Class 12 – Page 7 – Maharashtra Board Solutions

12th Chemistry Chapter 15 Exercise Introduction to Polymer Chemistry Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 15 Introduction to Polymer Chemistry Textbook Exercise Questions and Answers.

Introduction to Polymer Chemistry Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 15 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 15 Exercise Solutions

1. Choose the correct option from the given alternatives.

Question i.
Nylon fibers are …………………………………..
A. Semisynthetic fibres
B. Polyamide fibres
C. Polyester fibres
D. Cellulose fibres
Answer:
B. polyamide fibres

Question ii.
Which of the following is naturally occurring polymer?
A. Telfon
B. Polyethylene
C. PVC
D. Protein
Answer:
D. Protein

Question iii.
Silk is a kind of …………………………………. fibre
A. Semisynthetic
B. Synthetic
C. Animal
D. Vegetable
Answer:
C. an animal

Question iv.
Dacron is another name of …………………………………. .
A. Nylon 6
B. Orlon
C. Novolac
D. Terylene
Answer:
D. Terylene

Question v.
Which of the following is made up of polyamides?
A. Dacron
B. Rayon
C. Nylon
D. Jute
Answer:
C. Nylon

Question vi.
The number of carbon atoms present in the ring of ε – caprolactam is
A. Five
B. Two
C. Seven
D. Six
Answer:
D. Six

Question vii.
Terylene is …………………………………. .
A. Polyamide fibre
B. Polyester fibre
C. Vegetable fibre
D. Protein fibre
Answer:
B. Polyester fibre

Question viii.
PET is formed by …………………………………. .
A. Addition
B. Condensation
C. Alkylation
D. Hydration
Answer:
D. Hydration

Question ix.
Chemically pure cotton is …………………………………. .
A. Acetate rayon
B. Viscose rayon
C. Cellulose nitrate
D. Cellulose
Answer:
D. Cellulose

Question x.
Teflon is chemically inert, due to presence of …………………………………. .
A. C-H bond
B. C-F bond
C. H- bond
D. C=C bond
Answer:
A. C-H bond

2. Answer the following in one sentence each.

Question i.
Identify ‘A’ and ‘B’ in the following reaction …………………………………. .

Answer:

Question ii.
Complete the following statements
a. Caprolactam is used to prepare …………………………………. .
b. Novolak is a copolymer of …………………………………. and …………………………………. .
c. Terylene is ………………………………….. polymer of terephthalic acid and ethylene glycol.
d. Benzoyl peroxide used in addtion polymerisation acts as …………………………………. .
e. Polyethene consists of polymerised …………………………………. .
Answer:
a. Nylon-6
b. Phenol, formaldehyde
c. polyester
d. initiator (catalyst)
e. linear or branched-chain

Question iii.
Draw the flow chart diagram to show the classification of polymers based on type of polymerisation.
Answer:

Question iv.
Write examples of Addition polymers and condensation polymers.
Answer:
Addition polymers : Polyvinyl chloride, polythene
Condensation polymers : Bakelite, terylene, Nylon-66

Question v.
Name some chain-growth polymers.
Answer:
Chain growth polymers : Polythene, polyacrylonitrile and polyvinyl chloride.

Question vi.
Define the terms :
1) Monomer
2) Vulcanisation
3) Synthetic fibres
Answer:
1. Monomer is a small and simple molecule and has a capacity to form two chemical bonds with other monomers. Examples : Ethene, Propylene.
2. The process by which a network of cross-links is introduced into an elastomer is called vulcanisation or it can also be defined as the process of heating natural rubber with sulphur to increase the tensile strength, toughness and elasticity of natural rubber is known as vulcanization of rubber.
3. The man-made fibres prepared by polymerization of one monomer or copolymerization of two or more monomers are called synthetic fibres.

Question vii.
What type of intermolecular force leads to high-density polymer?
Answer:
High density polymers have low degree of branching along the hydrocarbon chain. The molecules are closely packed together during crystallization. This closer packing means that the van der Waals attraction between the chains are greater and so the plastic (high density polymer) is stronger and has a melting point.

Question viii.
Give one example each of copolymer and homopolymer.
Answer:
Homopolymer : PVC, Nylon-6
Copolymer : Terylene, Buna-S

Question ix.
Identify Thermoplastic and Thermosetting Plastics from the following …………………………………. .
1. PET
2. Urea-formaldehyde resin
3. Polythene
4. Phenol formaldehyde
Answer:
Thermoplastic plastics : PET, Polythene
Thermosetting plastics : Urea formaldehyde resin, Phenol formaldehyde

3. Answer the following.

Question i.
Write the names of classes of polymers formed according to intermolecular forces and describe briefly their structural characteristics.
Answer:
Molecular forces bind the polymer chains either by hydrogen bonds or Vander Waal’s forces. These forces are called intermolecular forces. On the basis of magnitude of intcrmolccular forces, polymers are further classified as ebstomers, fibres, thermoplastic polymers. thermosetting polymers.

(1) Elastomers: Weak van der Waals type of intermolecular forces of attraction between the polymer chains are observed in cbstomcrs. When polymer is stretched, the polymer chain stretches and when the strain is relieved the chain returns to its odginal position, Thus, polymer shows elasticity and is called elastomers. Elastomers, the elastic polymers, have weak van der Waals type of intermolecular forces which permit the polymer to be stretched. Lilastorners are soft and stretchy and used in making rubber bands. E.g.. neoprene, vulcanized rubber, buna.S, buna-N.

(2) Fibres : It consists of strong intermolecular forces of’ attraction due to hydrogen bonding and strong dipole-dipole forces. These polymers possess high tensile strength. Due to these strong intermolecular forces the fibres are crystalline in nature. They are used in textile industries, strung tyres. etc.. e.g., nylon, terylene.

(3) Thermoplastic polymers: These polymer possess moderately strong intermolecular forces of attraction between those of elastomers and fibres. These polymers arc called thermoplastic because they become soft on heating and hard on cooling. They are either linear or branched chain polymers. They can be remoulded and recycled. E.g. polyethenc, PVC, polystyrene.

(4) Thermosetting polymers: These polymers are cross linked or branched molecules and are rigid polymers. During their formation they have property of being shaped on heating. but they get hardened while hot. Once hardened these become infusible, cannot be softened by heating and therefore, cannot be remoulded and recycled.
This shows extensive cross linking by covalent bonds formed in the moulds during hardening/setting process while hot. E.g. Bakelite, urea formaldehyde resin.

Question ii.
Write reactions of formation of :
a. Nylon 6
b. Terylene
Answer:
Terylene is polyester fibre formed by the polymerization of terephthalic acid and ethylene glycol.

Terylene is obtained by condensation polymerization of ethylene glycol and terephthalic acid in presence of catalyst zinc acetate and antimony trioxide at high temperature.

Properties :

Terylene has relatively high melting point (265 °C)
It is resistant to chemicals and water.

Uses :

It is used for making wrinkle free fabrics by blending with cotton (terycot) and wool (terywool), and also as glass reinforcing materials in safety helmets.
PET is the most common thermoplastic which is another trade name of the polyester polyethylenetereph- thalate.
It is used for making many articles like bottles, jams, packaging containers.

Question iii.
Write the structure of natural rubber and neoprene rubber along with the name and structure of thier monomers.
Answer:

Question iv.
Name the polymer type in which the following linkage is present.
Answer:
The linkage is present in terylene or dacron polymer.

Question v.
Write the structural formula of the following synthetic rubbers :
a. SBR rubber
b. Buna-N rubber
c. Neoprene rubber
Answer:

Question vi.
Match the following pairs :
Name of polymer – Monomer
1. Teflon – a. CH₂= CH₂
2. PVC – b. CF₂ = CF₂
3. Polyester – c. CH₂= CHCl
4. Polythene – d. C₆H₅OH and HCHO
5. Bakelite – e. Dicarboxylic acid and polyhydoxyglycol
Answer:

Teflon – CF₂ = CF₂
PVC – CH₂ = CHCI
Polyester-Dicarboxylic acid and polyhydoxyglycol
Polythene – CH₂ = CH₂
Bakelite – C₆H₅OH and HCHO

Question vii.
Draw the structures of polymers formed from the following monomers
1. Adipic acid + Hexamethylenediamine
2. e – Aminocaproic acid + Glycine
Answer:

Question viii.
Name and draw the structure of the repeating unit in natural rubber.
Answer:
Repeating unit of natural rubber (Basic unit : isoprene)

Question ix.
Classify the following polymers as natural and synthetic polymers
a. Cellulose
b. Polystyrene
c. Terylene
d. Starch
e. Protein
f. Silicones
g. Orlon (Polyacrylonitrile)
h. Phenol-formaldehyde resins
Answer:

Natural Polymers	1. Cellulose 4. Starch 5. Protein
Synthetic Polymers	2. Polystyrene 3. Terylene 6. Silicones 7. Orion (Polyacrylonitrile) 8. phenol-formaldehyde resin

Question x.
What are synthetic resins? Name some natural and synthetic resins.
Answer:
Synthetic resins are artificially synthesised high molecular weight polymers. They are the basic raw material of plastic. The main properties of plastic depend on the synthetic resin it is made from.

Examples of natural resins : Rosin, Damar, Copal, Sandarac, Amber, Manila
Examples of synthetic resins : Polyester resin, Phenolic resin, Alkyl resin, Polycarbonate resin, Polyamide resin, Polyurethane resin, silicone resin, Epoxy resin, Acrylic resin.

Question xi.
Distinguish between thermosetting and thermoplastic resins. Write example of both the classes.
Answer:

Thermosetting resin	Thermoplastic resin
(1) They harden when heated. Once hardened it no longer melts.	(1) They soften when heated and harden again when cooled.
(2) They cannot be re-shaped.	(2) They can be reshaped
(3) They are strong, hard.	(3) They are weak, soft.
(4) Thermosetting resin show cross-linking. Examples : Melamine resin Epoxy resins, Bake-lite.	(4) Thermoplastic molecules do not cross link, hence are flexible. Examples : Polythene, polypropylene, nylon, polyester.

Question xii.
Write name and formula of raw material from which bakelite is made.
Answer:
The raw material or monomers used to prepare bakelite are o-hydroxymethyl phenol and formaldehyde (HCHO)

4. Attempt the following :

Question i.
Identify condensation polymers and addition polymers from the following.

Answer:

Question ii.
Write the chemical reactions involved in the manufacture of Nylon 6, 6
Answer:
Nylon-6, 6 is a linear polyamide polymer formed by the condensation polymerisation reaction. The monomers used in the preparation of Nylon-6, 6 are :
(1) Adipic acid : HOOC-(CH₂)₄-COOH
(2) Hexamethylene diamine : H₂N-(CH₂)₆-NH₂

When equimolar aqueous solutions of adipic acid and hexamethylene diamine are mixed and heated, there is neutralization to form a nylon salt. During polymerisation at 553 k nylon salt loses a water molecule to form nylon 6, 6 polymer. Both monomers (hexamethylene diamine and adipic acid) contain six carbon atoms each, hence the polymer is termed as Nylon-6,6.

Properties and uses : Nylon 6,6 is high molecular mass (12000 – 50000 u) linear condensation polymer. It possesses high tensile strength. It does not soak in water. It is used for making sheets, bristles for brushes, surgical sutures, textile fabrics, etc.

Question iii.
Explain the vulcanisation of rubber. Which vulcanizing agents are used for the following synthetic rubber.
a. Neoprene
b. Buna-N
Answer:
The process by which a network of cross links is introduced into an elastomer is called vulcanization.

Vulcanization enhances the properties of natural rubber like tensile strength, stiffness, elasticity, toughness etc. Sulphur forms cross links between polyisoprene chains which results in improved properties of rubber.

For neoprene vulcanizing agent is MgO.
For Buna-N vulcanizing agent is sulphur.

Question iv.
Write reactions involved in the formation of …………………………………. .
1) Teflon
2) Bakelite
Answer:
The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.

Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.

Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

2. Teflon is polytetrafluoroethylene. The monomer used in preparation of teflon is tetrafluoroethylene, (CF₂ = CF₂) which is a gas at room temperature. Tetrafluoroethylene is polymerized by using free radical initiators such as hydrogen peroxide or ammonium persulphate at high pressure.

Properties:

Telflon is tough, chemically inert and resistant to heat and attack by corrosive reagents.
C – F bond is very difficult to break and remains unaffected by corrosive alkali, organic solvents.
Uses : Telflon is used in making non-stick cookware, oil seals, gaskets, etc.

Question v.
What is meant by LDP and HDP? Mention the basic difference between the same with suitable examples.
Answer:

LDP is low density polyethylene and HDP is high density polyethylene.
LDP is a branched polymer with low density due to chains are loosely held and HDP is a linear polymer with density due to close packing.
HDP is much stiffer than LDP and has high tensile strength and hardness.

LDP is mainly used in preparation of pipes for agriculture, irrigation and domestic water line connections. HDP is used in manufacture of toys and other household articles like bucket, bottles, etc.

Question vi.
Write preparation, properties and uses of Teflon.
Answer:
Teflon is polytetrafluoroethylene. The monomer used in preparation of teflon is tetrafluoroethylene, (CF₂ = CF₂) which is a gas at room temperature. Tetrafluoroethylene is polymerized by using free radical initiators such as hydrogen peroxide or ammonium persulphate at high pressure.

Properties:

Telflon is tough, chemically inert and resistant to heat and attack by corrosive reagents.
C – F bond is very difficult to break and remains unaffected by corrosive alkali, organic solvents.
Uses : Telflon is used in making non-stick cookware, oil seals, gaskets, etc.

Question vii.
Classify the following polymers as straight-chain, branched-chain and cross-linked polymers.

Answer:

5. Answer the following.

Question i.
How is polythene manufactured? Give their properties and uses.
Answer:
LDP means low density polyethylene. LDP is obtained by polymerization of ethylene under high pressure (1000 – 2000 atm) and temperature (350 – 570 K) in presence of traces of O₂ or peroxide as initiator.

The mechanism of this reaction involves free radical addition and H-atom abstraction. The latter results in branching. As a result the chains are loosely held and the polymer has low density.

Properties of LDP :

LDP films are extremely flexible, but tough chemically inert and moisture resistant.
It is poor conductor of electricity with melting point 110 °C.

Uses of LDP :

LDP is mainly used in preparation of pipes for agriculture, irrigation, domestic water line connections as well as insulation to electric cables.
It is also used in submarine cable insulation.
It is used in producing extruded films, sheets, mainly for packaging and household uses like in preparation of squeeze bottles, attractive containers, etc.

HDP means high density polyethylene. It is a linear polymer with high density due to close packing.

HDP is obtained by polymerization of ethene in presence of Zieglar-Natta catalyst which is a combination of triethyl aluminium with titanium tetrachloride at a temperature of 333K to 343K and a pressure of 6-7 atm.

Properties of HDP :

HDP is crystalline, melting point in the range of 144 – 150 °C.
It is much stiffer than LDP and has high tensile strength and hardness.
It is more resistant to chemicals than LDP.

Uses of HDP :

HDP is used in manufacture of toys and other household articles like buckets, dustbins, bottles, pipes, etc.
It is used to prepare laboratory wares and other objects where high tensile strength and stiffness is required.

Question ii.
Is synthetic rubber better than natural rubber? If so, in what respect?
Answer:
Yes. Synthetic rubber is more resistant to abrasion than natural rubber and is also superior in resistance to heat and the effects of aging (lasts longer). Many types of synthetic rubber are flame-resistant, so they can be used in insulation for electrical devices.

It also remains flexible at low temperatures and is resistant to grease and oil. It is resistant to heat, light and certain chemicals.

Question iii.
Write main specialities of Buna-S, Neoprene rubber?
Answer:
Buna-S is an elastomer and it is copolymer of styrene with butadiene. Its trade name is SBR. Buna-S is superior to natural rubber, because of its mechanical strength and abrasion resistance. It is used in tyre industry. It is vulcanized with sulphur. Neoprene is a synthetic rubber and it is a condensation polymer of chloroprene (2-chloro-l, 3-butadiene). Vulcanization of neoprene takes place in presence of MgO. It is resistant to petroleum, vegetable oils. Neoprene is used in making hose pipes for transport of gasoline and making gaskets.

Question iv.
Write the structure of isoprene and the polymer obtained from it.
Answer:

Question v.
Explain in detail the free radical mechanism involved during the preparation of the addition polymer.
Answer:
Polymerisation of ethylene is carried out at high temperature and at high pressure in presence of small amount of acetyl peroxide as initiator.

(1) Formation of free radicals : The first step involves clevage of acetyl peroxide to form two carboxy radicals. These carboxy radicals immediately undergo decarboxylation to give methyl initiator free radicals.

(2) Chain initiating step : The methyl radical thus formed adds to ethylene to form a new larger free radical.

(3) Chain propagation step : The larger free radical formed in the chain initiating step reacts with another molecule of ethene to form another big size free radicals and chain grows. This is called chain propagation step.

The chain reaction continues till thousands of ethylene molecules are added.

(4) Chain terminating step : The continuous chain reaction can be terminated by the combination of free radicals to form polyethene.

Activity :
i. Collect the information of the process like extrusion and moulding in Textile Industries.
ii. Make a list of polymers used to make the following articles
a. Photographic film
b. Frames of spectacles
c. Fountain pens
d. Moulded plastic chains
e. Terywool or Terycot fabric
iii. Prepare a report on factors responsible for degradation of polymers giving suitable example.
iv. Search and make a chart/note on silicones with reference to monomers, structure, properties and uses.
v. Collect the information and data about Rubber industry, plastic industry and synthetic fibre (rayon) industries running in India.

12th Chemistry Digest Chapter 15 Introduction to Polymer Chemistry Intext Questions and Answers

Use your brain power! (Textbook Page No 323)

Question 1.
Differentiate between natural and synthetic polymers.
Answer:

Natural polymers	Synthetic polymers
(1) The polymers are obtained either from plants or animals.	(1) The man made fibres prepared by polymerization of monomer or copolymerization of two or more monomers.
(2) They are further divided into two types : (i) plant polymers (ii) Animal polymers.Examples: Cotton, linen, latex	(2) They are further divided into three subtypes : (i) fibres (ii) synthetic rubbers (iii) plastics.Examples : Nylon, terylene Buna-S

Use your brain power! (Textbook Page No 325)

Question 1.
What is the type of polymerization in the following examples?

Answer:
(i) Addition polymerization
(ii) Condensation polymerization

Problem 15.1 : (Textbook Page No 326)

Question 1.
Refer to the following table listing for different polymers formed from respective monomers. Identify from the list whether it is copolymer or homopolymer.

Monomer	Polymers
Ethylene	Polyethene
Vinyl chloride	Polyvinyl chloride
Isobutylene	Polyisobutylene
Acrylonitrile	Polyacrylonitrile
Caprolactam	Nylon 6
Hexamethylene diammonium adipate	Nylon 6, 6
Butadiene + styrene	Buna-S

Solution :
In each of first five cases, there is only one monomer which gives corresponding homopolymer. In the sixth case hexamethylene diamine reacts with adipic acid to form the salt hexamethylene diammonium adipate which undergoes condensation to form Nylon 6, 6. Hence nylon 6, 6 is homopolymer. The polymer Buna-S is formed by polymerization of the monomers butadiene and styrene in presence of each other. The repeating units corresponding to the monomers butadiene and styrene are randomly arranged in the polymer. Hence Buna-S is copolymer.

Use your brain power! (Textbook Page No 328)

Question 1.
(1) From the cis-polyisoprene structure of natural rubber explain the low strength of van der Waals forces in it.
(2) Explain how the vulcanization of natural rubber improves its elasticity. (Hint : consider the intermolecular links.)
Answer:
(1) (i) Natural rubber is cis-polyisoprene. It is obtained by polymerization of isoprene units at 1, 4 positions. In rubber molecule, double bonds are located between C₂ and C₃ of each isoprene unit. These cis-double bonds do not allow the polymer chain to come closer. Therefore, only weak vander Waals’ forces are present. Since the chains are not linear, they can be stretched just like springs and exhibit elastic properties.

(ii) Cis-1, 4 polyisoprene (Natural rubber), due to this cis configuration about the double bonds, has the adjacent chain that do not fit together well (there is no close packing of adjacent chains). The only force that interact is the weak or low strength of van der Waals’ forces.

(iii) Cis-polyisoprene has a coiled structure in which the various polymer chains are held together by weak van der Waals’ forces.

(2) (i) Vulcanization of rubber is a process of improvement of the rubber elasticity and strength by heating it in the presence of sulphur, which results in three dimensional cross-linking of the chain rubber molecules (polyisoprene) bonded to each other by sulphur atoms.

(ii) Vulcanisation makes rubber more elastic and more stiffer. On vulcanisation, sulphur forms cross links at the reactive sites of double bonds and thus rubber get stiffened.

(iii) The improved properties of vulcanised rubber are (i) high elasticity (ii) low water absorption tendency,

(iii) resistance to oxidation.

Use your brain power! (Textbook Page No 334)

Question 1.
Write structural formulae of styrene and polybutadiene.
Answer:

(1) Classify the following polymers as addition or condensation.
(i) PVC (ii) Polyamides
(iii) Polystyrene
(iv) Polycarbonates
(v) Novolac
Answer:
Addition polymers: PVC, Polystyrene
(ondensatlon polymers: Polyamides. Polycarbonates, Novolac

Question 2.
Completed the following table :
Answer:

Use your brain power! (Textbook Page No 335)

(1) Represent the copolymerization reaction between glycine and e aminocaproic acid to form the copolymer nylon 2-nylon 6.
(2) What is the origin of the numbers 2 and 6 in the name of this polymer?
Answer:
(1) It is a copolymer and has polyamide linkages. The monomers glycine and e-amino caproic acid undergo condensation polymerisation to form nylon-2-nylon-6.

Nylon-2-nylon-6 is used in orthopaedic devices and implants.

(2) Monomer glycine contains two carbon atoms and e amino caproic acid contains six carbon atoms, hence the polymer is termed as nylon-2-nylon-6.

12th Std Chemistry Questions And Answers:

12th Biology Chapter 12 Exercise Biotechnology Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 12 Biotechnology Textbook Exercise Questions and Answers.

