12th Biology Chapter 14 Exercise Ecosystems and Energy Flow Solutions Maharashtra Board

Class 12 Biology Chapter 14

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 14 Ecosystems and Energy Flow Textbook Exercise Questions and Answers.

Ecosystems and Energy Flow Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 14 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 14 Exercise Solutions

1. Multiple choice questions

Question 1.
Which one of the following has the largest population in a food chain?
(a) Producers
(b) Primary consumers
(c) Secondary consumers
(d) Decomposers
Answer:
(a) Producers

Maharashtra Board Class 12 Biology Solutions Chapter 14 Ecosystems and Energy Flow

Question 2.
The second trophic level in a lake is ……………………
(a) Phytoplankton
(b) Zooplankton
(c) Benthos
(d) Fishes
Answer:
(b) Zooplankton

Question 3.
Secondary consumers are …………………….
(a) Herbivores
(b) Producers
(c) Carnivores
(d) Autotrophs
Answer:
(c) Carnivores

Question 4.
What is the % of photosynthetically active radiation in the incident solar radiation?
(a) 100%
(b) 50%
(c) 1-5%
(d) 2-10%
Answer:
(b) 50%

Question 5.
Give the term used to express a community in its final stage of succession?
(a) End community
(b) Final community
(c) Climax community
(d) Dark community
Answer:
(c) Climax community

Question 6.
After landslide which of the following type of succession occurs?
(a) Primary
(b) Secondary
(c) Tertiary
(d) Climax
Answer:
(a) Primary

Question 7.
Which of the following is most often a limiting factor of the primary productivity in any ecosystem?
(a) Carbon
(b) Nitrogen
(c) Phosphorus
(d) Sulphur
Answer:
(c) Phosphorus

2. Very short answer question.

Question 1.
Give an example of ecosystem which shows inverted pyramid of numbers.
Answer:
Number of insects dependent on a single tree, is an example of ecosystem having inverted pyramid of numbers.

Question 2.
Give an example of ecosystem which shows inverted pyramid of biomass.
Answer:
Oceanic ecosystem has inverted pyramid of biomass.

Question 3.
Which mineral acts as limiting factor for productivity in an aquatic ecosystem?
Answer:
Phosphorus acts as limiting factor for productivity in an aquatic ecosystem.

Maharashtra Board Class 12 Biology Solutions Chapter 14 Ecosystems and Energy Flow

Question 4.
Name the reservoir and sink of carbon in carbon cycle.
Answer:
Atmosphere is the reservoir of carbon cycle, while fossil fuels embedded in ocean and oceanic waters are the sink of carbon in carbon cycle.

3. Short answer questions.

Question 1.
Upright and inverted pyramid of biomass.
Answer:

Upright pyramid Inverted pyramid
1. In upright pyramid, the number and biomass of the organisms which are at first trophic level of producers is high. 1. In inverted pyramid, the number and biomass of organisms at first trophic levels of producers is lowest.
2. The biomass goes on decreasing at each trophic level. 2. The biomass foes on increasing at each trophic level.
3. The base of the pyramid is always in large number of producers. 3. The base of pyramid is always in small numbers of producers.
4. Pyramid is always upright. 4. Pyramid is always inverted.

Question 2.
Food chain and Food web.
Answer:

Food chain Food web
1. Food chain is the linear sequence of organisms for feeding purpose. 1. Food web is interconnections between many small food chains.
2. In food chain the flow of energy is through a single straight pathway from the lower trophic level to the higher trophic level. 2. In food web, the energy flow is interconnected through numerous food chains in the ecosystem.
3. In a food chain, members present at higher trophic level feeds on only single type of organisms. 3. In a food web, one organism can feed on multiple types of organisms.
4. Energy flow can be easily calculated in food chain. 4. Energy flow is difficult to calculate in a food web.
5. In food chain there is increased instability due to increasing number of separate and confined food chains. 5. In food web there is increased stability due to the presence of the complex food chains.
6. The whole food chain gets affected even if one group of an organism is disturbed. 6. The food web does not get disturbed by the removal of one group of organisms.
7. Member of higher trophic level depends or feed upon the single type of organisms of the lower trophic level. 7. The members of higher trophic level depend or feed upon many different types of the organism of the lower trophic level.
8. Food chain consists of only 4-6 trophic levels of different species. 8. Food web contains numerous trophic levels and also of different populations of species.
9. Competition is seen in members of same trophic level. 9. Competition is seen in members of same as well as different trophic levels.
10. Food chains are of two types:

1. Grazing food chain 2. Detritus food chain.

10. In food web there are no types.

4. Long answer questions

Question 1.
Define ecological pyramids and describe with examples, pyramids of number and biomass.
Answer:
1. Ecological Pyramids : Ecological Pyramids are the representation of relationships between different components of ecosystem at successive trophic levels.

2. Pyramid of numbers:

  • Pyramid of numbers is the diagrammatic representation which shows the relationship between producers, herbivores and carnivores at successive trophic levels in terms of their numbers.
  • As we go up the trophic levels, the interdependent organisms keep on reducing in their numbers.
  • For example, the number of grasses are more than the number of herbivores which eat them. The number of herbivores such as rabbits would be lesser than grass but greater than the carnivores that are dependent upon the population of rabbits.
  • Thus, the producers would be more than primary consumers and primary consumers would be more than secondary consumers. The top level consumers would be least in their numbers. This pyramid shows upright nature.

Maharashtra Board Class 12 Biology Solutions Chapter 14 Ecosystems and Energy Flow

3. Pyramid of biomass:
(1) Pyramid of biomass are constructed by taking into consideration the different biomass in every successive trophic level.
(2) Pyramid of biomass in seas in inverted as the biomass of fishes is more than the biomass of phytoplankton.
Maharashtra Board Class 12 Biology Solutions Chapter 14 Ecosystems and Energy Flow 1

Question 2.
What is primary productivity? Give brief description of factors that affect primary productivity.
Answer:
(1) Primary Productivity : The rate of generation of biomass in an ecosystem which is expressed in units of mass per unit surface (or volume) per unit time, for instance grams per square metre per day (g/m²/day) is called primary productivity.

(2) Primary productivity is described as gross primary productivity (GPP) and net primary productivity (NPP).

(3) The rate of production of organic matter during photosynthesis is called gross primary productivity of an ecosystem. Of this the amount of energy lost through respiration of plants is called respiratory losses.

(4) Gross primary productivity minus respiratory losses gives the net primary productivity (NPP).

(5) Net primary productivity is the available biomass for the consumption to heterotrophs (herbivores, carnivores and decomposers).

(6) Factors affecting primary productivity: Gross primary productivity (GPP) depends on the following factors:

  • Plant species inhabiting a particular area.
  • Variety of environmental factors such as temperature, sunlight, salinity, oxygen and carbon dioxide content, etc.
  • Availability of nutrients and
  • Photosynthetic capacity of plants.

Question 3.
Define decomposition and describe the processes and products of decomposition.
Answer:

  1. Decomposition is the process carried out by the decomposer organisms.
  2. Most of the bacteria, actinomycetes and fungi are decomposers. They convert the dead and decaying organic matter into simpler compounds. These simpler inorganic substances return back to the environment.
  3. Decomposition takes place through detritus food chain. It starts from the dead organic matter. Detritus eating organisms called detritivores like earthworm, etc. breakdown the detritus into smaller fragments. Therefore, this first step of decomposition is called fragmentation.
  4. Water soluble inorganic nutrients seep into the soil after fragmentation. These nutrients get precipitated as salts. Therefore, this second step of decomposition is called leaching.
  5. The third step of decomposition is called catabolism. In this step, fungal and bacterial enzymes degrade the detritus into simple inorganic substances.
  6. The partially decomposed organic matter is called humus which is formed by the process of humification. Humus is a dark coloured amorphous substance which is the reservoir of nutrients.
  7. Humus too undergoes decomposition by bacterial action at a very slow rate and ultimately releases inorganic matter. This process is therefore called mineralization.
  8. Decomposition requires oxygen in greater amount. The rate of decomposition is dependent upon the temperature and the humidity of the environment.

Question 4.
Write important features of a sedimentary cycle in an ecosystem.
Answer:

  1. Reservoir of sedimentary cycles is earth’s crust.
  2. The nutrients such as phosphorus which show sedimentary cycle, moves through hydrosphere, lithosphere and biosphere.
  3. There is no respiratory release of nutrients into the atmosphere which show sedimentary cycle.
  4. Natural reservoir of such nutrients are usually in the form of rocks. The rocks upon weathering release such nutrients into circulation.
  5. Sedimentary cycles are very slow in their reactions.

Question 5.
Describe carbon cycle and add a note on the impact of human activities on carbon cycle.
Answer:
I. Carbon cycle:
(1) The entire carbon cycle has following basic processes viz. Photosynthesis, Respiration, Decomposition, Sedimentation and Combustion.

(2) Carbon is an important element as it forms 49% of the dry weight of all organisms. 71% of global carbon is present in the oceans. Therefore, ocean is the major reservoir of carbon. Carbon is also present in all fossil fuels. This is long term storage places or sinks for carbon which is in the form of coal, natural gas, etc.

Maharashtra Board Class 12 Biology Solutions Chapter 14 Ecosystems and Energy Flow

(3) Respiration and photosynthesis are the two events that keep the carbon in cyclic circulation. During respiration, oxygen is used for combustion of carbohydrates as a result of which carbon dioxide and water are formed with the release of energy. The process of photosynthesis utilizes carbon dioxide and water vapour liberating oxygen and producing carbohydrates at the same time.

(4) Solar energy is stored in the carbon-carbon bonds of carbohydrates during photosynthesis whereas respiration releases the same stored energy.

(5) The main reservoirs for carbon dioxide are in the oceans and in rocks. Carbon dioxide is highly soluble in water and forms mild carbonic acid upon dissolving. This dissolved carbon dioxide precipitate as a solid rock or limestone which is calcium carbonate. This reaction in the seas is aided by corals and algae which in turn builds the coral reefs made up of limestone.

