when he started-to sow the seeds and shoots on the eroded island covered with sand and silt. Thirty-six years later he had converted the once unproductive land into a forest.

Payeng’s forest is now home to five Royal Bengal tigers, over a hundred deer, wild boar, vultures, and several species of birds. For his remarkable initiative, the Jawaharlal Nehru University invited Payeng in 2012 on Earth Day and honoured him with the title of the ‘Forest Man of India’.

Question 29.
Discuss briefly the following:
(a) Catalytic converter
(b) Greenhouse gases
(c) Ecosan
Answer:
(a) A catalytic converter is a device that converts toxic gases and pollutants into less toxic pollutants by catalyzing redox reactions.

(b) Gases that trap the heat within the atmosphere are called Greenhouse gases. E.g: CO₂ and CO

(c) About 150 liters of wastewater at an average is generated by an Indian individual daily, and a large amount of it is generated from toilets. Ecological sanitation (EcoSan) is a sustainable system for handling human excreta by using dry composting toilets. EcoSan toilets not only reduce wastewater generation but also generate natural fertilizer from recycled human excreta, which forms an excellent substitute for chemical fertilizers. This method is based on the principle of recovery and recycling of nutrients from excreta to create a valuable supply for agriculture.

Question 30.
What are some solutions to toxic dumping in our oceans?
Answer:
Reducing energy usage, Eat sustainable seafood, use reusable plastic products, properly dispose of hazardous materials. Avoid littering on the beach and buying ocean-friendly products.

Question 31.
Describe how deforestation might contribute to global warming.
Answer:
Deforestation refers to the cutting down of trees. Trees are one of the major sources of CO₂ uptake. CO₂ is a major greenhouse gas. If there is large-scale destruction of forests, the level of CO₂ will be increased leading to global warming.

Question 32.
How does forest conservation help to reduce air pollution?
Forests decrease the CO₂ level and increase the oxygen level of the atmosphere. Particulate matters in the air will settle down on leaves. Thus they help to reduce water pollution.

12th Bio Zoology Guide Environmental Issues Additional Important Questions and Answers

12th Bio Zoology Guide Environmental Issues One Mark Questions and Answers

Question 1.
The gaseous envelope which surrounds the Earth is called…………………
(a) Stratosphere
(b) Atmosphere
(c) Troposphere
(d) Ozonosphere
Answer:
(b) Atmosphere

Question 2.
…………….. are the major causes of CO pollution in large cities and towns.
(a) Fossil fuels
(b) Ocean
(c) Deforestation
(d) Automobiles
Answer:
(d) Automobiles

Question 3
…………… and ………… are the major cause of acid rain
(a) Sulphur dioxide and Hydrogen peroxide
(b) Hydrogen peroxide and Sulphuric acid
(c) Hydrochloride and sulphur dioxide
(d) Sulphur dioxide and Nitrogen oxide
Answer:
(d) Sulphur dioxide and Nitrogen oxide.

Question 4.
What is the name of the app published by the Central Pollution Control Board that provides updates on AQI
(a) Hamear
(b) Jhoan
(c) Sameer
(d) Industan
Answer:
(c) Sameer

Question 5.
PAN stands for ……………..
(a) Peroxy acetic nitrogen
(b) Perchloro acetate
(c) Peroxyacetyl nitrate
(d) Peractyl nitroxide
Answer:
(d) Peroxyacetyl nitrate

Question 6.
Average human consumption of oxygen per day is ……………….
(a) 280 L
(b) 550 L
(c) 620L
(d) 730L
Answer:
(b) 550 L

Question 7.
Name the famous flagship programme of Central Government launched to achieve pollution abatement and rejuvenation of River Ganga.
Answer:
Namami Gange

Question 8.
The intensity of noise is measured in…………..
(a) Dobson
(b) Hertz
(c) Decibel
(d) Frequency
Answer:
(c) Decibel

Question 9.
According to noise pollution rules 2000, the permissible level of noise in the commercial area is………….. during the day and …………. during nitght.
(a) 55 db, 65 db
(b) 65 db, 55 db
(c) 70 db, 60 db
(d) 75 db, 65 db
Answer:
(b) 65 db, 55 db

Question 10.
Which is not a physical method of wastewater treatment.
(i) Floatation
(ii) Filtration
(iii) Phydro remediation
(iv) Oxidation
(a) i and iii
(b) ii and iv
(c) i and ii
(d) iii and iv
Answer:
(d) iii and iv

Question 11.
Match List I with List II

Answer:
a – iv, b – i, c – ii, d – iii

Question 12.
Which is not a method of disposal of radioactive waste.
(a) Dilute and dispense
(b) Delay and decay
(c) Recycle and reuse
(d) Limit generation
Answer:
(e) Recycle and reuse

Question 13.
E – Wastes are basically…………………..
(a) Poly iodinated biphenyl based compounds
(b) Polychlorinated biphenyl based compounds
(c) Polyhydroxy biphenyl based compounds
(d) Poly acetyl biphenyl based compounds
Answer:
(c) Polyhydroxy biphenyl based compounds

Question 14.
Identify the correct statement indicating 4Rs of treating water.
(a) Regenerate, Reduce, Reuse and Recycle
(b) Refuse, Reduce, Rejenuvate, and Reuse
(c) Redeem, Refuse, Rejenuvate and Reduce
(d) Refuse, Reduce, Rescue and Recycle.
Answer:
(d) Refuse, Reduce, Rescue and Recycle.

Question 15.
UN conference on Sustainable development in 2012 was held at ……………….
(a) Ruanda
(b) Rio de Janeiro
(c) Geneva
(d) Stockholm
Answer:
(b) Rio de Janeiro

Question 16.
The molecular formula for ozone is………………
(a) O₂
(b) O₄
(c) O₃
(d) O₇
Answer:
(c) O₃

Question 17.
World Ozone Day was observed on……………..
(a) September 16^th
(b) October 12^th
(e) December 1^st
(d) August 18^th
Answer:
(a) September 16th

Question 18.
Who is the leader of Chipko Movement?
Answer:
Sunderlal Bahuguna

Question 19.
I am called the Forest Man of India? Who am I?
Answer:
Jadav Payeng

Question 20.
Identify the incorrect statement.
(i) EcoSan toilets is a sustainable way of handling human excreta by using dry composting toilets
(ii) It reduces wastewater generation
(iii) It is based on the recovery and recycling of nutrients from excreta
(iv) EcoSan toilets are used in several parts of India and Srilanka.
(a) i and ii only
(b) iii and iv only
(c) all the above
(d) none of the above
Answer:
(d) none of the above

Question 21.
What is the name of the action plan for sustainable development framed at the Rio conference in 1992?
(a) Action 21
(b) Declaration 21
(c) Protocol 21
(d) Agenda 21
Answer:
(d) Agenda 21

Question 22.
Eutrophication is a result of…………..
(a) Agricultural and sewage waste
(b) Vehicle emission
(c) Pesticides
(d) Industrial effluent
Answer:
(c) Pesticides

Question 23.
BOD stands for………………
(a) Biological Oxidation Demand
(b) Biotic Oxygen Deficient
(c) Biological Oxygen Deficit
(d) Biochemical Oxidation Deficit
Answer:
(a) Biological Oxidation Demand

Question 24.
The stratosphere is mainly depleted by…………..
(a) Excess CO
(b) CFC’_s
(c) Ozone
(cl) Excess CO₂
Answer:
(b) CFC’_s

Question 25.
Treatment of sewage involves
(a) Floatation, Filtration and Sedimentation of suspended particles
(b) Aerating it for bacterial action
(c) Removal of nitrates and phosphates
(d) All of the above
Answer:
(d) All of the above

Question 26.
Assertion (A): The ozone layer protects the UV rays entering the Earth.
Reason (R): UV rays may cause melanoma
(a) A is right R is wrong
(b) A is the wrong R is right
(c) Both A and R are correct
(d) R explains A
Answer:
(d) R explains A

Question 27.
Assertion (A): Evolution of Greenhouse gases leads to Global warming
Reason (R): The energy released by the greenhouse gases move away from the atmospheric surface
(a) A is right R is wrong
(b) A is wrong R is right
(c) Both A and R are correct
(d) R explains A
Answer:
(a) A is right R is wrong

Question 28.
Statement (1): Incomplete combination of fossil fuels releases CO
Statement (2): CO is a GHG
(a) Statement 1 is true, statement 2 is false
(b) Statement 1 is false, statement 2 is true
(c) Both Statements 1 and 2 are true
(d) Both statements 1 and 2 are false
Answer:
(c) Both statements 1 and 2 are true.

Question 29.
Statement (1): The intensity of noise is measured in the dobson (dB) unit.
Statement (2): Noise provides immense bliss.
(a) Statement 1 is true, statement 2 is false
(b) Statement 1 is false, statement 2 is true
(c) Both Statements 1 and 2 are true
(d) Both statements 1 and 2 are false
Answer:
(d) Both statements 1 and 2 are false.

Question 30.
The threshold of noise level is ………………
Answer:
120db

Question 31
Chemicals used in agriculture as fertilizer and pesticides are generally called as
Answer:
Agrochemicals

12th Bio Zoology Guide Environmental Issues Two MarkS Questions and Answers

Question 1.
Define pollution.
Answer:
Pollution is any undesirable change in the physical, chemical and biological characteristics of the environment due to natural causes and human activities.

Question 2.
What is air pollution?
Answer:
The alterations or changes in the composition of the Earth’s atmosphere by natural or human activities (anthropogenic factors) are referred as Air Pollution.

Question 3.
Classify air pollutants.
Answer:
Air pollutants can be

• discharge of dusts or particulate matter (PM: 2.5 and 10)
• discharge of gases (SO_x, NO_x, CO and CO₂)

Question 4.
What are particulate matters? Give examples.
Answer:
Particulate matters are tiny particles of solid matter suspended in a gas or liquid. Combustion of fossil fuels, fly ash produced in thermal power plants, forest fires, asbestos mining units and cement factories are the main sources of particulate matter pollution.

Question 5.
Name the secondary pollutant in photochemical smog. Mention its adverse effect.
Answer:
Peroxyacetyl nitrate (PAN) is a secondary pollutant present in photochemical smog. It is thermally unstable and decomposes into phenoxyethanol radicals and nitrogen dioxide gas causing eye irritation.

Question 6.
How ozone hole is developed?
Answer:
Ozone depletion: Thinning of the stratospheric ozone layer is known as ozone depletion. Such depletion causes the ‘ozone hole’, resulting in the poor screening of the harmful UV rays and an increase in incidences of skin cancer. Some of the common agents that deplete ozone are CFC_s.

Question 7.
List out the greenhouse gases. (Any four)
Answer:
a) CO₂
b) CFC_s
c) Methane
d) Nitrous oxide

Question 8.
What is Acid rain?
Answer:
Acid rain is a form of precipitation that contains acidic components, such as sulfuric acid or nitric acid. It damages trees, crops and harms marine animals (coral reefs), and induces corrosion.

Question 9.
Point out any two non-point sources of water pollution.
Answer:

• Agriculture chemical runoff.
• Dumping of plastics in water bodies.

Question 10.
How will you define noise pollution?
Answer:
Sound that is unwanted and undesirable or can disrupt one’s quality of life is called noise. When there is a lot of ‘noise’ in the environment, it is termed Noise Pollution.

Question 11.
‘Expand USEPA and MOEFCC.
Answer:
USEPA: the United States Environmental Protection Agency.
MOEFCC: Ministry of Environment, Forest, and Climate Change.

Question 12.
What are agrochemicals?
Answer:
Chemicals which are used in agriculture for the growth of plants and pest control are called agrochemicals or agrichemicals.

Question 13.
Which organism is affected by colony collapse syndrome? What will be the impact?
Answer:
Colony collapse syndrome in Honey bees due to pesticides/herbicides can lead to the destruction of hives and lower agricultural productivity.

Question 14.
Define the term eutrophication.
Answer:
When run-off from land containing nutrients reaches water bodies like lakes, it results in a dense growth of plant life. This phenomenon is called Eutrophication.

Question 15.
What is meant by accelerated eutrophication?
Answer:
Pollutants from anthropogenic activities like effluents from the industries and homes can radically accelerate the aging process. This phenomenon is known as Cultural or Accelerated Eutrophication.

Question 16.
List out the chemical methods of wastewater treatment.
Answer:
Chemical methods of wastewater treatment include:

• Generation of insoluble solids.
• Produce an insoluble gas.
• Produce biologically degradable substances from a non-biodegradable substance.
• Oxidize or reduce to produce a non-objectionable substance.

Question 17.
Expand RZWT and DEWATS.
Answer:
RZWT: Root Zone Wastewater Treatment
DEWATS: Decentralized Wastewater Treatment System

Question 18.
What are the medical wastes?
Answer:
Any kind of waste that contains infectious material generated by hospitals, laboratories, medical research centers, Pharmaceutical companies, and Veterinary clinics is called medical wastes.

Question 19.
What does 4R stand for?
Answer:
‘4R’ refers to Refuse, Reduce, Reuse, and Recycle methods to manage plastic waste.

Question 20.
Define deforestation.
Answer:
Deforestation is the destruction of forests in order to clear the land and make it available for other uses.

12th Bio Zoology Guide Environmental Issues Three Marks Questions and Answers

Question 21.
Classify degradable pollutants.
Answer:
Based on the time taken to breakdown into their ingredients, degradable pollutants are classified as rapidly degradable (non-persistent) and slowly degradable (persistent).

a) Rapidly degradable or non-persistent pollutants: These can be broken down by natural processes. Domestic sewage and vegetable waste are examples of such pollutants.

b) Slowly degradable or persistent pollutants: These are pollutants that remain in the environment for many years in an unchanged condition and take decades or longer to degrade, as in the case of DDT.

Question 22.
What are the main sources of air pollution?
Answer:
The main sources of air pollution are:

• Transport sources – cars, buses, airplanes, trucks, and trains.
• Stationary sources – power plants, incinerators, oil refineries, industrial facilities, and factories.
• Area sources – agricultural – wood/stubble burning and fireplaces.
• Natural sources – wind-blown dust, wildfires, and volcanoes.

Question 23.
Define AQI. What will be the air quality if AQI is between 0-50?
Answer:
Air Quality Index (AQI) is a number used by government agencies to communicate to the public how polluted the air is at a given time. If AQI is between 0-50 then the quality of air is good.

Question 24.
Explain the three main sources of water pollution.
Answer:
There are three main types of sources: point sources, non-point sources, leaks, and spills. Point sources: Discharge of pollutants at specific locations through pipelines or sewers into the water body. Factory effluents, sewage, underground mines, oil wells, oil tankers, and agriculture are common point sources.

Non-point sources: Sources that cannot be traced to a single site of discharge like acid rain, dumping of the plastics in water bodies, and agricultural chemical runoff are common examples.

Leaks and Spills: This occurs mostly due to shipping collisions, offshore oil rigs, oil leakages, and discharges into the sea.

Question 25.
What will be the impact of water pollution on the ecosystem?
Answer:
Destruction of ecosystems: Ecosystems, especially aquatic systems, can be severely affected or destroyed by water pollution. Water pollutants affect existing niches and habitats and the survival of organisms. Soil fertility is affected and the system becomes uninhabitable.

Question 26.
Suggest a few ways to control water pollution.
Answer:

• Regulate or control pollutant(s) discharge at the point of generation.
• Wastewater can be pretreated by scientific methods before discharge to municipal treatment sources.
• Setting up of Sewage Treatment Plants (STP) and Effluent Treatment Plants (ETP).
• Regulate or restrict the use of synthetic fertilizers and pesticides.
• Public awareness and peoples’ involvement is essential.

Question 27.
Write a note on the ‘Namami Ganga’ programme.
Answer:
Namami Gange (National Mission for Clean Ganga) Programme is an Integrated Conservation Mission approved as the ‘Flagship Programme’ of the Union Government in June 2014 with a budget outlay of 20,000 crores to accomplish the twin objectives of effective abatement of pollution, conservation, and rejuvenation of River Ganga.

Question 28.
List out the sources of noise pollution.
Answer:
Vehicle engines, air horns, audio-video systems, trains, low flying aircraft, factory machines, sirens, motors, drillers and crushers, compressor machines, crackers, explosives, and modem supersonic transports are the common sources of noise pollution.

Question 29.
State the role of Government in controlled Noise pollution.
Answer:
Article 48-A and Article 51-A of the Constitution of India, Noise Pollution (Regulation and Control) Rules 2000, and Tamil Nadu State Environment Policy 2017 are some of the legal relief from noise pollution.

According to Noise Pollution (Regulation and Control) Rules, 2000, the permissible limit of noise in areas categorized as commercial is 65 decibels (dB) during the day and 55 dB during the night.

Question 30.
Comment on Bio-magnification of DDT.
Answer:
When non-degradable substances enter the food chain, they do not get metabolized or broken down or expelled and instead get transferred up the trophic levels of the food chain. During this process, they show an increase in concentration which is referred to as biomagnification. This results in increased toxicity and may even be lethal. This phenomenon is well established for mercury and DDT.

Question 31.
Explain the biological methods of wastewater treatment.
Answer:
Biological methods of Wastewater treatment:

• Bioremediation of wastewater includes aerobic treatment (oxidation ponds and aeration lagoons) and anaerobic treatment (anaerobic bioreactors and anaerobic lagoons).
• Phytoremediation of wastewater includes constructed wetlands, Root Zone Wastewater Treatment (RZWT), and Decentralized Waste Water Treatment System (DEWATS).

Question 32.
Write a note on organic farming.
Answer:
Organic farming is a method of a farming system which primarily aims at cultivating the land and raising crops in such a way, so as to keep the soil alive and in good health by use of organic wastes (crop, animal and farm wastes, aquatic wastes) and other biological materials along with beneficial microbes (biofertilizers) to release nutrients to crops for increased sustainable production in an eco-friendly pollution-free environment.

Question 33.
How radioactive wastes are generated?
Answer:
Radioactive wastes are generated during various operations of the nuclear power plant. Radioactive waste can be in gas, liquid, or solid form, and its level of radioactivity can vary. The waste can remain radioactive for a few hours or several months or even hundreds of thousands of years. Depending on the level and nature of radioactivity, radioactive wastes can be classified as Exempt Waste, Low and Intermediate Level Waste, and High-Level Waste.

Question 34.
What are geological repositories?
Answer:
Geological Repositories – A deep geological repository is a nuclear waste repository excavated deep within a stable geologic environment. It is suited to provide a high level of long-term isolation and containment without future maintenance. In India at Tarapur and Kalpakkam, a wet storage facility of Spent Fuel is the main mode of storage.

Question 35.
Give an account of E-waste.
Answer:
Electronic waste or e-waste describes discarded electrical electronic devices as well as any refuse created by discarded electronic devices and components and substances involved in their manufacture or use. Their disposal is a growing problem because electronic equipment frequently contains hazardous substances.

In a personal computer, for example, there may be lead (Pb) in the cathode ray tube (CRT) and soldering compound, mercury (Hg) in switches and housing, and cobalt (Co) in steel components, among other equally toxic substances. E-wastes are basically PCB (Polychlorinated biphenyl) based, which are non-degradable.

Question 36.
What will be the impact of global warming?
Answer:
Large-scale global warming will have a significant impact on people and nature. As global average temperatures rise, precipitation patterns could be, affected. Extreme wet and dry conditions can be expected (flooding and desertification). Coastal areas shall become more vulnerable to storm surges as sea level rises. Plant and animal species will migrate or disappear in response to climate change.

Global warming can directly affect the flora and fauna. This could also result in a shortage of food and even lead to a food crisis, and affect the health of the people and organisms.

Question 37.
Explain the process of Ozone formation in the atmosphere.
Answer:
Ozone is found in the layer of the atmosphere called the Stratosphere. It acts as a protective covering that absorbs ultraviolet (UV) radiation from the Sun. The ozone molecule (O₃) consists of three oxygen atoms. It is formed when atmospheric oxygen (O₂) on exposure to solar radiation breaks into two oxygen atoms; each atom then joins up with a single oxygen atom. The ozone molecule is unstable. It soon decays again to form molecular oxygen. This cycle is a continuous process in the upper reaches of the stratosphere.

Question 38.
How Ozone depletion can be controlled?
Answer:
Ozone layer depletion can be controlled by

•  Phase down or ban the use of CFCs (CFC free refrigerants).
• Minimizing the use of chemicals such as halons and halocarbons.
• Creating awareness about ozone-depleting agents.

Question 39.
What will be the effect of Ozone depletion on Earth?
Answer:
UV rays may penetrate deep into the skin and can lead to premature skin aging and wrinkling of the skin; suppression of the immune system, skin cancer (melanoma), and chronic effects leading to eye damage. DNA damage can result from free radicals and reactive oxygen and photons can damage the DNA itself.

Question 40.
Write a brief note on the Chipko movement.
Answer:
The Bishnois, who are known conservators of their forest, were an inspiration to many people’s participatory movements for Environmental protection in India. The Chipko movement resisted the destruction of forests of India in the 1970s. Sunderlal Bahuguna was the leader of this movement. People in the movement hugged the trees and prevented the felling of trees by contractors.

Question 41.
What do you mean by Ecosan Toilets?
Answer:
About 150 liters of wastewater at an average is generated by an Indian individual daily, and a large amount of it is generated from toilets. Ecological sanitation (EcoSan) is a sustainable system for handling human excreta by using dry composting toilets. clean toilets not only reduce wastewater generation but also generate natural fertilizer from recycled human excreta, which forms an excellent substitute for chemical fertilizers. This method is based on the principle of recovery and recycling of nutrients from excreta to create a valuable supply for agriculture. ‘EcoSan’ toilets are being used in several parts of India and Sri Lanka.

12th Bio Zoology Guide Environmental Issues Five Marks Questions and Answers

Question 42.
List out the effects of air pollution.
Answer:

• Affects all organisms as they depend on the atmosphere for respiration.
• Causes irritation in the throat, nose, lungs, and eyes. It causes breathing problems and aggravates existing health conditions such as emphysema and asthma.
• Contaminated air reduces the body’s defense mechanism and decreases the body’s capacity to fight other infections in the respiratory system.
• Frequent exposure to polluted air increases the risk of cardiovascular diseases. Breathing air that is filled with fine particulate matter can induce hardening of the arteries, triggering cardiac arrhythmia, or even a heart attack.’
• People who exercise outdoors can sometimes be susceptible to the adverse effects of air pollution because it involves deeper and faster breathing. Hence it is advisable to walk or jog in the mornings in places with ample tree cover.
• Gas leaks can be lethal or affect the quality of air in the affected area.
• CO in the atmosphere interferes with O₂ transport since hemoglobin has a greater affinity for carbon monoxide. At low concentration, it causes headaches and blurred vision. In higher concentration, it can lead to coma and death.

Question 43.
How air pollution can be controlled?
Answer:
Certain measures help to remove pollutants, reduce their presence or prevent their entry into the atmosphere.

• Trees are the best remedy for urban particulate and gaseous pollution.
• Forests act as carbon sinks and lungs of the planet.
• Catalytic converters in vehicles help to reduce polluting gases drastically.
• Diesel exhaust filters in automobiles cut particulates.
• Electrostatic precipitators reduce the release of industrial pollutants.
• Cost-effective air pollution treatment systems like indoor plants and high-performance biofilters can improve indoor air quality.

Question 44.
Point out the effects of Noise pollution.
Answer:

• According to the USEPA (the United States Environmental Protection Agency), there are direct links between noise and health. Heart disease, high blood pressure, stress-related illness, sleep disruption, hearing loss (deafness), and productivity loss are the problems related to noise pollution.
• Increased stress and tension, nervousness, irritability, anxiety, depression, and panic attacks.
• Peptic ulcer, severe headache, and memory loss.
• Marine animals are affected by noise pollution from offshore activities and port activities.
• Firecrackers frighten animals. Birds are often affected by increased air traffic.

Question 45.
How agrochemicals affect the ecosystem?
Answer:

• May kill beneficial bacteria and soil organisms.
• Can cause eutrophication in water bodies.
• Fect aquatic animals and their productivity.
• Pesticide containing water, even in trace quantities is unfit for human consumption.
• Particles (aerosols) and residues of these chemicals cause air pollution.
• Inhalation of contaminated air can cause respiratory problems.
• Consumption can lead to poisoning, side effects, and after-effects.
• Chemicals can cause skin rashes and irritation of the eyes.
• Many of these chemicals are reported to be carcinogenic.
• They can trigger hormonal disorders and neurotoxicity.
• Beneficial insects and animals can be affected.

Question 46.
Categorize the solid waste and its sources.
Answer:

Question 47.
How radioactive waste is managed? Suggest a few methods of disposal of radioactive ‘ waste.
Answer:
Radioactive waste management involves the treatment, storage, and disposal of liquid, airborne, and solid effluents from the nuclear industry.
Methods of disposal of radioactive wastes are:

• Limit generation – Limiting the generation of waste is the first and most important consideration in managing radioactive wastes.
• Dilute and disperse – For wastes having low radioactivity, dilution, and dispersion are adopted.
• Delay and decay – Delay and decay is frequently an important strategy because much of the radioactivity in nuclear reactors and accelerators are very short-lived.

Concentrate and confine process – Concentrating and containing is the objective of treatment activities for longer-lived radioactivity. The waste is contained in corrosion-resistant containers and transported to disposal sites. Leaching of heavy metals and radionuclides from these sites is a problem of growing concern.

Higher Order Thinking Skills (HOTs) Questions
Question 1.
Complete the following:
(a) Smoke + ______ = Smog
Answer:
Fog

Question 2.
What is the unit of measuring ozone thickness?
Answer:
Dobson Unit

Question 3.
List any four adverse effects of noise.
Answer:
High blood pressure Stress-related ailments Sleep disruption Hearing impairment

Question 4.
Name the scientist who discovered the ozone layer
Answer
Charles Fabry and Henri Buisson

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 6 Evolution Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 6 Evolution

12th Bio Zoology Guide Evolution Text Book Back Questions and Answers

Question 1.
The first life on Earth originated …………………..
(a) in air
(b) on land
(c) in water
(d) on mountain
Answer:
(c) in water

Question 2.
Who published the book “Origin of species by Natural Selection” in 1859?
(a) Charles Darwin
(b) Lamarck
(c) Weismann
(d) Hugo de Vries
Answer:
(a) Charles Darwin

Question 3.
Which of the following was the contribution of Hugo de Vries?
(a) Theory of mutation
(b) Theory of natural Selection
(c) Theory of inheritance of acquired characters
(d) Germplasm theory
Answer:
(a) Theory of mutation

Question 4.
The wings of birds and butterflies is an example of………………………………..
(a) Adaptive radiation
(b) convergent evolution
(c) divergent evolution
(d) variation
Answer:
(b) Convergent evolutions

Question 5.
The phenomenon of “Industrial Melanism” demonstrates……………………….
(a) Natural selection
(b) induced mutation
(c) reproductive isolation
(d) geographical isolation
Answer:
(a) Natural selections

Question 6.
Darwin’s finches are an excellent example of……………………….
(a) connecting links
(b) seasonal migration
(c) adaptive radiation
(d) parasitism
Answer:
(c) Adaptive radiations

Question 7.
Who proposed the Germplasm theory?
(a) Darwin
(b) August Weismann
(c) Lamarck
(d) Alfred Wallace
Answer:
(b) August Weismann

Question 8.
The age of fossils can be determined by
(a) electron microscope
(b) weighing the fossils
(c) carbon dating
(d) analysis of bones
Answer:
(c) Carbon dating

Question 9.
Fossils are generally found in
(a) igneous rocks
(b) metamorphic rocks
(c) volcanic rocks
(d) sedimentary rocks
Answer:
(d) Sedimentary rocks

Question 10.
Evolutionary history of an organism is called……………………….
(a) ancestry
(b) ontogeny
(c) phylogeny
Answer:
(c) Phylogeny

Question 11.
The golden age of reptiles was……………………….
(a) Mesozoic era
(b) Cenozoic era
(c) Paleozoic era
(d) Proterozoic era
Answer:
(a) Mesozoic era

Question 12.
Which period was called “Age of fishes”?
(a) Permian
(b) Triassic
(c) Devonian
(d) Ordovician
Answer:
(c) Devonian

Question 13.
Modem man belongs to which period?
(a) Quaternary
(b) Cretaceous
(c) Silurian
(d) Cambrian
Answer:
(a) Quaternary

Question 14.
The Neanderthal man had the brain capacity of………………………..
(a) 650 – 800cc
(b) 1200cc
(c) 900cc
(d) 1400cc
Answer:
(d) 1400cc

Question 15.
List out the major gases seems to fie found in the primitive Earth.
Answer:
CO₂, NH₃, UV and Water vapour

Question 16.
Explain the three major categories in which fossilization occur.
Answer:
(i) Actual remains is the most common method of fossilization. When marine animals die, their hard parts such as bones and shells, etc. are covered with sediments and are protected from further deterioration. They get preserved as such as they are preserved in vast ocean the salinity in them prevents decay. The sediments become hardened to form definite layers or strata. For example, Woolly Mammoth that lived 22 thousand years ago were preserved in the frozen coast of Siberia as such. Several human beings and animals living in die ancient city of Pompeii were preserved intact by volcanic ash which gushed out from Mount Vesuvius.

(ii) Petrifaction – When animals die the original portion of their body may be replaced molecule for a molecule by minerals and the original substance being lost through disintegration. This method of fossilization is called petrifaction. The principal minerals involved in this type of fossilization are iron pyrites, silica, calcium carbonate, and bicarbonates of calcium and magnesium.

(iii) Natural moulds and casts – Even after disintegration, the body of an animal might leave indelible impression on the soft mud which later becomes hardened into stones. Such impressions are called moulds. The cavities of the moulds may get filled up by hard minerals and get fossilized, which are called casts. Hardened faecal matter termed as coprolites, occur as tiny pellets. Analysis of the coprolites enables us to understand the nature of diet, the prehistoric animals thrived.

Question 17.
Differentiate between divergent evolution and convergent evolution with one example for each.
Answer:
Divergent Evolution :
Divergent evolution is a result of homology. Erg: The wings of bird and the forelimbs of human both are homologous structures modified according to functions. In birds, it is used for flight and in humans used for writing and other purposes.

Convergent Evolution :
Convergent evolution is a result of analogy, E.g: Root modification in sweet potato, and stem modification in potato are analogous structures both performing same function i.e., storage,

Question 18.
How does Hardy-Weinberg’s expression (p² + 2pq + q² = 1) explain that genetic equilibrium is maintained in a population? List any four factors that can disturb the genetic equilibrium.
Answer:
The allele frequencies in a population are stable and are constant from generation to generation in the absence of gene flow, genetic drift, mutation, recombination and natural selection. If a population is in a state of Hardy Weinberg equilibrium, the frequencies of alleles and genotypes or sets of alleles in that population will remain same over generations. Evolution is a change in the allele frequencies in a population over time. Hence population in Hardy Weinberg is not evolving.

Suppose we have a large population of beetles, (infinitely large) and appear in two colours ’ dark grey (black) and light grey, and their colour is determined by ‘A’ gene. ‘AA’ and ‘Aa’ beetles are dark grey and ‘aa’ beetles are light grey. In a population let’s say that ‘ A’ allele has frequency (p) of 0.3 and ‘a’ allele has a frequency (q ) of 0.7. Then p+q= 1.

If a population is in Hardy Weinberg equilibrium, the genotype’frequencycan be estimated by Hardy Weinberg equation.

(p + q)² = p² + 2pq + q²
p² = frequency of AA
2pq= frequency of Aa
q² = frequency of aa
p = 0.3, q = 0.7 then,
p² = (0.3)2 = 0.09 = 9 %AA
2pq = 2(0.3) (0.7) = 0.42 = 42 % Aa
q² = (0.7)2 0.49 = 49 % aa’

Hence the beetle population appears to be in Hardy- Weinberg equilibrium. When the beetles in Hardy- Weinberg equilibrium reproduce the allele and genotype frequency in the next generation would be: Let’s assume that the frequency of ‘A’ and ‘a’ allele in the pool of gametes that make the next generation would be the same, then there would be no variation in the progeny. The genotype frequencies of the parent appears in the next generation.
(i.e. 9% AA, 42% Aa and 49% aa).

If we assume that the beetles mate randomly (selection of male gamete and female gamete in the pool of gametes), the probability of getting the offspring genotype depends on the genotype of the combining parental gametes.

Question 19.
Explain how mutations, natural selection and genetic drift affect Hardy Weinberg equilibrium.
Answer:
Natural selection occurs when one allele (or combination of alleles of differences) makes an organism more or less fit to survive and reproduce in a given environment. If an allele reduces fitness, its frequencies tend to drop from one generation to the next.

The evolutionary path of a given gene (i.e) how its allele’s change in frequency in the population across generation, may result from several evolutionary mechanisms acting at once. For example, one gene’s allele frequencies might be modified by both gene flow and genetic drift, for another gene, mutation may produce a new allele, that is favoured by natural selection.

Genetic drift / Sewall Wright Effect is a mechanism of evolution in which allele frequencies of a population change over generation due to chance (sampling error). Genetic drift occurs in all population sizes, but its effects are strong in a small population. It may result in a loss of some alleles (including beneficial ones) and fixation of other alleles. Genetic drift can have major effects, when the population is reduced in size by natural disaster due to bottle neck effect or when a small group of population splits from the main population to form a new colony due to founder’s effect.

Although mutation is the original source of all genetic variation, mutation rate for most organisms is low. Hence new mutations on allele frequencies from one generation to the next is usually not large.

Question 20.
How did Darwin explain fitness of organisms?
Answer:
Organisms struggle for food, space and mate. As these become a limiting factor, competition exists among the members of the population. Darwin denoted struggle for existence in three ways Intra specific struggle between the same species for food, space and mate.
Interspecific struggle with different species for food and space.

Struggle with the environment to cope with the climatic variations, flood, earthquakes and drought, etc.
According to Darwin, nature is the most powerful selective force. He compared origin of species by natural selection to a small isolated group. Darwin believed that the struggle for existence resulted in the survival of the fittest. Such organisms become better adapted to the changing environment.

Question 21.
Mention the main objections to Darwinism.
Answer:
Some objections raised against Darwinism were Darwin failed to explain the mechanism of variation.

1. Darwinism explains the survival of the fittest but not the arrival of the fittest.
2. He focused on small fluctuating variations that are mostly non-heritable.
3. He did not distinguish between somatic and germinal variations.
4. He could not explain the occurrence of vestigial organs, overspecialization of some organs like large tusks in extinct mammoths and oversized antlers in the extinct Irish deer, etc.

Question 22.
Taking the example of Peppered moth, explain the action of natural selection. What do you call the above phenomenon?
Answer:
Natural selection can be explained clearly through industrial melanism. Industrial melanism is a classical case of Natural selection exhibited by the peppered moth, Bistort betularia. These were available in two colours, white and black. Before industrialization peppered moth both white and black coloured were common in England. Pre-industrialization witnessed white colpured background of the wall of the buildings hence the white coloured moths escaped from their predators.

Post industrialization, the tree trunks became dark due to smoke and soot let out from the industries. The black moths camouflaged on the dark bark of the trees and the white moths were easily identified by their predators. Hence the dark coloured moth population was selected and their number increased when compared to the white moths. Nature offered positive selection pressure to the black coloured moths. The above proof shows that in a population, organisms that can adapt will survive and produce more progenies resulting in increase in population through natural selection.

Question 23.
Darwin’s finches and Australian marsupials are suitable examples of adaptive radiation – Justify the statement.
Answer:
Darwin’s finches are the birds whose common ancestor arrived on the Galapagos about 2 million years ago. During that time, Darwin’s finches have evolved into 14 recognized species differing in body size, beak shape and feeding behavior. Changes in the size and form of the beak have enabled different species to utilize different food resources such as insects, seeds, nectar from cactus flowers and blood from iguanas, all driven by Natural selection. Genetic variation in the ALX1 gene in the DNA of Darwin finches is associated with variation in the beak shape. Mild mutation in the ALX1 gene leads to phenotypic change in the shape of the beak of the Darwin finches.

Marsupials in Australia and placental mammals in North America are two subclasses of mammals they have adapted in similar way to a particular food resource, locomotory skill or climate. They were separated from the common ancestor more than 100 million years ago and each lineage continued to evolve independently.

Despite temporal and geographical separation, marsupials in Australia and placental mammals in North America have produced varieties of species living in similar habitats with similar ways of life. Their overall resemblance in shape, locomotory mode, feeding and foraging are superimposed upon different modes of reproduction. This feature reflects their distinctive evolutionary relationships.

Over 200 species of marsupials live in Australia along with many fewer species of placental mammals. The marsupials have undergone adaptive radiation to occupy the diverse habitats in Australia, just as the placental mammals have radiated across North America.

Question 24.
Who disproved Lamarck’s Theory of acquired characters? How?
Answer:
Lamarck’s “Theory of Acquired characters” was disproved by August Weismann who conducted experiments on mice for twenty generations by cutting their tails and breeding them. All mice bom were with tail. Weismann proved that change in the somatoplasm will not be transferred to the next generation but changes in the germplasm will be inherited.

Question 25.
How does the Mutation Theory of De Vries differ from Lamarck and Darwin’s view on the origin of new species?
Answer:
According to de Vries, sudden and large variations were responsible for the origin of new species, whereas Lamarck and Darwin believed in gradual accumulation of all variations as the causative factors in the origin of new species.

Question 26.
Explain stabilizing, directional, and disruptive selection with examples.
Answer:
i. Stabilising selection (centripetal selection): This type of selection operates in a stable environment as shown in fig. The organisms with average phenotypes survive whereas the extreme individuals from both ends are eliminated. There is no speciation but the phenotypic stability is maintained within the population over a generation. For example, measurements of sparrows that survived the storm clustered around the mean, and the sparrows that failed to survive the storm clustered around the extremes of the variation showing stabilizing selection.

ii. Directional Selection: The environment” which undergoes gradual change is subjected to directional selection, as shown in fig. This type of selection removes the individuals from one end towards the other end of phenotypic distribution. For example, size differences between male and female sparrows. Both male and female look alike externally but differ in body weight. Females show directional selection in relation to body weight.

iii. Disruptive selection: (centrifugal selection) When homogenous environment changes into heterogenous environment this type of selection is operational as shown in fig. The organisms of both the extreme phenotypes are selected, whereas individuals with average phenotype are eliminated. This results in splitting of the population into subpopulation/species. This _ is a rare form of selection but leads to formation of two or more different species. It is also ( called adaptive radiation. (E.g:) Darwin’s finches beak size in relation to seed size inhabiting Galapagos islands. Group selection and sexual selection are other types of selection. The two major group selections are Altrusim and Kin selection.

Question 27.
Rearrange the descent in human evolution.
Answer:
Australopithecus → Homo erectus → Homo sapiens → Ramapithecus → Homo habilis
Ramapithecus → Australopithecus → Homo habilis → Homo erectus → Homo sapiens

Question 28.
Differentiate between the eating habit and brain size of Australopithecus and Ramapithecus.
Answer:

Question 29.
How does the Neanderthal man differ from the modern man in appearance?
Answer:
Neanderthal man differ from the modem human in having semierect posture, flat cranium, sloping forehead, thin large orbits, heavy brow ridges, protruding jaws and no chin.

Question 30.
Mention any three similarities found common in Neanderthal man and Homo sapiens.
Answer:
Common characters showed by Neanderthal man and Homo sapiens are:

• Usage of Fire
• Burying of dead bodies
• Protecting themselves from predators

Question 31.
According to Darwin, organic evolution is due to………………….
(а) Intraspecific competition
(b) Interspecific competition
(c) Competition within closely related species. ‘
(d) Reduced feeding efficiency in one species due to the presence of interfering species.
Answer:
(d) Reduced feeding efficiency in one species due to the presence of interfering species.

Question 32.
A population will not exist in Hardy – Weinberg equilibrium if……………….
(a) Individuals mate selectively
(b) There are no mutations
(c) There is no migration
(d) The population is large
Answer:
(a) Individuals mate selectively

12th Bio Zoology Guide Evolution Additional Important Questions and Answers

12th Bio Zoology Guide Evolution One Mark Questions and Answers

Question 1.
Identify the incorrect statement in concern with Neanderthals.
(a) Neanderthal human were found in Germany.
(b) They possessed flat cranium.
(c) They used to bury their dead.
(d) Their brain size is of 650 – 800 cc.
Answer:
(d) Their brain size is of 650 – 800 cc

Question 2.
Which of the following statement does not satisfy Hardy Weinberg principle?
(a) A population undergoing random mating
(b) Small-sized population
(c) Population where there is no mutation or gene flow
(d) Absence of natural selection
Answer:
(b) Small-sized population

Question 3.
Match column I with column II

(a) a – iii b – ii c – iv d – i
(b) a – iv b – iii c – i d – ii .
(c) a – iii b – iv c – i d – ii
(d) a – ii b – iii c – i d – iv
Answer:
(a) a – iii b-ii c-iv d-i

Question 4.
Placental mammals develop during……..
(a) Eocene
(b) Oligocene
(c) Pliocene
(d) Paleocene
Answer:
(d) Paleocene

Question 5.
Identify the correct sequence from oldest to youngest
(а) Cambrian → Permian → Devonian → Silurian → Ordovician
(б) Permian → Silurian → Devonian → Ordovician → Cambrian
(c) Permian → Devonian → Silurian → Cambrian → Ordovician
(d) Cambrian → Ordovician → Silurian → Devonian → Permian
Answer:
(d) Cambrian → Ordovician → Silurian Devonian → Permian

Question 6.
Match the scientists with their terminologies used

(a) a – iii b-iv c-ii d- i
(b) a – ii b-iv, c -i d- Hi
(c) a – Hi b – i c- iv d- ii
(d) a-i b-iv c-iii d-ii
Answer:
(b) a – ii b-iv c-i d – iii

Question 7.
Anatomical structures that have similar functions but not similar structures are called
(a) Homologous structures
(b) Vestigial structures
(c) Analogous structures
(d) Generalized structures
Answer:
(c) Analogous structures

Question 8.
Who propounded the theory of recapitulation?
(a) Ernst Von Haeckel
(b) Charles Darwin
(c) Thomas Huxley
(d) Oparin
Answer:
(c) Ernst Von Haeckel

Question 9.
Mammal in human male is…………….
(a) Atavistic organ
(b) Rudimentary Organ
(c) Vestigial organ
(d) Homologous structure
Answer:
(c) Vestigial organ

Question 10.
Which of the following is/are not examples of analogous structure
(a) Wings of Birds and Bats
(b) Wings of Birds and Insects
(c) Thom of Bougainvillea and Tendril of cururbita
(d) Flippers of Penguins and Dolphins
(i) a, b, c (ii) a and c (iii) b and d (iv) All the above
Answer:
(ii) a and c

Question 11.
identify the mismatched pairs
(a) Thom of Bougainvillea and Tenrdril of cucurbita – Analogy
(b) Forelimbs of whale and cat – Analogy
(c) Octopus eye & Mammalian eye – Homology
(d) Root of sweet potato & stem of potato – Homology
Answer:
(a) Thorn of Bougainvillea & Terdril of crucurbita – Analog

Question 12.
Witnesses for evolution are found in ……………….
(a) Rocks
(b) Ocean beds
(c) Fossils
(d) Desert
Answer:
(c) Fossils

Question 13.
Assertion (A): Oparin used the term coacervates
Reason (R): Coacervates are colloidal particles in an aqueous environment
(a) Both A and Rare incorrect
(b) Both A and R are correct
(c) Both A and R are correct. R explains A.
(d) A is correct R is incorrect
Answer:
(c) Both A and R are correct. R explains A.

Question 14.
According to the theory of spontaneous generation, life originated from…………….
(a) Cosmic particles
(b) Non-living materials
(c) Coacervates
(d) Sea
Answer:
(b) Non-living materials

Question 15.
Assertion (A): Hardy – Weinberg principle states that allelic frequency of a population remains constant
Reason (R) : Constancy is maintained through natural selection and mutation
(a) A is true R is false
(b) A is false R is true
(c) Both A and R are true
(d) R explains A
Answer:
(a) A is true R is false

Question 16.
Calculate the allelic frequency of Aa. frequency of 0.7
(a) 0.67
(b) 0.42
(c) 0.36
(d) 0.59
Answer:
(b) 0.42

Question 17.
Match the following Evolutionary Human

(d) a – iv b-i c – ii d – iii
ib) a – ii b – iv c – iii d – i
(c) a – ii b – iii c – iv d – i
{d) a – iii b-i c – ii d – iv
Answer:
(a) a-iv b-i c-ii d-iii

Question 18.
Genetic drift leads to………………
(a) Mutation
(b) Bottleneck effect
(c) Immigration
(d) Isolation
Answer:
(b) Bottleneck effect

Question 19.
Atavism refers to…………….
(a) Inheritance of trial by mother
(b) Inheritance of trial by father
(c) Criss-cross inheritance
(d) Inheritance of characters not shown by parents
Answer:
(d) Inheritance of characters not shown by parents

12th Bio Zoology Guide Evolution Two Marks Questions and Answers

Question 1.
State the theory of spontaneous generation.
Answer:
According to the theory of spontaneous generation or Abiogenesis, living organisms originated from non-living materials and occurred through stepwise chemical and molecular evolution over millions of years. Thomas Huxley coined the term abiogenesis.

Question 2.
List the four eras of geological time scale.
Answer:

1. Precambrian era
2. Paleozoic era
3. Mesozoic era
4. Cenozoic era

Question 3.
Which periods of the Paleozoic era are referred to as

• Age of fishes
• Invertebrates

Answer:

• Age of fishes – Devonian period
• Age of invertebrates – Cambrian period

Question 4.
Point out the epochs of Carboniferous period.
Answer:

•  Pennsylvanian
• Mississippian

Question 5.
Compare relative dating with absolute dating.
Answer:
Relative dating is used to determine a fossil by comparing it to similar rocks and fossils of known age. Absolute dating is used to determine the precise age of a fossil by using radiometric dating to measure the decay of isotopes.

Question 6.
Wing of a cockroach and the wing of parrot. What do you infer from this statement with reference to evolution?
Answer:
Both the wings of cockroach and bird are different in structure but similar in their function. Thus, they are the analogous structure that brings about convergent evolution.

Question 7.
Name the scientists who propounded the following theories.

1. Mutation theory
2. Chemical theory of evolution

Answer:

1. Mutation theory was propounded by Hugo de Vries.
2. The chemical theory of evolution was propounded by Oparin and Haldane

Question 8.
Define fossilization and mention its types.
Answer:
‘ Fossilization is the process by which plant and animal remains are preserved in sedimentary rocks. It is of three major types,

• Actual remains
• Petrifaction
• Natural moulds and casts.

Question 9.
Name the principle minerals involved in petrifaction.
Answer:
Iron pyrites, silica, calcium carbonate and bicarbonates of calcium and magnesium.

Question 10.
What is meant by petrifaction?
Answer:
When animals die the original portion of their body may be replaced molecule for molecule by minerals and the original substance being lost through disintegration. This method of fossilization is called petrifaction. The principle minerals involved in this type fossilization are iron pyrites, silica, calcium carbonate and bicarbonates of calcium and magnesium.

Question 11.
Define analogous organ with an example.
Answer:
Organisms having different structural patterns but similar function are termed as analogous structures. For example, the wings of birds and insects are different structurally but perform the same function of flight that brings about convergent evolution.

Question 12.
Mention any four organs homologous to human hand.
Answer:
Flippers of whale, wings of bat, wings of bird and forelimb of horse.

Question 13.
Thorn of Bougainvillea and tendrils of Pisum sativum represent homology. How?
Answer:
The thorn of Bougainvillea and the tendrils of Curcurbita and Pisum sativum represent homology. The thorn in former is used as a defence mechanism from grazing animals and the tendrils of latter is used as a support for climbing.

Question 14.
Which type of evolution is brought out by homologous structures and analogous structures?
Answer:
Homologous structures brings about divergent evolution. Analogous structures brings about convergent evolution.

Question 15.
What are vestigial organs? Give example.
Answer:
Structures that are of no use to the possessor, and are not necessary for their existence are called vestigial organs. Vestigial organs may be considered as remnants of structures which were well developed and functional in the ancestors, but disappeared in course of evolution due to their non-utilization. E.g: Human appendix.

Question 16.
Human appendix is a vestige. Give reason.
Answer:
Human appendix is the remnant of caecum which is functional in the digestive tract of
herbivorous animals like rabbit. Cellulose digestion takes place in the caecum of these .
animals. Due to change in the diet containing less cellulose, caecum in human became functionless and is reduced to a vermiform appendix, which is vestigial.

Question 17.
What are connecting link? Give example.
Answer:
The organisms which possess the characters of two different groups (transitional stage) are called connecting links. Example Peripatus (connecting link between Annelida and Arthropoda) Archaeopteryx (connecting link between Reptiles and Aves).

Question 18.
Name one fossilised connecting link between reptiles and Aves also one living connecting link between Annelida and Arthropoda.
Answer:
Archaeopteryx – connecting link between Reptiles and Aves.
Peripatus – Connecting link between Annelida and Arthropoda.

Question 19.
Why it is considered as a connecting link?
Answer:
Peripatus is a worm that shown the characters of both Annelidia and Arthropoda. Hence it is a connecting link between Annelida and Arthropoda. ‘

Question 20.
Atavistic organs – comment.
Answer:
Sudden appearance of vestigial organs in highly evolved organisms is called atavistic organs. Example, presence of tail in human baby is an atavistic organ.

Question 21.
Define Ontogeny and Phytogeny.
Answer:
Ontogeny refers to the life history of an individual.
Phytogeny refers to the evolutionary history of a race. ‘

Question 22.
Who proposed the theory of recapitulation? State the theory.
Answer:
Ernst Von Haeckel proposed the theory of recapitulation, which states that life history of an individual briefly repeats the evolutionary history of the race. ,

Question 23.
Name few Neo – Lamarckists.
Answer:
Cope, Osborn, Packard and Spencer.

Question 24.
Who proposed the theory of acquired characters? Also mention the scientist who disproved it.
Answer:
The theory of acquired characters was proposed by Jean Baptise de Lamarck and it was disproved by August Weismann.

Question 25.
Point out the basic principles of Darwin’s theory of evolution.
Answer:
Over production, struggle for existence, Universal occurence of variation, Survival of fittest and Natural selection.

Question 26.
Name any four Neo – Darwinists.
Answer:
Gregor Mendel, August Weismann, Russel Wallace and Heinrich.

Question 27.
Enumerate the salient features of mutation theory.
Answer:

• Mutations or discontinuous variation are transmitted to other generations.
• In naturally breeding populations, mutations occur from time to time.
• There are no intermediate forms, as they are fully fledged.
• They are strictly subjected to natural selection.

Question 28.
Who proposed Mutation theory? Name the organism on which the experiment was carried out.
Answer:
Mutation theory was put forth by Hugo de Vries. Based on the experiments in Oenothera lamarckiana (The evening primrose plant).

Question 29.
What are the basic factors of modern synthetic theory that leads to evolution?
Answer:
Gene mutation, Chromosomal mutation, Genetic recombination, Natural selection and Reproductive isolation.

Question 30.
Name the scientists who supported modern synthetic theory.
Answer:
Sewell Wright, Dobzhansky, Huxley and Simpson.

Question 31.
Define point mutation.
Answer:
Gene mutation refers to the changes in the structure of the gene. It is also called gene / point mutation. It alters the phenotype of an organism and produces variations in their offsprings.

Question 32.
Point out the factors that alters allelic frequency of a population.
Answer:
Natural selection, Genetic drift, Mutation and Geneflow

Question 33.
Mention any two differences between Homo habilis and Homo erectus
Answer:
Homo habilis :

1. The brain capacity was between 650-800 cc.
2. They were probably vegetarians.

Homo erectus :

1. The brain capacity was around 900 cc.
2. They probably ate meat

Question 34.
Write a brief note on Homo sapiens with respect to evolution.
Answer:
Homo sapiens or modem human arose in Africa some 25,000 years ago and moved to other continents and developed into distinct races. They had a brain capacity of 1300 – 1600 cc. “fhey started cultivating crops and domesticating animals.

Question 35.
Define evolution.
Answer:
The term evolution describes heritable changes in one or more characteristics of a population of species from one generation to the other.

12th Bio Zoology Guide Evolution Three Marks Questions and Answers

Question 36.
Write a short note on Big Bang theory.
Answer:
Big bang theory explains the origin of universe as a singular huge explosion in physical terms. The primitive Earth had no proper atmosphere, but consisted of ammonia, methane, hydrogen and water vapour. The climate of the Earth was extremely high. UV rays from the Sun split up water molecules into hydrogen and oxygen. Gradually the temperature cooled and the water vapour condensed to form rain. Rain water filled all the depressions to form water bodies. Ammonia and methane in the atmosphere combined with oxygen to form carbon dioxide and other gases.

Question 37.
Theory of chemical evolution states that organisms have evolved from inorganic substances. If so, what was the atmospheric condition that favoured evolution?
Answer:
The atmosphere was devoid of O₂, and with high level of CO₂ NH₀3 and UV radiations.

Question 38.
Name the periods of Mesozoic era. Also mention the flora and fauna dominates during that periods.
Answer:
Mesozoic era is divided into three periods namely Triassic, Jurassic and Cretaceous.
Dominating Fauna : Reptiles and Dinosaurs Dominating Flora : Conifers, Ferns and Ginkgon.

Question 39.
Which era is referred as Age of Mammals? What are the periods of that era? And also mention the fauna during the periods.
Answer:
Cenozoic era is called as Age of Mammals.
Tertiary and Quaternary are the two periods of Cenozoic era.
Tertiary periods marks the abundance of mammalian fauna. Quaternary period marks the beginning of human social life.

Question 40.
Write a short note on Cenozoic era.
Answer:
Cenozoic era (Age of mammals) is subdivided into two periods namely Tertiary and Quaternary. Tertiary period is characterized by abundant mammalian fauna. This period is subdivided into five epochs namely, Paleocene (placental mammals, Eocene (Monotremes except duck billed Platypus and Echidna, hoofed mammals and carnivores), Oligocene (higher placental mammals appeared), Miocene (origin of first man like apes) and Pliocene (origin of man from man like apes). Quaternary period witnesses decline of mammals and beginning of human social life.

Question 41.
Name the gaseous mixture used in Urey – Miller’s experiment. Which type of physical force is applied to generate amino acids?
Answer:
Ammonia, Methane, Hydrogen, Water vapour are the gaseous mixture allowed to circulate over electric discharge from a tungsten electrode.

Question 42.
Which is the most common methods of fossilization? Explain how it occurs.
Answer:
Actual remains – The original hard parts such as bones, teeth or shells are preserved as such in the Earth’s atmosphere. This is the most commpn method of fossilization. When marine animals die, their hard parts such as bones and shells, etc., are covered with sediments and are protected from further deterioration. They get preserved as such as they are preserved in vast ocean; the salinity in them prevents decay. The sediments become hardened to form definite layers or strata. For example, Woolly Mammoth that lived 22 thousand years ago were preserved in the frozen coast of Siberia as such. Several human beings and animals living in the ancient city of Pompeii were preserved intact by volcanic ash which gushed out from Mount Vesuvius. ,

Question 43.
What are coprolites? Mention its role in phytogeny.
Answer:
Coprolites are the hardened faecal matters occurs as small pieces. Analysing the coprolites helps to understand the nature of diet of pre-historic animals.

Question 44.
What are moulds and casts?
Answer:
Even after disintegration, the body of an animal might leave indelible impression on the soft mud which later becomes hardened into stones. Such impressions are called moulds. The cavities of the moulds may get filled up by hard minerals and get fossilized, which are called casts.

Question 45.
How will you compute the age of fossil?
Answer:
The age of fossils can be determined using two methods namely, relative dating and absolute dating. Relative dating is used to determine a fossil by comparing it to similar rocks and fossils of known age. Absolute dating is used to determine the precise age of a fossil by using radiometric dating to measure the decay of isotopes.

Question 46.
“Ontogeny recapitulates phylogeny” – comment on the statement with example.
Answer:
The embryonic stages of a higher animal resemble the adult stage of its ancestors. Appearance of pharyngeal gill slits, yolk sac and the appearances of tail in human embryos are some of the examples.

Question 47.
Biogenetic law is not universal – justify.
Answer:
The biogenetic law is not universal and it is now thought that animals do not recapitulate the adult stage of any ancestors. The human embryo recapitulates the embryonic history and not the adult history of the organisms.

Question 48.
How macro molecules like DNA and RNA play their crucial role in evolutionary history?
Answer:
Molecular evolution is the process of change in the sequence composition of molecules such as DNA, RNA and proteins across generations. It uses principles of evolutionary biology and population genetics to explain patterns in the changes of molecules.

One of the most useful advancement in the development of molecular biology is proteins and other molecules that control life processes are conserved among species. A slight change that occurs over time in these conserved molecules (DNA, RNA and protein) are often called molecular clocks. Molecules that have been used to study evolution are cytochrome c (respiratory pathway) and rRNA (protein synthesis).

Question 49.
Explain the principles of Lamarckian theory.
Answer:

i. The theory of use and disuse – Organs that are used often will increase in size and those that are not used will degenerate. Neck in giraffe is an example of use and absence of limbs in snakes is an example for disuse theory.
ii. The theory of inheritance of acquired characters – Characters that are developed during ’ the life time of an organism are called acquired characters and these are then inherited.

Question 50.
Write a note on Mutation theory.
Answer:
Hugo de Vries put forth the Mutation theory. Mutations are sudden random changes that occur in an organism that is not heritable. De Vries carried out his experiments in the Evening Primrose plant {Oenothera lamarckiana) and observed variations in them due to mutation. According to de Vries, sudden and large variations were responsible for the origin of new species whereas Lamarck and Darwin believed in gradual accumulation of all variations as the causative factors in the origin of new species.

Question 51.
What do you mean by “adaptive radiation”? Give example.
Answer:
The evolutionary process which produces new species diverged from a single ancestral form becomes adapted to newly invaded habitats is called adaptive radiation. Adaptive radiations are best exemplified in closely related groups that have evolved in relatively short time. Darwin’s finches and Australian marsupials are best examples for adaptive radiation.

Question 52.
Darwins finches are the classical examples studied for adaptive radation. Explain.
Answer:
Darwin’s finches are the birds whose common ancestor arrived on the Galapagos about 2million year ago. During that time, Darwin’s finches have evolved into 14 recognized species differing in body size, beak shape and feeding behavior. Changes in the size and form of the beak have enabled different species to utilize different food resources such as insects, seeds and nectar from cactus flowers and blood from iguanas, all driven by Natural selection. Genetic variation in the ALX1 gene in the DNA of Darwin finches is associated with variation in the beak shape. Mild mutation in the ALX1 gene leads to phenotypic change in the shape of the beak of the Darwin finches.

Question 53.
What is micro evolution?
Answer:
Microevolution (evolution on a small scale) refers to the changes in allele frequencies within a population. Allele frequencies in a population may change due to four fundamental forces of evolution such as natural selection, genetic drift, mutation and gene flow.

Question 54.
Name the major types of Natural Selection.
Answer:

1. Stabilising Selection
2. Directional Selection
3. Disruptive Selection

Question 55.
What do you mean by gene flow?
Answer:
Movement of genes through gametes or movement of individuals in (immigration) and out (emigration) of a population is referred to as gene flow. Organisms and gametes that enter the population may have new alleles or may bring in existing alleles but in different proportions than those already in the population. Gene flow can be a strong agent of evolution.

Question 56.
Give an account on Genetic drift. Mention its impact over a population.
Answer:
Genetic drift is a mechanism of evolution in which allele frequencies of a population change over generation due to chance (sampling error). Genetic drift occurs in all population sizes, but its effects are strong in a small population. It may result in a loss of some alleles (including beneficial ones) and fixation of other alleles. Genetic drift can have major effects, when the population is reduced in size by natural disaster due to bottle neck effect or when a small group of population splits from the main population to form a new colony due to founder’s effect.

Question 57.
State Hardy – Weinberg equilibrium.
Answer:
The allele frequencies in a population are stable and are constant from generation to generation in the absence of gene flow, genetic drift, mutation, recombination and natural selection.

Question 58.
Write in brief about the characters of Australian ape man.
Answer:
Australopithecus lived in East African grasslands about 5 mya and was called the Australian ape man. He was about 1.5 meters tall with bipedal locomotion, omnivorous, semi erect, and lived in caves. Low forehead, brow ridges over the eyes, protruding face, lack of chin, low brain capacity of about 350 – 450 cc, human like dentition, lumbar curve in the vertebral column were his distinguishing features.

Question 59.
Who is Cro-Magnon?
Answer:
Cro-Magnon was one of the most talked forms of modem human found from the rocks of Cro-Magnon, France and is considered as the ancestor of modem Europeans. They were not only adapted to various environmental conditions, but were also known for their cave paintings, figures on floors and walls.

12th Bio Zoology Guide Evolution Five Marks Questions and Answers

Question 60.
Explain Oparin – Haldane hypothesis on evolution.
Answer:
According to the theory of chemical evolution primitive organisms in the primordial environment of the Earth evolved spontaneously from inorganic substances and physical forces such as lightning, UV radiations, volcanic activities, etc. Oparin (1924) suggested that the organic compounds could have undergone a series of reactions leading to more complex molecules.

He proposed that the molecules formed colloidal aggregates or ‘coacervates’ in an aqueous environment. The coacervates were able to absorb and assimilate organic compounds from the environment. Haldane (1929) proposed that the primordial sea served as a vast chemical laboratory powered by solar energy. The atmosphere was oxygen free and the combination of CO₂, NH₃ and UV radiations gave rise to organic compounds.

The sea became a ‘hot’ dilute soup containing large populations of organic monomers and polymers. They envisaged that groups of monomers and polymers acquired lipid membranes and further developed into the first living cell. Haldane coined the term prebiotic soup and this became the powerful symbol of the Oparin-Haldane view on the origin of life (1924-1929). Oparin and Haldane independently suggested that if the primitive atmosphere was reducing – and if there was appropriate supply of energy such as lightning or UV light then a wide range of organic compounds can be synthesized.

Question 61.
How Urey – Miller’s experiment supports the origin of life?
Answer:
Urey and Miller (1953), paved way for understanding the possible
synthesis of organic compounds that led to the appearance of living organisms is depicted in the Fjgure In their experiment, a mixture of gases was allowed to circulate over electric discharge from an tungsten electrode. A small flask was kept boiling and the steam emanating from it was made to mix with the mixture of

gases (ammonia, methane and hydrogen) in the large chamber that was connected to the boiling Fig. 6.1 Diagrammatic representation of Urey-Miller’s water. The steam condensed to form water which ran down the ‘U’ tube. Experiment was conducted continuously for a week and the liquid was analysed. Glycine, alanine, beta alanine and aspartic acid were identified. Thus Miller’s experiments had an insight as to the possibility of abiogenetic synthesis of large amount of variety of organic compounds in nature from a mixture of sample gases in which the only source of carbon was methane. Later in similar experiments, formation of all types of amino acids, and nitrogen bases were noticed.

Question 62.
Give a detailed account of Modern Synthetic Theory.
Answer:
Sewell Wright, Fisher, Mayer, Huxley, Dobzhansky, Simpson and Haeckel explained Natural Selection in the light of Post-Darwinian discoveries. According to this theory ‘ gene mutations, chromosomal mutations, genetic recombinations, natural selection and reproductive isolation are the five basic factors involved in the process of organic evolution.

• Gene mutation refers to the changes in the structure of the gene. It is also called gene/ point mutation. It alters the phenotype of an organism and produces variations in their offsprings.
• Chromosomal mutation refers to the changes in the structure of chromosomes due to deletion, addition, duplication, inversion or translocation. This too alters the phenotype of an organism and produces variations in their offspring.
• Genetic recombination is due to crossing over of genes during meiosis. This brings about genetic variations in the individuals of the same species and leads to heritable variations.
• Natural selection does not produce any genetic variations but once such variations occur » it favours some genetic changes while rejecting others (driving force of evolution).
• Reproductive isolation helps in preventing interbreeding between related organisms.

Higher Order Thinking Skills (HO’ts) Questions

Question 1.
Name the connecting link for the following groups of organisms.
(a) Annelida and Arthropoda
(b) Reptiles and Aves
(c) Pisces and Amphibians
(d) Reptiles and Mammals
Answer:
(a) Peripatus
(b) Archeopteryx
(c) Lung fish
(d) Platypus

Question 2.
Point out any four condition under which Hardy Weinberg’s equilibrium is not attained.
Answer:

•  Selected mating
• Flow of genes (either by immigration or emmigration)
• Occurance of mutation
• Definite population size

Question 3.
Why are analogous structures a result of convergent evolution?
Answer:
Analogous structures are not anatomically similar though they perform same function.

Question 4.
Organs which are of no use to the organism is called as vestige. Name any four vestigal organs that can be noticed in your body.
Answer:
Wisdom teeth, Mammae in male, Body hair and Coccyx.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th English Guide Pdf Supplementary Chapter 6 The Never – Never Nest Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 11th English Solutions Supplementary Chapter 6 The Never – Never Nest

11th English Guide The Never – Never Nest Text Book Back Questions and Answers

B. Answer the following questions in about a paragraph of 100-150 words each:

Question 1.
Why is there a double negative in the title: The never – Never Nest? Elucidate with reasons from the play.
Answer:
Never – Never Nest is the title. It is absolutely justified because Jack and Jill, were living on a limited earning of just six pounds a month. The seed money given by Jane, as wedding gift was squandered by them in making advance payment for the house, fridge, piano etc. They continued to make EMI payments for all the items. The furniture, car and even the baby’s delivery fees was, running on EMI. There was nothing in the home they could call as completely their own.

Jack called himself the owner of the home but the EMI for the housing loan was going on and he had to continue it for many years. They are glad to be freed of the drudgery of paying rent. But they are entangled in paying EMI for the house, car, piano the bed, cot and the cozy furniture. It is very doubtful if ever the “nest” would be called a real nest. Would Jack and Jill ever become the real owners of the house and often gadgets at home is a pretty disturbing questions because Jack is borrowing every month to pay back his EMIs. Living beyond the means can never help a person to settle down in life. Such a person will have, insecurity every month.

Question 2.
Bring out the humorous elements in the play.
Answer:
“The Never – Never Nest” is a comic one-act play about a young couple. They make full use of the buy – now-pay. – later marketing system. This comedy is very relevant today, because we can. buy almost anything now on the instalment basis. The author uses humour elements.

When Jane asked about the car, Jill replied that they owned steering wheel of a car and one of the tyres and about two of the cylinders belong to them. It means the car does not belong to them. When Jane was asked to lie down by Jane She replied that she was going to trust herself in a bed that belongs to Mr Sage or Marks and Spencer or somebody.

Here the author brings out the humour element at the same time makes Jack realize his mistakes. At the end of the play, humour takes on wings when we hear that the couple had their first baby in instalment.

Question 3.
How does the play “The Never-Never Nest’ expose the harsh reality of modern living?
Answer:
In modem times, plastic currency has become popular in India, as a country known for saving for future needs; a country which is proud of the adage “save for a rainy day” has undergone drastic changes. Consumer culture has eroded into every common man’s home. Credit card has swindled the younger generation of their capacity to spend hard cash. Their future earnings are pledged for purchase of luxurious things. Modem man buys things which are heavily advertised and which are often caused by jealousy. Supermarkets, Amazon, Flipkart and other online commercial organizations encourage purchase of everything ranging to laptop, electronic items and from home appliances to undergarments and shoes using credit cards.

Popular malls, Hire purchase corporate giants like Vasanth & co, Rathna Fan house offer costly consumables with a down payment of just one rupee and the rest in easy equated monthly installments. Tempted by such fabulous offers, modem men and women lose their heads and go on a spending spree. They, like Jack and Jill, spend beyond their means.

Many of them eat into their future earnings. They buy house loans and when corporate giants like Sathyam sacks young Engineers out of jobs, they end up as chain snatchers and vehicle robbers unable to payback EMIs Or credit card monthly payments. Spending on future earnings is like issuing a post-dated cheque on a crashing bank. One must be very cautious. The best way out would be to avoid immediate gratification but save money and wait until enough money is there to purchase what one wants.

Question 4.
Jill said that they owned the steering wheel of a car, one of the tyres, two of the cylinders and leg of the sofa. What does this convey?
Answer:
Both Jack and Jill show their instant gratification for luxuries and had bought them on instalments without saving any money. Their life is based on buy – now – pay – later marketing system, they are not secure at all. Jill said that they owned the steering wheel of a car, one of the tyres, two of the sofa temporarily belong to them.

This situation tells that if anytime they would be unable to pay the instalments they might have to leave the house, which simply shows the insecurity of the luxuries of their life.

ஆசிரியரைப் பற்றி:

செட்ரிக் மவுண்ட் இங்கிலாந்து நாட்டைச் சார்ந்த நாடக ஆசிரியர். சிந்தனைகளை தூண்டக்கூடிய பல நாடகங்க ளை எழுதியுள்ளார். Twentieth“Century Lullaby”,“Tocutalongshort story short”,”Nature ablorsavacuum” என பலவற்றை படைத்துள்ளார்.

இவரின் ஓரங்க நாடகங்கள், நையாண்டி செய்வனவாகவும், அறிவார்ந்ததாகவும் இருக்கும். இந்த நாடகங்கள் வாழ்வின் பொய்மையை வெளிப்படுத்துகின்றது. அதை கண்டிக்கவும் செய்கிறது.

கதையைப் பற்றி:

சுலப தவனைகளில் பொருள் வாங்குவதனாலும், கடன் வாங்குவதனாலும் நடுத்தர மக்கள் சந்திக்கக் கூடிய இன்னல்களை இந்த கதை நமக்கு விளக்குகிறது. தேவையின் நிமிர்த்தம் கடன் பெறுவதும், தேவையில்லாத நிலையிலும் கடன் பெறுவதிலும் மாறுதல்கள் இருக்கிறது. இதைப்பற்றி தெளிவாக இக்கதையில் காண்போம்.

The Never – Never Nest Summary in Tamil

ஜாக், ஜில், அத்தை ஜேன், செவிலி (JACK, JILL, Aunt Jane, Nurse)

நியூ ஹம்ஸ்டெட் ஊரில் உள்ள ஜாக் – ஜில் ஆகியோரின் வீடு, வீட்டில் ஒரு மேசை அதன் மேல் எழுத்து பொருட்கள், இரண்டு நாற்காலிகள், திரை உயரும் போது, ஓய்வெடுக்கும் அறை காலியாக உள்ளது. ஜாக், ஜில் உள்ளே வருகிறார்கள். அவர்களின் பின்னே அத்தை Jane வருகிறார்.

JILL : ….. அப்புறமா இது தான் ஓய்வெடுக்குற இடம்….
அத்தை JANE : பிரமாதம்! பிரமாதம்! எவ்வளவு சின்ன நல்ல ரூம் அதோட நல்ல அழகான மேசை, நாற்காலி.
JACK : (நளினமாக) இது எங்களுக்கு பிடிச்சிருக்கு உட்கார நல்ல இடம். ரேடியோவும் கேட்டுக்கலாம்.
அத்தை Jane : நீங்க ரேடியோ, கார், அத்தோட பியானோ இதையெல்லாம் வாங்கிட்டீங்களா?
JACK : ஆமா அத்தை இப்பெல்லாம் ஒரு ரேடியோ கண்டிப்பா வெச்சுருக்கணும்.
JILL : ஜாக் வேலைக்கு போயிட்டா, இது கேட்க நல்லா இருக்கு, அவர் கிட்ட சொல்லி இதை சமையலறைக்குள்ள கொண்டு வரச் சொல்வேன் சமையல் செய்யும் போது, ரேடியோ கேட்பேன்.

JACK : உட்காருங்க, அத்தை. நீங்க சோர்வா இருப்பீங்க நாங்க எங்க வீடு எல்லாம் சுத்திக் காட்டிட்டோம்.
JILL : எங்க சின்ன கூட்டைப்பத்தி என்ன நினைக்கிறீங்க அத்தை.
AUNT JANE : இது ரொம்ப அழகாக இருக்கு., இந்த மேசை, நாற்காலி. அப்புறம். கார்… அப்புறமா பியோனா, பிரிட்ஜ், ரேடியோ அது என்ன…? ரொம்ப பிரமாதம்!
JACK : இது எல்லாம் எங்களுக்கு வந்தது உங்களால் தான்.
AUNT JANE : ஆமா, ஜாக். அதுதான் எனக்கு கவலையா இருக்கு.
JACK : என்ன அத்தை, கவலையா இருக்கா?
AUNT JANE : நான் கல்யாணத்தன்னிக்கு உங்களுக்கு அன்பளிப்பா, ஒரு செக் கொடுத்தேன் பாருங்க. அது இருநூறு பவுண்டு தான்! இல்லையா? நான், அதுல ரெண்டாயிரம் பவுண்டுன்னு எழுதலையே!
JILL : இல்ல ஜேன்! எப்படி உங்களுக்கு அப்படி ஒரு சந்தேகம் வந்துச்சு?

AUNT JANE : பரவாயில்ல, இருக்கட்டும். ஆனா, இன்னமும் எனக்கு ஒண்ணு புரியல. இந்த வீடு, இது ரொம்ப நல்லா இருக்கு… ஆனா, இதுக்கான வாடகை ரொம்ப அதிகமா இருக்குமே!
JACK : வாடகையா? இல்ல.. இல்ல.. நாங்க வாடகை கொடுக்கிறது இல்லை.
AUNT JANE : ஆனாஜாக், நீவாடகைகுடுக்கலேன்னா, உன்னையதெருவுக்குதள்ளிவிட்டுருவாங்களே! அது சரியில்ல. இப்ப, உனக்கு ஜில், அதோட ஒரு குழந்தை .. அதை நீ மனசுல வெச்சுக்கணும்.
JACK : இல்ல.. இல்ல.. அத்தை … நீங்க என்னைய தவறா புரிஞ்சுக்கிட்டீங்க. நாங்க வாடகை எதுவும் கொடுப்பதில்லை. ஏன்னா, இந்த வீடு எங்களுடையது.
AUNT JANE : உங்களோடதா?
JILL : ஆமா, பத்து பவுண்டு பணம் கட்டுனா, இந்த வீடு என்னோடது தான்.
JACK : இங்க பாருங்க அத்தை பத்து பவுண்டு பணம் கட்டுனா, நமக்கு சொந்தமா, ஒரு வீட்டையே வாங்குற போது, வருசா .. வருசம் … வாடகை கட்டிக்கிட்டு இருக்கறது சிக்கனமானது. இல்லை இதை நாங்க புரிஞ்சிக்கிட்டோம். அத்தோட, கால் வருட தொகை கட்டணும். அதான் பார்த்தேன். வாடகைக்காரனா குடியிருக்கறதை விட, ஓனரா இருக்கலாமே!
AUNT JANE : சரி. அதுல ஏதோ இருக்கு. இருந்தாலும் நீங்க நல்ல சம்பாதிச்சாத்தான் இப்படி ஒரு இடத்துல இருக்க முடியும்.
JILL : ஓ, ஆமா அத்தை. போன வருஷம் தான் இவருக்கு அஞ்சு ஷில்லிங் சம்பளம் கூட்டினார்கள் இல்லையா, ஜாக்?

JACK : (நளினமாக) ஆமா, அது ஒண்ணுமில்ல. இந்த கிறிஸ்மஸ்ல எனக்கு பத்து ஷில்லிங் சம்பளம் கூட்டித்தருவாங்க.
AUNT JANE : திடீரென) ஜாக்! இப்பதான் அதைப்பத்தி யோசிச்தேன் அந்த கார். அது உண்மையில உன்னோடது தானா?
JILL : ஆமா, அது என்னோடது தான்.
AUNT JANE : எல்லா காருமா?
JACK : அது வந்து, எல்லா காரும் இல்ல.
AUNT JANE : அப்ப அது எவ்வளவு?
JILL : உண்மையா சொல்லப்போனா, அந்த ஸ்டியரிங் அப்புறம் ஒரு டயர். அதுல இருக்கிற ரெண்டு சிலிண்டர். இவ்வளவு தான் எங்களுக்கு சொந்தம். ஆனா, அது ரொம்ப . அற்புதமானது இல்லையா?

JACK : உட்காருங்க, அத்தை. நீங்க சோர்வா இருப்பீங்க நாங்க எங்க வீடு எல்லாம் சுத்திக் காட்டிட்டோம்.
JILL : எங்க சின்ன கூட்டைப்பத்தி என்ன நினைக்கிறீங்க அத்தை.
AUNT JANE : இது ரொம்ப அழகாக இருக்கு., இந்த மேசை, நாற்காலி. அப்புறம். கார்… அப்புறமா பியோனா, பிரிட்ஜ், ரேடியோ அது என்ன…? ரொம்ப பிரமாதம்!
JACK : இது எல்லாம் எங்களுக்கு வந்தது உங்களால் தான்.
AUNT JANE : ஆமா, ஜாக். அதுதான் எனக்கு கவலையா இருக்கு.
JACK : என்ன அத்தை, கவலையா இருக்கா?
AUNT JANE : நான் கல்யாணத்தன்னிக்கு உங்களுக்கு அன்பளிப்பா, ஒரு செக் கொடுத்தேன் பாருங்க. அது இருநூறு பவுண்டு தான்! இல்லையா? நான், அதுல ரெண்டாயிரம் பவுண்டுன்னு எழுதலையே!
JILL : இல்ல ஜேன்! எப்படி உங்களுக்கு அப்படி ஒரு சந்தேகம் வந்துச்சு?
AUNT JANE : பரவாயில்ல, இருக்கட்டும். ஆனா, இன்னமும் எனக்கு ஒண்ணு புரியல. இந்த வீடு, இது ரொம்ப நல்லா இருக்கு… ஆனா, இதுக்கான வாடகை ரொம்ப அதிகமா இருக்குமே!

JACK : வாடகையா? இல்ல.. இல்ல.. நாங்க வாடகை கொடுக்கிறது இல்லை.
AUNT JANE : ஆனாஜாக், நீவாடகைகுடுக்கலேன்னா, உன்னையதெருவுக்குதள்ளிவிட்டுருவாங்களே! அது சரியில்ல. இப்ப, உனக்கு ஜில், அதோட ஒரு குழந்தை .. அதை நீ மனசுல வெச்சுக்கணும்.
JACK : இல்ல.. இல்ல.. அத்தை … நீங்க என்னைய தவறா புரிஞ்சுக்கிட்டீங்க. நாங்க வாடகை எதுவும் கொடுப்பதில்லை. ஏன்னா, இந்த வீடு எங்களுடையது.
AUNT JANE : உங்களோடதா?

JILL : ஆமா, பத்து பவுண்டு பணம் கட்டுனா, இந்த வீடு என்னோடது தான்.
JACK : இங்க பாருங்க அத்தை பத்து பவுண்டு பணம் கட்டுனா, நமக்கு சொந்தமா, ஒரு வீட்டையே வாங்குற போது, வருசா .. வருசம் … வாடகை கட்டிக்கிட்டு இருக்கறது சிக்கனமானது. இல்லை இதை நாங்க புரிஞ்சிக்கிட்டோம். அத்தோட, கால் வருட தொகை கட்டணும். அதான் பார்த்தேன். வாடகைக்காரனா குடியிருக்கறதை விட, ஓனரா இருக்கலாமே!
AUNT JANE : சரி. அதுல ஏதோ இருக்கு. இருந்தாலும் நீங்க நல்ல சம்பாதிச்சாத்தான் இப்படி ஒரு இடத்துல இருக்க முடியும்.
JILL : ஓ, ஆமா அத்தை. போன வருஷம் தான் இவருக்கு அஞ்சு ஷில்லிங் சம்பளம் கூட்டினார்கள் இல்லையா, ஜாக்?
AUNT JANE : (நளினமாக) ஆமா, அது ஒண்ணுமில்ல. இந்த கிறிஸ்மஸ்ல எனக்கு பத்து ஷில்லிங் சம்பளம் கூட்டித்தருவாங்க. திடீரென) ஜாக்! இப்பதான் அதைப்பத்தி யோசிச்தேன் அந்த கார். அது உண்மையில உன்னோடது தானா?
JILL : ஆமா, அது என்னோடது தான்.
AUNT JANE : எல்லா காருமா?
JACK : அது வந்து, எல்லா காரும் இல்ல.
AUNT JANE : அப்ப அது எவ்வளவு?
JILL : உண்மையா சொல்லப்போனா, அந்த ஸ்டியரிங் அப்புறம் ஒரு டயர். அதுல இருக்கிற ரெண்டு சிலிண்டர். இவ்வளவு தான் எங்களுக்கு சொந்தம். ஆனா, அது ரொம்ப. அற்புதமானது இல்லையா?

AUNT JANE : இதுல என்ன அற்புதம்ன்னு எனக்கு தெரியல.
JILL : ஆனா, அதுல ஒரு அற்புதம் இருக்கு. நாம ஒரு காரையே வாங்க முடியலைன்னாலும், அஞ்சு பவுண்டு குடுத்தா, அதை ஜாலியா ஓட்டலாம் இல்லையா?
AUNT JANE : மத்ததெல்லாம், சுலப தவணைகள், என்று நான் நினைக்கிறேன்.
JILL: சரியா சொன்னீங்க.
AUNT JANE : சரி, அந்த ரேடியோ எப்படி? அது என்ன!…..
JACK: அது…. அது……
AUNT JANE : அப்புறமா பியானோ.
JILL: ஆமா.
AUNT JANE : அப்புறம், மேசை நாற்காலி.
JACK : ஆமா… அதுவும், அப்படித்தான்…
JILL : நல்லது. அங்க ஒண்ணு இருக்கே … (எதையோ கை காட்டுகிறார்).
AUNT JANE : மத்ததெல்லாம், மிஸ்டர் சேஜ்க்கு சொந்தமானது இல்லையா?
JILL: ஆமா.

AUNT JANE : நல்லது மிஸ்டர் சேஜ்க்கு சொந்தமான எதிலேயும் நான் உக்கார மாட்டேன். (அவர் எழுந்து நிற்கிறார்). இப்ப சொல்லுங்க இந்த மொத்த தவணை எல்லாம் மொத்தம் எவ்வளவு வருது?
JACK : அது வந்து, உண்மையில.. (அவன் தன்னுடைய சட்டைப்பாக்கெட் நோட்டை எடுத்து, அதைப் பார்க்கிறான்).. ஒரு வாரத்துக்கு ஏழு பவுண்டு எட்டு ஷில்லிங், எட்டு பென்னி.
AUNT JANE : கடவுளே! நீ எவ்வளவு சம்பாதிக்கற?
JACK : சொல்லப்போனா, வந்து … ஆறு பவுண்டு.
AUNT JANE : ஆனா, இது ரொம்ப முட்டாள்தனம் ஆறு பவுண்டு சம்பளத்தை வைத்து எப்படி ஏழு பவுண்டு எட்டு ஷில்லிங், எட்டு பென்னி கடனை கட்டுவீங்க?
JACK : ஓ.. அது ரொம்ப ஈஸி, அதுல பாருங்க. நமக்கு அதிகமா தேவைப்படுற பணத்தை “த்ரிப்ட் அண்ட் ப்ரொவிடென்ஸ் டிரஸ்ட் கார்பொரேஷன்”ல கடனா வாங்கிக்க வேண்டியது தான்.
JILL : அவங்க எவ்வளவு கடன் கேட்டாலும், தர்றாங்க அங்க ப்ரோ – நோட் எழுதி தரவேண்டும்.
AUNT JANE : சரி இதை எப்படி திரும்பி கட்டப்போறீங்க?
JACK : ஓ, அதுவும் ரொம்ப ஈஸி, அதை தவணையில திரும்பி கட்டணும்.
AUNT JANE : தவணையா! (அவள் தனது தலையில் கையை வைத்துக் கொண்டு நாற்காலியில் உட்கார்ந்து விடுகிறாள். பிறகு, அது மிஸ்டர் சேஜ்க்கு சொந்தமானது என்பதை உணர்ந்து உடனே அலறிக்கொண்டு, கால்களால் குதித்து தரையில் நிற்கிறாள்))
JACK : அத்தை என்னாச்சு? நீங்க படுத்துக்க விரும்புறீங்களா!
AUNT JANE : இங்கயா?அந்த மிஸ்டர் சேஜ் அல்லது மார்க்ஸ் அல்லது ஸ்பென்ஸர்க்கு அல்லது வேற யாருக்காவது, சொந்தமான படுக்கையில் விழுந்து கிடப்பேன்னு நினைக்கிறீங்களா? இல்ல, நான் வீட்டுக்குப் போறேன்.
JILL : ஓ, நீங்க போயே ஆகணுமா?
AUNT JANE : அதுதான் நல்லதுன்னு நினைக்கிறேன்.
JACK : நான் உங்கள என் கார்ல கூட்டிக்கிட்டுப் போயி ஸ்டேஷன்ல விடுறேன்.
AUNT JANE : ஒரு டயர், ரெண்டு சாமான் மட்டுமே இருக்கிற கார்ல போகணுமா? வேணாம் நன்றி, நான் பஸ்ல போயிக்கிறேன்.
JACK : நல்லது, அதுதான் உங்களோட முடிவுன்னா சரி.

AUNT JANE : நான் கொஞ்சம் கடுமையா பேசினதுக்கு வருத்தப்படுறேன். (சிறிது வருத்தத்துடன்) நீங்க குடும்பம் நடத்துறவிதத்தை பாத்துட்டு, நான் அதிர்ச்சியடைஞ்சுட்டேன். என் வாழ்க்கையில ஒருத்தருக்கும் ஒரு பைசா கடன் கொடுக்கிற மாதிரி நான் இருக்கல. உடனடி ரொக்கம் அதுதான் என் கொள்கை நீங்களும் அதே மாதிரி இருக்கணும்னு நான் விரும்பறேன்.

(தனது கைப்பையைத் திறந்து) இங்க பாருங்க! உங்களுக்காக நான் கொடுக்கணும்னு “வெச்சுருந்த ஒரு சின்ன தொகைக்கான செக். (அதை ஜில்லிடம் கொடுத்து விட்டு இதை வெச்சுக்கிட்டு, நீங்க உங்களோட தவணையை கட்டி, குறைஞ்ச பட்சம் ஒரு சாமானாவது உங்களுக்கு சொந்தமாகுற மாதிரி செய்வீங்கன்னு நினைக்கிறேன்.

JILL : வந்து…. நன்றி அத்தை நீங்க செஞ்சது ரொம்ப நல்லதா இருக்கு.
AUNT JANE : சரி. நான் போகணும் (அவனது கையை தட்டிக் கொடுக்கிறார்)
JACK : உங்களை நான் பஸ் வரை வந்து வழியனுப்புகிறேன்.
JILL : குட்பை அத்தை, உங்க அன்பளிப்புக்கு நன்றி.

AUNT JANE : குட்பை கண்ணு! (அவளும் ஜாக்கும் வெளியே போகிறார்கள். ஜில், அந்த செக்கை பார்த்து விட்டு, காற்றில் பறக்கும் முத்தமிடுகிறாள். பிறகு “பத்து பவுண்டா!” என்று ஆச்சரியத்தில் கத்துகிறாள். பிறகு மேசைக்கு சென்று, ஒரு தபால் உறை எடுத்து, அதன்மேல், ஒரு விலாசத்தை எழுதுகிறாள். அந்த செக்கை வேறொருவருக்கு பெயர் மாற்றம் செய்கிறாள். அந்த செக்குடன் ஒரு ரசீதை இணைத்து அதை அந்த தபால் உறையினுள் போடுகிறாள். பிறகு மணியடிக்கிறாள். ஒரு செவிலி ஒருநொடியில் அங்கு கையில் குழந்தையுடன் வருகிறாள்.)

JILL : ஓ, நர்ஸ், உடனே ஓடிப்போய், இதை தபால்ல போட்டுட்டு வா, நீ போயிட்டு திரும்பும் வரை, நான் குழந்தையை பார்த்துக்கிறேன்.
NURSE : கண்டிப்பாக, மேடம் (தன் கையில் இருக்கும் குழந்தையை, அவள் ஜில்லிடம் தந்து விட்டு, கடிதத்தை வாங்கிக்கொண்டு போகிறாள்).
JACK : நல்லது அத்தை போயிட்டா! என்ன ஒரு எரிச்சலாக்குற ஆளு! இருந்தாலும், அவ கொஞ்சம் பணம் கொடுத்துவிட்டு போயிருக்கா. அது எவ்வளவு?
JILL : பத்து பவுண்டு
JACK : ஓ.. அது ரொம்ப அருமை…! நாம, அந்த காருக்கு அடுத்த ரெண்டு மாசத்துல கட்ட வேண்டிய தவணைக்கு இதை கட்டிவிடலாம் (விசில் அடிக்கிறான்)
JILL : நாம அதை அப்படி செய்ய முடியாது.
JACK : ஏன் முடியாது?

JILL : இங்க பாருங்க, அதை வேற ஒண்ணுக்காக கொடுத்து அனுப்பிட்டேன். நம்ம நர்ஸ் அதை எடுத்துக்கிட்டு தபால்ல போட போயிருக்காங்க.
JACK : நல்லது அது சரி தான்! யாருக்கு அதை அனுப்பிச்சுருக்க.
JILL : டாக்டர் மார்ட்டின்.
JACK : டாக்டர் மார்ட்டினா! நீ என்ன பிசாசு பிடிச்சு இப்படி செய்யுறியா?
JILL : (அழும் நிலையில்) பாருங்க! என் மேல கோபப்படுறீங்க!.
JACK : நான் கோபப்படல, ஆனா, ஏன் இவ்வளவு பணத்தை டாக்டருக்கு கொடுக்கிறீங்க? டாக்டர் பணம் குடுப்பாங்கன்னு எதிர்பார்த்து இருக்க மாட்டாங்க.
JILL : (சிறிது அழுகிறார்) ஆனா…. ஆனா…. நீங்க புரிஞ்சுக்க மாட்டுறீங்க!
JACK : என்ன புரிஞ்சுக்கல?
JILL : இன்னும்… ஒரே ஒரு தவணை தான் உள்ளது. அப்புறம் இந்த குழந்தை நமக்கு சொந்தம். (அவள், கொஞ்சம் பரிதாபமாக, குழந்தையை எடுத்து முன்னே நீட்டிக்காட்டுகிறாள். நாம் இருளடைகிறோம்)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 5 Molecular Genetics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

12th Bio Zoology Guide Molecular Genetics Text Book Back Questions and Answers

Question 1.
Hershey and Chase experiment with bacteriophage showed that
(a) Protein gets into the bacterial cells
(b) DNA is the genetic material
(c) DNA contains radioactive sulphur
(d) Viruses undergo transformation
Answer:
(b) DNA is the genetic material

Question 2.
DNA and RNA are similar with respect to
(a) Thymine as a nitrogen base
(b) A single-stranded helix shape
(c) Nucleotide containing sugars, nitrogen bases and phosphates
(d) The same sequence of nucleotides for the amino acid phenyl alanine
Answer:
(c) Nucleotide containing sugars, nitrogen bases and phosphates

Question 3.
A mRNA molecule is produced by
(a) Replication
(b) Transcription
(c) Duplication
(d) Translation
Answer:
(b) Transcription

Question 4.
The total number of nitrogenous bases in human genome is estimated to be about
(a) 3.5 million
(b) 35000
(c) 35 million
(d) 3.1 billion
Answer:
(d) 3.1 billion

Question 5.
E. coli cell grown on 15N medium are transferred to 14N medium and allowed to grow for two generations. DNA extracted from these cells is ultracentrifuged in a cesium chloride density gradient. What density distribution of DNA would you expect in this experiment?
(a) One high and one low density band
(b) One intermediate density band
(c) One high and one intermediate density band
(d) One low and one intermediate density band
Answer:
(d) One low and one intermediate density band

Question 6.
What is the basis for the difference in the synthesis of the leading and lagging strand of DNA molecules?
(a) Origin of replication occurs only at the 5’ end of the molecules
(b) DNA ligase works only in the 3’ → 5’ direction
(c) DNA polymerase can join new nucleotides only to the 3′ ends of the growing stand
(d) Helicases and single-strand binding proteins that work at the 5’ end
Answer:
(d) DNA polymerase can join new nucleotides only to the 3’ end of the growing stand

Question 7.
Which of the following is the correct sequence of event with reference to the central dogma?
(a) Transcription, Translation, Replication
(b) Transcription, Replication, Translation
(c) Duplication, Translation, Transcription
(d) Replication, Transcription, Translation
Answer:
(d) Replication, Transcription, Translation

Question 8.
Which of the following statements about DNA replication is not correct?
(a) Unwinding of DNA molecule occurs as hydrogen bonds break
(b) Replication occurs as each base is paired with another exactly like it
(c) Process is known as semiconservative replication because one old strand is conserved in the new molecule
(d) Complementary base pairs are held together with hydrogen bonds
Answer:
(b) Replication occurs as each base is paired with another exactly like it

Question 9.
which of the following statements is not true about DNA replication in eukaryotes?
(a) Replication begins at a single origin of replication.
(b) Replication is bidirectional from the origins.
(c) Replication occurs at about 1 million base pairs per minute.
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.
Answer:
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time

Question 10.
The first codon to be deciphered was ………………. which codes for ……………..
(a) AAA, proline
(b) GGG, alanine
(c) UUU, Phenylalanine
(d) TTT, arginine
Answer:
(c) UUU, Phenylalanine

Question 11.
Meselson and Stahl’s experiment proved
(a) Transduction
(b) Transformation
(c) DNA is the genetic material
(d) Semi-conservative nature of DNA replication
Answer:
(d) Semi-conservative nature of DNA replication

Question 12.
Ribosomes are composed of two subunits; the smaller subunit of a ribosome has a binding site for……………..and the larger subunit has two binding sites for two…………………
Answer:
mRNA, tRNA

Question 13.
Anoperonisa:
(a) Protein that suppresses gene expression
(b) Protein that accelerates gene expression
(c) Cluster of structural genes with related function
(d) Gene that switched other genes on or off
Answer:
(d) Cluster of structural genes with related function

Question 14.
When lactose is present in the culture medium:
(a) Transcription of lacy, lac z, lac a genes occurs
(b) Repressor is unable to bind to the operator
(c) Repressor is able to bind to the operator
(d) Both (a) and (b) are correct
Answer:
(d) Both (a) and (b) are correct

Question 15.
Give reasons: Genetic code is ‘universal’.
Answer:
The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial, and chloroplast genomes. However, similarities are more common than differences.

Question 16.
Name the parts marked ‘A’ and ‘B’ in the given transcription unit:

Answer:
A – Promoter site B – Structural gene

Question 17.
Differentiate – Leading strand and lagging strand
Answer:

Question 18.
Differentiate – Template strand and coding strand.
Answer:

1. Template Strand: During replication, DNA strands having the polarity 3’ → 5’ act as the template strand.
2. Coding Strand: During replication, a DNA strands having the polarity 5’ → 3’ act as a coding strand.

Question 19.
Mention any two ways in which single nucleotide polymorphism (SNPs) identified in human genome can bring revolutionary change in biological and medical science.
Answer:
Scientists have identified about 1.4 million locations, where single-base DNA differences (SNPs – Single nucleotide polymorphism – pronounced as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease-associated sequences and tracing human history.

Question 20.
State any three goals of the human genome project.
Answer:

• Identify all the genes (approximately 30000) in human DNA.
• Determine the sequence of the three billion chemical base pairs that make up the human DNA.
• To store this information in databases.

Question 21.
In E.coli, three enzymes 0- galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.
Answer:
In the absence of lactose, the repressor protein binds to the operator and prevents the transcription of structural gene by RNA polymerase, hence the enzymes are not produced. However, there will always be a minimal level of lac operon expression even in absence of lactose.

Question 22.
Distinguish between structural gene, regulatory gene and operator gene.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls the transcriptional, activity of the structural gene.

1. The structural gene codes for proteins, rRNA, and tRNA required by the cell.
2. Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
3. The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Question 23.
A low level of expression of lac operon occurs at all the time in E-coli. Justify the statement.
Answer:
One of the enzyme synthesized by lac operon is permease which is involved in the transport of lactose into the cells. If the lac operon gets inactivated, permease is not synthesized hence lactose cannot enter the cell. Lactose acts as a inducer, binding to the repressor protein and switch on the operator to initiate gene expression.

Question 24.
Why the human genome project is called a megaproject?
Answer:
The international human genome project was launched in the year 1990. It was a mega project and took 13 years to complete. The human genome is about 25 times larger than the genome of any organism sequenced to date and is the first vertebrate genome to be completed. Human genome is said to have approximately 3 >109 bp. HGP was closely associated with the rapid development of a new area in biology called bioinformatics.

Question 25.
From their examination of the structure of DNA, What did Watson and Crick infer about the probable mechanism of DNA replication, coding capability and mutation?
Answer:
Inference of Watson and Crick on DNA replication.
They concluded that each of the DNA strand in a helix act as template during DNA replication leading to formation of new daughter DNA molecules, which are complementary to parental strand, (i.e., Semi-conservative method of replication)

Inference on coding capability
During transcription, the genetic information in the DNA strand is coded to mRNA as complementary bases, (except for uracil in place of thymine in RNA)

Inference on mutation
Any changes in the nucleotide sequence of DNA leads to corresponding alteration in aminoacid sequence of specific protein thus confirming the validity of genetic code.

Question 26.
Why tRNA is called an adapter molecule?
Answer:
The transfer RNA, (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called an adapter molecule. This term was postulated by Francis Crick.

Question 27.
What are the three structural differences between RNA and DNA?
Answer:
DNA :

1. Sugar is deoxyribose sugar. .
2. Double-stranded structure.
3. Nitrogen bases are Adenine, Guanine, Cytosine and Thymine.

RNA :

1. Sugar is ribose sugar
2. Single-stranded molecule.
3. Nitrogen bases are Adenine, Guanine, Cytosine and Uracil.

Question 28.
Name the anticodon required to recognize the following codons: AAU, CGA, UAU, and GCA.
Answer:
UUA, GCU, AUA and CGU.

Question 29.
a) Identify the figure given below
b) Redraw the structure as a replicating fork and label the parts
c) Write the source of energy for this replication and name the enzyme involved in this process.
d) Mention the differences in the synthesis of protein, based on the polarity of the two template strands.
Answer:
a) Replication fork
b)

c) Deoxy nucleotide, triphosphate acts as a energy source for replication. DNA polymerase is used for replication
d) mRNA contacting information for protein synthesis will developed from DNA strand having polariy 5 ’ → 3 ’

Question 30.
If the coding sequence in a transcription unit is written as follows:
5’ TGCATGCATGCATGCATGCATGCATGC 3’
Write down the sequence of mRNA.
Answer:
mRNA sequence is 3’ACGUACGUACGUUCGUACGUACGUACG5’

Question 31.
How is the two-stage process of protein synthesis advantageous?
Answer:
The split gene feature of eukaryotic genes is almost entirely absent in prokaryotes. Originally each exon may have coded for a single polypeptide chain with a specific function. Since exon arrangement and intron removal are flexible, the exon coding for these polypeptide subunits act as domains combining in various ways to form new genes. Single genes can produce different functional proteins by arranging their exons in several different ways through alternate splicing patterns, a mechanism known to play an important role in generating both protein and functional diversity in animals.

Introns would have arose before or after the evolution of eukaryotic gene. If introns arose late how did they enter eukaryotic gene? Introns are mobile DNA sequences that can splice themselves out of, as well as into, specific ‘target sites’ acting like mobile transposon-like elements (that mediate transfer of genes between organisms – Horizontal Gene Transfer – HGT). HGT occurs between lineages of prokaryotic cells, or from prokaryotic to eukaryotic cells and between eukaryotic cells. HGT is now hypothesized to have played a major role in the evolution of life on Earth.

Question 32.
Why did Hershey and Chase use radioactively labelled phosphorous and sulphur only? Would they have got the same result if they use radiolabelled carbon and nitrogen?
Answer:
Generally proteins contain sulphur but not phosphorous and nucleic acid (DNA) contains , phosphorous but not sulphur. Hence Hershey – Chase used radioactive isotopes of sulphur (35S) and phosphorus (32P) to keep separate track of viral protein and nucleic acid in culture medium. The expected result cannot be achieved, if radioactive carbon and nitrogen is used, since these molecules are present in both DNA and proteins.

Question 33.
Explain the formation of a nucleosome.
Answer:
Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere.

The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix. The histone octameres are in close contact and DNA is coiled on the outside of nucleosome.

Question 34.
It is established that RNA is the first genetic material. Justify giving reasons.
Answer:
Three molecular biologists in the early 1980’s (Leslie Orgel, Francis Brick and Carl Woese) independently proposed the ‘RNA world’ as the first stage in the evolution of life, a stage when RNA catalysed all molecules necessary for survival and replication. The term ‘RNA world’ first used by Walter Gilbert in 1986, hypothesizes RNA as the first genetic material on Earth. There is now enough evidence to suggest that essential life processes (such as metabolism, translation and splicing etc.,) evolved around RNA.

RNA has the ability to act as both genetic material and catalyst. There are several biochemical reactions in living systems that are catalysed by RNA. This catalytic RNA is known as ribozyme. But, RNA being a catalyst was reactive and hence unstable. This led to evolution of a more stable form of DNA, with certain chemical modifications. Since DNA is a double-stranded molecule having complementary strand, it has resisted changes by evolving a process of repair. Some RNA molecules function as gene regulators by binding to DNA and affect gene expression. Some viruses use RNA as the genetic material. Andrew Fire and Craig Mellow (recipients of Nobel Prize in 2006) were of the opinion that RNA is an active ingredient in the chemistry of life.

12th Bio Zoology Guide Molecular Genetics Additional Important Questions and Answers

12th Bio Zoology Guide Molecular Genetics One Mark Questions and Answers

Question 1.
The term gene was coined by……………
Answer:
Wilhelm Johannsen

Question 2.
Whose experiment finally provided convincing evidence that DNA is the genetic material?
(a) Griffith experiment
(b) Avery, Macleod and McCarty’s experiment
(c) Hershey-Chase experiment
(d) Urey-Miller’s experiment
Answer:
(c) Hershey-Chase experiment

Question 3.
In Hershey – Chase experiment, the DNA of T₂ phase was made radioactive by using
(a) ³²P
(b) ³⁵S
(c) ³⁵P
(d) ³⁵S
Answer:
(a) ³²P

Question 4.
A nucleoside is composed of………………………………
(a) Sugar and Phosphate
(b) Nitrogen base and Phosphate
(c) Sugar and Nitrogen base
(d) Sugar, Phosphate and Nitrogenous base
Answer:
(c) Sugar and Nitrogen base

Question 5.
Identify the incorrect statement
(a) a base is a substance that accepts H+ ion
(b) Both DNA and RNA have four bases
(c) Purines have single carbon-nitrogen ring
(d) Thymine is unique for DNA
Answer:
(c) Purines have single carbon-nitrogen ring

Question 6.
Watson and Crick proposed their double helical DNA model based on the X-ray diffraction analysis of………………
(a) Erwin Chargaff
(b) Meselson and Stahl
(c) Wilkins and Franklin
(d) Griffith
Answer:
(c) Wilkins and Franklin

Question 7.
The term ‘RNA world’ was first used by………………
Answer:
Walter Gilbert

Question 8.
The distance between two consecutive base pairs in DNA is …………………
(a) 0.34 nm
(b) 3.4 nm
(c) 0.034 nm
(d) 34 nm
Answer:
(a) 0.34 nm

Question 9.
If the length of E. coli DNA is 1.36 mm, the number of base pairs is
(a) 0.36 x 10⁶m
(b) 4 x 10⁶m
(c) 0.34 x 10^-9nm
(d) 4 x 10^-9m
Answer:
(b) 4 x 10⁶m

Question 10.
Identify the proper sequence in the organisation of the eukaryotic chromosome.
(a) Nucleosome – Solenoid – Chromatid
(b) Chromatid – Nucleosome – Solenoid
(c) Solenoid – chromatin – DNA
(d) Nucleosome – solenoid – genophore
Answer:
(a) Nucleosome – Solenoid – Chromatid

Question 11.
Assertion (A) : Genophore is noticed in prokaryotes.
Reason (R) : Bacteria possess circular DNA without chromatin organisation.
(a) Both A and R are correct
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(c) R explains A

Question 12.
Assertion (A): Heterochromatin is transcriptionally active.
Reason (R): Tightly packed chromatin which stains dark.
(a) Both A and R are correct , (b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct .
(d) A is incorrect R is correct
Answer:
(d) A is incorrect R is correct

Question 13.
Assertion (A): the semi-conservative model was proposed by Hershey and Chase.
Reason (R) : The daughter DNA contains only new strands.
(a) Both A and R are incorrect
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(a) Both A and R are incorrect

Question 14.
Komberg enzyme is called as ……………..
Answer:
DNA polymerase I

Question 15.
Replication of DNA occurs at phase of cell cycle.
(a) M
(b) S
(c) G₁
(d) G₂
Answer:
(b) S

Question 16.
Semi-conservative model of replication was proved by
(a) Hershey and Chase
(b) Griffith
(c) Meselson and Stahl
(d) Macleod and McCarty
Answer:
(c) Meselson and Stahl

Question 17.
How many types of DNA polymerases does a eukaryotic cell possess?
(a) two
(b) three
(c) four
(d) five
Answer:
(d) Five

Question 18.
Identify the incorrect statement
(a) Replication occurs at ori – site of DNA
(b) Deoxy nucleotide triphosphate acts as a substrate
(c) Unwinding of DNA strand is carried out by topoisomerase
(d) DNA polymerase catalyses the polymerization at 3-OH
Answer:
(c) Unwinding of DNA strand is carried out by topoisomerase

Question 19.
The discontinuously synthesized fragments of lagging strand are called………………
Answer:
Okazaki fragments

Question 20.
Retroviruses possess ………………..as genetic material.
Answer:
RNA

Question 21.
Which is NOT a part of transcription unit?
(a) Promoter
(b) Operator
(c) Structural gene
(d) Terminator
Answer:
(b) Operator

Question 22.
Goldberg – Hogness box of eukaryotes is equivalent to of prokaryotes.
Answer:
Pribnow box

Question 23.
Okazaki fragments are joined by the enzyme during DNA replication.
Answer:
DNA ligase

Question 24.
Match the following:

(a) A – iv, B – C – ii, D – iii
(b) A – i, B – ii, C – iii, D – iv
(c) A – ii, B – iii, C – i, D – ii
(d) A – iii, B – ii, C – ii, D – i
Answer:
(a) A – iv, B – C – ii, D – iii

Question 25.
The RNA polymerase of prokaryotes binds with ………………….factor to initiate polymerization.
(a) rho
(b) theta
(c) sigma
(d) psi
Answer:
(c) sigma A

Question 26.
Precursor
(a) Capung
(b) Tailing
(c) Splicing
(d) Transcribing
Answer:
(c) Splicing

Question 27.
Which of the following feature is absent in prokaryotes?
(a) Prokaryotes possess three major types of RNAs
(b) Structural genes are polycistronic
(c) Initiation process of transcription requires ‘P’ factor
(d) Split gene feature
Answer:
(d) Split gene feature

Question 28.
Which of the following sequence has completely translated?
(i) AGA, UUU, UGU, AGU, UAG
(ii) AUG, UUU, AGA, UAC, UAA
(iii) AAA, UUU, UUG, UGU, UGA
(iv) AUG,AAU,AAC,UAU,UAG
(a) i and ii
(b) ii only
(c) i and iii
(d) ii and iv
Answer:
(d) ii and iv

Question 29.
Capping of mRNA occurs using
(a) Poly A residues
(b) Methyl guanosine triphosphate
(c) Deoxy ribonucleotide triphosphate
(d) Ribonucleotide triphosphate
Answer:
(b) Methyl guanosine triphosphate

Question 30.
One of the aspect is not a feature of genetic code?
(a) Specific
(b) Degenerate
(c) Universal
(d) Ambiguous
Answer:
(d) Ambiguous

Question 31.
Which of the triplet codon is not a code of proline?
(i) CCU (ii) CAU (iii) CCG (zv) CAA
(a) i only
(b) ii and iv
(c) iii only
(d) all the above
Answer:
(b) ii and iv

Question 32.
Coding sequences found in split genes are called.
(a) Operons
(b) Introns
(c) Exons
(d) Cistron
Answer:
(c) Exons

Question 33.
Which of the following mRNA yields 6 aminoacids after translation?
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU
(ii) UGA AGA UAG GAG CAU CCC UAC UAU GAU
(iii) GUC UGC UGG GCU GAU UAA AGG AGC AUU
(iv) AUG UAC CAU UGC UGA UGC AGG AGC CCG
Answer:
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU

Question 34.
The transcription termination factor associated with RNA polymerase in prokaryotes is
(a) ∑
(b) σ
(c) ρ
(d) ∑
Answer:
(c) p

Question 35.
In a DNA double-strand, if guanine is 30%, what will be the percentage of thymine?
(a) 100%
(b) 20%
(c) 10%
(d) 70%
Answer:
(b) 20%

Question 36.
Identify the triplet pairs that code for Tyrosine
(a) UUU,UUC
(b) UAU, UAU
(c) UGC, UGU
(d) CAU, CAC
Answer:
(b) UAU, UAU

Question 37.
Match the following:

(a) A – ii B – i C – iv D – iii
(b) A – iii B – ii C – i D – iv
(c) A – iv B – i C – ii D – iii
(d) A – ii B – iii C – i D – iv
Answer:
(a) A-ii B-i C-iv D-iii

Question 38.
AUG code is for……………….
(a) Arginine
(b) Tyrosine
(c) Tryptophan
(d) Methionine
Answer:
(d) Methionine

Question 39.
The sequence of bases in coding strand of DNA is GAGTTAGCAGGC, then the sequence of codons in primary transcript is ……………
(d) CUCAUACGCCCG
(b) CUCAAUCGUCCG
(c) UCAGAUCUGCGC
(d) UUCAAUCGUGCG
Answer:
(b) CUCAAUCGUCCG

Question 40.
The promoter region of eukaryote is …………..
(a) TATAA
(b) AUGUT
(c) UUUGA
(d) AAAAU
Answer:
(a) TATAA

Question 41.
Match the following:

(a) A – iii B – i C – iv D – ii
(b) A – iii B – ii C – i D – iv
(c) A – iv B – i C – ii D – iii
(d) A – ii B – iii C – iv D – i
Answer:
(a) A – iii B-i C-iv D-ii

Question 42.
……………….number of codons, codes for cystine.
Answer:
Two

Question 43.
In sickle cell anaemia, the ……………….codon of p – globin gene is modified.
(a) Eighth
(b) Seventh
(c) Sixth
(d) Ninth
Answer:
(c) Sixth

Question 44.
Pick out the incorrect statement.
(a) tRNA acts as a adapter molecule
(b) Stop codons do not have tRNA’s
(c) Addition of amino acid leads to hydrolysis of tRNA {d) tRNA has four major loops
Answer:
(c) Addition of amino acid leads to hydrolysis of tRNA

Question 45.
Which of the following antibiotic inhibits the interaction between tRNA and mRNA?
(a) Neomycin
(b) Streptomycin
(c) Tetracycline
(d) Chloramphenicol
Answer:
(a) Neomycin

Question 47.
The cluster of genes with related function is called …………….
(a) Cistron
(b) Operon
(c) Muton
(d) Recon
Answer:
(b) Operon

Question 48.
Repressor protein of Lac operon binds to ……………of operon
(a) Promoter region
(b) Operator region
(c) terminator region
(d) inducer region
Answer:
(b) Operator region

Question 49.
Lac Z gene codes for ………………..
(a) Permease
(b) transacetylase
(c) P-galactosidase
(d) Aminoacyl transferase
Answer:
(c) p-galactosidase

Question 50.
Lac operon model was proposed by ………………
Answer:
Jacob and Monod

Question 51.
Approximate count of base pair in human genome is …………….
Answer:
3 x 10⁹ bp

Question 52.
Automated DNA sequences are developed by.
Answer:
Frederick Sanger

Question 53.
Which of the chromosome has a higher gene density?
(a) Chromosome 20
(b) Chromosome 19
(c) Chromosome 13
(d) Chromosome Y
Answer:
(b) Chromosome 19

Question 54.
Number of genes located in chromosome Y is
(a) 2968
(b) 213
(c) 2869
(d) 231
Answer:
(d) 231

Question 55.
How many structural genes are located in the lac operon of E.Coli?
(a) 4
(b) 3
(c) 2
(d) 1
Answer:
(b) 3

Question 56.
DNA fingerprinting technique was developed by
(d) Jacob and Monod
(b) Alec Jeffreys
(c) Frederick Sanger
Answer:
(b) Alec Jeffreys

Question 57.
In DNA fingerprinting, separation of DNA fragments is done by………
(a) Centrifugation
(b) Electrophoresis
(c) X-ray diffraction
(d) denaturation
Answer:
(b) Electrophoresis

Question 58.
SNP stands for
(a) Single nucleotide Polymorphism
(b) Single Nucleoside Polypeptide
(c) Single nucleotide Polymorphism
(d) Single nucleotide polymer
Answer:
(a) Single nucleotide Polymorphism

Question 59.
Specific sequences of mRNA that are not translated are……………..
Answer:
Untranslated Regions (UTR)

Question 60.
Non-coding or intervening DNA sequence is called ……………….
Answer:
Intron

Question 61.
………………is the monomer of DNA.
Answer:
Nucleotide

Question 62.
Which one of the following is wrongly matched?
(a) Transcription – Copying information from DNA to RNA
(b) Translation – Decoding information from mRNA to protein
(c) Replication – Making of DNA copies
(d) Splicing – Joining of exons with introns
Answer:
(d) ‘ Splicing – Joining of exons with introns

12th Bio Zoology Guide Molecular Genetics Two Marks Questions and Answers

Question 1.
Who proposed One gene – One enzyme hypothesis? Define it.
Answer:
George Beadle and Edward Tatum proposed the One gene – One enzyme hypothesis which states that one gene controls the production of one enzyme.

Question 2.
Differentiate nucleoside from nucleotide.
Answer:
Nucleoside :
Nucleoside subunit is composed of nitrogenous bases linked to a pentose sugar molecule.

Nucleotide :
Nucleotide subunit is composed of nitrogenous bases, a pentose sugar and a phosphate group.

Question 3.
State the key differences between DNA and RNA.
Answer:
DNA

1. DNA is made of deoxyribose sugar.
2. Nitrogenous bases of DNA are Adenine, Guanine, Cytosine and Thymine.

RNA:

1. RNA is made of ribose sugar.
2. Nitrogenous bases of RNA are Adenine,Guanine, Cytosine and Uracil.

Question 4.
Point out the nitrogenous bases of RNA.
Answer:
Adenine, Guanine, Cytosine and Uracil.

Question 5.
What makes the DNA and RNA acidic molecules?
Answer:
The phosphate functional group (P04) gives DNA and RNA the property of an acid at physiological pH, hence the name nucleic acid.

Question 6.
Which type of bond is formed

1. between a purine and pyrimidine base?
2. between the pentose sugar and adjacent nucleotide?

Answer:

1. Purine and pyrimidine bases are linked by hydrogen bonds.
2. Pentose sugar is linked to adjacent nucleotide by phosphodiester bonds.

Question 7.
DNA acts as genetic material for majority of living organisms and not the RNA. Give reasons to support the statement.
Answer:

1. RNA was reactive and hence highly unstable.
2. Some RNA molecules acts as gene regulators by binding to DNA and affect gene expression.
3. Uracil of RNA is less stable than thymine of DNA.

Question 8.
Name any two viruses whose genetic material is RNA.
Answer:

1. Tobacco Mosaic Virus (TMV)
2. Bacteriophage (B)

Question 9.
What are the properties that a molecule must possess to act as genetic material?
Answer:

• Self-replication
• Information storage
• Stability
• Variation through mutation

Question 10.
How many base pairs are present in one complete turn of DNA helix? What is the distance between two consecutive base pairs?
Answer:
There are ten base pairs in each turn with a distance of 0.34 x 10‘9 m between two adjacent base pairs.

Question 11.
What is a genophore?
Answer:
In prokaryotes such as E. coli though they do not have defined nucleus, the DNA is not scattered throughout the cell. DNA (being negatively charged) is held with some proteins (that have positive charges) in a region called the nucleoid. The DNA as a nucleoid is organized into large loops held by protein. DNA of prokaryotes is almost circular and lacks chromatin organization, hence termed genophore.

Question 12.
What is nucleosome? How many base pairs are there in a typical nucleosome?
Answer:
The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix.

Question 13.
Expand and define NHC
Answer:
NHC : Non-histone Chromosomal protein.
In eukaryotes, apart from histone proteins, additional set of proteins are required for packing of chromatin at higher level and are referred as non – histone chromosomal proteins.

Question 14.
Differentiate between Heterochromatin and Euchromatin.
Answer:
Heterochromatin:

1. Region of nucleus where the chromatin are loosely packed and stains light are called Heterochromatin.
2. Transcriptionally inactive.

Euchromatin:

1. Region of the nucleus where the chromatin are tightly packed and stains dark are called Euchromatin.
2. Transcriptionally active.

Question 15.
Which is the widely accepted model of DNA replication? Who has proved it?
Answer:
Semi-conservative replication model. It was proved by Meselson and Stahl in 1958.

Question 16.
Name the chemical substance which is called by the name

1. Kornberg Enzyme
2. Ochoa’s enzyme

Answer:

1. DNA polymerase I is also known as Komberg enzyme.
2. Polynucleotide phosphorylase is also known as Ochoa’s enzyme.

Question 17.
Name the various types of prokaryotic DNA polymerase. State their role in replication process.
Answer:

Question 18.
What is the function of Deoxy nucleotide triphosphate in replication?
Answer:
Deoxy nucleotide triphosphate acts as substrate and also provides energy for polymerization reaction.

Question 19.
Given below are some events of eukaryotic replication. Name the enzymes involved in the process.

1. Unwinding of DNA
2. Joining of Okazaki fragments
3. Addition of nucleotides to a new strand
4. Correcting the repair

Answer:

1. Helicase
2. DNA ligase
3. DNA polymerase
4. Nuclease

Question 20.
Differentiate leading strand from lagging strand
Answer:
Leading strand :

1. Leading strand has the polarity 3’ →‘ 5’.
2. Replication is continuous.

Lagging strand :

1. Lagging strand has the polarity 5’ →‘ 3’
2. Replication is discontinuous.

Question 21.
What are Okazaki fragments?
Answer:
The discontinuously synthesized fragments of the lagging strand are called the Okazaki fragments are joined by the enzyme DNA ligase.

Question 22.
What is a replication fork?
Answer:
At the point of origin of replication, the helicases and topoisomerases (DNA gyrase) unwind and pull apart the strands, forming a Y-Shaped structure called the replication fork. There are two replication forks at each origin.

Question 23.
Apart from DNA polymerase, name any other four enzymes which were involved in DNA replication of eukaryotic cell.
Answer:
DNA ligase, Topoisomerase (DNA gyrase), Helicase and Nuclease.

Question 24.
Who proposed the central dogma? Write its concept.
Answer:

Question 25.
Define transcription and name the enzyme involved in this process.
Answer:
The process of copying genetic information from one strand of DNA into RNA is termed transcription. This process takes place in presence of DNA dependent RNA polymerase.

Question 26.
What is TATA box? State its function.
In eukaryotes, the promoter has AT-rich regions called TATA box or Goldberg-Hogness box. It acts as a binding site for RNA polymerase.

Question 27.
Structural gene of eukaryotes differ from prokaryotes. How?
In eukaryotes, the structural gene is monocistronic coding for only one protein whereas in prokaryotes the structural gene is polycistronic coding for many proteins.

Question 28.
What are the two major components of prokaryotic RNA polymerase? How do they act?
Answer:
Bacterial (prokaryotic) RNA polymerase consists of two major components, the core enzyme and the sigma subunit. The core enzyme (P1, p, and a) is responsible for RNA synthesis ” whereas a sigma subunit is responsible for recognition of the promoter.

Question 29.
Distinguish between exons and introns.
Answer:
Exons:
Expressed sequences (Coding sequences) of a eukaryotic gene

Introns:
Intervening sequences (non-coding sequences) of an eukaryotic gene

Question 30.
Define splicing.
Answer:
The process of removing introns from hnRNA is called splicing.

Question 31.
What is capping and tailing?
Answer:
In capping an unusual nucleotide, methyl guanosine triphosphate is added at the 5’ end of hnRNA, whereas adenylate residues (200-300) (Poly A) are added at the 3’ end in tailing.

Question 32.
If a double-stranded DNA has 20% of cytosine, calculate the percentage of adenine in DNA.
Answer:
Cytdsine = 20, hence Guanine = 20
As per ChargafFs rule (A+T) = (G+C) =100
Percent of Thymine + Adenine = 20 + 20 = 100
(T + A) = (20 + 20) = 100
(T + A) = 100 – (20 + 20)
T + A = 100 – 40
T + A = 60
Therefore the percent of Adenine will be 60/2 = 30%.

Question 33.
Mention the dual functions of AUG.
Answer:
AUG has dual functions. It acts as a initiator codon and also codes for the amino acid methionine.

Question 34.
How many codons are involved in termination of translation. Name them.
Answer:
Three codons terminate translation process. They are UAA, UAG and UGA.

Question 35.
Degeneracy of codon – comment.
Answer:
A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.

Question 36.
Point out the exceptional categories to universality of genetic code.
Answer:
Exceptions to the universal nature of genetic code are noticed in prokaryotic mitochondrial and chloroplast genomes.

Question 37.
What are non-sense codons?
Answer:
UGA, UAA, and UAG are the non-sense codons, which terminate translation.

Question 38.
Name the triplet codons that code for

1. Tyrosine
2. Histidine

Answer:

1. Tyrosine – UAU, UAC
2. Histidine – CAU, CAC

Question 39.
Why hnRNA has to undergo splicing?
Answer:
Since hnRNA contains both coding sequences (exons) and non-coding sequences (introns) it has to undergo splicing to remove introns.

Question 40.
State the role of following codons in the translation process

1. AUG
2. UAA

Answer:

1. AUG is the initiator codon and also codes for methionine.
2. UAA is a terminator codon.

Question 41.
Given below is mRNA sequence. Mention the aminoacids sequence that is formed after its translation.
3’AUGAAAGAUGGGUAA5’
Answer:
Methionine – Lysine – Aspartic acid – Glycine

Question 42.
Name the four codons that codes valine.
Answer:
GUU, GUC, GUA and GUG.

Question 43.
The base sequence in one of the DNA strand is TAGC ATGAT. Mention the base sequence in its complementary strand.
Answer:
The complementary strand has ATCGTACTA.

Question 44.
Why t-RNA is called as adapter molecule?
Answer:
The transfer RNA (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called as adapter molecule.

Question 45.
What do you mean by charging of tRNA? Name the enzyme involved in this process.
Answer:
The process of addition of amino acid to tRNA is known as aminoacylation or charging and the resultant product is called aminoacyl- tRNA (charged tRNA). Aminoacylation is catalyzed by an enzyme aminoacyl – tRNA synthetase.

Question 46.
What is UTR’s?
Answer:
mRNA also have some additional sequences that are not translated and are referred to as Untranslated Regions (UTR). UTRs are present at both 5’ end (before start codon) and at 3’ end (after stop codon).

Question 47.
What is the S – D sequence?
Answer:
The 5’ end of the mRNA of prokaryotes has a special sequence which precedes the initial AUG start codon of mRNA. This ribosome binding site is called the Shine – Dalgarno sequence or S-D sequence. This sequences base-pairs with a region of the 16Sr RNA of the small ribosomal subunit facilitating initiation.

Question 48.
Define translation unit.
Answer:
A translation unit in mRNA is the sequence of RNA that is flanked by the start codon on 5’ end and stop codon on 3’ end and codes of polypeptide.

Question 49.
Mention the inhibitory role of tetracycline and streptomycin in bacterial translation.
Answer:
Tetracycline inhibits binding between aminoacyl tRNA and mRNA.
Streptomycin inhibits initiation of translation and causes misreading.

Question 50.
At what stage, does the gene expression is regulated?
Answer:
Gene expression can be controlled or regulated at transcriptional or translational levels.

Question 51.
What is a operon? Give example.
Answer:
The cluster of genes with related functions is called operon.
E.g: lac operon in E.coli.

Question 52.
Considering the lac operon of E.coli, name the products of the following genes.

1. a) i gene
2. (b) lac Z gene
3. (c) lac Y gene
4. (d) lac a gene

Answer:

1. i gene – repressor protein
2. lac Z gene – fS-galactosidase
3. Lac Y gene – Permease
4. lac a gene – transacetylase

Question 53.
Expand

1. ETS
2. YAC.

Answer:

1. ETS: Expressed Sequence Tags
2. YAC: Yeast Artificial Chromosomes

Question 54.
Name the human chromosome that has

1. most number of genes
2. least number of genes

Answer:

1. Chromosome 1 has the maximum number of genes (2968 genes)
2. Chromosome Y has least genes (231 genes)

Question 55.
What are SNPs? Mention its uses.
Answer:
SNPs : Single nucleotide polymorphism. It helps to find chromosomal locations for disease-associated sequences and tracing human history.

Question 56.
Mention any four areas where DNA fingerprinting can be used.
Answer:

• Forensic analysis
• Pedigree analysis
• Conservation of wildlife
• Anthropological studies

12th Bio Zoology Guide Molecular Genetics Three Marks Questions and Answers

Question 57.
Classify nucleic acid-based on sugar molecules.
Answer:
There are two types of nucleic acids depending on the type of pentose sugar. Those containing deoxyribose sugar are called Deoxyribo Nucleic Acid (DNA) and those with ribose sugar are known as Ribonucleic Acid (RNA). The only difference between these two sugars is that there is one oxygen atom less in deoxyribose.

Question 58.
Both purines and pyrimidines are nitrogen bases yet they differ. How?
Answer:
Both purines and pyrimidines are nitrogen bases. The purine bases Adenine and Guanine have double carbon-nitrogen ring, whereas cytosine and thymine bases have single carbon-nitrogen ring.

Question 59.
How 5’ of DNA differ from its 3’?
Answer:
The 5’ of DNA refers to the carbon in the sugar to which phosphate (P04V) functional group is attached. The 3’ of DNA refers to the carbon in the sugar to which a hydroxyl (OH) group is attached.

Question 60.
State Chargaff’s rule.
Answer:
According to Erwin Chargaff,
a) Adenine pairs with Thymine with two hydrogen bonds.
b) Guanine pairs with Cytosine with three hydrogen bonds.

Question 61.
Chemically DNA is more stable than RNA – Justify.
Answer:
In DNA, the two strands being complementary, if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2’ OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

Question 62.
Write in simple about semi-conservative mode of DNA replication.
Answer:
Semi-conservative replication was proposed by Watson and Crick in 1953. This mechanism of replication is based on the DNA model. They suggested that the two polynucleotide strands of DNA molecule unwind and start separating at one end. During this process, covalent hydrogen bonds are broken. The separated single strand then acts as template for the synthesis of a new strand. Subsequently, each daughter double helix carries one polynucleotide strand from the parent molecule that acts as a template and the other strand is newly synthesised and complementary to the parent strand.

Question 63.
Draw a simplified diagram of nucleosome and label it.
Answer:

Question 64.
What is a primer?
Answer:
A primer is a short stretch of RNA. It initiates the formation of new strand. The primer produces 3’-OH end on the sequence of ribonucleotides, to which deoxyribonucleotides are added to form a new strand.

Question 65.
Both strands of DNA are not copied during transcription. Give reason.
Answer:
Both the strands of DNA are not copied during transcription for two reasons.

1. If both the strands act as a template, they would code for RNA with different sequences. This in turn would code for proteins with different amino acid sequences. This would result in one segment of DNA coding for two different proteins, hence complicate the genetic information transfer machinery.

2. If two RNA molecules were produced simultaneously, double-stranded RNA complementary to each other would be formed. This would prevent RNA from being translated into proteins.

Question 66.
What do you mean by a template strand and coding strand?
Answer:
DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→5’ acts as a template, and is called the template strand. The other strand which has the polarity 5’→3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called coding strand.

Question 67.
Name the factors that are responsible for initiation and termination of transcription in prokaryotes.
Answer:
Sigma factor is responsible for initiation of transcription.
Rho factor is responsible for termination of transcription.

Question 68.
Name the major RNA types of prokaryotes and mention their role.
Answer:
In prokaryotes, there are three major types of RNAs: mRNA, tRNA, and rRNA. All three \ RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA
brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role during translation.

Question 69.
Define genetic code.
Answer:
P The order of base pairs along DNA molecule controls the kind and order of amino acids found in the proteins of an organism. This specific order of base pairs is called genetic code.

Question 70.
Explain Wobble hypothesis.
Answer:
Wobble Hypothesis is proposed by Crick (1966) which states that tRNA anticodon has the t. ability to wobble at its 5’ end by pairing with even non-complementary baSe of mRNA codon.’ According to this hypothesis, in codon-anticodon pairing the third base may not be complementary. The third base of the codon is called wobble base and this position is called wobble position. The actual base pairing occurs at first two positions only. The importance of Wobbling hypothesis is that it reduces the number of tRNAs required for polypeptide synthesis and it overcomes the effect of code degeneracy.

Question 71.
Explain the nature of eukaryotic ribosomes.
Answer:
The ribosomes of eukaryotes (80 S) are larger, consisting of 60 S and 40 S subunits. ‘S’ denotes the sedimentation efficient which is expressed as Svedberg unit (S). The larger subunit in eukaryotes consist of a 23 S RNA and 5Sr RNA molecule and 31 ribosomal proteins. The smaller eukaryotic subunit consist of 18Sr RNA component and about 33 proteins.

Question 72.
Expand and define ORF.
Answer:
Any sequence of DNA or RNA, beginning with a start codon and which can be translated into a protein is known as an Open Reading Frame (ORF).

Question 73.
What are the components of the initiation complex of prokaryotic translation?
Answer:
Initiation of translation in E. coli begins with the formation of an initiation complex, consisting of the 30S subunits of the ribosome, a messenger RNA and the charged N-formyl methionine tRNA (f^met -t RNA f^met), three proteinaceous initiation factors (IF 1, IF2, IF3), GTP (Guanine Tri Phosphate) and Mg ²⁺.

Question 74.
Explain the components of operon.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls transcriptional activity of the structural gene.

• The structural gene codes for proteins, rRNA and tRNA required by the cell.
• Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
• The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

12th Bio Zoology Guide Molecular Genetics Five Marks Questions and Answers

Question 75.
Describe the Hershey and chase experiment. What is concluded by their experiment?
Answer:
Alfred Hershey and Martha Chase (1952) conducted experiments on bacteriophages that infect bacteria. Phage T₂ is a virus that infects the bacterium Escherichia coli. When phages (virus) are added to bacteria, they adsorb to the outer surface, some material enters the bacterium, and then later each bacterium lyses to release a large number of progeny phage. Hershey and Chase wanted to observe whether it was DNA or protein that entered the bacteria. All nucleic acids contain phosphorus, and contain sulphur (in the amino acid cysteine and methionine).

Hershey and Chase designed an experiment using radioactive _ isotopes of Sulphur (³⁵S) and phosphorus (³²P) to keep separate track of the viral protein and nucleic acids during the infection process. The phages were allowed to infect bacteria in culture medium which containing the radioactive isotopes ³⁵S or ³²P. The bacteriophage that grew in the presence of ³⁵S had labelled proteins and bacteriophages grown in the presence of ³²P had labelled DNA. The differential labelling thus enabled them to identity DNA and proteins of the phage. Hershey and Chase mixed the labelled phages with unlabeled E. coli and allowed bacteriophages to attack and inject their genetic material.

Soon after infection (before lysis of bacteria), the bacterial cells were gently agitated in a blender to loosen the adhering phase particles. It was observed that only ³²P was found associated with bacterial cells and ³⁵S was in the surrounding medium and not in the bacterial cells. When phage progeny was studied for radioactivity, it was found that it carried only ³²P and not ³⁵S. These results clearly indicate that only DNA and not protein coat entered the bacterial cells. Hershey and Chase thus conclusively proved that it was DNA, not protein, which carries the hereditary information from virus to bacteria.

Question 76.
Explain the properties of DNA that makes it an ideal genetic material.
Answer:
Self Replication: It should be able to replicate. According to the rule of base pairing and complementarity, both nucleic acids (DNA and RNA) have the ability to direct duplications. Proteins fail to fulfill this criteria.

Stability: It should he stable structurally and chemically. The genetic material should be stable enough not to change with different stages of life cycle, age or with change in physiology of the organism. Stability as one of property of genetic material was clearly evident in Griffith’s transforming principle. Heat which killed the bacteria did not destroy some of the properties of genetic material. In DNA the two strands being complementary, if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2’ OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

Information storage: It should be able to express itself in the form of ‘Mendelian characters’. RNA can directly code for protein synthesis and can easily express the characters. DNA, however depends on RNA for synthesis of proteins. Both DNA and RNA can act as a genetic material, but DNA being more stable stores the genetic information and RNA transfers the genetic information.

Variation through mutation: It should be able to mutate. Both DNA and RNA are able to mutate. RNA being unstable, mutates at a faster rate. Thus viruses having RNA genome with shorter life span can mutate and evolve faster. The above discussion indicates that both RNA and DNA can function as a genetic material. DNA is more stable, and is preferred for storage of genetic information.

Question 77.
How the DNA is packed in a eukaryotic cell? ft
Answer:
In eukaryotes, organization is more complex. Chromatin is formed by a series of repeating units called nucleosomes. Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere. The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical * nucleosome contains 200 bp of DNA helix.

The histone octameres are in close contact and DNA is coiled on the outside of nucleosome. Neighbouring nucleosomes are connected by linker DNA (HI) that is exposed to enzymes. The DNA makes two complete turns around the histone octameres and the two turns are sealed off by an HI molecule. Chromatin lacking HI has a beads-on-a-string appearance in which DNA inters and leaves the nucleosomes at random places.

HI of one nucleosome can interact with 33l of the neighbouring nucleosomes resulting in the further folding of the fibre. The chrof&atin fiber in interphase nuclei and mitotic chromosomes have a diameter that vary between 200-300 nm and represents inactive chromatin. 30 nm fibre arises from the folding of nucfeosbme, chains into a solenoid structure having six nucleosomes per turn. This structure is stabilized by interaction between different HI molecules. DNA is a solenoid and packed about,%)_folds.

The hierarchical nature of chromosome structure is illustrated. Additional set of^pteins are required for packing of chromatin at higher level and are referred to as non-histone chromosomal proteins (NHC). In*,a typical nucleus, some regions of chromatin are Ibosely packed (lightly stained) and are referred to as euchromatin. The chromatin that is,-tightly packed (stained darkly) is called heterochromatin. Euchromatin is transcriptionally active and heterochromatin is transcriptionally inactive.

Question 78.
Meselson and Stahl’s experiment proved the semi-coflBptervation mode of DNA replication. Explain.
Answer:
The mode of DNA replication was determined in 1958 by Meselson and Stahl. They designed an experiment to distinguish between semi-conservative, conservative and dispersive replications. In their experiment, they grew two cultures of E.coli for many generations in separate media. The ‘heavy’ culture was grown in a medium in which the nitrogen source (NH₄Cl) contained the heavy isotope ₁₅N and the ‘ light’ culture was grown in a medium in which the nitrogen source contained light isotope ₁₄N for many generations.

At the end of growth, they observed that the bacterial DNA in the heavy culture contained only ₁₅N and in the light culture only ₁₅N. The heavy DNA could be distinguished from light DNA (₁₅N from ₁₄N) with a technique called Cesium Chloride (CsCl) density gradient centrifugation. In this process, heavy and light DNA extracted from cells in thtytwo cultures settled into two distinct and separate bands (hybrid DNA). ;

The heavy culture (₁₅N) was then transferred into a medium that had only NH₄Cl, and took samples at various definite time intervals (20 minutes duration). After the first replication, they extracted DNA and subjected it to density gradient centrifugation. The DNA settled into a band that was intermediate in position between the previously determined heavy and light bands. After the second replication (40 minutes duration), they again extracted DNA samples, and this time found the DNA settling into two bands, one at the light band position and one at intermediate position. These results confirm Watson and Crick’s semi-conservative replication hypothesis.

Question 79.
Give a detailed account of a transcription unit.
Answer:
A transcriptional unit in DNA is defined by three regions, a promoter, the structural gene and a terminator. The promoter is located towards the 5 ’ end. It is a DNA sequence that provides binding site for RNA polymerase. The presence of promoter in a transcription unit, defines the template and coding strands. The terminator region located towards the 3’ end of the coding strand contains a DNA sequence that causes the RNA polymerase to stop transcribing.

In eukaryotes the promoter has AT-rich regions called TATA box (Goldberg- Hogness box) ‘ and in prokaryotes this region is called Pribnow box. Besides promoter, eukaryotes also require an enhancer.

The two strands of the DNA in the structural gene of a transcription unit have opposite polarity. DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→5’ acts as a template, and is called the template strand. The other strand which has the polarity 5 ’→3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called the coding strand (Fig. 5.7).

The structural gene may be monocistronic (eukaryotes) or polycistronic (prokaryotes). In eukaryotes, each mRNA carries only a single gene and encodes information for only a single protein and is called monocistronic mRNA. In prokaryotes, clusters of related genes, known as operon, often found next to each other on the chromosome are transcribed together to give a single mRNA and hence are polycistronic.

Question 80.
Explain the transcription process in prokaryotes with the needed diagram.
Answer:
In prokaryotes, there are three major types of RNAs: mRNA, tRNA, and rRNA.
All three RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play a structural and catalytic role during translation. There is a single DNA-dependent RNA polymerase that catalyses the transcription of all types of RNA. It binds to the promoter and initiates transcription (Initiation).

The polymerases binding sites are called promoters. It uses nucleoside triphosphate as substrate and polymerases in a template depended fashion following the rule of complementarity. After the initiation of transcription, the polymerase continues to elongate the RNA, adding one nucleotide after another to the growing RNA chain. Only a short stretch of RNA remains bound to the enzyme, when the polymerase reaches a terminator at the end of a gene, the nascent RNA falls off, so also the RNA polymerase.

The RNA polymerase is only capable of catalyzing the process of elongation. The RNA polymerase associates transiently with initiation factor sigma (a) and termination factor rho (p) to initiate and terminate the transcription, respectively. Association of RNA with these factors instructs the RNA polymerase either to initiate or terminate the process of transcription.

In bacteria, since the mRNA does not require any processing to become active and also since transcription and translation take place simultaneously in the same compartment (since there is no separation of cytosol and nucleus in bacteria), many times the translation can begin much before the mRNA is fully transcribed. This is because the genetic material is not ‘ separated from other cell organelles by a nuclear membrane consequently; transcription and translation can be coupled in bacteria.

Question 81.
Write the salient features of genetic code.
Answer:
The salient features of genetic code are as follows:

• The genetic codon is a triplet code and 61 codons code for amino acids and 3 codons do not code for any amino acid and function as stop codon (Termination).
• The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino
• acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial, and chloroplast genomes. However, similarities are more common than differences.
• A non-overlapping codon means that the same letter is not used for two different codons. For instance, the nucleotide sequence GUTJ and GUC represents only two codons.
• It is comma less, which means that the message would be read directly from one end to the other i.e., no punctuation are needed between two codes.
• A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.
• Non-ambiguous code means that one codon will code for one amino acid.
• The code is always read in a fixed direction i.e. from 5’ → 3’ direction called polarity.
• AUG has dual functions. It acts as an initiator codon and also codes for the amino acid methionine.
• UAA, UAG (tyrosine), and UGA (tryptophan) codons are designated as termination (stop) codons and also are known as “non-sense” codons.

Question 82.
Mutations on genetic code affect the phenotype. Describe with an example.
Answer:
The simplest type of mutation at the molecular level is a change in nucleotide that substitutes one base for another. Such changes are known as base substitutions which may occur spontaneously or due to the action of mutagens. A well studied example is sickle cell anaemia in humans which results from a point mutation of an allele of (3-haemoglobin gene (PHb).

A haemoglobin molecule consists of four polypeptide chains of two types, two a chains and two β-chains. Each chain has a heme group on its surface. The heme groups are involved in the binding of oxygen. The jruman blood disease, sickle cell anaemia is due to abnormal haemoglobin. This abnormality in haemoglobin is due to a single base substitution at the sixth codon of the beta globingene from GAG to GTG in β -chain of haemoglobin. It results in a change of amino acid gluconic acid to valine at the 6th position of the β -chain. This is the classical example of point mutation that results in the change of amino acids residue glutamic acid to valine. The mutant haemoglobin undergoes polymerisation under oxygen tension causing the change in the shape of the RBC from biconcave to a sickle shaped structure.

Question 83.
Explain the mechanism of AteArperon of the E-coli.
Answer:
The Lac (Lactose) operon: The metabolism of lactose in E.coli requires three enzymes – permease, β-galactosidase (P-gat) and transacetylase. The enzyme permease is needed for entry of lactose into the cell, Pjgglactosidase brings about hydrolysis of lactose to glucose and galactose, while transacetylqSSgtransfers acetyl group from acetyl Co A to P-galactosidase. The lac operon consists of one-regulator gene (T gene refers to inhibitor) promoter sites (p), and operator site (o). Besides these, it has three structural genes namely lac z, y and lac a. The lac ‘z’ gene codes for P-gaiaqtttsidase, lac ‘y’ gene codes for permease and ‘a’ gene codes for transacetylase.

Jacob and Monod proposed the classical model of Lac operon to explain gene expression and regulation in E.coli. In lac a polycistronic structural gene is regulated by a common promoter and regulatory genfc When the cell is using its normal energy source as glucose, the ‘i’ gene transcribes a repressor mRNA and after its translation, a repressor protein is produced. It binds to the operator region of the operon and prevents translation, as a result, β-galactosidase is not produced. In the absence of preferred carbon source such as glucose, if lactose is available as an energy source for the bacteria then lactose enters the cell as a result of permease enzyme. Lactose acts as an inducer and interacts with the repressor to inactivate it. The repressor protein binds to the operator of the operon and prevents RNA polymerase from transcribing the operon. In the presence of inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase to bind to the promotor site and transcribe the operon to produce lac mRNA which enables formation of all the required enzymes needed for lactose metabolism. This regulation of lac operon by the repressor is an example of negative control of transcription initiation.

Question 84.
What are the objectives of the Human Genome project?
Answer:

• The main goals of the Human Genome Project are as follows:
•  Identify all the genes (approximately 30000) in human DNA.
• Determine the sequence of the three billion chemical base pairs that make up the human DNA.
• To store this information in databases.
• Improve tools for data analysis.
• Transfer related technologies to other sectors, such as industries.
• Address the ethical, legal and social issues (ELSI) that may arise from the project.

Question 85.
Write the salient features of the Human Genome Project.
Answer:

• Although human genome contains 3 billion nucleotide bases, the DNA sequences that encode proteins make up only about 5% of the genome.
• An average gene consists of 3000 bases, the largest known human gene being dystrophin with 2.4 million bases.
• The function of 50% of the genome is derived from transposable elements such as LINE Molecular Genetics 315 and ALU sequence.
• Genes are distributed over 24 chromosomes. Chromosome 19 has the highest gene density. Chromosomes 13 and the Y chromosomes have the lowest gene densities.
• The chromosomal organization of human genes shows diversity.
• There may be 35000-40000 genes in the genome and almost 99.9 nucleotide bases are exactly the same in all people.
• Functions for over 50 percent of the discovered genes are unknown.
• Less than 2 percent of the genome codes for proteins.
• Repeated sequences make up very large portion of the human genome. Repetitive sequences have no direct coding functions but they shed light on chromosome structure, dynamics and evolution (genetic diversity).
• Chromosome 1 has 2968 genes, whereas chromosome ’Y’ has 231 genes.
• Scientists have identified about 1.4 million locations, where single-base DNA differences (SNPs – Single nucleotide polymorphism – pronounce as ‘snips’) occur in humans.
• Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease-associated sequences and tracing human history.

Question 86.
Describe the principle involved in the DNA fingerprinting technique.
Answer:
The DNA fingerprinting technique was first developed by Alec Jeffreys in 1985. The DNA of a person and fingerprints are unique. There are 23 pairs of human chromosomes with 1.5 million pairs of genes. It is a well known fact that genes are segments of DNA which differ in the sequence of their nucleotides. Not all segments of DNA code for proteins, some DNA .segments have a regulatory function, while others are intervening sequences (introns) and still others are repeated DNA sequences.

In DNA fingerprinting, short repetitive nucleotide sequences are specific for a person. These nucleotide sequences are called variable number tandem repeats (VNTR). The VNTRs of two persons generally show variations and are useful as genetic markers. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation. The bulk DNA forms a major peak and the other small peaks are referred to as satellite DNA.

Depending on base composition (A: T rich or G : C rich), length of segment and number of repetitive units, the satellite DNA is classified into many subcategories such as micro-satellites and mini¬satellites, etc. These sequences do not code for any proteins, but they form a large portion of the human genome. These sequences show a high degree of polymorphism and form the basis of DNA fingerprinting. DNA isolated from blood, hair, skin cells, or other genetic evidences left at the scene of a crime can be compared through VNTR patterns, with the DNA of a criminal suspect to determine guilt or innocence. VNTR patterns are also useful in establishing the identity of a homicide victim, either from DNA found as evidence or from the body itself.

Question 87.
Draw a flow chart depicting the steps of the DNA fingerprinting technique.
Answer:

Higher Order Thinking Skills (HOTs) Questions

Question 1.
An mRNA strand has a series of triplet codons of which the first three codons are given below
(a) AUG (b) UUU (c) UGC
(i) Name the amino acid encoded by these triplet codons.
(ii) Mention the DNA sequence from which these triplet codons would have transcribed?
Answer:
(i) AUG codes for Methionine UUU codes for Phenylalanine UGC codes for Cysteine
(ii) TAC sequence of DNA is transcribed to AUG
AAA sequence of DNA is transcribed to UUU
ACG sequence of DNA is transcribed to UGC

Question 2.
Given below are the structures of tRNA molecules which are involved in translation process. In one tRNA, codon is mentioned but not the amino acid. In another tRNA molecule, amino acid is named and not the codon. Complete the figure by mentioning the respective amino acids and codons.
Answer:

Question 3.
A DNA fragment possesses 32 adenine bases and 32 cytosine bases. How many total number of nucleotides does that DNA fragment contains? Explain.
Answer:
128 nucleotides. Adenine always pair Thymine base.
If there are 32 adenine bases then there must be 32 Thymine bases. Similarly cytosine pairs with guanine. If cytosine bases are 32 in number the guanine bases will be equal to cytosine. So it makes a total of 128 nucleotides.

Question 4.
Following is a DNA sequence representing a part of the gene TAC TCG CCC TAT UAA CCC AAA ACC TCT using this derive A.

1. The RNA transcript
2. The spliced mRNA (consider all the codons with two Aderine bases are introns)
3. The total number of amino acids coded by the mRNA

Answer:

1. RNA transcript: AUG UGC GGG AUA GGG UUU UGG AGA
2. Spliced mRNA: AUG UGC GGG GGG UUU UGG
3. 6 amino acids are coded by mRNA

Question 5.
Complete the molecular processes by naming them

1. DNA → DNA
2. mRNA → Protein
3. RNA transcript → mRNA

Answer:

1. Replication
2. Translation
3. Splicing

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 9 Limits and Continuity Ex 9.6 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.6

Choose the correct or the most suitable answer

Question 1.

(1) 1
(2) 0
(3) ∞
(4) -∞
Answer:
(2) 0

Explaination:

Question 2.

(1) 2
(2) 1
(3) -2
(4) 0
Answer:
(3) -2

Explaination:

Question 3.

(1) 0
(2) 1
(3) √2
(4) does not exist
Answer:
(3) √2

Explaination:

Question 4.

(1) 1
(2) -1
(3) 0
(4) 2
Answer:
(1) 1

Explaination:

Question 5.

(1) e⁴
(2) e²
(3) e³
(4) 1
Answer:
(1) e⁴

Explaination:

Question 6.

(1) 1
(2) 0
(3) -1
(4) \(\frac{1}{2}\)
Answer:
(4) \(\frac{1}{2}\)

Explaination:

Question 7.

(1) log ab
(2) log \(\left(\frac{\mathbf{a}}{\mathbf{b}}\right)\)
(3) log \(\left(\frac{\mathbf{b}}{\mathbf{a}}\right)\)
(4) \(\frac{\mathbf{a}}{\mathbf{b}}\)
Answer:
(2) log \(\left(\frac{\mathbf{a}}{\mathbf{b}}\right)\)

Explaination:

Question 8.

(1) 2 log 2
(2) 2(log 2)²
(3) log 2
(4) 3 log 2
Answer:
(2) 2(log 2)²

Explaination:


= (log 2) × (log 4)
= log 2 × log 2²
= log 2 × 2 log 2
= 2 (log 2)²

Question 9.

(1) -1
(2) 0
(3) 2
(4) 4
Answer:
(2) 0

Explaination:

Question 10.

(1) 2
(2) 3
(3) does not exist
(4) 0
Answer:
(3) does not exist

Explaination:

Question 11.
Let the function f be defined by

(1)
(2)
(3)
(4)
Answer:
(4)

Explaination:

Question 12.

(1) -2
(2) -1
(3) 0
(4) 1
Answer:
(3) 0

Explaination:

Question 13.

(1) 1
(2) 2
(3) 3
(4) 0
Answer:
(4) 0

Explaination:

Question 14.

(1) 6
(2) 9
(3) 12
(4) 4
Answer:
(3) 12

Explaination:

Question 15.

(1) \(\sqrt{2}\)
(2) \(\frac{1}{\sqrt{2}}\)
(3) 1
(4) 2
Answer:
(1) \(\sqrt{2}\)

Explaination:

Question 16.

(1) \(\frac{1}{2}\)
(2) 0
(3) 1
(4) ∞
Answer:
(1) \(\frac{1}{2}\)

Explaination:

Question 17.

(1) 1
(2) e
(3) \(\frac{1}{\mathrm{e}}\)
(4) 0
Answer:
(1) 1

Explaination:

Question 18.

(1) 1
(2) e
(3) \(\frac{1}{2}\)
(4) 0
Answer:
(1) 1

Explaination:


Put tan x – x = y
When x = 0, tan 0 – 0 = y
0 = y

= e⁰ × 1 = 1 × 1 = 1

Question 19.
The value of is
(1) 1
(2) -1
(3) 0
(4) ∞
Answer:
(1) 1

Explaination:

Question 20.
The value of , where k is an integer is
(1) -1
(2) 1
(3) 0
(4) 2
Answer:
(2) 1

Explaination:

Question 21.
At x = \(\frac{3}{2}\) the function f(x) = \(\frac{|2 x-3|}{2 x-3}\) is
(1) continuous
(2) discontinuous
(3) differentiable
(4) non zero
Answer:
(2) discontinuous

Explaination:
f(x) = \(\frac{|2 x-3|}{2 x-3}\)
f(x) is not defined at x = \(\frac{3}{2}\)
∴ f(x) is discontinuous at x = \(\frac{3}{2}\)

Question 22.
Let f : R → R be defined by

(1) discontinuous x = \(\frac{1}{2}\)
(2) continuous x = \(\frac{1}{2}\)
(3) continuous everywhere
(4) discontinuous everywhere
Answer:
(2) continuous x = \(\frac{1}{2}\)

Explaination:

Question 23.

(1) \(\frac{2}{3}\)
(2) –\(\frac{2}{3}\)
(3) 1
(4) 0
Answer:
(2) –\(\frac{2}{3}\)

Explaination:

At = -1, f(x) has a removable discontinuity Redefining f(x) as

Question 24.
Let f be a continuous function on [2, 5]. If f takes only rational values for all x and f(3) = 12, then f (4.5) is equal to
(1)
(2) 12
(3) 17.5
(4)
Answer:
(2) 12

Explaination:
Given f(3) = 12
f takes only rational values
f(x) = 12
f(3) = 12
f(4.5) = 12

Question 25.
Let a function f be defined by f(x) = \(\) for x ≠ 0 and f(0) = 2. Then f is
(1) continuous nowhere
(2) continuous everywhere
(3) continuous for all x except x = 1
(4) continuous for all x except x = 0
Answer:
(4) continuous for all x except x = 0

Explaination:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 9 Limits and Continuity Ex 9.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.5

Question 1.
Prove that f(x) = 2x² +3x – 5 is continuous at all points in R.
Answer:
f(x) = 2x² + 3x – 5
Clearly f(x) is defined for all points of R.
Let x₀ be an arbitrary point in R. Then
f(x₀) = 2x₀² + 3x₀ – 5 ——- (1)

Thus, f(x) is defined at all points of R limit of f(x) exist at all points of R and is equal to the value of the function f (x).
Thus f (x) is continuous at all points of R.

Question 2.
Examine the continuity of the following
(i) x + sin x
Answer:
Let f(x) = sin x
f (x) is defined at all points of R.
Let x₀ be an arbitrary point in R.

= x_o + sin x₀ …….. (1)
f (x_o) = x_o + sin x_o ……… (2)
From equations (1) and (2) we get

∴ At all points of R, the limit of f (x) exists and is equal to the value of the function.
Thus, f( x) satisfies ail conditions for continuity.
Therefore, f(x) is continuous at all points of f(x).

(ii) X² cos x
Answer:
Let f(x) = x² cos x
f (x) is defined at all points of R.
Let x₀ be an arbitrary point in R. Then

= x²₀ cos ₀
f(x₀) = x²₀ cos ₀
From equation (1) and (2), we have

∴ The limit at x = x₀ exist and is equal to the value of the function f(x) at x = x₀.
Since x₀ is arbitrary, the limit of the function exist and is equal to the value of the function for all points in R.
∴ f( x) satisfies all conditions for continuity. Hence
f (x) is a continuous function in R.

(iii) e^x tan x
Answer:
Let f(x) = e^x tan x
f (x) is defined at ail points of R.
except at (2n + 1)\(\frac{\pi}{2}\), n ∈ Z.
Let x₀ be an arbitrary point in R – (2n + 1)\(\frac{\pi}{2}\), n ∈ Z

∴ Limit at x = x₀ exist and is equal to the value of the function f(x) at x = x₀.
Since x₀ is arbitrary the limit of the function. f(x) exists at all points in R – (2n + 1)\(\frac{\pi}{2}\), n ∈ Z and is equal to the value of the function f (x) at that points.
∴ f (x) satisfies all conditions for continuity. Hence,
f(x) is continuous at all points of R – (2n + 1)\(\frac{\pi}{2}\), n ∈ Z

(iv) e^2x + x²
Answer:
Let f(x) = e^2x + x²
Clearly, f(x) is defined for all points in R.
Let x₀ be an arbitrary point in R.

From equations (1) and (2) we have,
The limit of the function f(x) exist at x = x₀ and is equal to the value of the function f(x) at x – x₀.
Since x₀ is an arbitrary point in R, the above is true for all points in R. Hence f (x) satisfies all conditions for continuity. Hence f (x) is continuous at all points of R.

(v) x . log x
Answer:
Let f(x) = x log x
The function f(x) is defined in the open interval (0 , ∞) since log x is defined for x > 0. Let x₀ be an arbitrary point in (0, ∞). Then

= x₀ log x₀
f(x₀) = x₀ log x₀
From equation (1) and (2) we have

∴ The limit of the function f(x) exists at x = x₀ and is equal to the value of the function .
Since x₀ is an arbitrary point the above is true for all points in (0, ∞).
∴ f(x) is continuous at all points of (0, ∞).

(vi) \(\frac{\sin x}{x^{2}}\)
Answer:
f(x) = \(\frac{\sin x}{x^{2}}\)
f(x) is not defined at x = 0
∴ f(x) is defined for all points of R – {0}
Let x₀ be an arbitrary point in R – {0}. Then

From equation (1) and (2) we have

∴ The limit of the function f(x) exist at x = x₀ and is equal to the value of the function f(x) at x = x₀.
Since x₀ is an arbitrary point in R – {0}, the above result is true for all points in R – {0}.
∴ f(x) is continuous at all points of R – {0}.

(vii) \(\frac{x^{2}-16}{x+4}\)
Answer:
Let f(x) = \(\frac{x^{2}-16}{x+4}\)
f(x) is not defined at x = – 4
∴ f(x) is defined for all points of R – {- 4}.
Let x₀ be an arbitrary point in R – {- 4}. Then

From equation (1) and (2) we have

Thus the limit of the function f (x) exist at x = x₀ and is equal to the value of the function f (x) at x = x₀.
Since x₀ is an arbitrary point in R – {- 4} the above result is true for all points in R – { – 4}.
∴ f(x) is continuous at all points of R – {- 4}.

(viii) |x + 2| + |x – 1|
Answer:
let f(x) = |x + 2| + |x – 1|
f( x) is defined for all points of R. Let x₀ be an arbitrary point in R. Then

From equation (1) and (2) we get

Thus the limit of the function f(x) exist at x = x₀ and is equal to the value of the function at x = x₀. Since x = x₀ is an arbitrary point in R, the above
result is true for all points in R. Hence f (x) is continuous at all points of R.

(ix) \(\frac{|x-2|}{|x+1|}\)
Answer:
Let f(x) = \(\frac{|x-2|}{|x+1|}\)
f(x) is defined for all points of R except at x = – 1.
∴ f (x) is defined for all points of R – { – 1 }.
Let x₀ be an arbitrary point in R – {- 1}.Then


From equation (1) and (2) we have

Hence the limit of the function f(x) at x = x₀ exists and is equal to the value of the function at x = x₀.
Since x = x₀ is an arbitrary point in R – { – 1 }, the above result is true for all points in R – {- 1).
∴ f(x) is continuous at all points of R – {- 1}.

(x) cot x + tan x
Answer:
Let f(x) = cot x + tan x

∴ The limit of the function f(x) exists at x = x₀ and is equal to the value of the function f (x) at x = x₀.
Since x₀ is an arbitrary point , the above result is true for all points of R – \(\left\{\frac{\mathrm{n} \pi}{2}\right\}\), n ∈ z.
∴ f(x) is continuous at all points of R – \(\left\{\frac{\mathrm{n} \pi}{2}\right\}\), n ∈ Z

Question 3.
Find the points of discontinuity of the function f, where
(i)
Answer:

(ii)
Answer:

For the point x₀ < 2, we have the limit of the function that exists and is equal to the value of the function at that point.
Since x₀ is an arbitrary point the above result is true for all x < 2.
∴ f(x) is continuous in (-∞, 2).
Let x₀ be an arbitrary point such that x₀ > 2 then

∴ For the point x₀ > 2, the limit of the function exists and is equal to the value of the function.
Since x₀ is an arbitrary point the above result is true for all x > 2.
∴ The function is continuous at all points of (2, ∞). Hence the given function is continuous at all points of R.

(iii)
Answer:
Clearly, the given function is defined at all points of R.
Case (i) At x = 2

Case (ii) for x < 2
Let x₀ be an arbitrary point in (- ∞, 2).

∴ f(x)is continuous at x = x₀ in (- ∞, 2).
Since x₀ is an arbitrary point in (- ∞, 2), f(x) is continuous at all points. f(-∞, 2).

Case (iii) for x > 2
Let y₀ be an arbitrary point in (2, ∞). Then


Hence, f(x) is continuous at x = y₀ in (2, ∞).
Since y_o is an arbitrary point of (2, ∞), f (x) is continuous at all points of (2, ∞).

∴ By case (i) case (ii) and case (iii) f(x) is continuous at all points of R.

(iv)
Answer:

Hence, f (x) is continuous at x = x₀. Since x₀ is an arbitrary point of \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\), f(x) is continuous at all points of \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\).
Hence, f (x) is continuous at all points \(\left[0, \frac{\pi}{2}\right)\),

Question 4.
At the given point x₀ discover whether the given function is continuous or discontinuous citing the reasons for your answer
(i)
Answer:

(ii)
Answer:

Question 5.
Show that the function

Answer:

Clearly, the given function f(x) is defined at all points of R.
Case (i) Let x₀ ∈ (- ∞, 1) then

f (x) is continuous at x = x₀.
Since x₀ is arbitrary f(x) is continuous at all points of (-∞, 1).

Case (ii) Let x₀ ∈ (1, ∞) then

f (x) is continuous at x = x₀.
Since x₀ is an arbitrary point of (1, ∞), f(x) is continuous at all points of (1, ∞).

Case (iii) Let x₀ = 1 then

Hence, f (x) is continuous at x = 1.
Using all the three cases, we have f (x) is continuous at all the points of R.

Question 6.

Answer:

Question 7.

Graph the function. Show that f(x) continuous on (- ∞, ∞)
Answer:


When x < 0 We have y = 0
When 0 ≤ x < 2 We have y = x²
When x ≥ 2 We have y = 4
Case (i) If x < 0 ie. (-∞, 0) then f (x) = 0
which is clearly continuous in (-∞, 0).

Case (ii) If 0 ≤ x < 2 je. [0 , 2)
Let x₀ be an arbitrary point in [0, 2)

Hence f(x) is continuous at x = x₀. Since x = x₀ is an arbitrary f(x) is continuous at all points of [0, 2).

Case (iii) x ≥ 2 je. [2, ∞)
f( x) = 4 which is clearly continuous in [2 , ∞)

Case (iv) at x = 2,

f(2) = 4
∴ f(x) is continuous at x = 2.
∴ Using case (j) case (ii) case (iii) and case (iv) we have f(x) is continuous at all points of R.

Question 8.
If f and g are continuous functions with f(3) = 5 and [2f(x) – g(x)] = 4, find g(3)
Answer:
Given f and g are continuous functions.

2 f(3) – g(3) = 4
2 × 5 – g(3) = 4
10 – 4 = g(3)
g(3) = 6

Question 9.
Find the points at which f is discontinuous. At which of these points f is continuous from the right, from the left, or neither? Sketch the graph of f.
(i)
Answer:

(ii)
Answer:

Question 10.
A function f is defined as follows:

is the function continuous?
Answer:

Question 11.
Which of the following functions f has a removable discontinuity at x = x₀? If the discontinuity is removable, find a function g that agrees with f for x ≠ x₀ and is continuous on R
(i)
Answer:
f(x) is not defined at x = -2

Redefine the function f(x) as

∴ f (x) has a removable discontinuity at x = -2.
Clearly, g (x) is continuous on R.

(ii)
Answer:
The function f(x) is not defined at x = -4.

Limit the function f (x) exist at x = -4.
∴ The function f (x) has a removable discontinuity at x = -4.
Redefine the function f (x) as

Clearly, the function g(x) is continuous on R.

(iii)
The function f(x) is not defined at x = 9.


∴ Limit of the function f(x) exists at x = 9.
Hence, the function f(x) has a removable discontinuity at x = 9. Redefine the function f(x) as

Clearly, g (x) is defined at all points of R and is continuous on R.

Question 12.
Find the constant b that makes g continuous on (-∞, ∞).

Answer:

Given g is continuous on R.
∴ g (x) is continuous at x = 4.

4² – b² = b × 4 + 20
16 – b² = 4b + 20
b² + 4b + 20 – 16 = 0
b² + 4b + 4 = 0
(b + 2)² = 0
b + 2 = 0 ⇒ b = -2

Question 13.
Consider the function f (x) = x sin \(\frac{\pi}{x}\) What value must we give f (0) in order to make the function continuous everywhere?
Answer:
f(x) = x sin \(\frac{\pi}{x}\)
Define f(x) on R as

∴ f(0) = 0. Then f(x) is continuous on R.

Question 14.
The function f(x) = \(\frac{x^{2}-1}{x^{3}-1}\) is not defined at x = 1. What value must we give f(1) in order to make f(x) continuous at x = 1 ?
Answer:


The function f (x) has a removable discontinuity at x = 1. Redefine f (x) as

∴ f(1) = \(\frac{2}{3}\). Then f(x) will be continuous at x = 1

Question 15.
State how continuity is destroyed at x = x₀ for each of the following graphs.
Answer:
(a)
The left – hand limit and right hand limit does not coincide at x = x₀

(b)
The function f(x) is not defined at x = x₀ and hence the continuity is destroyed at x = x₀

(c)
The limit of f(x) does not exist at x = x₀

(d)
The left hand limit and right – hand limit does not coincide at x = x₀

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 9 Limits and Continuity Ex 9.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.4

Evaluate the following limits:

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.

Answer:

Question 9.

Answer:

Question 10.

Answer:

Question 11.

Answer:

Put x² + a² = y put x² + b² = z
when x = 0 ⇒ y = a²
when x = 0 ⇒ z = b²

Question 12.

Answer:

Question 13.

Answer:

Question 14.

Answer:

Question 15.

Answer:

Question 16.

Answer:

Question 17.

Answer:

Question 18.

Answer:

Question 19.

Answer:

Question 20.

Answer:

Question 21.

Answer:

Question 22.

Answer:

Question 23.

Answer:

Question 24.

Answer:

Question 25.

Answer:

Question 26.

Answer:

Question 27.

Answer:

Question 28.

Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 11 Biodiversity and its Conservation Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 11 Biodiversity and its Conservation

12th Bio Zoology Guide Biodiversity and its Conservation Text Book Back Questions and Answers

Question 1.
Which of the following region has maximum biodiversity?
(a) Taiga
(b) Tropical forest
(c) Temperate rain forest
(d) Mangroves
Answer:
(b) Tropical forest

Question 2.
Conservation of biodiversity within their natural habitat is ……………….
(a) In-situ conservation
(b) Ex-situ conservation
(c) In vivo conservation
(d) In vitro conservation
Answer:
(a) In-situ conservation

Question 3.
Which one of the following is not coming under in-situ conservation?
(a) Sanctuaries
(b) Natural parks
(c) Zoological park
(d) Biosphere reserve
Answer:
(c) Zoological park

Question 4.
Which of the following is considered a hotspot of biodiversity in India?
(a) Western ghats
(b) Indo-gangetic plain
(c) Eastern Himalayas
(d) A and C
Answer:
(d) A and C

Question 5.
The organization which published the red list of species is ………………
(a) WWF
(b) IUCN
(c) ZSI
(d) UNEP
Answer:
(b) IUCN

Question 6.
Who introduced the term biodiversity?
(a) Edward Wilson
(b) Walter Rosen
(c) Norman Myers
(d) Alice Norman
Answer:
(b) Walter Rosen

Question 7.
Which of the following forests is known as the lungs of the planet Earth?
(a) Tundra forest
(b) Rain forest of north east India
(c) Taiga forest
(d) Amazon rain forest
Answer:
(d) Amazon rain forest

Question 8.
Which one of the following are at high risk extinction due to habitat destruction?
(a) Mammals
(b) Birds
(c) Amphibians
(d) Echinoderms
Answer:
(c) Amphibians

Question 9.
Assertion: The Environmental conditions of the tropics are favourable for speciation and diversity of organisms.
Reason: The climate seasons, temperature, humidity and photoperiod are more or less stable and congenial.
(a) Both Assertion and Reason are true and Reason explains Assertion correctly.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true, but Reason is false.
(d) Both Assertion and Reason are false.
Answers:
(a) Both Assertion and Reason are true and Reason explains Assertion correctly.

Question 10.
Define endemism.
Answer:
Endemism: A species or a taxon which is confined to a specific area
E..g: Nilgiri Thar

Question 11.
How many hotspots are there in India? Name them.
Answer:
India encloses 4 biodiversity hotspots. They are

• Himalayan
• Indo-Burma
• Western Ghats
• Sunderland

Question 12.
What are the three levels of biodiversity?
Answer:

• Genetic Diversity
• Species Diversity
• Community / Ecosystem Diversity

Question 13.
Name the active chemical found in the medicinal plant Rauwolfia vomitoria. What type of diversity does it belong to?
Answer:
Rauwolfia vomitoria can be cited as an example of genetic diversity. Reserpine is an active chemical extracted from Rauwolfia vomitoria.

Question 14.
“Amazon forest is considered to be the lungs of the planet”-Justify this statement.
Answer:
Amazon rain forest is known as the “lungs of the planet” since a great proportion of CO₂ released due to anthropogenic activities are uptaken by their dense tropical forests, which is _ homologous to the functioning of human beings except for the difference in gases.

Question 15.
‘Red data book’-What do you know about it?
Answer:
Red Data Book or Red list is a catalogue of taxa facing the risk of extinction. IUCN – International Union of Conservation of Nature and Natural Resources, which is renamed as WCU – World Conservation Union (Morges Switzerland) maintains the Red Databook. The concept of Red list was mooted in 1963. The purpose of preparation of Red List are:

• To create awareness on the degree of threat to biodiversity
• Identification and documentation of species at high risk of extinction
• Provide global index on declining biodiversity
• Preparing conservation priorities and help in conservation of action

Information on international agreements on conservation of biological diversity Red list has eight categories of species.

1. • Extinct
   • Extinct in wild
   • Critically Endangered
   • Endangered
   • Vulnerable
   • Lower risk
   • Data deficiency
   • Not evaluated

Question 16.
Extinction of a keystone species led to loss of biodiversity – Justify.
Answer:
A keystone species is an organism that helps define an entire ecosystem. Without the keystone species, a particular ecosystem would be dramatically disturbed or even ceased. Keystone species either directly or indirectly affects every species in a particular ecosystem. If a keystone species is lost or removed no other organism would compensate its ecological niche.

E.g. Jaguar is a keystone species. As a top predator, it plays a crucial role in the ecosystem. Without jaguar, there is an exponential increase in herbivoral population that would decimate the plants of the ecosystem. At one point even the herbivore populations also get declined due to the lack of vegetation. Thus jaguar acts a keystone species.

Question 17.
Compare and Contrast the insitu and exsitu conservation.
Answer:

Question 18.
What are called endangered species? Explain with examples.
Answer:
A species which has been categorized as very likely to become extinct. E.g., Bengal tiger, Polar bears.

Question 19.
Why do we find a decrease in biodiversity distribution, if we move from the tropics towards the poles?
Answer:
There is a decrease in biodiversity as we move from tropics towards poles due to a drop in temperature which makes the condition unfavourable for the majority of organisms to survive.

Question 20.
What are the factors that drive habitat loss?
Answer:
Natural habitats are destroyed for the purpose of settlement, agriculture, mining, industries, and construction of highways. As a result, species are forced to adapt to the changes in the environment or move to other places. If not, they become victim to predation, starvation, disease and eventually die or results in human animal conflict.

Question 21.
Where are biodiversity hotspots normally located? Why?
Answer:
Hotspots are areas characterized with high concentration of endemic species experiencing unusual rapid rate of habitat modification loss. Norman Myers defined hot spots as “regions that harbour a great diversity of endemic species and at the same time, have been significantly impacted and altered by human activities.”

Question 22.
Why is biodiversity so important and worthy of protection?
Answer:
Biodiversity reflects the number of different organisms and their relative frequencies in an ecological system and constitutes the most important functional component of a natural ecosystem. It helps to maintain ecological processes, create soil, recycle nutrients, influence climate, degrade waste, and control diseases. It provides an index of the health of an ecosystem. The survival of the human race depends on the existence and well-being of all life forms (plants and animals) in the biosphere.

Question 23.
Why do animals have greater diversification than plant diversity?
Answer:
The distribution of plants and animals is not uniform around the world. Organisms require different sets of conditions for their optimum metabolism and growth. Plants in general have the ability to withstand and overcome environmental fluctuations. Moreover, majority of plants are autotrophs so they can synthesize their own food, hence they show minimal modifications.
In case of animals, they have to tolerate climatic fluctuations, migrate to other areas in search of food, or adapt themselves or their body parts according food they consume thus varying from ancestors leading to evolution of new species. Hence it is understood that climatic condition food source, predation, competition and other natural forces lead to much diversification among animals than in plants.

Question 24.
Alien species invasion is a threat to endemic species – substantiate this statement.
Answer:
Exotic species are organisms often introduced unintentionally or deliberately for commercial purpose, as biological control agents and other uses. They often become invasive and drive away the local species and is considered as the second major cause for extinction of species. Tilapia fish (Jilabi kendai) (Oreochromis mosambicus) introduced from east coast of South Africa in 1952 for its high productivity into Kerala’s inland waters, became invasive, due to which the native species such as Puntius dubius and Labeo kontius face local extinction.
Amazon sailfin catfish is responsible for destroying the fish population in the wetlands of Kolkata. The introduction of the Nile Perch, a predatory fish into Lake Victoria in East Africa led to the extinction of an ecologically unique assemblage of more than 200 nature species of cichlid fish in the lake.

Question 25.
Mention the major threats to biodiversity caused by human activities. Explain.
Answer:
Human activities, both directly and indirectly are today’s main reason for habitat loss and biodiversity loss. Fragmentation and degradation due to agricultural practices, extraction (mining, fishing, logging and harvesting) and development (settlements, industrial and associated infrastructures) leads to habitat loss and fragmentation leads to the formation of isolated, small and scattered populations and as endangered species.

Some of the other threats include specialised diet, specialized habitat requirement, large size, small population size, limited geographic distribution and high economic or commercial value. Large mammals by virtue of their size require larger areas to obtain the necessities of life – food, cover and mates than do smaller mammals.

The individual home range of Lion can be . “ about 100 square Km. Mammals have specialized dietary needs such as carnivores, frugivores and the need to forage over much larger areas than general dietary herbivores and omnivores. Mammals also have low reproductive output other than small rodents.

Question 26.
What is mass extinction? Will you encounter one such extinction in the near future. Enumerate the steps to be taken to prevent it.
Answer:
The Earth has experienced quite a few mass extinctions due to environmental catastrophes. A mass extinction occurred about 225 million years ago during the Permian, where 90% of shallow-water marine invertebrates disappeared.

Question 27.
In northeastern states, the jhum culture is a major threat to biodiversity – substantiate.
Answer:
In shifting cultivation, plots of natural tree vegetation are burnt away and the cleared patches 1 are farmed for 2-3 seasons, after which their fertility reduces to a point where crop production is no longer profitable.

The farmer then abandons this patch and cuts down a new patch of forest trees elsewhere for crop production. This system is practiced in the north-eastern regions of India. When vast areas are cleared and burnt, it results in loss of forest cover, pollution, and discharge of CO₂, which in turn attributes to a loss of habitat and climate change which has an impact on the faunal diversity of that region.

Question 28.
List out the various causes for biodiversity losses.
Answer:
The major causes of biodiversity decline are:

• Habitat loss, fragmentation, and destruction (affects about 73% of all species),
• Pollution and pollutants (smog, pesticides, herbicides, oil slicks, and GHGs).
• Climate change.
• Introduction of alien/exotic species.
• Overexploitation of resources (poaching, indiscriminate cutting of trees, overfishing, hunting, and mining).
• Intensive agriculture and aquacultural practices.
• Hybridization between native and non-native species and loss of native species
• Natural disasters (Tsunami, forest fire, earthquake,s, and volcanoes).
• Industrialization, Urbanization, infrastructure development, Transport – Road and Shipping activity, communication towers, dam construction, unregulated tourism, and monoculture are a common area of specific threats.
• Co-extinction

Question 29.
How can we contribute to promoting biodiversity conservation?
Answer:

• identify and protect all threatened species
• identify and conserve in protected areas the wild relatives of all the economically important organisms
• identify and protect critical habitats for feeding, breeding, nursing, resting of each species
• resting, feeding, and breeding places of the organisms should be identified and protected.
• Air, water, and soil should be conserved on a priority basis
• Wildlife Protection Act should be implemented

Question 30.
‘Stability of a community depends upon its species diversity’ – Justify the statement.
Answer:
Species diversity leads to a stable community because an area with more species diversity always leads to higher productivity thus maintains a stable community.

Question 31.
Write a note on
(i) Protected areas
(ii) Wildlife sanctuaries
(iii) WWF.
Answer:
(i) Protected areas are biogeographical areas, where biological diversity along with natural and cultural resources is protected, maintained, and managed through legal measures. Protected areas include national parks, wildlife sanctuaries, community reserves, and biosphere reserves.

(ii) Any area other than the area comprised of any reserve forest or the territorial waters can be notified by the State Government to constitute as a sanctuary if such area is of adequate ecological, faunal, floral, geomorphological, natural or zoological significance. This is for the purpose of protecting, endangered factual species. Some restricted human activities are allowed inside the sanctuary area. Ecotourism is permitted, as long as animal life is undisturbed.

(iii) WWF stands for World Wide Fund for nature is an international NGO working in the field of wildlife conservation.

12th Bio Zoology Guide Biodiversity and its Conservation Additional Important Questions and Answers

12th Bio Zoology Guide Biodiversity and its Conservation One Mark Questions and Answers

Question 1.
Who coined the term Bio-diversity?
Answer:
Walter Rosen.

Question 2.
Which is not an indices of species diversity?
(a) Alpha diversity
(b) Beta diversity
(c) Delta diversity
(d) Gamma diversity
Answer:
(c) Delta diversity

Question 3.
The total number of mega biodiversity countries in the world is …………….
(a) Twelve
(b) Fifteen
(c) Seventeen
(d) Nineteen
Answer:
(c) Seventeen

Question 4.
How many biogeographic zones are there in India?
(a) Twelve
(b) Seventeen
(c) Ten
(d) Fifteen
Answer:
(c) Ten

Question 5.
The most important pattern of biodiversity is ……………………..
(a) Longitudinal gradient in diversity
(b) Latitudinal gradient in diversity
(c) Polar gradient diversity
(d) Equatorial gradient in diversity
Answer:
(b) Latitudinal gradient in diversity

Question 6.
Which of the following denotation is correct regarding increasing diversity?
(a) Poles < Equator
(b) Equator < Pole
(c) Pole = Equator
(d) Latitude = Longitude
Answer:
(a) Poles < Equator

Question 7.
Select the proper sequence indicating the increasing order of biodiversity.
(a) Polar, Temperate and Polar
(b) Tropics, Temperate and Polar
(c) Temperate, Tropic and Polar
(d) Polar, Tropic and Temperate
Answer:
(a) Polar, Temperate and Polar

Question 8.
Select the correct linear equation describing the species-area relationship.
(a) loc C = log S + Z log A
(b) Z log A = log S + log C
(c) log S = log C + Z log A
(d) log C = log S ± Z log A
Answer:
(c) log S = log C + Z log A

Question 9.
If meat-eating animals are called carnivores, how do you call the animals that thrive mostly on fruits?
Answer:
Frugivore

Question 10.
The wild ass is endemic to ………………..
(a) the Western Ghats
(b) Deccan Peninsula
(c) the Himalayas
(d) Indian desert
Answer:
(d) Indian desert

Question 11.
Which is considered as the Biogeographical Gateway of India?
(a) the Himalayas
(b) Andaman & Nicobar
(c) North – East India
(d) Mumbai
Answer:
(c) North – East India

Question 12.
Species introduced deliberately in an area are referred to as …………………
(a) Endemic species
(b) Vulnerable species
(c) Exotic species
(d) Extinct species
Answer:
(c) Exotic species

Question 13.
Tilapia fish (Oreochromis Mozambique) is an exotic breed from ……………………
(a) Mexico
(b) South Africa
(c) Canada
(d) Central America
Answer:
(b) South Africa

Question 14.
Mention the correct number of biodiversity hotspots identified throughout the world.
(a) 29
(b) 16
(c) 34
(d) 46
Answer:
(c) 34

Question 15.
Which is not an accepted biodiversity hotspot of India?
(a) Indian Himalayas
(b) Western Ghats
(c) Indo-Burma
(d) Deccan Plateau
Answer:
(d) Deccan Plateau

Question 16.
A species is considered extinct
(a) When its member is confined to a particular area
(b) When its member is maintained in a non-native area
(c) When none of its members is alive in the native area
(d) When none of its members alive anywhere in the world
Answer:
(d) When none of its members are alive anywhere in the world.

Question 17.
The concept of Red list was noted in …………………
(a) 1953
(b) 1963
(c) 1973
(d) 2003
Answer:
(6) 1963

Question 18.
Match the following
(a) Tiger reserves in India – (i) 4
(b) Hotspots in India – (ii) 104
(c) Biosphere reserves in India – (iii) 27
(d) National parks in India – (iv) 18
Answer:
a – iii, b – i, c – iv, d – ii

Question 19.
Statement 1: Biodiversity is the assemblage of different life form.
Statement 2: The term biodiversity was introduced by Edward Wilson.
(a) Statement 1 is correct, statement 2 in incorrect
(b) Statement 1 is incorrect, statement 2 in correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(fl) Statement 1 is correct, statement 2 is incorrect

Question 20.
Statement 1: India is the seventh-largest country in the world in terms of area.
Statement 2: It includes ten biogeographic areas.
(a) Statement 1 is correct, statement 2 is incorrect
(b) Statement 1 is incorrect, statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both the statements are correct

Question 21.
Statement 1: the Western Ghats extend from South Gujarat to Karnataka.
Statement 2: Wild ass is an endemic species of Western Ghats
(a) Statement 1 is correct, statement 2 is incorrect
(b) Statement 1 is incorrect, statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(d) Both the statements are incorrect

Question 22.
Statement 1: Exotic species are a non-native organisms.
Statement 2: Sailfin catfish is an exotic species in India.
(a) Statement 1 is correct, statement 2 in incorrect
(b) Statement 1 is incorrect, statement 2 in correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both the statements are correct

12th Bio Zoology Guide Biodiversity and its Conservation Two Marks Questions and Answers

Question 1.
Define biodiversity.
Answer:
Biodiversity is the variability among living organisms from all sources, including terrestrial, marine, and other aquatic ecosystems and the ecological complexes of which they are part. This includes diversity within species, between species and ecosystems of a region.

Question 2.
Who coined the term biodiversity? Name the levels of biodiversity.
Answer:
The term biodiversity was coined by Walter Rosen (1986). The levels of biodiversity are – Genetic diversity, Species diversity, and ecosystem diversity.

Question 3.
What is species richness? Give example.
Answer:
The number of species per unit area at a specific time is called species richness, which denotes the measure of species diversity.
E.g., Western ghats have more amphibian species than Eastern ghats.

Question 4.
Enlist the factors that determine the distribution of biodiversity.
Answer:
Temperature, precipitation, distance from the equator (latitudinal gradient), altitude from sea level (altitudinal gradient) are some of the factors that determine biodiversity distribution patterns.

Question 5.
What are the most important causes of biodiversity loss?
Answer:
Habitat loss, fragmentation, and destruction.

Question 6.
Name any two alien animal species introduced in India.
Answer:
(a) Tilapia fish (Oreochromis Mozambique)
(b) African Apple snail (Achatina Fulica)

Question 7.
Name any four biogeographic zones in India.
Answer:
(a) the Himalayas
(b) Indian deserts
(c) Trans-Himalayan region
(d) Western ghats.

Question 8.
What do you mean by the term co-extinction?
Answer:
Coextinction of a species is the loss of a species as a consequence of the extinction of another.
E.g., orchid bees and forest trees by cross-pollination.

Question 9.
What are the natural causes of biodiversity loss?
Answer:
Natural threats include spontaneous jungle fires, tree falls, landslides, defoliation by insects or locust attacks.

Question 10.
Define hotspots.
Answer:
Hotspots are areas characterized with high concentration of endemic species experiencing unusual rapid rate of habitat modification loss.

Question 11.
What will be the consequences of the loss of biodiversity?
Answer:
Loss of biodiversity has a immense impact on plant and animal life. The loss of diversity leads to,

• Extinction of species
• Dramatic influence on the food chain and food web in.
• It will lead to immediate danger for food necessity

Question 12.
Name the four biodiversity hotspots in India.
Answer:
(a) the Himalayas
(b) Western Ghats
(c) Indo-Burma region
(d) Sundaland

Question 13.
What does IUCN stand for?
Answer:
IUCN – International Union for Conservation of Nature and Natural resources.

Question 14.
Define species extinction. What makes a species to become extinct?
Answer:
A species is considered extinct when none of its members are alive anywhere in the world. Environmental changes and population characteristics are the two major reasons for species extinction.

Question 15.
When a species is considered as locally extinct?
Answer:
A species is considered to be locally extinct when it is no longer found in an area it once inhabited but is still found elsewhere in the wild.

Question 16.
State the mission of IUCN.
Answer:
IUCN’s mission is to influence, encourage, and assist societies throughout the world to conserve nature and to ensure that any use of natural resources is equitable and ecologically sustainable.

Question 17.
What is the Red list? How many categories of species are mentioned in the Red List?
Answer:
Red List or Red databook is a catalogue of taxa facing a risk of extinction. It has 8 categories of species.

Question 18.
Mention any four categories of species mentioned in the Red data book.
Answer:

• Extinct
• Endangered
• Extinct in wild
• Vulnerable

Question 19.
How a national park can be defined?
Answer:
The national park is a natural habitat that is notified by the state government to be constituted as a National Park due to its ecological, faunal, floral, geomorphological, or zoological association of importance. No human activity is permitted inside the national park except the activities permitted by the Chief Wildlife Warden of the state.

Question 20.
Name any two species that are extinct due to human activities.
Answer:

• Dodo of Mauritius
• Steller’s cow of Russia

Question 21.
Define in-situ conservation.
Answer:
Conservation of animals in their natural habitat is called in-site conservation.
E.g., National parks.

Question 22.
What is the goal of “Project Tiger”?
Answer:
The project ensures a viable population of Bengal tigers in their natural habitats, protecting them from extinction and preserving areas of biological importance as a natural heritage.

Question 23.
Give the number of national parks in India. Name any two of them in Tamil Nadu.
Answer:
India has 104 National Parks. Guindy National Park (Chennai) and Mudumalai National Park (Nilgiris) are located in Tamil Nadu.

Question 24.
State the role of Biosphere Reserve.
Answer:
Biosphere Reserves are designated to deal with the conservation of biodiversity, economic and social development, and maintenance of associated cultural values.

Question 25.
Name few endangered species protected in Arignar Anna Zoological Park.
Answer:
Royal Bengal Tiger, Lion Tailed Macaque, Nilgiri Langur, and Gray Wolf.

Question 26.
Give the names of two methods of in-situ conservation.
Answer:

• Wildlife Sanctuaries
• Biosphere reserve

12th Bio Zoology Guide Biodiversity and its Conservation Three Marks Questions and Answers

Question 27.
Point out the biosphere reserves in Tamil Nadu.
Answer:

• Nilgiris (Tamil Nadu – Kerala)
• Agasthya malai (Tamil Nadu – Kerala – Karnataka)
• Gulf of Mannar (Tamil Nadu)

Question 28.
Write a note on Sacred Groves.
Answer:
A sacred grove or sacred woods are any groves of trees that are of special religious importance to a particular culture. Sacred groves feature in various cultures throughout the world.

Question 29.
What is ex-situ conservation?
Answer:
Ex-situ conservation of selected rare plants/ animals in places outside their natural homes. It includes offsite collections and gene banks.

Question 30.
Why Red list is prepared periodically?
Answer:
The purpose of preparation of Red List are:

• To create awareness on the degree of threat to biodiversity
• Identification and documentation of species at high risk of extinction
• Provide global index on declining biodiversity
• Preparing conservation priorities and help in conservation of action
• Information on international agreements on the conservation of biological diversity

Question 31.
Name the types of extinctions.
Answer:

• Natural Extinction
• Mass Extinction
• Anthropogenic Extinction

Question 32.
Point out the human activities that threaten biodiversity.
Answer:
Direct and indirect human activities have a detrimental effect on biodiversity. Direct human, activities like change in local land use, species introduction or removal, harvesting, pollution, and climate change contribute a greater pressure on the loss of biodiversity. Indirect human drivers include demographic, economic, technological, cultural, and religious factors.

Question 33.
Extinction of Dodo bird led to the danger of Calvaria tree – Justify,
Answer:
Another example for co-extinction is the connection between Calvaria tree and the extinct bird of Mauritius Island, the Dodo. The Calvaria tree is dependent on the Dodo bird for completion of its life cycle. The mutualistic association is that the tough horny endocarp of the seeds of Calvaria tree are made permeable by the actions of the large stones in birds gizzard and digestive juices thereby facilitating easier germination. The extinction of the Dodo bird led to the imminent danger of the Calvaria tree coextinction.

Question 34.
Give an account on slash and burn agriculture.
Answer:
In shifting cultivation, plots of natural tree vegetation are burnt away and the cleared patches are farmed for 2-3 seasons, after which their fertility reduces to a point where crop production is no longer profitable. The farmer then abandons this patch and cuts down a new patch of forest trees elsewhere for crop production.

This system is practiced in the north-eastern regions of India. When vast areas are cleared and burnt, it results in loss of forest cover, pollution and discharge of CO₂ which in turn attributes to a loss of habitat and climate change which has an impact on the faunal diversity of that regions.

Question 35.
Impact of Industrialization on Biodiversity – Comment.
Answer:
Industrialization is a major contributor to climate change and a major threat to biodiversity. Energy drives our industries, which is provided by burning of fossil fuels. This increases the emission of CO₂, a GHG, leading to climate change. Due to large scale deforestation, the emitted CO₂ cannot be absorbed fully, and its concentration in the air increases.

Climate change increases land and ocean temperature changes precipitation patterns and raises the sea level. This, in turn, results in melting of glaciers, water inundation, less predictability of weather patterns, extreme weather conditions, outbreak of squalor diseases, migration of animals and loss of trees in forest. Thus, climate change is an imminent danger to the existing biodiversity.

Question 36.
What are exotic species? Explain with example.
Answer:
Exotic species are organisms often introduced unintentionally or deliberately for commercial purpose, as biological control agents and other uses. They often become invasive and drive away the local species and is considered as the second major cause for extinction of species. Exotic species have proved harmful to both aquatic and terrestrial ecosystems.
Tilapia fish (Jilabi kendai) (Oreochromis mosambicus) introduced from east coast of South Africa in 1952 for its high productivity into Kerala’s inland waters, became invasive, due to which the native species such as Puntius dubius and Labeo kontius face local extinction.

Question 37.
Write a brief note on Habitat fragmentation.
Answer:
Habitat fragmentation is the process where a large, continuous area of habitat is both, reduced in area and divided into two or more fragments. Fragmentation of habitats like forest land into croplands, orchard lands, plantations, urban areas, industrial estates, transport, and transit systems has resulted in the destruction of complex interactions amongst species, (food chain and webs) destruction of species in the cleared regions, annihilation of species restricted to these habitats (endemic) and decreased biodiversity in the habitat fragments.

Animals requiring large territories such as mammals and birds are severely affected. The elephant corridors and migratory routes are highly vulnerable. The dwindling of many well-known birds (sparrows) and animals can be attributed to this.

Question 38.
Write a note on the biogeographic area – the Gangetic plains.
Answer:
Gangetic Plains: These plains are relatively homogeneously defined by the Ganges river system and occupy about 11% of the country’s landmass. This region is very fertile and extends up to the Himalayan foothills. Fauna includes rhinoceros, elephant, buffalo, swamp deer, hog-deer.

Question 39.
Compare Alpha diversity with Beta diversity.
Answer:
(i) Alpha diversity: It is measured by counting the number of taxa (usually species) within a particular area, community or ecosystem.
(ii) Beta diversity: It is species diversity between two adjacent ecosystems and is obtaining by comparing the number of species unique to each of the ecosystem.

Question 40.
What is species diversity?
Answer:
Species diversity refers to the variety in number and richness of the species in any habitat. The number of species per unit area at a specific time is called species richness, which denotes the measure of species diversity. The Western Ghats have greater amphibian species diversity than the Eastern Ghats. The more the number of species in an area the more is the species richness. The three indices of diversity are – Alpha, Beta and Gamma diversity.

Question 41.
State the principle of Stockholm declaration – 1972.
Answer:
The natural resources of the Earth, including air, water, land, flora, and fauna of natural ecosystems must be safeguarded for the benefit of the present and future generations through careful planning and management, as appropriate – Principle of the Stockholm Declaration. 1972.

12th Bio Zoology Guide Biodiversity and its Conservation Five Marks Questions and Answers

Question 42.
Give an account on genetic diversity and community diversity.
Answer:
Genetic diversity refers to the differences in genetic make-up (number and types of genes) between distinct species and to the genetic variation within a single species; also covers genetic variation between distinct populations of the same species. Genetic diversity can be measured using a variety of molecular techniques. India has more than 50,000 genetic variants of Paddy and 1000 variants of Mango. Variation of genes of a species increases with diversity in size and habitat. It results in the formation of different races, varieties and subspecies. Rauwolfia vomitaria, a medicinal plant growing in different ranges of the Himalayas shows differences in the potency and concentration of the active ingredient reserpine due to genetic diversity. Genetic diversity helps in developing adaptations to changing environmental conditions.

Community/Ecosystem diversity is the variety of habitats, biotic communities, and ecological processes in the biosphere. It is the diversity at ecosystem level due to diversity of niches, trophic levels and ecological processes like nutrient cycles, food webs, energy flow, and several biotic interactions. India with its alpine meadows, rain forests, mangroves, coral reefs, grasslands, and deserts has one of the greatest ecosystem diversity on Earth.

Question 43.
Why Tropical regions are rich in biodiversity?
Answer:
The reasons for the richness of biodiversity in the Tropics are:

• Warm tropical regions between the tropic of Cancer and Capricorn on either side of the equator possess congenial habitats for living organisms.
• Environmental conditions of the tropics are favourable not only for speciation but also for supporting both variety and number of organisms.
• The temperatures vary between 25°C to 35°C, a range in which most metabolic activities of living organisms occur with ease and efficiency.
• The average rainfall is often more than 200 mm per year.
• Climate, seasons, temperature, humidity, photoperiods are more or less stable and encourage both variety and numbers.
• Rich resource and nutrient availability.

Question 44.
What is the significance of the slope of regression in a species-area relationship?
Answer:
German Naturalist and Geographer Alexander von Humboldt explored the wilderness of South American jungles and found that within a region the species richness increased with the increasing area but upto a certain limit. The relationship between species richness and area for a wide variety of taxa (angiosperm plants, birds, bats and freshwater fishes) turned out to be the rectangular hyperbola. On a logarithmic scale, the relationship is a straight line described by the equation.
log S = log C + Z log A
where S = Species richness
A = Area
Z = Slope of the line
(regression coefficient)
C = Y-intercept

Regression coefficient Z generally has a value of 0.1-0.2 regardless of taxonomic group or region. However, in case of the species-area relationship in very large areas like entire continents, the slope of the line appears to be much steeper (Z-value in the range of 0.6-1.2). For example, in the case of the fruit eating (frugivorous) birds and mammals in the tropical forests of different continents, the slope is found to be a steeper line.

Question 45.
Point out any 5 functional attributions of biodiversity.
Answer:
The major functional attributes are:

• continuity of nutrient cycles or biogeochemical cycles (N₂, C, H₂O, P, S cycles)
• soil formation, conditioning or maintenance of soil health (fertility) by soil microbial diversity along with the different trophic members
• increases ecosystem productivity and provide food resources
• act as water traps, filters, water flow regulators, and water purifiers (forest cover and vegetation)
• climate stability (forests are essential for rainfall, temperature regulation, C02 absorption, which in turn regulate the density and type of vegetation)
• forest resource management and sustainable development

Question 46.
Explain in detail about various types of extinctions.
Answer:
There are three types of Extinctions
i. Natural extinction: It is a slow process of replacement of existing species with better adapted species due to changes in environmental conditions, evolutionary changes, predators and diseases. A small population can get extinct sooner than a large population due to inbreeding depression (less adaptivity and variation)

ii. Mass extinction: The Earth has experienced quite a few mass extinctions due to environmental catastrophes. Amass extinction occurred about 225 million years ago during the Permian, where 90% of shallow-water marine invertebrates disappeared.

iii. Anthropogenic extinctions: These are abetted by human activities like hunting, habitat destruction, overexploitation, urbanization and industrialization. Some examples of extinctions are Dodo of Mauritius and Steller’s sea cow of Russia. Amphibians seem to be at higher risk of extinction because of habitat destruction. The most serious aspect of the loss of biodiversity is the extinction of species. The unique information contained in its genetic material (DNA) and the niche it possesses are lost forever.

Question 47.
Give a comparative account on ex-situ conservation.
Answer:
Ex-Situ Conservation: It is conservation of selected rare plants/ animals in places outside their natural homes. It includes offsite collections and gene banks.

Offsite Collections: They are live collections of wild and domesticated species in Botanical gardens, Zoological parks, Wildlife safari parks, Arborata (gardens with trees and shrubs). The organisms are well maintained for captive breeding programmes. As a result, many animals which have become extinct in the world continue to be maintained in Zoological Parks. As the number increases in captive breeding, the individuals are selectively released in the wild. In this way, the Indian crocodile and Gangetic dolphin have been saved from extinction.

Gene Banks: Gene banks are a type of biorepository which preserves genetic materials. Seeds of different genetic strains of commercially important plants can be stored for long periods in seed banks, gametes of threatened species can be preserved in viable and fertile condition for long periods using cryopreservation techniques. However, it is not economically feasible to conserve all biological wealth and all the ecosystems. The number of species required to be saved from extinction far exceeds the conservation efforts.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
What will be the result, if the relationship between species richness and area for wide variety of taxa are plotted on a long scale?
Answer:
A rectangular hyperbola.

Question 2.
What may be the reasons for the entry of wild .lives into the agriculatural lands or towns?
Answer:

• Habitat lose / Habitat fragmentation
• Lack of food or water source

Question 3.
When does a species is categorized as endangered?
Answer:
A species that has been categorized as very likely to become extinct is an endangered species.

Question 4.
Give any two examples of anthropogenic extinction.
Answer:
Dodo of Mauritius Steller’s cow of Russia

Question 5.
Mention any two species that had become extinct very recently.
Answer:

• George, the tree snail (Achatinella apexfulva)
• Sudan – Northern white rhinoceros (Ceratotherium simum)

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th English Guide Pdf Supplementary Chapter 5 The Singing Lesson Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 11th English Solutions Supplementary Chapter 5 The Singing Lesson

11th English Guide The Singing Lesson Text Book Back Questions and Answers

C. Answer the following questions in a paragraph of about 150 words:

Question 1.
Describe Miss Meadow’s mood before and after receiving the telegram How did it affect her class?
Answer:
Miss. Meadow was heart-broken. The letter written by Basil had pierced her heart and she was bleeding. Her hatred and anger became a knife and she carried it with her. Her icy cold response to science Miss demonstrates it. She is least bothered about the tender feelings of young children who look at her face all-time for a friendly nod or smile of approval. Her favourite pupil Mary Beazley is baffled at her treatment of the chrysanthemum she had brought with so much love. The choice of the song “A lament” perfectly jells well with her worst mood. She is in fact in her heart lamenting over the loss of love, trust and future hopes. She is unnecessarily severe with young children forcing them to redo the singing which drives them to despair, pain and tears they manage to stifle.

After she receives the telegram from Basil apologizing for his insane letter, her mood changes to joy. She takes the chrysanthemum and keeps it close to her lips to conceal her blush. She goads the children to sing a song of joy congratulating someone for success. She persuades them to show warmth in their voices. Her warm and lively voice dominates the tremulous voices of the young ones. The young ones now realize that Miss Meadow who was in a wax earlier is now in her elements.

“My moods don’t just swing – they bounce, pivot, recoil, rebound, oscillate, fluctuate, and occasionally PIROUETTE. ”

Question 2.
‘The only difference between a good day and a bad day is your attitude’ Relate this to a real-life experience you have to share your thought in class.
Answer:
The difference between ‘A Good day’ and `A Bad day’ depends upon our attitude. Actually, you can make that happen by simply changing your attitude. It really is just about our attitude that shifts one good feeling to a bad one. If we let the bad feeling overcome us, chances are no good will come out of it.

Once I got a scolding from my mother for getting low marks in the Quarterly Exam I didn’t take my breakfast. With a bad state of the mind, I went to school. I thought that the day would be a bad day for me. During the break time, I sat alone and thought about why my mother advised and scolded me.

I realized my mistakes. Then I started to study well, concentrate on my classwork and listened to the class with keen eyes I changed my attitude towards the right way In the evening, I went to my mother and asked her to forgive me. My mother hugged me. So, that day ended with happiness.

Question 3.
You are busy getting ready for school. You receive a What’s App message from your best friend, saying that he/she is very upset over the fight you had yesterday and doesn’t want to talk to you anymore. This distresses you as she sounds very firm. However, today is a big day at school with two tests lined up. What will be your state of mind? How will you handle this situation?
Answer:
I always remember an anecdote. Kannadasan has recounted this anecdote. A temple elephant was proceeding to the temple. It’s mahout had washed him and applied sandal paste and holy ash on his forehead. Passerby greeted him like a God. As he was walking majestically, he was followed by a she elephant. A pig crossing the male elephant told its wife, “You see how the elephant was scared of me and gave way”. Overhearing the arrogant words of the pig, “The ‘she-elephant’ asked the ‘male elephant’ if it was true. The gentle animal smiled wisely and said, “I always focus on my goal.

We are on our way to a holy place. Even if I stamp on the pig by mistake, he would die. But I need to return to the tank for another wash.” “In life we need to avoid confrontation to ensure continuous progress in the chosen path. When I am a student, academics is quite important. Friendship is also important.

If a friend gets upset and if she really values my friendship, I can always say sorry and bring her around after the examination. If she is pig-headed and refuses to give up arrogance or anger, I will tell her I shall pray for her and move on. I will definitely find someone worthy of my true friends. In reality, true friends, can’t be angry for long with each other. Realizing the value of true friends.

I’ll send a message wishing her the very best of luck for her exam and promise to sort out the issue in the evening. Nothing needs to be taken as a permanent failure in a relationship or even in the examination. I would like to remember the Chinese proverb “One can’t help birds of sorrow hover over one’s head. But one can prevent them from building a nest in one’s head”.

“Never leave a true relationship for a few faults. Nobody is perfect; Nobody is correct. In the end… Affection is always greater than perfection.”

ஆசிரியரைப் பற்றி:

கேத்ரீன் மேன்ஸ்பீ ல்டு முர்ரி (1888-1923) இவர் கேத்தரின் மேன்ஸ்பீ ல்டு என்ற புனைப்பெயரில் எழுதிய நியூசிலாந்தின் சிறுகதை எழுத்தாளர். இவர் புகழ்பெற்ற எழுத்தாளர்களான டி.எச்.

லாரண்ஸ், விர்ஜீனியா உல்ப் ஆகியோருடன் தொடர்பு கொண்டிருந்தார். “Bliss, The garden Party”ஆகியவை இவரின் சிறுகதைத் தொகுப்பாகும். இவரால் தொகுக்கப்பட்டட கடிதங்கள் இவருக்கு மிகச்சிறந்த வெற்றியை தந்துள்ளது.

கதையைப் பற்றி:

இந்த கதையில் ஒரு இசையாசிரியை தன்னுடைய வாழ்க்கை அனுபவங்களை சொல்வதாக அமைந்துள்ளது. இசை ஆசிரியரின் வாழ்வில் நடந்த துன்பங்களை இசையின் மூலம் கடந்து சென்றதை விளக்குகிறது.

தன் காதல் வாழ்விலும் , திருமண வாழ்விலும் அனுபவித்த கசப்பான அனுபவங்களைப் பற்றி மிஸ்மெடோஸ் நீள கவுன் உடையுடன் கையில் சிறிய குச்சியுடன் (இசை மீட்ட உதவும் குச்சி (பேட்டன்)) இசையின் மூலம் பகிர்ந்து கொள்கிறார். வார்க்கையில் குடும்பத்தை வெறுத்த இவரின் கதையை விரிவாகக் காண்போம்.

The Singing Lesson Summary in Tamil

“இசைப் பாடம்” என்ற இக்கதையில் ஒரு இசை ஆசிரியரின் மனநிலையானது நியாயமற்ற முறையினால் மன அமைதியிழந்து அவளின் மனநிலைக்கேற்ப எவ்வாறு கடுமையாக மாற்றத்திற்குள்ளாகிறது என்பதை வாசிப்போம்.

நம்பிக்கையற்று – இதயத்தின் ஆழம் வரை சென்று புதைந்து கிடக்கும் மோசமான கத்தியைப் போன்று, செல்வி, (Miss Meadows), நீண்ட தளர் உடை மற்றும் தொப்பி அணிந்தவளாய், கையில் இசைக்குழுத் தலைவர் இசையை வழிநடத்தக் கூடிய சிறு மெல்லிய குச்சியைத் தூக்கிக் கொண்டு, இசை கற்பிக்கக் கூடிய கூடத்திற்குச் செல்லும் குளிரான நடைக் கூடத்தின் வழியாக நடந்தாள்.

பல்வேறு வயதுடைய சிறுமிகள் பரபரப்பான மகிழ்ச்சியான சூழலில் வளமான வாய்ப்புகளுடைய மனநிலையுடன் இதமான இலையுதிர்காலத்தின் ஒரு காலை வேளையில் பள்ளிக்கு வரும் நிலையில் விரைந்தும், குதித்துத் தாவியும், படபடப்புடனும் கடந்து சென்றார்கள், தாழ்வான வகுப்பறை களிலிருந்து முரசு போன்ற குரலொலிகள் கேட்டன; மணி அடித்தது; பறவையின் குரலையொத்த ஒரு குரலானது மீயூரியல்’ என்று சப்தமாக ஒலித்தது.

மாடியிலிருந்து தடாலென்ற ஒரு மாபெரும் சப்தம் வந்தது. யாரோ தசைகளை வலுப்படுத்தப் பயன்படுத்தும் இருபுறமும் சம எடையுள்ள உடற்பயிற்சிக் கருவியை நழுவ விட்டிருந்தனர்.

அறிவியல் ஆசிரியை செல்வி மெடோஸை நிறுத்தினார்.

அவள் தன்னுடைய இனிமையான குரலில் மெல்ல இழுத்துப் பேசுவது போல “காலை வணக்கம்” என்று கத்தினாள். “கடும் குளிராக இல்லையா? இது குளிர்காலமாக இருக்கும்” என்றாள். செல்வி மெடோஸ், துயரத்தை அணைத்துக் காண்டு, அறிவியல் ஆசிரியை வெறுப்புணர்ச்சியுடன் வெறிக்கப் பார்த்தாள்.

அவளது பொருத்த வரையில் எல்லாமே தேனைப் போன்று இனிமையானது, வெளிறிய நிறமுடையது. சிக்கலான முறுக்கிய அவளது மஞ்சள் நிற முடியில் தேனீ கூட சிக்கிக் கொள்வதை நீங்கள் யாரும் ஆச்சர்யப்படாமலிருக்க முடியாது.

“மிகவும் கடுமையாக இருக்கிறது” என்று மிஸ் மெடோஸ் இருக்கமுடன் கூறினாள். மற்றவள் போலியான புன்னகையை உதிர்த்தாள்.

“இறுக்கமான நிலையில் காணப்படுகிறீர்கள்”. என்று அவள் கூறினாள். அவளுடைய நீல நிறக் கண்கள் அகலத்திறந்தன; அதிலிருந்து கேலி செய்கின்ற ஒரு ஒளி தென்பட்டது. (அவள் எதையோ கண்டுவிட்டாளோ?)

“ஓ அவ்வளவு மோசமாக இல்லை” என்றாள் மிஸ் மெடோஸ்; பதிலுக்கு அறிவியல் ஆசிரியையிடம், அவளது புன்னகைக்காக, வெறுப்பைத் தன் முகத்தில் காட்டிவிட்டுக் கடந்து சென்றாள்.

வகுப்புப் படிநிலை நான்கு, ஐந்து மற்றும் ஆறு இசை பயிலும் அரங்கத்தில் கூடியிருந்தன. செவிகளைப் பிளக்கும் அளவுக்குச் சப்தமாக இருந்தது. மேடையில் பியானோவுக்கருகில், மேரி பெஸ்ஸி, மிஸ் மெடோஸின், பின்னணியிசை வாசிக்கின்ற செல்லப் பிள்ளை நின்று கொண்டிருந்தாள். அவள் மிஸ் மெடோஸைக் கண்டவுடன் அமைதியாய் இருக்கும் – படி எச்சரிக்கைக் குரலெழுப்பினாள்.

மிஸ் மெடோஸ் தன் கைகளை சட்டைக் கைகளுக்குள் மறைத்துக் கொண்டும் இசைக்குழுவை வழி நடத்துகின்ற குச்சியை தன் அக்குள் பகுதியில் இடுக்கிக் கொண்டும், நடுவேயுள்ள நாற்காலி வரிசைப் பகுதிகளுக்கிடையேயான வழியில் நடந்து, படிகளிலேறி, திடீரெனத் திரும்பி, வெண்கல இசைத் தாங்கியை இழுத்து அவள் முன்னால் நிறுத்தி, அமைதியாய் இருக்கும் படி தனது கையிலுள்ள மெல்லிய குச்சியால் இருமுறை கடுமையாகத் தட்டினாள்.

“அமைதி, தயவுசெய்து! உடனடியாக!” என்றும் யாரையும் குறிப்பிட்டுப் பார்க்காமல், கடல் போன்ற ப்ளானல் சீருடையணிந்து அலைபோன்று அசைக்கின்ற சிவந்த முகங்கள் மற்றும் கைகள், தலைகளில் பட்டாம்பூச்சி தலையணிகள் அணிந்து அசைப் புத்தகங்களை திறந்து விரித்து வைத்துக் கொண்டு நின்ற குழந்தைகளை மேலோட்டமாக பார்த்தாள். அவர்கள் என்ன நினைத்துக் கொண்டிருக்கிறார்கள் என்று அவளுக்கு முழுமையாகத் தெரியும்.

“ இசையாசிரியை கோபமுடன் இருக்கிறாள்.” நன்று, அவர்கள் அப்படியே நினைக்கட்டும்; அவள் கண்கள் படபடவெனத் துடித்தன; அவர்களுக்குச் சவாலாகத் தன் தலையைத் திருப்பினாள். இதயத்தை ஊடுருவிக் குத்திய இவ்வாறான ஒரு கடிதத்தினால் இரத்தம் கசிந்து இறந்து கொண்டிருக்கின்ற ஒருவருக்கு இவ்வுயிரினங்களால் என்ன செய்துவிட முடியும்.

”நமது திருமணமானது தவறானதாகி விடும் என்று மேலும் மேலும் நான் உணர்கிறேன்”. நான் உன்னை நேசிக்கவில்லை என்பதற்காக இல்லை.

எந்தவொருப் பெண்ணையும் நேசிக்கக் கூடிய அளவுக்கு உன்னையும் நான் நேசிக்கிறேன் ஆனால் உண்மையைச் சொல்ல வேண்டுமென்றால், நான் திருமணம் செய்துகொள்கின்ற ஒரு நபரில்லை என்ற முடிவுக்கு வந்து விட்டேன், மற்றும் திருமணமாகி சௌகரியமாக இருக்க வேண்டும் என்பது ஒன்றுமில்லை”-மற்றும் “ஏற்கமுடியாததென்று’ என்ற அந்தச்சொல் லேசாக அடிக்கப்பட்டு “வருந்தத்தக்கது” என்ற அதன் மேலேயே எழுதப்பட்டிருந்தது.

பாசில்! என்று மிஸ் மெடோஸ் பியானோவுக்குப் பின் சென்றாள். மேலும் இந்த நேரத்திற்காகக் காத்துக் கொண்டிருந்த மேரி பிளேஸ்லி, முன்பக்கம் குனிந்து; அவளுடைய சுருள் போன்ற முடி கன்னங்களில் தவழ மெல்ல, “காலை வணக்கம், மிஸ் மெடோஸ்” என்று வாழ்த்தி, தன்னுடைய ஆசிரியருக்கு அழகிய மஞ்சள் வண்ணமுடைய சாமந்திப் பூ ஒன்றைக் கொடுத்தாள்.

இந்த சிறு சடங்கு சம்பிரதாயமானது பல காலமாக ஒன்றரைப் பருவமாக நடந்து கொண்டிருக்கிறது. அது பியானோ இசை வகுப்பைத் தொடங்குவதன் ஒரு பகுதியாக இருந்தது.

ஆனால் இன்று காலையிலோ, அதை ஏற்றுக் கொள்ளாமல், அதைத் தன் இடைப்பட்டையில் சொருகிக் கொள்ளாமல், வழக்கமாக மேரியை நோக்கி குனிந்து சொல்வாள், “நன்றி, மேரி. எவ்வளவு நன்றாக இருக்கிறது! பக்கம் முப்பத்தியிரண்டைத் திருப்புங்கள்”.

மேரியுடைய பயம் என்னவென்றால், மிஸ் மெடோஸ் சாமந்திப் பூ வழங்கிய போது முற்றிலுமாக கண்டுகொள்ளவில்லை தவிர்த்து விட்டாள், அவளுடைய வார்த்தைக்குப் பதிலளிக்கவில்லை. ஆனால் பனிக்கட்டி போன்ற குரலில், “பக்கம் பதினான்கு, தயவுசெய்து உச்சரிப்பை நன்றாகக் குறித்துக்கொள்ளுங்கள்” என்று கூறினாள்.

தடுமாற்றமான நேரம்! கண்கள் கலங்கி நிற்க மேரி முகம் சிவந்தாள். ஆனால் மிஸ்மெடோஸ் இசைத் தாங்கியினருகில் சென்றுவிட்டாள்; அவளுடைய குரல் இசை அரங்கம் முழுவதும் எதிரொலித்தது. “பக்கம் பதினான்கு, பக்கம் பதினான்கில் தொடங்குவோம், ‘ஒரு இரங்கற்பா’, இப்பொழுது, சிறுமியரே, இதை நீங்கள் தெரிந்து கொண்டிருக்க வேண்டும்.

எல்லாரும் சேர்ந்து இதைப் பாடுவோம்; தனிக் குழுக்களில் அல்ல, எல்லோரும் சேர்ந்து எந்தவொரு உணர்வையும் வெளிப்படுத்தாமல், உங்கள் இடது கையால் கால அளவைத் தட்டிக் கொண்டு, ஆனாலும் மிக எளிமையாகப் பாடுங்கள்”.

மெல்லகுச்சியை உயர்த்தி; இசைத்தாங்கியை இருமுறை தட்டினாள். மேரியும் தொடக்க ஒத்திசைச்சுரங்களை இசைக்க, அவர்களுடைய இடது கைகள் கீழேவர, காற்றிலசைத்துக்கொண்டு, ஒரே குரலாய் அந்த இளமையான சோகக் குரல்கள்

ஒலித்தன:- “வேகமாக! ஆ, மிகவும் வேகமாக மகிழ்ச்சியின் ரோஜாக்கள் வாடுகின்றன; விரைவிலேயே இலையுதிர்காலம் சலிப்பான குளிர்காலத்திற்கு வழிவிடுகிறது. கூட்டமாக! அ கூட்டமாக இசையின் ஈர்ப்பு அளவு கேட்கும் காதுகளிலிருந்து கடந்து செல்கிறது.”

அடகடவுளே, இந்த சோகப் பாடலைத் தவிர மோசமானது எதுவாக இருக்கமுடியும்! ஒவ்வொரு இசைக் குறியும் (குறியீடும்) ஒரு பெருமூச்சாகவும் அழுகையாகவும் ஆராத்துயரின் மோசமான துன்ப ஒலியாக இருக்கறிது. தனது அகண்ட மேலங்கியோடு மிஸ் மெடோஸ் தன் கைகளை உயர்த்தினாள், மற்றும் தனது இரு கைகளினாலும் வழிநடத்தத் தொடகினாள்.

“நமது திருமணம் தவறானதாகிவிடும் என்று நான் மேன்மேலும் உணர்கிறேன்” அவன் தட்டினாள். எல்லாக் குரல்களுமே ஒலித்தன; “கூட்டமாக! ஆ கூட்டமாக”.

இந்த மாதிரியான கடிதத்தை எழுதுவதற்கு அன் மனதில் என்ன குடி கொண்டிருக்கும்! அதற்கு என்ன வழி நடத்தியிருக்க வேண்டும்! ஒன்றுமில்லாமல் இது வரவில்லை.

அவனுடைய முந்தைய கடிதத்தில் எங்களுடைய புத்தகங்களை வைப்பதற்காக புகையுட்டப்பட்ட ஓக் மரத்தாலான புத்தக அலமாரியைப் பற்றியல்லாமல் கூறியிருந்தான் மற்றும் கூடத்தில் வைக்கக் கூடிய ஒரு சிறிய நேர்த்தியான, அவன் பார்த்திருந்த தாங்கி ஒன்றை, ஒரு தூய்மையான வேலைப்பாடுடைய ஆந்தை உருவம் அதன் மேற்புரத்திலிருக்க, அதன் கால்களிலிருக்கின்ற நகங்களில் மூன்று தொப்பித் தூரிகைகள்.

திடீரென்று கதவு திறந்தது. நீல நிற உடை அணிந்த ஒரு சிறு பெண் அரங்கின் நடை பாதையில் தலையை தாழ்த்தியபடி தனது உதடுகளை கடித்தபடி அவளது சிவந்த கைகளில் இருந்த வெள்ளி வளையலை திருகியபடியே பரபரப்பாக நடந்து வருகிறாள். அவள் படிகளில் ஏறி, மிஸ் மெடோசின் முன்பு நிற்கிறாள்.

என்ன மோனிகா, என்ன அது?

உங்களுக்கு முடியும்னா, இப்ப, மிஸ் வயட் உங்களை அவரோட அறைக்கு வந்து பார்க்க வேண்டும் என சொல்லியிருக்காங்க. அறையில் வைத்து பார்க்க வேண்டுமென்று விரும்புகிறார்கள்.

நல்லது என்கிறார் மிஸ் மெடோஸ். மிஸ் மெடோஸ் பெண் பிள்ளைகளை அழைத்து நான் இப்போது வெளியே பேபாகிறேன். நீங்கள் சத்தம் செய்யாமல் மெதுவாக பேசிக்கொள்ளலாம் என்கிறாள். ஆனால் பெண் பிள்ளைகள் எதுவுமே செய்யமுடியாதபடி ஒடுங்கியிருக்கின்றனர். அவர்களில் பெரும்பாலனவர்கள் தங்கள் மூக்கை சீந்தியபடி இருக்கின்றனர்.

வராந்தாக்கள் குளிர்ச்யிாகவும் அமைதியாகவும் இருக்கின்றன. அவை மிஸ் மெடோசின் காலடிகளை எதிரெபாலிக்கின்றன. தலைமையாசிரியை தனது மேசையில் அமர்ந்திருக்கிறார். ஒரு நொடிப்பொழுது அவர் அவளை நிமிர்ந்து கூட பார்க்கவில்லை அவள் தனது வழக்கப்படி கண் கண்ணாடிகளை கழற்றும்போது அவளது கழுத்தில் இருக்கும் துணியின் வார் இழையில் அது சிக்கிக் கொள்கிறது.

உக்காருங்க மிஸ் மெடோஸ் என்று அன்பாகச் சொன்னவள், மை ஒற்றும் பலகையிலிருந்து ஒரு இளஞ்சிவப்பு நிற உறையை எடுக்கிறாள். “ இந்த தந்தி உங்களுக்கு வந்திருக்கிறது. அதனால் உங்களை வரச்சொன்னேன்.”

மிஸ் வயட் எனக்கு தந்தியா?

பசில் அவன் தற்கொலை செய்து கொண்டுவிட்டான் என்று மிஸ் மெடோஸ் முடிவு செய்துவிட்டாள். அவளது கைகள் முன்னே சென்றன. ஆனால், மிஸ் வயட் கொஞ்சம் தன் கையை பின்னுக்கு இழுத்துக்கொண்டு, இது கெட்ட செய்தி இல்லை என்று நினைக்கிறேன். எப்போதும் இல்லாத அன்போடு சொன்னாள் அவள். மிஸ் மெடோஸ் அதைப்பிரித்துப் பார்த்தாள்.

அந்தப்கடிதத்தை பொருட்படுத்தாதே. அது ஏதோ பைத்தியக்காரத்தனமாக எழுதியது. ஒரு தொப்பிவைக்கும் ஸ்டாண்டு வாங்கினேன். என்று அவள் அதில் வாசித்தாள். அந்த தந்தியிலிருந்து அவளால் கண்களை எடுக்க முடியவில்லை.

அது ஏதோ ஒரு வருத்தப்படுகிற விசயம் இல்லை என்று நான் நினைக்கிறேன் என்றாள் மிஸ் வயட்.

ஓ, அதெல்லாம் இல்ல. நன்றி மிஸ் வயட், என்றாள் மிஸ் மெடோஸ், வெட்கப்பட்டபடி, அது ஒண்ணுமே இல்ல. அதன் பின்பு மன்னிப்பு கேட்கும் விதமாக ஒரு சின்ன சிரிப்பு சிரித்தாள். இது எனக்கு மாப்பிள்ளையாகப் போகிறவர் கிட்ட இருந்து வந்திருக்கு. அதுல என்ன இருக்குன்னா ….. ஒரு இடைவெளி….. அப்படியா அது சரி என்கிறாள் மிஸ் வயட். மீண்டும் ஒரு இடைவெளி. பிறகு உங்களுக்கு இன்னும் பதினைந்து நிமிடம் வகுப்பு இருக்கிறது இல்லையா மிஸ் மெடோஸ்.

ஆமா மிஸ் வயட். அவள் எழுந்தாள். அவள் கிட்டத்தட்ட கதவை நோக்கி ஓடினாள்.

ஒரு நிமிடம் மிஸ் மெடோஸ். உங்க கிட்ட ஒண்ணு சொல்லணும். டீச்சர்களுக்கு பள்ளிக்கூட நேரத்துல யாராவது தந்தி அனுப்பறதை நான் ஏத்துக்கறதில்ல. அது ரொம்ப கெட்ட செய்தியா இருந்தா மட்டுமே. அதாவது செத்துப் போன செய்தி அல்லது அது மாதிரி ஒண்ணு நல்ல செய்தி தான் எப்பயும் வேணும்.

இதைத் தெரிஞ்சுக்கோங்க என்று மிஸ் வயட் அவளிடம் விளக்கிளாள். நம்பிக்கை, நேசம், மகிழ்ச்சி ஆகிய சிறகுகளோடு இசை அரங்குக்கு விரைந்து சென்று அரங்கின் மையப் பாதையில் நடந்து சென்று படிகளில் ஏறி, பியானோவை நோக்கி சென்றாள் மிஸ் வயட்.

பக்கம் முப்பத்து ரெண்டு மேரி என்றாள் அவள். பக்கம் முப்பத்து ரெண்டு என்றபடி அருகிருந்த சாமந்திப் பூவை கையில் எடுத்துக் கொண்டாள். தனது புன்னகையை மறைப்பதற்காக அதை தன் உதடுகளின் அருகே கொண்டு சென்றாள். பிறகு அவள் பெண் பிள்ளைகளை நோக்கித் திரும்பி தனது (பேட்டன்) குச்சியால் தட்டியபடி பிள்ளைகளா, பக்கம் முப்பத்து ரெண்டு பக்கம் முப்பத்து ரெண்டு என்றாள்.

மலர்கள் அதிகம் சுமந்து வந்தோம் இன்றைக்கு நாமே. அதோடு கூடைகளில் கனிகளும், ரிப்பன்களும், வாழ்த்தவே.

நிறுத்து. ! நிறுத்து என்று கத்தினாள் மிஸ் மெடோஸ். இது ரொம்ப மோசம்படு மோசமா இருக்கு. அவள் பெண் பிள்ளைகளை நோக்கி உங்களுக்கு என்ன ஆச்சு. யோசிச்சுப்பாருங்க. நீங்க என்ன பாடுறோம்னு யோசிச்சு பாடுங்க. உங்க கற்பனையை பயன்படுத்து. மலர்கள் அதிகம் சுமந்து வந்தோம்.

அதோடு கூடைகளில் கனிகளும் ரிப்பன்களும் வாழ்த்தவே. மிஸ் மெடோஸ் காட்டமாகச் சொன்னாள். பிள்ளைகளா ரொம்ப சோகமா பாடாதீங்க. அது கொஞ்சம் இதமா இருக்கனும். மகிழ்ச்சியா ஆர்வமா வாழ்த்தவே மறுபடி ஒருமுறை வேகமா எல்லோரும் இப்ப வாங்க.

இந்தத் தடவை, மிஸ் மெடோசின் குரல் மற்ற எல்லாக் குரலைவிட அதிக ஒலியுடன் ஒலித்தது. முழுமையாய், ஆழமாய் உணர்வுகளோடு ஒளிரும்படியாய் ஒலித்தது.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 4 Principles of Inheritance and Variation Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

12th Bio Zoology Guide Principles of Inheritance and Variation Text Book Back Questions and Answers

Question 1.
Haemophilia is more common in males because it is a………………
(a) Recessive character carried by Y-chromosome
(b) Dominant character carried by Y-chromosome
(c) Dominant trait carried by X-chromosome
(d) Recessive trait carried by X-chromosome
Answer:
(d) Recessive trait carried by X-chromosome

Question 2.
ABO blood group in man is controlled by …………………….
(a) Multiple alleles
(b) Lethal genes
(c) Sex linked genes
(d) Y-linked genes
Answer:
(a) Multiple alleles

Question 3.
Three children of a family have blood groups A, AB and B. What could be the genotypes of their parents?
(a) I^AI^B and ii
(b) I^A1^O and I^BI^O
(c) I^B I^B and I^AI^A
(d) I^AI^A and ii
Answer:
(b) I^A1^O and I^BI^O

Question 4.
Which of the following is not correct?
(a) Three or more alleles of a trait in the population are called multiple alleles.
(b) A normal gene undergoes mutations to form many alleles
(c) Multiple alleles map at different loci of a chromosome
(d) A diploid organism has only two alleles out of many in the population
Answer:
(c) Multiple alleles map at different loci of a chromosome

Question 5.
Which of the following phenotypes in the progeny are possible from the parental combination
(a)AandB only
(b) A,B andAB only
(c) AB only
(d) A,B,AB and O
Answer:
(d) A,B,AB and O

Question 6.
Which of the following phenotypes is not possible in the progeny of the parental genotypic combination I^AI^O x l^Al^B ?
(a) AB
(b) O
(c) A
(d) B
Answer:
(b) O

Question 7.
Which of the following is true about Rh factor in the offspring of a parental combination DdXDd (both Rh positive)?
(a) All will be Rh positive
(b) Half will be Rh positive
(c) About 3/4 will be Rh negative
(d) About one fourth will be Rh negative
Answer:
(d) About one fourth will be Rh negative

Question 8.
What can be the blood group of offspring when both parents have AB blood group?
(a) AB only
(b) A, B and AB
(c) A, B, AB and O
(d) A and B only
Answer:
(b) A, B and AB

Question 9.
If the childs blood group is ‘O’ and fathers blood group is ‘A’ and mother’s blood group is ‘B’ the genotype of the parents will be …………….
(a) I^AI^A and I^BI^O
(b) I^AI^O and I^BI^O
(c) I^AI^O and I^OI^O
(d) I^OI^O and I^BI^B
Answer:
(b) I^AI^O and I^BI^O

Question 10.
XO type of sex determination and XY type of sex determination are examples of …………………
(a) Male heterogamety
(b) Female heterogamety
(c) Male homogamety
(d) Both (b) and (c)
Answer:
(a) Male heterogamety

Question 11.
In an accident there is great loss of blood and there is no time to analyse the blood group Question which blood can be safely transferred?
(a) ‘O’ and Rh negative
(b) ‘O’ and Rh positive
(c) ‘B’ and Rh negative
(d) ‘AB’ and Rh positive
Answer:
(a) ‘O’ and Rh negative

Question 12.
Father of a child is colourblind and mother is carrier for colourblindness, the probability of the child being colourblind is ………………
(a) 25%
(b) 50%
(c) 100%
(d) 75%
Answer:
(b) 50%

Question 13.
A marriage between a colourblind man and a normal woman produces
(a) All carrier daughters and normal sons
(b) 50% carrier daughters, 50% normal daughters
(c) 50% colourblind sons, 50% normal sons
(d) All carrier offsprings
Answer:
(a) All carrier daughters and normal sons

Question 14.
Mangolism is a genetic disorder which is caused by the presence of an extra chromosome number.
(a) 20
(b) 21
(c) 4
(d) 23
Answer:
(b) 21

Question 15.
Klinefelters’ syndrome is characterized by a karyotype of………………
(a) XYY
(b) XO
(c) XXX
(d) XXY
Answer:
(d) XXY

Question 16.
Females with Turners’syndrome have………………
(a) Small uterus
(b) Rudimentary ovaries
(c) Underdeveloped breasts
(d) All of these
Answer:
(d) All of these

Question 17.
Pataus’ syndrome is also referred to as………………
(a) 13-Trisomy
(b) 18-Trisomy
(c) 21-Trisomy
(d) None of these
Answer:
(a) 13-Trisomy

Question 18.
Who is the founder of Modem Eugenics movement?
(a) Mendel
(b) Darwin
(c) Fransis Galton
(d) Karl pearson
Answer:
(c) Fransis Galton

Question 19.
Improvement of human race by encouraging the healthy persons to marry early and produce large number of children is called………………
(a) Positive eugenics
(b) Negative eugenics
(c) Positive euthenics
(d) Positive euphenics
Answer:
(a) Positive eugenics

Question 20.
The ……………… deals with the control of several inherited human diseases especially inborn errors of metabolism.
(a) Euphenics
(b) Eugenics
(c) Euthenics
(d) All of these
Answer:
(a) Euphenics

Question 21.
“Universal Donor” and “Universal Recipients” blood group are ……………… and ………………  respectively.
(a) AB, O
(b) O, AB
(c) A, B
(d) B, A
Answer:
(b) O, AB

Question 22.
ZW-ZZ system of sex determination occurs in………………
(a) Fishes
(b) Reptiles
(c) Birds
(d) All of these
Answer:
(d) All of these

Question 23.
A co-dominant blood group is
(a) A
(b) AB
(c) B
(d) O
Answer:
(b) AB

Question 24.
Which of the following is incorrect regarding ZW-ZZ type of sex determination?
(a) It occurs in birds and some reptiles
(b) Females are homogametic and males are heterogametic
(e) Male produce two types of gametes
(d) It occurs in gypsy moth
Answer:
(b) Females are homogametic and males are heterogametic

Question 25.
What is haplodiploidy?
Answer:
In haplodiploidy, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid).

Question 26.
Distinguish between heterogametic and homogametic sex determination systems.
Answer:
Heterogametic Sex :

1. Organisms producing two different types of gametes.
2. Example: Human male.
   Sperm with X chromosome
   Sperm with Y chromosome

Homogametic Sex :

1. Organisms producing only one type of gametes.
2. Example: Human female.
   Every egg produced contain X chromosomes.

Question 27.
What is Lyonisation?
Answer:
Lyonisation is a process of inactivation of one of the X chromosomes in some females.

Question 28.
What is criss-cross inheritance?
Answer:
Inheritance of genes from a male parent to female child and then to male grandchild or female parent to male child and then to female grandchild. E.g., X-linked gene inheritance.

Question 29.
Why are sex-linked recessive characters more common in male human beings?
Answer:
Sex linked inherited traits are more common in males than females because, males are hemizygous and therefore express the trait when they inherit one mutant allele.

Question 30.
What are holandric genes?
Answer:
The genes present in the differential region of the Y chromosome are called Y- linked or holandric genes. The Y linked genes have no corresponding allele in X chromosome.

Question 31.
Mention the symptoms of Phenylketonuria.
Answer:
Severe mental retardation, light pigmentation of skin and hair. Phenylpyruvic acid is excreted in the urine.

Question 32.
Mention the symptoms of Down’s syndrome.
Answer:
Severe mental retardation, defective development of the central nervous system, increased separation between the eyes, flattened nose, ears are malformed, mouth is constantly open and the tongue protrudes.

Question 33.
Differentiate Intersexes from Supersexes.
Answer:Intersexes:
Intersexes refers to the individuals having the characteristics of both female and male sexes and their sexual anatomy does not seem to fit the typical definition of male or female.

Supersexes:
Supersexes ar formed as a result of an abnormal combination of sex chromosomes.
Example: Super males in humans human beings have 44+XYY chromosomes.

Question 34.
Explain the genetic basis of ABO blood grouping in man.
Answer:
Multiple allele inheritance of ABO blood groups
Blood differs chemically from person to person. When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells. Karl Landsteiner discovered two kinds of antigens called antigen ‘A’ and antigen ‘B’ on the surface of RBC’s of human blood. Based on the presence or absence of these antigens three kinds of blood groups, type ‘A’, type ‘B’, and type ‘O’ (universal donor) were recognized. The fourth and the rarest blood group ‘AB’ (universal recipient) was discovered in 1902 by two of Landsteiner’s students Von De Castelle and Sturli.

Bernstein in 1925 discovered that the inheritance of different blood groups in human beings is determined by a number of multiple allelic series. The three autosomal alleles located on chromosome 9 are concerned with the determination of blood group in any person. The gene controlling blood type has been labeled as ‘L’ (after the name of the discoverer, Landsteiner) or I (from isoagglutination). The I gene exists in three allelic forms, I^A, I^B and I^O. I^A specifies A antigen. I^B allele determines B antigen and I^O allele specifies no antigen. Individuals who possess these antigens in their fluids such as the saliva are called secretors.

Each allele (I^A and I^B) produces a transferase enzyme. I^A allele produces N-acetyl galactose transferase and can add N-acetyl galactosamine (NAG) and I^B allele encodes for the enzyme galactose transferase that adds galactose to the precursor (i.e. H substances). In the case of I^O/I^O allele no terminal transferase enzyme is produced and therefore called “null” allele and hence cannot add NAG or galactose to the precursor.

From the phenotypic combinations it is evident that the alleles I^A and I^B are dominant to 1°, but co-dominant to each other (I^A = I^B). Their dominance hierarchy can be given as (I^A=I^B> 1^O). A child receives one of three alleles from each parent, giving rise to six possible genotypes and four possible blood types (phenotypes). The genotypes are I^AI^A , I^AI^O, I^BI^B, I^BI^O, I^AI^B and I^OI^O.

Question 35.
How is sex determined in human
Answer:
Genes determining sex in human beings are located on two sex chromosomes, called allosomes. In mammals, sex determination is associated with chromosomal differences between the two sexes, typically XX females and XY males. 23 pairs of human chromosomes include 22 pairs of autosomes (44A) and one pair of sex chromosomes (XX or XY). Females are homogametic producing only one type of gametes (egg), each containing one X chromosome while the males are heterogametic producing two types of sperms with X and Y chromosomes. An independently evolved XX: XY system of sex chromosomes also exist in Drosophila.

Question 36.
Explain male heterogamety.
Answer:
Male heterogamety (XY males) is a type of sex determination in which males produce two different types of gametes. For example, human males produce two kinds of sperms that is sperm with X-chromosome and sperms with Y-chromosome.

Question 37.
Brief about female heterogamety.
Answer:
Female heterogamety (ZO females) refers to the condition, where female produces two types of egg cells. Some with Z chromosome and some without Z chromosome.

Question 38.
Give an account of genetic control of Rh factor?
Answer:
Genetic control of Rh factor
Fisher and Race hypothesis: Rh factor involves three different pairs of alleles located on three different closely linked loci on the chromosome pair. This system is more commonly in use today, and uses the ‘Cde’ nomenclature.
In the given figure, three pairs of Rh alleles (Cc, Dd and Ee) occur at 3 different loci on homologous chromosome pair-1. The possible genotypes will be one C or c, one D or d, one E or e from each chromosome. For e.g. CDE/cde; CdE/cDe; cde/cde; CDe/CdE etc. All genotypes carrying a dominant ‘D’ allele will produce Rh+positive phenotype and double recessive genotype ‘dd’ will give rise to Rh negative phenotype.

Wiener Hypothesis
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R² ,R⁰ ,R^z) will produce ‘Rh-positive’ ^phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 39.
Explain the mode of sex determination in honeybees.
Answer:
In hymenopteran insects such as honeybees, ants and wasps, a mechanism of sex determination called haplodiploidy mechanism of sex determination is common. In this system, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid), hence the name haplodiploid for this system of sex determination.

This mode of sex determination facilitates the evolution of sociality in which only one diploid female becomes a queen and lays the eggs for the colony. All other females which are diploid having developed from fertilized eggs help to raise the queen’s eggs and so contribute to the queen’s reproductive success and indirectly to their own, a phenomenon known as Kin Selection. The queen constructs their social environment by releasing a hormone that suppresses fertility of the workers.

Question 40.
Discuss the genic balance mechanism of sex determination with reference to Drosophila?
Answer:
XX-XY type (Lygaeus Type) sex determination is seen in Drosophila. The females are homogametic with XX chromosomes, while the males are heterogametic with X and Y chromosomes. Homogametic females produce only one kind of egg, each with one X chromosome, while the heterogametic males produce two kinds of sperms some with X chromosome and some with Y chromosome.

The sex of the embryo depends on the fertilizing sperm. An egg fertilized by an ‘X’ bearing sperm produces a female, if fertilized by a ‘Y’ bearing sperm, a male is produced.

Question 41.
What are the applications of Karyotyping?
Answer:

• Karyotyping helps in gender identification.
• It is used to detect the chromosomal aberrations like deletion, duplication, translocation, non-disjunction of chromosomes.
• It helps to identify the abnormalities of chromosomes like aneuploidy.
• It is also used in predicting the evolutionary relationships between species.
• Genetic diseases in human beings can be detected by this technique.

Question 42.
Explain the inheritance of sex linked characters in human being.
Answer:
Haemophilia is commonly known as bleeder’s disease, which is more common in men than women. This hereditary disease was first reported by John Cotto in 1803. Haemophilia is caused by a recessive X-linked gene. A person with a recessive gene for haemophilia lacks a normal clotting substance (thromboplastin) in blood, hence minor injuries cause continuous ’bleeding, leading to death. The females are carriers of the disease and would transmit the disease to 50% of their sons even if the male parent is normal. Haemophilia follows the characteristic criss-cross pattern of inheritaitce.

Question 43.
What is extra chromosomal inheritance? Explain with an example.
Answer:
The cytoplasmic extra nuclear genes have a characteristic pattern of inheritance which does not resemble genes of nuclear chromosomes and are known as Extrachromosomal/ Cytoplasmic inheritance.

Question 44.
Comment on the methods of Eugenics.
Answer:
Eugenics refers to the study of the possibility of improving the qualities of human population.

Methods of Eugenics:

• Sex-education in school and public forums.
• Promoting the uses of contraception.
• Compulsory sterilization for mentally retarded and criminals.
• Egg donation.
• Artificial insemination by donors.
• Prenatal diagnosis of genetic disorders and performing MTP
• Gene therapy
• Cloning
• Egg/sperm donation of healthy individuals.

12th Bio Zoology Guide Principles of Inheritance and Variation Additional Important Questions and Answers

12th Bio Zoology Guide Principles of Inheritance and Variation One Mark Questions and Answers

Question 1
If a colorblind female marries a normal male, their sons will be ………………
(a) All normal visioned
(b) All color blinded
(c) One half normal visioned other half colorblind
(d) Three fourth colorblind one fourth normal
Answer:
(c) One half normal visioned other half colorblind

Question 2
Excess hair growth on pinna is a feature noticed only in males because ……………
(a) Males produce more testosterone
(b) gene responsible for the character is located in Y-chromosome
(c) Estrogen suppresses the character in females
(d) females act only as a carriers for this character
Answer:
(b) gene responsible for the character is located in Y-chromosome

Question 3.
ABO blood group is a classical example for ………………..
(a) Multiple allelism
(b) Pleotropism
(c) Incomplete dominance
(d) Polygenic mechanism
Answer:
(a) Multiple allelism

Question 4
Unit of heredity is ……………….
(a) allele
(b) allelomorph
(c) trait
(d) gene
Answer:
(d) gene

Question 5.
Identify the proper dominance hierarchy.
(a) I^A = I^O > I^B
(b) I^A = I^B > O
(b) I^A = I^B > O
(d) I^B = I^A > O
Answer:

Question 6
Haemophilia is more common in human males than human females. The reason is due to
(а) X-linked dominant gene
(b) X-linked recessive gene
(c) Y-linked recessive gene
(d) Allosomal abnormality
Answer:
(b) X-linked recessive gene

Question 7.
Identify the correct statement.
(a) Homozygous sex chromosome (XX) produce males in Drosophila
(b) Homozygous sex chromosome (ZZ) determine female sex in birds
(c) Heterozygous sex chromosome (XO) determine male sex in grasshopper
(d) Heterozygous sex chromosome (ZW) determine male sex in gypsy moth
Answer:
(c) Heterozygous sex chromosome (XO) determine male sex in grasshopper

Question
Which blood group doesnot possess antibodies?
(a) I^AI^B
(b) I^OI^O
(c) I^AO
(d) I^BI^B
Answer:
(a) I^AI^B

Question 9.
Assertion (A): On diagnosis, Ramu is reported to have underdeveloped testis and gynecomastia.
Reason (R): His karyotype reveals XXY condition.
(а) A is right but R is wrong
(b) R explains A
(c) Both A and R are wrong
(d) Both and R are right but R is not the correct explanation of A
Answer:
(b) R explains A

Question 10.
Pick out the odd man.
(a) Klinefelter’s syndrome
(b) Turner’s syndrome
(c) Huntington’s chorea
(d) 13-Trisomy
Answer:
(c) Huntington’s chorea

Question 11.
Pick the odd one out regarding Mendelian disorder.
(a) Thalassemia
(b) phenylketonuria
(c) Albinism
(d) Huntington’s chorea
Answer:
(d) Huntington’s chorea

Question 12.
Match the following:

(a) A – iv, B – ii, C – i, D – iii
(b)A – ii, B – iv, C – iii, D – i
(a)A – iii, B – i, C – ii, D – iv
(c) A – i, B – iv, C – iii, D – iii
Answer:
(a) A – iv, B – ii, C – i, D – iii

Question 13.
Identify the proper ratio of normal visioned individuals against colorblind individuals, if colorblind carrier female marries a normal male.
(a) 1 : 1
(b) 3:1
(c) 1 : 3
(d) All four are normal visioned
Answer:
(c) 1 : 3

Question 14.
Pick out the correct statement.
(i) Karyotyping helps in gender identification
(ii) Holandric genes are located on X-chromosome
(iii) Trisomy-21 is an allosomal abnormality
(iv) Cooley’s anemia is an autosomal recessive disorder
(a) i, iii, iv are correct
(c) i and iv are correct
Answer:
(c) i and iv are correct

Question 15.
DOPA stands for ……………….
(a) 3,4- dihydroxy phenylacetate
(b) 3, 4 – dihydroxy phenylalanine
(c) 3,4- dihydroxy phenyl aspartate
(d) 3, 4- dihydroxy phenol aldehyde
Answer:
(b) 3,4 – dihydroxy phenylalanine

Question 16.
The type of antibody generated against Rh antigen is ….
(a)IgE
(b) IgG
(c) IgA
(d) IgB
Answer:
(b) IgG

Question 17.
Which of the following symbol is used in the pedigree analysis to represent unspecified sex?

Answer:

Question 18:
A colorblind man marries a woman with normal sight who has no history of color blindness in her family. What is the probability of their grandson being colorblind?
(a) 1/4
(b) 3/4
(c) 2/4
(d) 4/4
Answer:
(a) 1/4

Question 19.
Multiple alleles are located……………………
(a) at different loci on homologous chromosome
(b) at same locus on homologous chromosome
(c) at different loci on non-homologous chromosome
(d) at different chromosomes
Answer:
(b) at same locus on homologous chromosome

Question 20.
Identify the incorrect statement regarding haplodiploidy.
(g) Haplodiploidy is noticed in honeybees and drosophila
(b) Unfertilized eggs develop into drones
(c) Fertilized eggs develop into queen and worker bees
(d) Males have half the total chromosomal number
Answer:
(a) Haplodiploidy is noticed in honeybees and drosophila

Question 21.
I^A and I^B genes of ABO blood group are
(a) Co-dominant
(b) Pleotropic
(c) Dominant and recessive
(d) Epistatic
Answer:
(a) Co-dominant

Question 22.
Which one of the following crosses show 3 : 1 ratio of normal visioned versus carrier blind?
(a) X^CX^C x X⁺Y
(b) X⁺ X^C x X^C Y^–
(c) X⁺X^C x X⁺Y^–
(d) X⁺X⁺ x X^CY^–
Answer:
(c) X⁺X^C x X⁺Y^–

12th Bio Zoology Principles of Inheritance and Variation Two Marks Questions and Answers

Question 1.
Define multiple allelism.
Answer:
When three or more alleles of a gene that control a particular trait occupy the same locus on the homologous chromosome of an organism, they are called multiple alleles and their inheritance is called multiple allelism.

Question 2.
Name the discoverers of antigen A, B and AB.
Answer:
Antigens A and Antigen B was discovered by Karl Landsteiner. Antigen AB was discovered by Von De Castelle and Sturli.

Question 3.
What happens if type A blood is injected to a person having B blood group? Explain the reason.
Answer:
When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells.

Question 4.
State the allelic forms of I gene and mention its chromosomal location.
Answer:
The I gene exists in three forms: I^A, I^B and I^O. The alleles are located on chromosome 9.

Question 5.
Write the possible genotypes for a person having a B-blood group.
Answer:
The possible genotypes of a B-blood group person are I^BI^B or I^BI^O.

Question 6.
State Wiener Hypothesis on Rh-factor.
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R² ,R⁰ ,R^z) will produce ‘Rh-positive’ phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 7.
Distinguish between homogametic and heterogametic condition with example.
Answer:
Homogametic organism:

1. Organism producing only one type of gametes.
2. e.g. Human female (Only X)

Heterogametic organism :

1. Organism producing two different types of gametes.
2. e.g. Human Male (X and Y)

Question 8.
Name any four organism expressing ZW-ZZ type of sex determination.
Answer:
Gypsy moth, fishes, reptiles and birds.

Question 9.
Expand (a) SRY (b) TDF
Answer:
SRY – Sex Determining region Y
TDF – Testes Determining Factor

Question 10.
Define Barr body.
Answer:
In 1949, Barr and Bertram first observed a condensed body in the nerve cells of female cat which was absent in the male. This condensed body was called sex chromatin by them and was later referred as Barr body.

Question 11.
Based on Lyon’s hypothesis, mention the number of Barr bodies in XXY males, XO females.
Answer:
XXY males – One Barr body.
XO females – No Barr body.

Question 12.
State Lyon’s hypothesis.
Answer:
Lyon’s hypothesis states that in mammals the necessary dosage compensation is accomplished by the inactivation of one of the X chromosomes in females so that both males and females have only one functional X chromosome per cell.
Mary Lyon suggested that Barr bodies represented an inactive chromosome, which in females becomes tightly coiled into a heterochromatin, a condensed and visible form of chromatin (Lyon’s hypothesis). The number of Barr bodies observed in cell was one less than the number of X-Chromosome. XO females have no Barr body, whereas XXY males have one Barr body.

Question 13.
Mention few X-linked inherited diseases.
Answer:
Red-green colour blindness or daltonism, haemophilia and Duchenne’s muscular dystrophy.

Question 14.
Define Karyotyping.
Answer:
Karyotyping is a technique through which a complete set of chromosomes is separated from a cell and the chromosomes are arranged in pairs. An idiogram refers to a diagrammatic representation of chromosomes.

Question 15.
Explain the inheritance pattern of Y-linked genes for example.
Answer:
Genes in the non-homologous region of the Y-chromosome are inherited directly from male to male. In humans, the Y-linked or holandric genes for hypertrichosis (excessive development of hairs on pinna of the ear) are transmitted directly from father to son, because males inherit the Y chromosome from the father. Female inherits only X chromosome from the father and are not affected.

Question 16.
Observe the symbol used in pedigree analysis and give the proper terms they represent.
Answer:

Question 17.
Write a brief note on pedigree analysis.
Answer:
Pedigree is a “family tree”, drawn with standard genetic symbols, showing the inheritance pathway for specific phenotypic characters. Pedigree analysis is the study of traits as they have appeared in a given family line for several past generations.

Question 18.
What do you mean by ‘Mendelian disorder’.
Answer:
Alteration or mutation in a single gene causes Mendelian disorders. These disorders are transmitted to the offsprings on the same line as the Mendelian pattern of inheritance. E.g., Thalassemia.

Question 19.
Name any four Mendelian disorders.
Answer:
(a) Thalassemia (b) Albinism (c) sickle cell anaemia (d) Huntington’s chorea

Question 20.
What is the phenotype of (a) I^AI^O (b) I^OI^O
Answer:
(a) I^AI^O – A blood group person
(b) I^OI^O – O blood group person

Question 21.
On which chromosomes does HBA1 gene and HBB genes are located?
Answer:
HBA1 gene is located on chromosome 16.
HBB gene is located on chromosome 11.

Question 22.
Complete the equation.

Answer:
(a) A = Phenylalanine hydroxylase
(b) B = Tyrosinase

Question 23.
Write a note on Huntington’s chorea.
Answer:
Huntington’s chorea is inherited as an autosomal dominant lethal gene in man. It is characterized by involuntary jerking of the body and progressive degeneration of the nervous system, accompanied by gradual mental and physical deterioration. The patients with this disease usually die between the age of 35 and 40.

Question 24.
Comment on Trisomy-21.
Answer:
Trisomic condition of chromosome – 21 results in Down’s syndrome. It is characterized by severe mental retardation, defective development of the central nervous system, increased separation between the eyes, flattened nose, ears are malformed, mouth is constantly open and the tongue protrudes.

Question 25.
Mention the genetic makeup of Turner’s syndrome person and Klinefelter’s syndrome , person.
Answer:
Klinefelter’s syndrome – 44AA+XXY
Turner’s syndrome – 44AA+XO

Question 26.
List out any four clinical symptoms of Klinefelter’s syndrome.
Answer:
Gynaecomastia, high pitched voice, under developed genetalia and tall with long limbs.

12th Bio Zoology Principles of Inheritance and Variation Three Marks Questions and Answers

Question 27.
Write the types of sex-determination mechanisms does the following crosses as shown. Give an example for each.
(a) Female XX with Male XO (6) Female ZW with Male ZZ
Answer:
(a) Male heterogamety. e.g., Human beings.
(b) Female heterogamety. e.g., Birds.

Question 28.
What are the enzymes encoded by the alleles I^A, I^B and I^O?
Answer:
IA allele produces N-acetyl galactose transferase and can add N-acetyl galactosamine (NAG) and I^B allele encodes for the enzyme galactose transferase that adds galactose to the precursor (i.e. H substances). In the case of I^O/I^O allele no terminal transferase enzyme is produced and therefore called “null” allele and hence cannot add NAG or galactose to the precursor.

Question 29.
Draw a tabular column representing various types of blood group in human beings, their phenotypes, genotypes, antigens and respective antibodies.
Answer:
Genetic basis of the human ABO blood groups:

Question 30.
Give an account on Rhesus factor.
Answer:
Rhesus or Rh – Factor: The Rh factor or Rh antigen is found on the surface of erythrocytes. It was discovered in 1940 by Karl Landsteiner and Alexander Wiener in the blood of rhesus monkey, Macaca rhesus and later in human beings. The term ‘Rh factor’ refers to “immunogenic D antigen of the Rh blood group system. An individual having D antigen are Rh D positive (Rh+) and those without D antigen are Rh D negative (Rh”)”. Rhesus factor in the blood is inherited as a dominant trait.

Naturally occurring Anti D antibodies are absent in the plasma of any normal individual. However if an Rh” (Rh negative) person is exposed to Rh+ (Rh positive) blood cells (erythrocytes) for the first time, anti D antibodies are formed in the blood of that individual. On the other hand, when an Rh positive person receives Rh-negative blood no effect is seen.

Question 31.
How Erythroblastosis foetalis can be prevented?
Answer:
If thefmother is Rh negative and foetus is Rh positive, anti D antibodies should be administered to the mother at 28th and 34th week of gestation as a prophylactic measure. If the Rh-negative mother delivers Rh positive child then anti D antibodies should be administered to the mother soon after delivery. This develops passive immunity and prevents the formation of anti D antibodies in the mothers blood by destroying the Rh foetal RBC before the mother’s immune system is sensitized. This has to be done whenever the woman attains pregnancy.

Question 32.
Explain XX-XO type of sex determination.
Answer:
XX-XO method of sex determination is seen in bugs, some insects such as cockroaches and grasshoppers. Pi The female with two X chromosomes are homogametic Gametes (XX) while the males with only one X chromosome are heterogametic (XO). The presence of unpaired X chromosomes determines the male sex. The males PI Generation with unpaired ‘X’ chromosome produce two types of sperms, one half with X chromosome and other half without X chromosome. The sex of the offspring depends upon the sperm that fertilizes the egg.

Question 33.
Name the type of sex-determination mechanism of the following organisms.
(a) Gypsy moth (A) Human beings (c) Butterflies
Answer:
(a) Gypsy moth -ZW – ZZ type (ZW-females, ZZ – males)
(b) Human beings – XX – XY type (XX-females, XY – males)
(c) Butterflies – ZO – ZZ type (ZO-females, ZZ – males)

Question 34.
Complete the following cross.

Answer:

Question 35.
Role of Y- chromosome is crucial for maleness – Justify.
Answer:
Current analysis of Y chromosomes has revealed numerous genes and regions with potential genetic function; some genes with or without homologous counterparts are seen on the X. Present at both ends of the Y chromosome are the pseudoautosomal regions (PARs) that are similar with regions on the X chromosome which synapse and recombine during meiosis.

The remaining 95% of the Y chromosome is referred as the Non-combining Region of the Y (NRY). The NRY is divided equally into functional genes (euchromatic) and non-functional genes (heterochromatic). Within the euchromatin regions, is a gene called Sex determining region Y (SRY). In humans, absence of Y chromosome inevitably leads to female development and this SRY gene is absent in X chromosome. The gene product of SRY is the testes determining factor (TDF) present in the adult male testis.

Question 36.
Color blindness is a perfect example for criss-cross of inheritance – Justify the statement.
Answer:
A marriage between a colour blind man and a normal visioned woman will produce normal visioned male and female individuals in F1 generation but the females are carriers. The marriage between a F1 normal visioned carrier woman and a normal visioned male will produce one normal visioned female, one carrier female, one normal visioned male and one colour blind male. Colour blind trait is inherited from the male parent to his grandson through carrier daughter, which is an example of criss-cross pattern of inheritance.

Question 37.
How the Karyotype of lymphocytes was prepared by Tjio and Levan?
Answer:
Preparation of Karyotype Tjio and Levan (1960) described a simple method of culturing lymphocytes from the human blood. Mitosis is induced followed by addition of colchicine to arrest cell division at metaphase stage and the suitable spread of metaphase chromosomes is photographed. The individual chromosomes are cut from the photograph and are arranged in an orderly fashion in homologous pairs. This arrangement is called a karyotype. Chromosome banding permits structural definitions and differentiation of chromosomes.

Question 38.
What is a genetic disorder? Mention its types?
Answer:
A genetic disorder is a disease or syndrome that is caused by an abnormality in an individual -DNA. Abnormalities can range from a small mutation in a single gene to the addition or subtraction of an entire chromosome or even a set of chromosomes. Genetic disorders are of two types namely, Mendelian disorders and chromosomal disorders.

Question 39.
Explain the genetic basis of Phenylketonuria.
Answer:
Phenylketonuria is an inborn error of Phenylalanine metabolism caused due to a pair of autosomal recessive genes. It is caused due to mutation in the gene PAH (phenylalanine hydroxylase gene) located on chromosome 12 for the hepatic enzyme “phenylalanine hydroxylase”.

This enzyme is essential for the conversion of phenylalanine to tyrosine. Affected individual lacks this enzyme, so phenylalanine accumulates and gets converted to phenylpyruvic acid and other derivatives. It is characterized by severe mental retardation, light pigmentation of skin and hair. Phenylpyruvic acid is excreted in the urine.

Question 40.
Give an account of Patau’s syndrome.
Answer:
Trisomic condition of chromosome 13 results in Patau’s syndrome. Meiotic non-disjunction is thought to be the cause for this chromosomal abnormality. It is characterized by multiple and severe body malformations as well as profound mental deficiency. Small head with small eyes, cleft palate, malformation of the brain and internal organs are some of the symptoms of this syndrome.

Question 41.
Define aneuploidy.
Answer:
Failure of chromatids to segregate during cell division resulting in the gain or loss of one or more chromosomes is called aneuploidy. It is caused by the non-disjunction of chromosomes.

12th Bio Zoology Principles of Inheritance and Variation Five Marks Questions and Answers

Question 42.
What do you mean by “syndrome”? Give two examples.
Answer:
Group of signs and symptoms that occur together and characterize a particular abnormality is called a syndrome, e.g., Down’s syndrome and Turner’s syndrome.

Question 42.
Explain in detail about Erythroblastosis foetalis.
Answer:
Rh incompatability has great significance in childbirth. If a woman is Rh-negative and the man is Rh positive, the foetus may be Rh positive having inherited the factor from its father. The Rh negative mother becomes sensitized by carrying Rh positive foetus within her body. Due to damage of blood vessels, during child birth, the mother’s immune system recognizes the Rh antigens and gets sensitized. The sensitized mother produces Rh antibodies. The antibodies are IgG type which are small and can cross placenta and enter the foetal circulation. By the time the mother gets sensitized and produce anti ‘D’ antibodies, the child is delivered.

Usually no effects are associated with exposure of the mother to Rh positive antigen during the first child birth, subsequent Rh positive children carried by the same mother, may be exposed to antibodies produced by the mother against Rh antigen, which are carried across the placenta into the foetal blood circulation. This causes haemolysis of foetal RBCs resulting in haemolytic jaundice and anaemia. This condition is known as Erythroblastosis foetalis or Haemolytic disease of the new bom (HDN).

Question 43.
Decribe female heterogamy and its types.
Answer:
Heterogametic Females:
In this method of sex determination, the homogametic male possesses two ‘X’ chromosomes as in certain insects and certain vertebrates like fishes, reptiles and birds producing a single type of gamete; while females produce dissimilar gametes. The female sex consists of a single ‘X’ chromosome or one ‘X’ and one ‘Y’ chromosome. Thus the females are heterogametic and produce two types of eggs. Heterogametic females are of two types, ZO-ZZ type and ZW-ZZ type. ,

ZO-ZZ Type
This method of sex determination is seen in certain moths, butterflies and domestic chickens. In this type, the female possesses single ‘Z’ chromosome in its body cells and is heterogametic (ZO) producing two kinds of eggs some with ‘Z’ chromosome and some without ‘Z’ chromosome, while the male possesses two ‘Z’ chromosomes and is homogametic (ZZ).

ZW-ZZ type
This method of sex determination occurs in certain insects (gypsy moth) and in vertebrates such as fishes, reptiles and birds. In this method the female has one ‘Z’ and one ‘ W’ chromosome (ZW) producing two types of eggs, some carrying the Z chromosomes and some carry the W chromosome. The male sex has two ‘Z’ chromosomes and is homogametic (ZZ) producing a single type of sperm.

Question 44.
Write elaborately about the following Mendelian disorders.
(a) Thalassemia (b) Albinism
Answer:
(a) Thalassemia
Thalassemia is an autosomal recessive disorder. It is caused by gene mutation resulting in excessive destruction of RBC’s due to the formation of abnormal haemoglobin molecules. Normally haemoglobin is composed of four polypeptide chains, two alpha and two beta globin chains. Thalassemia patients have defects in either the alpha or beta globin chain causing the production of abnormal haemoglobin molecules resulting in anaemia.

Thalassemia is classified into alpha and beta based on which chain of haemoglobin molecule _ is affected. It is controlled by two closely linked genes HBA1 and HBA2 on chromosome 16. Mutation or deletion of one or more of the four alpha gene alleles causes Alpha Thalassemia. In Beta Thalassemia, production of beta globin chain is affected. It is controlled by a single gene (HBB) on chromosome 11. It is the most common type of Thalassemia and is also known as Cooley’s anaemia. In this disorder, the alpha chain production is increased and damages the membranes of RBC.

(b) Albinism
Albinism is an inborn error of metabolism, caused due to an autosomal recessive gene. Melanin pigment is responsible for skin colour. Absence of melanin results in a condition called albinism. A person with the recessive allele lacks the tyrosinase enzyme system, which is required for the conversion of dihydroxyphenylalanine (DOPA) into melanin pigment inside the melanocytes. In an albino, melanocytes are present in normal numbers in their skin, hair, iris, etc., but lack melanin pigment.

Question 45.
Discuss any two Allosomal anomalies in human.
Answer:
Allosomal abnormalities in human beings
Mitotic or meiotic non-disjunction of sex chromosomes causes allosomal abnormalities. Several sex chromosomal abnormalities have been detected. E.g. Klinefelter’s syndrome and Turner’s syndrome.

1. Klinefelter’s Syndrome (XXY Males)
This genetic disorder is due to the presence of an additional copy of the X chromosome resulting in a karyotype of 47,XXY. Persons with this syndrome have 47 chromosomes (44AA+XXY). They are usually sterile males, tall, obese, with long limbs, high pitched voice, under developed genetalia and have feeble breast (gynaecomastia) development.

2. Turner’s Syndrome (XO Females)
This genetic disorder is due to the loss of a X chromosome resulting in a karyotype of 45,X. Persons with this syndrome have 45 chromosomes (44 autosomes and one X chromosome) (44AA+XO) and are sterile females. Low stature, webbed neck, underdeveloped breast, rudimentary gonads lack of menstrual cycle during puberty, are the main symptoms of this syndrome.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
On analysis, a person’s karyotype reveals an extra one chromosome of the twenty-first pair. What does this condition represent? which type of symptoms can be noticed in the person?
Answer:
Trisomy-21 or Down’s syndrome.
Symptoms – Mental retardation, malformed ears, protruded tongue, mouth is constantly open etc.

Question 2.
A female whose blood group is AB^– got conceived and later it is diagnoised that her – foetus possess B+. What measures would be taken to prevent the foetus from Haemolytic disease of Newborn (HDN)
Answer:
If the mother is Rh-negative and foetus is Rh-positive, anti D antibodies should be administered to the mother at 28th and 34th week of gestation as a prophylactic measure. If the Rh-negative mother delivers a Rh-positive child then anti D antibodies should be administered to the mother soon after delivery. This develops passive immunity and prevents the formation of anti D antibodies in the mother’s blood by destroying the Rh foetal RBC before the mother’s immune system is sensitized. This has to be done whenever the woman attains pregnancy.

Question 3.
The following table shows the genotypes for ABO blood grouping and these phenotypes. Complete the table by filling the gaps.

Answer:
2) I^AI^O
3) I^AI^B
4) O

Question 4.
Give one example for each of the following group of drugs, (a) Stimulants (b) Analgesic (c) Hallucinogens
Answer:
(a) Stimulants – Eg.: Nicotine
(b) Analgesic – Eg.: Opium
(c) Hallucinogens – Phencyclidine