Biotechnology Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 12 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 12 Exercise Solutions

1. Multiple choice questions

Question 1.
MU The bacterium which causes a plant disease called crown gall is ………………..
(a) Helicobacter pylori
(b) Agrobacterium tumifaciens
(c) Thermophilus aquaticus
(d) Bacillus thuringienesis
Answer:
(b) Agrobacterium tumtfaciens

Question 2.
The enzyme nuclease hydrolyses ……………….. of polynucleotide chain of DNA.
(a) hydrogen bonds
(b) phosphodiester bonds
(c) glycosidic bonds
(d) peptide bonds
Answer:
(b) phosphodiester bonds

Question 3.
In vitro amplification of DNA or RNA segment is known as ………………..
(a) chromatography
(b) southern blotting
(c) polymerase chain reaction
(d) gel electrophoresis
Answer:
(c) polymerase chain reaction

Question 4.
Which of the following is the correct recognition sequence of restriction enzyme hind III.
(a) 5′ —A-A-G-C-T-T— 3′
3′ —T-T-C-G-A-A—5′
(b) 5′ — G-A-A-T-T-C—3′
3′ — C-T-T-A-A-G—5′
(c) 5′ — C-G-A-T-T-C—3′
3′ — G-C-T-A-A-G—5′
(d) 5′ — G-G-C-C—3′
3′ — C-C-G-G—5′
Answer:
(a) 5’ —A-A-G-C-T-T—3’
3’ —T-T-C-G-A-A—5’

Question 5.
Recombinant protein ……………….. is used to dissolve blood clots present in the body.
(a) insulin
(b) tissue plasminogen activator
(c) relaxin
(d) erythropoietin
Answer:
(b) tissue plasminogen activator

Question 6.
Recognition sequence of restriction enzymes are generally ……………….. nucleotide long.
(a) 2 to 4
(b) 4 to 8
(c) 8 to 10
(d) 14 to 18
Answer:
(b) 4 to 8

2. Very short answer questions

Question 1.
Name the vector which is used in production of human insulin through recombinant DNA technology.
Answer:
PBR 322

Question 2.
Which cells from Langerhans of pancreas do produce a peptide hormone insulin?
Answer:
cells of islets of Langerhans of a peptide hormone insulin.

Question 3.
Give the role of Ca⁺⁺ ions in the transfer of recombinant vector into bacterial host cell.
Answer:
Ca⁺⁺ ions promotes binding of plasmid DNA to lipo polysaccharides on bacterial cell surface. Then plasmid can enter the cell on heat shock.

Question 4.
Expand the following acronyms which are used in the held of biotechnology:

YAC
RE
dNTP
PCR
GMO
MAC
CCMB.

Answer:

YAC : Yeast Artificial chromosome
RE : Restriction Endonuclease
dNTP : Deoxyribonucleoside triphosphates
PCR : Polymerase Chain Reaction
GMO : Genetically Modified Organisms
MAC : Mammalian Artificial Chromosome
CCMB : Centre for Cellular and Molecular Biology

Question 5.
Fill in the blanks and complete the chart.

GMO	Purpose
(i) Bt cotton	———–
(ii) ———-	Delay the softening of tomato during ripening
(iii) Golden rice	———–
(iv) Holstein cow	———–

Answer:

GMO	Purpose
(i) Bt cotton	Insect resistance
(ii) Flavr savr Tomato	Delay the softening of tomato during ripening
(iii) Golden rice	Rich in vitamin A
(iv) Holstein cow	High milk productivity

3. Short answer type questions.

Question 1.
Explain the properties of a good or ideal cloning vector for r-DNA technology.
Answer:
Desired characteristics of ideal cloning vector are as follows:

Vector should be able to replicate independenly (through ori gene), so that as vector replicates, multiple copies of the DNA insert are also produced.
It should be able to easily transferred into host cells.
It should have suitable control elements like promoter, operator, ribosomal binding sites, etc.
It should have marker genes for antibiotic resistance and restriction enzyme recognition sites within them.

Question 2.
A PCR machine can rise temperature up to 100 °C but after that it is not able to lower the temperature below 70 °C automatically. Which step of PCR will be hampered first in this faulty machine? Explain why?
Answer:

If the faulty machine is not able to lower the temperature below 70 °C, then the primer annealing step will be hampered first.
Each primer has a specific annealing temperature, depending upon its A, T, G, C content.
For most of the primers annealing temperature is about 40-60 °C.
Hence, if temperature is more than primers annealing temperature, it will be able to pair with its complementary sequence in ssDNA.

Question 3.
In the process of r-DNA technology, if two separate restriction enzymes are used to cut vector and donor DNA then which problem will arise in the formation of r-DNA or chimeric DNA? Explain.
Answer:
In the process of r-DNA technology, if two separate restriction enzymes are used to cut vector and donor DNA, then it will result in fragments with different sticky ends which will not be complementary to each other.

Question 4.

Recombinent protein	Its use in or for
(1) Platelet derived growth factor	(a) Anemia
(2) a-antitrypsin	(b) Cystic fibrosis
(3) Relaxin	(c) Haemophilia A
(4) Eryhthropoietin	(d) Diabetes
(5) Factor VIII	(e) Emphysema
(6) DNA ase	(f) Parturition
	(g) Atherosclerosis

Answer:

Recombinent protein	Its use in or for
(1) Platelet derived growth factor	(g) Atherosclerosis
(2) a-antitrypsin	(e) Emphysema
(3) Relaxin	(f) Parturition
(4) Eryhthropoietin	(a) Anemia
(5) Factor VIII	(c) Haemophilia A
(6) DNA ase	(b) Cystic fibrosis

4. Long answer type questions.

Question 1.
(i) Define and explain the terms Bioethics.
Answer:

Bioethics is the study of moral vision, decision and policies of human behaviour in relation to biological phenomena or events.
Bioethics deals with wide range of reactions on new developments like cloning, transgenic, gene therapy, eugenics, r-DNA technology, in vitro fertilization, sperm bank, gene therapy, euthanasia, death, maintaining those who are in comatose state, prenatal genetic selection, etc.
Bioethics also includes the discussion on subjects like what should and should not be done in using recombinant DNA techniques.

Ethical aspects pertaining to the use of biotechnology are:

Use of animals cause great sufferings to them.
Violation of integration of species caused due to transgenosis.
Transfer of human genes into animals and vice versa.
Indiscriminate use of biotechnology pose risk to the environment, health and biodiversity.
The effects of GMO on non-target organisms, insect resistance crops, gene flow, the loss of diversity.
Modification process disrupting the natural process of biological entities.

(ii) Define and explain the term Biopiracy.
Answer:

Biopiracy is defined as ‘theft of various natural products and then selling them by getting patent without giving any benefits or compensation back to the host country’.
It is unauthorized misappropriation of any biological resource and traditional knowledge.
It is bio-patenting of bio-resource or traditional knowledge of another nation without proper permission of the concerned nation or unlawful exploitation and use of bioresources without giving compensation.

Following are the examples of biopiracy:
(a) Patenting of Neem (Azadirachta indica):

Pirating India’s traditional knowledge about the properties and uses of neem, the USDA and an American MNC W.R. Grace sought a patent from the European Patent Office (EPO) on the “method for controlling on plants by the aid of hydrophobic extracted neem oil,” in the early 90s.
The patenting of the fungicidal properties of Neem, was an example of biopiracy.

(b) Patenting of Basmati:

Texmati is a trade name of “Basmati rice line and grains” for which Texas based American company Rice Tec Inc was awarded a patent by the US Patent and Trademark Office (USPTO) in 1997.
This is a case of biopiracy as Basmati is a long-grained, aromatic variety of rice indigenous to the Indian subcontinent.
Very broad claims about “Inventing” the said rice was the basis of patent application.
The UPSTO has rejected all the claims due to people movement against Rice Tec in March 2001.

A patent claim about the healing properties of Haldi was made by two American researchers of Indian origin of the University of Mississippi Medical Center, to the US Patent and Trademark Office.
They were granted a patent in March 1995.
This is an example of biopiracy because healing properties of Haldi is not a new discovery, but it is a traditional knowledge in ayurvedas for centuries.
The Council of Scientific and Industrial Research (CSIR) applied to the US Patent Office for a reexamination and they realized the mistake and cancelled the patent.

(iii) Define and explain the term Biopatent.
Answer:

Biopatent is a biological patent awarded for strains of microorganisms, cell lines, genetically modified strains, DNA sequences, biotechnological processes, product processes, product and product applications.
It allows the patent holder to exclude others from making, using, selling or importing protected invention for a limited period of time.
Duration of biopatentis five years from the date of the grant or seven years from the date of filing the patent application, whichever is less.
Awarding biopatents provides encouragement to innovations and promote development of scientific culture in society. It also emphasizes the role of biology in shaping human society.
First biopatent was awarded for genetically engineered bacterium ‘Pseudomonas’ used for clearing oils spills.
Patent jointly issued by Delta and Pineland company and US department of agriculture having title ‘control of plant gene expression’, is based on a gene that produces a protein toxic to plant and thus prevents seed germination.

This patent was not granted by Indian government. Such a patent is considered morally unacceptable and fundamentally unequitable. Such patents would pose a threat to global food security as financially powerful corporations would acquire monopoly over biotechnological process.

Question 2.
Explain the steps in process of r-DNA technology with suitable diagrams.
Answer:

The steps involved in gene cloning are as follows:
(1) Isolation of DNA (gene) from the donor organism:

To obtain the desired gene to be cloned, the cells of the donor organism are sheared with the blender and treated with suitable detergent. Genetic material is then isolated and purified.
Isolated purified DNA is then cleaved using restriction Endonucleases.
Restriction fragment containing desired gene is isolated and selected for cloning. This is now called foreign DNA or passanger DNA.
A desired gene can also be obtained directly from genomic library or c-DNA library.

(2) Insertion of desired foreign gene into a cloning vector (vehicle DNA):

The foreign DNA or passanger DNA is inserted into a cloning vector (vehicle DNA) like bacterial plasmids and the bacteriophages like lamda phage and M13. The most commonly used plasmid is pBR 322.
Plasmids are isolated from the bacteria and are cleaved by using same RE which is used in the isolation of the desired gene from the donor.
Enzyme DNA ligase is used to join foreign DNA and the plasmid DNA.
Plasmid DNA containing foreign DNA is called recombinant DNA (r-DNA) or chimeric DNA.

(3) Transfer of r-DNA into suitable competent host or cloning organism:

The r-DNA is introduced into a competent host cell, which is mostly a bacterium.
Host cell takes up naked r-DNA by process of ‘transformation’ and incorporates it into its own chromosomal DNA which finally expresses the trait controlled by passenger DNA.
The transfer of r-DNA into a bacterial cell is assisted by divalent Ca⁺⁺.
The cloning organisms are E.coli and Agrobacterium tumifaciens.
The competent host cells which have taken up r-DNA are called transformed cells.
By using techniques like electroporation, microinjection, lipofection, shot gun, ultrasonification, biolistic method, etc. Foreign DNA can also be transferred directly into the naked cell or protoplast of the competent host cell, without using vector.
In plant biotechnology the transformation is through Ti plasmids of A. tumifaciens.

(4) Selection of the transformed host cell:

For isolation of recombinant cell from non-recombinant cell, marker gene of plasmid vector is employed.
For example, pBR322 plasmid vector contains different marker genes like ampicillin resistant gene and tetracycline resistant gene. When pstl RE is used, it knocks out ampicillin resistant gene from the plasmid, so that the recombinant cells become sensitive to ampicillin.

(5) Multiplication of transformed host cell:

The transformed host cells are introduced into fresh culture media where they divide.
The recombinant DNA carried by them also multiplies.

(6) Expression of gene to obtain desired product. Then desired products like enzymes, antibiotiocs etc. separated and purified through down stream processing using bioreactors.

Question 3.
Explain the gene therapy. Give two types of it.
Answer:
Gene therapy is the treatment of genetic disorders by replacing, altering or supplementing a gene that is absent or abnormal and whose absence or abnormality is responsible for the disease.
Types of gene therapy:
(a) Germ line gene therapy:

In this germ cells are modified genetically to correct a genetic defect.
Normal gene is introduced into germ cells like sperms, eggs, early embryos.
It allows transmission of the modified genetic information to the next generation.
Although it is highly effective in treatment of the genetic disorders, its use is not preferred in human beings because of various technical and ethical reasons.

(b) Somatic cell gene therapy:

In this somatic cells are modified genetically to correct a genetic defect.
Healthy genes are introduced in somatic cells like bone marrow cells, hepatic cells, fibroblasts endothelium and pulmonary epithelial cells, central nervous system, endocrine cells and smooth muscle cells of blood vessel walls.
Modification of somatic cells only affects the person being treated and the modified chromosomes cannot be passed on the future generations.
Somatic cell gene therapy is the only feasible option and the clinical trials have already employed for the treatment of disorders like cancer, rheumatoid arthritis, SCID, Gaucher’s disease, familial hypercholesterolemia, haemophilia, phenylketonuria, cystic fibrosis, sickle-cell anaemia, Duchenne muscular dystrophy, emphysema, thalassemia, etc.

Question 4.
How are the transgenic mice used in cancer research?
Answer:

Transgenic mice are used in various research areas of cancer research.
Transgenic mice containing a particular oncogene (cancer causing gene) develop specific cancer.
They are used to study the relationship between oncogenes and cancer development, cancer treatment and prevention of malignancy.
The transgenic mouse model for the investigation of the breast cancer was developed in the laboratory of Philip Leder in Harvard (USA).
Transgenic mice containing oncogenes myc and ras were analyzed to find out role of these genes in the development of breast cancer.

Question 5.
Give the steps in PCR or polymerase chain reaction with suitable diagrams.
Answer:

(1) The DNA segment and excess of two primer molecules, four types of dNTPs, the thermostable DNA polymerase are mixed together in ‘eppendorf tube’.

(2) One PCR cycle is of 3-4 minutes duration and it involves following steps:

Denaturation : The reaction mixture is heated at 90-98°C. Due to this hydrogen bonds in the DNA break and two strands of DNA separate. This is called denaturation.
Annealing of primer : When the reaction mixture is cooled to 40-60°C, the primer pairs with its complementary sequences in ssDNA. This is called annealing.
Extension of primer : In this step, the temperature is increased to 70-75°C. At this temperature thermostable Taq DNA polymerase adds nucleotides to 3’end of primer using single-stranded DNA as template. This is called primer extension. Duration of this step is about two minutes.

(3) In an automatic thermal cycler, the above three steps are automatically repeated 20-30 times.
(4) Thus, at the end of ‘n’ cycles 2n copies of DNA segments, get synthesized.

Question 6.
What is a vaccine? Give advantages of oral vaccines or edible vaccines.
Answer:

A vaccine is a biological preparation that provides active acquired immunity against a certain disease.
Vaccine is often made from a weakened or killed form of the microorganism, its toxins or one of its surface protein antigens.
Edible vaccine is an edible plant part engineered to produce an immunogenic protein, which when consumed gets recognized by immune system.
Immunogenic protein of certain pathogens are active when’administered orally.
When animals or mainly humans consume these plant parts, they get vaccinated against certain pathogen.
Oral or edible vaccines have low cost, they are easy to administer and store.

Question 7.
Enlist different types of restriction enzymes commonly used in r-DNA technology? Write on their role.
Answer:

Different restriction enzymes commonly used in r-DNA technology are Alu I, Bam HI, Eco RI, Hind II, Hind III, Pst I, Sal I, Taq I, Mbo II, Hpa I, Bgl I, Not I, Kpn I, etc.
They are the molecular scissors which recognize and cut the phosphodiester back bone of DNA on both strands, at highly specific sequences.
The sites recognized by them are called recognition sequences or recognition sites.
Different restriction enzymes found in different organisms recognize different nucleotide sequences and therefore cut DNA at different sites.
Restriction cutting may result in DNA fragments with blunt ends or cohesive or sticky ends or staggered ends (having short, single stranded projections).
Restriction endonucleases like Bam HI and EcoRI produce fragments with sticky ends.
Restriction endonucleases like Alu I, Hind III produce fragments with blunt ends.
Type I restriction endonucleases fuction simultaneously as endonuclease and methylase e.g. EcoK.
Type II restriction endonucleases have separate cleaving and methylation activities. They are more stable and are used in r-DNA technology e.g. EcoRI, Bgll. They cut DNA at specific sites within the palindrome.
Type III restriction endonucleases cut DNA at specific non-palindromic sequences e.g. Hpal, MboII.
In bacterial cells, REs destroy various viral DNAs that might enter the cell, thus restricting the potential growth of the virus.

Question 8.
Enlist and write in brief about the different biological tools required in r-DNA technology.
Answer:
The biological tools used in r-DNA technology are various enzymes, cloning vectors and competent hosts.
(1) Enzymes:

Enzymes like lysozymes, nucleases (exonucleases and endonucleases), DNA ligase, reverse transcriptase, DNA polymerase, alkaline phosphatases, etc. are used in r-DNA technology.
The restriction endonucleases are used as biological or molecular scissors. They are able to cut a DNA molecule at a specific recognition site.

(2) Vectors:

Vectors are DNA molecules which carry foreign DNA segment and replicate inside the host cell.
Vectors may be plasmids, bacteriophages (M13, lambda virus), cosmid, phagemids, BAC (bacterial artificial chromosome), YAC (yeast artificial chromosome), transposons, baculoviruses and mammalian artificial chromosomes (MACs).
Most commonly used vectors are plasmid vectors (pBR 322, pUC, Ti plasmid) and bacteriophages (lamda phage, M13 phage).

(3) Competent host cells:

They are bacteria like Bacillus haemophilus, Helicobacter pyroliand E. coli.
Mostly E. coli is used for the transformation with recombinant DNA.

12th Std Biology Questions And Answers:

12th Biology Chapter 4 Exercise Molecular Basis of Inheritance Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 4 Molecular Basis of Inheritance Textbook Exercise Questions and Answers.

Molecular Basis of Inheritance Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 4 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 4 Exercise Solutions

1. Multiple Choice Questions

Question 1.
Griffith worked on ………………..
(a) Bacteriophage
(b) Drosophila
(c) Frog eggs
(d) Streptococci
Answer:
(d) Streptococci

Question 2.
The molecular knives of DNA are ………………..
(a) Ligases
(b) Polymerases
(c) Endonucleases
(d) Transcriptase
Answer:
(c) Endonucleases

Question 3.
Translation occurs in the ………………..
(a) Nucleus
(b) Cytoplasm
(c) Nucleolus
(d) Lysosomes
Answer:
(b) Cytoplasm

Question 4.
The enzyme required for transcription is ………………..
(a) DNA polymerase
(b) RNApolymerase
(c) Restriction enzyme
(d) RNase
Answer:
(b) RNA polymerase

Question 5.
Transcription is the transfer of genetic information from ………………..
(a) DNA to RNA
(b) t-RNA to m-RNA
(c) DNA to m-RNA
(d) m-RNA to t-RNA
Answer:
(a) DNA to RNA

Question 6.
Which of the following is NOT part of protein synthesis?
(a) Replication
(b) Translation
(c) Transcription
(d) All of these
Answer:
(a) Replication

Question 7.
In the RNA molecule, which nitrogen base is found in place of thymine?
(a) Guanine
(b) Cytosine
(c) Thymine
(d) Uracil
Answer:
(d) Uracil

Question 8.
How many codons are needed to specify three amino acids?
(a) 3
(b) 6
(c) 9
(d) 12
Answer:
(a) 3

Question 9.
Which out of the following is NOT an example of inducible operon?
(a) Lactose operon
(b) Histidine operon
(c) Arabinose operon
(d) Tryptophan operon
Answer:
(d) Tryptophan operon

Question 10.
Place the following event of translation in the correct sequence ………………..
i. Binding of met-t-RNA to the start codon.
ii. Covalent bonding between two amino acids.
iii. Binding of second t-RNA.
iv. Joining of small and large ribosome subunits.
(a) iii, iv, i, ii
(b) i, iv, iii, ii
(c) iv, iii, ii, i
(d) ii, iii, iv, i
Answer:
(b) i, iv, iii, ii

2. Very Short Answer Questions

Question 1.
What is the function of an RNA primer during protein synthesis?
Answer:
During DNA replication, RNA primer provides 3’ OH to which DNA polymerase enzyme can add nucleotides to synthesize new strand using parental strand of DNA as template.
[Note : RNA primer has no direct role in protein synthesis.]

Question 2.
Why is the genetic code considered as commaless?
Answer:
The triplet codon are arranged one after the other on m-RNA molecule without any gap or space and therefore genetic code is considered as commaless.

Question 3
Genome
Answer:
Genome is the total genetic constitution of an organism or a complete copy of genetic information (DNA) or one complete set of chromosomes (monoploid or haploid) of an organism.

Question 4.
Which enzyme does remove supercoils from replicating DNA?
Answer:
Super-helix relaxing enzyme (Topoisomerase) removes supercoils from replicating DNA.

Question 5.
Why are Okazaki fragments formed on lagging strand only?
Answer:
Okazaki fragments are formed only on lagging template as only short stretch of lagging template becomes available for replication at one time.

Question 6.
When does DNA replication take place?
Answer:
In eukaryotes DNA-replication takes place during S-phase of interphase of cell cycle and in prokaryotes. DNA replicates prior to cell division.

Question 7.
Define term Codogen and Codon
Answer:
Codogen is a triplet of nucleotides present on the DNA which specifies one particular amino acid.
Codon is a triplet of nucleotides present on the m-RNA which specifies one particular amino acid.

Question 8.
What is degeneracy of genetic code?
Answer:
Genetic code is degenerate as 61 codons code for 20 amino acids, that is two or more codons can specify the same amino acid. E.g. Cysteine has two codons, while isoleucine has three codons.

Question 9.
Which are the nucleosomal ‘core’ histones?
Answer:
Two molecules each of histone proteins, viz. H₂A. H₂B, H₃ and H₄ are the nucleosomal ‘core’ histones.

3. Short Answer Questions

Question 1.
DNA packaging in eukaryotic cell.
Answer:

In eukaryotic cells, DNA (2.2 metres) is condensed, coiled and supercoiled to be packaged efficiently in the nucleus (10^-16 m).
DNA is associated with histone and non-histone proteins.
Histones are a set of positively charged, basic proteins, rich in basic amino acid residues lysine and arginine.
Nucleosome consists of nucleosome core (two molecules of each of histone proteins viz. H₂A, H₂B, H₃ and H₄ forming histone octamer) and negatively charged DNA (146 bps) that wraps around the histone octamer by 1 3/4 turns.
H₁ protein binds the DNA thread where it enters and leaves the nucleosome.
Adjacent nucleosomes are linked with linker DNA (varies in length from 8 to 114 bp, average length of linker DNA is about 54 bp).
Each nucleosome contains 200 bp of DNA.
Packaging involves formation of – Beads on string (10 nm diameter), Solenoid fibre (looks like coiled telephone wire, 30 nm diameter/300Å), Chromatin fibre and Chromosome.
Non-Histone Chromosomal Proteins (NHC) contribute to the packaging of chromatin at a higher level.