(6) Carbon moves through food chains. Autotrophic green plants on land and in water take up carbon dioxide and manufacture carbohydrates by the process of photosynthesis. The carbon stored in plants has three different fates, viz. liberation into atmosphere, consumption by animals upon feeding, storage in the plant till the plant dies.

(7) Animals get their carbon requirement through their food. When autotrophs are consumed, the heterotrophs obtain carbon. Carbon in animals also has three fates, viz. release back into the atmosphere in the process of respiration, release of stored carbon from the body by the action of decomposers or conversion into fossil fuels if buried intact.

(8) Fossil fuels such as coal, oil, natural gas, etc. can be mined and burned for energy purposes. This burning releases carbon dioxide back into the atmosphere.

(9) Carbon from limestone can also be released if pushed to the surfaces and slowly weathered away. Subducting and volcanic eruptions can also release the stored carbon from sediments.
Maharashtra Board Class 12 Biology Solutions Chapter 14 Ecosystems and Energy Flow 2

II. Impact of human activities on carbon cycle:
(1) Excessive burning of fossils fuels for power plants, industrial processes and vehicular traffic, adds excessive carbon dioxide into atmosphere. When fossil fuels burn to run factories, power plants, motor vehicles, most of the carbon quickly enters the atmosphere as carbon dioxide gas.

(2) Each year, 5.5 billion tonnes of carbon is released through combustion of fossil fuels. Of this massive amount, 3.3 billion tonnes stays in the atmosphere.

(3) Rapid deforestation also increases carbon dioxide. Since plants absorb carbon dioxide for their photosynthesis, they always reduce the concentration of CO2. But deforestation upsets this balance.

(4) Massive burning of fossil fuel for energy and transport, have significantly increased the rate of release of carbon dioxide into the atmosphere which is causing global warming and resultant climate change.

12th Std Biology Questions And Answers:

12th Chemistry Chapter 11 Exercise Alcohols, Phenols and Ethers Solutions Maharashtra Board

Class 12 Chemistry Chapter 11

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 11 Alcohols, Phenols and Ethers Textbook Exercise Questions and Answers.

Alcohols, Phenols and Ethers Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 11 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 11 Exercise Solutions

1. Choose the correct option.

Question i.
Which of the following represents the increasing order of boiling points of (1), (2) and (3)?
(1) CH3 – CH2 – CH2 – CH2 – OH
(2) (CH3)2 CH – O – CH3
(3) (CH3)3COH
A. (1) < (2) < (3)
B. (2) < (1) < (3)
C. (3) < (2) < (1)
D. (2) < (3) < (1)
Answer:
(a) (1) < (2) < (3)

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Question ii.
Which is the best reagent for carrying out following conversion ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 272
A. LiAlH4
B. Conc. H2SO4, H2O
C. H2/Pd
D. B2H6, H2O2 – NaOH
Answer:
B. Conc. H2SO4, H2O

Question iii.
Which of the following reaction will give ionic organic product on reaction ?
A. CH3 – CH2 – OH + Na
B. CH3 – CH2 – OH + SOCl2
C. CH3 – CH2 – OH + PCl5
D. CH3 – CH2 – OH + H2SO4
Answer:
C. CH3 – CH2 – OH + PCl5

Question iv.
Which is the most resistant alcohol towards oxidation reaction among the follwoing ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 273
Answer:
(c)

Question v.
Resorcinol on distillation with zinc dust gives
A. Cyclohexane
B. Benzene
C. Toluene
D. Benzene-1, 3-diol
Answer:
(b) Benzene

Question vi.
Anisole on heating with concerntrated HI gives
A. Iodobenzene
B. Phenol + Methanol
C. Phenol + Iodomethane
D. Iodobenzene + methanol
Answer:
B. Phenol + Methanol

Question vii.
Which of the following is the least acidic compound ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 274
Answer:
(b)

Question viii.
The compound incapable of hydrogen bonding with water is ……
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 275
Answer:
(b)

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Question ix.
Ethers are kept in air tight brown bottles because
A. Ethers absorb moisture
B. Ethers evaporate readily
C. Ethers oxidise to explosive peroxide
D. Ethers are inert
Answer:
C. Ethers oxidise to explosive peroxide

Question x.
Ethers reacts with cold and concentrated H2SO4 to form
A. oxonium salt
B. alkene
C. alkoxides
D. alcohols
Answer:
A. oxonium salt

2. Answer in one sentence/ word.

Question i.
Hydroboration-oxidation of propene gives…..
Answer:
n-propyl alcohol (CH3 – CH2 – CH2 – OH)

Question ii.
Write the IUPAC name of alcohol having molecular formula C4H10O which is resistant towards oxidation.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 244

Question iii.
Write the structure of optically active alcohol having molecular formula C4H10O
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 245

Question iv.
Write name of the electrophile used in Kolbe’s Reaction.
Answer:
Electrophile : Carbon dioxide (O = C = O)

3. Answer in brief.

Question i.
Why phenol is more acidic than ethyl alcohol ?
Answer:
(1) In ethyl alcohol, the -OH group is attached to sp3 – hybridised carbon while in phenols, it is attached to sp2 – hybridised carbon.

(2) Due to higher electronegativity of sp2 – hybridised carbon, electron density on oxygen decreases. This increases the polarity of O-H bond and results in more ionization of phenol than that of alcohols.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 117

(3) Electron donating inductive effect (+1 effect) of the alkyl group destabilizes alkoxide ion. As a result alcohol does not ionize much in water, therefore alcohol is neutral compound in aqueous medium.

(4) In alkoxide ion, the negative charge is localized on oxygen, while in phenoxide ion the negative charge is delocalized. The delocalization of the negative charge (structure I to V) makes phenoxide ion more stable than that of phenol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 118

The delocalization of charge in phenol (structures VI to X), the resonating structures have charge separation (where oxygen atom of OH group to be positive and delocalization of negative charge over the ortho and para positions of aromatic ring) due to which phenol molecule is less stable than phenoxide ion. This favours ionization of phenol. Thus phenols are more acidic than ethyl alcohol.

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Question ii.
Why p-nitrophenol is a stronger acid than phenol ?
Answer:
(1) In p-nitrophenol, nitro group (NO2) is an electron withdrawing group present at para position which enhances the acidic strength (-1 effect). The O-H bond is under strain and release of proton (H+) becomes easy. Further p-nitrophenoxide ion is more stabilised due to resonance.

(2) Since the absence of electron withdrawing group (like – NO2) in phenol at ortho and para position, the acidic strength of phenol is less than that of p-nitrophenol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 119

Question iii.
Write two points of difference between properties of phenol and ethyl alcohol.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 122

Question iv.
Give the reagents and conditions necessary to prepare phenol from
a. Chlorobenzene
b. Benzene sulfonic acid.
Answer:
(1) From chlorobenzene : Reagents required : NaOH and dil. HC1 Temperature : 623 K, Pressure : 150 atm
(2) From Benzene sulphonic acid : Reagents required : aq NaOH, caustic soda, dil. HC1 Temperature : 573 K

Question v.
Give the equations of the reactions for the preparation of phenol from isopropyl benzene.
Answer:
Preparation of phenol from cumene (isopropylbenzene) : This is the commercial method of preparation of phenol. When a stream of air is passed through cumene (isopropylbenzene) suspended in aqueous Na2CO3 solution in the presence of cobalt naphthenate catalyst, isopropyl benzene hydroperoxide or cumene hydroperoxide is formed. Isopropylbenzene hydroperoxide on warming with dil. H2SO4 gives phenol and acetone. Acetone is an important by-product of the reaction and is separated by distillation. The reaction is called auto oxidation.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 111

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Question vi.
Give a simple chemical test to distinguish between ethanol and ethyl bromide.
Answer:
When ethyl bromide is heated with aq NaOH; ethyl alcohol is formed whereas ethanol does not react with aq NaOH
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 123

4. An ether (A), C5H12O, when heated with excess of hot HI produce two alkyl halides which on hydrolysis form compound (B)and (C), oxidation of (B) gave and acid (D), whereas oxidation of (C) gave a ketone (E). Deduce the structural formula of (A), (B), (C), (D) and (E).
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 243

5. Write structural formulae for

a. 3-Methoxyhexane
b. Methyl vinyl ether
c. 1-Ethylcyclohexanol
d. Pentane-1,4-diol
e. Cyclohex-2-en-1-ol
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 35

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

6. Write IUPAC names of the following

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 276
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 36
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 37

Activity :
• Collect information about production of ethanol as byproduct in sugar industry and its importance in fuel economy.
• Collect information about phenols used as antiseptics and polyphenols having antioxidant activity.

12th Chemistry Digest Chapter 11 Alcohols, Phenols and Ethers Intext Questions and Answers

Use your brain power! (Textbook Page No 235)

Question 1.
Classify the following alcohols as l0/2°/3° and allylic/benzylic
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 7
Answer:
(1) Ally lie alcohol (primary)
(2) Allylic alcohol (secondary)
(3) Allylic alcohol (tertiary)
(4) Benzylic alcohol (primary)
(5) Benzylic alcohol (secondary)

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Use your brain power ….. (Textbook Page No 236)

Question 1.
Name t-butyl alcohol using carbinol system of nomenclature.
Answer:
Trimethyl carbinol.

Problem 11.1 (Textbook Page No 238)

Question 1.
Draw structures of following compounds:
(i) 2,5-DiethIphenoI
(ii) Prop-2-en-I-oI
(iii) 2-methoxypropane
(iv) Phenylmethanol
Solution :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 32

Try this ….. (Textbook Page No 238)

Write IUPAC names ol (he following compounds.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 33
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 34

Do you know (Textbook Page No 238)

Question 1.
The mechanism of hydration of ethylcnc to ethyl alcohol.
Answer:
The mechanism of hydration of ethylene involves three steps:

Step 1: Ethylene gets protonated to form carbocation by electrophilic attack of H3O (Formation of carbocation intermediate).
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 59
Step 2 : Nucleophilic attack of water on carbocation
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 60
Step 3 : Deprotonation to form an alcohol
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 61

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Problem 11.2 : (Textbook Page No 239)

Question 1.
Predict the products for the following reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 76
Solution:
The substrate (A) contains an isolated Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 77 and an aldehyde group. H2/Ni can reduce both these functional groups while LiAlH4 can reduce only – CHO of the two, Hence
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 78

Try this ….. (Textbook page 240)

Question 1.
Arrange O – H, C – H and N – H bonds in increasing order of their bond polarity.
Answer:
Increasing order of polarity :C – H, N – H, O – H

Problem 11.3 : (Textbook Page No 241)

Question 1.
The boiling point of n-butyl alcohol, isobutyl alcohol, sec-butyl alcohol and tert-butyl alcohol are 118 °C, 108 °C. 99 °C and 82 °C respectively. Explain.
Solution:
As branching increases, intermolecular van der Waal’s force become weaker and the boiling point decreases. Therefore, n-butyl alcohol has highest boiling point 118 °C and tert-butyl alcohol has lowest boiling point 83 °C. Isobutyl alcohol is a primary alcohol and hence its boiling point is higher than that of sec-butyl alcohol.

Problem 11.4 : (Textbook Page No 242)

The solubility of o-nitrophenol and p-nitrophenol is 0.2 g and 1.7 g/100 g of H2O respectively. Explain the difference.
Solution :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 115
p-Nitrophenol has strong intermolecular hydrogen bonding with solvent water. On the other hand, o-nitrophenol has strong intramolecular hydrogen bonding and therefore the intermolecular attraction towards solvent water is weak. The stronger the intermolecular attraction between solute and solvent higher is the solubility. Hence p-nitrophenol has higher solubility in water than that of o-nitrophenol.

Problem 11.5 : (Textbook Page No 243 & 244)

Question 1.
Arrange the following compounds in decreasing order of acid strength and justify.
(1) CH3 – CH2 – OH
(2) (CH3)3 C – OH
(3) C6H5 – OH
(4) p-NO2 – C6H4 – OH
Solution :
Compounds (3) and (4) are phenols and therefore are more acidic than the alcohols (1) and (2). The acidic strengths of compounds depend upon stabilization of the corresponding conjugate bases. Hence let us compare electronic effects in the conjugate bases of these compounds :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 133

The conjugate base of the alcohol (1) is destabilized by + 1 effect of one alkyl group, whereas conjugate base of the alcohol (2) is destabilized by +1 effect of three alkyl groups. Hence (2) is weaker acid than (1)
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 134

Phenols : The conjugate base of p-nitrophenol (4) is better resonance stabilized due to six resonance structures compared to the five resonance structure of conjugate base of phenol (3). The resonance structure VI has – ve charge on only electronegative oxygens. Hence the phenol (4) is stronger acid than (3). Thus the decreasing order of acid strength is (4), (3), (1), (2).

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Use your brain power (Textbook Page No 244)

Question 1.
What are the electronic effects exerted by – OCH3 and – Cl? Predict the acid strength of
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 135
Answer:
The electronic effects exerted by – Cl and – O CH3 are as follows :
(1) Cl being more electronegative atom it pulls the bonding electrons towards itself. This is known as negative inductive effect (- I).

(2) – OCH3 is less electronegative group which repels the bonding electrons away from it. This is known as positive inductive effect ( + I).

(3) The relative to parent phenol, Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 136 is more acidic than Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 137.

Problem 11.6 : (Textbook Page No 245)

Question 1.
Mechanism of acid catalyzed dehydration of ethanol to give ethene.
Answer:
The mechanism of dehydration of ethanol involves the following order :
Step 1 : Formation of protonated alcohols : Initially ethyl alcohol gets protonated to form ethyl oxonium ion.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 164
Step 2 : Formation of carbocation : It is the slowest step and hence, the rate determining step of the reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 165
Steps 3: Formation of ethene: Removal of a proton (H+) from carbocation.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 166

The acidused in step I is released in step 3, the equilibrium is shifted to the right, ethene is removed as it is formed.

Problem 11.6 : (Textbook Page No 245)

Question 1.
Write the reaction showing major and minor products formed on heating butan-2-ol with concentrrated sulphuric acid.
Solution :
In the reaction described butan-2-ol undergoes dehydration to give but-2-ene (major) and but-l-ene (minor) in accordance with Saytzeff rule.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 167

Problem 11.7 : (Textbook Page No 246)

Question 1.
Write and explain reactions to convert propan-l-ol into propan-2-ol.
Solution :
The dehydration of propane-l-ol to propene is the first step. Markownikoff hydration of propene is the second step to get the product propan-2-ol. This is brought about by reaction with concemtrated H2SO4 followed by hydrolysis.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 168

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Problem 11.8 : (Textbook Page No 246)

Question 1.
An organic compound gives hydrogen on reaction with sodium metal. It forms an aldehyde having molecular formula C2H4O on oxidation with pyridinium chlorochromate. Name the compounds and give equations of these reactions.
Solution :
The given molecular formula C2H4O of aldehyde is written as CH3 – CHO. Hence the formula of alcohol from which this is obtained by oxidation must be CH3 – CH2 – OH. The two reactions can, therefore, be represented as follows :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 178

(Do you know? Textbook Page No 248)

Question 90.
Write the mechanism of dehydration of alcohol to give ether.
Answer:
Dehydration of alcohols to form ether is SN2 reaction. The mechanism of dehydration of ethanol involves the following steps.

Step 1 (Protonation) : Initially ethyl alcohol gets protonated in the presence of acid to form ethyl oxonium ion.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 191
Step 2 (SN2 mechanism) : Protonated alcohol species undergoes a backside attack by second molecule of alcohol is a slow step.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 192

Step 3 (Deprotonation) : Formation of diethyl ether by elimination of proton
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 193

Problem 11.9 : (Textbook Page No 249)

Question 1.
Ethyl isopropyl ether does not form on reaction of sodium ethoxide and isopropyl chloride.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 202
(i) What would be the main product of this reaction?
(ii) Write another reaction suitable for the preparation of ethyl isopropyl ether.
Solution :
(i) Isopropyl chloride is a secondary chloride. On treating with sodium ethoxide it gives elimination reaction to form propene as the main product.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 203
(ii) Ethyl isopropyl ether can be prepared as follows using ethyl chloride (10 chloride) as substrate.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 204

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Do you know? (Textbook Page No 250)

Question 1.
The mechanism of the reaction of HI with methoxy ethane.
Answer:
The reaction mechanism takes place as follows :
Step 1 : Protonation of ether Initially the ether molecule (methoxy ethane) protonated by cone. HI to form oxonium ion.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 229

Step 2 : Iodide is a good nucleophile. It attacks the least substituted carbon of the oxonium ion formed in step 1 and displaces an alcohol molecule by SN2 mechanism.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 230

For example :
• Use of excess HI converts the alcohol into alkyl iodide.
• In case of ether having one tertiary alkyl group the reaction with hot HI follows SN1 mechanism, and tertiary iodide is formed rather than tertiary alcohol.

Step 1 :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 231
Step 2 :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 232

12th Std Chemistry Questions And Answers:

12th Biology Chapter 10 Exercise Human Health and Diseases Solutions Maharashtra Board

Class 12 Biology Chapter 10

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 10 Human Health and Diseases Textbook Exercise Questions and Answers.

Human Health and Diseases Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 10 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 10 Exercise Solutions

1. Multiple Choice Questions

Question 1.
Which of the following is NOT caused by unsterilized needles?
(a) Elephantiasis
(b) AIDS
(e) Malaria
(d) Hepatitis
Answer:
(a) Elephantiasis

Question 2.
Opium derivative is …………………
(a) Codeine
(b) Caffeine
(c) Heroin
(d) Psilocybin
Answer:
(c) Heroin

Maharashtra Board Class 12 Biology Solutions Chapter 10 Human Health and Diseases

Question 3.
The stimulant present in tea is …………………
(a) tannin
(b) cocaine
(C) caffeine
(d) crack
Answer:
(c) caffeine

Question 4.
WhIch of the following Is caused by smoking?
(a) Liver cirrhosis
(b) Pulmonary tuberculosis
(c) Emphysema
(d) Malaria
Answer:
(c) Emphysema

Question 5.
An antibody is …………………
(a) molecuic that binds specifically an antigen
(b) WBC which invades bacteria
(c) secretion of mammalian RBC
(d) cellular component of blood
Answer:
(a) molecule that binds specifically an antigen

Question 6.
The antiviral proteins released by a virus-infected cell are called …………………
(a) histamines
(b) interferons
(c) pyrogens
(d) allergens
Answer:
(b) interferons

Question 7.
Both B-cells and T-cells are derived from …………………
(a) lymph nodes
(b) thymus glands
(c) liver
(d) stem cells in bone marrow
Answer:
(b) thymus glands

Question 8.
Which of the following diseases can be contracted by droplet infection?
(a) Malaria
(b) Chicken pox
(c) Pneumonia
(d) Rabies
Answer:
(c) Pneumonia

Question 9.
Confirmatory test used for detecting HIV infection is …………………
(a) ELISA
(b) Western blot
(c) Widal test
(d) Eastern blot
Answer:
(b) Western blot

Question 10.
Elephantiasis is caused by …………………
(a) W. barterofti
(b) P. vivax
(c) Bedbug
(d) Elephant
Answer:
(a) W. bancrofti

Question 11.
Innate immunity is provided by …………………
(a) phagocytes
(b) antibody
(c) T-lymphocytes
(d) B-lymphocytes
Answer:
(c) T-lymphocytes

2. Short Answer Questions

Question 1.
What is the source of cocaine?
Answer:
Source of cocaine is coca plant – Erythroxylum coca.

Question 2.
Name one disease caused by smoking.
Answer:
Emphysema. (Damaged and enlarged lungs causing breathlessness)

Question 3.
Which cells stimulate B-cells to form antibodies ?
Answer:
Helper T-cells stimulate B-cells to form antibodies.

Question 4.
What does the abbreviation AIDS stand for?
Answer:
AIDS stands for Acquired Immuno Deficiency Syndrome.

Question 5.
Name the causative agent of typhoid fever.
Answer:
Salmonella typhi

Question 6.
What is Rh factor?
Answer:
Antigen ‘D’ present on the surface of RBCs is known as Rh factor.

Question 7
What is schizont?
Answer:
Schizont is a ring-like form produced from merozoites inside the erythrocytes of human beings, infected by Plasmodium, which again forms new merozoites.

Question 8.
Name the addicting component found in tobacco.
Answer:
Nicotine

Maharashtra Board Class 12 Biology Solutions Chapter 10 Human Health and Diseases

Question 9.
Name the pathogen causing Malaria.
Answer:
Plasmodium vivax

Question 10.
Name the vector of Filariasis.
Answer:
Female Culex mosquito

Question 11.
Name of the causative agent of ringworm.
Answer:
Trichophyton

Question 12.
Health
Answer:
Health is defined as the state of complete physical, mental and social well¬being and not merely the absence of disease or infirmity.

3. Short Answer Questions

Question 1.
What are acquired diseases?
Answer:
Diseases which are developed after the birth of an individual are called acquired diseases. These are of two types, viz. (a) Communicable or infectious diseases and (b) Non- communicable or Non-infectious diseases. Communicable or infectious diseases are transmitted from infected person to another healthy person either directly or indirectly. They are caused due to pathogens like viruses, bacteria, fungi, helminth worms, etc. Non-communicable or Non-infectious diseases cannot be transmitted from infected person to another healthy one either directly or indirectly.

Question 2.
Antigen and antibody.
Answer:

Antigen Antibody
1. Antigens are foreign proteins which are capable of producing infection. 1. Antibodies are immunoglobulins produced by the body to act against the antigens.
2. The structure of antigens is variable dependent upon the type of pathogen. 2. The structure of antibody is Y-shaped.
3. The antigen is the ‘non-self’ molecule. 3. The antibody is ‘self’ molecule.
4. The antigens have epitope sites which bind with the antibody molecule. 4. The antibodies have paratope sites which bind with the antigen molecule.

Question 3.
Name the infective stage of Plasmodium. Give Symptoms of malaria
Answer:
Sporozoite
I. Symptoms of malaria:

  1. Fever accompanied by shivering.
  2. Joint pain or arthralgia.
  3. Vomiting.
  4. Anaemia caused due to rupture of RBCs or haemolysis.
  5. Haemoglobinuria.
  6. Retinal damage.
  7. Convulsions.
  8. Cyclical occurrence of sudden coldness followed by rigor and then fever and sweating lasting for four to six hours. This is called a classic symptom of malaria.
  9. Splenomegaly or enlarged spleen, severe headache, cerebral ischemia, hepatomegaly, i. e. enlarged liver, hypoglycaemia and haemoglobinuria with renal failure may occur in severe infections.

II. Spread / Transmission of malaria:

  1. Malaria parasite is transmitted through the female Anopheles mosquito and hence it is known as mosquito-borne disease. Mosquito acts as a vector.
  2. There are four species of Plasmodium, viz., P. vivax, P. falciparum, P. ovale and P. malariae which transmit malaria.

Question 4.
Explain the mode of infection and cause of elephantiasis.
Answer:
Mode of infection, i.e. transmission:

  1. The parasite Wuchereria bancrofti is transmitted from a patient to other normal human being by female Culex mosquito.
  2. The filarial larvae leave mosquito body and arrive on the human skin where they penetrate the skin and enter inside.
  3. They undergo two moultings to become adults. Later they settle in the lymphatic system. They incubate for about 8-16 months.
  4. When they settle in lymphatic system, this infection is called lymphatic filariasis.
  5. The worms start infecting lymphatic circulation resulting into enlargement of lymph vessels and lymph nodes. The extremities like legs or limbs become swollen which resembles elephant legs. Therefore it is called elephantiasis.
  6. This condition is lymphoedema, i.e. accumulation of lymph fluid in tissue causing swelling.

Question 5.
Why is smoking a bad habit?
Answer:

  1. Smoking involves inhaling the cigarette smoke which contains nicotine and other toxic substances like N-nitrosodimethlene. There is some amount of carbon monoxide.
    All these substances affect the normal respiratory health.
  2. Smoking invites problems like asthma, hypertension, heart disease, stroke, lung damage.
  3. The worst impact is that these substances are carcinogenic and hence can cause cancer of larynx, trachea, lung, etc.
  4. Smoking not only affects the smokers but also has bad effect on others due to passive smokers.
  5. In women, smoking is still hazardous as their ovaries can undergo mutations due to mutagenic chemicals found in smoke.
  6. Therefore, smoking is a very bad habit.

Question 6.
What do the abbreviations AMIS and CMIS denote?
Answer:
AMIS is Antibody-mediated immune system or humoral immunity and CMIS is cell- mediated immune system.

Question 7.
What is a carcinogen? Name one chemical carcinogen with its target tissue.
Answer:

  1. Carcinogen is the substance or agent that causes cancer.
  2. Urinary bladder cancer caused by 2-naphthylamine and 4-aminobiphenyl.

Question 8.
Active immunity and passive immunity.
Answer:

Active immunity Passive immunity
1. Active immunity is produced in response to entry of pathogens and their antigenic stimuli. 1. Passive immunity is produced due to antibodies that are transferred to the body.
2. Active immunity is the long lasting immunity. 2. Passive immunity is short-lived immunity.
3. In active immunity, the body produces its own antibodies. 3. In passive immunity, antibodies are given to the body from outside.
4. Natural acquired active immunity is obtained due to infections by pathogens. 4. Natural acquired passive immunity is obtained through antibodies of mother transmitted- to baby by placenta or colostrum.
5. Artificial acquired active immunity is obtained through vaccinations. These vaccines contain dead or live but attenuated pathogens. 5. Artificial acquired passive immunity is also obtained through vaccinations, but here the vaccines contain the readymade antibodies which are prepared with the help of other animals such as horses.

4. Short Answer Questions

Question 1.
B-cells and T-cells.
Answer:

B-cells T-cells.
1. B-cells are type of lymphocyte whose origin is in bone marrow but maturation is in blood. 1. T-cells are type of lymphocytes which originate in bone marrow but maturation occurs in thymus.
2. B-cells Eire type of lymphocytes which are involved in humoral mediated immunity. 2. T-cells are type of lymphocytes which are involved in cell-mediated immunity.
3. 20% of lymphocytes present in the blood are B-cells. 3. 80% of lymphocytes present in the blood are T-cells.
4. Two types of B-cells are Memory cells and Plasma cells. 4. T-cells are of following subtypes : Cytotoxic T-cells, helper T-cells, suppressor T-cells.
5. They are involved in antibody mediated immunity. (AMI) 5. They are involved in cell-mediated immunity (CMI).
6. B-cells produced antibodies with which they fight against pathogens. 6. T-cells do not produce antibodies.
7. B-cells have membrane bound immunoglobulins located on the surface. 7. There is a presence of T cell receptors on the T-cell surface.

Maharashtra Board Class 12 Biology Solutions Chapter 10 Human Health and Diseases

Question 2.
What are the symptoms of malaria? How does malaria spread?
Answer:
Symptoms of malaria:

  1. Fever accompanied by shivering.
  2. Joint pain or arthralgia.
  3. Vomiting.
  4. Anaemia caused due to rupture of RBCs or haemolysis.
  5. Haemoglobinuria.

Question 3.
AIDS.
Answer:
(1) AIDS or the acquired immuno deficiency syndrome, is fatal viral disease caused by a retrovirus (ss RNA) known as the human immuno deficiency virus (HIV) which weakens the body’s immune system. It is called a modern pandemic.

(2) The HIV attacks the immune system which in turn causes many opportunistic infections, neurological disorders and unusual malignancies ultimately leading to death.

(3) AIDS was first noticed in USA in 1981 whereas in India, first confirmed case of AIDS was in April 1986 from Tamil Nadu.

(4) HIV is transmitted through body fluids such as saliva, tears, nervous system tissue, spinal fluid, blood, semen, vaginal fluid and breast milk. However, only blood, semen, vaginal secretions and breast milk generally transmit infection to others.

(5) The transmission of HIV occurs by sexual contact, through blood and blood products and by contaminated syringes, needles, etc. There is also transplacental transmission or through breast milk at the time of nursing.

(6) Accidental needle injury, artificial insemination with infected donated semen and transplantation with infected organs are some of the rare occasions of transmission of HIV.

(7) HIV infection is not spread by casual contact such as hugging, bite of mosquitoes or using other objects touched by a patient.

(8) Acute HIV infection progresses over time to asymptomatic HIV infection and then to early symptomatic HIV infection. Later, it progresses to full blown AIDS when patient shows advanced HIV infection with CD4 T-cell count below 200 cells/mm.

Question 4.
Give the symptoms of cancer.
Answer:
Symptoms of cancer:

  1. Presence of lump or tumour.
  2. White patches in the mouth.
  3. Change in a wart or mole on the skin.
  4. Swollen or enlarged lymph nodes.
  5. Vertigo, headaches or seizures if cancer affect the brain.
  6. Coughing and shortness of breath if lungs are affected due to cancer.

Question 5.
Antigens on blood cells.
Answer:

  1. There are about 30 known antigens on the surface of human red blood cells. They decide the type of blood group such as ABO, Rh, Duffy, Kidd, Lewis, P MNS, Bombay.
  2. The different blood groups are determined genetically due to presence of a particular antigen.
  3. Landsteiner found two antigens or agglutinogens on the surface of human red blood cells which are named as antigen A and antigen B.
  4. There is another antigen called Antigen D which decides the Rh status of the blood. If Antigen D is present, the person is said to be RH positive and when it is lacking, the person is Rh negative.
  5. These antigens are responsible for types of blood group and the specific transfusions.
  6. Antigens present on the RBCs and antibodies present in the serum can cause agglutination reactions if they are non-compatible. Therefore, at the time of transfusion blood groups are checked properly.

Question 6.
Antigen-antibody complex:
Img 1
Answer:

  1. Between antigen and antibody there is specificity.
  2. Each antibody is specific for a particular antigen.
  3. On the antigens there are combining sites which are called antigenic determinants or epitopes.
  4. Epitopes react with the corresponding antigen binding sites of antibodies which are called paratopes.
  5. The antigen binding sites are located on the variable regions of the antibody. Variable regions have small variations which make each antibody highly specific for a particular antigen.
  6. Owing to variable region the antibody can recognize the specific antigen.
  7. Antibody thus binds to specific antigen in a lock and key manner, forming an antigen- antibody complex.

Question 7.
What are the various public health measures, which you would suggest as safeguard against infectious diseases?
Answer:
Infectious diseases spread through pathogens, therefore, it is an important duty of each person to decrease the risk of infecting our own self or others. This can be achieved by

  1. Washing hands often, especially whenever, we are in contact with food and water. Before and after preparing food, before eating and after using the toilet, hand wash is a must.
  2. Vaccinations : Immunization helps us to protect against contracting many diseases. Therefore, timely vaccination should be taken. Especially at the time of epidemic, one must keep distance from infected area or get vaccinated.
  3. One must be at home if there are signs and symptoms of an infection. By going out, we may infect other healthy persons.
  4. Proper diet and exercise should be followed to improve one’s own immunity.
  5. Hygiene should be utmost in the kitchen and dining area. One must take care while eating uncovered and leftover food.
  6. Bathroom and toilet should be cleaned daily as there can be a high concentration of bacteria or other infectious agents in these areas.
  7. One should have responsible sexual behaviour to avoid sexually transmitted diseases.
  8. Personal items such as toothbrush, comb, towel, undergarments or razor blade should never be shared.
  9. Travelling should be avoided because we may infect other passengers during travel. Moreover, our illness can be aggravated. Some special immunizations are needed during certain travels, such as anti-cholera vaccine while going to Pandharpur during Ashadhi.

Question 8.
How does the transmission of each of the following diseases take place?
(a) Amoebiasis:
Answer:
Amoebiasis is usually transmitted by the following ways:

  1. The faecal-oral route.
  2. Through contact with dirty hands or objects.
  3. By anal-oral contact.
  4. Through contaminated food and water.

(b) Malaria:
Answer:
Symptoms of malaria:

  1. Fever accompanied by shivering.
  2. Joint pain or arthralgia.
  3. Vomiting.
  4. Anaemia caused due to rupture of RBCs or haemolysis.
  5. Haemoglobinuria.

(c) Ascariasis:
Answer:

  1. Unsafe and unhygienic food and drinks contaminated with the eggs of Ascaris are the main mode of transmission.
  2. Eggs hatch inside the intestine of the new host.
  3. The larvae pass through various organs and settle as adults in the digestive system.

(d) Pneumonia:
Answer:

  1. Pneumonia usually spreads by direct person to person contact.
  2. It is also spread via droplet infection, i.e. droplets released by infected person.
  3. Using clothes and utensils of the patient.

Question 9.
What measures would you take to prevent water-borne diseases?
Answer:

  1. To prevent water-borne diseases, use of safe, clean and potable water is a must. Water should be filtered, then boiled and stored in covered container. If possible water purifier systems should be installed at home.
  2. One should preferably use bottled water or carry our own water container while travelling.
  3. Cleaning of water containers and maintaining personal hygiene near water storage is a must.
  4. Megacities offer chlorinated and purified water for citizens. But villages and smaller rural set ups use river water which may be highly contaminated with pathogens. Such water should be purified before consumption to prevent water-borne diseases.

Maharashtra Board Class 12 Biology Solutions Chapter 10 Human Health and Diseases

Question 10.
Typhoid.
Answer:
Typhoid is an infective disease caused by Gram-ve bacterium, Salmonella typhi.
(1) It is food and water-borne infection. In the intestinal lumen of infected person this bacteria is found.

(2) The bacterium has “O” – antigen, which is a lipopolysaccharide (LPS), present on surface coat and its flagella has “H” – antigen. Thus it becomes pathogenic.

(3) Signs and Symptoms of typhoid are as follows:
Prolonged and high fever with nausea, fatigue, headache.
Abdominal pain, constipation or diarrhoea. In severe cases rose-coloured rash is seen on skin. Tongue shows white coating and there is cough. Anorexia or loss of appetite is seen. In chronic cases there is breathlessness, irregular heartbeats and haemorrhage.

(4) Poor hygiene habits and poor sanitation and insects like houseflies and cockroaches spread typhoid.

(5) Typhoid is diagnosed by Widal test.

(6) Antibiotics like Chloromycetin can cure typhoid. Preventive vaccines such as oral Ty21a vaccine and injectable typhim vi and typherix against typhoid are also available. Chronic cases need surgical removal of gall bladder.

5. Match the following.

Column I Column II
(a) AIDS (i) Antibody production
(b) Lysozyme (ii) Activation of B-cells
(c) B-cells (iii) Immunoglobulin
(d) T-helper cells (iv) Tears
(e) Antibody (v) Immuno deficiency

Answer:

Column I Column II
(a) AIDS (v) Immuno deficiency
(b) Lysozyme (iv) Tears
(c) B-cells (i) Antibody production
(d) T-helper cells (ii) Activation of B-cells
(e) Antibody (iii) Immunoglobulin

6. Long Answer Questions

Question 1.
Describe the structure of antibody.
Answer:
Img 2

  1. Antibodies are highly specific to specific antigens. They are glycoprotein called immunoglobulins (Igs.).
  2. They are produced by plasma cells. Plasma cells are in turn formed by B-lymphocytes.
  3. About 2000 molecules of antibodies are formed per second by the plasma cells.
  4. Antibody is a ‘Y’-shaped molecule. It has four polypeptide chains, two heavy or H-chains and two light or L-chains.
  5. Disulfide bonds (-s-s-) hold the polypeptide chains together to form a ‘Y’-shaped structure.
  6. The region holding arms and stem of antibody is termed as hinge. Each chain of the antibody has two distinct regions, the variable region and the constant region.
  7. Variable regions have a paratope which is an antigen-binding site. This part of antibody recognizes and binds to the specific antigen forming an antigen-antibody complex.
  8. Antibodies are called bivalent as they carry two antigen binding sites.

Question 2.
Vaccination.
Answer:

  1. Vaccines are prepared from inactivated pathogen, in the form of protein or sugar from pathogen or dead form of pathogen or toxoid from pathogens or attenuated pathogen.
  2. These when they are administered to a person to protect against a particular pathogen, it is called vaccination.
  3. Vaccination ’teaches’ the immune system to recognize and eliminate pathogenic organism. Because, already in the body the vaccine is injected and body has made antibodies in response to it. Thus, body is prepared before the attack, if at all it is exposed to pathogen.
  4. Thus, it is an important form of primary prevention, which reduces the chances of illness by protecting people. It works by exposing the pathogen in a safe form.
  5. Vaccinations control spread of diseases like measles, polio, tetanus and whooping cough that once threatened many lives.
  6. Vaccination controls the epidemic outbreak of diseases, if all the people Eire pre-vaccinated.
  7. Some hazardous diseases like small, pox and polio have been completely eradicated by the vaccination.

Question 3.
What is cancer? Differentiate between benign tumour and malignant tumour. The main five types of cancer
Answer:
I. Cancer : Cancer is a disease caused by uncontrolled cell division due to disturbed cell cycle.

II. Difference between benign tumour and malignant tumour:

Benign tumour malignant tumour
1. Benign tumour is localized and it does not spread to neighbouring areas. 1. Malignant tumour starts as local but spreads rapidly to neighbouring areas.
2. Benign tumour is enclosed in connective tissue sheath. 2. Malignant tumour is not enclosed in connective tissue sheath.
3. Benign tumour compresses the surrounding normal tissue. 3. Malignant tumour invades and destroys the surrounding tissue.
4. Benign tumours can be removed surgically. 4. Malignant tumours need further treatment after removal.
5. Except for brain tumour, benign tumours are usually not fatal. 5. Malignant tumours are fatal.
6. Benign tumours do not show metastasis. 6. Malignant tumours show metastasis.
7. Benign tumours are well differentiated. 7. Malignant tumours are poorly differentiated.
8. Benign tumours show slow and progressive growth. 8. Malignant tumours show rapid and erratic growth.

III. The main five types of cancer:

Types of Cancer : According to the tissue affected, the cancers are classified into five main types. These are as follows:

  1. Carcinoma : Cancer of epithelial tissue covering or lining the body organs is known as carcinoma. E.g. breast cancer, lung cancer, cancer of stomach, skin cancer, etc.
  2. Sarcoma : Cancer of connective tissue is called sarcoma. Following are the types of sarcoma osteosarcoma (bone cancer), myosarcoma (muscle cancer),
    chondrosarcoma (cancer of cartilage) and liposarcoma (cancer of adipose tissue).
  3. Lymphoma : Cancer of lymphatic tissue is called lymphoma. Lymphatic nodes, spleen and tissues of immune system are affected due to lymphoma.
  4. Leukaemia : Leukaemia is blood cancer. In this condition, excessive formation of leucocytes take place in the bone marrow. There are millions of abnormal immature leucocytes which cannot fight infections. Monocytic leukaemia, lymphoblastic leukaemia, etc. are the types of leukaemia.
  5. Adenocarcinoma : Cancer of glandular tissues such as thyroid, pituitary, adrenal, etc. is called adenocarcinoma.

Maharashtra Board Class 12 Biology Solutions Chapter 10 Human Health and Diseases

Question 4.
Describe the different type of immunity.
Answer: There are two basic types of immunity, viz. innate immunity and acquired immunity.
(A) Innate immunity:

  1. Innate immunity is natural, inborn immunity, which helps the body to fight against the invasion of microorganisms.
  2. Innate immunity is non-specific because it does not depend on previous exposure to foreign substances.
  3. Innate immunity mechanisms consist of various types of barriers such as anatomical barriers, physiological barriers, phagocytic barriers and inflammatory barriers. They prevent entry of foreign agents into the body.

(B) Acquired immunity:

  1. The immunity that an individual acquires during his life is called acquired immunity or adaptive immunity or specific immunity. It helps the body to adapt by fighting against specific antigens hence it is called adaptive immunity. Since it is produced specifically against an antigen, it is called specific immunity.
  2. Acquired immunity takes long time for its activation.
  3. This type of immunity is seen only in vertebrates.
  4. Due to acquired immunity, the body is able to defend against any invading foreign agent.

Question 5.
Describe the ill-effects of alcoholism on health.
Answer:

  1. Alcohol in any form is toxic for the body. Hence as soon as alcohol is consumed, the liver tries to detoxify it.
  2. In low doses it acts as a stimulant but in high dose, it acts on central nervous system, especially the cerebrum and cerebellum. Still higher dose can induce a comatose condition.
  3. Alcohol affect the gastrointestinal tract by causing inflammation and damage to gastric 4 mucosa. Ulceration and painful condition arises in alcoholics.
  4. Excessive doses of alcohol induce vomiting.
  5. The worst effect of alcohol is on liver causing diseases like cirrhosis.
  6. Alcohol induces hypertension and cardiac problems.
  7. Apart from physical effect, it causes deterioration of mental health and emotional well-being.
  8. Alcoholic person cannot think due to numbness in his/her cerebrum.
  9. The social health is greatly affected as the alcoholic can cause problems to his family, friends and society in general.

Question 6.
In your view, what motivates the youngsters to take to alcohol or drugs and how can this be avoided?
Answer:
I. Taking drugs or alcohol:

  1. Youngsters are at the vulnerable age, where they lack the planning about their future.
  2. If they fall into bad company or are facing parental neglect, they get hooked on to alcohol or drugs.
  3. Some common causes for addiction among youngsters are insufficient parental supervision and monitoring or excessive pressure and expectations from them. Lack of communication between an adolescent and parents.
  4. Poorly defined rules for the family. Continuous family conflicts.
  5. Favourable parental attitudes towards alcohol and drug uses. Many a times, at home children are exposed to such habits.
  6. Inability to cope up with present and hence switching to the addictions. Risk taking behaviour which is common among youngsters.

II. Methods/measures to avoid drug abuse:

  1. There should be complete acceptance for the child, because the adolescent phase is the most crucial phase when the children should be treated with love, care and respect.
  2. Many physical, hormonal and psychological transformations are taking place in this phase. Therefore child suffers from stressful situation.
  3. Wrong company and bad influence of peer group can trap the child in bad addictive habits. Thus, family should be supportive and communicative to help such children.
  4. The sexual thoughts should be sublimed by channelizing energy into healthy pursuits like sports, reading, music, yoga and other extracurricular activities.
  5. Ill-effects of drugs or alcohol should be told to youngsters.
  6. Education and counselling can control the children from getting hooked on to the addictions.

Maharashtra Board Class 12 Biology Solutions Chapter 10 Human Health and Diseases

Question 7.
Do you think that friends can influence one to take alcohol/drugs? If yes, how may one protect himself/herself from such an influence?
Answer:
Friends can influence one to take alcohol and drugs, if a boy or girl is timid and non-communicative with his or her parents and teachers. It also depends on the personality of the indtvidual. In the adolescent age, many fall in trap due to such peer pressure. The confusion in the mind and role of hormones playing on the psyche and thought process makes one unable to understand the hazards of such habits. Also there is curiosity to do these experimentations due to bad influence of media.

If there is complete trust and friendship with sensible parents, then such influence does not work. One should protect himself or herself by a strong denial. Communicating such incidents to an elder in whom a boy or girl can confide, is very important. One should tell his or her friends about the ill-effects of alcohol and drugs. He should be made aware of these aspects that he or she has learnt in this lesson.

12th Std Biology Questions And Answers:

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

Class 12 Chemistry Chapter 15

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

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

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

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

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 …………………………………. .
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 1
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 70

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:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 71

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.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

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.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 26

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:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 27

Question iv.
Name the polymer type in which the following linkage is present.
Answer:
The Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 74 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:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 41

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

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:

  1. Teflon – CF2 = CF2
  2. PVC – CH2 = CHCI
  3. Polyester-Dicarboxylic acid and polyhydoxyglycol
  4. Polythene – CH2 = CH2
  5. 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:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 32

Question viii.
Name and draw the structure of the repeating unit in natural rubber.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 14Repeating 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.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

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 Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 35 and formaldehyde (HCHO)

4. Attempt the following :

Question i.
Identify condensation polymers and addition polymers from the following.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 2
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 68

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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 24

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.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 33
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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 34
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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 22

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.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 22

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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 3
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 8

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

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.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question iv.
Write the structure of isoprene and the polymer obtained from it.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 82

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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 15

(2) Chain initiating step : The methyl radical thus formed adds to ethylene to form a new larger free radical.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 16

(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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 17

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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 18

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.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

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?
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 11
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.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

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:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 43

(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:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 44

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 46
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

Class 12 Biology Chapter 12

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

Maharashtra Board Class 12 Biology Solutions Chapter 12 Biotechnology

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:

  1. YAC
  2. RE
  3. dNTP
  4. PCR
  5. GMO
  6. MAC
  7. CCMB.

Answer:

  1. YAC : Yeast Artificial chromosome
  2. RE : Restriction Endonuclease
  3. dNTP : Deoxyribonucleoside triphosphates
  4. PCR : Polymerase Chain Reaction
  5. GMO : Genetically Modified Organisms
  6. MAC : Mammalian Artificial Chromosome
  7. 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

Maharashtra Board Class 12 Biology Solutions Chapter 12 Biotechnology

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:

  1. Vector should be able to replicate independenly (through ori gene), so that as vector replicates, multiple copies of the DNA insert are also produced.
  2. It should be able to easily transferred into host cells.
  3. It should have suitable control elements like promoter, operator, ribosomal binding sites, etc.
  4. 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:

  1. If the faulty machine is not able to lower the temperature below 70 °C, then the primer annealing step will be hampered first.
  2. Each primer has a specific annealing temperature, depending upon its A, T, G, C content.
  3. For most of the primers annealing temperature is about 40-60 °C.
  4. 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:

  1. Bioethics is the study of moral vision, decision and policies of human behaviour in relation to biological phenomena or events.
  2. 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.
  3. 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:

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

Maharashtra Board Class 12 Biology Solutions Chapter 12 Biotechnology

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

  1. 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’.
  2. It is unauthorized misappropriation of any biological resource and traditional knowledge.
  3. 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):

  1. 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.
  2. The patenting of the fungicidal properties of Neem, was an example of biopiracy.

(b) Patenting of Basmati:

  1. 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.
  2. This is a case of biopiracy as Basmati is a long-grained, aromatic variety of rice indigenous to the Indian subcontinent.
  3. Very broad claims about “Inventing” the said rice was the basis of patent application.
  4. The UPSTO has rejected all the claims due to people movement against Rice Tec in March 2001.

(c) Haldi (Turmeric) Biopiracy:

  1. 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.
  2. They were granted a patent in March 1995.
  3. 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.
  4. 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:

  1. 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.
  2. It allows the patent holder to exclude others from making, using, selling or importing protected invention for a limited period of time.
  3. 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.
  4. 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.
  5. First biopatent was awarded for genetically engineered bacterium ‘Pseudomonas’ used for clearing oils spills.
  6. 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:
Maharashtra Board Class 12 Biology Solutions Chapter 12 Biotechnology 1
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.

Maharashtra Board Class 12 Biology Solutions Chapter 12 Biotechnology

(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:

  1. In this germ cells are modified genetically to correct a genetic defect.
  2. Normal gene is introduced into germ cells like sperms, eggs, early embryos.
  3. It allows transmission of the modified genetic information to the next generation.
  4. 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:

  1. In this somatic cells are modified genetically to correct a genetic defect.
  2. 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.
  3. Modification of somatic cells only affects the person being treated and the modified chromosomes cannot be passed on the future generations.
  4. 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:

  1. Transgenic mice are used in various research areas of cancer research.
  2. Transgenic mice containing a particular oncogene (cancer causing gene) develop specific cancer.
  3. They are used to study the relationship between oncogenes and cancer development, cancer treatment and prevention of malignancy.
  4. The transgenic mouse model for the investigation of the breast cancer was developed in the laboratory of Philip Leder in Harvard (USA).
  5. 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:
Maharashtra Board Class 12 Biology Solutions Chapter 12 Biotechnology 2

(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:

  1. A vaccine is a biological preparation that provides active acquired immunity against a certain disease.
  2. Vaccine is often made from a weakened or killed form of the microorganism, its toxins or one of its surface protein antigens.
  3. Edible vaccine is an edible plant part engineered to produce an immunogenic protein, which when consumed gets recognized by immune system.
  4. Immunogenic protein of certain pathogens are active when’administered orally.
  5. When animals or mainly humans consume these plant parts, they get vaccinated against certain pathogen.
  6. 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:

  1. 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.
  2. They are the molecular scissors which recognize and cut the phosphodiester back bone of DNA on both strands, at highly specific sequences.
  3. The sites recognized by them are called recognition sequences or recognition sites.
  4. Different restriction enzymes found in different organisms recognize different nucleotide sequences and therefore cut DNA at different sites.
  5. Restriction cutting may result in DNA fragments with blunt ends or cohesive or sticky ends or staggered ends (having short, single stranded projections).
  6. Restriction endonucleases like Bam HI and EcoRI produce fragments with sticky ends.
  7. Restriction endonucleases like Alu I, Hind III produce fragments with blunt ends.
  8. Type I restriction endonucleases fuction simultaneously as endonuclease and methylase e.g. EcoK.
  9. 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.
  10. Type III restriction endonucleases cut DNA at specific non-palindromic sequences e.g. Hpal, MboII.
  11. In bacterial cells, REs destroy various viral DNAs that might enter the cell, thus restricting the potential growth of the virus.

Maharashtra Board Class 12 Biology Solutions Chapter 12 Biotechnology

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:

  1. They are bacteria like Bacillus haemophilus, Helicobacter pyroliand E. coli.
  2. 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

Class 12 Biology Chapter 4

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

Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance

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.

Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance

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:

  1. In eukaryotic cells, DNA (2.2 metres) is condensed, coiled and supercoiled to be packaged efficiently in the nucleus (10-16 m).
  2. DNA is associated with histone and non-histone proteins.
  3. Histones are a set of positively charged, basic proteins, rich in basic amino acid residues lysine and arginine.
  4. 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.
  5. H1 protein binds the DNA thread where it enters and leaves the nucleosome.
  6. Adjacent nucleosomes are linked with linker DNA (varies in length from 8 to 114 bp, average length of linker DNA is about 54 bp).
  7. Each nucleosome contains 200 bp of DNA.
  8. 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.
  9. 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

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

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

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

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

  1. 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.
  2. Lactose binds to repressor and inactivates it.
  3. Repressor – lactose complex cannot bind with the operator gene, which is then turned on.
  4. RNA polymerase transcribes all the structural genes to produce lac m-RNA which is then translated to produce all enzymes.
  5. Thus, lactose acts as an inducer.
  6. 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:

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

Question 2.
Describe the structure of operon.
Answer:

  1. An operon is a unit of gene expression and regulation.
  2. It includes the structural genes and their control elements. Control elements are promoters and operators.
  3. The structural genes code for proteins, r-RNA and t-RNA that are necessary for all the cells.
  4. Promoters are signal sequences in DNA. They start the RNA synthesis. They also act as sites where the RNA polymerases are bound during transcription.
  5. Operators are present between the promoters and structural genes.
  6. There is repressor protein that binds to the operator region of the operon.
  7. 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
Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance 1
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.
Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance 2
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.

Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance

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:
Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance 3
DNA replication is semi-conservative replication. It involves following steps:
Activation of Nucleotides:

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

Point of Origin or Initiation point:

  1. Replication begins at specific point ‘O- Origin and terminates at point ‘T’.
  2. 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:

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

Replicating fork:

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

Synthesis of new strands:

  1. Each separated strand acts as a template for the synthesis of new complementary strand.
  2. 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.
  3. 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).
  4. The phosphodiester bonds are formed between nucleotides of new strand to form a polynucleotide strand.
  5. The enzyme DNA polymerase catalyses synthesis of new complementary strand always in 5′ – 3′ direction.

Leading and Lagging strand:

  1. The template strand with free 3′ is called the leading template.
  2. The template strand with free 5′ end is called the lagging template.
  3. The replication always starts at C-3 end of template strand and proceeds towards C-5 end.
  4. New strands are always formed in 5′ → 3′ direction.
  5. The new strand which develops continuously towards replicating fork is called the leading strand.
  6. The new strand which develops discontinuously away from the replicating fork is called the lagging strand.
  7. Maturation of Okazaki fragments : The lagging strand is synthesized in the form of small Okazaki fragments which are joined by enzyme DNA ligase.
  8. 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.
  9. Gaps formed are filled by complementary DNA sequence with the help of DNA polymerase-I in prokaryotes and DNA polymerase-a in eukaryotes.
  10. Finally, DNA gyrase (topoisomerase) enzyme forms double helix to form daughter DNA molecules.

Formation of two daughter DNA molecules:

  1. In each daughter DNA molecule, one strand is parental and the other one is newly synthesized.
  2. 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:
Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance 4
Transcription involves three stages, viz. Initiation, Elongation and Termination.
(1) Initiation:

  1. RNA polymerase binds to promoter site.
  2. It then moves along the DNA and causes local unwinding of DNA duplex into two strands in the region of the gene.
  3. 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:
Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance 5
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.

Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance

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.

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

iii. Translocation.

  1. 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.
  2. Due to this A’-site becomes vacant to receive next charged t-RNA molecule.
  3. The events like arrival of t-RNA – amino acid complex, formation of peptide bond, ribosomal translocation and release of previous t-RNA, are repeated.
  4. 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:
Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance 6
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:

  1. DNA as the genetic material has to be chemically and structurally stable.
  2. It should be able to generate its replica.
  3. Sugar-phosphate backbone and complementary base pairing between the two strands, give stability to DNA.
  4. Both the strands of DNA act as template for synthesis of their complementary strands. This allows accurate replication of DNA.
  5. 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:

  1. 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.
  2. Ribosome has one binding site for m-RNA. It orients m-RNA molecule in such a way that all the codons are properly read.
  3. 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).
  4. t-RNAs place the required amino acids in correct sequence and translate the coded message of RNA.
  5. 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.
  6. Thus ribosome plays an important role in translation.

Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance

(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

Class 12 Biology Chapter 15

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.
Maharashtra Board Class 12 Biology Solutions Chapter 15 Biodiversity, Conservation and Environmental Issues 1
(a) Line A represents, S = CA²
(b) Line B represents, log C = log A + Z log S
(c) Line A represents, S = CAZ
(d) Line B represents, log S = log Z + C log A
Answer:
(c) Line A represents, S = CAZ

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

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:

  1. Molasses
  2. Bagasse.

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

  1. Tissue culture
  2. In vitro fertilization of eggs
  3. 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.

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

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:

  1. Tropical regions are bestowed by thicker vegetation and ample food due to available sunlight and humidity.
  2. Polar regions are covered over with snow, with almost no vegetation.
  3. Only handful species of animals can survive here due to their adaptations.
  4. 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:

  1. Genetic diversity develops the capability of the species to adapt to the varying changes in the environment.
  2. 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.
  3. Some individuals have, a better capacity to endure the increasing pollution in the environment whereas some do not have it.
  4. Those that do not have show infertility or even death from the same conditions.
  5. Those who are able to endure and adapt to this change survive and live in a better way.
  6. This is called natural selection which leads to a loss of genetic diversity in particular habitats.
  7. 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.

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

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

  1. Carbon monoxide is tasteless, colourless and odourless gas, therefore its presence goes unnoticed.
  2. It can inhibit the blood’s ability to carry oxygen to body tissues.
  3. Supply of oxygen to vital organs such as.the heart and brain is affected due to presence of CO.
  4. 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.
  5. 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:

  1. Montreal Protocol was an international treaty signed at Montreal in Canada in 1987.
  2. Later many more efforts have been made and protocols have laid down definite roadmaps separately for developing and developed countries.
  3. All these efforts were for reducing emission of CFCs and other ozone depleting chemicals.
  4. 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.
  5. Montreal Protocol was a very positive move because after 1987, there have been much better condition of ozone layer.

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

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:

  1. BS emission standards changed over the time due to changing city life and more vehicular traffic on the road, especially in the megacities.
  2. 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.
  3. These norms were set to reduce sulphur and aromatic content of petrol and diesel. Also the vehicular engines were upgraded.
  4. 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.
  5. Since population of Delhi was to be saved, in 2001, Bharat stage II emission norms were set for CNG and LPG vehicles.
  6. 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.
  7. 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:

  1. The toilets deployed at Pandharpur at the time of vari are of the Ecosan type.
  2. Ecosan toilet is a closed system without water and it is an alternative to leach pit toilets.
  3. When the pit of an Ecosan toilet fills up after some time, then it is closed and sealed for about 8-9 months.
  4. In this time the faeces get completely composted to organic manure. In this way the organic waste can be disposed.
  5. It is a practical, efficient and cost-effective solution for human waste disposal.
  6. 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.

  1. 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.
  2. Vatapournima festival is worshipping a banyan tree.
  3. Various other festivals teach the value of plants and animals surrounding us. Even the cattle are worshipped on a particular day as a tradition.
  4. Jain religion strongly advocates protection of all animals through vegetarianism.

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

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:

  1. Humans always reap material benefits from biodiversity in the form of resources for basic needs such as food, clothes, shelter.
  2. Industrial products like resins, tannins, perfume base, etc. are also obtained through biodiversity resources.
  3. For making ornaments or artefacts for aesthetic purpose, again biodiversity is sacrificed.
  4. Many medicines are also obtained through biodiversity resources which shares 25% of global medicine market.
  5. Around 25000 species are used for traditional medicines by tribal population worldwide.
  6. 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:

  1. 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.
  2. Insects carry out pollination and seed dispersal.
  3. If insects do not carry out pollination and seed dispersal, man would go hungry without crops and fruits.
  4. 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

Class 12 Biology Chapter 3

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

Maharashtra Board Class 12 Biology Solutions Chapter 3 Inheritance and Variation

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.

Maharashtra Board Class 12 Biology Solutions Chapter 3 Inheritance and Variation

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:

  1. Tall habit versus dwarf habit (Height of the plant).
  2. Purple flowers versus white flowers. (Colour of flowers)
  3. Yellow seeds versus green seeds. (Colour of seeds)
  4. Round seeds versus wrinkled seeds. (Shape of seeds)
  5. Green pods versus yellow pods. (Colour of pods)
  6. Inflated pods versus constricted pods. (Shape of pods)
  7. 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:
Maharashtra Board Class 12 Biology Solutions Chapter 3 Inheritance and Variation 1

Question 3.
Pleiotropy.
Answer:

  1. 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.
  2. The pleiotropic ratio is always 1 : 2 instead . of normal 3 : 1.
  3. 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.
  4. 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.
  5. Thus a single gene produces two different expressions.
  6. 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.

Maharashtra Board Class 12 Biology Solutions Chapter 3 Inheritance and Variation

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

  1. Mendel planned his experiments carefully and these experiments consisted of large sample.
  2. He always recorded the results of number of plants of each type and their ratios.
  3. The contrasting characters that he chose were easily recognizable.
  4. 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.
  5. 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:

  1. 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.
  2. Mother, on the other hand, is homogametic, producing all similar types of ova, i.e 22 + X chromosomal combination.
  3. If 22+X bearing sperm fertilise an egg, female child is formed while if Y bearing sperm fertilizes an egg, male child is formed.
  4. 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:

  1. 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.
  2. Humans have 23 linkage groups because they have 23 pairs of chromosomes.
  3. Maize plant has 10 linkage groups because they have 10 pairs of chromosomes.

Question 7.
PKU.
Answer:

  1. PKU means phenylketonuria which is an autosomal recessive inborn error.
  2. In this disorder the metabolism of phenylalanine does not occur due to deficiency of phenylalanine hydroxylase (PAH) enzyme.
  3. This enzyme is necessary to metabolize the amino acid phenylalanine to the amino acid tyrosine.
  4. 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.
  5. PKU is caused by mutations in the PAH gene on chromosome no. 12.
  6. 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:

  1. Chromosomal theory identifies chromosomes as the carrier of genetic material.
  2. 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.
  3. Chromosomes are found in pairs in somatic or diploid cells.
  4. 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.
  5. 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
Maharashtra Board Class 12 Biology Solutions Chapter 3 Inheritance and Variation 2
(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:

  1. First Generation : Carrier woman marrying a sufferer man. Their three children are in following birth order.
  2. Second generation : First son is normal, second daughter is carrier and third daughter is sufferer.
  3. 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

Maharashtra Board Class 12 Biology Solutions Chapter 3 Inheritance and Variation

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:
Maharashtra Board Class 12 Biology Solutions Chapter 3 Inheritance and Variation 3
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
Maharashtra Board Class 12 Biology Solutions Chapter 3 Inheritance and Variation 4
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.

Maharashtra Board Class 12 Biology Solutions Chapter 3 Inheritance and Variation

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:

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

Maharashtra Board Class 12 Biology Solutions Chapter 3 Inheritance and Variation 5

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:
Maharashtra Board Class 12 Biology Solutions Chapter 3 Inheritance and Variation 6
(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 : XC
  • Gene for colour blindness : Xc
  • Normal female : XCXC
  • Normal male : XCY
  • Colour blind female : XcXc
  • Carrier female : XCXc
  • Colour blind male : Xc Y

II. Crosses showing the inheritance of colour blindness:
(i) A cross between normal female and colour-blind male.
Maharashtra Board Class 12 Biology Solutions Chapter 3 Inheritance and Variation 7

(ii) A cross of carrier female with normal male.
Maharashtra Board Class 12 Biology Solutions Chapter 3 Inheritance and Variation 8

(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.

Maharashtra Board Class 12 Biology Solutions Chapter 3 Inheritance and Variation

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

Class 12 Biology Chapter 11

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

Maharashtra Board Class 12 Biology Solutions Chapter 11 Enhancement of Food Production

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.

Maharashtra Board Class 12 Biology Solutions Chapter 11 Enhancement of Food Production

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:

  1. Many food preparations made at home involves the use of microorganisms.
  2. The microbes Lactobacilli are used in the preparation of dhokla from gram flour and buttermilk by the process of fermentation.
  3. Dosa and idlis are prepared by using batter of rice and black gram which is fermented by air borne Leuconostoc and Streptococcus bacteria.
  4. 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.
  5. 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.
  6. 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:

  1. Microbial source of Chloromycetin. – Streptomyces venezuelae
  2. Microbial source of Erythromycin. – Streptomyces erythreus
  3. Microbial source of Penicillin. – Penicillium chrysogenum
  4. Microbial source of Streptomycin. – Streptomyces griseus
  5. Microbial source of Griseofulvin. – Penicillium griseojulvum
  6. Microbial source of Bacitracin. – Bacillus licheniformis
  7. Microbial source of Oxytetracyclin / Terramycin. – Streptomyces aurifaciens
  8. The enzyme produced by Streptococcus bacterium. – Streptokinase
  9. Microbial source of Invertase. – Saccharomyces cerevisiae
  10. Microbial source of Pectinase. – Sclerotinia libertine, Aspergillus niger
  11. Microbial source of Lipase. – Candida lipolytica
  12. Microbial source of Cellulase. – Trichoderma konigii

Maharashtra Board Class 12 Biology Solutions Chapter 11 Enhancement of Food Production

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

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

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

  1. Fish is a nutritious food and thus is a source of many vitamins, minerals and nutrients.
  2. Commercial products such as fish oil, fish meal and fertilizers, fish guano, fish glue, isinglass are prepared from fish.
  3. These by-products are used in paints, soaps, oils and medicines.
  4. Some organisms like prawns and lobsters have high export value and market price.
  5. Fish farming and other fishery trades provide job opportunity and self-employment
  6. 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.

Maharashtra Board Class 12 Biology Solutions Chapter 11 Enhancement of Food Production

(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:

  1. 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.
  2. Lac is a complex substance having resin, sugar, water, minerals and alkaline substances.
  3. Lac insect is colonial in habit and it feeds on succulent twigs like ber, peepal, palas, kusum, babool,
  4. These plants are artificially inoculated in order to get better and regular supply of good quality and quantity of lac.
  5. Natural lac is always contaminated and pure form of lac obtained by washing and filtering is called as shellac.
  6. Lac is used to make bangles, toys, woodwork, inks, mirrors, etc.
  7. India’s share is 85% of total lac produced in the world.

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

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

The different methods of fish preservation are as follows:

  1. Chilling : This involves covering the fish with layers of ice. Ice is effective for short term preservation. It inhibits the activity of autolytic enzymes.
  2. 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.
  3. 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.
  4. Sun drying : This inhibits the growth of microorganisms that spoil the fish.
  5. 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.
  6. Salting : Salt removes the moisture from the fish tissues by osmosis. High salt concentration destroys autolytic enzymes and halts bacterial activity.
  7. 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:

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

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

  1. Mutations are sudden heritable changes in the genotype.
  2. Natural mutations occur at a very slow rate.
  3. Natural physical mutagens include exposure to high temperature, high concentration of C02, X-rays, UV rays.
  4. Mutations can be induced by using various mutagens.
  5. Mutagens cause gene mutations and chromosomal aberrations.
  6. Chemical mutagens include nitrous acid, EMS (Ethyl – Methyl – Sulphonate), mustard gas, colchicine, etc.
  7. Seedlings or seeds are irradiated by using CO60 or UV bulbs or X-ray machines.
  8. The mutated seedlings are then screened for resistance to diseases/pests, high yield, etc.
  9. 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).

Maharashtra Board Class 12 Biology Solutions Chapter 11 Enhancement of Food Production

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

Class 12 Chemistry Chapter 7

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) 104 -1 cm2 mol-1
(c) 10-2 S cm2 mol-1
(d) 102-1 cm2 mol-1
Answer:
(b) 104-1 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

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 1
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
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 2
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 ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 3
Answer:
(b) \(2 \lambda_{\mathrm{Al}^{3+}}^{0}+3 \lambda_{\mathrm{SO}_{4}^{2-}}^{0}\)

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 4
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:
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 5

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.

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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 : E0Ni2+/Ni = -0.25 V
E0Al3+/Al = – 1.66 V; E0cell = ?
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,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 6
(2) The part of equation namely,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 7
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,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 8

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,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 9
Single vertical line represents separation of two phases, solid Zn(s) and solution of ions.
(ii) Double vertical lines represent a salt bridge.

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 10
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 → H2(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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 11
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,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 12
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

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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:
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 13

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, \(K=\frac{[\text { Products }]}{[\text { Reactants }]}\)
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 14
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 = -nFE0cell
By comparing above equations,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 15

Question viii.
It is impossible to measure the potential of a single electrode. Comment.
Answer:
(1)
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 16
Fig 5.12(a) : Measurement of single electrode potential
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 17
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.
Ecell = E2 – E1
where E1 and E2 are reduction potentials of two electrodes.

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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 :
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 18

(2) Net cell reaction :
(i) Thus, the overall cell reaction during discharging is
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 19
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 :
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 20
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)

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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, BxAy 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,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 21

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,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 22
where λ0CH3COO and λ0H+ 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,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 23
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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 24
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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 25
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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 26
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.

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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 : WCu = 2.4 g; t = 1 hr = 1 × 60 × 60 s
MCu = 63.5 g mol-1; I = ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 27
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
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 28

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
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 29
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 30
= 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
E0Zn = – 0.76 V, E0Cr = – 0.74 V
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 31
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 32

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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 33
In SI units it is expected as m-1.

The resistance of an electrolytic solution is measured by using a conductivity cell and Wheatstone
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 52
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 Rsolution is the resistance of KCl solution and Rx 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 = κKcl × Rsolution.

Step II : Determination of conductivity of the given solution :
KCl solution is replaced by the given electrolytic solution and its resistance (Rs) 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,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 53
Since the concentration of the solution is known, ∧m can be calculated.

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 34

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 :
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 35
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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 36
(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.
2H2(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.
O2(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.
2H2(g) + O2(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 → 2NH3(aq) + H2(g) (Reduction half reaction)
2MnO2(s) + H2 → Mn2O3(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
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 37
(Note : Actually Br2 acts as an oxidising agent but not Br.)

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 38
(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:
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 39

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.
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 40
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
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 41
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.)

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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 × 10-3-1 cm-1 at 25 °C ?
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 42
∴ 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,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 43
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
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 44
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 = \(\lambda_{+}^{0}\) + \(\lambda_{-}^{0}\).

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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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 (fc) 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 = κKCl × RKCl
For example, the conductivity of 0.01 M KCl is 0.00141 Ω-1 cm-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 :
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 45
Adding equations (i) and (ii) and subtracting equation (iii) we get equation (I).
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 46

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,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 47

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:
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 48

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, PH2) | Pt
Reduction reaction : H+(aq) + e → \(\frac {1}{2}\)H2(g)
(ii) Chlorine gas electrode :
Cl(aq) | Cl2(g, PCl2) | Pt
Reduction reaction : \(\frac {1}{2}\)Cl2(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++ – MnO4, 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)
SnCl2(aq), SnCl4(aq) | Pt
Reduction reaction : Sn4+(aq) + 2e →Sn2+(aq)

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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

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

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

Electrochemical cell (Galvanic cell or Voltaic cell):

  1. This device is used to produce electrical energy by a spontaneous chemical reaction.
  2. It is used to generate electricity.
  3. In this cell, chemical energy is converted into electrical energy.
  4. In this cell, the cathode is positive and the anode is negative.
  5. Electrochernical cells are reversible.
  6. Oxidation takes place at the negative electrode and reduction at the positive electrode.
  7. The electrons move from anode to cathode in the external circuit.
  8. 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,
\(E_{\text {cell }}^{0}=\frac{0.0592}{n} \log _{10} K\)

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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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, ZnSO4(aq) for an electric contact. The vessel is filled with mercury, a paste of Hg and Hg2Cl2 (calomel) and saturated KCl solution.
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 49
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) | ZnSO4(aq) || KCl(aq) | Hg2Cl2(s) | Hg
OR Zn(s) | ZnSO4(aq) || Calomel electrode
Reduction reaction for calomel electrode :
Hg2Cl2(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).

Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

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,
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 50
Maharashtra Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry 51

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