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Short Answer Type Questions:

Q1.How does the unequal distribution of heat over the planet earth in space and time cause variations in weather and climate?

Answer

The earth receives almost all of its energy from the sun. The earth in turn radiates back to space the energy received from the sun. As a result, the earth neither warms up nor does it get cooled over a period of time. Thus, the amount of heat received by different parts of the earth is not the same. This variation causes pressure differences in the atmosphere. This leads to transfer of heat from one region to the other by winds. Thus, the unequal distribution of heat over the planet earth in space and time cause variations in weather and climate.

Q2.What are the factors that control temperature distribution on the surface of the earth?
Answer

The factors that control temperature distribution on the surface of the earth are:
• The latitude of the place• The altitude of the place• Distance from the sea, the airmass circulation• The presence of warm and cold ocean currents• Local aspects
Q3. In India, why is the day temperature maximum in May and why not after the summer solstice?

Answer
The day temperature maximum in May because of the summer solstice. At that time, sun’s rays are overhead the tropic of cancer (23.5°N). Tropic of Cancer passes through the middle of India. It remains till the end of May in India. Before summer solstice i.e., 21st June, monsoon starts in India which brings a cooling effect to the climate of India. This is why, India experiences high temperature before summer solstice.

Q4. Why is the annual range of temperature high in the Siberian plains?

Answer

The mean January temperature between 80°N and 50°N is minus 20°C and the temperature in July is more than 10°C. That is why annual range of temperature is very high.

Q5. How do the latitude and the tilt in the axis of rotation of the earth affect the amount of radiation received at the earth’s surface?

Answer

The amount of insolation received is the angle of inclination of the rays. This depends on the latitude of a place. The higher the latitude the less is the angle they make with the surface of the earth resulting in slant sun rays. The area covered by vertical rays is always less than the slant rays. If more area is covered, the energy gets distributed and the net energy received per unit area decreases. Moreover, the slant rays are required to pass through greater depth of the atmosphere resulting in more absorption, scattering and diffusion. Thus, the high latitudinal areas get less isolation an vice versa.
Sunrays fall vertically on equator throughout the year. The sun rays keep changing from 0° to 23.5° north and south. The sun is in the southern hemisphere and its rays fall vertically on tropic of cancer from 1st March to 21st March. The sun is in the northern hemisphere and its rays fall vertically on tropic of Capricorn from 23rd September to 22nd September. As we towards the poles, temperature keeps on decreasing. After 66 1⁄2 ° north and south there is cold zone. Here, the temperature remains low throughout the year because the sun’s rays fall tilted on it. Thus, the tilt in the axis of rotation of the earth affect the amount of radiation received at the earth’s surface.

Q6. Discuss the processes through which the earth-atmosphere system maintains heat balance.

Answer

The energy of sun reaches earth through radiation and circulates through various processes. The earth as a whole does not accumulate or loose heat. It maintains its temperature. This can happen only if the amount of heat received in the form of insolation equals the amount lost by the earth through terrestrial radiation. 
• Of the 100% energy radiated by Sun. While passing through the atmosphere some amount of energy is reflected, scattered and absorbed. 
• Only the remaining part reaches the earth surface. Roughly 35 units are reflected back to space even before reaching the earth’s surface. 
• Of these, 27 units are reflected back from the top of the clouds and 2 units from the snow and ice-covered areas of the earth. 
• The remaining 65 units are absorbed, 14 units within the atmosphere and 51 units by the earth’s surface. The earth radiates back 51 units in the form of terrestrial radiation.• Of these, 17 units are radiated to space directly and the remaining 34 units are absorbed by the atmosphere. 48 units absorbed by the atmosphere are also radiated back into space. 
• Thus, the total radiation returning from the earth and the atmosphere respectively is 17+48=65 units which balance the total of 65 units received from the sun. 
This is termed the heat budget or heat balance of the earth which explains the earth neither warms
up nor cools down despite the huge transfer of heat that takes place.

Q7.Compare the global distribution of temperature in January over the northern and the southern hemisphere of the earth.

Answer

The isotherms are generally parallel to the latitude. In the northern hemisphere the land surface area is much larger than in the southern hemisphere. Hence, the effects of land mass and the ocean currents are well pronounced. In January the isotherms deviate to the north over the ocean and to the south over the continent. This can be seen on the North Atlantic Ocean. The presence of warm ocean currents, Gulf Stream and North Atlantic drift, make the Northern Atlantic Ocean warmer and the isotherms bend towards the north. Over the land the temperature decreases sharply and the isotherms bend towards south in Europe. The effect of the ocean is well pronounced in the southern hemisphere. Here the isotherms are more or less parallel to the latitudes and the variation in temperature is more gradual than in the northern hemisphere. The isotherm of 20° C, 10° C, and 0° C runs parallel to 35° S, 45° S and 60° S latitudes respectively.

Long Answer Type Questions:

Q1.Explain about inversion of temperature.
Answer:
At times, the situations are reversed and the normal lapse rate is inverted. It is called inversion of temperature. Inversion is usually of short duration but quite common nonetheless. A long winter night with clear skies and still air is ideal situation for inversion. The heat of the day is radiated off during the night, and by early morning hours, the earth is cooler than the air above.

Over polar areas, temperature inversion is normal throughout the year. Surface inversion promotes stability in the lower layers of the atmosphere. Smoke and dust particles get collected beneath the inversion layer and spread horizontally to fill the lower strata of the atmosphere. Dense fogs in mornings are common occurrences especially during winter season. This inversion commonly lasts for few hours until the sun comes up and beings to warm the earth. The inversion takes place in hills and mountains due to air drainage.

Q2.Explain the heating and the cooling mechanism of atmosphere.
Or
Discuss the process through which earth and the atmosphere system maintain heat balance.
Answer:
(a) Conduction:

• The earth after being heated by insolation transmits the heat to the atmospheric layers near to the earth in long wave form. The air in contact with the land gets heated slowly and the upper layers in contact with the lower layers also get heated.
• Conduction takes place when two bodies of unequal temperature are in contact with one another, there is a flow of energy from the warmer to cooler body. The transfer of heat continues until both the bodies attain the same temperature or the contact is broken. Conduction is important in heating the lower layers of the atmosphere.

(b) Convection:

• The air in contact with the earth rises vertically on heating in the form of currents and further transmits the heat of the atmosphere. This vertical heating of atmosphere is known as convection.
• The convection transfer of energy is confined only to the troposphere.

(c) Advection:

• The transfer of heat through horizontal movement of air is called advection. Horizontal movement of the air is relatively more important than the vertical movement.
• In tropical regions particularly in northern India during summer season local winds called ‘loo’ is the outcome of advection process

Short Answer Type Questions:

Q1.What do you understand by atmosphere?
Answer
Atmosphere is a mixture of different gases and it envelopes the earth all round. It contains life-giving gases like oxygen for humans and animals and carbon dioxide for plants.
Q2.What are the elements of weather and climate?
Answer
The elements of weather and climate are temperature, pressure, winds, humidity, clouds and precipitation. These elements are subject to change and which influence human life on earth.
Q3.Describe the composition of atmosphere.
Answer
The atmosphere is composed of gases, water vapour and dust particles. Nitrogen constitutes 78.8%, oxygen constitutes 20.94% and argon constitutes 0.93%. Both gases together constitute 99% of the atmosphere. Other gases include are Carbon dioxide, Neon, Helium, Krypto, Xenon and Hydrogen. 
Q4.Why is troposphere the most important of all the layers of the atmosphere?
Answer
Troposphere is the most important of all the layers of the atmosphere:→ All changes in climate and weather take place in this layer.→  This layer contains dust particles and water vapour→ All biological activities take place in this layer.

Q5.Describe the composition of the atmosphere.
Answer
The atmosphere is composed of gases, water vapour and dust particles. The given table specifies the constituent of atmosphere with their volume.

Constituent	Formulae	% by Volume
Nitrogen	N2	78.08
Oxygen	O2	20.95
Argon	Ar	0.93
Carbon-dioxide	CO2	0.036
Neon	Ne	0.002
Helium	He	0.0005
Krypton	Kr	0.001
Xenon	Xe	0.0009
Hydrogen	H2	0.0005

Nitrogen and Oxygen gases together constitute 99% of the atmosphere. Other gases include are Carbon dioxide, Neon, Helium, Krypto, Xenon and Hydrogen. The proportion of gases changes in the higher layers of the atmosphere in such a way that oxygen will be almost in negligible quantity at the height of 120 km. Similarly, carbon dioxide and water vapour are found only up to 90 km from the surface of the earth. Carbon dioxide absorbs a part of terrestrial radiation and reflects back some part of it towards the earth’s surface. It is largely responsible for the green house effect. Ozone is another important component of the atmosphere which absorbs the ultra-violet rays radiating from the sun and prevents them from reaching the surface of the earth.

Q6.Draw a suitable diagram for the structure of the atmosphere and label it and describe it.

Answer

The atmosphere consists of different layers with varying density and temperature. The column of atmosphere is divided into five different layers depending upon the temperature condition. They are: troposphere, stratosphere, mesosphere, thermosphere and exosphere.

• Troposphere: It is the lowermost layer of the atmosphere. Its average height is 13 km and extends roughly to a height of 8 km near the poles and about 18 km at the equator. This layer contains dust particles and water vapour. All changes in climate and weather take place in this layer. The temperature in this layer decreases at the rate of 1 C for every 165m of height. This is the most important layer for all biological activity.

→ Tropopause: The zone separating the tropsophere from stratosphere is known as the tropopause. The air temperature at the tropopause is about minus 80°C over the equator and about minus 45°C over the poles. The temperature here is nearly constant, and hence, it is called the tropopause.

• Stratosphere: It is found above the tropopause and extends up to a height of 50 km. It contains the ozone layer. This layer absorbs ultra-violet radiation and shields life on the earth from intense, harmful form of energy.

• Mesosphere: It lies above the stratosphere, which extends up to a height of 80 km. In this layer, once again, temperature starts decreasing with the increase in altitude and reaches up to minus 100 C at the height of 80 km. The upper limit of mesosphere is known as the mesopause.

• Thermososphere: It is the outermost layer of the atmosphere. It is located from 80 km with no definite upper limit. The air in this layer is very hot because heat coming from the sun strikes the thermosphere first.

→ Ionosphere: The lower layer of thermosphere is called ionosphere. It is located between 80 and 400 km above the mesopause. It contains electrically charged particles known as ions, and hence, it is known as ionosphere. Radio waves transmitted from the earth are reflected back to the earth by this layer. Temperature here starts increasing with height.

• Exosphere: The uppermost layer of the atmosphere above the thermosphere is known as the exosphere. This is the highest layer but very little is known about it. Whatever contents are there, these are extremely rarefied in this layer, and it gradually merges with the outer space.

Long Answer Type Questions :

Q1.Write about elements of weather and climate in detail.
Answer:
The main elements of atmosphere which are subject to change and which influence human life on earth are temperature, pressure, winds, humidity, clouds and precipitation. These elements act and react on each other. These elements determine the direction and speed of wind, amount of sunlight received, cloud formation and amount of rainfall. These in turn affect weather and climate. These factors behave differently in different places. All these elements are affected by a number of factors in turn. For example, temperature is affected by latitude and height; humidity is affected by distance from the sun and pressure is affected by height from sea level.

Q2.Write about the structure of atmosphere in detail.
Answer:
Structure of Atmosphere: The layers of atmosphere differ from one another with respect to density and temperature. On the basis of chemical composition the atmosphere is mainly divided into

1. Homosphere
2. Hetrosphere

1. Homosphere:

• It extends upto 90 km.
• It is uniform in chemical composition.
•  It consists of three layers
  • Troposphere
  • Stratosphere
  • Mesosphere

Troposphere:

• Lower most layer of atmosphere
• Average height is 13 km although it is roughly 8 km.
• The thickness of troposphere is greater at equation due to upward transportations of heat by conventional currents. This layer consists of dust particles and water vapours.
• The temperature decrease with height in this layer at a rate ldegree for every 165 m. this is known as Normal Lapse Rate.
• It is layer is important for all biological activities besides that all climatic and weather conditions takes place in this layer.

Tropopause:

• The upper limit of troposphere separating it from stratosphere is called tropopause. It is very unstable at a thin layer and very thin layers of 1.5 km thickness.
• The temperature of tropopause is -80degree centigrade censius at equator and -40 degree centigrade at poles.
• The jet planes at the other activities occur in this layer.

Stratosphere:

• It extends upto 50km.
• It is thicker at poles then at equator.
• The temperature is almost constant in its lower portion upto 20 km and their it gradually increases upto 50 km due to the presence of Ozone which absorbs UV rays.
• The temperature rises in the upper limits of the stratosphere as there are no clouds, no conventional currents, no dust particles and the air moves in the horizontal direction. The upper’ limit of stratosphere is called stratosphere which has concentration of Ozone gas.

Mesosphere:

• It extends from 50* to 90 km.
• Temperature decreases with height in this layer and false upto minus 100 degree centigrade at a height of 80-90 km. this is due to the clouds in high latitudes.
• The upper limit of Mesosphere is called as Mesopause.

2. Hetrosphere:

• It has heterogeneous chemical.
• It consist of two layers
  • Ionosphere
  • Exosphere

Ionosphere

•  It extends from 80 to 400 km above the mesopause.
• It contains electrically charged particles known as ions.

Exosphere

• It is the uppermost layer of the atmosphere above the thermosphere.

Short Answer Type Questions:

Q1. What do incised meanders in rocks and meanders in plains of alluvium indicate?

Answer
The incised meanders in rocks and meanders in plains of alluvium indicates the status of original land surfaces over which streams have developed.

Q2. Explain the evolution of valley sinks or uvalas.
Answer
Generally, the surface run-off simply goes down swallow and sink holes and flow as underground streams and re-emerge at a distance downstream through a cave opening. When sink holes and dolines join together because of slumping of materials along their margins or due to roof collapse of caves, long, narrow to wide trenches called valley sinks or Uvalas form.

Q3.Underground flow of water is more common than surface run-off in limestone areas. Why?
Answer
Underground flow of water is more common than surface run-off in limestone areas because limestone is rich in calcium carbonate, the surface water as well as groundwater through the chemical process of solution and precipitation deposition, develop varieties of landforms. These two processes of solution and precipitation are active in limestones occurring either exclusively or interbedded with other rocks.
Q4.Glacial valleys show up many linear depositional forms. Give their locations and names.
Answer

Glacial valleys show up many linear depositional forms:
• Terminal moraines: formed at the end (toe) of the glaciers.
• Lateral moraines – formed along the sides parallel to the glacial valleys
• Ground moraines – many valley glaciers retreating rapidly leave an irregular sheet of till over their valley floors.
• Eskers – flow over the ground with ice forming its banks.
• Outwash Plains – The plains at the foot of the glacial mountains or beyond the limits of continental ice sheets.
• Drumlins – form beneath heavily loaded ice through fissures in the glacier.
Q5.How does wind perform its task in desert areas? Is it the only agent responsible for the erosional features in the deserts?

Answer
Winds also move along the desert floors with great speed and the obstructions in their path create turbulence. Winds cause deflation, abrasion and impact. Deflation includes lifting and removal of dust and smaller particles from the surface of rocks. In the transportation process sand and silt act as effective tools to abrade the land surface. The impact is simply sheer force of momentum which occurs when sand is blown into or against a rock surface. The wind action creates a number of interesting erosional and depositional features in the deserts. Winds are not the only agent responsible for the erosional features in the deserts. The rain or sheet wash is also important.

Q6. Running water is by far the most dominating geomorphic agent in shaping the earth’s surface in humid as well as in arid climates. Explain.

Answer

In humid regions, There are two components of running water. One is overland flow on general land surface as a sheet. Another is linear flow as streams and rivers in valleys. Most of the erosional landforms made by running water are associated with vigorous and youthful rivers flowing along gradients. With time, stream channels over steep gradients turn gentler due to continued erosion, and as a consequence, lose their velocity, facilitating active deposition. Overland flow causes sheet erosion. Depending upon irregularities of the land surface, the overland flow may concentrate into narrow to wide paths. In the early stages, down-cutting dominates during which irregularities such as waterfalls and cascades will be removed. In the middle stages, streams cut their beds slower, and lateral erosion of valley sides becomes severe. During their terminal stages, the running water makes deltas.In arid regions, though rain is scarce in deserts, it comes down torrentially in a short period oftime. The desert rocks devoid of vegetation, exposed to mechanical and chemical weathering processes due to drastic diurnal temperature changes, decay faster and the torrential rains help in removing the weathered materials easily. The weathered debris in deserts is moved by not only windbut also by rain/sheet wash.Thus, Running water is by far the most dominating geomorphic agent in shaping the earth’s surface in humid as well as in arid climates.
Q7. Limestones behave differently in humid and arid climates. Why? What is the dominant and almost exclusive geomorphic process in limestone areas and what are its results?

Answer

Limestones are permeable, thinly bedded and highly jointed and cracked therefore, the surface water
percolates well. After vertically going down to some depth, the water under the ground flows horizontally through the bedding planes, joints or through the materials themselves. This downward
and horizontal movement of water which causes the rocks to erode. Physical or mechanical removal of materials by moving groundwater is insignificant in developing landforms.
In arid climates, water table is below the surface therefore, there is less amount of surface water.
The amount of water differ in these two areas, therefore, limestones behave differently in humid and arid climates.
The dominant and almost exclusive geomorphic process in limestone is the processes of solution and deposition by the action of the groundwater. Many depositional forms develop within the
limestone caves. The depositional landforms in limestone areas by the action of ground water are stalctites, stalagmites and pillars.

Q8. How do glaciers accomplish the work of reducing high mountains into low hills and plains?

Answer

Masses of ice moving as sheets over the land or as linear flows down the slopes of mountains in broad trough-like valleys are called glaciers. The movement of glaciers is slow unlike water flow. The movement could be a few centimetres to a few metres a day or even less or more. Glaciers move basically because of the force of gravity.
Erosion by glaciers is tremendous because of friction caused by sheer weight of the ice. The material plucked from the land by glaciers get dragged along the floors or sides of the valleys and cause great damage through abrasion and plucking. Glaciers can cause significant damage to even un-weathered rocks and can reduce high mountains into low hills and plains.
As glaciers continue to move, debris gets removed, divides get lowered and eventually the slope is reduced to such an extent that glaciers will stop moving leaving only a mass of low hills and vast outwash plains along with other depositional features.

Long Answer Type Questions :

Q1.Explain the landforms that are seen in upper part of the river.
Answer:
In upper part of the river, many beautiful and attractive landforms are formed. Some of them are as follows:

•  V-shaped valleys: Valleys start as small and narrow rills; the rills will gradually develop into long and wide gullies; the gullies will further deepen, widen and lengthen to give rise to valleys. Depending upon dimensions and shape, many types of valleys like V-shaped valley, gorge, canyon, etc. can be recognised.
• Gorge: A gorge is a deep valley with very steep to straight sides.
• Canyon: A canyon is characterised by steep step-like side slopes and may be as deep as a gorge. A gorge is almost equal in width at its top as well as its bottom. In contrast, a canyon is wider at its top than at its bottom. In fact, a canyon is a variant of gorge.
• Waterfall: When the rivers start falling in pits in mountainous regions, it makes waterfall.
• Plunge pools: Once a small and shallow depression forms, pebbles and boulders get collected in those depressions and get rotated by flowing water and consequently the depressions grow in dimensions. A series of such depressions eventually join and the stream valley gets deepened. At the foot of waterfalls also, large potholes, quite deep and wide, form because of the sheer impact of water and rotation of boulders. Such large and deep holes at the base of waterfalls are called plunge pools.

Q2.Explain the landforms made by erosion caused by groundwater.
Answer:
Important landforms made by erosion are as follows:

1. Pools: These are conical shaped pits whose depth is three to nine metres. The width of the mouth is more than one metre. Due to solubility in water, when cracks in limestone increase, then pools take birth.

2.Swallow holes: Small to medium sized round to sub-rounded shallow depressions called swallow holes form on the surface of limestones through soil.

3. Sinkholes: A sinkhole is an opening more or less circular at the top and funnel -shaped towards the bottom with sizes varying in area from a few square metre to a hectare and with depth from a less than half a metre to thirty metres or more.

4. Uvalas: When sinkholes and dolines join together because of slumping of materials along their margins or due to roof collapse of caves, long, narrow to wide trenches called uvalas are formed.

5. Collapse sinks: If the bottom of the sinkholes forms the roof of a void or cave underground it might collapse leaving a large hole opening into a cave or a collapse sinks.

6. Lapies: Gradually, most of the surface of the limestone is eaten away by these pits and trenches, leaving it extremely irregular with a maze of points, grooves and ridges or lapies. Especially, these
ridges or lapies form due to differential solution activity along parallel to sub¬parallel joints. The lapie field may eventually turn into somewhat smooth limestone pavements.