Question 2.
Enlist the characteristics of genetic code.
Answer:
The characteristics of genetic code are

Genetic code is triplet, commaless and non-overlapping.
It is degenerate and non-ambiguous.
It is universal
It has polarity.

Question 3.
Applications of DNA fingerprinting.
Answer:
Applications of DNA fingerprinting are as follows:

In forensic science to solve rape and murder cases.
Finds out the biological father or mother or both, of the child, in case of disputed parentage.
Used in pedigree analysis in cats, dogs, horses and humans.

Question 4.
Explain the role of lactose in ‘Lac Operon’.
Answer:

A small amount of beta-galactoside permease enzyme is present in cell even when Lac operon is switched off and it allows a few molecules of lactose to enter into the cell.
Lactose binds to repressor and inactivates it.
Repressor – lactose complex cannot bind with the operator gene, which is then turned on.
RNA polymerase transcribes all the structural genes to produce lac m-RNA which is then translated to produce all enzymes.
Thus, lactose acts as an inducer.
When the inducer level falls, the operator is blocked again by repressor and structural genes are repressed again. This is negative feedback.

4. Short Answer Questions

Question 1.
Human genome project.
Answer:
1. Human Genome Project (HGP) was initiated in 1990 under the International administration of the Human Genome Organization (HUGO) and it was completed r in 2003.

2. The main aims:

To sequence 3 billion base pairs of DNA in human genome and to map an estimated 33000 genes.
To store the information collected from the project in databases.
To develop tools and techniques for analysis of the data.
Transfer of the related technologies to the private sectors, such as industries.
Taking care of the legal, ethical and social issues which may arise from project.
To sequence the genomes of several other organisms such as bacteria e.g. E.coli, Caenorhabditis elegans, Saccharomyces cerevisiae, Drosophil, rice, Arabidopsis), Mus musculus, etc.

3. Significance:

HGP has a major impact in the fields like Medicine, Biotechnology, Bioinformatics and the Life sciences.
More understanding of functions of genes, proteins and human evolution.

Question 2.
Describe the structure of operon.
Answer:

An operon is a unit of gene expression and regulation.
It includes the structural genes and their control elements. Control elements are promoters and operators.
The structural genes code for proteins, r-RNA and t-RNA that are necessary for all the cells.
Promoters are signal sequences in DNA. They start the RNA synthesis. They also act as sites where the RNA polymerases are bound during transcription.
Operators are present between the promoters and structural genes.
There is repressor protein that binds to the operator region of the operon.
There are regulatory genes which are responsible for the formation of repressors which interact with operators.

Question 3.
In the figure below A, B and C are three types of

Answer:
Answer: A, B and C are A : m-RNA, B : r-RNA, C : t-RNA

Question 4.
Identify the labelled structures on the following diagram of translation.

Part A is the ………………
Part B is the ………………
Part C is the ………………
Answer:
Part A is the anti-codon.
Part B is the amino acid.
Part C is the larger subunit of ribosome.

Question 5.
Match the entries in Column I with those of Column II and choose the correct answer.

Column I	Column II
A. Alkali treatment	i. Separation of DNA fragments on gel slab
B. Southern blotting	ii. Splits DNA fragments into single strands
C. Electrophoresis	iii. DNA transferred to nitrocellulose sheet
D. PCR	iv. X-ray photography
E. Autoradiography	v. Produce fragments different sizes
F. DNA treated with REN	vi. DNA amplification

Answer:

Column I	Column II
A. Alkali treatment	ii. Splits DNA fragments into single strands
B. Southern blotting	iii. DNA transferred to nitrocellulose sheet
C. Electrophoresis	i. Separation of DNA fragments on gel slab
D. PCR	vi. DNA amplification
E. Autoradiography	iv. X-ray photography
F. DNA treated with REN	v. Produce fragments different sizes

5. Long Answer Questions

Question 1.
Explain the process of DNA replication.
Answer:

DNA replication is semi-conservative replication. It involves following steps:
Activation of Nucleotides:

Nucleotides (dAMP dGMR dCMP and dTMP) present in the nucleoplasm, are activated by ATP in presence of an enzyme phosphorylase.
This phosphorylation results in the formation of deoxyribonucleotide triphosphates i.e. dATE dGTR dCTP and dTTE

Point of Origin or Initiation point:

Replication begins at specific point ‘O- Origin and terminates at point ‘T’.
At the point ‘O’, enzyme endonuclease nicks (breaks the sugar-phosphate backbone or the phosphodiester bond) one of the strands of DNA, temporarily.

Unwinding of DNA molecule:

Enzyme DNA helices breaks weak hydrogen bonds in the vicinity of ‘O’.
The strands of DNA separate and unwind. This unwinding is bidirectional.
SSBP (Single strand binding proteins) remains attached to both the separated strands and prevent them from recoiling (rejoining).

Replicating fork:

Y-shape replication fork is formed due to unwinding and separation of two strands.
The unwinding of strands results in strain which is released by super-helix relaxing enzyme.

Synthesis of new strands:

Each separated strand acts as a template for the synthesis of new complementary strand.
A small RNA primer (synthesized by activity of enzyme RNA primase) get attached to the 3′ end of template strand and attracts complementary nucleotides from surrounding nucleoplasm.
These nucleotides bind to the complementary nucleotides on the template strand by hydrogen bonds (i.e. A = T or T = A; G = C or C = G, CEG).
The phosphodiester bonds are formed between nucleotides of new strand to form a polynucleotide strand.
The enzyme DNA polymerase catalyses synthesis of new complementary strand always in 5′ – 3′ direction.

Leading and Lagging strand:

The template strand with free 3′ is called the leading template.
The template strand with free 5′ end is called the lagging template.
The replication always starts at C-3 end of template strand and proceeds towards C-5 end.
New strands are always formed in 5′ → 3′ direction.
The new strand which develops continuously towards replicating fork is called the leading strand.
The new strand which develops discontinuously away from the replicating fork is called the lagging strand.
Maturation of Okazaki fragments : The lagging strand is synthesized in the form of small Okazaki fragments which are joined by enzyme DNA ligase.
Later RNA primers are removed by the combined action of RNase H, an enzyme that degrades the RNA strand of RNA-DNA hybrids, and polymerase I.
Gaps formed are filled by complementary DNA sequence with the help of DNA polymerase-I in prokaryotes and DNA polymerase-a in eukaryotes.
Finally, DNA gyrase (topoisomerase) enzyme forms double helix to form daughter DNA molecules.

Formation of two daughter DNA molecules:

In each daughter DNA molecule, one strand is parental and the other one is newly synthesized.
Thus, 50% part (i.e. one strand of the helix) is contributed by mother DNA. Hence, it is described as semiconservative replication.

Question 2.
Describe the process of transcription in protein synthesis.
Answer:

Transcription involves three stages, viz. Initiation, Elongation and Termination.
(1) Initiation:

RNA polymerase binds to promoter site.
It then moves along the DNA and causes local unwinding of DNA duplex into two strands in the region of the gene.
Only antisense strand functions as template.

(2) Elongation:

The complementary ribonucleoside tri-phosphates get attached to exposed bases of DNA template chain.
As transcription proceeds, the hybrid DNA-RNA molecule dissociates and makes m-RNA molecule free.
As the m-RNA grows, the transcribed region of DNA molecule becomes spirally coiled and regains double helical form.

(3) Termination:
When RNA polymerase reaches the terminator site on the DNA, both enzyme and newly formed m-RNA (primary transcript) gets released.

Question 3.
Describe the process of translation in protein synthesis.
Answer:

Translation involves the following steps:
1. Activation of amino acids and formation of charged t-RNA (t-RNA – amino acid complex):
i. In the presence of an enzyme amino acyl t-RNA synthetase, the amino acid is activated and then attached to the specific t-RNA molecule at 3’ end to form charged t-RNA (t-RNA – amino acid complex).

ii. ATP is essential for the reaction.

2. Initiation of Polypeptide chain:

Small subunit of ribosome binds to the m-RNA at 5’ end.
Start codon is positioned properly at P-site.
Initiator t-RNA, (carrying amino acid methionine in eukaryotes or formyl methionine in prokaryotes) binds with initiation codon (AUG) of m-RNA, by its anticodon (UAC) through hydrogen bonds.
The large subunit of ribosome joins with the smaller subunit in the presence of Mg⁺⁺.
Thus, initiator charged t-RNA occupies the P-site and A – site is vacant.

3. Elongations of polypeptide chain:
Addition of amino acid occurs in 3 Step cycle-
i. Codon recognition.
Anticodon of second (and subsequent) amino acyl t-RNA molecule recognizes and binds with codon at A-site by hydrogen bonds.

ii. Peptide bond formation.

Ribozyme catalyzes the peptide bond formation between amino acids on the initiator t-RNA at P-site and t-RNA at A-site.
It takes less than 0.1 second for formation of peptide bond.
Initiator t-RNA at ‘P’ site is then released from E-site.

iii. Translocation.

Translocation is the process in which sequence of codons on m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.
Due to this A’-site becomes vacant to receive next charged t-RNA molecule.
The events like arrival of t-RNA – amino acid complex, formation of peptide bond, ribosomal translocation and release of previous t-RNA, are repeated.
As ribosome move over the m-RNA, all the codons on m-RNA are exposed oiie by one for translation.

4. Termination and release of polypeptide:
When stop codon (UAA, UAG, UGA) gets exposed at the A-site, the release factor binds to the stop codon, thereby terminating the translation process
The polypeptide gets released in the cytoplasm.
Two subunits of ribosome dissociate and last t-RNA and m-RNA are released in the cytoplasm.
m-RNA gets denatured by nucleases immediately.

Question 4.
Describe Lac ‘Operon’.
Answer:

Lac operon consists of the following components:
(1) Regulator gene:

Regulator gene precedes the promoter gene.
It may not be present immediately adjacent to operator gene.
Regulator gene codes for a repressor protein which binds with operator gene and represses (stops) its action.

(2) Promoter gene:

It precedes the operator gene.
It is present adjacent to operator gene.
RNA Polymerase enzyme binds at promoter site.
Promoter gene base sequence determines which strand of DNA acts a template.

(3) Operator gene:

It precedes the structural genes.
When operator gene is turned on by an inducer, the structural genes get transcribed to form m-RNA.

(4) Structural gene:

There are 3 structural genes in the sequence lac-Z, lac-Y and lac-A.
Enzymes produced are β-galactosidase, β-galactoside permease and transacetylase respectively.
Inducer Allolactose acts as an inducer. It inactivates the repressor by binding with it.

Question 5.
Justify the statements. If the answer is false, change the underlined word(s) to make the statement true.
(i) The DNA molecule is double stranded and the RNA molecule is single stranded.
Answer:

DNA as the genetic material has to be chemically and structurally stable.
It should be able to generate its replica.
Sugar-phosphate backbone and complementary base pairing between the two strands, give stability to DNA.
Both the strands of DNA act as template for synthesis of their complementary strands. This allows accurate replication of DNA.
Single stranded RNA can be folded to form complex structures and perform specific functions such as synthesis of proteins.

(ii) The process of translation occurs at the ribosome.
Answer:

Translation is the process in which sequence of codons of m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.
Ribosome has one binding site for m-RNA. It orients m-RNA molecule in such a way that all the codons are properly read.
Ribosome has three binding sites for t-RNA : P-site (peptidyl t-RNA-site), A-site (aminoacyl t-RNA-site) and E-site (exit site).
t-RNAs place the required amino acids in correct sequence and translate the coded message of RNA.
In eukaryotes, a groove which is present between two subunits of ribosomes, protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.
Thus ribosome plays an important role in translation.

(iii) The job of m-RNA is to pick up amino acids and transport them to the ribosomes.
Answer:
The job of t-RNA is to pick up amino acids and transport them to ribosomes. t-RNA is an adapter molecule. It reads the codons of m-RNA and also simultaneously transfer specific amino acid to m-RNA Ribosome complex. It binds with amino acid at its 3′ end.

(iv) Transcription must occur before translation may occur.
Answer:
In prokaryotes, translation can start before transcription is complete, as both these processes occur in the same compartment, i.e. cytoplasm. But in eukaryotes, transcription and processing of hnRNA occurs in nucleus. hnRNA then comes out of the nucleus through nuclear pores and then it is translated at ribosomes in the cytoplasm.

Question 6.
Guess
(i) the possible locations of DNA on the collected evidence from a crime scene and
(ii) the possible sources of DNA.

Evidence	Possible location of DNA on the evidence	Sources of DNA
e.g. Eyeglasses	e.g. Earpieces	e.g. Sweat, Skin
Bottle, Can, Glass	Sides, mouthpiece	—————-
————–	Handle	Sweat, skin, blood
Used cigarette	Cigarette butt	—————–
Bite mark	—————–	Saliva
————-	Surface area	Hair, semen, sweat, urine

Answer:

Evidence	Possible location of DNA on the evidence	Sources of DNA
e.g. Eyeglasses	e.g. Earpieces	e.g. Sweat, Skin
Bottle, Can, Glass	Sides, mouthpiece	Saliva
Door	Handle	Sweat, skin, blood
Used cigarette	Cigarette butt	Saliva
Bite mark	Teeth impression	Saliva
Clothes	Surface area	Hair, semen, sweat, urine

12th Std Biology Questions And Answers:

12th Biology Chapter 15 Exercise Biodiversity, Conservation and Environmental Issues Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 15 Biodiversity, Conservation and Environmental Issues Textbook Exercise Questions and Answers.

Biodiversity, Conservation and Environmental Issues Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 15 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 15 Exercise Solutions

1. Multiple choice questions

Question 1.
Observe the graph and select correct option.

(a) Line A represents, S = CA²
(b) Line B represents, log C = log A + Z log S
(c) Line A represents, S = CA^Z
(d) Line B represents, log S = log Z + C log A
Answer:
(c) Line A represents, S = CA^Z

Question 2.
Select odd one out on the basis of Ex situ conservation.
(a) Zoological park
(b) Tissue culture
(c) Sacred groves
(d) Cryopreservation
Answer:
(a) Zoological park

Question 3.
Which of the following factors will favour species diversity?
(a) Invasive species
(b) Glaciation
(c) Forest canopy
(d) Co-extinction
Answer:
(a) Invasive species

Question 4.
The term “terror of Bengal’ is used for
(a) algal bloom
(b) water hyacinth
(c) increased BOD
(d) eutrophication
Answer:
(b) water hyacinth

Question 5.
CFC are air polluting agents which are produced by
(a) Diesel trucks
(b) Jet planes
(c) Rice fields
(d) Industries
Answer:
(b) Jet planes

2. Very short answer type questions.

Question 1.
Give two examples of biodegradable materials released from sugar industry.
Answer:

Molasses
Bagasse.

Question 2.
Name any two modern techniques of protection of endangered species.
OR
Two modern methods of ex-situ conservation of species
Answer:

Tissue culture
In vitro fertilization of eggs
Cryopreservation.

Question 3.
Where was ozone hole discovered?
Answer:
Ozone hole was discovered in Antarctica.

Question 4.
Give one example of natural pollutant.
Answer:
Volcanic ash is a natural pollutant.

Question 5.
What do you understand by EW category of living being?
Answer:
A species which becomes extinct in the wild (EW) is called EW category, their members are seen only in captivity or as a naturalized population outside its historic range due to massive habitat loss.

3. Short answer type questions.

Question 1.
Dandiya raas is not allowed after 10.00 pm. Why?
Answer:
Dandiya rass involves blaring loudspeakers which cause noise pollution. It is undesired loud sound which could be hazardous for ears and general health. In India, the Air (Prevention and Control of Pollution) Act 1981, Amendment 1987, includes noise as an air pollutant. As per law noise after 10 pm is not allowed as many people may be resting. Therefore, Dandiya Raas is not allowed after 10 pm.

Question 2.
Tropical regions exhibit species richness as compared to polar regions. Justify.
Answer:

Tropical regions are bestowed by thicker vegetation and ample food due to available sunlight and humidity.
Polar regions are covered over with snow, with almost no vegetation.
Only handful species of animals can survive here due to their adaptations.
Species richness always shows latitudinal gradient for many plants and animal species. It is high at lower latitudes and there is a steady decline towards the poles. Therefore, tropical regions show more species richness.

Question 3.
How does genetic diversity affect sustenance of a species?
Answer:

Genetic diversity develops the capability of the species to adapt to the varying changes in the environment.
The large variation of the different gene sets allows an individual or the whole population to have the capacity to endure environmental stress in any form.
Some individuals have, a better capacity to endure the increasing pollution in the environment whereas some do not have it.
Those that do not have show infertility or even death from the same conditions.
Those who are able to endure and adapt to this change survive and live in a better way.
This is called natural selection which leads to a loss of genetic diversity in particular habitats.
Thus, due to genetic diversity can affect sustenance of some species.

Question 4.
Greenhouse effect is boon or bane? Give your opinion.
Answer:
(1) The natural greenhouse effect is good, it is a boon but human enhanced greenhouse effect is a bane.

(2) In the absence of an atmosphere, Earth’s surface temperature would be about -18 °C, or 0 °F, which is too cold for sustaining life.

(3) Earth is habitable because of the natural greenhouse effect. Heating of Earth’s atmosphere due to the presence of greenhouse gases such as water vapour, carbon dioxide (CO₂), methane (CH₄) and oxides of nitrogen (NO₂).

(4) Greenhouse gases have just the right molecular structure to absorb infrared radiation that the Earth emits. It re-emits most of that infrared energy in all directions, warming the atmosphere to its comfortable average temperature of 15 °C (60 °F). So, the greenhouse effect was a boon in olden days before industrialization and invention of automobiles.

(5) However, due to human impact, the proportion of greenhouse gases has increased tremendously causing global warming. Thus, now greenhouse effect has become a bane.

Question 5.
State the effects of CO in human body.
OR
How does CO cause giddiness and exhaustion?
Answer:
Effects of Carbon monoxide:

Carbon monoxide is tasteless, colourless and odourless gas, therefore its presence goes unnoticed.
It can inhibit the blood’s ability to carry oxygen to body tissues.
Supply of oxygen to vital organs such as.the heart and brain is affected due to presence of CO.
When CO is inhaled, it combines with the oxygen carrying haemoglobin of the blood to form carboxyhaemoglobin. Once combined with the haemoglobin, that haemoglobin is no longer available for transporting oxygen.
The symptoms of CO poisoning are headache, nausea, giddiness, etc.

Question 6.
Name two types of particulate pollutants found in air. Add a note on ill effects of the same on human health.
OR
Describe any 2 particulate and gaseous pollutants.
Answer:
I. Types of gaseous pollutants include CO₂, CO, SO₂, NO, NO₂, etc.
(1) Carbon dioxide : It is a greenhouse gas. It is produced in excess due to human activities such as burning of fossil fuels. It is also rising due to increasing deforestation. The natural cycle of Carbon dioxide is disturbed due to human interference. Otherwise, the process of photosynthesis can balance CO₂ : O₂ ratio of the air. Aeroplane traffic such as a jet plane also emits lots of CO₂.

(2) Carbon monoxide (CO) : CO is produced due to incomplete combustion of fuels. It is a toxic gas. Vehicular exhausts produce lot of CO.

II. Types of particulate pollutants are mist, dust, fume and smoke particles, smog, pesticides, heavy metals and radioactive elements, etc.
(1) Dust are fine particles which enter the respiratory passage and can cause damage to delicate tissues in the lungs. Various processes such as construction work, demolition of buildings and traffic can cause dust pollution. There are natural causes of release of dust too, through wind or volcanic eruption.

(2) Smoke and smog are worst type of particulate air pollutants which can cause many respiratory problems like emphysema or asthma.

4. Long answer type questions.

Question 1.
Montreal Protocol is an essential step. Why is it so?
Answer:

Montreal Protocol was an international treaty signed at Montreal in Canada in 1987.
Later many more efforts have been made and protocols have laid down definite roadmaps separately for developing and developed countries.
All these efforts were for reducing emission of CFCs and other ozone depleting chemicals.
All nations realized that ozone depletion can cause penetration of harmful UV radiations to the earth’s surface. This is very hazardous, for flora, fauna and for mainly human beings. Therefore, urgent action was needed to combat this effect.
Montreal Protocol was a very positive move because after 1987, there have been much better condition of ozone layer.

Question 2.
Name any 2 personalities who have contributed to control deforestation in our country. Elaborate on importance of their work.
Answer:
Two personalities who have contributed to control deforestation in our country are:
Saalumara Thimmakka from Karnataka and Moirangthem Loiya from Manipur.
1. Saalumara Thimmakka :

Saalumara Thimmakka is the best example of peoples’ participation in reforestation.
She is an Indian environmentalist from Karnataka. She has taken up work of planting and tending to 385 banyan trees along a 4 km stretch of highway between Hulikal and Kudur. Other 800 trees are also planted by her.
She is honoured with the National Citizens Award of India and Padma Shri in 2019.

2. Moirangthem Loiya :

Moirangthem Loiya is from Manipur who has restored Punshilok forest. For last 17 years he is planting trees after leaving his job.
He brought the lost glory back for the 300 acres forest land. He planted a variety of trees like, bamboo, oak Ficus, teak, jackfruit and Magnolia.
This forest now has over 250 varieties of plants including 25 varieties of bamboo along with many animals making the forest rich in biodiversity.

Question 3.
How BS emission standards changed over time? Why is it essential?
Answer:

BS emission standards changed over the time due to changing city life and more vehicular traffic on the road, especially in the megacities.
Since capital city of Delhi was declared as worst polluted city as far as its air quality is concerned, various measures were taken by the Government of India. There was new fuel policy declared, in which Bharat stage emission standards (BS) were set.
These norms were set to reduce sulphur and aromatic content of petrol and diesel. Also the vehicular engines were upgraded.
Bharat stage emission standards (BS) are standards which are equivalent to Euro norms and have evolved on similar lines as Bharat Stage II (BS II) to BS VI from 2001 to 2017.
Since population of Delhi was to be saved, in 2001, Bharat stage II emission norms were set for CNG and LPG vehicles.
This helped in reduced emission of sulphur which was controlled at 50 ppm in diesel and 150 ppm in petrol. Also aromatic hydrocarbons were reduced at 42% in concerned fuel according to norms.
Because, in spite of all the efforts, Delhi was declared as worst air-polluted city in the world in 2016, therefore, Government of India directly adapted BS VI in the year 2018, skipping BS V These efforts decreased the levels of CO₂ and SO₂ in Delhi.

Question 4.
During large public gatherings like Pandharpur vari, mobile toilets are deployed by the government. Explain how this organic waste is disposed.
Answer:

The toilets deployed at Pandharpur at the time of vari are of the Ecosan type.
Ecosan toilet is a closed system without water and it is an alternative to leach pit toilets.
When the pit of an Ecosan toilet fills up after some time, then it is closed and sealed for about 8-9 months.
In this time the faeces get completely composted to organic manure. In this way the organic waste can be disposed.
It is a practical, efficient and cost-effective solution for human waste disposal.
Also, open-air defecation is prohibited which can cause health problems. Therefore, during large public gatherings like Pandharpur vari mobile toilets like Ecosan are deployed by the government.

Question 5.
How Indian culture and traditions helped in bio-diversity conservation? Give importance of conservation in terms of utilitarian reasons.
Answer:
In Indian culture and traditions in different religions, biodiversity is protected and conserved. Few examples of worship of animals and plants can be given here.

Nagpanchami festival is towards the respect of snakes. They are worshipped on that day and the local people are aware of their role in ecosystem of control of rat population.
Vatapournima festival is worshipping a banyan tree.
Various other festivals teach the value of plants and animals surrounding us. Even the cattle are worshipped on a particular day as a tradition.
Jain religion strongly advocates protection of all animals through vegetarianism.

Conservation in terms of utilitarian reasons:
The conservation of biodiversity can be done in utilitarian way or for ethical reasons. Utilitarian reasons are further classified into narrowly utilitarian and broadly utilitarian reasons:

I. Narrowly utilitarian reasons:

Humans always reap material benefits from biodiversity in the form of resources for basic needs such as food, clothes, shelter.
Industrial products like resins, tannins, perfume base, etc. are also obtained through biodiversity resources.
For making ornaments or artefacts for aesthetic purpose, again biodiversity is sacrificed.
Many medicines are also obtained through biodiversity resources which shares 25% of global medicine market.
Around 25000 species are used for traditional medicines by tribal population worldwide.
Bioprospecting which is a systematic search for development of new sources of chemical compounds, genes, microorganisms, macroorganisms, and other valuable products from nature which is of economically important species is also due to biodiversity.

II. Broadly utilitarian reasons:

Production of oxygen done by all green plants helps human beings to thrive. Amazon forest alone gives 25% of the oxygen to the entire world.
Insects carry out pollination and seed dispersal.
If insects do not carry out pollination and seed dispersal, man would go hungry without crops and fruits.
Biodiversity also is useful in recreation of human beings.

III. Taking all these aspects in consideration, conservation of biodiversity becomes essential. Therefore, to protect and conserve our rich biodiversity on the planet, we have to remember all the utilitarian reasons.

12th Std Biology Questions And Answers:

12th Biology Chapter 3 Exercise Inheritance and Variation Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 3 Inheritance and Variation Textbook Exercise Questions and Answers.

Inheritance and Variation Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 3 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 3 Exercise Solutions

1. Multiple Choice Questions

Question 1.
Phenotypic ratio of incomplete dominance in Mirabilis jalapa.
(a) 2 : 1 : 1
(b) 1 : 2 : 1
(c) 3 : 1
(d) 2 : 2
Answer:
(b) 1 : 2 : 1

Question 2.
In dihybrid cross, F₂ generation offspring show four different phenotypes while the genotypes are ……………….
(a) six
(b) nine
(c) eight
(d) sixteen
Answer:
(b) nine

Question 3.
A cross between an individual with unknown genotype for a trait with recessive plant for that trait is ……………….
(a) back cross
(b) reciprocal cross
(C) monohybrid cross
(d) test cross
Answer:
(d) test cross

Question 4.
When phenotypic and genotypic ratios are the same, then it is an example of ……………….
(a) incomplete dominance
(b) complete dominance
(c) multiple alleles
(d) cytoplasmic inheritance
Answer:
(a) incomplete dominance

Question 5.
If the centromere is situated near the end of the chromosome, the chromosome is called ……………….
(a) Metacentric
(b) Acrocentric
(c) Sub-Metacentric
(d) Telocentric
Answer:
(d) Telocentric

Question 6.
Chromosomal theory of inheritance was proposed by ……………….
(a) Sutton and Boveri
(b) Watson and Crick
(c) Miller and Urey
(d) Oparin and Halden
Answer:
(a) Sutton and Boveri

Question 7.
If the genes are located in a chromosome as p-q-r-s-t, which of the following gene pairs will have least probability of being inherited together ?
(a) p and q
(b) r and s
(c) s and t
(d) p and s
Answer:
(d) p and s

Question 8.
Find the mismatched pair:
(a) Down’s syndrome = 44 + XY
(b) Turner’s syndrome = 44 + XO
(c) Klinefelter’s syndrome = 44 + XXY
(d) Super female = 44 + XXX
Answer:
(a) Down’s syndrome = 44 + XY

Question 9.
A colourblind man marries a woman, who is homozygous for normal colour vision, the probability of their son being colour blind is ……………….
(a) 0%
(b) 25%
(c) 50%
(d) 100%
Answer:
(a) 0%

2. Very Short Answer Questions

Question 1.
Explain the statements
a. Test cross is back cross but back cross is not necessarily a test cross.
b. Law of dominance is not universal.
Answer:
a. (1) Test cross is the cross between F₁ hybrid and its homozygous recessive parent.
(2) Back cross is the cross of offspring with any one of the parents, either dominant or recessive.
(3) Therefore, test cross can be a back cross – but back cross cannot be a test cross.

b. (1) There are many traits in many organisms which show dominance. For example, widow’s peak in human beings is a dominant trait. Yellow seed colour and round seed shape are dominant traits in pea plant.
(2) However, there are characters which are either co-dominant, such as genes for human blood group A and B or incompletely dominant as in flower colour of Mirabilis jalapa.
(3) Therefore the law of dominance is not universally applicable.

Question 2.
Define the following terms:
a. Dihybrid cross
b. Homozygous
c. Heterozygous
d. Test cross
Answer:
a. A cross between parents differing in two heritable traits is called dihybrid cross.
b. An individual possessing identical alleles for a particular trait is called homozygous or pure for that trait. E.g. TT for tallness and tt for dwarfness.
c. An individual possessing contrasting allele for a particular trait is called heterozygous. E.g. Tt showing tallness.
d. The cross of F₁ progeny with homozygous recessive parent is called a test cross.

Question 3.
What are allosomes?
Answer:
Allosomes are the chromosomes which decide the sex of an organism.

Question 4.
What is crossing over?
Answer:
Crossing over is the process of forming new recombinations by interchanging and exchanging non-sister chromatid arms of the homologous chromosomes.

Question 5.
Give one example of autosomal recessive disorder.
Answer:
Thalassemia is an example of autosomal recessive disorder.

Question 6.
What are X-linked genes?
Answer:
Genes located on the non-homologous region of X chromosome are called X-linked genes.

Question 7.
What are holandric traits?
Answer:
Genes located on the non-homologous region of Y chromosome are called Y-linked genes. The traits due to such genes are called holandric traits which are seen only in male sex.

Question 8.
Give an example of chromosomal disorder caused due to non-disjunction of autosomes.
Answer:
Down’s syndrome is an example of chromosomal disorder caused due to non-disjunction of autosomes.

Question 9.
Give one example of complete sex linkage.
Answer:
Sex linkage can be complete X linkage and complete Y linkage. X linkage is haemophilia and Y linkage is hypertrichosis.

3. Short Answer Questions

Question 1.
Enlist seven traits of pea plant selected / studied by Mendel.
Answer:
Seven traits in pea selected by Mendel:

Tall habit versus dwarf habit (Height of the plant).
Purple flowers versus white flowers. (Colour of flowers)
Yellow seeds versus green seeds. (Colour of seeds)
Round seeds versus wrinkled seeds. (Shape of seeds)
Green pods versus yellow pods. (Colour of pods)
Inflated pods versus constricted pods. (Shape of pods)
Axial flower versus terminal flower. (Position of a flower)

Question 2.
Why law of segregation is also called the law of purity of gametes?
Answer:
(1) Mendel’s law of segregation is also called Law of purity of gametes because, during formation of gametes, the alleles separate/ segregate from each other and only one allele enters a gamete.

(2) The separation of one allele does not affect other. Since single allele enters a gamete means gametes will be pure for a trait.
E.g. The contrasting characters such as tall (T) and dwarf (t) present in F₁ hybrid (Tt) segregate during the formation of gametes.

(3) Owing to this, two types of gametes i.e. T and t are formed which are pure for the characters which they carry.
(4) Thus for example:

Question 3.
Pleiotropy.
Answer:

When a single gene controls two or more different traits, it is called a pleiotropic gene and the phenomenon is known as pleiotropy or pleiotropism.
The pleiotropic ratio is always 1 : 2 instead . of normal 3 : 1.
Sickle-cell anaemia is caused by the gene HbS. The healthy or normal gene which is dominant is HbA. The heterozygotes or carriers i.e., HbA/Hbs show anaemia as there is deficiency of haemoglobin due to sickling of RBCs. Abnormally low concentration of oxygen can cause sickling of RBCs.
The homozygotes possessing the recessive gene HbS die because of fatal anaemia because the gene for sickle-cell anaemia is lethal in homozygous condition and causes sickle-cell trait in heterozygous carrier.
Thus a single gene produces two different expressions.
When two carriers are married they will produce normal carriers and Sickle-cell anaemic children in the ratio of 1 : 2 : 1. Out of these three children sickle-cell anaemic child will die leaving the ratio 1 : 2 instead of 3 : 1.

Question 4.
What are the reasons of Mendel’s success?
Answer:
Reasons for Mendel’s success:

Mendel planned his experiments carefully and these experiments consisted of large sample.
He always recorded the results of number of plants of each type and their ratios.
The contrasting characters that he chose were easily recognizable.
The seven pairs of contrasting characters that he selected were under control of a single factor each. They were present on separate chromosomes and were transmitted from one generation to the next.
Mendel studied and introduced concept of dominance and recessiveness.

Question 5.
“Father is responsible for determination of sex of child and not the mother”. Justify.
Answer:

Human made is heterogame tic, i.e. he produces two different types of sperms. One is bearing X chromosome along with 22 autosomes and the other is Y bearing sperm with 22 autosomes.
Mother, on the other hand, is homogametic, producing all similar types of ova, i.e 22 + X chromosomal combination.
If 22+X bearing sperm fertilise an egg, female child is formed while if Y bearing sperm fertilizes an egg, male child is formed.
Thus the sex of the child is dependent upon type of sperm that father gives, therefore, it is said that father is responsible for determination of sex of a child and not the mother.

Question 6.
What is linkage? How many linkage groups do occur in human being and maize?
Answer:

Linkage is defined as the tendency of the genes to be inherited together because they are present in the same chromosome. Linkage group is group of genes situated on a chromosome.
Humans have 23 linkage groups because they have 23 pairs of chromosomes.
Maize plant has 10 linkage groups because they have 10 pairs of chromosomes.

Question 7.
PKU.
Answer:

PKU means phenylketonuria which is an autosomal recessive inborn error.
In this disorder the metabolism of phenylalanine does not occur due to deficiency of phenylalanine hydroxylase (PAH) enzyme.
This enzyme is necessary to metabolize the amino acid phenylalanine to the amino acid tyrosine.
When PAH activity is reduced, phenylalanine accumulates in blood and cerebrospinal fluid and is converted into phenylpyruvate or phenyl-ketone which is a toxic compound. This may cause mental retardation. Excess phenylalanine is excreted in urine, hence this disease is called phenylketonuria.
PKU is caused by mutations in the PAH gene on chromosome no. 12.
Untreated PKU causes abnormal phenotype which includes growth failure, poor skin pigmentation, microcephaly, seizures, global developmental delay and severe intellectual impairment. However, at birth if an infant is checked for PKU, the further abnormalities can be avoided.

Question 8.
Compare X-chromosome and Y-chromosome.
Answer:

X-chromosome	Y-chromosome
1. X-chromosome is straight, rod like and longer 1. than Y chromosome. It is metacentric.	1. Y-chromosome is shorter chromosome which is acrocentric.
2. X-chromosome has large amount of euchromatin and small amount of heterochromatin.	2. Y-chromosome has small amount of euchromatin and large amount of heterochromatin.
3. X-chromosome has large amount of DNA, hence it is genetically active due to more genes.	3. Y-chromosome has less amount of DNA, hence it is genetically less active or inert due to lesser genes.
4. Non-homologous region of X-chromosome is longer and contains more genes.	4. Non-homologous region of Y-chromosome is shorter and contains lesser genes.
5. Contains X-linked genes on non-homologous region.	5. Contains Y-linked genes on non-homologous region.
6. X-chromosome is present in men as well as women.	6. Y-chromosome is present only in men.

Question 9.
Explain the chromosomal theory of inheritance.
Answer:
Chromosomal theory of inheritance was put forth by Sutton and Boveri after studying paraillel behaviour of genes and chromosomes during meiotic division. This theory states following points:

Chromosomal theory identifies chromosomes as the carrier of genetic material.
All the hereditary characters are transmitted by gametes. Nucleus of gametes, i.e. sperms and ova of the parents contain chromosomes which transmit the heredity to offspring.
Chromosomes are found in pairs in somatic or diploid cells.
During gamete formation, homologous chromosomes pair and segregate independently at meiosis. The diploid condition is converted into haploid condition. Thus each gamete contains only one chromosome of a pair.
During fertilization, the union of sperm and egg restores the diploid number of chromosomes.

Question 10.
Observe the given pedigree chart and answer the following questions

(a) Identify whether the trait is sex-linked or autosomal.
(b) Give an example of a trait in human beings which shows such a pattern of inheritance.
Answer:
Pedigree given above shows:

First Generation : Carrier woman marrying a sufferer man. Their three children are in following birth order.
Second generation : First son is normal, second daughter is carrier and third daughter is sufferer.
Third generation : The sufferer daughter marries a normal man. Her children are normal daughter and sufferer son.

(a) The above pedigree show sex-linked (X-linked) trait. Since criss-cross inheritance is seen in the trait, it must be sex-linked inheritance.
(b) Such trait and its inheritance can be seen in colour blindness.

4. Match the Columns

rewrite the matching pairs.

Column I	Column II
(1) 21 trisomy	(a) Turner’s syndrome
(2) X-monosomy	(b) Klinefelter’s syndrome
(3) Holandric traits	(c) Down’s syndrome
(4) Feminized male	(d) Hypertrichosis

Answer:

Column I	Column II
(1) 21 trisomy	(c) Down’s syndrome
(2) X-monosomy	(a) Turner’s syndrome
(3) Holandric traits	(d) Hypertrichosis
(4) Feminized male	(b) Klinefelter’s syndrome

5. Long Answer Questions

Question 1.
What is dihybrid cross? Explain with suitable example and checker board method.
Answer:
1. A cross which involves two pairs of alleles is called a dihybrid cross. A phenotypic ratio of 9 : 3 : 3 : 1 obtained in the F₂ generation of a dihybrid cross is called a dihybrid ratio.

(2) Thus for example, when we cross a true breeding pea plant bearing round and yellow seeds with a true breeding pea plant bearing wrinkled and green seeds we get pea plants bearing round and yellow seeds in the F₁ generation.

(3) When F₁ plants are selfed, we get a ratio of 9 : 3 : 3 : 1 in the F₂ generation, where 9 plants bear yellow round seeds, 3 plants bear yellow wrinkled seeds, 3 plants bear green round seeds and 1 plant bears green wrinkled seeds.

(4) Parents (P₁) : RRYY × rryy
Gametes of P₁ RY and ry
F₁ generation : RrYy(Yellow round)
On selfing F₁ : RrYy × RrYy
Gametes of F₁ : RY, Ry, rY, ry

P₂ generation:

Round Yellow : 9 Round green : 3 Wrinkled yellow : 3 Wrinkled green : 1
Phenotypic ratio : 9 : 3 : 3 : 1
Genotypic ratio : 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1

Question 2.
Explain with suitable example an independent assortment.
Answer:
(1) The law of independent assortment states that when hybrid possessing two or more pairs of contrasting characters bearing alleles form gametes, the alleles in each pair segregate independently of the other pair. Therefore, the inheritance of one pair of characters is independent of that of the other pair of characters.
(2) For example, when we cross a pea plant which is tall and having purple flowers with dwarf plant having white flowers we obtain all tall plants with purple flowers in F₁ generation. When F₁ generation are selfed, 9 : 3 : 3 : 1 ratio was obtained in F₂ generation with 9 tall and purple flower, 3 tall with white flowers, 3 dwarf with purple flowers and 1 which was dwarf and white. Tallness and purple colour are dominant traits while dwarfness and white colour are recessive traits.

(i) Homozygous tall purple – TTPP
(ii) Homozygous dwarf white – ttpp

Tall purple = 9. Tall white = 3
Dwarf purple = 3, Dwarf white = 1,
Phenotypic ratio = 9 : 3 : 3 : 1
Results : The offspring of F₁ generation will be in the proportion of 9 : 3 : 3 : 1, where 9 are tall purple, 3 are tall white, 3 are dwarf purple and 1 is dwarf white.

Question 3.
Define test cross and explain its significance.
Answer:
1. Definition of test cross : A cross between F₁ offspring and its homozygous recessive parent is called a test cross.
2. Significance of test cross:

Test cross can be used to find out the genotype of any plant which shows dominant characters.
Whether the plant is homozygous or heterozygous can be understood by performing test cross.
Test cross is used to introduce useful recessive traits in the hybrids of self- pollinated plants.
Test cross is quicker method to improve the variety of crop plants and thus it is useful for breeders and geneticists.
Test cross can be used for verifying the laws of inheritance.

Question 4.
What is parthenogenesis? Explain the haplodiploid method of sex determination in honey bee.
Answer:
I. Parthenogenesis is a natural form of asexual reproduction in which growth and development of embryos occur without fertilization by sperm. In some insects like honey bees, parthenogenesis means development of an embryo from an unfertilized egg cell.

II. In honey bee:

Sex determination is by haplodiploid system.
Sex is determined by the number of sets of chromosomes received by an individual.
The egg which is fertilized by sperm, becomes diploid and develops into female.
The egg which is not fertilized develops by parthenogenesis and develops into a male.
The queen and worker bee therefore contain 32 chromosomes. The drone, i.e. male bears 16 chromosomes.
The sperms are produced by mitosis while eggs are produced by meiosis.

Question 5.
In the answer for inheritance of X-linked. genes, Madhav had shown carrier male. His answer was marked incorrect. Madhav was wondering why his marks were cut. Explain the reason.
Answer:
Males can never be carriers. They have single X and other Y chromosome. In X linked inheritance, the genes are present on the non-homologous region of X chromosome. Males do not have other X and hence if the genes are present on his X chromosome, they will not be suppressed in them. The Y chromosome does not have dominant gene to hide this expression as there is no homolorous region too. But in case of females, there are double X chromosomes and hence if X-linked gene is recessive, the other X can hide the expression of such X-linked gene.

Thus she becomes a carrier without showing any physical characters. She is physically normal and does not suffer from such X-linked recessive disorder. Thus, Madhav will get his answer wrong due to incorrect concept.

Question 6.
With the help of neat labelled diagram, describe the structure of chromosome.
Answer:

(1) A chromosome is best visible during metaphase, when it is highly condensed.

(2) Chromosome shows two identical halves, called sister chromatids. Chromatids are held together at centromere which is also called primary constriction.

(3) Primary constriction has disc shaped plate called kinetochore. This plate is useful for attachment of spindle fibres at the time of cell division.

(4) Additional narrow areas called secondary constrictions are seen in some chromosomes which are known as nucleolar organizers. They help in the formation of nucleolus. At secondary constriction (i) there is nucleolar organising region. Secondary constriction (ii) shows attachment of satellite body or SAT body.

(5) Each chromatid is made up of sub¬chromatids called chromonemata. Each chromonema consists of a long, unbranched, slender, highly coiled DNA thread. This double stranded DNA molecule extends throughout the length of the chromosome.

(6) The ends of the chromatid arms are called telomeres.

Question 7.
What is criss-cross inheritance? Explain with suitable example.
Answer:
Criss-cross inheritance is the type of inheritance in which the genes are passed on from father to daughter and then to her son, i.e. from male to female and from female to male (grandson). In other words, it is also said that the transmission is from the grandfather to his grandson through his daughter.

I. Inheritance of Colour blindness show criss-cross pattern.
(1) Colour blindness is a sex-linked disorder in which the person concerned cannot distinguish between red and green colours.

(2) It is recessively X-linked disorder, which is expressed in males. It is rarely seen in females.

(3) The genes for normal vision are dominant whereas those for colour blindness are recessive.

(4)

Gene for normal vision : X^C
Gene for colour blindness : X^c
Normal female : X^CX^C
Normal male : X^CY
Colour blind female : X^cX^c
Carrier female : X^CX^c
Colour blind male : X^c Y

II. Crosses showing the inheritance of colour blindness:
(i) A cross between normal female and colour-blind male.

(ii) A cross of carrier female with normal male.

(1) Normal female with Colour blind male. Such cross produces 50% carrier daughters and 50% normal sons.

(2) Carrier female with normal male. Such a cross produces 25% normal daughters, 25% normal sons, 25% carrier daughters and 25% colour blind sons.

(3) Colour blind father transmits the disorder to his grandson through his carrier daughter. The inheritance of characters from the father to his grandson through his daughter is called criss-cross inheritance.

Question 8.
Describe the different types of chromosomes.
Answer:
I. Chromosomes are classified into the following four types according to the position of the centromere in them:
(1) Metacentric : In metacentric chromosome, the centromere is situated in the middle of the chromosome. The two arms of the chromosome are nearly equal. It appears ‘V’-shaped during anaphase.

(2) Sub-metacentric : In sub-metacentric chromosome, the centromere is situated some distance away from the middle. Due to this, one arm of the chromosome is shorter than the other. It appears T-shaped during anaphase.

(3) Acrocentric : In acrocentric chromosome, the centromere is situated near the end of the chromosome. One arm of the acrocentric chromosome is very short while the other is long making it appear like ‘J’-shaped during anaphase.

(4) Telocentric : In telocentric chromosome, the centromere is situated at the tip of the chromosome. Telocentric chromosome has only one arm thus it appears rod-shaped.

II. Based on the functions, chromosomes are divided into autosomes and allosomes. Autosomes are somatic chromosomes which decide the body characters. Allosomes are sex chromosomes which decide the sex of the individual.

12th Std Biology Questions And Answers:

12th Biology Chapter 11 Exercise Enhancement of Food Production Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 11 Enhancement of Food Production Textbook Exercise Questions and Answers.

Enhancement of Food Production Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 11 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 11 Exercise Solutions

1. Multiple choice questions

Question 1.
Antibiotic Chloromycetin is obtained from ………………….
(a) Streptomyces erythreus
(b) Penicillium chrysogenum
(c) Streptomyces venezuelae
(d) Streptomyces griseus
Answer:
(c) Streptomyces venezuelae

Question 2.
Removal of large pieces of floating debris, oily substances, etc. during sewage treatment is called ………………….
(a) primary treatment
(b) secondary treatment
(c) final treatment
(d) amplification
Answer:
(a) primary treatment

Question 3.
Which one of the following is free living bacterial biofertilizer?
(a) Azotobacter
(b) Rhizobium
(c) Nostoc
(d) Bacillus thuringiensis
Answer:
(a) Azotobacter

Question 4.
Most commonly used substrate for industrial production of beer is ………………….
(a) barley
(b) wheat
(c) corn
(d) sugar cane molasses
Answer:
(a) barley

Question 5.
Ethanol is commercially produced through a particular species of ………………….
(a) Aspergillus
(b) Saccharomyces
(c) Clostridium
(d) Trichoderma
Answer:
(b) Saccharomyces

Question 6.
One of the free-living anaerobic nitrogen- fixers is ………………….
(a) Azotobacter
(b) Beijerinckia
(c) Rhodospirillum
(d) Rhizobium
Answer:
(c) Rhodospirillum

Question 7.
Microorganisms also help in production of food like ………………….
(a) bread
(b) alcoholic beverages
(c) vegetables
(d) pulses
Answer:
(a) bread

Question 8.
MOET technique is used for ………………….
(a) production of hybrids
(b) inbreeding
(c) outbreeding
(d) outcrossing
Answer:
(a) production of hybrids

Question 9.
Mule is the outcome of ………………….
(a) inbreeding
(b) artificial insemination
(c) interspecific hybridization
(d) outbreeding
Answer:
(c) interspecific hybridization

2. Very Short Answer Questions

Question 1.
What makes idlis puffy?
Answer:
During preparation of idlis, rice and black gram flour is fermented by air borne Leuconostoc and Streptococcus bacteria. CO₂ produced during fermentation makes them puffy.

Question 2.
Bacterial biofertilizers.
Answer:
Rhizobium, Frankia, Pseudomonas striata, Bacillus polymyxa, Agrobacterium, Microccocus, Azotobacter, Costridium, Beijerinkia, Klebsiella.

Question 3.
What is the microbial source of vitamin B₁₂?
Answer:
The microbial source of vitamin B₁₂ is Pseudomonas denitrificans.

Question 4.
What is the microbial source of enzyme invertase?
Answer:
The microbial source of enzyme invertase is Saccharomyces cerevisiae.

Question 5.
Milk starts to coagulate when Lactic Acid Bacteria (LAB) are added to warm milk as a starter. Mention any two other benefits of LAB.
Answer:
Lactic Acid Bacteria (LAB) check the growth of disease causing microbes and produce vitamin B.

Question 6.
Name the enzyme produced by Streptococcus bacterium. Explain its importance in medical sciences.
Answer:
The enzyme produced by Streptococcus spp. is streptokinase. It is used as a ‘clot buster’ for clearing blood clots in the blood vessels in heart patients.

Question 7.
What is breed?
Answer:
Breed is a group of animals related by descent and similar in most characters like general appearance, features, size, configuration, etc.

Question 8.
Estuary
Answer:
Estuary is a place where river meets the sea.

Question 9.
What is shellac?
Answer:
Shellac is the pure form of lac obtained by washing and filtering.

3. Short Answer Questions.

Question 1.
Many microbes are used at home during preparation of food items. Comment on such useful ones with examples.
Answer:

Many food preparations made at home involves the use of microorganisms.
The microbes Lactobacilli are used in the preparation of dhokla from gram flour and buttermilk by the process of fermentation.
Dosa and idlis are prepared by using batter of rice and black gram which is fermented by air borne Leuconostoc and Streptococcus bacteria.
Large, fleshy fruiting bodies of some mushrooms and truffles are directly used as food. It is sugar free, fat free food rich in proteins, vitamins, minerals and amino acids. It is food with low calories.
Curd is prepared by inoculating milk with Lactobacillus acidophilus. Lactic acid produced during fermentation causes coagulation and partial digestion of milk protein casein and milk turns into curd.
Buttermilk is the acidulated liquid left after churning of butter from curd, is called buttermilk.

Question 2.
What is biogas? Write in brief about the production process.
Answer:
Biogas is a mixture of methane CH₄ (50-60%), CO₂ (30-40%), H₂S (0-3%) and other gases (CO, N₂, H₂) in traces.

Biogas production process:
a. A typical biogas plant consists of digester (made up of concrete bricks and cement or steel and is partly buried in the soil) and gas holder (a cylindrical gas tank to collect gases).
b. Raw materials like cow dung is mixed with water in equal proportion to make slurry which is fed into the digester’ through a side opening (charge pit).

Anaerobic digestion involves following processes:
i. Hydrolysis or solubilization:
Anaerobic hydrolyzing bacteria like Clostridium and Pseudomonas hydrolyse carbohydrates into simple sugars, proteins into amino acids and lipids into fatty acids.

ii. Acidogenesis:
Facultative and obligate anaerobic, acidogenic bacteria convert simple organic substances into acids like formic acid, acetic acid, H₂ and CO₂

iii. Methanogenesis:
Anaerobic methanogenic bacteria like Methanobacterium, Methanococcus convert acetate, H₂ and CO₂ into Methane, CO₂ and H₂O and other products.
12 mol CH₃COOH → 12CH₄ + 12CO₂ 4mol H.COOH → CH₄ + 3CO₂ + 2H₂O CO₂ + 4H₂ → CH₄ + 2H₂O

Question 3.
Biocontrol agents.
Answer:
(1) Biocontrol agents are the organisms like (bacteria, fungi, viruses and protozoans) act which are employed for controlling pathogens, pests and weeds.

(2) They cause the disease to the pest or compete or kill them.

(3) The use of biocontrol measures greatly reduces use of toxic chemicals and pesticides that are harmful to human beings and also pollute our environment.

(4) Biocontrol agents and their hosts.

Bacteria (Bacillus thuringiensis, B. papilliae and B. lentimorbus Hosts : Caterpillars, cabbage worms, adult beetles
Fungi (Beauveria bassiana, Entomophothora, pallidaroseum, Zoophthora radicans) Host : Aphid crocci, A. unguicilata, mealy bugs, mites, white flies, etc.
Protozoans (Nosema locustae) Host: Grasshoppers, caterpillars, crickets
Viruses (Nucleopolyhedro virus-NPV, Granulovirus-GV) Host : Caterpillars, Gypsy moth, ants and beetles.

(5) Some examples:

Bacillus thuringiensis (Bt) is a microbiai pesticide used to get rid of butterfly, caterpillars.
Trichoderma fungus is an effective biocontrol agent against soil borne fungal plant pathogens which infect roots and rhizomes.
Phytophthora palmiuora is a mycoherbicide that controls milk weed in orchards.
Pseudomonas spp. is a bacterial herbicide that attacks several weeds.
Tyrea moth controls the weed Senecio jacobeac.

Question 4.
Name any two enzymes and antibiotics with their microbial source.
Answer:

Microbial source of Chloromycetin. – Streptomyces venezuelae
Microbial source of Erythromycin. – Streptomyces erythreus
Microbial source of Penicillin. – Penicillium chrysogenum
Microbial source of Streptomycin. – Streptomyces griseus
Microbial source of Griseofulvin. – Penicillium griseojulvum
Microbial source of Bacitracin. – Bacillus licheniformis
Microbial source of Oxytetracyclin / Terramycin. – Streptomyces aurifaciens
The enzyme produced by Streptococcus bacterium. – Streptokinase
Microbial source of Invertase. – Saccharomyces cerevisiae
Microbial source of Pectinase. – Sclerotinia libertine, Aspergillus niger
Microbial source of Lipase. – Candida lipolytica
Microbial source of Cellulase. – Trichoderma konigii

Question 5.
Write the principles of farm management.
Answer:
The principles of farm management are as follows:

Selection of high-yielding breeds.
Understanding the feed requirements of farm animals.
Supply of adequate nutritional sources for the animals.
Maintaining the cleanliness of environment.
Maintenance of health with the help of veterinary supervision.
Undertaking vaccination programmes.
Development of high-yielding cross-bred varieties.
Making various products and their preservation.
Distribution and marketing of the farm produce.

Question 6.
Give the economic importance of fisheries.
Answer:
Economic importance of fisheries is as follows:

Fish is a nutritious food and thus is a source of many vitamins, minerals and nutrients.
Commercial products such as fish oil, fish meal and fertilizers, fish guano, fish glue, isinglass are prepared from fish.
These by-products are used in paints, soaps, oils and medicines.
Some organisms like prawns and lobsters have high export value and market price.
Fish farming and other fishery trades provide job opportunity and self-employment
Productivity and national economy is improved through fishery practices.

Question 7.
Enlist the species of honey bee mentioning their specific uses.
Answer:
(1) The four species of honey bees commonly found in India : Apis dorsata (rock bee, or wild bee), Apis jlorea (little bee), Apis mellifera (European bee) and Apis indica (Indian bee).

(2) Uses:

Rock bee : They produce 36 kg of honey per comb per year. They produce bee wax.
Little bee : They produce half kg of honey per hive per year.
European bee : The average production per colony per year is 25 to 40 kg.
Indian bee : The average production per colony per year is 6 to 8 kg.

Question 8.
What are A, B, C, D in the table given below.

Types of microbe	Name	Commercial Product
Fungus	A	Penicillin
Bacterium	Acetobacter aceti	B
C	Aspergillus niger	Citric acid
Yeast	D	Ethanol

Answer:
A : Penicillium chrysogenum
B : Vinegar (Acetic acid)
C : Fungus
D : Sachharomyces cerevisiae var. ellipsoidis

4. Long Answer Questions.

Question 1.
Explain the process of sewage water treatment before it can be discharged into natural bodies. Why is this treatment essential?
Answer:
Sewage treatment includes following steps:
(1) Preliminary Treatment:

Screening: The larger suspended or floating objects are filtered and removed in screening chambers by passing the sewage through screens or net in the chambers.
Grit Chamber : Filtered sewage is passed into series of grit chambers which contain large stones (pebbles) and brick-ballast. Coarse particles which settle down by gravity are removed.

(2) Primary treatment (physical treatment):

The sewage water is pumped into the primary sedimentation tank where 50-70% of the suspended solid or organic matter get sedimented and about 30-40% (in number) of coliform organisms are removed.
The organic matter which is settled down is called primary sludge.
Primary sludge is removed by mechanically operated devices.
Dissolved organic matter and micro-organisms in the supernatant (effluent) are then removed by the secondary treatment.

(3) Secondary treatment (biological treatment):

The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it.
The mesh like masses of aerobic bacteria, slime and fungal hyphae, known as floes are formed.
Aerobic microbes consume most of the organic matter and this reduces BOD (Biochemical Oxygen Demand) of the effluent.

(4) Tertiary treatment:

Once the BOD is sufficiently reduced, waste water is passed into a settling tank where the floes are allowed to sediment.
The sediment is called activated sludge.
Small part of activated sludge is transferred to aeration tank and the major part is pumped in to large anaerobic sludge digesters.
In these tanks, anaerobic bacteria grow and digest the bacteria and fungi in the sludge and gases like methane, hydrogen sulphide, CO₂, etc. are released.
Effluents from these digesters are released in natural water bodies like rivers and streams after chlorination which kills pathogenic bacteria.
Digested sludge is then disposed.

Question 2.
Lac culture.
Answer:

Lac is a pink coloured resin secreted by dermal glands of female lac insect (Trachardia lacca) that hardens on coming in contact with air forming lac.
Lac is a complex substance having resin, sugar, water, minerals and alkaline substances.
Lac insect is colonial in habit and it feeds on succulent twigs like ber, peepal, palas, kusum, babool,
These plants are artificially inoculated in order to get better and regular supply of good quality and quantity of lac.
Natural lac is always contaminated and pure form of lac obtained by washing and filtering is called as shellac.
Lac is used to make bangles, toys, woodwork, inks, mirrors, etc.
India’s share is 85% of total lac produced in the world.

Question 3.
Describe various methods of fish preservation.
Answer:

Fish is a highly perishable commodity.
After catching the fish it immediately starts spoilage process.
In order to prevent this process, the fish preservation is done.

The different methods of fish preservation are as follows:

Chilling : This involves covering the fish with layers of ice. Ice is effective for short term preservation. It inhibits the activity of autolytic enzymes.
Freezing : It is a long duration preservation method. Fish are freezed at 0°C to -20°C. This also inhibit autolytic enzyme activities and slows down bacterial growth.
Freeze drying : The deep frozen -fish at -20°C are dried by direct sublimation of ice to water vapour with any melting into liquid water. This is achieved by exposing the frozen fish to 140°C in a vacuum chamber. The fish is then packed or canned in dried condition.
Sun drying : This inhibits the growth of microorganisms that spoil the fish.
Smoke drying : Smoke is prepared by burning woods with less resinous matter. Bacteria are destroyed by the acid content of the smoke. Smoking also give the characteristic colour, taste and odour to fish.
Salting : Salt removes the moisture from the fish tissues by osmosis. High salt concentration destroys autolytic enzymes and halts bacterial activity.
Canning : Canning involves sealing the food in a container, heat ‘sterilising’ the sealed unit and cooling it to ambient temperature for subsequent storage.

Question 4.
Give an account of poultry diseases.
Answer:
Various poultry diseases are as follows:

Viral diseases : Ranikhet, Bronchitis, Avian influenza (bird flu), etc. Bird flu had serious impact on poultry farming and also caused human infection.
Bacterial diseases : Pullorum, Cholera, Typhoid, TB, CRD (chronic respiratory disease), Enteritis, etc.
Fungal diseases : Aspergillosis, Favus and Thrush.
Parasitic diseases : Lice infection, round worm, caecal worm infections, etc.
Protozoan diseases : Coccidiosis.

Question 5.
Give an account of mutation breeding with examples.
Answer:

Mutations are sudden heritable changes in the genotype.
Natural mutations occur at a very slow rate.
Natural physical mutagens include exposure to high temperature, high concentration of C02, X-rays, UV rays.
Mutations can be induced by using various mutagens.
Mutagens cause gene mutations and chromosomal aberrations.
Chemical mutagens include nitrous acid, EMS (Ethyl – Methyl – Sulphonate), mustard gas, colchicine, etc.
Seedlings or seeds are irradiated by using CO60 or UV bulbs or X-ray machines.
The mutated seedlings are then screened for resistance to diseases/pests, high yield, etc.
Examples of mutant varieties in different crops are Jagannath (rice), NP 836 (rust resistant wheat variety), Indore-2 (cotton variety resistant to bollworm), Regina-II (cabbage variety resistant to bacterial rot).

Question 6.
Describe briefly various steps of plant breeding methods.
Answer:
The main steps of the plant breeding program (Hybridization) are as follows:

(1) Collection of variability:

Germplasm collection is the entire collection of all the diverse alleles for all genes in a given crop.
Wild species and relatives of the cultivated species having desired traits are collected and preserved.
Forests and natural reserves are the means of in situ conservation of germplasm.
Botanical gardens, seed banks, etc. are means of ex situ conservation of germplasm.

(2) Evaluation and selection of parents:

The collected germplasm is evaluated to identify healthy and vigorous plants with desirable and complementary characters.
Selected parents are selfed for three to four generations to increase homozygosity.
Only pure lines are selected, multiplied and used in the hybridization.

(3) Hybridization:

The variety showing maximum desirable features is selected as female (recurrent) parent and the other variety which lacks good characters found in recurrent parent is selected as male parent (donor).
The pollen grains from anthers of male parent are artificially dusted over stigmas of emasculated flowers of female parent.
Hybrid seeds are collected and sown to grow F₁ geneartion.

(4) Selection and Testing of Superior Recombinants:

The F₁ hybrid plants which are superior to both the parents and having high hybrid vigour, are selected and selfed for few generations to make them homozygous for the said desirable characters.
This ensures that there is no further segregation of the characters.

(5) Testing, release and commercialization of new cultivars:

The newly selected lines are evaluated for the productivity and desirable features like disease resistance, pest resistance, quality, etc.
They are initially grown under controlled conditions of water, fertilizers, etc. and their performance is recorded.
The selected lines are then grown for at least three generations in natural field, in different agroclimatic zones.
Finally variety is released as new variety for use by the farmers.

12th Std Biology Questions And Answers:

12th Chemistry Chapter 5 Exercise Electrochemistry Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 5 Electrochemistry Textbook Exercise Questions and Answers.

Electrochemistry Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 5 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 5 Exercise Solutions

1. Choose the most correct option.

Question i.
Two solutions have the ratio of their concentrations 0.4 and ratio of their conductivities 0.216. The ratio of their molar conductivities will be
(a) 0.54
(b) 11.574
(c) 0.0864
(d) 1.852
Answer:
(a) 0.54

Question ii.
On diluting the solution of an electrolyte,
(a) both ∧ and κ increase
(b) both ∧ and κ decrease
(c) ∧ increases and κ decreases
(d) ∧ decreases and κ increases
Answer:
(c) ∧ increases and κ decreases

Question iii.
1 S m² mol^-1 is equal to
(a) 10^-4 S m² mol^-1
(b) 10⁴Ω^-1 cm² mol^-1
(c) 10^-2 S cm² mol^-1
(d) 10² Ω^-1 cm² mol^-1
Answer:
(b) 10⁴Ω^-1 cm² mol^-1

Question iv.
The standard potential of the cell in which the following reaction occurs
H₂₊ (g, 1 atm) + Cu²⁺ (1 M) → 2H (1 M) + Cu_(s), (\(E_{\mathrm{Cu}}^{0}\) = 0.34 V) is
(a) – 0.34 V
(b) 0.34 V
(c) 0.17 V
(d) -0.17 V
Answer:
(b) 0.34 V

Question v.
For the cell, Pb_(s)|Pb²⁺ (1 M)|| Ag⁺ (1 M)|Ag_(s), if concentration of an ion in the anode compartment is increased by a factor of 10, the emf of the cell will
(a) increase by 10 V
(b) increase by 0.0296 V
(c) decrease by 10 V
(d) decrease by 0.0296 V
Answer:
(d) decrease by 0.0296 V

Question vi.
Consider the half reactions with standard potentials

The strongest oxidising and reducing agents respectively are
(a) Ag and Fe²⁺
(b) Ag⁺ and Fe
(c) Pb²⁺ and I^–
(d) I₂ and Fe²⁺
Answer:
(b) Ag⁺ and Fe

Question vii.
For the reaction
Ni_(s) + Cu²⁺ (1 M) → Ni²⁺ (1 M) + Cu_(s), \(E_{\text {cell }}^{0}\) = 0.57 V. Hence ΔG⁰ of the reaction is
(a) 110 kJ
(b) -110 kJ
(c) 55 kJ
(d) -55 kJ
Answer:
(b) -110 kJ

Question viii.
Which of the following is not correct ?
(a) Gibbs energy is an extensive property
(b) Electrode potential or cell potential is an intensive property.
(c) Electrical work = -ΔG
(d) If half reaction is multiplied by a numerical factor, the corresponding E⁰ value is also multiplied by the same factor.
Answer:
(d) If half reaction is multiplied by a numerical factor, the corresponding E⁰ value is also multiplied by the same factor.

Question ix.
The oxidation reaction that takes place in lead storage battery during discharge is

Answer:
(c) \(\mathrm{Pb}_{(\mathrm{s})}+\mathrm{SO}_{4(\mathrm{aq})}{ }^{2-} \longrightarrow \mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{e}^{-}\)

Question x.
Which of the following expressions represent molar conductivity of Al₂(SO₄)₃ ?

Answer:
(b) \(2 \lambda_{\mathrm{Al}^{3+}}^{0}+3 \lambda_{\mathrm{SO}_{4}^{2-}}^{0}\)

2. Answer the following in one or two sentences.

Question i.
What is a cell constant ?
Answer:
(A) Cell constant of a conductivity cell is defined as the ratio of the distance between the electrodes divided by the area of cross section of the electrodes.

In SI units it is expressed as m-1.

Question ii.
Write the relationship between conductivity and molar conductivity and hence unit of molar conductivity.
Answer:
If k is conductivity and ∧_m is molar conductivity then, ∧_m = \(\frac{\kappa \times 1000}{C}\)
Unit of molar conductivity is, Ω^-1 cm² mol^-1 or S cm² mol^-1.

Question iii.
Write the electrode reactions during electrolysis of molten KCl.
Answer:

Question iv.
Write any two functions of salt bridge.
Answer:
The functions of a salt bridge are :

It maintains the electrical contact between the two electrode solutions of the half cells.
It prevents the mixing of electrode solutions.
It maintains the electrical neutrality in both the solutions of two half cells by a flow of ions.
It eliminates the liquid junction potential.

Question v.
What is standard cell potential for the reaction
3Ni_(s) + 2Al³⁺ (1M) → 3NI²⁺(1M) + 2Al_(s)
if \(\boldsymbol{E}_{\mathrm{Ni}}^{0}\) = – 0.25 V and \(\boldsymbol{E}_{\mathrm{Al}}^{0}\) = -1.66V?
Solution :
Given : E⁰_Ni²⁺/Ni = -0.25 V
E⁰_Al³⁺/Al = – 1.66 V; E⁰_cell = ?
Since Ni is oxidised and Al³⁺ is reduced,
\(E_{\text {cell }}^{0}=E_{\mathrm{Al}^{3+} / \mathrm{Al}}^{0}-E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{0}\)
= – 1.66 – (-0.25)
= – 1.41 V
Ans. \(E_{\text {cell }}^{0}\) = -1.41 V
[Note : Since \(E_{\text {cell }}^{0}\) is negative, the given reaction is not possible but reverse reaction is possible.]

Question vi.
Write Nerst equation. What part of it represents the correction factor for nonstandard state conditions ?
Answer:
(1) Nernst equation for cell potential is,

(2) The part of equation namely,

represents the correction factor for nonstandard state conditions.

Question vii.
Under what conditions the cell potential is called standard cell potential ?
Answer:
In the standard cell, the active masses of the substances taking part in the electrochemical reaction have unit value, i.e., 1 M solution or ions and 1 atm gas.

Question viii.
Formulate a cell from the following electrode reactions :
\(\mathbf{A u}_{(\mathrm{aq})}^{3+}+\mathbf{3 e}^{-} \longrightarrow \mathbf{A} \mathbf{u}_{(\mathrm{s})}\)
\(\mathbf{M g}_{(\mathbf{s})} \longrightarrow \mathbf{M g}_{(\mathrm{aq})}^{2+}+\mathbf{2 e}^{-}\)
Answer:
An electrochemical cell from above electrode reactions is,

Question ix.
How many electrons would have a total charge of 1 coulomb ?
Answer:
Given : 1 Faraday = charge on 1 mol of electrons
= 6.022 × 10²³ electrons and 1 Faraday = 96500 C
∵ 96500 C = 6.022 × 10²³ electrons 6 022 × 10²³
∴ 1 C ≡ \(\frac{6.022 \times 10^{23}}{96500}\) = 6.24 × 10¹⁸ electrons
Ans. Number of electrons = 6.24 × 10¹⁸

Question x.
What is the significance of the single vertical line and double vertical line in the formulation galvanic cell.
Answer:
(i) Consider representation of Daniell cell,

Single vertical line represents separation of two phases, solid Zn_(s) and solution of ions.
(ii) Double vertical lines represent a salt bridge.

3. Answer the following in brief

Question i.
Explain the effect of dilution of solution on conductivity ?
Answer:

The conductance of a solution is due to the presence of ions in the solution. More the ions, higher is the conductance of the solution.
Conductivity or the specific conductance is the conductance of unit volume (1 cm³) of the electrolytic solution.
The conductivity of the electrolytic solution always decreases with the decrease in the concentration of the electrolyte or the increase in dilution of the solution.
On dilution, the concentration of the solution decreases, hence the number of (current carrying) ions per unit volume decreases. Therefore the conductivity of the solution decreases, with the decrease concentration or increase in dilution. (It is to be noted here that, molar conductivity increases with dilution.)

Question ii.
What is a salt bridge ?
Answer:
A salt bridge is a U-shaped glass tube containing a saturated solution of a strong electrolyte, like KCl, NH₄NO₃, Na₂SO₄ in a solidified agar-agar gel. A hot saturated solution of these electrolytes in 5% agar solution is filled in the U-shaped tube and allowed it to cool and solidify forming a gel.

Fig. 5.9 : Salt bridge
It is used to connect two half cells or electrodes forming a galvanic or voltaic cell.

Question iii.
Write electrode reactions for the electrolysis of aqueous NaCl.
Answer:
Reactions in electrolytic cell :
(i) Reduction half reaction at cathode : There are Na⁺ and H⁺ions but since H⁺ are more reducible than Na⁺, they undergo reduction liberating hydrogen and Na⁺ are left in the solution.
2H₂O_(l) + 2e^– → H_2(g) + 2OH^–_(aq) (reduction) E⁰ = -0.83 V

(ii) Oxidation half reaction at anode : At anode there are Cl^– and OH^–. But Cl^– ions are preferably oxidised due to less decomposition potential.

Net cell reaction : Since two electrons are gained at cathode and two electrons are released at anode for each redox step, the electrical neutrality is maintained. Hence we can write,

Since Na⁺ and OH^– are left in the solution, they form NaOH_(aq).

Question iv.
How many moles of electrons are passed when 0.8 ampere current is passed for 1 hour through molten CaCl₂ ?
Answer:
Given : I = 0.8 A; t = 1 × 60 × 60 = 3600 s
Number of moles of electrons = ?
Q = I × t
= 0.8 × 3600
= 2880 C
1 Faraday = 1 mol electrons
1 Faraday = 96500 C
∵ 96500 C = 1 mol electrons
∴ 2880 C ≡ \(\frac{2880}{96500}\)
= 0.02984 mol electrons
Ans. Number of moles of electrons = 0.02984

Question v.
Construct a galvanic cell from the electrodes Co³⁺|Co and Mn²⁺|Mn. \(\boldsymbol{E}_{\mathrm{Co}}^{0}\) = 1.82 V,
\(\boldsymbol{E}_{\mathrm{Mn}}^{0}\) = – 1.18V. Calculate \(\boldsymbol{E}_{\text {cell }}^{0}\).
Answer:

Question vi.
Using the relationsip between ∆G⁰ of cell reaction and the standard potential associated with it, how will you show that the electrical potential is an intensive property ?
Answer:
(1) For an electrochemical cell involving n number of electrons in the overall cell reaction,
ΔG⁰ = -nF\(E_{\text {cell }}^{0}\)
where ΔG⁰ is standard Gibbs energy change and \(E_{\text {cell }}^{0}\) is a standard cell potential.
(2) ∴ \(E_{\mathrm{cell}}^{0}=\frac{-\Delta G^{0}}{n F}\)
Since ΔG⁰ changes according to number of moles of electrons involved in the cell reaction, the ratio, ΔG⁰/nF remains constant.
(3) Therefore \(E_{\text {cell }}^{0}\) is independent of the amount of substance and it represents the intensive property.

Question vii.
Derive the relationship between standard cell potential and equilibrium constant of cell reaction.
Answer:
For any galvanic cell, the overall cell reaction at equilibrium can be represented as,
Reactants ⇌ Products.
[For example for Daniell cell,
\(\mathrm{Zn}_{(s)}+\mathrm{Cu}_{(\mathrm{aq})}^{2+} \rightleftharpoons \mathrm{Zn}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(\mathrm{s})}\) ]
The equilibrium constant, K for the reversible reaction will be, \(K=\frac{[\text { Products }]}{[\text { Reactants }]}\)

The equilibrium constant is related to the stan-dard free energy change Δ G⁰, as follows,
ΔG⁰ = -RTlnK
If \(E_{\text {cell }}^{0}\) is the standard cell potential (or emf) of the galvanic cell, then ΔG⁰= -nFE⁰_cell
By comparing above equations,

Question viii.
It is impossible to measure the potential of a single electrode. Comment.
Answer:
(1)

Fig 5.12(a) : Measurement of single electrode potential

Fig 5.12(b) : Measurement of cell potential
According to Nemst theory, electrode potential is the potential difference between the metal and ionic layer around it at equilibrium, i.e. the potential across the electric double layer.

(2) For measuring the single electrode potential, one part of the double layer, that is metallic layer can be connected to the potentiometer but not the ionic layer. Hence, single electrode potential can’t be measured experimentally.

(3) When an electrochemical cell is developed by combining two half cells or electrodes, they can be connected to the potentiometer and the potential difference or cell potential can be measured.
E_cell = E₂ – E₁
where E₁ and E₂ are reduction potentials of two electrodes.

Question ix.
Why do the cell potential of lead accumulators decrease when it generates electricity ? How the cell potential can be increased ?
Answer:
Working of a lead accumulator :
(1) Discharging : When the electric current is withdrawn from lead accumulator, the following reactions take place :

(2) Net cell reaction :
(i) Thus, the overall cell reaction during discharging is

OR
Pb_(s) + PbO₂_(s) + 2H₂SO₄_(aq) → 2PbSO₄_(s) + 2H₂O_(l)
The cell potential or emf of the cell depends upon the concentration of sulphuric acid. During the operation, the acid is consumed and its concentration decreases and specific gravity decreases from 1.28 to 1.17. As a result, the emf of the cell decreases. The emf of a fully charged cell is about 2.0 V.

(ii) Recharging of the cell : When the discharged battery is connected to external electric source and a higher external potential is applied the cell reaction gets reversed generating H₂SO₄.
Reduction at the -ve electrode or cathode :
PbSO4_(s) + 2e^– → Pb_(s) + \(\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
Oxidation at the + ve electrode or anode :

The emf of the accumulator depends only on the concentration of H₂SO₄.

Question x.
Write the electrode reactions and net cell reaction in NICAD battery.
Answer:
Reactions in the cell :
(i) Oxidation at cadmium anode :
Cd_(s) + 2OH^–_(aq) → Cd(OH)₂_(s) + 2e^–
(ii) Reduction at NiO₂_(s) cathode :
NiO₂_(s) + 2H₂O_(l) + 2e^– → Ni(OH)₂_(s) + 2OH^–_(aq)
The overall cell reaction is the combination of above two reactions.
Cd_(s) + NiO₂_(s) + 2H₂O_(l) → Cd(OH)₂_(s) + Ni(OH)₂_(s)

4. Answer the following :

Question i.
What is Kohrausch law of independent migration of ions? How is it useful in obtaining molar conductivity at zero concentration of a weak electrolyte ? Explain with an example.
Answer:
(A) Statement of Kohlrausch’s law : This states that at infinite dilution of the solution, each ion of an electrolyte migrates independently of its co-ions and contributes independently to the total molar conductivity of the electrolyte, irrespective of the nature of other ions present in the solution.

(B) Explanation : Both the ions, cation and anion of the electrolyte make a definite contribution to the molar conductivity of the electrolyte at infinite dilution or zero concentration (∧₀).
If \(\lambda_{+}^{0}\) and \(\lambda_{-}^{0}\) are the molar ionic conductivities of cation and anion respectively at infinite dilution, then
∧₀ = \(\lambda_{+}^{0}\) + \(\lambda_{-}^{0}\).
This is known as Kohlrausch’s law of independent migration of ions.
For an electrolyte, B_xA_y giving x number of cations and y number of anions,
∧₀ = x\(\lambda_{+}^{0}\) + y\(\lambda_{-}^{0}\).

(C) Applications of Kohlrausch’s law :
(1) With this law, the molar conductivity of a strong electrolyte at zero concentration can be determined. For example,
\(\wedge_{0(\mathrm{KCl})}=\lambda_{\mathrm{K}^{+}}^{0}-\lambda_{\mathrm{Cl}^{-}}^{0}\)
(2) ∧₀ values of weak electrolyte with those of strong electrolytes can be obtained. For example,

Molar conductivity of a weak electrolyte at infinite dilution or zero concentration cannot be measured experimentally.
Consider the molar conductivity (∧₀) of a weak acid, CH₃COOH at zero concentration. By Kohlrausch s law,

where λ⁰CH₃COO^– and λ⁰_H⁺ are the molar ionic conductivities of CH₃COO^– and H⁺ ions respectively.
If ∧_0CH3COONa, ∧_0HCl and ∧_0NaCl are molar conductivities of CH₃COONa, HCl and NaCl respectively at zero concentration, then by
Kohlrausch’s law,

Hence, from ∧₀ values of strong electrolytes, ∧₀ of a weak electrolyte CH₃COOH, at infinite dilution can be calculated.

Question ii.
Explain electrolysis of molten NaCl.
Answer:
(1) Construction of an electrolytic cell : It consists of a vessel containing molten (fused) NaCl. Two graphite (carbon) inert electrodes are dipped in it, and connected to an external source of direct electric current (battery). The electrode connected to a negative terminal of the battery is a cathode and that connected to a positive terminal is an anode.

(2) Working of the cell :
(A) In the external circuit, the electrons flow through the wires from anode to cathode of the cell.
(B) The fused NaCl dissociates to form cations (Na⁺) and anions (Cl^–).
\(\mathrm{NaCl}_{\text {(fused) }} \longrightarrow \mathrm{Na}_{(\mathrm{l})}^{+}+\mathrm{Cl}_{(\mathrm{l})}^{-}\)
Na⁺ migrate towards cathode and Cl^– migrate towards anode.

Fig. 5.7 : Electrolysis of fused sodium chloride

(C) Reactions in electrolytic cell :
(i) Reduction half reaction at cathode : The Na⁺ ions get reduced by accepting electrons from a cathode supplied by a battery and form metallic sodium.
\(\mathrm{Na}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Na}_{(\mathrm{s})} \text { (reduction) }\)

(ii) Oxidation half reaction at anode : The Cl^– ions get oxidised by giving up electrons to the anode forming neutral Cl atoms in the primary process, and these Cl atoms combine forming Cl₂ gas in the secondary process.

The released electrons in the anodic oxidation half reaction return to battery through the metallic wires.

Net cell reaction : In order to maintain the electrical neutrality, the number of electrons gained at cathode must be equal to the number of electrons released at anode. Hence the reduction half reaction is multiplied by 2 and both reactions, oxidation half reaction and reduction half reaction are added to obtain a net cell reaction.

Results of electrolysis :

A molten silvery white Na is formed at cathode which floats on the surface of molten NaCl.
A pale green Cl₂ gas is liberated at anode.

Question iii.
What current strength in amperes will be required to produce 2.4g of Cu from CuSO₄ solution in 1 hour ? Molar mass of Cu = 63.5 g mol^-1.
Answer:
Given : W_Cu = 2.4 g; t = 1 hr = 1 × 60 × 60 s
M_Cu = 63.5 g mol^-1; I = ?

Ans. Current strength = I = 2.026 A

Question iv.
Equilibrium constant of the reaction,
2Cu⁺_(aq) → Cu²⁺_(aq) + Cu_(s)
is 1.2 × 10⁶. What is the standard potential of the cell in which the reaction takes place ?
Answer:
For the cell reaction, n = 1

Question v.
Calculate emf of the cell
Zn_(s)|Zn²⁺(0.2M)||H⁺(1.6M)|H₂(g, 1.8 atm)|Pt at 25°C.
Answer:
Given : Zn_(s)|Zn²⁺(0.2M)||H⁺(1.6M)|H₂(g, 1.8 atm)|Pt

= 0.763 – 0.0296 × (- 0.8521)
= 0.763 + 0.02522
= 0.7882
Ans. \(E_{\text {cell }}^{0}\) = 0.7882 V

Question vi.
Calculate emf of the following cell at 25°C.
Zn_(s)| Zn²⁺(0.08M)||Cr³⁺(0.1M)|Cr
E⁰_Zn = – 0.76 V, E⁰_Cr = – 0.74 V
Answer:

Question vii.
What is a cell constant ? What are its units? How is it determined experimentally?
Answer:
(A) Cell constant of a conductivity cell is defined as the ratio of the distance between the electrodes divided by the area of cross section of the electrodes.

In SI units it is expected as m^-1.

The resistance of an electrolytic solution is measured by using a conductivity cell and Wheatstone

Fig. 5.6 : Measurement of conductance
The measurement of molar conductivity of a solution involves two steps as follows :
Step I : Determination of cell constant of the conductivity cell :
KCl solution (0.01 M) whose conductivity is accurately known (κ = 0.00141 Ω^-1 cm^-1) is taken in a beaker and the conductivity cell is dipped. The two electrodes of the cell are connected to one arm while the variable known resistance (R) is placed in another arm of Wheatstone bridge.

A current detector D’ which is a head phone or a magic eye is used. J is the sliding jockey (contact) that slides on the arm AB which is a wire of uniform cross section. A source of A.C. power (alternating power) is used to avoid electrolysis of the solution.

By sliding the jockey on wire AB, a balance point (null point) is obtained at C. Let AC and BC be the lengths of wire.

If R_solution is the resistance of KCl solution and R_x is the known resistance then by Wheatstone’s bridge principle,
\(\frac{R_{\text {solution }}}{\mathrm{BC}}=\frac{R_{x}}{\mathrm{AC}}\)
∴ \(R_{\text {solution }}=\mathrm{BC} \times \frac{R_{x}}{\mathrm{AC}}\)
Then the cell constant ‘ b ’ of the conductivity cell is obtained by, b = κ_Kcl × R_solution.

Step II : Determination of conductivity of the given solution :
KCl solution is replaced by the given electrolytic solution and its resistance (R_s) is measured by Wheatstone bridge method by similar manner by obtaining a null point at D.
The conductivity (κ) of the given solution is,
κ = \(\frac{\text { cell constant }}{R_{\mathrm{s}}}=\frac{b}{R_{\mathrm{s}}}\)

Step III: Calculation of molar conductivity :
The molar conductivity (∧_m) is given by,

Since the concentration of the solution is known, ∧_m can be calculated.

Question viii.
How will you calculate the moles of electrons passed and mass of the substance produced during electrolysis of a salt solution using reaction stoichiometry.
Answer:
Calculation of moles of electrons passed : The charge carried by one mole of electrons is referred to as one faraday (F). If total charge passed is Q C, then moles of electrons passed = \(\frac{Q(\mathrm{C})}{F\left(\mathrm{C} / \mathrm{mol} \mathrm{e}^{-}\right)}\)

Calculation of mass of product : Mass, W of product formed is given by,
W = moles of product × molar mass of product (M)
= \(\frac{Q}{96500}\) × mole ratio × M
= \(\frac{I \times t}{96500}\) × mole ratio × M 96500
When two electrolytic cells containing different electrolytes are connected in series so that same quantity of electricity is passed through them, then the masses W₁ and W₂ of products produced are given by,

Question ix.
Write the electrode reactions when lead storage cell generates electricity. What are the anode and cathode and the electrode reactions during its recharging?
Answer:
Recharging of the cell : When the discharged battery is connected to external electric source and a higher external potential is applied the cell reaction gets reversed generating H₂SO₄.
Reduction at the – ve electrode or cathode :
\(\mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}^{(s)}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
Oxidation at the + ve electrode or anode :

The emf of the accumulator depends only on the concentration of H₂SO₄.

Question x.
What are anode and cathode of H₂-O₂ fuel cell ? Name the electrolyte used in it. Write electrode reactions and net cell reaction taking place in the fuel cell.
Answer:
Construction :
(i) In fuel cell the anode and cathode are porous electrodes with suitable catalyst like finely divided platinum.

(iii) H₂ is continuously bubbled through anode while O, gas is bubbled through cathode.

Working (cell reactions) :
(i) Oxidation at anode : At anode, hydrogen gas is oxidised to H₂O.
2H_2(g) + 4OH^–_(aq) → 4H₂O_(l) + 4e^– (oxidation half reaction)
(ii) Reduction at cathode : The electrons released at anode travel to cathode through external circuit and reduce oxygen gas to OH^–.
O_2(g) + 2H₂O_(l) + 4e^– → 4OH^–_(aq) (reduction half reaction)

(iii) Net cell reaction: Addition of both the above reactions at anode and cathode gives a net cell reaction.
2H_2(g) + O_2(g) → 2H₂O_(l) (overall cell reaction)

Question xi.
What are anode and cathode for Leclanche’ dry cell ? Write electrode reactions and overall cell reaction when it generates electricity.
Answer:
A dry cell has zinc vessel as anode and graphite rod as cathode and moist paste of ZnCl₂, MnO₂, NH₄Cl as electrolytes.
At anode :
Zn_(s) → \(\mathrm{Zn}_{(\mathrm{aq})}^{2+}\) + 2e^– (Oxidation half reaction)
At graphite (c) cathode :
\(2 \mathrm{NH}_{4(\mathrm{e})}^{+}\) + 2e^– → 2NH_3(aq) + H_2(g) (Reduction half reaction)
2MnO_2(s) + H₂ → Mn₂O_3(s) + H₂O_(l)
There is a side reaction inside the cell, between Zn²⁺ ions and aqueous NH₃.
\(\mathrm{Zn}_{(\mathrm{aq})}^{2+}+4 \mathrm{NH}_{3(\mathrm{aq})} \longrightarrow\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]_{(\mathrm{aq})}^{2+}\)

Question xii.
Identify oxidising agents and arrange them in order of increasing strength under standard state conditions. The standard potentials are given in parenthesis.
Al(- 1.66 V), Cl₂ (1.36 V), Cd²⁺ (-0.4 V), Fe (-0.44 V), I₂ (0.54 V), Br^– (1.09 V).
Answer:
The oxidising agents are I₂, Br^– and Cl₂. The increasing strength is

(Note : Actually Br₂ acts as an oxidising agent but not Br^–.)

Question xiii.
Which of the following species are reducing agents? Arrange them in order of increasing strength under standard state conditions. The standard potentials are given in parenthesis.
K (-2.93V), Br₂(1.09V), Mg(-2.36V), Co³⁺(1.61V), Ti²⁺(-0.37V), Ag⁺(0.8V), Ni (-0.23V).
Answer:
Lower the standard reduction potential, higher is reducing power. The reducing agents are Ni, Mg and K. Their increasing strength is,

(Note : Cations don’t act as reducing agent since they are already in oxidised state.)

Question xiv.
Predict whether the following
reactions would occur spontaneously
under standard state conditions.
a. Ca_(s) + Cd²⁺_(aq) → Ca²⁺_(aq) + Cd_(s)
b. 2 Br-_(s) + Sn²⁺(aq) → Br₂_(l) + Sn_(s)
c. 2Ag_(s) + Ni²⁺_(aq) → 2 Ag⁺_(aq) + Ni_(s)
(use information of Table 5.1)
Answer:

12th Chemistry Digest Chapter 5 Electrochemistry Intext Questions and Answers

Question 1.
How does electrical resistance depend on the dimensions of an electronic (metallic) conductor?
Answer:
The electrical resistance of an electronic conductor is linearly proportional to its length (l) and inversely proportional to its cross section area a.

Fig. 5.3 : Electronic conductor
Thus, R ∝ l; R ∝ \(\frac{1}{a}\)
∴ R ∝ \(\frac{l}{a}\) or R = ρ × \(\frac{l}{a}\)
where the proportionality constant p is called specific resistance. IUPAC recommends the term resistivity for specific resistance.

Question 2.
What are the units of resistivity ?
Answer:
For an electronic conductor of length l, and cross section area a, the resistance R is represented as
R = ρ × \(\frac{l}{a}\)
where ρ is the resistivity of the conductor.
∴ ρ = R × \(\frac{a}{l}\)
If l = 1 m, a = 1 m², ρ = R

Hence, resistivity is the resistance of a conductor of volume of 1 m³.
(In C.G.S. units, the units of ρ are ohm cm. Hence, ρ is the resistance of a conductor of unit volume or 1 cm³.)

Question 3.
Define resistivity. What are its units ?
Answer:
Resistivity (or specific resistance) : It is the resistance of a conductor that is 1 m in length and 1 m² in cross section area in SI units. (In C.G.S. units, it is the resistance of a conductor that is 1 cm in length and 1 cm² in cross section area.) Hence, the resistivity is the resistance of a conductor of unit volume. (In case of electrolytic solution, ρ is the resistivity i.e., resistance of a solution of unit volume.)
It has SI units, ohm m and C.G.S. units, ohm cm.

Question 4.
Why is alternating current used in the measurement of conductivity of the solution ?
Answer:
If direct current (D.C.) by battery is used, there will be electrolysis and the concentration of the solution is changed. Hence alternating current (A.C.) with high frequency is used.

Try this… (Textbook page No. 93)

Question 1.
What must be the concentration of a solution of silver nitrate to have the molar conductivity of 121.4 Ω^-1 cm² mol^-1 and the conductivity of 2.428 × 10^-3 Ω^-1 cm^-1 at 25 °C ?
Answer:

∴ Concentration of a Solution = 0.02 M

Try this… (Textbook page No. 96)

Question 1.
Obtain the expression for dissociation constant in terms of ∧_c and ∧₀ using Ostwald’s dilution law.
Answer:
Consider a solution of a weak electrolyte, BA having concentration C mol dm^-3. If α is the degree of dissociation, then by Ostwald’s theory of weak electrolytes,

If K is the dissociation constant of the weak electrolyte, then by Ostwald’s dilution law,
K = \(\frac{\alpha^{2} C}{(1-\alpha)}\)
If ∧_m is the molar conductivity of the electrolyte BA at the concentration C and ∧₀ is the molar conductivity at zero concentration or infinite dilution, then

Hence by measuring ∧_m at the concentration C and knowing ∧₀, the dissociation constant can be calculated.
If \(\lambda_{+}^{0}\) and \(\lambda_{-}^{0}\) are the ionic conductivities, then by Kohlrauseh’s law, ∧₀ = \(\lambda_{+}^{0}\) + \(\lambda_{-}^{0}\).

Learn this as well…

Question 1.
How is the cell constant of a conductivity cell determined?
Answer:
The cell constant of a given conductivity cell is obtained by measuring the resistance (R) (or the conductance) of a standard solution whose conductivity (f_c) is accurately known by using Wheatstone’s bridge (discussed in Q. 37). For this purpose, KCl solution of accurately known conductivity is used.
\(\kappa_{\mathrm{KCl}}=\frac{1}{R_{\mathrm{KCl}}} \times \frac{l}{a}\) where \(\frac{l}{a}\) is a cell constant, represented by b.
∴ \(\kappa_{\mathrm{KCl}}=\frac{b}{R_{\mathrm{KCl}}}\)
or b = κ_KCl × R_KCl
For example, the conductivity of 0.01 M KCl is 0.00141 Ω^-1cm^-1 (S cm^-1). Hence by measuring R KCl the cell constant b can be obtained.

Try this… (Textbook page No. 95)

Question 1.
Calculate ∧₀ (CH₂ClCOOH) if ∧₀ values for HCl, KCl and CH₂ClCOOK are respectively, 4.261, 1.499 and 1.132 Ω^-1 m² mol^-1.
Solution :

Adding equations (i) and (ii) and subtracting equation (iii) we get equation (I).

Can you tell ? (Textbook page No. 103)

Question 1.
You have learnt Daniel cell in XIth standard. Write notations for anode and cathode. Write the cell formula.
Answer:
Daniel cell is represented as,

Try this… (Textbook page No. 104)

Question 1.
Write electrode reactions and overall cell reaction for Daniel cell you learnt in standard XI.
Answer:
Reactions for Daniell cell:

Question 1.
Describe different types of reversible electrodes with examples. (1 mark for each type)
Answer:
A reversible electrochemical cell or a galvanic cell consists of two reversible half cells or electrodes. There are four types of reversible electrodes according to their compositions.
(1) Metal-metal ion electrode : This electrode is set up by dipping a metal in a solution containing its own ions, e.g. Zn rod dipped into ZnSO₄ solution containing Zn⁺⁺ ions of concentration C.
It is represented as,
\(\mathrm{Zn}^{2+}{ }_{(\mathrm{aq})} \mid \mathrm{Zn}_{(\mathrm{s})}\)
The reduction reaction at the electrode is,
Zn⁺⁺_(aq) + 2e^– → Zn_(s)

(2) Metal-sparingly soluble salt electrode : This electrode consists of a metal coated with one of its sparingly soluble salts and immersed in a solution containing an electrolyte having a common anion as that of the salt. For example, silver electrode coated with sparingly soluble AgCl dipped in KCl solution with common anion Cl^–. This electrode is represented as,
Cl^–_(aq) | AgCl_(s) | Ag_(s)
The reduction reaction is,
AgCl_(s) + e → Ag_(s) + Cl^–_(aq)

(3) Gas electrode : This is developed by bubbling pure and dry gas around a platinised platinum foil dipped in the solution containing ions (of the gas) reversible with respect to the gas bubbled.
The gas is adsorbed on the surface of platinum foil and establishes an equilibrium with its ions in the solution. Pt electrode provides electrical contact and also acts as a catalyst.
Some of the gas electrodes are represented as follows :
(i) Hydrogen gas electrode :
H⁺_(aq) | H₂(g, P_H₂) | Pt
Reduction reaction : H⁺_(aq) + e^– → \(\frac {1}{2}\)H_2(g)
(ii) Chlorine gas electrode :
Cl^–_(aq) | Cl₂(g, PCl₂) | Pt
Reduction reaction : \(\frac {1}{2}\)Cl_2(g) + e- → Cl^–_(aq)

(4) Redox electrode (Oxidation reduction electrode) : This electrode consists of a platinum wire dipped in a solution containing the ions of the same metal (or a substance) in two different oxidation states, like Fe²⁺ – Fe³⁺, Sn²⁺ – Sn⁴⁺, Mn⁺⁺ – MnO^–₄, etc.
A platinum electrode which provides an electrical contact and acts as catalyst aquires an equilibrium between two ions in the solution, due to their tendency to undergo a change from one oxidation state to another. The electrodes are represented as,
Fe²⁺_(aq), Fe³⁺_(aq) | Pt
Reduction reaction : Fe³⁺_(aq) + e^– → Fe²⁺_(aq)
SnCl_2(aq), SnCl_4(aq) | Pt
Reduction reaction : Sn⁴⁺_(aq) + 2e^– →Sn²⁺_(aq)

Use your brain power! (Textbook page No. 98)

Question 1.
Distinguish between electrolytic and galvanic cells.
Answer:
Electrolytic cell:

This device is used to bring about a non-spontaneous chemical reaction by passing an electric current.
It is used to bring about a chemical reaction generally for the dissociation (electrolysis) of compounds.
In this cell, electrical energy is converted into chemical energy.
In this cell, the cathode is negative and the anode is positive.
Electrolytic cells are irreversible.
Oxidation takes place at the positive electrode and reduction at the negative electrode.
The electrons are supplied by the external source and enter through cathode and come out through anode.
It is used for electroplating, electrorefining, etc.

Electrochemical cell (Galvanic cell or Voltaic cell):

This device is used to produce electrical energy by a spontaneous chemical reaction.
It is used to generate electricity.
In this cell, chemical energy is converted into electrical energy.
In this cell, the cathode is positive and the anode is negative.
Electrochernical cells are reversible.
Oxidation takes place at the negative electrode and reduction at the positive electrode.
The electrons move from anode to cathode in the external circuit.
It is used as a source of electric current.

Try this… (Textbook page No. 107)

Question 1.
Write expressions to calculate equilibrium constant from
i. Concentration data
ii. Thermochemical data
iii. Electrochemical data
Answer:
(i) Consider following a reversible cell reaction.
aA + bB ⇌ cC + dD
If [A], [B], [C] and [D] represent concentrations of reactants and products then the equilibrium constant K is,
K = \(\frac{[\mathrm{C}]^{c} \times[\mathrm{D}]^{d}}{[\mathrm{~A}]^{a} \times[\mathrm{B}]^{b}}\)
(ii) If ΔG⁰ is the standard Gibbs free energy change at temperature T then,
ΔG⁰ = – RTlnK = – 2.303 RTlog₁₀K
(iii) From electrochemical data,
if \(E_{\text {cell }}^{0}\) is the standard cell potential and K is the equilibrium constant for the cell reaction at a temperature T, then,
\(E_{\text {cell }}^{0}=\frac{0.0592}{n} \log _{10} K\)

Learn this as well…

Question 1.
The construction and working of the calomel electrode.
Answer:
(1) Since standard hydrogen electrode (SHE) is not convenient for experimental use, a secondary reference electrode like calomel electrode is used.
(2) Construction : It consists of a glass vessel with side arm B for dipping in a desired solution of another electrode like, ZnSO_4(aq) for an electric contact. The vessel is filled with mercury, a paste of Hg and Hg₂Cl₂ (calomel) and saturated KCl solution.

Fig. 5.15 : Determination of standard electrode potential using calomel electrode
(3) The potential developed depends upon the concentration of Cl^– or KCl solution. When saturated KCl solution is used, its reduction potential is 0.242 V.
(4) Consider following cell :
Zn_(s) | ZnSO_4(aq) || KCl_(aq) | Hg₂Cl_2(s) | Hg
OR Zn_(s) | ZnSO_4(aq) || Calomel electrode
Reduction reaction for calomel electrode :
Hg₂Cl_2(s) + 2e^– → 2Hg_(l) + 2Cl^–_(aq)
Hence potential of calomel electrode depends on the concentration of Cl^– or KCl solution.

Can you tell ? (Textbook page No. 114)

Question 1.
In what ways are fuel cells and galvanic cells similar and in what ways are they different ?
Answer:
Similarity between fuel cells and galvanic cells :

In both the cells, there is oxidation at anode and j reduction at cathode.
The cell potential is developed due to net redox reactions.
Both are galvanic cells.

Difference in fuel cells and galvanic cells :

Fuel cells involve electrodes with large surface area while galvanic cells involve electrodes with j compact surface area.
Fuel cells involve gaseous materials on a large scale while galvanic cells involve gaseous materials at a definite pressures along with electrolytes or there may not be gases.
In fuel cells, the cell potential is developed due to exothermic combustion reactions while in galvanic cell, cell potential is developed due to normal redox reactions.
In fuel cells gaseous electrode materials are continuously supplied from outside while in galvanic cells electrode materials have constant concentration or may change due to reactions.

Use your brain power (Textbook page No. 114)

Question 1.
Indentify the strongest and the weakest oxidizing agents from the electrochemical series.
Answer:
From the electrochemical series,
(a) The strongest oxidising agent is fluorine since it has the highest standard reduction potential (\(E_{\mathrm{F}_{2} / \mathrm{F}^{-}}^{0}\) = + 2.87 V).
(b) The weakest oxidising agent (or the strongest reducing agent) is lithium since it has the lowest standard reduction potential, (\(E_{\mathrm{Li}^{+} / \mathrm{Li}}^{0}\) = -3.045 V).

Use your brain power (Textbook page No. 115)

Question 1.
Identify the strongest and the weakest reducing agents from the electrochemical series.
Answer:
(a) From the electrochemical series, the strongest reducing agent is lithium since it has the lowest standard reduction potential (\(E_{\mathrm{Li}^{+} / \mathrm{Li}}^{0}\) = -3.045 V).
(b) The weakest reducing agent is fluorine since it has the highest standard reduction potential,
(\(E_{\mathrm{F}_{2} / \mathrm{F}^{-}}^{0}\) = +2.87 V).

Question 2.
From E° values given in Table 5.1, predict whether Sn can reduce I₂ or Ni²⁺.
Answer:
From electrochemical series,

12th Std Chemistry Questions And Answers:

12th Biology Chapter 1 Exercise Reproduction in Lower and Higher Plants Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 1 Reproduction in Lower and Higher Plants Textbook Exercise Questions and Answers.

Reproduction in Lower and Higher Plants Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 1 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 1 Exercise Solutions

1. Multiple Choice Questions

Question 1.
Insect pollinated flowers usually possess ………………
(a) sticky pollen with a rough surface
(b) large quantities of pollens
(c) dry pollens with a smooth surface
(d) light colored pollens
Answer:
(a) sticky pollen with a rough surface

Question 2.
In ovule, meiosis occurs in ………………
(a) Integument
(b) Nucellus
(c) Megaspore
(d) Megaspore mother cell
Answer:
(d) Megaspore mother cell

Question 3.
The ploidy level is NOT the same in ………………
(a) Integuments and nucellus
(b) Root tip and shoot tip
(c) Secondary nucleus and endosperm
(d) Antipodals and synergids
Answer:
(c) Secondary nucleus and endosperm

Question 4.
Which of the following types require pollination but result is genetically similar to autogamy?
(a) Geitonogamy
(b) Xenogamy
(c) Apogamy
(d) Cleistogamy
Answer:
(a) Geitonogamy

Question 5.
If diploid chromosome number in a flowering plant is 12, then which one of the following will have 6 chromosomes?
(a) Endosperm
(b) Leaf cells
(c) Cotyledons
(d) Synergids
Answer:
(d) Synergids

Question 6.
In angiosperms, endosperm is formed by/ due to ………………
(a) free nuclear divisions of megaspore
(b) polar nuclei
(c) polar nuclei and male gamete
(d) synergids and male gametes
Answer:
(c) polar nuclei and male gamete

Question 7.
Point out the odd one.
(a) Nucellus
(b) Embryo sac
(c) Micropyle
(d) Pollen grain
Answer:
(d) Pollen grain

2. Very Short Answer Questions

Question 1.
The part of gynoecium that determines the compatible nature of pollen grain.
Answer:
Stigmatic surface.

Question 2.
How many haploid cells are present in a mature embryo sac?
Answer:
6 cells, 2 synergids, 1 egg cell, 3 antipodals.

Question 3.
Even though each pollen grain has 2 male gametes why at least 20 pollen grains are required to fertilize 20 ovules in a particular carpel?
Answer:
Angiosperms have phenomenon of double fertilization in which both the male gametes are utilized, one for fusion with egg cell to form zygote and other for fusion with secondary nucleus to form endosperm.

Question 4.
Megasporogenesis
Answer:
It is the process of formation of haploid megaspores from diploid megaspore mother cell.

Question 5.
What is hydrophily?
Answer:
Transfer of pollen grains in pollination process through agency of water is known as hydrophily.

Question 6.
The layer which supplies nourishment to the developing pollen grains.
Answer:
Tapetum

Question 7.
Parthenocarpy
Answer:
The condition in which fruit is developed without the process of fertilization is called parthenocarpy.

Question 8.
Are pollination and fertilization necessary in apomixis?
Answer:
Apomixis is formation of embryos without formation of gametes hence there is no need of pollination and fertilization.

Question 9.
The part of pistil which develops into fruit and seed.
Answer:
Ovary develops into fruit and ovules into seed.

Question 10.
What is the function of filiform apparatus ?
Answer:
Filiform apparatus guides the pollen tube towards egg cell.

3. Short Answer Questions

Question 1.
How polyembryony can be commercially exploited?
Answer:

Polyembryony is the development of more than one embryo inside the seed.
When such polyembryonic seed germinate we get multiple seedlings from it.
This condition increases the chances of survival of new plants.
Nucellar embryos are genetically identical to parent plants hence we get uniform plants.
In horticulture we can utilize these as rootstock for grafting, hence they have significant role in fruit breeding programmes e.g. Citrus, Mango.

Question 2.
Pollination and seeds formation are very crucial for the fruit formation, Justify.
Answer:

After fertilization, ovary is transformed into fruit, where ovary wall becomes fruit wall, i.e pericarp.
Mature ovules are transformed into seeds after fertilization.
Fertilization is a process where male gametes unites with female gamete to form zygote which develops into embryo.
In pollination process pollen grains carrying non-motile male gamete are transferred on stigma.
Seeds have embryo which germinate into new plant hence the goal of reproduction to create offspring for next generation is achieved. Hence these are the crucial events for fruit formation.

Question 3.
Incompatibility is a natural barrier in the fusion of gametes. How will you explain this statement?
Answer:

Self incompatibility or self-sterility is a genetic mechanism that prevents germination of pollen on stigma of same flower. This favours cross pollination. E.g. Tobacco.
In pollen-pistil interaction, when pollen grain is deposited on stigma, pistil has the ability to recognize and allow germination of right type of pollen.
Special type of proteins on stigmatic surface determine compatibility or incompatibility.
A physiological mechanism operates to ensure successful germination of compatible pollen.
Compatible pollen absorbs water and nutrients from stigmatic surface that are absent in pollen and then pollen tube emerges which grow-s through style.

Question 4.
Describe three devices by which cross pollination is encouraged in Angiosperms by avoiding self-pollination?
Answer:

Unisexuality, dichogamy, prepotency, heteromorphy and herkogamy are the outbreeding devices.
Unisexuality : The plants bear either male or female flowers. Due to unisexual nature, self-pollination is avoided. Plants are either dioecious, e.g. Papaya or monoecious, e.g. maize.
Heteromorphy : In same plants different types of flowers are produced. In these flowers, stigmas and anthers are situated at different levels. There is heterostyly and heteroanthy. This prevents self-pollination e.g. Primrose.
Herkogamy : In bisexual flowers we may come across mechanical device to prevent self-pollination. Natural physical barrier avoids contact of pollens with stigma. E.g. Calotropis where pollinia are situated below the stigma.

4. Long Answer Questions

Question 1.
Describe the process of double fertilization.
Answer:
Double fertilization:
(1) Out of the two male gametes produced by the male gametophyte in angiosperms, one unites with the female gamete and the other with the secondary nucleus. Since both the male gametes take part in fertilization and fertilization occurs twice, it is called double fertilization.

(2) During double fertilization, the pollen tube on reaching the ovule enters the embryo sac through micropyle and bursts in one of the synergids. Owing to this, the two male gametes contained in the pollen tube, are set free.

(3) Out of the two male gametes, one unites with the egg or female gamete and the other unites with the secondary nucleus of the embryo sac, forming a triploid or triple fusion nucleus, called the primary endosperm nucleus. The process involving the fusion of one of the male gametes with the egg nucleus, resulting in the formation of a diploid zygote is called syngamy.

(4) The reproductive process in which non-motile male nuclei are carried to the egg cell through a pollen tube is called siphonogamy.

(5) After fertilization, zygote develops into an embryo. Certain changes take place in the ovule leading to the development of a seed.

Question 2.
Explain the stages involved in the maturation of microspore into male gametophyte.
OR
Describe the development of male gametophyte before pollination in angiosperms.
OR
Sketch and label male gametophyte in angiosperm.
Answer:

Microspore or pollen grain is first cell of male gametophyte.
The protoplast of pollen grain divides mitotically to form two unequal cells – a small thin walled generative cell and a large naked vegetative or tube cell.
The generative cell possesses thin cytoplasm and a nucleus. It separates and floats in the cytoplasm of vegetative cell.
The vegetative, possesses thick cytoplasm, irregular shaped nucleus and the reserved food.
In majority of the angiosperms, the pollen grains are liberated at two-celled stage after the dehiscence of the anther.
The generative cell of the pollen grain divides by mitosis to form two male non-motile gametes.

Question 3.
Explain the development of dicot embryo.
Answer:
Development of embryo (dicot) in angio- sperm:

The oospore undergoes a transverse division to form a large basal cell towards the micropyle and a small apical or terminal cell towards the chalaza of the embryo sac. This two celled structure is called proembryo. The basal cell or suspensor initial undergoes repeated transverse divisions to form a multicellular structure called suspensor. The suspensor pushes the embryo towards the endosperm to draw its nutrition.

The development of embryo from a zygote is called embryogenesis.
The fusion of male gamete and an egg cell during fertilization results in the formation of a diploid zygote. The zygote develops a wall around it and is converted into oospore.
The apical cell or embryonal initial of the proembryo undergoes a transverse division followed by two vertical divisions at right angles to form an octant stage.
From octant, the lower four cells form hypocotyl and radicle while four cells of upper side form plumule with two cotyledons.
The lowermost cell of suspensor is hypophysis and by its further division forms part of radicle and root cap.
The cells from upper side of octant divide repeatedly to form heart shaped which elongated further to form two lateral cotyledons.
Enlargement of hypocotyl and cotyledon results into curved embryo which appears horse shoe shaped.

Question 4.
Draw a diagram of the L.S of anatropous ovule and list the components of embryo sac and mention their fate after fertilization.
Answer:
Components of Embryo sac.

Mature embryo sac is 7-celled and 8 nucleate.
Egg apparatus at micropylar end – with 2 synergids and egg cell.
Central cell with secondary nucleus formed by 2 polar nuclei
Antipodal cells at chalazal end – 3 cells.
Pollen tube enters the synergids, Synergids guide the growth of pollen tube towards egg.
Male gamete fuses with female gamete, i.e. syngamy to form zygote which develops into embryo.
One male gamete fuses with secondarynucleus to form primary endosperm nucleus (PEN) which forms endosperm, nutritive tissue for embryo.

5. Fill in the Blanks

Question 1.
The ……………… collects the pollen grains.
Answer:
biotic agents

Question 2.
The male whorl, called the ……………… produces ………………
Answer:
androecium, pollen grains

Question 3.
The pollen grains represent the ………………
Answer:
male

Question 4.
The ……………… contains the egg or ovum.
Answer:
embryo sac

Question 5.
…………….. takes place when one male gamete and the egg fuse together. The fertilized egg grows vs into seed from which the new plants can grow.
Answer:
Fertilization

Question 6.
The ……………… is the base of the flower to which other floral parts are attached.
Answer:
thalamus

Question 7.
……………… is the transfer of pollen grains from anther of the flower to the stigma of the same or a different flower.
Answer:
Pollination

Question 8.
Once the pollen reaches the stigma, pollen tube traverses down the ……………… to the ovary where fertilization occurs.
Answer:
style

Question 9.
The ……………… are coloured to attract the insects that carry the pollen. Some flowers also produce ……………… or ……………… that attracts insects.
Answer:
petals, fragrance, nectar

Question 10.
The whorl ……………… is green that protects the flower until it opens.
Answer:
Calyx.

6. Label the Parts of seed.

Answer:

7. Match the following

Column I (Structure Before seed formation)	Column II (Structure After seed formation)
A. Funiculus	i. Hilum
B. Scar of Ovule	ii. Tegmen
C. Zygote	iii. Testa
D. Inner Integument	iv. Stalk of Seed
	v. Embryo

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

12th Std Biology Questions And Answers:

12th Biology Chapter 7 Exercise Plant Growth and Mineral Nutrition Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 7 Plant Growth and Mineral Nutrition Textbook Exercise Questions and Answers.

Plant Growth and Mineral Nutrition Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 7 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 7 Exercise Solutions

1. Multiple choice questions

Question 1.
Which of the hormone can replace vernalization ?
(a) Auxin
(b) Cytokinin
(c) Gibberellins
(d) Ethylene
Answer:
(c) Gibberellins

Question 2.
The principle pathway of water translocation in angiosperms is ………………..
(a) Sieve cells
(b) Sieve tube elements
(c) Xylem
(d) Xylem and phloem
Answer:
(c) Xylem

Question 3.
Abscissic acid controls ………………..
(a) cell division
(b) leaf fall and dormancy
(c) shoot elongation
(d) cell elongation and wall formation
Answer:
(b) leaf fall and dormancy

Question 4.
Which is employed for artificial ripening of banana fruits?
(a) Auxin
(b) Ethylene
(c) Cytokinin
(d) Gibberellin
Answer:
(b) Ethylene

Question 5.
Which of the following is required for stimulation of flowering in plants?
(a) Adequate oxygen
(b) Definite photoperiod
(c) Adequate water
(d) Water and minerals
Answer:
(b) Definite photoperiod

Question 6.
For short day plants, the critical period is ………………..
(a) light
(b) dark/night
(c) UV rays
(d) Both (a) and (c)
Answer:
(b) dark/night

Question 7.
Which of the following is NOT day neutral plant?
(a) Tomato
(b) Cotton
(c) Sunflower
(d) Soybean
Answer:
(d) Soybean

Question 8.
Essential macro elements are ………………..
(a) manufactured during photosynthesis
(b) produced by enzymes
(c) obtained from soil
(d) produced by growth hormones
Answer:
(c) obtained from soil

Question 9.
Function of Zinc is ………………..
(a) closing of stomata
(b) biosynthesis of 3-IAA
(c) synthesis of chlorophyll
(d) oxidation of carbohydrates
Answer:
(b) biosynthesis of 3-LAA

Question 10.
Necrosis means ………………..
(a) yellow spot on the leaves
(b) death of tissue
(c) darkening of green colour in leaves
(d) wilting of leaves
Answer:
(b) death of tissue

Question 11.
Conversion of nitrates to nitrogen is called ………………..
(a) ammonification
(b) nitrification
(c) nitrogen fixation
(d) denitrification
Answer:
(d) denitrification

Question 12.
How many molecules of ATP are required to fix one molecule of nitrogen?
(a) 12
(b) 20
(c) 6
(d) 16
Answer:
(d) 16

2. Very short answer questions

Question 1.
Enlist the phases of growth in plants.
Answer:
The three phases of growth are phase of cell division, phase of cell enlargement and phase of cell maturation.

Question 2.
Give full form of IAA.
Answer:
Full form is Indole Acetic Acid.

Question 3.
What does it mean by ‘open growth’?
Answer:
In plants the growth is indeterminate and takes place throughout the life at specific regions having meristems.

Question 4.
Plant stress hormone.
Answer:
Abscissic acid.

Question 5.
What is denitrification?
Answer:
Anaerobic bacteria can convert nitrates of soil back into nitrogen gas. That process performed by denitrifying bacteria is denitrification.

Question 6.
Bacteria responsible for conversion of nitrite to nitrate.
Answer:
Nitrobacter.

Question 7.
What is the role of gibberellins in rosette plants?
Answer:
In rosette plants like beet and cabbage, bolting, i.e. elongation of internodes before flowering is observed due to effect of gibberellins.

Question 8.
Vernalization
Answer:
The response of plant to the influence of low temperature on flowering in plants is called vernalization.

Question 9.
Photoperiodism
Answer:
The response of plant to the influence of light for initiation of flowering is known as photoperiodism.

Question 10.
What is grand period of growth?
Answer:
There are three phases of growth and the total time required for all phases to occur is called grand period of growth.

3. Short Answer Questions

Question 1.
(i) Differentiation
Answer:

It is a process of maturation of cells derived from apical meristems.
Differentiation is a permanent change in structure and function of cells that leads to its maturation.
Cell undergoes major anatomical and physiological change during differentiation process.
In hydrophytic plants parenchyma cells develop large schizogenous cavities which help them in aeration, buoyancy and mechanical support.

(ii) Redifferentiation
Answer:

It is a process in which cells produced by de-differentiation lose their capacity of division and become mature.
The cells mature to perform specific function.
Interfascicular cambium is formed by process of dedifferentiation loses its capacity to divide.
Secondary xylem and secondary phloem is formed form this cambium in vascular cylinder.

Question 2.
Arithmetic growth and Geometric growth
Answer:

Arithmetic growth	Geometric growth
1. In arithmetic growth only one daughter cell continues to divide, while the other undergoes differentiation and maturation.	1. In geometric growth both the daughter cells continue to divide and redivide again and again.
2. Rate of growth is constant.	2 Rate growth is initially slow but later on rapid rate.
3. Linear curve is obtained.	3. Exponential curve is obtained.
4. Mathematical expression is L_t = Lo + rt whereL_t = length of time ‘t’ L_o = Length at time zero r_t = growth rate, t = time of growth	4. Mathematical expression is W_t = Wo^{e rt} where, W_t = final size, Wo = initial size, r = growth rate, t = time of growth E = base of natural logarithm
5. e.g. Elongation of root	5. e.g. Divisions of zygote during embryo development.

Question 3.
Enlist the role and deficiency symptoms of: (a) nitrogen (b) phosphorus (c) potassium.
Answer:
(a) Nitrogen:
Role : Constituent of proteins as amino acids, nucleic acids, vitamins, hormones, coenzymes, ATP and chlorophyll molecule.
Deficiency symptoms : stunted growth and chlorosis.

(b) Phosphorus:
Role : Constituent of cell membrane, certain proteins, nucleic acids and nucleotides, required for all phosphorylation reactions.
Deficiency symptoms : Poor growth, leaves dull green

(c) Potassium :
Role : Determination of anion – cation balance in cell, necessary for protein synthesis, involved in formation of cell membrane, opening and closing of stomata, activates enzymes, helps in maintenance of turgidity of cells.
Deficiency symptom : Yellow edges in leaves, premature death.

Question 4.
What is short day plant? Give any two examples.
Answer:
The plants which flower when the day length or light period is shorter than the critical photoperiod are called short day plants or SDP
SDPs usually flower during winter and late summer.
Examples – Dahlia, Aster, Tobacco, Chrysanthemum, Soybean (Glycine max) and Cocklebur (Xanthium).

Question 5.
What is vernalization? Give its significance.
Answer:
A low temperature or chilling treatment that induces early flowering in plants is known as vernalization.

Significance:

Due to chilling treatment crops can be produced earlier.
Crops can be grown in areas where they do not grow naturally.

4. Long answer questions

Question 1.
Explain sigmoid growth curve with the help of diagram.
Answer:

When growth occurs in plants three distinct phases of growth are noticed.
Phase of cell formation is first phase where meristematic cells divide and new cells added.
In phase of cell enlargement newly formed cells elongate and with turgidity there is cell enlargement.
In phase of cell maturation cells get differentiated.
When we compare the growth rate it differs in these three phases.
In first phase or lag phase it is slow, while in log phase or exponential phase, growth rate accelerates and it reaches maximum.
In stationary phase of maturation growth rate slows down and comes to steady state.
When this changing rate of growth is plotted against time duration in a graph a sigmoid or S-shaped growth curve is obtained.

Question 2.
Describe the types of plants on the basis of photoperiod required, with the help of suitable examples.
Answer:

Effect of light duration on flowering of plants is known as photoperiodism.
Depending on photoperiodic response, plants are categorised into three types – Short day plants, long day plants and day neutral plants.

1. Short day plants : Plants that flower under short day length conditions are called short day plants. Plants such as Dahlia, Xanthium, Soybean, Aster, Tobacco and Chrysanthemum are short day plants or SDR. Short day plants require a long uninterrupted dark period for flowering. Therefore, they are also called long night plants.

2. Long day plants : Plants that flower only when they are exposed to light period longer than their critical photoperiod are called long day plants or LDP Long day plants require a short dark or night period for flowering. Hence, they are also called short night plants. Plants such as radish, spinach, wheat, poppy, cabbage, pea, sugar beet, etc. are long day plants.

3. Day neutral plants : Plants in which the flowering is not affected by the day length period are called day neutral plants or DNP or photoneutral plants. Plants such as cucumber, sunflower, cotton, balsam, maize, tomato, etc. are day neutral plants.

Question 3.
Explain biological nitrogen fixation with example.
Answer:

Conversion of atmospheric nitrogen into nitrogenous salts to make it available to plants for its update is described as nitrogen fixation.
When living organisms are involved in nitrogen fixation process it is known as biological nitrogen fixation.
The process is mainly carried out by prokaryotic organisms, i.e. different kinds of bacteria present in soil.
The nitrogen fixing organisms are known as diazotrophs or nitrogen fixers and about 70% nitrogen is fixed by them.
The nitrogen fixers are either free living bacteria or symbiotic associated with other higher plants e.g. Rhizobium.
The cyanobacteria have specialized cells heterocysts which help in process of nitrogen fixation.
Nitrogen fixation is high energy requiring process and 16 ATP molecules are needed for fixation of one molecule of nitrogen to ammonia.
Soil bacteria like Nitrosomonas, Nitrosocyccus convert ammonia to nitrate and the Nitrobacter convert nitrite to nitrate. This is known as nitrification, biological oxidation.
These bacteria are chemoautotrophic and utilize these processes for their metabolism.
Fabaceae plants like pea, bean have root nodules which harbour symbiotic bacterium Rhizobium which fixes nitrogen. It is host specific, soil bacterium, Nitrogen is made available to host plant.

Question 4.
Write on macro and micro nutrients required for plant growth.
Answer:

Plants absorb mineral nutrients from their surroundings.
For a proper growth of plants about 35 to 40 different elements are required.
Plants absorb these nutrients in ionic or dissolved form from soil with their root system e.g. Phosphorus as PO₄^–, Sulphur as SO₄^2- etc.
Based on their requirement in quantity, they are classified as major nutrients or macronutrients and those needed in small amounts Eire minor or micronutrients.
Macroelements are required in large amounts, as they play nutritive and structural roles e.g. C, H, O, R Mg, N, K, S and Ca. – Ca pectate cell wall component, Mg component of chlorophyll.
C, H, O are non-mineral major elements obtained from air and water e.g. CO₂ is source of carbon, Hydrogen from water.
Microelements are required in traces as they mainly have catalytic role as co-factors or activators of enzymes.
Microelements may be needed for certain activity in life cycle of plant e.g. B for pollen germination, Si has protective role during stress conditions and fungal attacks, Al enhances availability of phosphorus.
The important micronutrients for plant growth are Mn, B, Cu, Zn, Cl.

12th Std Biology Questions And Answers:

12th Biology Chapter 13 Exercise Organisms and Populations Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 13 Organisms and Populations Textbook Exercise Questions and Answers.

Organisms and Populations Class 13 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 13 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 13 Exercise Solutions

1. Multiple choice questions

Question 1.
Which factor of an ecosystem includes plants, animals and microorganisms?
(a) Biotic factor
(b) Abiotic factor
(c) Direct factor
(d) Indirect factor
Answer:
(a) Biotic factor

Question 2.
An assemblage of individuals of different species living in the same habitat and having functional interactions is ……………….
(a) Biotic community
(b) Ecological niche
(c) Population
(d) Ecosystem
Answer:
(a) Biotic community

Question 3.
Association between sea anemone and Hermit crab in gastropod shell is that of ………………..
(a) Mutualism
(b) Commensalism
(c) Parasitism
(d) Amensalism
Answer:
(b) Commensalism

Question 4.
Select the statement which explains best parasitism.
(a) One species is benefited.
(b) Both the species are benefited.
(c) One species is benefited, other is not affected.
(d) One species is benefited, other is harmed.
Answer:
(d) One species is benefited, other is harmed.

Question 5.
Growth of bacteria in a newly inoculated agar plate shows ………………….
(a) exponential growth
(b) logistic growth
(c) Verhulst-Pearl logistic growth
(d) zero growth
Answer:
(c) Verhulst-Pearl logistic growth

2. Very short answer questions.

Question 1.
Define the following terms
a. Commensalism
Answer:
The interaction between two species in which one species gets benefits and the other is neither harmed nor benefited is called commensalism.

b. Parasitism
Answer:
The interaction between two species in which one parasitic species derives benefit from the other host species by harming it is called parasitism.

c. Camouflage
Answer:
Camouflage is the disguising colouration or behaviour to merge with the surrounding so that prey or predator can remain hidden.

Question 2.
Give one example for each
a. Mutualism
b. Interspecific competition
Answer:
a. Lichen is composed of alga (cyanobacteria) and fungus. They cannot survive independently. Their association is mutualistic alga synthesises food by photosynthesis and fungus does the absorption of moisture.

b. Leopard and lion competing for a same prey. Sheep and cow competing for grazing in the same land.

Question 3.
Name the type of association:
a. Clown fish and sea anemone
b. Crow feeding the hatchling of Koel
c. Humming birds and host flowering plants
Answer:
a. Commensalism
b. Brood parasitism
c. Mutualism

Question 4.
What is the ecological process behind the biological control method of managing with pest insects?
Answer:

Pest insects act as prey to predator birds or frogs.
The biological control method consists of releasing the predators in the farms so that they can control the pest population in the natural way.
This also eliminates the use of chemical pesticides.
Frogs are natural predators of locust, therefore the population of this hazardous insect is controlled by frogs and the produce from agricultural farm can be saved.

Protocooperation:

Protocooperation is a type of population interaction where two species interact with each other.
Both are benefited but they have no need to interact with each other.
They can survive and grow even in the absence of other species.
Therefore this interaction is purely for the gain that they receive in such type of interaction.
The interaction that occurs can be between different kingdoms.

3. Short answer questions.

Question 1.
How is the dormancy of seeds different from hibernation in animals?
Answer:
In dormancy seed is not showing any metabolic activities. It can come back to life if and only if it gets suitable moisture and sunlight. Hibernation is suspended state, in which metabolic reactions do take place but at a very reduced pace. Animal arouses on its own after the winter sleep is over. This arousal is spontaneous and depends upon the ambient temperature. Dormant seed does not show such change unless it is planted or thrown in to moist place.

Question 2.
If a marine fish is placed in a fresh water aquarium, will it be able to survive? Give reason.
Answer:
Marine fish has its own osmoregulation which is different from the osmoregulation seen in fresh water fish. In marine water, the ambient salinity is more than the concentration of ions in the body. But in fresh water reverse is the case. Therefore, marine fish has different machinery to cope up with high saline environment. Therefore, it cannot survive in fresh water as its osmoregulation is not possible in less saline waters.

Question 3.
How is the dormancy of seeds different from hibernation in animals?
Answer:
In dormancy seed is not showing any metabolic activities. It can come back to life if and only if it gets suitable moisture and sunlight. Hibernation is suspended state, in which metabolic reactions do take place but at a very reduced pace. Animal arouses on its own after the winter sleep is over. This arousal is spontaneous and depends upon the ambient temperature. Dormant seed does not show such change unless it is planted or thrown into moist place.

Question 4.
An orchid plant is growing on the branch of mango tree. How do you describe this interaction between the orchid and the mango tree?
Answer:

Orchid is an epiphyte. It gets the support from the mango tree. But it does not cause any harm to the mango tree.
Mango tree does not derive any benefit from this association. Therefore, this interaction is of type of commensalism.

Question 5.
Distinguish between the following:
a. Hibernation and Aestivation
Answer:

Hibernation	Aestivation
1. Hibernation is winter sleep shown by some warm-blooded and some cold-blooded animals.	1. Aestivation is the type of summer sleep, shown by cold-blooded animals.
2. It is for the whole winter.	2. It is of short duration.
3. The animals look out for the warmer place to enter into hibernation.	3. Animals search for the moist, shady and cool place to sleep.
4. Metabolic activities of hibernators slowdown in this dormant stage.	4. Metabolic activities of aestivators remain low during aestivation period.
5. Hibernation helps in maintaining the body temperature and prevents any internal body damage due to low temperatures. E.g. Bats, birds, mammals, insects, etc. show hibernation.	5. Aestivation helps in maintaining the body temperature by avoiding the excessive water loss and thus prevents any internal body damaged due to high temperatures. E.g. Bees, snails, earthworms, salamanders, frogs, earthworms, crocodiles, tortoise, etc. show aestivation.

b. Ectotherms and Endotherms
Answer:

Ectotherms	Endotherms
1. Ectotherms do not have ability to generate heat in the body.	1. Endotherms possess the ability to generate their own body heat.
2. Ectotherms depend on the environmental sources to heat their bodies. E.g sunlight.	2. Endotherms do not depend upon outside sources to generate heat.
3. Most ectotherms are confined to warmer parts of the world.	3. Endotherms inhabit coldest parts of the earth.
4. Body temperature of ectotherms fluctuate according to ambient temperature.	4. Body temperatures of endotherms remain constant and do not show fluctuations as per ambient temperatures.
5. Metabolic rate of ectotherms is low. E.g. Amphibians and reptiles.	5. Metabolic rate of endotherms is high. E.g. Mammals and birds

c. Parasitism and Mutualism
Answer:

Parasitism	Mutualism
1. Parasitism is the relationship where only one organism receive benefits, while the other is harmed in return.	1. Mutualism is the relationship where both the organisms of distinct species are benefited.
2. Parasite cannot survive without host but if the host is overexploited then parasite too dies.	2. Both the species are dependent on each other for their benefits and survival.
3. Parasitism can be facultative or obligatory.	3. Mutualism is obligatory relationship.
4. Parasitism is a negative interaction.	4. Mutualism is a positive interaction.

Question 6.
Write a short note on
a. Adaptations of desert animals
Answer:

Animals which are well-adapted to live in deserts are called xerocoles. These animals show adaptations for water conservation or heat tolerance.
These animals show low basal metabolic rate. They obtain moisture from succulent plants and rarely drink water. E.g Gazella and Oryx.
Desert animals like camel produce concentrated urine and dry dung.
Many other hot desert animals are nocturnal, seeking out shade during the day or dwelling underground in burrows.
Smaller animals from desert, emerge from their burrows at night.
Mammals living in cold deserts have developed greater insulation through warmer body fur and insulating layers of fat beneath the skin.
Few adaptations to desert life are unable to cool themselves by sweating so they shelter during the heat of the day. Many desert reptiles are ambush predators and often bury themselves in the sand, waiting for prey to come within range.
Other animals have bodies designed to save water. Scorpions and wolf spiders have a thick outer covering which reduces moisture loss. The kidneys of desert animals concentrate urine, so that they excrete less water.

b. Adaptations of plants to water scarcity
Or
Adaptations in desert plants.
Answer:

Thick cuticle on their leaf surfaces
Stomata of desert plants is sunken that is it is in deep pits to minimize loss of water through transpiration.
Desert plants also have a special photosynthetic pathway (CAM -Crassulacean acid metabolism) that enables their stomata to remain closed during daytime.
Some desert plants like Opuntia, have their leaves reduced or they are modified to spines. Loss of leaf surface helps in prevention of transpiration.
Photosynthetic function is taken over by the flattened stems called as phylloclade.

c. Behavioural adaptations in animals
Answer:

Behavioural responses to cope with variations in their environment are shown by few animals.
Desert lizards manage to keep their body temperature fairly constant by behavioural adaptations. They bask in the sun and absorb heat, when their body temperature drops below the comfort zone, but move into shade, when the ambient temperature starts increasing. Even snakes also show basking during winter months.
Since they are ectothermic, this kind of behaviour saves them from extreme temperatures.
Many smaller animals show burrowing behaviour to adapt to the temperature extremes.
Some species burrow into the sand to hide and escape from the heat.
Migrations shown by the birds and mammals are also behavioural responses for adapting to severe winter temperatures.

Question 7.
Define Population and Community.
Answer:
Population:
Group of organisms belonging to same species that can potentially interbreed with each other and live together in a well-defined geographical area by sharing or competing for similar resources, is called population.

Community:
Several populations of different species in a particular area makes a community.

4. Long answer questions.

Question 1.
With the help of suitable diagram, describe the logistic population growth curve.
Answer:

Naturally all populations of any species always have limited resources to permit exponential growth. Due to this there is always competition between individuals for limited resources. The most fit organisms succeed by survival and reproduction.
A given habitat has enough resources to support a maximum possible number, but beyond a particular limit the further growth is impossible.
This limit is called nature’s carrying capacity (K) for that species in that habitat.
A population growing in a habitat with limited resources show following phases in a sequential manner, (a) A lag phase (b) Phase of acceleration (c) Phase of deceleration (d) An asymptote, when the population density reaches the carrying capacity.
A plot of N in relation to time (t) results in a sigmoid curve. This type of population growth is called Verhulst-Pearl Logistic Growth.
Since resources for growth for most animal populations are finite and become limiting sooner or later, the logistic growth model is considered as a more realistic one.
Logistic growth thus always shows sigmoid curve.

Question 2.
Enlist and explain the important characteristics of a population.
Answer:
Important characteristics of a population are as follows:
1. Natality:

Natality is the birth rate of a population. Due to increased natality the population density rises.
Natality is a crude birth rate or specific birth rate.
Crude birth rate : Number of births per 1000 population/year gives crude birth rate. Crude birth rate is helpful in calculating population size.
Specific birth rate : Crude birth rate is relative to a specific criterion such as age. E.g. If in a pond, there were 200 carp fish and their population rises to 800. Then, taking the current population to 1000, the birth rate becomes 800/200 = 4 offspring per carp per year. This is specific birth rate.
Absolute Natality : The number of births under ideal conditions when there is no competition and the resources such as food and water are abundant, then it give absolute natality.
Realized Natality : The number of births under different environmental pressures give realized natality. Absolute natality will be always more than realized natality.

2. Mortality:

Mortality is the death rate of a population. It gives a measure of the number of deaths in a particular population, in proportion to the size of that population, per unit of time.
Mortality rate is typically expressed in deaths per 1,000 individuals per year.
A mortality rate of 9.5 (out of 1,000) in a population of 1,000 would mean 9.5 deaths per year in that entire population or 0.95% out of the total.
Absolute Mortality : The number of deaths under ideal conditions when there is no competition, and all the resources such as food and water are abundant, then it gives absolute mortality.
Realized Mortality : The number of deaths under environmental pressures come into play gives realized mortality.
It must be remembered that absolute mortality will always be less than realized mortality.

3. Density:
The density of a population in a given habitat during a given period fluctuates due to changes in four basic processes, viz.

Natality i.e. birth rate (The number of births during a given period in the population that are added to the initial density).
Mortality i.e. death rate (The number of deaths in the population during a given period).
Immigration i.e. number of individuals of the same species that have come into the habitat from elsewhere during the time period under consideration.
Emigration i.e. the number of individuals of the population who left the habitat and gone elsewhere during the time period under consideration.
Natality and immigration increase in population density whereas mortality and emigration decrease it.

4. Sex ratio : Ratio of the number of individuals of one sex (male) to that of the other sex (female) is called sex ratio. In nature male, female ratio is always 1 : 1. This 1 : 1 ratio is called evolutionary stable strategy of ESS for each population.

5. Age distribution and age pyramid : This parameter is important for human population. Each population is composed of individuals of different ages. The age distribution is plotted for the population, the resulting structure is called an age pyramid. For making the age pyramid, the entire population is divided into three age groups as Pre-Reproductive (age 0-14 years), Reproductive (age 15-44 years) and Post-reproductive (age 45 -85+ years).

6. Growth : Growth of a population causes rise in its density. The size and density are dynamic parameters as they keep on changing with time, and various factors including food, predation pressure and adverse weather. From the density, one comes to know if the population is flourishing or declining.

12th Std Biology Questions And Answers: