Yearly Archives: 2022

In This Post we are  providing Chapter- 15 BIODIVERSITY AND CONSERVATION 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 BIODIVERSITY AND CONSERVATION

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1. Hot spots are the regions of exceptionally high biodiversity . But they have become regions of accidental habitat loss too. Name the three hot spots of our country. Why are they called ‘Hot spot’?
Ans. Western Ghats and Sri lanka; Indo-Burma; Himalaya called ¡¥biodiversity hot spots¡¦ as they show
(i) High level of species richness
(ii) High degree of

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2. Study the diagram of the earth given below . Give the name of the pattern of biodiversity therein. Suggest any two reasons for this type of occurance.

Ans. Latitudinal gradients
(i) More solar energy available in tropics, more productivity.
(ii) Tropical environments are less seasonal, so more predictable.

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3. What is so special about tropics that might account for their greater biological diversity?
Ans.a) Speciation is a function of time, unlike temperate regions subjected to frequent glaciations in the past, tropical latitude have remained relatively undisturbed for million of years and thus had long evolutionary time for species diversification
b) Tropical environment are less seasonal, more const ant and predictable
c) More solar energy awailable in the tropics contributing to high productivity leading to greater diversity .

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4.What do you mean by biodiversity? What are the different types of Biodiversity?
Ans. Biodiversity can be defined as the totality of genes species & ecosystem of a given region.
Three important components of Biodiversity are:-
i) Genetic Biodiversity:- It refers to the diversity of genes within a species, Greater the genetic diversity amongorganisms of a species. More sustenance it has against environmental perteburations whereas geneticallyuniform populations are highly prone to diseases or harsh environment
ii) Species Biodiversity:- It refers to variety of species within a region. It has two important measures :-
a) Species richness:- i.e. number of species per unit area.
b) Species evenness:- i.e. abundance with which each species is represented in an area.
iii) Ecosystem Biodiversity:- It refers to variation of habitats, community types & abiotic environment present in an area. It is further of three types:-
a) αα – diversity- It refers to number of species in a given community.
b) ββ– diversity – biodiversity which appears in range of communities due to replacement ofspecies with change in community is called ββ– diversity.
c) γγ– diversity – It refers to diversity of habitats over the total geographical area.

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5.What do you mean by latitudinal gradient? What could be the possible reasons for diversity between tropic & temperate region?
Ans. Latitudinal gradient in diversity means that species diversity usually decreases as we move away fromequator towards the poles, Tropic area of latitudinal range 23.50c harbor more species than temperate orpolar area. Three hypothesis have been proposed to explain this difference:-
i) Speciation is a function of time, which temperate regions were subjected to frequent glaciations in the past,the tropics have remain unchanged & hence evolved more species diversity.
ii) As compared to temperate region, tropical environment are less seasonal, relatively more constant &predictable; such constant environment have promotes niche specialization & greater species diversity.
iii) There is more solar radiation avail able in tropical region this contributes directly to greater productivity &indirectly to greater species diversity.

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6.Why is it necessary to conserve biodiversity?
Ans.The reasons for conserving biodiversity can be grouped into three categories.
i) Narrow utilitarian reasons:- Human beings derive a number of economic benefits like food, fibre,firewood, industrial product & medicinal products.
ii) Broad utilitarian reasons:- Biodiversity plays a major role in providing ecosystem services like :-
a) production of oxygen
b) Pollination of flowers, without which seeds or fruits are not produced.
c) Aesthetic pleasures like bird watching, watching spring flowers, walking through thick forest, workingup to bulbul’s song etc.
iii) Ethical reasons :- Every species has an intrinsic value even if it is not of any economic value to us-wehave a moral duty to care for their well-being & pass on the biological legacy in a proper from to ourfuture generation.

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7.What is the relation between species richness & area? What is the significance of slope of regression?
Ans.Alexander Von Humboldt has observed that within a region, species richness increased withincrease explored area but only upto a limit thus the relationship between species richness & area for anumber of taxa is found to be a rectangular hyperbola. On a log scale, the relation ship becomes linear & is described layequation
Log S = log C + Z log A
The values of slope of regression are identical regardless of the taxonomic group or the region. When such analysis is made among very large areas, the slope ofregression would be much steeper.

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8.What are the different approaches for biodiversity conservation in India?
Ans.There are two major approaches for conservation of biodiversity:-
i) In-situconservation :- It is the process of protecting the endangered species of plant or animal in thenatural habitat lay either protecting or cleaning up the habitat or by defending species from predators Itincludes:-
a) Biosphere Reserves:- There are 425 biosphere reserve in the world of which 14 are in India. Hotspotshave been identified for maximum protection to endemic or endangered species.
b) National park or wildlife Sanctuaries:- India has about 90 national parks & 448 wildlife sanctuaries.
c) Sacred forests:- These are undisturbed forests without any human intervention & are surrounded by highly degraded landscapes.
ii) Ex- situ Conservation:- It is the process of protecting the endangered species of plants or animals by removing it from threatened habitat & placing them under care of humans. It includes :-
a) Botanical garden, zoological park and arboreta are conventional methods of ex-situ conservation
b) Cryopreservation to the storage of materials at ultra low temperature either by rapid cooling or by gradual cooling & simultaneous dehydration at low temperature.

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9.Give an account of Biodiversity in India?
Ans.India is one of the 10th mega biodiversity countries of the world because of the presence of variety ofclimatic conditions prevailing on different ecological habitat ranging from tropical, subtropical,temperate, So far as biodiversity of India is concerned, it comprises about 47,000 plants & 81,000 animalspecies. India occupies 2-4% of total land area of would but in terms of biodiversity, India contributeabout 10-35% of global diversity.
A large number of species is native of India. About 5000 species of flowering plants belonging to 141genera& 47 families had a birth in India. There are 62% of amphibian species& 50% of lizards endemicto our country with large number in Western Ghats. India is an origin place of 166 species of crop plants& 320 species of wild relatives of cultivated crop. India is rich in marine biodiversity lying along coastlineof 7500 km. There are two hotspots located in India out of 25 in would – These are Western Ghats&Eastern Himalayas.

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10.What is the significance of Biodiversity to Human beings?
Ans.Biodiversity provide numerous direct or indirect services to human beings. These are-
i) Source of food & improved varieties:- Biodiversity directly or indirectly adds as the source of food,cloth& shelter.
ii) Fats & Oils:- A variety of plants are used to extract different kinds of oils.
iii) Fibres:- A variety of plants eg. cotton, hemp, jute are chief sources of fibres.
iv) Resins:- Resins are sticky exudation of plants.
v) Gums, Timber, Paper, Tannins, Dyes:- Plants species provide variety of useful products eg. gums,raisins, dyes, similarly animal species provide leather, fur, honey, silk, pearl etc.
vi) Drugs & Medicines:- Hiving organism also contain number of therapeutically useful substances.
vii) Stability of Ecosystem:- The food web, food chain energy flow in various tropic level & biochemicalcycles occurs in natural ways without any hindrance if there is proper availability of diversified species
viii) Aesthetic, Scientific & Recreational values :- Indian people grow many plants because they regardthem as sacred.

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In This Post we are  providing Chapter- 10 MICROBES IN HUMAN WELFARE 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 MICROBES IN HUMAN WELFARE

Question 1.

Name the toxin produced by B. Thuringiensis.
Answer:
∝-exotoxin, β-exotoxin, γ-exotoxin, and louse factor

(ii) Nitrogen fixers are available on a commercial basis in the market? Also, name the beneficial crop and microbes used in the following table.

Product	Microbe used	Beneficial crop
1. NitraginTM	(A)	Soybean
2. Rhizocote	Rhizobium	(B)
3. Nodosit	Rhizobium	(C)

Answer:
A. Rhizobium B. Legume C. Legume

(iii) Expand BOD and COD
Answer:
BOD- Biological Oxygen Demand COD- Chemical Oxygen Demand

Question 2.
By a flow chart showing the stages in anaerobic digestion during the production of biogas.
Answer:

Stages in Anaerobic Digestion during biogas formation

Question 3.
Given below is a list of six microorganisms. State their usefulness to humans.
(i) Nucleopolyhedrovirus
(ii) Saccharomyces cerevisiae
(iii) Monascus Purpureus
(iv) Trichoderma polypore
(v) Penicillium Notatum
(vi) Propionibacterium shamanic.
Answer:

Name of Micro-organisms	Uses
(i) Nucleopolyhedrovirus	Used in biocontrol of insects
(ii) Saccharomyces cerevisiae	Bread making, Ethanol making
(iii) Monascus Purpureus	Produces Statin used as blood cholesterol-lowering agent
(iv) Trichoderma polypore	Preparation of cyclosporin having antifungal, anti-inflammatory, immuno-suppressive properties
(v) Penicillium Notatum	Production of antibiotic, Penicillin
(vi) Propionibacterium shamanic	Preparation of large-holed swiss cheese.

Question 4.
Explain the different steps involved in the secondary treatment of sewage. (CBSE Sample paper 2018—19)
Or
Secondary treatment of sewage is also called biological treatment. Justify this statement and explain the process. (CBSE 2018)
Answer:

1. Secondary treatment of sewage is a biological process that employs the heterotrophic bacteria naturally present in the sewage.
2. The effluent from the primary treatment is passed into large aeration tanks, where it is constantly agitated and the air is pumped into it.
3. This allows the rapid growth of aerobic microbes into ‘floes’ which consume the organic matter of the sewage and reduce the biological oxygen demand (BOD). The greater is the BOD of wastewater, the more is its polluting potential.
4. When the BOD of sewage is reduced significantly, the effluent is passed into a settling tank, where the ‘floes’ are allowed to sediment forming the activated sludge.
5. A small part of the activated sludge is pumped back into the aeration tanks.
6. The remaining major part of the sludge is pumped into anaerobic sludge digesters, where the anaerobic bacteria digest the bacteria and fungi in the sludge-producing methane, hydrogen sulfide, and carbon dioxide,
   i. e. biogas. This is why secondary treatment of sewage is also called biological treatment.
7. The effluent after secondary treatment is released into water-bodies like streams or rivers.

Question 5.
Microbes can be used to decrease the use of chemical fertilizers. Explain how this can be accomplished. (CBSE Delhi 2019)
Answer:

1. Rhizobium bacteria present in the root nodules of leguminous plants (pea family) forms a symbiotic association and fixes atmospheric nitrogen into organic forms as nitrates/nitrites which are used by the plant as nutrient.
2. Free-living bacteria in the soil Azospirillum and Azotobacter can fix atmospheric nitrogen thus enriching the nitrogen content of the soil.
3. Many members of the genus Glomus (Fungi) form mycorrhizal symbiotic associations with higher plants. In these, the fungal symbiont absorbs phosphorus from soil and passes it to the plant.

Question 6.
Organic farmers prefer biological control of diseases and pests to the use of chemicals for the same purpose. Justify.
Answer:
Chemical methods often kill both useful and harmful living beings indiscriminately. The organic farmer holds the view that the eradication of the creatures that are often described as pests is not only possible but also undesirable, for without them the beneficial predatory and parasitic insects which depend upon them as food or hosts would not be able to survive. Thus, the use of biocontrol measures will greatly reduce our dependence on toxic chemicals and pesticides.

(ii) Give an example of a bacterium, a fungus, and an insect that are used as biocontrol agents. (CBSE 2018)
Answer:
Insects = Ladybird and Dragonflies. Bacteria = Bacillus thuringiensis. Fungus = Trichoderma

Question 7.
The three microbes are listed below. Name the product produced by each one of them and mention their use.
(i) Aspergillus niger
(ii) Trichoderma polypore
(iii) Monascus Purpureus (CBSE Delhi 2018C)
Or
(i) A patient had suffered myocardial infarction and clots were found in his blood vessels. Name a ‘clot buster’ that can be used to dissolve clots and the microorganism from which it is obtained.
(ii) A woman had just undergone a kidney transplant. A bioactive molecular drug is administered to oppose kidney rejection by the body. What is the bioactive molecule? Name the microbe from which this is extracted.
(iii) What do doctors prescribe to lower the blood cholesterol level in patients with high blood cholesterol? Name the source organism from which this drug can be obtained.
Answer:
(i) Aspergillus niger produces citric acid. Citric acid is used as a flavoring agent and as a food preservative.
(ii) Trichoderma Polysporum produces a bioactive molecule cyclosporin A. It is used as an immunosuppressive agent in organ transplant patients.
(iii) Monascus Purpureus produces statins. Statins are capable of competitive inhibition of enzymes required for cholesterol synthesis. Hence, it is used as blood cholesterol-lowering agents.
Or
(i) Streptokinase-‘Clot buster’ can be used to dissolve clots. It is obtained from the bacteria Streptococcus.
(ii) The bioactive molecule is Cyclosporin A which is used as an immunosuppressive agent in organ transplantation. It is produced by the fungus Trichoderma Polysporum.
(iii) Doctors prescribe Statins to lower blood cholesterol. It is obtained from the fungus Monascus Purpureus.

Question 8.
Baculoviruses are good examples of biocontrol agents. Justify giving reasons. (CBSE Delhi 2018C)
Answer:
Baculoviruses kill insects and other arthropods, hence they are used as biocontrol agents especially Nucleopolyhedrovirus.

Reasons for their use:

• These viruses are species-specific and have narrow spectrum insecticidal applications.
• They do not harm non-target organisms like other harmless insects, birds, animals, etc.
• It is very useful in integrated pest management programs or treatment of ecologically sensitive areas.

Question 20.
Describe the primary and secondary treatment of domestic sewage before it is released for reuse. (CBSE, 2014)
Answer:
Treatment of domestic sewage. The municipal wastewaters are treated in Effluent Treatment Plant (ETP) prior to disposal in water bodies.

It consists of 3 steps: primary, secondary, and tertiary.
1. Primary treatment. It includes physical processes, such as sedimentation, floatation, shredding (fragmenting and filtering). These processes remove most of the large debris.

2. Secondary treatment. It is a biological method. Activated sludge method. Sewage, after primary treatment, is pumped into aeration tanks or oxidation ponds. Here, it is mixed with air and sludge containing algae and bacteria. Bacteria consume organic matter. The process results in the release of C02 and the formation of sludge or biosolid. Algae produce oxygen for the bacteria. The water, which is now almost clear of organic matter, is chlorinated to kill microorganisms.

3. Tertiary treatment. It involves. removal of nitrates and phosphates. The water, after the above treatment, is then released. It can be reused.

Question 9.
Explain biological control of pests and plant pathogens with examples.
Answer:
The very familiar beetle with red and black markings the Ladybird, and Dragonflies are useful to get rid of aphids and mosquitoes, respectively.

Role of Bacillus Thuringinesis:
Bt Coming to microbial biocontrol agents that can be introduced in order to control butterfly caterpillars is the bacteria Bacillus thuringiensis (often written as Bt). These are available in sachets of dried spores which are mixed with water and sprayed onto vulnerable plants such as Brassica and fruit trees, where these are eaten by the insect larvae. In the gut of the larvae, the toxin is released and the larvae get killed.

The bacterial disease will kill the caterpillars, but leave other insects unharmed. Because of the development of the methods of genetic engineering in the last decade or so, scientists have introduced B. thuringiensis toxin genes into plants. Such plants are resistant to attack by insect pests. Bt-cotton is one such example which is being cultivated in some states of our country.

Biological control of plant pathogens: A biological control developed for use in the treatment of plant disease is the fungus Trichoderma. Trichoderma sp. are free-living fungi that are very common in soil and root ecosystems. They are effective biocontrol agents of several plant pathogens.

Baculoviruses are pathogens that attack insects and other arthropods. The majority of baculoviruses used as biological control agents are in the genus Nucleopolyhedrovirus. These viruses are excellent candidates for species-specific, narrow spectrum insecticidal applications.

They have been shown to have no negative impacts on plants, mammals, birds, fish, or even on non-target insects. This is especially desirable when beneficial insects are being conserved to aid in an overall IPM (integrated pest management) program, or when an ecologically sensitive area is being treated.

Question 22.
How do biofertilizers enrich the soil?
Answer:
Biofertilizers play a vital role to solve the problems of soil fertility and soil productivity.

1. Anabaena azollae, a cyanobacterium, lives in symbiotic association with the free-floating water fern, Azolla. The symbiotic system Azolla-Anabaena complex is known to contribute 40-60 mg N ha-1 per rice crop. In addition to this, cyanobacteria add organic matter, secretes growth-promoting substances like auxins and vitamins, mobilizes insoluble phosphate, and thus improves the physical and chemical nature of the soil.
2. Rhizobium Leguminoserum and Azospirillum fix atmospheric nitrogen as nitrates and nitrites.
3. Mycorrhizae formed by an association of bacteria and roots of higher plants increase soil fertility.

Question 10.
Discuss the role of Microbes as Biofertilizers. (CBSE Delhi 2011, 2015, 2019)
Answer:
Role of microbes as biofertilizers:
Bacteria, cyanobacteria, and fungi (mycorrhiza) are the three groups of organisms used as biofertilizers.
1. Bacteria:
(a) Symbiotic bacteria Rhizobium.
(b) Free-living bacteria Azospirillum and Azotobacter.
(c) They fix the atmospheric nitrogen and enrich soil nutrients.

2.  Cyanobacteria, e.g. Anabaena, Nostoc, Aulosira, Oscillatoria, etc.
(a) They function as biofertilizers by fixing atmospheric nitrogen and
(b) Increasing the organic matter of the soil through their photosynthetic activity.

3. Fungi/mycorrhizae:
(a) Fungi form a symbiotic association with roots of higher plants (mycorrhizae), e.g. Glomus.
(b) The fungus absorbs phosphorus and passes it on to the plant.
(c) Other benefits of mycorrhizae are :

• resistance to root-borne pathogens.
• tolerance to salinity.
• tolerance to drought.
• the overall increase in the plant growth and development

In This Post we are  providing Chapter-11 BIOTECHNOLOGY: PRINCIPLES AND PROCESSESS 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 BIOTECHNOLOGY: PRINCIPLES AND PROCESSESS

1. Observe the given sequence of nitrogenous bases on a DNA fragment and answer the following question –
5´ – CAGAATTCTTA – 3´
3´ – GTCTTAAGAAT – 5´
(a) Name a restriction enzme which can recognise this DNA sequence.
(b) Write the sequence after digestion.
(c) Why are the ends generated after digestion called sticky ends?
Ans. (a) EcoRI

(c) These are named sticky ends, because they form hydrogen bonds with their complementary cut parts.

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2. A selectable marker is used in the section of recombinants on the basis of their ability to produce colour in presence of chromogenic substrate.
(a) Mention the name of mechanism involved.
(b) Which enzyme is involved in production of colour?
(c) How is it advantageous over using antibiotic resistant gene as a selectable marker?
Ans. (a) Insertional inactivation
(b) b-galactosidase.
(c) Selection of recombinants due to inactivation of antibiotics requires simultaneous plating on two plates having different antibiotics.

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3.Mention the important properties which a good vector must possess?
Ans.The important properties which a good vector must possess are :-
i) Size :- The vector must have small size so that it is easier to purify & isolate.
ii) Origin of replication :- This is a sequence of base pairs where replication starts. Any piece of DNA linked to this sequence can be made to replicate within its host cell & thus, controls the copy number of linked DNA.
iii) Selectable Marker :- A marker is a gene which helps in selecting those host cells which contain the vector & eliminating the non – transformants Common Selectable marker include gene encoding resistance to antibiotics.
iv) Cloning Sites :- The vector Should have a few or at least one unique recognition site to link the foreign / alien DNA. Presence of a particular recognition site enables the particular restriction enzyme to cut the vector.

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4.Describe any three vectors less method of introducing the rDNA into a competent host cell?
Ans. i) Transformation :- In order to force bacteria to take up the plasmid, the bacterial cell must first be made competent to take up DNA. This is done by treating them with specific concentration of divalent cationeg. Ca2+ which increases the efficiency with which DNA enters the bacterium through pores in its cell wall Recombinant DNA can then be forced into such cells by incubating the cells with recombinant DNA on ice, followed by placing them at 420 C& then putting them back into ice. This enables the bacteria to take up the recombinant DNA.
ii) Microinjection :- recombinant DNA is directly injected into the nucleus of an animal cell using a micro – needle of tip with diameter (~ 4mm)
iii) Biolistics / Gene gun :- cells are bombarded with high velocity micro – particles of gold or tungsten coated with DNA.

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5.Why is Agrobacterium mediated genetic transformation described as Natural Genetic engineering in plants?
Ans.Agrobacterium tumefacien, anatural pathogen of Several dicot plants is able to deliver a piece of DNA known as “t – DNA” to transform normal plant cell into a tumor & direct gene transfer transform tumor cells to produce chemicals required by pathogen . The tumor inducing (Ti) plasmid of
Agrobacterium tumefacien has now been modified into a cloning vector which is no more pathogenic to plant but is still able to use the mechanism to deliver genes of our interest into a variety of plants Since Agrobacterium tumefacien has the natural ability to donate a part of its DNA to the plant during infection. This property of Agrobacterium is exploited and a gene of interest is ligated into T-DNA so that it automatically gets transformed into plant cell thus, Agrobacterium tumefacien is known as “Natural Genetic Engineer” of plants.

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6.Mention the important tools required for genetic engineering technology?
Ans.The process of genetic engineering is accomplished only when we have following key tools :-
a)Restriction enzymes:- Restriction enzymes are a group of endonucleases which cut the DNA at Specific position anywhere in its length. Each restriction endonuclease functions by inspecting the length of DNA & binds to DNA at the recognition Sequence.
b)Cloning Vector:- The DNA molecule which carry the desired DNA Segment of an organism & transfer it to cell or DNA of another organism is called cloning vector.
c)Desired foreign DNA:- The segment of DNA containing genes having desired characters & which are being transferred into genome of another cell with the help of vector is called foreign DNA.

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7. In the given figure, one cycle of polymerase chain reaction (PCR) is shown-

(a) Name the steps A, B and C.
(b) Give the purpose of each of these steps.
(c) State the contribution of bacterium Thermus aquaticus in this process.
Ans. (a) Denaturation – Heat denatures DNA to separate complementary strands.
(b) Annealing : Primers hybridises to the denatured DNA strands.
(c) Extension : Extension of primers resulting in synthesis of copies of target DNA sequence. Enzyme Tag polymerase is isolated from the bacterium Thermusaquaticus. This enzyme induces denaturation of double stranded DNA at high temperature.

8.Describe the various steps involved in Recombinant DNA technology with the help of a well labeled. Diagram?
Ans. i) Identification of DNA with desirable Genes:- Other molecules in the target cell can beremovedby appropriate treatment & purified DNA ultimately precipitates out after addition ofchilled ethanol.
ii) Cutting the DNA at specific location :- After having cut the source DNA as well as vector DNA with Specific restriction enzyme, the cut out “gene of interest” from the source DNA & the cut vector with space are mixed & ligase is added.
iii)Insertion of Recombinant DNA into host cell :- Recipient cells after making them competent to receive takes up DNA in its surrounding. Recombinant DNA is introduced into suitable host cell by vector – based or vector – less method.

iv)Selection & Screening :- If a recombinant DNA bearing gene for resistance to an antibiotic is trAns.ferred into E-coli the host – cell become trAns.formed into ampicillin – resistant cells. Due to this amp gene one is able to select a trAns.formed cell in the presence of ampicillin. This amp r gene is called selectable marker.
v)Obtaining the foreign Gene product :- After having cloned the gene of interest & having optimized the conditions to induce expression of the target protein, one has to consider producing it on large scale.

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9.Expand PCR? Describe the different Steps involved in this technique?
Ans. PCR stands for polymerase chain reaction.It is a technique for amplification of gene of interest
or to obtain multiple copies of DNA of interest.
The PCR requires primers, taq polymerase, target sequence, DNA sample &deoxyribonucleotides.
PCR includes number of cycles for amplifying DNA of interest invitro. Each cycle has three steps :-
a)DENATURATION:- The first step is denaturation of SNA sample in a reaction mixture to 940 c. During this step, DNA strand gets separated.
b)RENATURATION / ANNEALING:- The temperature is allowed to cool down to 500c to allow two oligo-nucleotide primers to anneal to complementary sequence in DNA molecule.
c)EXTENSION:- The temperature is raised to 750c. At this temperature, taq – polymerase initiates DNA Synthesis at 3-OH end of primer.

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10.What are Restriction enzyme? Why do bacteria have these restriction enzymes. Show diagrammatically a restriction enzyme its recognition & the product it produces?
Ans. Restriction enzymes are endonucleases which recognize a specific sequence within DNA and cut the DNA within that sequence at a specific point. In bacteria, these restriction enzymes operates modification restriction system which modifies & cuts the foreign DNA entering into the bacterial cell& thus, provides immunity to bacterial cell.
Name of Restriction enzyme- EcoRI Substrate DNA on which it acts

In This Post we are  providing Chapter- 12 BIOTECHNOLOGY AND ITS APPLICATION 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 BIOTECHNOLOGY AND ITS APPLICATION

1.Describe how nematode – resistant transgenic plants have been obtained?
Ans.A nematode Meloidogyne incognita infects tobacco plant &reduces its yield. The specific genes fromparasite are introduced into plant using Agrobacterium. The genes are introduced in such a way that bothsense& Antisense RNA are produced. Since these two RNAs are complementary, they form a doublestranded RNA (ds RNA). This neutralizes the specific RNA of nematode by a process called RNAinterference as a result, the parasite cannot live in transgenic host & plant is protected from the pest.

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2.What are Cry proteins? Name an organism that produces it. How has man exploited this protein to his benefit?
Ans.The soil bacterium Bacillus thuringiensis produces crystal proteins called cry proteins that are toxicto larvae of insects like tobacco budworm, beetles & mosquitoes. The cry proteins exist as inactiveprotoxin& gets converted into active toxin when ingested lay the insect, as the alkaline pH of gutSolublises the crystal. The activated toxia binds to surface of epithelial cells ofmidgut& create pores thiscauseslysis of cells leading to death of insects.The genes encoding this protein are isolated from bacterium & incorporated into crop-plant to make them insect – resistant.

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3.Write an account on the production of human insulin in transgenic organisms.
Ans. Human insulin consists of two short polypeptide chains: chain A & B linked by disulfide bonds.Insulin is secreted as prohormone which has to be processed before it becomes a mature & functionalhormone. The prohormone contains another polypeptides called C-peptide which is removed during
maturation.Using genetic engineering, the two DNA sequences coding for chains A & B of human insulin areintroduced into plasmid of E – coli – to produce insulin. The two chains produced are extracted &combinedby creating disulfide bridges.

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4.Compare & contrast the advantages & disadvantage of production of Genetically modified organisms?
Ans. ADVANTAGES OF PRODUCING GMOS.

1. GM crops produce desired phenotypic traits in crop plants.
2. The genes responsible for production of specific proteins are inserted into GM crops. These crops then produce that specific protein.
3. Transgenic crops synthesizes new end product of specific biochemical pathway.
4. These crops also help in preventing expression of existing native Gene.

DISADVANTAGES OF PRODUCING GMOS:

1. Transgenic crops may endanger wild & native species.
2. GM crops may cause health problems by supplying allergens.
3. GM crops may damage to the natural environment.

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5.What is RNA Silencing? How is this strategy used to create pest – resistant plants?
Ans.RNA silencing is a technique which involves silencing or disabling of specific mRNA due tocomplementary ds RNA molecule that binds to & prevent translation of mRNA. This strategy is used toprevent infection of roots of tobacco plants lay nematode Meloidegyne incognita. In this strategy, complementary ds RNA is produced against specific mRNA. The source of this complementary RNA couldbe from an infection by viruses having RNA genomes. Using Agrobacterium vector nematode specificgenes were introduced into host plant. The introduction of DNA was such that it produced both sense &anti-dense RNA in the host cell. These two RNA’s being complementary to each other formed a doublestrand RNA that initiated RNAi& thus silenced specific mRNA of the nematode. The consequence wasthat parasite could not survive in transgenic host.

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6.What are the steps involved in synthesis of genetically engineered insulin.
Ans. Steps involved in Insulin production are :-

1. for synthesis of Insulin, RNA is extracted from β−β−cells of islets of Langehans of pancreas.
2. With the help of enzyme Reverse transcriptase, single stranded DNA complementary to mRNA is synthesized second strand of DNA complementary to first is synthesized with enzyme DNA polymerase.
3. The two strands of copy DNA is joined to plasmid by using an enzyme called terminal transferase.
4. The two ends of DNA get annealed by enzyme called ligase thus ends of inserted DNA & plasmid are sealed & a new circular plasmid is formed. This is a molecule of recombinant DNA.
5. This recombinant DNA is then inoculated in a new bacterial cell of E-coli & inserted in a bacterial gene after having cut by restriction enzyme.
6. After proper expression of genes the bacterial cells of both cultures are lysed with appropriate chemicals. The fragments of insulin are then separated from enzyme by cyanogen bromide.

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7. The clinical gene therapy is given to a 4 years old patient for an enzyme which is crucial for the immune system to function.

Observe the therapeutical flow chart and give the answer of the following:
(a) Complete the missing steps (B) and (D)
(b) Identify the disease to be cured.
(c) Why the above method is not a complete solution to the problem?
(d) Scientists have developed a method to cure this disease permanently. How?
Ans. (a) Step (B) : Lymphocytes are grown in culture medium. Step (D) : Infusion of genetically engineered lymphocytes into patients.
(b) Adenosine deaminase (ADA) deficiency.
(c) As genetically engineered lymphocytes are not immortal, the patient requires periodic
infusion of cells.
(d) If the gene isolated from bone marrow cells producing ADA is introduced into cells at early embryonic stages, it could be a permanent cure.

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8. In the given figure, Agrobacterium is utilized for the production of a transgenic crop. Explain the steps a, b, c, d and e shown in the figure.

Ans. Step (a) Plasmid is removed and cut open with restriction endonuclease.
Step (b) Gene of interest is isolated from another organism and amplified using PCR
Step (c) New gene is inserted into plasmid
Step (d) Plasmid is put back into Agrobacterium
Step (e) Agrobacterium based transformation.

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9. In the given figure, Form (A) and Form (B) represents different forms of a proteinaceous hormone secreted by pancreas in mammals.

(a) What type of bonding is present between chains of this hormone?
(b) What are these form (A) and form (B). How these forms differ from each other?
(c) Explain how was this hormone produced by Eli Lilly, an American company, using rDNA technology.
Ans. (a) Disulphide bonds
(b) Form (A) – Proinsulin
Form (B) – Mature insulin.Proinsulin contains an extra stretch called C – peptide which is absent in mature insulin.
(c) Eli Lilly company prepared two DNA sequences corresponding to A and B peptide chains of human insulin and introduced them in plasmid E. coli to produce insulin chains. Chains A and B were producedseparately, extracted and combined by creating disulphide bonds to form insulin.

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10.What is Gene therapy – Illustrate using example of Adenosine deaminase deficiency?
Ans.Gene therapy is a collection of methods that allows correction of a gene defect. In this method,genes are inserted into the cells & tissues of an individual to correct certain hereditary diseases. Itinvolves delivery of a normal gene into the individual or embryo to replace the defective mutant allele of the gene. Viruses which attack the host cell & introduce genetic material into host are usedas vectors.
For example Adenosine deaminase (ADA) deficiency can be cured by bone marrow transplantation in some children but is not curative for Gene therapy, lymphocytes are grown in a culture & functional ADA, cDNA is introduced into these lymphocytes. These lymphocytes are then transferred into body of patient the patient requires infusion of such genetically engineered lymphocytes.

In This Post we are  providing Chapter- 8 HUMAN HEALTH AND DISEASE 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 HEALTH AND DISEASE

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1.Describe the ill – effects of drug abuse in males & females. Also mention the preventive measures that is to be taken to reduce such effects.
Ans.1) ILL – DEFECTS IN MALES :- acne, increased aggressiveness , mood swing depression reduction of size of testicles, decreased Sperm production, kidney & liver dysfunction, premature baldness.
2) ILL – EFFECTS IN FEMALES :-masculinisation, increased aggressiveness, mood swings, depression abnormal menstrual cycle, excessive hair growth on face & body & deepening of voice.
The following measures are need to be taken to prevent such problems :-

1. EDUCATION & COUNSELLING :- to face problem or stress, to accept failure as part of life & to channelize child’s energy to some health promoting activities.
2. AVOID UNDUE PEER PRESSURE :- to pressurize a child to perform beyond his capabilities
3. SEEKING HELP FROM PARENTS &PEERS :- to share the feeling of anxiety & guilty.
4. SEEKING PROFESSIONAL FOR MEDICAL HELP :- help available in the form of highly qualified psychiatrist, psychologist etc.

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2.What is vaccination? What type of immunity is provided by vaccination?
Ans.Vaccination is a process of development of immunity with administration of vaccines in the body, here weakened pathogen are infected into the body to produce immunity against a particular pathogen. This pathogen stimulates the body to produce antibodies. The antibodies produced against these antigens would neutralize the pathogenic agent. The vaccine also generates memory B – and T – cells that recognize pathogen quickly on subsequent exposure & overwhelm the invaders with massive production of antibodies.
The type of immunity is ACTIVE IMMUNITY.
However, if a person is infected with some deadly microbe to which quick immune response is required, we need to directly injected to patient’s body, This type of immunization is called PASSSIVE IMMUNISATION.

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3. (i) Differentiate between communicable & non – communicable diseases?
(ii) Name the body part & the host in which following events takes place in life cycle of plasmodium.
(a) fertilization
(b) Development of Gametophyte :-
(c) Release of sporozoites :-
(d) Asexual Reproduction.
Ans. (i) Communicable diseases are caused lay biological agents & can spread from one person to another or one place to another through air, water, physical contact etc.
Non – communicable diseases are confined to a person & do not easily spread from one person to another.
(ii) (a) in the gut of female anopheles.
(b) in RBCS of Human beings.
(c) Salivary gland of female anopheles.
(d) liver cells of human beings.

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4. Answer the following with respect to Caner.
(a) How does a cancerous cell differ from a normal cell?
(b) Benign tumor is less dangerous than malignant tumor. Why
(c) Describe causes of cancer.
(d) mention two methods of treatment of the disease.
Ans. (a) In normal cells, growth and differentiation is highly controlled and regulated (contact inhibition). The cancerous cells have lost the property of contact inhibition, hence continue to divide giving rise to masses of cells (tumors).
(b) The benign tumor remains confined in the organ affected as it is enclosed in a connective tissue sheath and does not enter the metastatic stage.
(c) Cancer may be caused due to carcinogens which are physical (radiations), chemicals (Nicotine, Aflatoxin, Cadmium oxide, Asbestos) and biological (viral oncogens).
(d) Surgery, radiotherapy, Chemotherapy

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5. The pathogen of a disease depends on RBCs of human for grwoth and reproduction. The person with this pathogen suffers with chill and high fever.
(a) Identify the disease.
(b) Name the pathogen.
(c) What is the cause of fever?
(d) Represent the life cycle of the pathogen diagrammatically.
Ans. (a) Malaria
(b) Different species of Plasmodium viz P. vivax, P. Malariae and P. falciparum.
(c) Malaria is caused by the toxins (haemozoin) produced in the human body by the malarial parasite. This toxin is released by the rupturing of RBCs.
(d) Life cycle of Plasmodium : Fig. 8.1 Page 148, NCERT book, Biology – XII

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6. The immune system of a person is supressed. He was found positive for a pathogen in the diagnostic test ELISA.
(a) Name the disease, the patient is suffering from.
(b) Which pathogen is identified by ELISA test?
(c) Which cells of the body are attacked by the pathogen?
(d) Suggest preventive measure of the infection.
Ans. (i) AIDS (Acquired Immuno Deficiency Syndrome)
(ii) HIV (Human Immunodeficiency Virus)
(iii) Helper T-cells, macrophages, B-lymphocytes.
(iv) Preventive measures :
(a) People should be educated about AIDS transmission.
(b) Disposable needles and syringes should be used
(c) Sexual habits should be changed immediately
(d) High-risk groups should be discouraged from donating blood.
(e) Routine screening may be done.

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7.Discuss the role of lymphoid organs in the immune response. Explain the different types of lymphoid organs giving two examples of each type in humans.
Ans.Lymphoid organs are organs where origin or maturation & proliferation of lymphocytes occurs. These lymphoid organs are of two types:-
1.PRIMARY LYMPHOID ORGAN: – where immature lymphocytes differentiate into antigen – sensitive lymphocytes. It includes :-

(a)BONEMARROW :– It is the main lymphoid organ present in the thigh region where all types of blood cells including lymphocytes are formed. It provides micro – environment for the development & maturation of B – cells.
(b)THYMUS :- It is located beneath the chest bone near heart. It provides microenvironment for the development & maturation of T – lymphocytes.
2.SECONDAR LYMPHOID ORGAN :- They provide the site for interaction of lymphocytes with antigen which then proliferate to become effector cells. It includes.
(a)SPLEEN :- It is large bean shaped organ & contains mainly lymphocytes & phagocytes. It acts as a filter of blood by trapping blood – bound micro – organism.
(b)LYMPHNODE :- They are small – solid structure located at different points along lymphatic system. It serves to trap antigen which happens to get into lymph & tissue fluid. Antigen trapped in lymph nodes are responsible for activation of lymphocytes,

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8.With the help of a well – labelled diagram, Describe the life cycle of malarial parasite.
Ans.Malaria is caused by plasmodium vivax. It has two hosts – female anopheles is the vector of plasmodium while the primary host is man where the parasite maintains an amoeboid stage in RBCS & later produces gametophyte.
Life cycle of plasmodium involves following steps:-

1. The sporozoites enters the human body, reach the liver through blood & multiply within the liver cells such liver cells burst & release the parasites into blood.
2. They attack RBCS, multiply & cause their rupture.

The rupturing of RBCS is associated with the release of a toxin called haemozoin, which is responsible for recurring fever & the chill / shivering.

1. Gametophytes are developed in RBCS.
2. When a female anopheles mosquito bites an infected person, these parasites enter the mosquito’s body & undergo further development. These parasites multiply within then in the stomach & develop a cyst.
3. The cyst produces sporozoites which reach salivary gland of mosquito. When such infected Anopheles sucks blood of a healthy person, it transfers. Sporozoites to repeat amoeboid stage again.

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9. What do you mean by “Out – breeding”. What are the different methods employed for out breeding.
Ans.Breeding between the unrelated ale & female animals is called Outbreeding. It can be done in following ways:-

1. OUT CROSS :- The mating of animals within the same breed but do not have any common ancestor on either side of their pedigree for 4-6 generation is called an out cross. It is the best method of breeding of animals that are below average in milk production, growth rate of beef cattle etc.
2. CROSS–BREEDING :- It is a cross between superior males of one breed & Superior females of another breed. It allows the desirable qualities of two different breeds to be combined & are used for commercial production eg .Hisardale, a new breed of sheep is developed by crossing bikaneri ewes & Marino rams.
3. INTERSPECIFIC HYBRIDISATION :- male & female animals of two different related species are mated so, that progeny may combine desirable features of both parents eg. mule is produced by crossly donkey & a female house.

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10. What is somatic hybridization – Explain the steps involved in the production of somatic hybrids?
Ans.Somatic hybridization is the process of fusing protoplasts of somatic cells derived from two different varieties or species of a plant on a suitable nutrient culture medium under sterile condition. One example of somatic hybrid is topato produced by fusion of protoplast of tomato & potato.
Somatic hybridization involves the following steps:-

1. Isolation of protoplast from two different varieties of plants – each having a desirable character,
2. Fusion of cytoplasm of two protoplast results in coalescence of cytoplasm. The nuclei of two protoplasts may or may not fuse together even after fusion of cytoplasm, fusion of protoplast requires a suitable agent called fusogeneg. PEG or polyethylene glycol.
3. Under favourable conditions, hybrid protoplast synthesise new cell wall around it. Hybrid cell functions as a single cell & then undergo sustained division to form callus.
4. The regenerated callus is transferred to a new culture plates containing suitable culture media.Here callus divide & form root & shoot after organogenesis.

In This Post we are  providing Chapter- 9 STRATEGIES FOR ENHANCEMENT FOR FOOD PRODUCTION 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 STRATEGIES FOR ENHANCEMENT FOR FOOD PRODUCTION

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1.What measures would you undertake to improve the quality & quantity of milk production?
Ans.The quality & quantity of milk production depends on three factors :-

1. Genetic makeup.
2. Nutrition &
3. Environment

Thus, the following steps should be taken to improve management of livestock :-
a.SHEDS :- Sheds should be neat & clean, well – ventilated with pucca floor & will drained channel.
b.BALANCED DIET :- a balanced feed consists of appropriate quantities of carbohydrates, proteins, vitamins, minerals & water. The feed consists of two main components :-
i.Roughage – include fodder, hay, straw & Silage.
ii.Concentrates – broken forge crops, grams, cereals, mullets, cotton, seeds.
a.CLEAN WATER :
b.HEALTH CARE:- It requires regular inspection with proper record keeping.

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2.What is “tissue culture”. What are the steps involved in tissue culture?
Ans.“Tissue culture is an experimental process through which a mass of cells (callus) is produced froman explant tissue & used directly to regenerate plant It invoices following steps :-

1. Selection of an elite plant
2. Preparation of suitable culture media
3. Sterilisation of an explant & inoculation on culture media under controlled temp ~ 250 c in light
4. Callus induction in explant.
5. Organogenesis :- a high cytokine : auxin ratio induce Shoot formation while high auxin : cytokinin ratio induce root formation.
6. Acclamatization :- test tube rooted plantlets are first subjected to acclimatization in green house & then transferred to the field.

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3.What are the measures that need to be taken for effective poultry farm management?
Ans.i) It requires a crowd – free, rainproof, well ventilated & protected brood house.
ii).Brood house should be clean & disinfected.
iii).Good drainage system.
iv).Proper fed & clean & fresh drinking water.
v).Proper light management for optimum egg production.
vi).Poultry are more sensitive to heat so, measures should be adopted to overcome heat shock.
a).Sheds should be covered with grass or low vegetation.
b).Provide sprinklers on roof.
c).Maximum Ventilation.
vii)Disease – free & suitable breeds should be selected for breeding.

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4.The steps in a programme are :-
Collection of germplasm, crossbreeding the selected parents, selection superior recombinant progeny & Testing, releasing & marketing new cultivars?
i) What is this programme related to?
ii) Name two special qualities as the basis of selection of progeny.
iii) What was the outcome of the programme?
iv) What is the popular term given to this outcome? Also name the India Scientist who is credited with chalking out of this programme.
v) Among the above – mentioned step which is the most crucial step of this programme& why?
Ans. i). Plant breeding.
ii). Disease resistance & yield.
iii). Production of improved varieties.
iv). The popular term give to this outcome is HYBRID. Dr. S. Swamminathan is credited with chalking out of this programme.
v). Selection of superior progeny is the most crucial step of this programme because it yields plants thatare superior to both parents & are then self – pollinated for several generations.

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5. What is apiculture? What are the requirements to consider for bee–keeping?
Ans.The culturing of honey bees for the production of honey or beewax is called Apiculture.
Bee – keeping can be practised in any area where there is sufficient bee pastures of some wild shrubs, fruits orchards & cultivated crops. The following points are important for successful bee – keeping :-

1. Knowledge of nature & habits of bee.
2. Selection of suitable location of keeping beehives.
3. Catching & hiving of swarms.
4. Management of beehives during different seasons.
5. Handling & collection of honey &beewax.

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6.What are the major steps involved in Plant breeding?
Ans.The major steps involved in plant breeding are :-
i). Collection of varieties :- collection & preservation of all the different wild varieties, species & relatives ofthe cultivates species.
ii). Evaluation & Selection of Parents :- Germplasm collected is evaluated to identify plants with desirablecharacter. The selected plants are multiplied & used.
iii). Hybridisation of Selected Parents :- The selected parents are hybridized so that the traits in them can becombined in the hybrid progeny.
iv). Selection & Testing of Superior Recombinants :- Individuals with desired combination of characters haveto be selected from among the progeny. Such hybrids are superior to both the parents.
v). Testing, Release & commercialization of New cultivars :-
Evaluation is done by growing these plants in the research field & recording their performance under ideal conditions of irrigation, fertilizers & other crop practices. The selected plants are then tested in the farmer’s field for at least three growing seasons. The material thus selected is certified & released as a variety.

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7. Does apiculture offer multiple advantages to farmers? List its advantages, if it is located near a place of commercial flower cultivation. Name the most common species of bee which is reared in India.
Ans. Apiculture or Bee-Keeping is the maintenance of hives of honeybees for the production of honey. Apiculture is beneficial for farmers in many ways. Honey bee also produces beewax which is used in industries, such as in preparation of cosmetics and polishes of various kinds. If Bee keeping is practiced in any area the commercial flowers are cultivated, it will be beneficial in the following ways.
(i) Bees are pollinators of many crop species including flowering crops such as sunflower.
(ii) It improves the honey yield, because honeybees collect the nectar from flowers for making honey.
Apis indica is the msot common species whch is reared in India.

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8. What is somatic hybridisation? Describe the various steps in producing somatic hybrids from protoplasts. Mention any two uses of somatic hybridisation.
Ans. Somatic Hybridisation : The process of fusing protoplasts of Somatic cells derived from different varieties or species of plants to produce a hybrid.
Steps :
(i) Removal of cell wall of fusing cells by digestion with a combination of pectinase and cellulase to form protoplasts.
(ii) Fusion between protoplasts of selected parents is induced by the use of poly ethylene glycol (PEG).
(iii) The resulted product is cultured on a suitable medium to regenerate cell walls.
(iv) The cells obtained begin to divide to produce plantlets called somatic hybrids.
Uses/Applications :
(i) Somaclonal variations can be created
(ii) Lines or varieties/species of plants which can not be sexually hybridised, they can be hybridised.
(iii) Allopolyploids can be raised by the method.

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9.What do you mean by “Out – breeding”. What are the different methods employed for out breeding.
Ans. Breeding between the unrelated ale & female animals is called Outbreeding. It can be done in following ways:-
i)OUT CROSS :- The mating of animals within the same breed but do not have any common ancestor on either side of their pedigree for 4-6 generation is called an out cross. It is the best method of breeding of animals that are below average in milk production, growth rate of beef cattle etc.
ii)CROSS–BREEDING :- It is a cross between superior males of one breed & Superior females of another breed. It allows the desirable qualities of two different breeds to be combined & are used for commercial production eg . Hisardale, a new breed of sheep is developed by crossing bikaneri ewes & Marino rams.
iii)INTERSPECIFIC HYBRIDISATION :- male & female animals of two different related species are mated so, that progeny may combine desirable features of both parents eg. mule is produced by crossly donkey & a female house.

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10.What is somatic hybridization – Explain the steps involved in the production of somatic hybrids?
Ans. Somatic hybridization is the process of fusing protoplasts of somatic cells derived from twodifferent varieties or species of a plant on a suitable nutrient culture medium under sterilecondition. One example of somatic hybrid is topato produced by fusion of protoplast of tomato &potato.
Somatic hybridization involves the following steps:-
I).Isolation of protoplast from two different varieties of plants – each having a desirable character,
II). Fusion of cytoplasm of two protoplast results in coalescence of cytoplasm. The nuclei of twoprotoplasts may or may not fuse together even after fusion of cytoplasm, fusion of protoplast requires a suitable agent called fusogen eg. PEG or polyethylene glycol.
III). Under favourable conditions, hybrid protoplast synthesise new cell wall around it. Hybrid cellfunctions as a single cell & then undergo sustained division to form callus.
IV). The regenerated callus is transferred to a new culture plates containing suitable culture

Short Answer Type Question:

Q1. What is the unit used in measuring pressure? Why is the pressure measured at station level reduced to the sea level in preparation of weather maps?
Answer
Millibar or Pascal is the unit used in measuring pressure. It is reduced to the sea level in the preparation of weather maps as the gravity of air at the surface is denser and hence has higher pressure.

Q2. While the pressure gradient force is from north to south, i.e. from the subtropical high pressure to the equator in the northern hemisphere, why are the winds north easterlies in the tropics?
Answer
The Coriolis force acts perpendicular to the pressure gradient force. The pressure gradient force is perpendicular to an isobar. The higher the pressure gradient force, the more is the velocity of the wind and the larger is the deflection in the direction of wind. As a result of these two forces operating perpendicular to each other, in the low-pressure areas the wind blows around it. The winds blow from high pressure to the low pressure, so this pressure gradient force is from north to south. Therefore the north easterlies blow from north east to south west.

Q3.What are the geostrophic winds?

Answer

When isobars are straight and when there is no friction, the pressure gradient force is balanced by the Coriolis force and the resultant wind blows parallel to the isobar. This wind is known as the geostrophic wind.
Q4.Explain the land and sea breezes.

Answer
During the day the land heats up faster and becomes warmer than the sea. Therefore, over the land the air rises giving rise to a low pressure area, whereas the sea is relatively cool and the pressure over sea is relatively high. Thus, pressure gradient from sea to land is created and the wind blows from the sea to the land as the sea breeze. In the night the reversal of condition takes place. The land loses heat faster and is cooler than the sea. The pressure gradient is from the land to the sea. This is known as land breeze.

Q5.Discuss the factors affecting the speed and direction of wind.
Answer
The speed and direction o the wind is controlled by the combined effects of three forces:
• Pressure Gradient Force: The differences in atmospheric pressure produces a force. The rate of change of pressure with respect to distance is the pressure gradient. The pressure gradient is strong where the isobars are close to each other and is weak where the isobars are apart.

• Frictional Force: It affects the speed of the wind. It is greatest at the surface and its influence generally extends upto an elevation of 1 – 3 km. Over the sea surface the friction is minimal.• Coriolis Force: The rotation of the earth about its axis affects the direction of the wind. This force is called the Coriolis force after the French physicist who described it in 1844. It deflects the wind to the right direction in the northern hemisphere and to the left in the southern hemisphere.

Q6.Draw a simplified diagram to show the general circulation of the atmosphere over the globe. What are the possible reasons for the formation of subtropical high pressure over 30°N and S latitudes?
Answer

The general circulation of the atmosphere also sets in motion the ocean water circulation which influences the earth’s climate. The air at the Inter Tropical Convergence Zone (ITCZ) rises because of convection caused by high insolation and a low pressure is created. The winds from the tropics converge at this low pressure zone. The converged air rises along with the convective cell. It reaches
the top of the troposphere up to an altitude of 14 km. and moves towards the poles. This causes accumulation of air at about 30°N and S. Part of the accumulated air sinks to the ground and forms a subtropical high. Another reason for sinking is the cooling of air when it reaches 30°N and S latitudes.

Q7.Why does tropical cyclone originate over the seas? In which part of the tropical cyclone do torrential rains and high velocity winds blow and why?
Answer
Tropical cyclones originate and intensify over warm tropical oceans. The conditions favourable for the formation and intensification of tropical storms are:
(i) Large sea surface with temperature higher than 27° C.(ii) Presence of the Coriolis force.(iii) Small variations in the vertical wind speed.(iv) A pre-existing weaklow-pressure area or low-level-cyclonic circulation.(v) Upper divergence above the sea level system.
The torrential rains and high velocity winds, occur in the region of the ‘eye-wall’. The eye is a region of calm with subsiding air. Around the eye is the eye wall, where there is a strong spiralling ascent of air to greater height reaching the tropopause. The wind reaches maximum velocity in this region, reaching as high as 250 km per hour.

Long Answer Type Questions:

Q1.Write a detailed note on tornado.
Answer:
Meaning: A thunderstorm is a well- grown cumulonimbus cloud producing thunder and lightening. When the clouds extend to heights where sub-zero temperature prevails, hails are formed and they come down as hailstorm. If there is insufficient moisture, a thunderstorm can generate dust- storms. A thunderstorm is characterised by intense updraft of rising warm air, which causes the clouds to grow bigger and rise to greater height.

This causes precipitation. Later, downdraft brings down to earth the cool air and the rain. From severe thunderstorms sometimes spiralling wind descends like a trunk of an elephant with great force, with very low pressure at the centre, causing massive destruction on its way. Such a phenomenon is called a tornado. Features:

• Tornadoes generally occur in middle latitudes.
• The tornado over the sea is called water spouts.
• These violent storms are the manifestation of the atmosphere’s adjustments to varying energy distribution.
• The potential and heat energies are converted into kinetic energy in these storms and the restless atmosphere again returns to its stable state.

Q2.Write a detailed note on Coriolis Force.
Answer:
The rotation of the earth about its axis affects the direction of the wind. This force is called the Coriolis force after the French physicist who described it in 1844. It deflects the wind to the right direction in the northern hemisphere and to the left in the southern hemisphere. The deflection is more when the wind velocity is high. The Coriolis force is directly proportional to the angle of latitude. It is maximum at the poles and is absent at the equator.

The Coriolis force acts perpendicular to the pressure gradient force. The pressure gradient force is perpendicular to an isobar. The higher the pressure gradient force, the more is the velocity of the wind and the larger is the deflection in the direction of wind. As a result of these two forces operating perpendicular to each other, in the low-pressure areas the wind blows around it. At the equator, the Coriolis force is zero and the wind blows perpendicular to the isobars. The low pressure gets filled instead of getting intensified.
Differentiate between vertical variation of pressure and horizontal distribution of atmospheric pressure on earth.

Vertical variation of pressure: In the lower atmosphere the pressure decreases rapidly with height. The decrease amounts to about 1 mb for each 10 m increase in elevation. It does not always decrease at the same rate. Horizontal Distribution of Pressure: Small differences in pressure are highly significant in terms of the wind direction and purposes of comparison. The sea level pressure distribution is shown on weather maps. Low- pressure system is enclosed by one or more isobars with the lowest pressure in the centre. High- pressure system is also enclosed by one or more isobars with the highest pressure in the centre.
The vertical pressure gradient force is much larger than that of the horizontal pressure gradient. But, it is generally balanced by a nearly equal but opposite gravitational force. Hence, we do not experience strong upward winds.

Q3.What factors affect direction and velocity of winds?
Answer:
Air is set in motion due to the differences in atmospheric pressure. The air in motion is called wind. The wind blows from high pressure to low pressure. The wind at the surface experiences friction. Following factors affect the direction and velocity of winds.
1. Pressure gradient force: The differences in atmospheric pressure produces a force. The rate of change of pressure with respect to distance is the pressure gradient.

2. Frictional force: It affects the speed of the wind. It is greatest at the surface and its influence generally extends upto an elevation of 1 – 3 km. Over the sea surface the friction is minimal.

3.Coriolis force: The rotation of the earth about its axis affects the direction of the wind. This force is called the Coriolis force after the French physicist who described it in 1844. In addition, rotation of the earth also affects the wind movement. The force exerted by the rotation of the earth is known as the Coriolis force.

4. Pressure and wind: The velocity and direction of the wind are the net result of the wind generating forces. The winds in the upper atmosphere, 2-3 km above the surface, are free from frictional effect of the surface and are controlled mainly by the pressure gradient and the Coriolis force. When isobars are straight and when there is no friction, the pressure gradient force is balanced by the Coriolis force and the resultant wind blows parallel to the isobar. This wind is known as the geostrophic wind.

Short Answer Type Questions:

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

Answer

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

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

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

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

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

Answer

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

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

Answer

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

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

Answer

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

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

Answer

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

Long Answer Type Questions:

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

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

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

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

(b) Convection:

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

(c) Advection:

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

Short Answer Type Questions:

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

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

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

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

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

Answer

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

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

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

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

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

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

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

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

Long Answer Type Questions :

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

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

1. Homosphere
2. Hetrosphere

1. Homosphere:

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

Troposphere:

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

Tropopause:

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

Stratosphere:

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

Mesosphere:

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

2. Hetrosphere:

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

Ionosphere

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

Exosphere

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

Short Answer Type Questions:

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

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

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

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

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

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

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

Answer

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

Answer

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

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

Answer

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

Long Answer Type Questions :

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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