7. Caves: In areas where there are alternating beds of rocks (shales, sandstones, quartzites) with limestones or dolomites in between or in areas where limestones are dense, massive and occurring as thick beds, cave formation is prominent. Water percolates down either through the materials or through cracks and joints and moves horizontally along bedding planes. It is along these bedding planes that the limestone dissolves and long and narrow to wide gaps called caves result. There can be a maze of caves at different elevations depending upon the limestone beds and intervening rocks. Caves normally have an opening through which cave streams are discharged. Caves having openings at both the ends are called tunnels.

Q3.Explain the depositional landforms made by rivers.
Answer:
Depositional Landfoi, made by rivers:

1. Alluvial Fans: Alluvia ms are formed when streams flowing from higher levels break into foot slope plains of low gradient. Normally very coarse load is carried by streams flowing over mountain slopes. This load becomes too heavy for the streams to be carried over gentler gradients and gets dumped and spread as a broad low to high cone shaped deposit called alluvial fan. Usually, the streams which flow over fans are not confined to their original channels for long and shift their position across the fan forming many channels called distributaries. Alluvial fans in humid areas show normally low cones with gentle slope from head to toe.

2. Deltas: Delta is like alluvial fans but develop at a different location. The load carried by the rivers is dumped and spread into the sea. If this load is not carried away far into the sea or distributed along the coast, it spreads and accumulates. Such areas over flood plains built up by abandoned or cut-off channels contain coarse deposits. The flood deposits of spilled waters carry relatively finer materials like silt and clay. The flood plains in a delta are called delta plains.

3. Floodplains: Floodplain is a major landform of river deposition. Large sized materials are deposited first when stream channel breaks into a gentle slope. Thus, normally, fine sized materials like sand, silt and clay are carried by relatively slow moving waters in gentler channels usually found in the plains and deposited over the bed and when the waters spill over the banks during flooding above the bed.

4. Natural Levees: Natural levees are found along the banks of large rivers. They are low, linear and parallel ridges of coarse deposits along the banks of rivers, quite often cut into individual mounds. During flooding as the water spills over the bank, the velocity of the water comes down and large sized and high specific gravity materials get dumped in the immediate vicinity of the bank as ridges. They are high nearer the banks and slope gently away from the river. The levee deposits are coarser than the deposits spread by flood waters away from the river. When rivers shift laterally, a series of natural levees can form.

5. Point Bars: Point bars are also known as meander bars. They are found on the convex side of meanders of large rivers and are sediments deposited in a linear fashion by flowing waters along the bank. They are almost uniform in profile and in width and contain mixed sizes of sediments. If there more than one ridge, narrow and elongated depressions are found in between the point bars.

Q4.Explain the erosional landforms created by waves and currents.
Answer:
Cliffs, Terraces, Caves and Stacks are important landforms created by erosion caused by waves and currents.

• Wave-cut cliffs: Almost all sea cliffs are steep and may range from a few m to 30 m or even more. At the foot of such cliffs there may be a flat or gently sloping platform covered by rock debris derived from the sea cliff behind. Such platforms occurring at elevations above the average height of waves is called a wave-cut terrace.
• Terraces: The lashing of waves against the base of the cliff and the rock debris that gets smashed against the cliff along with lashing waves create hollows and these hollows get widened and deepened to form sea caves. The roofs of caves collapse and the sea cliffs recede further inland.
• Sea stacks: Retreat of the cliff may leave some remnants of rock standing isolated as small islands just off the shore. Such resistant masses of rock, originally parts of a cliff or hill are called sea stacks.

Like all other features, sea stacks are also temporary and eventually coastal hills and cliffs will disappear because of wave erosion giving rise to narrow coastal plains, and with onrush of deposits from over the land behind m ay get covered up by alluvium or may get covered up by shingle or sand to form a wide beach.

Q5.Explain the different stages of a river.
Answer:
A river passes through three stages like a human being: youth, mature and old.

1. Youth Stage: Youth streams are less in number. In this stage with poor integration and flow over original slopes showing shallow V-shaped valleys with no floodplains or with very narrow floodplains along trunk streams. Streams divides are broad and flat with marshes, swrnmp and lakes. If meanders are present, they develop over these broad upland surfaces. These meanders may eventually entrench themselves into the uplands. Waterfalls and rapids may exist where local hard rock bodies are exposed.

2. Mature Stage: During this stage streams are plenty with good integration. The valleys are still V-shaped but deep; trunk streams are broad enough to have wider floodplains within which streams may flow in meanders confined within the valley. The flat and broad inter stream areas and swamps and marshes of youth disappear and the stream divides turn sharp. Waterfalls and rapids disappear.

3. Old Stage: Smaller tributaries during old age are few with gentle gradients. Streams meander freely over vast floodplains showing natural levees, oxbow lakes, etc. Divides are broad and flat with lakes, swamps and marshes. Most of the landscape is at or slightly above sea level.

Q1.It is weathering that is responsible for bio-diversity on the earth. How?
Answer
Weathering processes are responsible for the bio-diversity on the earth. Biomes and biodiversity
is basically a result of forests or vegetation. The forests depend upon the depth of weathering mantles.

Q2.What are mass movements that are real rapid and perceptible? List.

Answer

Mass movements transfer the mass of rock debris down the slopes under the direct influence of gravity. No geomorphic agent like running water, glaciers, wind, waves and currents participate
in the process of mass movements.The mass movements that are real rapid and perceptible are:• Earth flow• Mud flow• Landslide
Q3.What are the various mobile and mighty exogenic geomorphic agents and what is the prime job they perform?

Answer

Weathring, mass movements, erosion, transportation and deposition are the various mobile and mighty exogenic geomorphic agents. These agents bring the geomorphic changes on the surface of the earth.

Q4. Is weathering essential as a pre-requisite in the formation of soils? Why?

Answer

Yes, weathering is an essential as a pre-requisite in the formation of soils. Weathering processes are responsible for breaking down the rocks into smaller fragments and prepare the for formation of soils.

Q5.“Our earth is a playfield for two opposing groups of geomorphic processes.” Discuss.

Answer

It is right to say that our earth is a playfield for two opposing groups of geomorphic processes. The earth’s crust is dynamic, it has moved and moves vertically and horizontally. The differences in the internal forces operating from within the earth which built up the crust have been responsible for the variations in the outer surface of the crust. The earth’s surface is being continuously subjected to external forces induced basically by energy (sunlight). Also, the internal forces are still active though with different intensities. That means, the earth’s surface is being continuously subjected to by external forces originating within the earth’s atmosphere and by internal forces from within the earth. The external forces are known as exogenic forces and the internal forces are known as endogenic forces.The actions of exogenic forces result in wearing down of relief or elevations and filling up of basins/depressions, on the earth’s surface. The endogenic forces continuously elevate or build up parts of the earth’s surface and hence the exogenic processes fail to even out the relief variations
of the surface of the earth. So, variations remain as long as the opposing actions of exogenic and
endogenic forces continue.In general terms, the endogenic forces are mainly land building forces and the exogenic processes are mainly land wearing forces.
Q6. Exogenic geomorphic processes derive their ultimate energy from the sun’s heat. Explain.

Answer

The exogenic processes derive their energy from atmosphere determined by the ultimate energy from the sun and also the gradients created by tectonic factors. All the exogenic geomorphic processes arecovered under a general term, denudation. Weathering, mass movements, erosion and transportation are included in denudation.
• Weathering: It is action of elements of weather and climate over earth materials. The components of of weather and climate are temperature, pressure, winds, humidity and precipitation. All these components directly or indirectly derive their energy from the sun.

• Mass Movement: These movements transfer the mass of rock debris down the slopes under the direct influence of gravity. Weathering is not a pre-requisite for a mass movement. However, weathering aids in mass movement.
• Erosion and deposition: Erosion involves acquisition and transportation of rock debris. The erosion and transportation of earth materials are brought about by the wind, running water, glaciers, waves and ground water. Of these, the first three agents are controlled by climatic conditions while climate is decided by the energy of the sun.
Thus, All exogenic geomorphic processes derive their ultimate energy from the sun’s heat. However, the gravitational force of earth aids in all exogenic geomorphic processes because gravity makes mobility possible.
Q7.Are physical and chemical weathering processes independent of each other? If not, why? Explain with examples.

Answer

Physical and chemical weathering processes are not independent of each other. Physical weathering processes depend on some applied forces. The applied forces could be gravitational forces such as overburden pressure, load and shearing stress; expansion forces due to temperature changes, crystal growth or animal activity;  water pressures controlled by wetting and drying cycles. While in chemical weathering processes such as solution, carbonation, hydration, oxidation and reduction act on the rocks to decompose, dissolve or reduce them to a fine clastic state through chemical reactions by oxygen, surface and/or soil water and other acids.

Chemical Weathering process depends on the work of physical weathering process. The agents of physical weathering such as temperature change and freezing break the rocks and provide easy passage to chemical weathering process to work on. The chemical weathering processes make rocks decayed and decomposed which can be easily broken down by the physical weathring processes.

Q8.How do you distinguish between the process of soil formation and soil forming factors? What is the role of climate and biological activity as two important control factors in the formation of soils?

Answer

The process of soil formation starts with weathring. The weathring mantle provide the basic input for soil to form. First, the weathered material or transported deposits are colonised by bacteria and other inferior plant bodies like mosses and lichens. Also, several minor organisms may take shelter within the mantle and deposits. The dead remains of organisms and plants help in humus accumulation. Minor grasses and ferns may grow; later, bushes and trees will start growing through seeds brought in by birds and wind. Plant roots penetrate down, burrowing animals bring up particles, mass of material becomes porous and sponge like with a capacity to retain water and to permit the passage of air and finally a mature soil, a complex mixture of mineral and organic products forms.

Soil formaing factors control the formation of soils. These are five in number: (i) parent material; (ii) topography; (iii) climate; (iv) biological activity; (v) time. Soil forming factors act in union and affect the action of one another.
The climate and biological activity play very important role. The climatic elements involved in soil
development are moisture and temperature. Precipitation gives soil its moisture content which makes the chemical and biological activities possible. Excess of water helps in the downward transportation of soil components through the soil (eluviation) and deposits the same down below (illuviation). Temperature acts in two ways — increasing or reducing chemical and biological activity. Chemical activity is increased in higher temperatures, reduced in cooler temperatures (with an exception of carbonation) and stops in freezing conditions.

Biological Activity includes the effects of vegetative cover, organisms and bacteria. The vegetative cover and organisms help in adding organic matter, moisture retention, nitrogen etc. Dead plants provide humus, the finely divided organic matter of the soil.  With the increase in temorature, biological activity increases. In humid tropical and equatorial climates, bacterial growth and action is intense.

Long Answer Type Questions:

Q1.What are different types of mass movements?
Answer:
There are three types of mass movements: Slow Movements: Creep is one type under this category which can occur on moderately steep, soil covered slopes. Movement of materials is extremely slow and imperceptible except through extended observation. Materials involved can be soil or rock debris. Soil creep, talus creep, rock creep, rock- glacier creep etc can be identified. It also includes solifluction which involves slow downslope flowing soil mass or fine grained rock debris saturated or lubricated with water. This process is quite common in moist temperate areas where surface melting of deeply frozen ground and long continued rain respectively, occur frequently. When the upper portions get saturated and when the lower parts are impervious to water percolation, flowing occurs in the upper parts.

Rapid Movements: These movements are mostly prevalent in humid climate regions and occur over gentle to steep slopes. Movements of water- saturated clayey or silty earth materials down low angle terraces or hill slides is known as earth flow. When slopes are steeper ever the bedrock especially of soft sedimentary rocks like shale or deeply weathering igneous rock may slide downslope. With heavy rainfall, thick layers of weathered
materials get saturated with water and either slowly or rapidly flow down along definite channels. It looks like a stream of mud within a valley.

Landslides: The types of landslides.

• Slumps: The slipping of one or several units of rock debris with a backward rotation with respect to the slope over which the movement takes place.
• Debris slide: rapid rolling or sliding of earth debris without backward rotation of mass is known as Debris slide.
• Rockslide: Sliding of individual rock masses down bedding, joint or fault surface is rockslide.
• Rock fall: Rock fall is free falling of rock blocks over any steep slope keeping itself away from the slope. Rock falls occurs from the superficial layers of the rock face.

Q2.Explain different types of chemical weathering.
Answer:
Different types of chemical weathering includes:

1. Oxidation and Reduction: Oxidation is the effect of oxygen in air and water on the rocks. The atmospheric oxygen in rainwater unites with minerals in rocks specially with iron compounds. When oxidised minerals are placed in an environment where oxygen is absent, reduction takes place. It exists normally below water table, in area of stagnant water in more hot and humid climates.

2. Carbonation: When the carbon dioxide in atmosphere dissolves in water it form carbonic acid that affects the rocks, it is carbonation. It has acidic affect and dissolves calcium carbonates and magnesium carbonates such as gypsum, marble, limestone.

3.  Hydration: When the hydrogen of water dissolves in rocks hydration occurs. Certain minerals in rocks increase their volume and become heavy when observe water contains hydrogen. They break due to its increased pressure and the colour also changes.

4. Solution: Rainwater is able to dissolve certain minerals and leaching of the soil occurs. Normally solids are also removed during leaching. For e.g.: gypsum, rock salt, etc. undergo solution.

Q3.Explain different types of physical weathering.
Answer:
Different types of physical weathering includes:

• Exfoliation: Due to differential heating and resulting expansion and contraction of surface layers and their subsequent exfoliation from the surface results in smooth rounded surfaces in rocks. In rocks like granites, smooth surfaced and rounded small to big boulders called tors form due to such exfoliation.
• Frost: It is an active agent in cold climatic regions in high altitudes and the cracks are filled with water during the day time, this water is frozen at night when temperature falls below freezing point.
• Pressure: Many igneous and metamorphic rocks crystallize deep in the interior under the combine influence of high pressure and temperature. The salt near surface pores cause splitting of the grains within the rocks which eventually falls off, this result into granules disintegration.

Q4.Explain about erosion and deposition.
Answer:
Erosion involves acquisition and transportation of rock debris. When massive rocks break into smaller fragments through weathering and any other process, erosional geomorphic agents like running water, groundwater, glaciers, wind and waves remove and transport it to other places depending upon the dynamics of each of these agents. Abrasion by rock debris carried by these geomorphic agents also aids greatly in erosion. By erosion, relief degrades, i.e., the landscape is worn down. Weathering aids erosion it is not a pre-condition for erosion to take place. Weathering, mass-wasting and erosion are degradational processes. It erosion that is largely responsible for continuous changes that the earth’s surface is undergoing. The erosion and transportation of earth materials is brought about by wind, running water, glaciers, waves and ground water.

Deposition is a consequence of erosion. The erosional agents loose their velocity and hence energy on gentler slopes and the materials carried by them start to settle themselves. In other words, deposition is not actually the work of any agent. The coarser materials get deposited first and finer ones later. By deposition depressions get filled up. The same erosional agents viz., running water, glaciers, wind, waves and groundwater act as aggradational or depositional agents also. What happens to the surface of the earth due to erosion and deposition is elaborated in the next chapter on landforms and their evolution. There is a shift of materials in mass movements as well as in erosion from one place to the other.

In This Post we are  providing Chapter-MOLECULAR BASIS INHERITANCE NCERT MOST IMPORTANT QUESTIONS for Class 12 BIOLOGY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON MOLECULAR BASIS INHERITANCE

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1.Two claimant fathers filed a case against a lady claiming to be the father of her only daughter. How could this case be settled identifying the real biological father?
Ans.This case to identify the real biological father could lee settled lay DNA – finger printingtechnique. In this technique :-

1. first of all, DNA of the two claimants who has to be tested is isolated.
2. Isolated DNA is then digested with suitable restriction enzyme & digest is subjected to gelelectrophoresis.
3. The fragments of ds DNA are denatured to produce ss DNA by alkali treatment.
4. The electrophoresed DNA is then transferred from get into a nitrocellulose filter paper where itis fixed.
5. A known sequence of DNA is prepared called probe – DNA & is labelled with radioactive esotope32p & then probe is added to nitrocellulose paper.
6. The nitrocellulose paper is photographed on X –ray film through auto radiography. The film isanalysed to determine the presence of hybrid nucleic acid.

Then, the DNA fingerprints of the two claimants is compared with the DNA fingerprint of the lady &her daughter, whosoever matches with each other would be declared as biological father of herdaughter.

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2.The length of DNA in an eukaryotic cell is N 2.2 m How can such a huge DNA be packaged in a nucleus of micrometer in diameter.

Ans.In eukaryotes, the DNA is wrapped around positivelycharged histone octamer into a structure called nucleosome. Atypical nucleosome consists of 200bp of DNA helix. Thenucleosomes are the repeating units that form chromatin fibres.
These chromatin fibres condense at metaphase stage of celldivision to form chromosomes. The packaging of chromatin at higher level requires additional set ofproteins called non-histone chromosomal proteins thus in nucleus, certain regions of the chromatinare loosely packed & they Stain lighter than the other region, these are called euchromatin. Theother region are lightly packed & they stain darker & are called heterochromatin

3.Describe the continuous & discontinuous Synthesis of DNA?

Ans.Synthesis of new strand of DNA takes place lay additionof fresh nucleotides to the 3 – OH group of the last nucleotideof the primer. This synthesis takes place in 5 direction&enzyme that catalyses this is DNA – polymerase
∴∴ synthesis of strand called leading strand iscontinuous.
The replication of second strand of the DNA molecule is
DISCONTINOUS on strand called lagging strand.
Primase initiates primer synthesis on strand near the fork. The RNA – primer thus formedprovides free for replication of single stranded region on lagging strand the newcomplementary strand is formed in small fragments of DNA called Okazaki fragments. It is calleddiscontinuous because it has to be initiated several times & every time an Okazaki fragment isproduced.

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4.What are the three types of RNA & Mention their role in protein Synthesis?
Ans. There are three types of RNA :

1. Messenger RNA (mRNA) :- It is a single – stranded RNA which brings the genetic information ofDNA transcribed on it for protein synthesis.
2. Transfer RNA (tRNA) :- It has a clover leaf like structure which acts as an adapter moleculewhich contains an “anticodon loop” on one end that reads the code on one hand &” an amino acid acceptor end which binds to the specific amino acid on other hand.
3. Ribosomal RNA (rRNA) :- Ribosomes provides the site for synthesis of protein &catalyse theformation of peptide bond.

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5. Define bacterial transformation? Who proved it experimentally & how?
Ans. The transformation is a mode of exchange or transfer of genetic information betweenorganism or from one organism to another.
Fredrick Griffith tested the virulence of two strains of Diplococci to show transformation in thefollowing steps :-

1. When S-III strains of bacteria are injected into mice. It developed pneumonia & died.
2. When R-II strains are infected into mice, they did not develop pneumonia & survive.
3. When heat – killed S-III strains of bacteria are injected into mice, No symptoms of pneumoniadevelops& mice remain healthy.
4. When a mixture of heat – killed S-III strain & lives R-II strain is injected into mice, theydeveloped pneumonia & died.

From these results, Griffith concluded that the presence of heat – killed S-III bacteria must convertliving R-II type bacteria to type S-III so as to restore them the capacity for capsule formation. Thiswas called “BACTERIAL TRANSFORM ATION”
S strain →→ Inject into mice →→ Mice die
R strain →→Injct into mice →→ Mice live
S strain (heat-killed)→→Inject into mice →→ Mice live
S strain (heat-killed) + R strain (live) →→ Inject into mice →→Mice die

6. What is meant by semi conservative replication? How did Meselson and Stahl prove it experimentally?
Ans .Meselson and Stahl, performed an experiment using E.coli to prove that DNA replication is semi conservative.

• They grew E.coli in a medium containing 15NH4Cl15NH4Cl.
• Then separated heavy DNA from normal (14N) by centrifugation in CsCl density gradient.
• The DNA extracted, after one generation of transfer from 15N medium to 14N medium, had an intermediate density.

-The DNA extracted after two generations consisted of equal amounts of light and hybrid DNA.
-They proved that DNA replicates in a semiconservative manner.

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7. What does the lac operon consist of? How is the operator switch turned on and off in the expression of genes in this operon? Explain.
Ans.Lac Operon consists of the following :

• Structural genes : z, y, a which transcribe a polycistronic mRNA.
• gene ‘z’ codes for b-galactosidase
• gene‘y’ codes for permease.
• gene‘a’ codes for transacetylase.
• Promotor : The site where RNA polymerase binds for transcription.
• Operator : acts as a switch for the operon
• Repressor : It binds to the operator and prevents the RNAPolymerase from transcribing.
• Inducer : Lactose is the inducer that inactivates the repressor by binding to it.
• Allows an access for the RNA polymerase to the structural gene andtranscription.

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8. What is an operon? Describe the major steps involved in an operon?
Ans.Operon is a group of controller & structural genes which controls the catabolism of the cell geneticallyeg lactose operon / lac operon.
(i) When inducer or lactose is absent :-
The lac regulator gene synthesize a repressor protein by transcription & translation. This repressor protein binds with operator site of lac operon & blocks RNA polymerase. Thus, RNA polymerase unable to transcribe mRNA & structural gene unable to translate enzyme B-galactosidase.
(ii) When inducer or lactose is present :_
The lac regulator gene transcribe mRNA &synthesise active lac repressor protein & at the same time lactose is converted into isomer allolactose. Allolactose binds to active lac repressor due to which it is converted to inactive repressor. This inactive repressor is released from operator site of lac operon & RNA polymerase binds to promoter & starts to transcribe mRNA & forms β-galactosidase are which converts lactose into glucose &galactose.
Thus, presence of lactose determines whether or not lac. Repressor is bound to operator & genes are expressed on not.

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9.Where do transcription & translation takes place in a prokaryotic cell? Describe the three stepsinvolved in translation?
Ans.In a prokaryotic cell both transcription & translation occurs in cytoplasm. It consist offollowing steps :-
(i)ACTIVATION OF AMINO ACIDS :- amino acids are activated in the presence of ATP lay enzaminoacyltRNASynthetase.
(ii)BINDING OF ACTIVATED AMINOACID WITH tRNA :- Activated amino acids binds with specific tRNA to form charged tRNA .
(iii)INITIATION OF POLYPEPTIDE CHAIN :- Initiation codon is AUG which codes for methionine. Initiation codon of mRNA binds to p-site of ribosome with the help of initiation factors.
(iv)ELONGATION OF POLYPEPTIDE CHAIN :-
(a)Second activated amino acid along itstRNA reaches the ‘A’ site & binds to mRNA codon next to AUG.
(b)A peptide bond is formed betweentwo amino acid by peptidyl transferase.
(c) Ribosomes translocation mRNA in -direction due to which free tRNA slips away &peptidyltRNA reaches at P – site. Now third amino acid reaches at A – site & process continues.
(d)TERMINATION OF POLYPEPTIDE CHAIN :- When a termination codon (UAA, UAG, UGA) reaches at A- site translation terminates Since there is no specific tRNA for these codons.
(i)

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10.Who performed the blender experiment? What does this experiment prove? Describe the steps followed in this experiment?
Ans.The proof for DNA as the genetic material came from the experiments of Harshey& chase whoworked with bacteriophage.
The bacteriophage on infection injects only the DNA into the bacterial cell & not the protein coat.
Bacterial cell treats the viral DNA as its own & subsequently manufactures more virus particles.
They grew some viruses on a medium that ‘contained radioactive Phosphorus & some other on medium that contained radioactive sulphur. Virus grown in the presence of radioactive phosphorus contained radioactive DNA but not proteins because DNA contains phosphorus. Similarly virus grown on radioactive sulfur contained radioactive protein because DNA does not contain sulfur.
Radioactive phages are allowed to infect E. coli bacteria & soon after infection the cultures weregently agitated in a blender to separate the adhering protein coat of virus from bacterial cell.It was found that when phage containing radioactive DNA was used to infect the bacteria itsradioactivity was found in bacterial cells indicating that DNA has been injected into bacterial cell so,the DNA is the genetic material & not proteins

In This Post we are  providing Chapter-7 EVOLUTION NCERT MOST IMPORTANT QUESTIONS for Class 12 BIOLOGY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON EVOLUTION

Question 1.
Using the Internet and discussing with your teacher, trace the evolutionary stages of any one animal, say, a horse.
Answer:
The major evolutionary trend of horses:

1. General increase (with occasional decrease) in size.
2. The progressive loss of toes.
3. Lengthening of toes that are retained.
4. Lengthening of limbs in general.
5. Enlargement of the brain (especially cerebral hemisphere).
6. Increase in height.
7. Increase in the complexity of molar teeth and an enlargement of the last two and, eventually, the last three premolars until they came to resemble molars.

Question 2.
Summarise Milter’s simulation experiment for organic synthesis. Comment on its efficacy. (CBSE Delhi 2012)
Answer:
Miller’s experiment. Milter (1953) made the first successful simulation experiment to assess the validity of the claim for the origin of organic molecules. Miller sealed in a spark chamber a mixture of water, methane, ammonia, and hydrogen gas. He made arrangements for boiling water.

The trap in turn was connected with the flask for boiling water. After 18 days, a significant amount of simple major organic compounds, such as amino acids, such as glycine, alanine, and aspartic acid, and peptide chains, began to appear. Simple sugars, urea, and short-chain fatty acids were also formed. In the atmosphere, this spark is provided by U.V. light or other energy sources.

Stanley Miller’s Experiment in the artificial production of organic compounds.

Question 3.
With the help of an algebraic equation, how did Hardy-Weinberg explain that in a given population the frequency of occurrence of alleles of a gene is supposed to remain the same through generations? (CBSE Delhi 2018)
Or
Explain Hardy-Weinberg’s principle. (CBSE Delhi 2019 C)
Answer:
In a given population, one can find out the frequency of occurrence of alleles of a gene or a locus. This frequency is supposed to remain fixed and even remain the same through generations. Hardy-Weinberg’s principle stated it using algebraic equations. According to this principle, allele frequencies in a population are stable and are constant from generation to generation. The gene pool (total genes and their alleles in a population) remains constant. This is called genetic equilibrium.

Sum total of all the allelic frequencies is 1. Individual frequencies, for example, can be named as p, q, etc. In a diploid, p and q represent the frequency of allele A and allele a, respectively. The frequency of AA individuals in a population is simply p². This is simply stated in another way, i.e. the probability that an allele A with a frequency of p appears on both the chromosomes of a diploid individual is simply the product of the probabilities, i.e. p². Similarly of aa is q², of Aa 2pq. Hence, p² + 2pq + q² = 1. This is a binomial expansion of (p + q)². When the frequency measured is different from expected values, the difference (direction) indicates the extent of evolutionary change.

Disturbance in genetic equilibrium, or Hardy-Weinberg equilibrium, i.e. change of frequency of alleles in a population, would then be interpreted as resulting in evolution.

Question 4.
(i) Differentiate between analogous and homologous structures.
Answer:

Analogous organs	Homologous organs
(i) Organs that are structurally dissimilar but functionally similar are called analogous organs. Example: wings of birds and insects.	(i) Organs that are structurally similar but functional dissimilar are called homologous organs. Example: forelimbs of frog, lizard, bird, bat, horse, man, etc.
(ii) They lead to convergent evolution.	(ii) They lead to divergent evolution.

(ii) Select and write analogous structures from the list given below:
(o) Wings of butterfly and birds
(b) Vertebrate hearts
(c) Tendrils of Cucurbita and thorns of Bougainvillea
(d) Tubers of sweet potato and potato (CBSE Delhi 2018)
Answer:
(a) Wings of butterflies and birds.
(b) Tubers of sweet potato and potato.

Question 5 .
Write thecharacteristicsofRamapithecus, Dryopithecus, and Neanderthal man. (CBSE Delhi 2017)
Answer:
Characteristics of Ramapithecus:

• They evolved around 15 mya.
• They were more man-like, walked more erect, and had teeth like modern men.

Characteristics of Dryopithecus:

• They evolved around 5 mya.
• They were ape-like, having hairy arms and legs of the same length, large brains. They used to eat soft fruits and leaves and walked like gorillas and chimpanzees.

Characteristics of Neanderthal Man:

• They evolved around 1,00,000-40,000 years ago.
• Fossil found in east and central Asia had brain size 1400 cc. They used hides to protect their body. They buried their dead.

Question 6.
How does the process of natural selection affect Hardy-Weinberg equilibrium? List the other four factors that disturb the equilibrium.
Or
Write Hardy-Weinberg principle.
Or
How can Hardy-Weinberg equilibrium be affected? Explain giving three reasons. (CBSE Delhi 2018C)
Answer:
Hardy-Weinberg Principle states that the sum of allelic frequencies in a population is stable and is constant from generation to generation, i.e. the gene pool (total genes and their alleles) in a population remains constant. This is called genetic equilibrium. The sum total of all the allelic frequencies is

Hardy-Weinberg’s Equilibrium p²+ q² + 2pq =

Five factors that influence these values are:
The five factors which affect Hardy- Weinberg’s equilibrium is as follows:

1. Gene migration: When some individuals of a population migrate to other populations or when certain individuals come into a population (i.e. migration and immigration), some genes are lost in the first case and added in the second.
2. Genetic drift: Random changes in the allele frequencies of a population occurring only by chance constitute genetic drift. The change in allele frequency may become so drastically different that they form a new species.
3. Mutations: The mutations are random and directionless. They are sufficient to create a considerable genetic variation for speciation to occur.
4. Recombination: New combinations of genes occur due to crossing over in meiosis during gametic formation.
5. Natural selection: It is the most critical evolutionary process that leads to changes in allele frequencies
   and favors adaptation as a product of evolution.

Question 7.
Define genetic drift. How does it produce the founder effect and genetic bottleneck?
Or
How does the original drifted population become a founder? (CBSE 2019 C)
Answer:
Genetic drift: Random change occurring in the allele frequency by chance alone is called genetic drift. It is due to habitat fragmentation, isolation, natural calamities, or any epidemics.

Founder effect: When a section of the population gets separated from the original population, then this section becomes genetically different from the original population due to a change in alleles frequency. The original population becomes the founder of the new population. This is called the founder effect which is the result of genetic drift, i.e. by chance. Genetic bottleneck.

When in a season one population died leaving few individuals of the population which become the founder of the new population, then it will produce only a few genes by selection only, i.e. by chance new population is emerged and it is similar to a bottle in which only certain population is allowed to flow as in the neck of a bottle.

Bottleneck effect

Question 8.
How does Darwin’s theory of natural selection explain new forms of life on earth? (CBSE 2008, 2016)
Answer:
Darwin’s Theory of evolution may be summed up as follows:
Darwin’s Theory of natural selection. Charles Darwin (1809 – 1882), a naturalist, proposed a theory to explain the process of evolution. His theory was published in his famous book “Origin of Species” published in 1858.

His theory of natural selection is termed Darwinism:

• Rapid multiplication
• Struggle for existence
• Variations
• Natural selection or survival of the fittest
• Inheritance of useful variations
• Origin of new species.

Evidence in favor of Darwin’s theory: Darwin’s theory is supported by natural selection, phenomena of mimicry and protective coloration, and the correlation between nectaries of flowers and proboscis of pollinating insects.

Darwin’s theory fails to explain the perpetuation of vestigial organs and over-specialization of organs.
Darwin’s theory has since been modified in the light of progress in genetics.

Question 9.
Describe the present-day concept of evolution.
Answer:
1. Modern concept of evolution: The modern concept of evolution is a modified form of Darwin’s theory of natural selection and is often called Neo-Darwinism. It comprises genetic variation, natural selection, and isolation.
(a) Mutations: These have been recognized as the ultimate source of biological changes and hence the raw material of evolution. The mutation in chromosomes may be due to changes in structure, number, or gene.

(b) Gene Recombination takes place during crossing over in meiosis. New combinations of genes produce new phenotypes.

(c) Hybridisation is the intermingling of the genes of the members of closely related species.

(d) Genetic drift is the elimination of the genes of some original characteristics of a species by extreme reduction due to different reasons.

In Monoparental reproduction, only chromosomal and gene mutation are sources of genetic variation,

2. Natural Selection: If differential reproduction (i.e. some individuals produce abundant offspring, some only a few, and some organisms none) continues for many generations, genes of the individuals which produce more offspring will become predominant in the gene pool of the population. Thus natural selection occurs through differential reproduction in successive generations. The migration of individuals from one to another population is an accessory factor for speciation (origin of new species).

3. Isolation: By selecting the most suitable genotypes, natural selection guides different populations into different adaptive channels. The reproductive isolation between the populations due to certain physical barriers or others leads to the formation of new species. Isolation plays a significant role in evolution.

Question 10.
(i) Name the primates that lived about 15 million years ago. List their characteristic features.
Answer:
Primates Dryopithecus and Ramapithecus lived about 15 mya.

Features:
(a) Hairy and walked like gorillas and chimpanzees. Height up to 4 feet but walked upright.

(ii) (a) Where was the first man-like animal found?
Answer:
Ethiopia and Tanzania

(b) Write the order in which Neanderthals, Homo habilis, and Homo erectus appeared on the earth. State the brain capacity of each one of them.
Answer:

• Homo habilis – 700 cc
• Homo erectus – 900 cc
• Neanderthals man – 1300-1600 cc

(c) When did modern Homo sapiens appear on this planet ? (CBSE Delhi 2011)
Answer:
Homo sapiens appeared about 34000 years ago.

Very Importance Figures:

(A) Foretimbs of vertebrates as homologous organs.
(B) AnaLogous organs. Wings of insect and bird.

Darwin finches

Adaptive radiations of Australian marsupials

Kinds of selection

In This Post we are  providing Chapter-5 PRINCIPLE OF INHERITANCE AND VARIATION NCERT MOST IMPORTANT QUESTIONS for Class 12 BIOLOGY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON PRINCIPLE OF INHERITANCE AND VARIATION

1. A woman with O blood group marries a man with AB blood group
(i) work out all the possible phenotypes and genotypes of the progeny.
(ii) Discuss the kind of dominance in the parents and the progeny in this case.
Ans. (i) Blood group AB has alleles as I­­A, IB and O group has ii which on cross gives the both blood groups A and B while the genotype of progeny will be IAi and IBi.
(ii) IA and IB are equally dominant (co-dominant). In multiple allelism, the gene I exists in 3 allelic forms, IA, IB and i.

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2. Explain the cause of Klinefelter’s syndrome. Give any four symptoms shown by sufferer of this syndrome.
Ans. Cause : Presence of an extra chromosome in male i.e., XXY. Symptoms : Development of breast, Female type pubic hair pattern, poor beard growth, under developed testes and tall stature with Feminized physique.

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3. In Mendels breeding experiment on garden pea, the offspring of F2 generation are obtained in the ratio of 25% pure yellow pod, 50% hybrid green pods and 25% green pods State (i) which pod colour is dominant (ii) The Phenotypes of the individuals of F1 generation. (iii) Workout the cross.
Ans. (i) Green pod colour is dominant
(ii) Green pod colour

Phenotypic ratio 3 : 1
Genotypic ratio 1 : 2 : 1

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4. In Antirrhinum majus a plant with red flowers was crossed with a plant with white flowers. Work out all the possible genotypes & phenotypes of F1 & F2 generations comment on the pattern of inheritance in this case?
Ans. The inheritance of flower colour in snapdragon or Antirrhinum majus is an example of incomplete dominance. When a cross was made between a red flowered plant & a white flowered plant, the F1hybrid was pink i-e-an intermediate between red & white which means that both red & white are incompletely dominant. When F1 individuals was self – pollinated, the F2 generation consists of red, pink & white flower appears in ratio 1:2:1 respectively.

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5. A red eyed male fruitfly is crossed with white eyed female fruitfly. Work out the possible genotype & phenotype of F1 & F2 generation. Comment on the pattern of inheritance in this cross?
Ans. When a red eyed is crossed with white eyed female fruitfly, offspring will have both white eyed male & red eyed female in 1:1 ration in F1 generation. In F2 generation, 50% females will be red – eyed & 50% will be white eyed, similarly, in males 50% will be red eyed & 50% will be white eyed. This result indicates that in sex-linked genes, males transmit their sex-linked characters to their grandson through their daughter; such type of inheritance is called criss-cross inheritance –

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6. A man with AB blood group marries a woman with O group blood.

(i) Work out all the possible phenotypes & genotypes of the progeny.
(ii) Discuss the kind of domination in parents & progeny in this case?
Ans. (i) Half the progeny will have blood group A with genotype IA IO & half the progeny will have blood group B with genotype IB IO.
(ii) IA & IB both the genes are dominant over IO gene hence progeny shows either blood group A or B while in parents since both the dominant genes are present together man will have blood group AB & this phenomena is called co-dominance.

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7. In an cross made between a hybrid tall & red plant (TtRr) with dwarf & white flower (ttrr). What will be the genotype of plants in F1 generation?
Ans.

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8. In dogs, barking trait is dominant over silent trait & erect ears are dominant over drooping ears. What is the expected phenotypic ratio of offspring when dogs heterozygous for both the traits are crossed?
Ans.

Ration :- Barking & erect = 9
Barking & drooping =3
Silent & erect = 3
Silent & drooping =1
Phenotypic ratio = 9 : 3 : 3 : 1

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9. Differentiate between dominance, co-dominance & Incomplete dominance with one example each.

Ans. (i) Dominance :- When a cross is made between true – breeding tall pea plant & true – breeding dwarf pea plant, all the plants in F1 generation are tall this sows that tall character is dominant over dwarf

(ii) Co-dominance :- If the two equally dominant genes are present together, both of them will be equally expressed, this phenomena is called co-dominance eg alleles of blood group IA & IB ore dominant over IO but when both the alleles are present together, both of them will equally express & forms a phenotype AB.

(iii) In complete dominance :- When a cross is made between two characters of which none of them is completely dominant then an intermediate character develops in the progeny eg. when a cross is made between red flower & white flower in snapdragon flower an intermediate pink colour appears in the progeny

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10. A dihybrid heterozygous tall & yellow pea plant was crossed with double recessive plant.
(i) What type of cross is this?
(ii) Work out the genotype & phenotype of progeny
(iii) What principle of Mendel is illustrated through result of this cross?
Ans. (i) Test cross.
(ii)

(iii) Principle of Independent Assortment – Acc to which, in the inheritance of contrasting characters the factors of each pair of character segregate independently of the factors of the other pair of characters.Search for:

In This Post we are  providing Chapter-4 REPRODUCTIVE HEALTH NCERT MOST IMPORTANT QUESTIONS for Class 12 BIOLOGY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON REPRODUCTIVE HEALTH

Question 1.
Define population. What are the aims of the population study?

Answer:
The population is defined as the total number of individuals of a species present in a particular area. The members of a population have some common characteristics, share a common gene pool, and are capable of interbreeding among themselves to produce fertile offsprings. Aims of Population Study. An alarming rise in the human population has created many serious problems. Therefore, population education has been introduced into the school and college curricula.

Population education is aimed at making the students aware of the:

• consequences of uncontrolled population growth such as environmental pollution, depletion of natural resources, extinction of species, etc ;
• benefits of lowering population growth rate to the biosphere ;
• advantages of a small family to humans ;
• growth, distribution, and density of population ;
• relation of population to the standards of life.

Question 2.
Define birth rate, death rate, and fertility rate.

Answer:
1. Birth or natality rate: It is generally expressed as the number of births per 1,000 individuals of a population per year. It increases the population size (total number of individuals of a population) and population density.

The national average birth rate in India is about 28.6 per 1,000 per year. Among Indian states, Kerala has the lowest birth rate of 18 per 1,000, while U.P. has the highest of 34.8 per 1,000.

2. Death or mortality rate: It is the opposite of the natality rate. It is commonly expressed as the number of deaths per 1,000 individuals of a population per year.

3. Fertility rate: It is the number of live births per unit time per unit number of fertile females. Fertility Rate
h

Question 3.
What is family planning? List the ways of family planning:

Answer:
Family planning: The main objective of family planning or family welfare program is to prevent the fertilization of the ovum by the male sperm and stop the increase in population growth by various methods, such as contraceptives, intrauterine devices, vasectomy, and tubectomy. The contraceptives (Anirudh) for males and intra-uterine devices, loop for females are used to avoid pregnancy.

Vasectomy is the method of sterilizing males by surgical operation of sperm duct or vas deferens. Tubectomy is the method of sterilizing females by the surgical operation of fallopian tubes. Whatever the method employed, it must take care of the health of the persons concerned.

Because of the family planning methods, the birth rate in India is reduced to some extent. The government gives incentives to those who adopt family planning.

Ways of family planning:

• Late marriage for young persons.
• Increase in the sources of recreation so as to divert the attention from sex.
• The couple should not mate between 8-18th days from the start of the menstrual cycle.
• Use of contraceptives.
• Sterilization.
• Use of drugs.
• Abortion.
• Restrict the family to two children.

Question 4.
Suggest the various measures of population control:

Answer:
Population control: Population explosion can be checked by two methods-population education and birth control.

A. Population education: The knowledge about the relationship of population size and the availability of resources for the welfare of the society is called population education.

1. The students should be convinced about the relationship between overpopulation and unemployment.
2. The citizens should be told how the large size of the population is eating away the resources of the state and the reasons for the limited availability of healthcare, education facilities, and other welfare schemes.
3. People should be made aware of how a large number of children eat away the meager resources of the family with nothing left for bad days, how large families rely on indebtedness to meet emergencies, how child bread earners do not improve the conditions of the family, how uneducated children remain a burden on the society, etc. They should be convinced that a small family can live comfortably even with meager resources.

B. Birth control:

1. Mass media of communication. Radio, television, newspapers, magazines, hoardings, and posters should be employed to spread the message of family planning and birth control and its advantages. The future of mankind depends on the stabilization of the human population at a level that ensures basic necessities of life, employment, and happiness,
2. The law about marriageable age should be widely published and strictly enforced (21 years for boys and 18 years for girls). In developed countries, women marry at the age of 25-35 years.
3. As far as possible, stress should be laid on raising the social status of women. Women having higher social status prefer smaller families. Such women generally marry late.
4. Remove the superstitions and wrong beliefs in the society about a higher number of children being God’s gift connected with earthly or heavenly prosperity.

Question 5.
What is amniocentesis? Write its procedure and significance:

Answer:
Amniocentesis is a fetal sex determination test based upon the chromosomal pattern in the amniotic fluid surrounding the developing embryo. It should be legally banned throughout the country as such a ban shall check increasing female foeticide cases and maintain a normal sex ratio in the country.

Procedure:

1. The fetus bathes in the amniotic fluid that fills the amniotic cavity. At an early stage of pregnancy (14th or 15th week), the location of the fetus and placenta is determined by sonography (use of high-frequency sound waves).
2. Then a small amount of amniotic fluid is drawn by passing a special surgical syringe needle through the abdominal wall and uterine wall into the amniotic sac containing the amniotic fluid.
3. Celts that have sloughed from the fetus’s skin or respiratory tract into the fluid are thus sucked into the syringe.

Significance:

1. These cells can be examined for chromosomal abnormalities, such as Down’s syndrome, Klinefelter’s syndrome, Turner’s syndrome, etc resulting from non-disjunction during cell division.
2. The cells can also be cultured and in about a fortnight enough cells become available for test. The cells and fluid are also tested for metabolic disorders such as phenylketonuria, sickle-cell anemia, etc.

Question 6.
Write a note on test-tube babies:

Answer:
Test-Tube Babies: In some women normal conception is not possible because of blocked oviducts or spermicidal secretions in the vagina or the low sperm count of the husband. In such cases, her ovum is removed, fertilized by her husband’s sperm in a laboratory dish, checked that development has begun, and a morula (up to 32 cell stage) replaced or implanted in her uterus.

The entire operation is carried out under sterilized conditions. With proper medical care, she will give birth to a normal child on the completion of gestation. The baby produced in this manner (conceived out of and nursed in the uterus) is called a test-tube baby. The baby is not reared in the test tube. A scientific term for this procedure is in vitro (“in glass”) fertilization

The success rate of the technique is less than 20%. To increase the chances of success, the prospective mother is given fertility drugs which cause many ovarian follicles to mature at the same time. This releases many eggs simultaneously, thereby increasing the chances of success.

Question 7.
Reproductive and Child Healthcare (RCH) programs are currently in operation. One of the major tasks of these programs is to create awareness amongst people about the wide range of reproduction-related aspects. This is important and essential for building a reproductive health society.
1. “Providing sex education in schools is one of the ways to meet this goal.” Give four points in support of your opinion regarding this statement.

Answer:
Sex education is important in schools:
(a) to provide the right information about myths and misconceptions.
(b) to create awareness about reproduction.
(c) to provide knowledge about the growth of reproductive organs and sexually transmitted diseases (STDs)
(d) to guide the students about social evils such as sex abuse, sex-related crimes, etc.

(ii) List any two indicators that indicate a reproductively healthy society.
Answer:
Indicators about a reproductively healthy society.
(a) Low infant mortality rate (IMR)
(b) Low maternal mortality rate (MMR)

• Increased number of couples with small families.
• Better detection and cure of STDs.

Question 8.
Give a brief account of Assisted Reproductive Technologies (ART).

Answer:
Where corrective treatments are not available, there are special techniques called Assisted Reproductive Technologies (ART) to help the couple produce children; they are as follows:
1. Test-Tube baby programs:
(a) In this method, ovum from the wife or a donor female and the sperms from the husband or a donor is allowed to fuse under simulated conditions (as that of the body) in the laboratory; it is called in vitro fertilization (IVF).

2. The zygote or early embryo is transferred into the uterus or fallopian tube for further development; this process is called Embryo Transfer (ET) and can be done in the following ways:
(a) The zygote or embryo up to eight blastomeres is transferred into the fallopian tube; it is called Zygote Intra Fallopian Transfer (ZIFT).
(b) Embryos with more than eight blastomeres are transferred into the uterus. It is called Intrauterine Transfer (IUT).

3. Gamete Intra Fallopian Transfer (GIFT): This method involves the transfer of an ovum collected from a donor female into another female, who cannot produce ova, but can provide suitable conditions for fertilization and further development of the fetus up to parturition.

4. Intra Cytoplasmic Sperm Injection (ICSI): In this method, the sperm is directly injected into the ovum to form an embryo in the laboratory, and then embryo transfer is carried out.

5. Artificial insemination:
(a) In this method, the semen collected from the husband or a healthy donor is artificially introduced into the vagina or into the uterus (intrauterine insemination).
(b) This method is used in cases where infertility is due to the inability of the male partner to inseminate the female or due to very low sperm counts in the ejaculates.

Question 9.
It is commonly observed that parents feel embarrassed to discuss freely with their adolescent children about sexuality and reproduction. The result of this parental inhibition is that the children go astray sometimes.
(i) Explain the reasons that you feel are behind such embarrassment amongst some parents to freely discuss such issues with their growing children.

Answer:
Parents feel embarrassed because of the following reasons:
(a) Indian society is not that broad-minded. So parents feel shy talking openly about these matters to their children.
(b) Improper communication and age gap are the reasons behind such embarrassment.

(ii) By taking one example of a local plant and animal, how would you help these parents to overcome such inhibitions about reproduction and sexuality?
Answer:
Parents can take the example of China rose to explain the process of sexual reproduction. They can also take an example of the male honeybee and orchid Ophrys flower.

It is evident that sexual attraction is a natural phenomenon. The honeybee is attracted to an Ophrys flower and assumes its one petal as its female partner and pseudo copulate with it. So it is a natural phenomenon and parents should talk regarding this matter to their children.

Question 10.
(a) Explain one application of each one of the following:
(A) Amniocentesis:

Answer:

• Detection of a genetic disorder
• Detection of chromosomal disorder
• Sex determination
• Karyotyping (used for detecting chromosomal aberrations)

(B) Lactational amenorrhea:
Answer:
It is a kind of natural contraception to prevent pregnancy. When the women breastfeed regularly her menstrual cycle stops for some period and thus can’t have a baby.

(C) ZIFT:
Answer:
Application of ZIFT (Zygote Intrafallopian Transfer)- In vitro fertilization, the zygote or early embryos at eight blastomeres stage are transferred to the fallopian tale to complete its further development inside the body of the mother. Hence this method is very helpful for infertile couples.

(b) Prepare a poster for the school program depicting the objectives of the “Reproductive and Child Health Care Programme”.
Answer:
Reproductive and Child Health Care:

Objectives of RCH:

• Creating awareness about various reproduction-related problems.
• Providing facilities and support for building up a reproductively healthy society.
• Providing audio-visual and print, media support, to various government and non-government organizations.
• Educating the people and providing the right information to save them from myths and misconceptions.
• Providing proper education regarding reproductive organs, adolescence and related changes, safe and hygienic sexual practices.
• Providing information regarding the danger of sexually transmitted diseases, AIDS, etc.
• Awareness regarding that gender selection and detection is punishable.

Example – Hum do hamare do, Beti bachao beti padhao, Do boond zindasi ke etc.

In This Post we are  providing Chapter-3 HUMAN REPRODUCTION NCERT MOST IMPORTANT QUESTIONS for Class 12 BIOLOGY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON HUMAN REPRODUCTION

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1. Differentiate between spermatogenesis and oogenesis.
Ans.

Spermatogenesis	Oogenesis
1. It occurs inside the testes.	1. It occurs inside the ovary.
2. All the stages are completed inside the testes.	2. Majority occurs inside the ovary but last stages occur in the oviduct.
3. Spermatogonia develop from the germinal epithelium lining in the seminiferous tubules.	3. Oogonia develop from the germinal epithelium overlying the ovary.
4. All spermatogonia give rise to spermatocytes.	4. Only few oogonia give rise to oocytes.
5. Primary spermatocytes divide by meiosis I to give rise to two secondary spematocytes	5. Primary oocyte undergoes meiosis I to give rise to one secondary oocyte and a polar body.
6. Secondary spermatocyte divides by meiosisII to give rise to two spermatids.	6. Secondary oocyte divides by meiosisII to form the ovum and the second polar body.
7. Each spermatid differentiates into spermatozoan or sperm.	7. No differentiation is required after meiosisII.
8. The sperms formed are motile.	8. The ovum or egg is non- motile.

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2. ‘A fertilized egg is the blue print of future development’. Explain
Ans. The sperm carries the genetic information from the father in form of 23 chromosomes (including the male sex chromosome X or Y) while the egg bears the genetic information from the mother (including the female sex chromosome X). Thus during fertilization the fusion of the male and the female gametes produce new genetic combination which introduces variation in the progeny. The zygote or the fertilized egg contain the genetic information which accordingly controls the development of the embryo.

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3. Briefly describe the stages of spermatogenesis in human?
Ans. Spermatogenesis consists of two phases:-
I. FORMATIDN OF SPERMATIDS :- It further consist of 3 phases

1. Multiplication phase :- undifferentiated germ cells undergo repeated division to produce sperm mother cell or spermatogonia.
2. Growth phase :- Spermatogonia increase in volume & is now called PRIMARY SPERMATOCYTES.
3. Maturation phase: – primary spermatocyte undergoes meiosis I to produce small size haploid secondary spermatocyte secondary spermatocyte divides by meiosis – II & forms haploid Spermatids.

II.FORMATION OF SPERMS :- The tramsformation or differentiation of spermatids into spermatozoa or sperm is called spermiogenesis & occurs under the influence of FSH

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4. Describe the hormonal control of human male reproduction system with the help of a flow chart & highlight the inhibitory & stimulatory directions in it?
Ans. i) Spermatogenesis is initiated due to an increase in the secretion of Gonadotropin releasing hormone from hypothalamus at the age of puberty.

1. The increased levels of GnRH act on anterior pituitary& stimulate the secretion of two gonadotropins i-e. leuteinizing hormone (LH) & follicle stimulating hormone (FSH)
2. LH acts on leydig cells & stimulate them to secrete testosterone
3. FSH acts on sertoli cells & stimulate secretion of some factors help in spermiogenesis

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5. A sperm has just fertilized a human egg in the fallopian tube. Trace the events that the fertilized eggs will undergoes upto implantation of blastocyst in the uterus.
Ans. 1. CLEAVAGE :-Fertilized egg starts dividing lay specific mitotic divisions called cleavage. The zygotes undergoes mitotic division in the isthmus of oviduct to form daughter cell the cells formed as a result of cleavage called blastomere
2. BLASTOCYST :- 3-4 days after fertilization, the morula twins into large mass of cells called blastocyst Outer peripheral cells enlarge & flatten further & form trophoblast. Trophoblast cells secretes a fluid into interior & form a cavity called blastocoel. The embryonic stage with blastocoels is called blastula.

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6. Where oogenesis does takes place. Describe the stages of this process?
Ans. The process of formation & maturation of ovum is called oogenesis. It takes place in ovary & is initiated during embryonic development of female foetus. It consists of 3 phases :–

1. Multiplication phase :- The primordial germ cells divide by meiosis to produce oogonia. These oogonia divide lay repeated mitotic divisions forming clusters. In each cluster only one of them enters into growth phase & is called primary oocyte.
2. Growth phase :- Growth phase occurs only after attainment of puberty. It involves – increase in size of oocyte to many folds & synthesis of you.
3. Maturation phase :- The first division is meiotic as a result two haploid (n) cells are produced. In this division, cytobinesis is unequal, large daughter cell with almost all cytoplasm is called secondary oocyte & smaller me with less cytoplasm is called polar body. The secondary oocyte then undergoes second meiotic division to form an ovum & second polar body.

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7.

Ans. (i) ’D’ Spermatids = undergo spermiogenesis
(ii) ‘A’= Spermatogonium; B = Primary spermatocyte
(iii) ‘B’ = Diploid E = Haploid
(iv) ‘F’ = Sertoli cells – Nutrition to germ cells
(v) Mitosis in Cell ‘A’, Meiosis in cell ‘B’

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8. Explain the development of human embryo with diagrams.
Ans. The Fusion of the sperm and the egg in humans result into formation of the diploid structure called zygote. The zygote starts dividing mitotically as it moves through the oviduct into the uterus to form 2,4,8,16 daughter cells called blastomeres. The stage is called morula. The Morula divides further and differentiates into blastocysts. The outer layer of blastomeres called trophoblast gets attached to the endometrial layer of the uterus.
The uterine wall divides and encloses the blastocysts and this is referred to as implantation.
The inner layer of blastomeres in the blastocysts gives rise to the embryo.

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9. What is menstruation? What are the specific actions of FSH, LH, estrogen & progesterone in menstrual cycle?
Ans. During menstrual phase of menstrual cycle which starts on 28th day the endometrial lining of female genital tract break down due to lack of progesterone As a result bleeding occurs. This monthly flow of blood is caller menstruation.
During menstrual cycles, the various changes occurs in the ovary under the influence of various hormones :-

1. Menstrual phase :- The levels of hormones LH ,FDH estrogen & progesterone is very less which results in breakdown of endometrial lining of uterus.
2. Follicular phase :- In this phase , the levels of pituitary hormones FSH & LH increases which causes ovarian hormone estrogen to release,. FSH controls the follicular phase , it stimulates the growth of follicles. Both FSH & LH reach their peak level in middle of cycle (14th day)
3. OVULATORY PHASE :- The level of LH hormones reaches its peak (called LH swing) induces the ruptures of mature Graffian follicle & there by release of ovum
4. Luteal phase :- The LH & FSH hormones begins to decline. After ovulation, the follicle becomes to ruptures & is transformed into corpus Luteum which secretes large quantities of progesterone

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10. A woman has conceived & implantation has occurred within her uterus. Discuss the sequence of changes up to parturition which will take place within her body under the influence of various hormones.
Ans. The following changes takes place in the body of women after implantation :-

1. The trophoblast differentiates into two layers outer layer secretes enzymes to dissolve the endometrium of uterus.
2. The inner layer grows out as finger – like projections called chorionic villi into uterine stoma. The chorionic villi & the uterine tissue become inter digitated to form structural & functional unit called placenta.
3. Placenta secretes hormones like HCG, HPL , estrogen & progesterone that are necessary to maintain pregnancy
4. Umbilical cord, the structure that connects the placenta with the foetus is formed.
5. Simultaneously, inner cell mass differentiates into outer layer called ectoderm & inner layer called endoderm. & a middle layer called mesoderm appears between ectoderm & endoderm.
6. The primary germ layers give rise to all the tissues & organs of the adults e.g. after one month heart is formed & after second month digits & limbs are formed.
7. By the end of ninth month of pregnancy, foetus is completely developed & is ready for delivery.
8. During parturition, ovary secretes a hormone called relaxin that facilitates parturition which softens the connective tissue. Mild contraction called foetal ejection reflex is induced. This triggers release of oxytocin from posterior pituitary. Oxytocin induces stronger leads to expulsion of baby from uterus, through birth canal.

In This Post we are  providing Chapter-1 REPRODUCTION IN ORGANISM NCERT MOST IMPORTANT QUESTIONS for Class 112BIOLOGY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON REPRODUCTION IN ORGANISM

Question 1.
Define:
(i) juvenile phase,

Answer:
Juvenile phase. The period of growth in the life of organisms before they start reproducing sexually and attain a level of maturity is called juvenile phase. It is followed by the reproductive phase.

(ii) reproductive phase
Answer:
Reproductive phase. The period of active reproductive behaviour, when the organisms show marked morphological and physiological changes is called reproductive phase. It is followed by senescence phase.

(iii) senescence phase.
Answer:
Senescence phase. The period when the reproductive phase ends and concomitant changes occur in the body such as slowing of metabolism is called senescence phase. It is followed by death.

Question 2.
Distinguish between asexual and sexual reproduction. Why is vegetative reproduction also considered as a type of asexual reproduction?

Answer:
1. Differences between asexual reproduction and sexual reproduction.

Asexual Reproduction	Sexual Reproduction
1. The process involves only one cell or one parent.	1. This process involves two cells or gametes belonging to either the same or different parents.
2. The whole body of the parent may act as a reproductive unit or it can be a single cell or a bud.	2. The reproductive unit is called gamete which is unicellular and haploid.
3. The offspring are genetically similar to the parent.	3. The offspring differ from the parents.
4. Only mitotic division takes place.	4. Meiosis and mitosis both take place.
5. No formation of sex organs.	5. Formation of sex organs is essential.
6. No evolutionary significance.	6. It introduces variation; hence it is of evolutionary significance.

2. Vegetative reproduction is also considered a type of asexual reproduction because it does not involve meiotic division and there is no formation and fusion of gametes.

Question 3.
How does an encysted Amoeba reproduce on the return of favourable conditions?

Answer:
Multiple fission in encysted Amoeba:

• Amoeba withdraws pseudopodia and secretes a cyst wall around itself. This phenomenon is called encystation.
• Amoeba divides by multiple fission.
• It produces a large number of pseudo- conidiospores.
• The cyst wall breakdown.
• The spores are liberated and settle down on suitable substrates and grow as amoebae. This process is also called sporulation.

Question 4.
Discuss the advantages and disadvantages of asexual reproduction.

Answer:
Advantages of asexual reproduction:

1. It involves simple mitotic division in single-parent and it may produce a large number of young ones.
2. Young ones produced by asexual methods are genetically similar to the parent.
3. It helps in the dispersal of offspring to far off places.

Disadvantages of asexual reproduction.

1. The young ones thus produced do not possess much capacity to adapt rapidly to the environmental changes taking place in quick succession.
2. No genetic recombination occurs; thus no variation occurs.

Question 5.
Discuss the advantages and disadvantages of sexual reproduction.

Answer:
Advantages of sexual reproduction:

1. Genetic recombination, interaction, etc. take place which causes variations in the offspring, thus also form raw materials for evolution.
2. The offspring adapt more comfortably and quickly to the change in environmental conditions and have better chances of survival.

Disadvantages of sexual reproduction. Usually, two parents of opposite sexes are required (except in hermaphrodite).

Question 6.
List various methods of natural vegetative propagation. Give examples:

Answer:

1. Vegetative propagation by stems, e.g.Grasses, Turmeric, Onion, Colocasia, Potato, Gladiolus and Crocus.
2. Vegetative propagation by roots, e.g. Murraya sp., Albizzia Lebbac, Dalbergia sissoo, Tuberous roots of sweet potato, Asparagus, Tapioca, Dahlia and Yams (Dioscorea).
3. Vegetative propagation from reproductive organs. Flower buds of century plant (Agave sp.) develop into bulbils.

Question 7.
Define external fertilisation. Mention its disadvantages:

Answer:
The fertilisation in which the fusion of gametes occurs outside the body of the female in an external medium, i.e. water, is called external fertilisation.

Examples. Bony fishes, amphibians, etc. Organisms that exhibit external fertilisation show great synchrony between the sexes in order to liberate the gametes at the same time.

Disadvantages of external fertilisation:

1. A large number of gametes are produced to ensure fertilisation, thus there is wastage.
2. The offspring formed are extremely vulnerable to predators, thus threatening their survival up to adulthood.

Question 8.
Explain the process of budding in yeast.

Answer:
Budding in yeast. It is a common type of vegetative reproduction. In a medium which is abundantly supplied with sugar, yeast cytoplasm forms a bud-like outgrowth. The growth soon enlarges and a part of the nucleus protrudes into the bud and breaks off. The bud then begins to grow and then separates from the mother cell. Often it will itself form a bud before it breaks away, and straight or branched chains are produced.

Thus, as a result, branched or unbranched chains of cells called pseudo my cilium are produced. The cells are loosely held together. Sooner or later they become independent.

Question 9.
Describe the importance of vegetative propagation.

Answer:
Merits of vegetative propagation:

1. Plants produced by vegetative propagation are genetically similar and constitute a uniform population called a clone.
2. Plants with reduced power of sexual reproduction, long dormant period of seed, poor viability, etc. are multiplied by vegetative methods.
3. Some fruit trees like banana and pineapple do not produce viable seeds. So these are propagated by only vegetative methods.
4. It is a more rapid and easier method of propagation.
5. Good characters are preserved by vegetative propagation.
6. Some plants such as doob grass (Cynodon dactylon) which produce only a small quantity of seed are mostly propagated by vegetative propagation.
7. Grafting helps in getting an economically important plant having useful characteristics of two different individuals in a short time.

Question 10.
Write a note on sexuality in plants.
Or
Coconut palm is monoecious while date palm is dioecious. Why are they called so?

Answer:
Sexuality in organisms: Sexual reproduction in organisms generally involves the coming together of gametes from two different individuals. But this is not always true.

Sexuality in Plants: Plants may have both male and female reproductive structures in the same plant (bisexual) or on different plants (unisexual). In several fungi and plants, terms such as homothallic and monoecious are used to denote the bisexual condition, and heterothallic and dioecious are used to describe the unisexual condition.

In flowering plants, the unisexual male flower is staminate, i.e. bearing stamens, while the female is pistillate or bearing pistils. In some flowering plants, both male and female flowers may be present on the same individual (monoecious) or on separate individuals (dioecious). Some examples of monoecious plants are cucurbits and coconuts and dioecious plants are papaya and date palm.

Very Important Figures: