Yearly Archives: 2022

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:

In This Post we are  providing Chapter-2 SEXUAL REPRODUCTION IN FLOWERING PLANTS 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 SEXUAL REPRODUCTION IN FLOWERING PLANTS

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1. Trace the development of microsporocyte into mature pollen grains.
Ans. i) When the anther is young, the microsporangium contains compactly arranged homogenous cells forming the Sporogeneous tissues.

1. Every cell of the sporogenous tissue is a potential Pollen mother cell (PMC) & give rise to microspore tetrad or Pollen grains.
2. But Some of them forego this Potential & become differentiated into pollen or microspore mother cell (MMC)
3. Each microspore mother cell undergoes meiosis to form a cluster of four haploid cells called microspore tetrad.
4. As the anther matures, microspores dissociate from tetrad & develop into pollen grains.
5. The nucleus of microspore undergoes mitosis to form large vegetative cell & small generative cell. They develop a two layered wall – outer exine made up of sporopollenin & inner intine made up of cellulose & pectin. Usually Pollen grains are liberated at two celled stage.

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2. i) Explain the structure of a maize grain with the help of a diagram
ii) Why cannot we use the term maize seeds for maize grains?
Ans. (i) In grass family ( eg. Maize ) fruit is single seeded where pericarp & seed coat are fused together to form the husk. Just below husk, there is a layer of cells called aleurone layer, with stores proteins. There is a large endosperm that stores starch. The embryo lies on one side of endosperm & consists of a single cotyledon called scutellum & embryonal axis. The region of embryonal axis that points down ward from point of attachment of cotyledons is radicle & is covered by protective sheath called coleorhiza. The region of embryonal axis that points upward from point of attachment of cotyledon is plumule, it is covered by foliaceous sheath called coleoptite

(ii) We cannot use the term seeds for maize grain because seed is not completely developed from embryo but retains a part of endosperm.

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3. Trace the development of megasporocyte into mature ovule.
Ans. i. A single Megaspore mother cell is differentiated in the micropylar region of nucleus of an ovule & undergoes meiosis & forms a cluster of haploid cells called megaspore tetrad. Of these, soon three degenerates & only one megaspore becomes functional
ii. Functional megaspore enlarges to form embryo sac. Its nucleus undergoes mitotic division & two nuclei move to opposite poles forming 2-nucleate embryo Sac.
iii. Two successive mitotic divisions in each of these two nuclei results in formation of 8-nucleate embryo sac.
iv. Three cells are grouped together at micropylar end to form egg apparatus. consisting of two synergids & a female egg cell .
v. Three cells are grouped together at the chalazal end, they are called antipodal cells.
vi. The remaining two nuclei are called Polar nuclei, they move to centre of embryo sac & fuse to form Secondary nucleus.
Thus a typical angiospermic embryo sac is 8-nucleate 7-celled

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4. “Incompatibility is the natural barrier in fusion of gamete”. Justify this statement.
Ans. Pollen grains of a plant species cannot germinate on stigma of other unrelated species because both the species are incompatible & process is called pollen – pistil incompatibility. In many angiospermic plants, it is seen that pollen grains germinate on stigma of unrelated species but male gametes produced in pollen tube cannot fertilize egg. This is called gametic incompatibility Self incompatibility can be achieved by any of the following ways :-

1. Pollen Stigma interaction: – In this phenomenon, pollen grains fails to germinate on Stigma because of incompatibility.
2. Pollen tube style interaction: – In this phenomena, pollen grains become able to germinate on stigma & pollen tube penetrate stigmatic surface but due to incompatibility growth of pollen tube within stigma & style is inhibited.
3. Pollen – ovule interaction: – pollen tube successfully pierces & grows within style & its growth is inhibited at micropyle of ovule.

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5. How dose pollination takes place in salivia. List any four adaptations required for such type of pollination.
Ans. In salivia, entomophily or pollination lay insects occurs. The flowers of salivia are bilipped. Its upper lip consists of two petals & lower lip consists of three petals. The lower lip functions as sitting pad for insects. In normal conditions, the connective remains upright. When insect enters the tube of corolla towards nectar sitting on lower lip, it pushes sterile anther lobe which automatically brings about fertile anther to touch the back of insects gets the blow of fertile lobe. Pollen grains are dusted on back feather & legs of insects.
ADAPTAIONS EOR ENTOMOPHILY :-

1. Flowers are brightly coloured.
2. Flowers possess nectar glands.
3. pollen grains are usually sticky & spiny
4. flowers are large – sized & stout

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6. Explain the formation of an embryo sac with diagrams.
Ans.

• The functional megaspore grows in size.
• The nucleus divides mitotically to form two nuclei which move to opposite poles.
• Each nucleus at the poles undergoes two mitotic divisions to form four nuclei in each pole or a total of 8 nuclei.
• two nuclei from each pole move to the centre to form the polar nuclei.
• the other nuclei, three at each pole get surrounded by bit of cytoplasm to form cells.
• the female gametophyte or the embryo sac thus has 7 cells and eight nuclei.

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7. Explain the development of embryo in a dicotyledonous plant with neatly labeled diagrams.
Ans. The embryo develops at the micropylar end where the zygote is located. The zygote starts developing only after certain amount of endosperm is formed to assure nutrition to the embryo. The zygote divides mitotically to form various stages including pro- embryo, globular, heart shaped and finally the mature embryo

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8.Describe the post-fertilization changes taking place in a flowering plant?

Ans. The major events taking place in a flowering plant after fertilization:-
(i) DEVELOPMENT OF ENDOSPERM:- Endosperm development proceeds embryo development . The most common method of endosperm development is nuclear type where triploid endosperm (PEN) undergoes repeated mitotic divisions without cytokinesis – Subsequently cell wall formation occurs from periphery & endosperm store food materials which is later used up by embryo.
(ii) DEVELOPMENT OF EMBRYO :- The zygote divides lay mitosis to for a pro-embryo first . Later development results in formation of globular & heart shaped embryo & that ultimately become horseshoe – shaped embryo with one or more cotyledons. In dicot embryo, the portion of embryonal axis about the level of attachment is epicotyl & it terminates into plumule while portion of embryonal axis below the level of attachment is hypocotyl & terminates into radicle.

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9.Trace the events that would take place in flower from the time of Pollen grain of species fall on stigma up To completion of fertilization.

Ans. GERMINATION OF POLLEN GRAINS ON STIGMA
The pollen grains absorb fluid present on stigma & swell up. The exine ruptures at the place of germ pore & intine comes out in the form of tube with its internal contents. This small tubular structure is called pollen tube & process is called pollen germination.

ii) Entry of pollen tube into Ovule: – The entry of pollen tube into ovule occurs through micropyle or chalaza or through lateral sides of ovule. Only one pollen tube enters inside the embryo sac of an ovule. Normal two synergids are destroyed while entry of pollen tube into embryo sac.
iii) Discharge of Mate Gametes :- After enter of pollen tube both the male gametes discharged into embryo sac by either forming two pores into pollen tube & each male gamete is discharged through every pore or sometime pollen tube may burst & release the male gametes into embryo sac.

iv) Fertilization:- The fusion of first male gamete (n) with egg (n) is called fertilization. It results in formation of a diploid zygote (2n). The second male gamete fuses with secondary nucleus (2n) to form triploid endosperm nucleus (3n). This fusion between second male gamete & secondary nuclei is triple fusion. Since process of fertilization occurs twice. It is called double fertilization.

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10. i)Why is zygotes dominant for sometime in fertilized ovule.
ii) What is polyembryony? Give an example.
iii) In fruits, what is formed from following parts :-
a) Ovary wall
b) Outer integument
c) Inner integument
d) zygote
e) primary endosperm
f) Ovary
g) Nucellus
Ans. (i) Zygote remain dominant for sometime in a fertilized ovule because embryo develops after formation of endosperm therefore zygote wants for formation of endosperm which supplies food material for developing embryo
(ii) The presence of more than one embryo in a seed is called polyembryony eg. Sometimes more than one embryo is formed within an embryo sac either by cleavage or splitting of egg, synergid, antipodal or endosperm.
(iii) In fruits, the following is formed from given parts:-

a)	Ovary wall	Per carp
b)	Outer integument	Testa
c)	Inner integument	Tegmen
d)	zygote	embryo
e)	primary endosperm	endosperm
f)	Ovary	fruit
g)	Nucellus	perisperm.

Short Answer Type Questions:

Q1. What do you mean by rocks? Name the three major classes of rocks.

Answer

A rock is an aggregate of one or more minerals. The earth’s crust is composed of rocks.The three major classes of rocks are:• Igneous rocks• Sedimentary rocks• Metamorphic rocks

Q2. What is an igneous rock? Describe the method of formation and characteristics of igneous rock.

Answer

Igneous rocks are those rocks which are formed through the cooling and solidification of magma or lava. These rocks are formed when magma in its upward movement cools and turns into solid form. The process of cooling and solidification can happen in the earth’s crust or on the surface of the earth.
Characteristics of Igneous rocks:
• These rocks are consist of crystals.
• These rocks are extremely hard in nature.
• Fossils are not present in these rocks.

Q3. What is meant by sedimentary rock? Describe the mode of formation of sedimentary rock.

Answer

Sedimentary rocks are those rocks which are formed by the deposition at favorable sites by agents of denudation such as wind, river and sea waves. These deposits gradually turn into rocks.
The rocks of the earth’s surface are exposed to denudational agents, and are broken up into various sizes of fragments. Such fragments are transported by different exogenous agencies and deposited. These deposits through compaction turn into rocks. This process is called lithification.

Q4. What relationship explained by rock cycle between the major type of rock?

Answer

Rock cycle is a continuous process through which old rocks are transformed into new ones.
• Igneous rocks are primary rocks and other rocks (sedimentary and metamorphic) form from these primary rocks.
• Igneous rocks can be changed into sedimentary rocks or metamorphic rocks.
• The fragments derived out of igneous and metamorphic rocks form into sedimentary rocks.
• Sedimentary rock can change into metamorphic rock or into igneous rock.
• Metamorphic rock can change into igneous or sedimentary rock.

Q5.Define the term ‘mineral’ and name the major classes of minerals with their physical characteristics.

Answer

A mineral is a naturally occurring inorganic substance, having an orderly atomic structure and a definite chemical composition and physical properties. A mineral is composed of two or more
elements. But, sometimes single element minerals like sulphur, copper, silver, gold, graphite etc. are found.
Major minerals with their physical properties:• Feldspar: It has light cream to salmon pink colour. It is used in ceramics and glass making. Half of the earth’s crust is composed of feldspar. • Quartz: It consists of silica. It is a hard mineral virtually insoluble in water. It is white or colourless and used in radio and radar.
• Pyroxene : It consists of calcium, aluminium, magnesium, iron and silica. It forms 10% of the earth’s crust. It is commonly found in meteorites.
• Amphibole : Its major elements are silica, aluminium, calcium and iron. They form the 7% of the earth’s crust. It is used in the asbestos industry. It is black or green in colour.
• Mica : It comprises potassium, aluminium, magnesium, iron, silica etc. It forms 4% of the earth’s crust. It is uses in the electrical instruments.
• Olivine : Magnesium, iron and silica are the major elements of olivine. It is used in jewellery. It is usually a greenish crystal often found in basaltic rocks.

Q6. Describe the nature and mode of origin of the chief types of rock at the earth’s crust. How will you distinguish them?

Answer

There are three chief types of rock at the earth’s crust:

• Igneous Rocks:  These are formed when magma cools and solidifies. The process of cooling and solidification can happen in the earth’s crust or on the surface of the earth.
→ These are extremely hard in nature.
→ Texture depends upon size and arrangement of grains or other physical conditions of the materials. If molten material is cooled slowly at great depths, mineral grains may be very large. Sudden cooling results in small and smooth grains.

• Sedimentary Rocks: Rocks (igneous, sedimentary and metamorphic) of the earth’s surface are exposed to denudational agents, and are broken up into various sizes of fragments. Such fragments are transported by different exogenous agencies and deposited. These deposits through compaction turn into rocks. The deposited material is known as sediment and the rocks formed are called sedimentary rocks.→ The sedimentary rocks are soft in nature.→ These rocks have many layers of varying thickness.
• Metamorphic Rocks: These rocks form under the action of pressure, volume and temperature (PVT) changes. Metamorphism is a process by which already consolidated rocks undergo recrystallisation and reorganisation of materials within original
rocks.
→ These rocks are crystalline in nature.
→ The materials of rocks chemically alter and recrystallise due to thermal metamorphism.

Q7.What are metamorphic rocks? Describe the types of metamorphic rock and how are they formed?
Answer

Metamophic rocks are formed by the physical or chemical alteration by heat and pressure of an existing igneous or sedimentary rocks. Metamorphism occurs when rocks are forced down to lower levels by tectonic processes or when molten magma rising through the crust comes in contact with the crustal rocks or the underlying rocks are subjected to great amounts of pressure by overlying rocks.
Metamorphism is a process by which already consolidated rocks undergo recrystallisation and reorganisation of materials within original rocks.

Types of Metamorphic rocks:

• Foliated rocks: These rocks are formed in the inerior of the earth under extremely high pressures that are unequal occurring when the pressure is greater in one direction than in the others. These rocks develop a platy or sheet-like structure. Slate, schist are examples.

• Non-foliated rocks: These rocks are formed around igneous intrusions where the temperatures are high but the pressures are relatively low and equal in all directions. These are not flat or elongate. Marble, quartzite are examples.

Long Answer Type Questions:

Q1.Explain any six physical characteristics?
Answer:
Brief information about some important minerals in terms of their nature and physical characteristics is given below :

• External crystal form: Determined by internal arrangement of the molecules- cubes octahedrons, hexagonal prism, etc.
• Fracture: Internal molecular arrangement so complex there are two planes of molecules; the crystal will break in an irregular manner, not along planes of cleavage.
• Lustre: Appearance of a material without regard to color; each mineral has a distinctive lustre like metallic, silky, glossy, etc.
• Streak: Colour of the ground powder of any mineral. It may be of the same colour as the mineral or may differ malachite is green and gives green streak.
• Structure: Particular arrangement of the individual crystal; fine, medium or coarse grained; fibrous separable, divergent and radiating.
• Specific gravity: The ratio between the weight of a given object and the weight of an equal volume of water; object weighted in air and then weighed in water and divide weight in air by the difference of the two weights.

Q2.Explain some important minerals and their characteristics.
Answer:
Some major minerals and their characteristics

• Feldspar: Silicon and oxygen are common elements in all types of feldspar and sodium, potassium, calcium, aluminium etc. are found in specific feldspar variety.
• Question uartz: It is one of the most important components of sand and granite. It consists of silica. It is a hard mineral virtually insoluble in water.
• Pyroxene: Pyroxene consists of cak ium, aluminum, magnesium, iron and silica. Pyroxene forms 10 percent of the earth’s crust.
• Amphibole: Aluminium, calcium, silica, iron, magnesium are the major elements of amphiboles. They form 7 per cent of the earth’s crust.
• Mica: It comprises of potassium, aluminium, magnesium, iron, silica etc. It forms 4 per cent of the earth’s crust.
• Olivine: Magnesium, iron and silica are major elements of olivine. It is used in jewellery. It is usually a greenish crystal, often found in basaltic rocks.

Q3.Rocks do not remain in their original form for long but may undergo transformation. Explain.
Answer:
Rocks do not remain in their original form for long but may undergo transformation. Rock cycle is a continuous process through which old rocks are transformed into new ones. Igneous rocks are primary rocks and other rocks form from these primary rocks. Igneous rocks can be changed into metamorphic rocks. The fragments derived out of igneous and metamorphic rocks form into sedimentary rocks. Sedimentary rocks themselves can turn into fragments and the fragments can be a source for formation of sedimentary rocks. The crustal rocks (igneous, metamorphic and sedimentary) once formed may be carried down into the mantle through subduction process and the same melt down due to increase in temperature in the interior and turn into molten magma, the original source for igneous rocks

Short Answer Type Question:

Q1. What were the forces suggested by Wegener for the movement of the continents?

Answer

Wegener suggested that the movement responsible for the drifting of the continents was caused by pole-fleeing force and tidal force. The polar-fleeing force relates to the rotation of the earth. The earth is not a perfect sphere; it has a bulge at the equator. This bulge is due to the rotation of the earth. The second force, the tidal force is due to the attraction of the moon and the sun that develops tides in oceanic waters. Wegener believed that these forces would become effective when applied over many million years.

Q2.How are the convectional currents in the mantle initiated and maintained?
Answer

The convectional currents in the mantle are generated due to radioactive elements causing
thermal differences in the mantle portion. The hot material rises up from greater depths and comparatively cold material goes down from above. The occurrence of this process repetitively give birth to convectional currents. 

Q3. What is the major difference between the transform boundary and the convergent or divergent boundaries of plates?

Answer

Transform Boundary	Convergent Boundary	Divergent Boundary
Where the crust is neither produced nor destroyed as the plates slide horizontally past each other.	Where the crust is destroyed as one plate dived under another.	Where new crust is generated as the plates pull away from each other.

Q4.What was the location of the Indian landmass during the formation of the Deccan Traps?

Answer

The Deccan Traps were formed during the movement of the Indian plate towards the Asiatic plate. This started somewhere around 60 million years ago and continued for a long period of time. At that time, Indian landmass was located in the south of the equator.

Q5.What are the evidences in support of the continental drift theory?

Answer

The evidences in support of the continental drift theory are:

• The Matching of Continents (Jig-Saw-Fit): The shorelines of Africa and South America facing each other have a remarkable and unmistakable match.
• Rocks of Same Age Across the Oceans: The belt of ancient rocks of 2,000 million years from Brazil coast matches with those from western Africa. The earliest marine deposits along the coastline of South America and Africa are of the Jurassic age. This suggests that the ocean did not exist prior to that time.
• Tillite: It is the sedimentary rock formed out of deposits of glaciers. The Gondawana system
of sediments from India is known to have its counter parts in six different landmasses of the Southern Hemisphere. The glacial tillite provides unambiguous evidence of palaeoclimates and also of drifting of continents.
• Placer Deposits: The occurrence of rich placer deposits of gold in the Ghana coast and the absolute absence of source rock in the region is an amazing fact. The gold bearing veins are in Brazil and it is
obvious that the gold deposits of the Ghana are derived from the Brazil plateau when the two continents lay side by side.
• Distribution of Fossils: When identical species of plants and animals adapted to living on land or in fresh water are found on either side of the marine barriers, a problem arises regarding accounting for such distribution. The observations that Lemurs occur in India, Madagascar and Africa led some
to consider a contiguous landmass “Lemuria” linking these three landmasses.

Q6. Bring about the basic difference between the drift theory and Plate tectonics.

Answer

Drift Theory	Plate tectonics
Drift theory suggested that in the past, there was a super continent called Pangaea. Over time, this super continent split apart to form the seven continents we have today.	Plate Tectonics is a theory of global tectonics which proposes that the earth’s lithosphere is divided into seven major and some minor plates.
It only covers the movement of continents.	It covers movements of both Continents and Oceans.
The evidences in support of drift theory are Jig-Saw-Fit, Placer Deposits, Distribution of Fossils and others.	It is based on scientific analysis of the processes inside the earth’s surface.
It does not give any future prediction.	This theory says that all plates shall continue to move in the future period as well.

Q7. What were the major post-drift discoveries that rejuvenated the interest of scientists in the study of distribution of oceans and continents?

Answer

A number of post-drift discoveries during the post-war period added new information to geological literature. Particularly, the information collected from the ocean floor mapping provided new dimensions for the study of distribution of oceans and continents.

• All along the midoceanic ridges, volcanic eruptions are common and they bring huge amounts of
lava to the surface in this area.

• The rocks equidistant on either sides of the crest of mid-oceanic ridges show remarkable similarities in terms of period of formation, chemical compositions and magnetic properties. Rocks closer to the
mid-oceanic ridges are normal polarity and are the youngest. The age of the rocks increases as one moves away from the crest.

• The ocean crust rocks are much younger than the continental rocks. The age of rocks in the oceanic crust is nowhere more than 200 million years old.

• The sediments on the ocean floor are unexpectedly very thin.

• The deep trenches have deep-seated earthquake occurrences while in the midoceanic ridge areas, the quake foci have shallow depths.

Long Answer Type Questions:

Q1.According to tectonic plates theory in how many plates has the earth been divided? Explain.
Answer:
The theory of plate tectonics proposes that the earth’s lithosphere is divided into seven major and some minor plates. The major plates are as follows:

• Antarctica and the surrounding oceanic plate
• North American plate
• South American plate
• Pacific plate
• India-Australia-New Zealand plate
• Africa with the eastern Atlantic floor plate
• Eurasia and the adjacent oceanic plate. Some important minor plates are:
  • Cocos plate: It is between Central America and Pacific plate
  • Nazca plate: It is between South America and Pacific plate
  • Arabian plate: It includes mostly the Saudi Arabian landmass
  • Philippine plate: It is between the Asiatic and Pacific plate
  • Caroline plate: It is between the Philippine and Indian plate (North of New Guinea)
  • Fuji plate: It includes North-east of Australia.
• Pacific plate is largely an oceanic plate whereas the Eurasian plate may be called a continental plate. Plates are not static. Plates may converge or diverge. Plates may break as well.

Q2.Explain tectonic plate theory and its j working.
Answer:
These plates have been constantly moving over the globe throughout the history of the earth.

• The theory of plate tectonics was introduced by Mckenzie, Parker and , Morgan in 1967.
• A tectonic plate is also called as lithosphere plate.
• It is a massive irregularly shaped slab of solid rock.
• Consists of oceanic and continental sphere.
• Plates move horizontally over the asthenosphere.
• Average thickness is 100 km of oceanic part and 200 km of continental part.
• It may be oceanic or continental.
• Pacific plate is largest oceanic plate whereas Eurasian plate is the largest continental plate.
• These plates are moving constantly throughout geological time not the continent, believed by Wegener.
  It creates three types of boundaries.

1. Divergent boundaries
   • New crust is generated
   • Plates move away from each other
   • These are called spreading sites
   • Ex. Mid atlantic ridge
2. Convergent boundaries |
   • Crust is destroyed
   • Sinking of plate is called “subduction zone”. There are three ways in which subduction occurs (i) between an oceanic and continental plates; (ii) between two oceanic plates; and (iii) between two continental plates.
   • Transform boundaries: Where the crust is neither produced nor destroyed as the plates slide horizontally past each other.

Q3.Explain important theories associated with the movement of continents.
Answer:
Continental drift: Abraham Ortelius a Dutch map maker in 1596 first proposed the possibility of joining the continents such as America with Europe and Africa. Antonio Pellegrini drew a map showing the three continents together. Alfred Wegener, a German meteorologist put forth the continental drift theory. According to him, all continents formed a single continental mass called Pangaea.

All oceans formed a single universal ocean called Panthalassa around 200 million years ago. The Pangaea began to split into two large continental masses called Laurasia and Gondwanaland. By further splitting Laurasia formed northern continents and Gondwanaland formed southern continents.

Sea Floor Spreading: The deep trenches have deep-seated earthquake occurrences while in the mid- oceanic ridge areas, the quake foci have shallow depths. These facts and a detailed analysis of magnetic properties of the rocks on either sides of the mid-oceanic ridge led Hess in 1961 to propose his hypothesis. It was called the “sea floor spreading”. Hess argued that constant eruptions at the crest of oceanic ridges cause the rupture of the oceanic crust forces and the new lava wedges into it, pushing the oceanic crust on either side. The ocean floor, thus spreads. Two facts made Hess think about the consumption of the oceanic crust.

• The younger age of the oceanic crust.
• The spreading of one ocean does not cause the shrinking of the other.

He further maintained that the ocean floor that gets pushed due to volcanic eruptions at the crest, sinks down at the oceanic trenches and gets consumed. Plate Tectonics: It was in 1967, McKenzie and Parker and also Morgan, independently collected the available ideas and came out with another concept termed Plate Tectonics. The theory of plate tectonics proposes that the earth’s lithosphere is divided into seven major and some minor plates. These plates have been constantly moving over the globe throughout the history of the earth. It is not the continent that moves as believed by Wegener. Continents are part of a plate and what moves is the plate. All the plates, without exception, have moved in the geological past, and shall continue to move in the future as well.

Q4.The ocean floor may be segmented into how many divisions based on the depth as well as the forms of relief?
Answer:
The ocean floor may be segmented into three major divisions based on the depth as well as the forms of relief. These divisions are:

(i) Continental margins,
(ii) Abyssal plains and
(iii) Mid-oceanic ridges.

1. Continental margins
   • Form transitional zone between continental shore and deep sea basins
   • They include continental slope , shelf, continental rise and deep oceanic trenches
2. Abyssal plains
   • Extensive plains
   • Found between continental margin and mid oceanic ridge
   • Continental sediments get deposited
3. Mid-oceanic ridges
   • Distribution of volcanoes and earthquakes
   • All volcanoes and earthquakes are parallel to the coast
   • This line also co-incides with mid- Atlantic ridge and Alpine Himalayan system
   • Around the Pacific Ocean it is called ring of fire mid oceanic ridges.