Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 8 Differentials and Partial Derivatives Ex 8.2 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.2

Question 1.
Find the differential dy for each of the following functions.
(i) y = \(\frac { (1-2x)^3 }{ 3-4x }\)
(ii) y = (3 + sin2x)^2/3
(iii) y = e^{x2 – 5x +7} cos(x² – 1)
Solution:

(ii) y = (3 + sin2x)^2/3

Question 2.
Find df for f(x) = x² + 3x and evaluate it for
(i) x = 2 and dx = 0.1
(ii) x = 3 and dx = 0.02
Solution:
y = f(x) = x² + 3x
dy = (2x + 3) dx
(i) dy {when x = 2 and ate = 0.1} = [2(2) + 3] (0.1)
= 7(0.1) = 0.7

(ii) dy {when x = 3 and dx = 0.02} = [2(3) + 3] (0.0.2)
= 9(0.02) = 0.18

Question 3.
Find Δf and df for the function f for the indicated values of x, Δx and compare:
(i) f(x) = x³ – 2x², x = 2, Δx = dx = 0.5
(ii) f(x) = x² + 2x + 3, x = -0.5, Δx = dx = 0.1
Solution:
(i) y = f(x) = x³ – 2x²
dy = (3x² – 4x) dx
dy (when x = 2 and dx = 0.5) = [3(2²) – 4(2)] (0.5)
= (12 – 8)(0.5) = 4(0.5) = 2
(i.e.,) df = 2
Now ∆f = f(x + ∆x) – f(x)
Here x = 2 and ∆x = 0.5
f(x) = x³ – 2x²
So f(x + ∆x) = f(2 + 0.5) = f(2.5) = (2.5)³ – 2 (2.5)² = (2.5)² [2.5 – 2] = 6.25 (0.5) = 3.125
f(x) = f(2) = 2³ – 2(2²) = 8 – 8 = 0
So ∆f = 3.125 – 0 = 3.125

(ii) y = f(x) = x² + 2x + 3
dy = (2x + 2) dx
dy (when x = – 0.5 and dx = 0.1)
= [2(-0.5) + 2] (0.1)
= (-1 + 2) (0.1) = (1) (0.1) = 0.1
(i.e.,) df = 0.1
Now ∆f = f(x + ∆x) – f(x)
Here x = -0.5 and ∆x = 0.1
x² + 2x + 3
f(x + ∆x) = f(-0.5 + 0.1) = f(-0.4)
= (-0.4)² + 2(-0.4) + 3
= 0.16 – 0.8 + 3 = 3.16 – 0.8 = 2.36
f(x) = f(-0.5) = (-0.5)² + 2(-0.5) + 3
= 0.25 – 1 + 3 = 3.25 – 1 = 2.25
So ∆ f = f(x + ∆x) – f(x) = 2.36 – 2.25 = 0.11

Question 4.
Assuming log₁₀ e = 0.4343, find an approximate value of Iog₁₀ 1003.
Solution:
Let f(x) = log ₁₀ x then
f ‘(x) = \(\frac { 1 }{ x }\) log₁₀ e (log₁₀ x = log₁₀ e log_e x)
f(x + Δx) – f(x) = f ‘(x) Δx
f(1003) – f(1000) = \(\frac { 0.4344 }{ 1000 }\) × 3
log₁₀ 1003 – log₁₀ 1000 = 0.0013029
log₁₀ 1003 = log₁₀ 10³ + 0.0013029
= 3 + 0.0013029
= 3.0013029
Approximate value of log₁₀ 1003 = 3.0013029

Question 5.
The trunk of a tree has a diameter of 30 cm. During the following year, the circumference grew 6 cm.
(i) Approximately how much did the tree diameter grow?
(ii) What is the percentage increase in the area of the cross-section of the tree?
Solution:
(i) Diameter of the trunk of the tree
D = 30 cm
Rate of change of circumference
ds = 6 cm per year
Circumference S = πD
dS = πdD
6 = πdD
\(\frac { 6 }{ π }\) = dD
Rate of increasing diameter = \(\frac { 6 }{ π }\) cm

Question 6.
An egg of a particular bird is very nearly spherical. If the radius to the inside of the shell is 5 mm and the radius to the outside of the shell is 5.3 mm, find the volume of the shell approximately.
Solution:
Radius of the inside shell = 5 mm
Radius of the outside shell = 5.3 mm
Volume V = \(\frac { 4 }{ 3 }\) πr³
dV = \(\frac { 4 }{ 3 }\) π3r²dr
= 4 π 5 × 5 × 0.3
= 100 π × 0.3
= 30 π
Approximate volume of the shell = 30 mm³

Question 7.
Assume that the cross-section of the artery of human is circular. A drug is given to a patient to dilate his arteries. If the radius of an artery is increased from 2 mm to 2.1 mm, how much is the cross-sectional area increased approximately?
Solution:
The radius of an artery section = 2 mm
dr = 2.1 – 2
= 0.1
Area A = πr²
dA = 2πrdr
= 2 × π × 2 × 0.1
= 0.4 π
Increased area = 0.4 π mm²

Question 8.
In a newly developed city, it is estimated that the voting population (in thousands) will increase according to V(t) =30 + 12t² – t³, 0 ≤ t ≤ 8 where t is the time in years. Find the approximate change in voters for the time change from 4 to 4 1/6 years.
Solution:
V(t) = 30 + 12t² – t³; dt = 4 \(\frac { 1 }{ 6 }\) – 4 = \(\frac { 1 }{ 6 }\)
V’(t) = (24t – 3t²)dt
= (24(4)-3 (4)²) × \(\frac { 1 }{ 6 }\)
= (96 – 48) × \(\frac { 1 }{ 6 }\)
= 48 × \(\frac { 1 }{ 6 }\)
= 8
Voters in thousands
∴ Approximate change of voters = 8 × 1000 = 8000

Question 9.
The relation between the number of words y a person learns in x hours is given by y = 52√x, 0 ≤ x ≤ 9. What is the approximate number of words learned when x changes from
(i) 1 to 1.1 hours?
(ii) 4 to 4.1 hours?
Solution:
y = 52 √x
dy = 52 × \(\frac { 1 }{ 2 }\) × x^-1/2 dx
x = 1, dx = 0.1
\(\frac { 26 }{ √x }\) × 0.1 = 26 × 0.1
= 2.6
≅ 3 words

(ii) y = 52 √y
dy = 52 × \(\frac { 1 }{ 2 }\) × x^-1/2 dx
x = 4, dx = 0.1
\(\frac { 26 }{ √4 }\) × 0.1 = 13 × 0.1
= 1.3
≅ 1 word

Question 10.
A circular plate expands uniformly under the influence of heat. If its radius increases from 10.5 cm to 10.75 cm, then find an approximate change in the area and the approximate percentage change in the area.
Solution:
Area of the circular plate A = πr²
= π × 10.5 × 105
= 110.25 π
dA = 2πrdr
= 2π × 10.5 × 0.251
= 5.25 π
Approximate percentage change in the area
= \(\frac { dA }{ A }\) × 100
= \(\frac { 5.25π }{ 110.25π }\) × 100
= 0.04761 × 100
= 4.76%

Question 11.
A coat of paint of thickness 0.2 cm is applied to the faces of cube whose edge is 10 cm. Use the differentials to find approximately how many cubic centimeters of paint is used to paint this cube. Also calculate the exact amount of paint used to paint this cube.
Solution:
v = a³
so dv = a² da
dv (when) a = 10 cm and da = 0.20 cm
= 3(10²) (0.2)
300 × 0.2 = 60 cm³

Actual paint used = v at x + ∆x = 10.2 and x = 10 cm
= a³ at x + ∆x = 10.2 and x = 10
= (10.2)³ – (10) = 61.2 cm³

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Samacheer Kalvi 12th Accountancy Book Solutions Answers Guide

Samacheer Kalvi 11th Accountancy Book Back Answers

Tamilnadu State Board Samacheer Kalvi 12th Accountancy Book Back Answers Solutions Guide.

• Chapter 1 Accounts from Incomplete Records
• Chapter 2 Accounts of Not-For-Profit Organisation
• Chapter 3 Accounts of Partnership Firms-Fundamentals
• Chapter 4 Goodwill in Partnership Accounts
• Chapter 5 Admission of a Partner
• Chapter 6 Retirement and Death of a Partner
• Chapter 7 Company Accounts
• Chapter 8 Financial Statement Analysis
• Chapter 9 Ratio Analysis
• Chapter 10 Computerised Accounting System-Tally

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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 8 Microbes in Human Welfare Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 8 Microbes in Human Welfare

12th Bio Zoology Guide Microbes in Human Welfare Text Book Back Questions and Answers

Question 1.
Which of the following microorganism is used for production of citric acid in industries?
(a) Lactobacillus bulgaris
(b) Penicillium citrinum
(c) Aspergillus niger
(d) Rhizopus nigricans
Answer:
(c) Aspergillus niger

Question 2.
Which of the following pair is correctly matched for the product produced by them?
(a) Acetobacter aceti – Antibiotics
(b) Methanobacterium – Lactic acid
(c) Penicilium notatum – Acetic acid
(d) Saccharomyces cerevisiae – Ethanol
Answer:
(d) Saccharomyces cerevisiae – Ethanol

Question 3.
The most common substrate used in distilleries for the production of ethanol is
(a) Soyameal
(b) Groundgram
(c) Molasses
(d) Com meal
Answer:
(c) Molasses

Question 4.
Cry toxins obtained from Bacillus thuringiensis are effective against for ………….
{a) Mosquitoes
(b) Flies
(c) Nematodes
(d) Bollworms
Answer:
(d) Bollworms

Question 5.
Cyclosporin – A is an immunosuppressive drug produced from ………….
(a) Aspergillus niger
(b) Manascus purpureus
(c) Penicillium notatum
(d) Trichoderma polysporum
Answer:
(d) Trichoderma polysporum

Question 6.
Which of the following bacteria is used extensively as a bio-pesticide?
(a) Bacillus thurigiensis
(b) Bacillus subtilis
(c) Lactobacillus acidophilus
(d) Streptococcus lactis
Answer:
(a) Bacillus thurigiensis

Question 7.
Which of the following is not involved in nitrogen fixation?
(a) Pseudomonas
(b) Azotobacter
(c) Anabaena
(d) Nostac
Answer:
(a) Pseudomonas

Question 8.
CO₂ is not released during
(a) Alcoholic fermentation
(b) Lactate fermentation
(c) Aerobic respiration in animals
(d) Aerobic respiration in plants
Answer:
(b) Lactate fermentation

Question 9.
The purpose of biological treatment of waste water is to ………….
(a) Reduce BOD
(b) Increase BOD
(c) Reduce sedimentation
(d) Increase sedimentation
Answer:
(a) Reduce BOD

Question 10.
The gases produced in anaerobic sludge digesters are ………….
(a) Methane, oxygen and hydrogen sulphide.
(b) Hydrogen sulphide, methane and sulphur dioxide.
(c) Hydrogen sulphide, nitrogen and methane.
(d) Methane, hydrogen sulphide and CO₂
Answer:
(d) Methane, hydrogen sulphide and CO₂

Question 11.
How is milk converted into curd? Explain the process of curd formation?
Answer:
The LAB bacteria grows in milk and convert it into curd, thereby digesting the milk protein casein. A small amount of curd added to fresh milk as a starter or inoculum contains millions of Lactobacilli, which under suitable temperature (< 40°C) multiply and convert milk into curd. Curd is more nutritious than milk as it contains a number of organic acids and vitamins.

Question 12.
Give any two bioactive molecules produced by microbes and state their uses.
Answer:

Question 13.
What is biological oxygen demand?
Answer:
The BOD (Biochemical oxygen demand or Biological oxygen demand). BOD refers to the amount of the oxygen that would be consumed, if all the organic matter in one litre of water were oxidized by bacteria. The sewage water is treated till the BOD is reduced. The greater the BOD of the waste water more is its polluting potential.

Question 14.
Explain the role of cry-genes in genetically modified crops.
Answer:

Bacillus thuringiensis is a soil-dwelling bacterium which is commonly used as a biopesticide and contains a toxin called cry toxin. Scientists have introduced this toxin-producing gene into cotton and have raised genetically engineered insect resistant cotton plants.

During sporulation Bacillus thuringiensis produces crystal proteins called Delta-endotoxin which is encoded by cry genes. Delta-endotoxins have specific activities against the insects of the orders Lepidoptera, Diptera, Coleoptera and Hymenoptera. When the insects ingest the toxin crystals their alkaline digestive tract denatures the insoluble crystals making them soluble. The cry toxin then gets inserted into the gut cell membrance and paralyzes the digestive tract. The insect then stops eating and starves to death.

Question 15.
Write the key features of organic farming.
Answer:

• Protecting soil quality using organic materials and encouraging biological activity.
• Indirect provision of crop nutrients using soil microorganisms.
• Nitrogen fixation in soils using legumes.
• Weed and pest control based on methods like crop rotation, biological diversity, natural predators, organic manures and suitable chemical, thermal and biological interventions.

Question 16.
Justify the role of microbes as a bio-fertilizer.
Answer:
Biofertilisers are formulation of living microorganisms that enrich the nutrient quality of the soil. They increase physico – chemical properties of soils such as soil structure, texture, water holding capacity, cation exchange capacity and pH by providing several nutrients and sufficient organic matter. The main sources of biofertilisers are bacteria, fungi and cyanobacteria. Rhizobium is a classical example for symbiotic nitrogen fixing bacteria. This bacterium infects the root nodules of leguminous plants and fixes atmospheric nitrogen into organic forms. Azospirillum and Azotobacter are free living bacteria that fix atmospheric nitrogen and enrich the nitrogen content of soil.

A symbiotic association between a fungus and the roots of the plants is called mycorrhiza. The fungal symbiont in these associations absorbs the phosphorus from soil and transfers to the plant. Plants having such association show other benefits such as resistance to rootbome pathogens, tolerance to salinity, drought, enhances plant growth and developments.

For example, many members of the genus Glomus form mycorrhiza. Cyanobacteria (or) blue green algae (BGA) are prokaryotic free-living organisms which can fix nitrogen. Oscillatoria, Nostoc, Anabaena, Tolypothrix are well known nitrogen fixing cyanobacteria. Their importance is realized in the water logged paddy fields where Cyanobacteria multiply and fix molecular nitrogen. Cyanobacteria secrete growth promoting substances like indole-3-acetic acid, indole-3 – butyric acid, naphthalene acetic acid, amino acids, proteins, vitamins which promotes plant growth and production.

Biofertilisers are commonly used in organic farming methods. Organic farming is a technique, which involves cultivation of plants and rearing of animals in natural ways. This process involves the use of biological materials, avoiding synthetic substances to maintain soil fertility and ecological balance thereby minimizing pollution and wastage.

Key features of organic farming

• Protecting soil quality using organic materials and encouraging biological activity.
• Indirect provision of crop nutrients using soil microorganisms.
• Nitrogen fixation in soils using legumes.
• Weed and pest control based on methods like crop rotation, biological diversity, natural predators, organic manures and suitable chemical, thermal and biological interventions.

Question 17.
Write short notes on the following.
(a) Brewer’s yeast
(b) Ideonella sakaiensis
(c) Microbial fuel cells
Answer:
(a) Brewer’s yeast – Saccharomyces cerevisiae is a widely used fungal species in preparation & softening of bakery products like dough.

(b) Ideonella sakaiensis is a bacterium is used to recycle PET plastics. The enzyme PETase and MHETase in the bacterium breakdown the PET plastics into terephthalic acid & ethylene glycol.

(c) A microbial fuel cell is a bio-electrochemical system that drives an electric current by using bacteria and mimicking bacterial interaction found in nature. Microbial fuel cells work by allowing bacteria to oxidize and reduce organic molecules. Bacterial respiration is basically one big redox reaction in which electrons are being moved around. A MFC consists of an anode and a cathode separated by a proton exchange membrane. Microbes at the anode oxidize the organic fuel generating protons which pass through the membrane to the cathode and the electrons pass through the anode to the external circuit to generate current.

Question 18.
List the advantages of biogas plants in rural areas.
Answer:
Biogas is used for cooking and lighting. The technology of biogas production was developed in India mainly due to the efforts of Indian Agricultural Research Institute (IARI) and Khadi and Village Industries Commission (KVIC).

Question 19.
When does antibiotic resistance develop?
Answer:
Antibiotic resistance occurs when bacteria develop the ability to defeat the drug designed to kill or inhibit their growth. It is one of the most acute threat to public health. Antibiotic resistance is accelerated by the misuse and over use of antibiotics, as well as poor infection prevention control. Antibiotics should be used only when prescribed by a certified health professional. When the bacteria become resistant, antibiotics cannot fight against them and the bacteria multiply. Narrow spectrum antibiotics are preferred over broad spectrum antibiotics. They effectively and accurately target specific pathogenic organisms and are less likely to cause resistance.

Question 20.
What is the key difference between primary and secondary sewage treatment?
Answer:
Primary Treatment:

1. Physical foam
2. Solid materials and particulate organic and inorganic materials are removed from sewage.
3. Principles involved are setting, sedimentation, and filtration.

Secondary Treatment:

1. Biological foam
2. Organic matters are removed by using microbes.
3. Principles involved in aeration and aerobic microbial degradation.

12th Bio Zoology Guide Microbes in Human Welfare Additional Important Questions and Answers

12th Bio Zoology Guide Microbes in Human Welfare One Mark Questions and Answers

Question 1.
The leavering of the dough during fermentation is due to
(a) Formation of ethyl alchohol
(b) Formation of CO₂
(c) Formation of oxygen
(d) Action of zymase enzyme
Answer:
(b) Formation of CO₂

Question 2.
Match list I with list II

Answer:
a – iv, b – iii, c – ii, d – i

Question 3.
Name the person who was the first to use the term antibiotic.
(a) Selman Waksman
(b) Alexander Flemming
(c) Earnest Chain
(d) Howard Florey
Answer:
(a) Selman Waksman

Question 4
……………. is commonly referred as the Queen of drugs.
Answer:
Penicillin

Question 5.
Tetracycline is a …………………..
(i) bactericidal antibiotic
(ii) bacteriastatic antibiotic
(iii) narrow spectrum antibiotic
(iv) Broad spectrum antibiotic

(a) i and iii
(b) ii and iii
(c) i and iv
(d) ii and iv
Answer:
(d) ii and iv

Question 6.
Chlortetracycline was isolated from the culture ……………….
(a) Streptomyces aureofaciens
(b) Streptomyces griseus
(c) Streptococcus lactis
(d) Aspergillus niger
Answer:
(a) Streptomyces aureofaciens

Question 7.
Identify the name and the formula of industrial alcohol.
(a) Butanol, C₄H₉OH
(b) Propanol, C₃H₇OH
(c) Ethanol, C₂H₅OH
(d) Methanol, CH₃OH
Answer:
(c) Ethanol, C₂H₅OH

Question 8.
Pick the bacterial species which is not used in ethanol formation.
(a) Zymomonas mobilis
(b) Sarcina ventriculi
(c) Saccharomyces cerevisiae
(d) Streptomyces aureofaciens
Answer:
(d) Streptomyces aureofaciens

Question 9.
Match the Microbes with the respective organic acids

Answer:
a-ii, b -i, c- iv, d-iii

Question 10.
Human insulin is being commercially produced from a transgenic species of…………………
(a) Escherichia
(b) Mycobacterium
(c) Streptococcus
(d) Penicillin
Answer:
(a) Escherichia

Question 11.
Select the correct statement from the following.
(a) Primary treatment of sewage involves biological oxidation
(b) Excreta of cattle is commonly called Gobur
(c) Delta endotoxin of Bacillus thuringiensis is encoded by pen-genes.
(d) Trichoderma is a free-timing bacteria very common in root ecosystem.
Answer:
(a) Escherichia

Question 12.
Oil strains in laundry can be removed using ……………….
(a) Peptidane
(b) Protease
(c) Amylase
(d) Lipase
Answer:
(d) Lipase

Question 13.
Find the odd sentence out.
(i) Biogas primarily consists of methane with CO₂ and hydrogen
(ii) The greater the BOD of wastewater the more its polluting potential
(iii) World biofuel day is observed on 10th August
(iv) Cyclosporin A is obtained from Trichoderma polysporum.

(a) i and iii
(b) ii and iv
(c) i and iv
(d) none of the above
Answer:
(d) none of the above

Question 14.
Match with correct pair
(a) Biofertilizer (i) Bacillus thuringiensis
(b) Bioremediation (ii) Rhizobium
(c) Biopesticide (iii) Pseudomonas
(d) Bioherbicide (iv) Phytophthora palmivora
Answer:
a – ii, b – iii, c-i, d-iv

Question 15.
Name the genus of virus which are used as effective Biocontrol agent.
Answer:
Nucleopolyhedrovirus

Question 16.
Statement 1: Prebiotics are the compounds in food that induce the growth of beneficial microbes
Statement 2: LAB is a probiotic
(a) Statement 1 is correct. Statement 2 is incorrect.
(b) Statement 1 is incorrect. Statement 2 is correct.
(c) Both statements 1 and 2 are incorrect.
(d) Both statements 1 and 2 are correct.
Answer:
(d) Both statements 1 and 2 are correct.

Question 17.
‘Statement 1: Saccharomyces cerevisiae is commonly called as Baker’s yeast.
Statement 2: Yogurt is produced by the fermentation of milk by saccharomyces cerevisiae
(a) Statement 1 is correct. Statement 2 is incorrect.
(b) Statement 1 is incorrect. Statement 2 is correct.
(c) Both statements 1 and 2 are incorrect.
(d) Both statements 1 and 2 are correct.
Answer:
(a) Statement 1 is correct. Statement 2 is incorrect.

Question 18.
The flavour in the yogurt is due to ………………..
(a) Formaldehyde
(b) Lactate
(c) Acetaldehyde
(d) Caesin
Answer:
(c) Acetaldehyde

Question 19.
Assertion (A): Streptomycin is an antibiotic.
Reason (R): Antibiotic are microbial chemicals inhibits the growth of pathogenic microbe.
(a) A is right R is wrong.
(b) R explains A.
(c) A and R are wrong.
(d) A and R are right. R cannot explain A. ‘
Answer:
(b) R explains A

Question 20.
Assertion (A): Oenology deals with study of wine and its preparation.
Reason (R): Zymology deals with the biochemical process of fermentation and its uses.
(a) A is right R is wrong.
(b) R explains A.
(c) A and R are wrong.
(d) A and R are right. R cannot explain A.
Answer:
(d) A and R are right. R can not explain A.

Question 21.
In primary sewage treatment, the floating debris are removed by………………….
(a) Distillation
(b) Sedimentation
(c) Sequential filtration
(d) Biological oxidation
Answer:
(c) Sequential filtration

Question 22.
Yamuna Action Plan was a bilateral project signed between ……………..
(a) India and Pakistan Government
(b) India and Japan Government
(c) India and China Government
(d) India and Srilanka Government
Answer:
(b) India and Japan Government

Question 23.
Select the correct option denoting the proper sequence of sewage tratment.
(a) Filtration, Sedimentation, Aeration, Biological oxidation and UV radiation
(b) Sedimentation, Filtration, Biological oxidation, Aeration and UV radiation.
(c) Filtration, Aeration, Biological oxidation, Sedimentation and UV radiation.
(d) UV radiation, Sedimentation, Filtration, Biological oxidation and Aeration.
Answer:
(a) Filtration, Sedimentation, Aeration, Biological oxidation and UV radiation.

Question 24.
The Ganga action plan was launched on ………………..
Answer:
14th January 1986.

Question 25.
Which of the following plant species is the most suitable oilseed for biodirect production?
(a) Ground nut
(b) Areca nut
(c) Jatropha curcas
(d) Phyllanthus anarus
Answer:
(c) Jatropha curcas

Question 26.
Match the following medicinal products with their microbial source.
(a) Humulin (z) Streptococcus
(b) Statins (ii) Trichoderma polysporum
(c) Cyclosporin A (z’z’z) Monascus purpureus
(d) Streptokinase (z’v) E-coli
Answer:
a – iv, b – iii, c – ii, d-i

12th Bio Zoology Guide Microbes in Human Welfare Two Marks Questions and Answers

Question 1.
What does LAB stands for? Give two examples.
Answer:

1. LAB stands for Lactic Acid Bacteria
2. Examples for LAB: Lactobacillus lactis, Streptococcus lactis.

Question 2.
Define LAB.
Answer:
Lactic Acid bacteria (LAB) are the probiotics that check the growth of pathogenic microbes in the stomach and other digestive parts. .

Question 3.
Given below are the food products obtained by microbial action. Name the respective organism responsible for their production, (a) Swiss Cheese (6) Bread dough
Answer:
(a) Swiss Cheese – Propionibacterium shermanii
(b) Bread dough – Saccharomyces cerevisiae

Question 4.
What is SCP?
Answer:
Single cell protein refers to edible unicellular microorganisms like Spirulina. Protein extracts from pure or mixed cultures of algae, yeasts, fungi or bacteria may be used as ingredient or as a substitute for protein rich foods and is suitable for human consumption or as animal feed.

Question 5.
What is fermentor?
Answer:
A fermentor (bioreactor) is a closed vessel with adequate arrangement for aeration, agitation, temperature, pH control and drain or overflow vent to remove the waste biomass of cultured microorganisms along-with their products.

Question 6.
What are antibiotics?
Answer:
Antibiotics are chemical substances produced by microorganisms which can kill or retard the growth of other disease causing microbes even in low concentration. Antibiotic means “against life”.

Question 7.
Give reason, (a) Flavour in Yogurt (6) Large holes in Swiss Cheese
Answer:
(a) The flavour in yogurt is due to acetaldehyde formation.
(b) Large holes in Swiss Cheese is due to more C02 production by propionibacterium shermanii.

Question 8.
Name few industrial microbial products.
Answer:
Beverages, antibiotics, organic acids, amino acids, vitamins, biofuels, single-cell protein, enzymes, steroids, vaccines, pharmaceutical drugs.

Question 9.
Name few home-made micorbial products.
Answer:
Yogurt, Paneer, Curd, Idli batter, Bread dough.

Question 10.
Who derived the term antibiotic? Name the antibiotic which he discovered.
Answer:
The term antibiotic was used first by Selman Waksman in 1943. He discovered Streptomycin.

Question 11.
The first antibiotic was extracted from fungus. Who had done it? And also mention the fungal species used.
Answer:
Alexander Flemming, discovered the first antibiotic from the fungus Penicillium notatum and Pencillium Chrysogenum.

Question 12.
Write as example for (a) Bacteriostatic antibiotic (b) Bactericidal antibiotic
Answer:
Bacteriostatic antibiotic – Eg: Tetracycline.
Bactericidal antibiotic – Eg: Streptomycin.

Question 13.
Name any four antibiotics.
Answer:
Erythromycin, Chloromycetin, Neomycin, Kenamycin.

Question 14.
Compare Broad-spectrum antibiotics with narrow-spectrum antibiotics.
Answer:
Broad-spectrum antibiotics act against a wide range of disease-causing bacteria. Narrow-spectrum antibiotics are active against a selected group of bacterial types.

Question 15.
Define (a) Antibiosis (A) Superbugs
Answer:
(a) Antibiosis: Property of antibiotics to. kill microbes.
(b) Superbugs: Bacterial strains gained resistance against antibiotics.

Question 16.
What is studied under zymology and oenology?
Answer:

• Zymology deals with the biochemical process of fermentation and its application.
• Oenology deals with study of wine and winemaking.

Question 17.
Saccharomyces cerevisiae is called as brewer’s yeast. Justify.
Answer:
Saccharomyces cerevisiae commonly called brewer’s yeast is used for fermenting malted cereals and fruit juices to produce various alcoholic beverages. Wine and beer are produced without distillation, whereas whisky, brandy and rum are obtained by fermentation and distillation.

Question 18.
How wine is made? Which organism is involved in the process.
Answer:

• Wine is made by the fermentation of grape juice.
• Grape juice is fermented by different strains of Saccharomyces cerevisiae into alcohol.

Question 19.
Name the 2 types of grape wine. How they differ from one another?
Answer:
Grape wine is of two types, red wine and white wine. For red wine, black grapes are used including skins and sometimes the stems also are used. In contrast white wine is produced only from the juice of either white or red grapes without their skin and stems.

Question 20.
Saccharomyces Cerevisiae is a widely used fungus in making beverages. Considering this complete the table by mentioning the raw material and respective product.

Answer:
A – Wine, B – Germinated barley, C – Whisky, D — Rum

Question 21.
Ethanol is a industrial alcohol – Justify.
Answer:
Saccharomyces cerevisiae is the major producer of ethanol (C2H5OH). It is used for industrial, laboratory and fuel purposes. So ethanol is referred to as industrial alcohol.

Question 22.
Name the bacterial species involved in ethanol production.
Answer:
Zymomonas mobilis and Sarcina ventriculi.

Question 23.
What are the substrates used in producing industrial alcohol.
Answer:
Molasses, Com, Potatoes, Wood wastes.

Question 24.
Write the equation for alcoholic fermentation.
Answer:

Question 25.
Complete the table by filling with respective organic acid and microbes.

Answer:
W- Acetobacter aceti
X – Citric acid,
Y – Rhizopus oryzae,
Z – Butyric acid

Question 26.
Genetically engineered Streptococci is used in medicine. How?
Answer:
Genetically engineered Streptococci are used as “clot buster” for, removing clots from the blood vessels of patients who have undergone myocardial infarction.

Question 27.
Write a brief note on Cyclosporin A.
Answer:
Cyclosporin A, an immunosuppressant used in organ transplantation is produced from the fungus Trichoderma polysporum. It is also used for its anti-inflammatory, anti-fungal and anti-parasitic properties.

Question 28.
State the source of statins and its medical role.
Answer:

• Statins is obtained from Monascus Purpureus (yeast fungus).
• It competitively inhibits the enzyme responsible for cholesterol synthesis there by reducing the blood cholesterol level.

Question 29.
Expand and define BOD.
Answer:
BOD – Biological Oxygen Demand (or) Biochemical Oxygen Demand
BOD refers to the amount of the oxygen that would be consumed, if all the organic matter in one litre of water were oxidized by bacteria.

Question 30.
Define activated sludge.
Answer:
Once the BOD of sewage water is reduced significantly, the effluent is then passed into a settling tank where the bacterial “floes” are allowed to sediment. This sediment is called activated sludge.

Question 31.
Give a brief account on Ganga action plan.
Answer:
The Ganga action plan was launched on 14th January 1986. The main objective of the programme is to improve the water quality of River Ganges by interception, diversion and treatment of domestic sewage and to identify grossly polluting units to prevent pollution.

Question 32.
Name the biocontrol agents used to get rid of Aphids & mosquito larvae.
Answer:
Lady bird beetle & Dragon flies.

Question 33.
How insect resistant plants are developed using Bacillus thuringiensisl
Answer:
Bacillus thuringiensis is a soil dwelling bacterium which is commonly used as a biopesticide and contains a toxin called cry toxin. Scientists have introduced this toxin producing genes into cotton and have raised genetically engineered insect resistant cotton plants.

Question 34.
Which group of insects can be controlled by using delta-endotoxins of Bt-crops?
Answer:
Lepidoptera, Diptera, Coleoptera and Hymenoptera.

Question 35.
Name the two viruses that acts as biocontrol agents.
Answer:
Buculoviruses and Nucleopolyhedrovirus

Question 36.
Rhizobium is a bacteria but acts as bio-fertilization. How?
Answer:
Rhizobium is symbiotic nitrogen-fixing bacteria that infects the root nodules of leguminous plants and fixes free atmospheric nitrogen into organic forms.

Question 37.
Give any two examples for free-living nitrogen-fixing bacteria.
Answer:
Azospirillum and Azotobacter

Question 38.
Name a few blue-green algae (BGA).
Answer:
Oscillatoria, Nostoc, Anabaena, Tolypothrix

Question 39.
Define organic farming.
Answer:
Organic farming is a technique, which involves the cultivation of plants and rearing of animals in natural ways. This process involves the use of biological materials, avoiding synthetic substances to maintain soil fertility and ecological balance thereby minimizing pollution and wastage.

Question 40.
Pseudomonas putida and pollution abatement – comment.
Answer:
Pseudomonas putida is a genetically engineered, multi plasmid hydrocarbon-degrading bacterium. These bacteria can digest the hydrocarbons in the oil spills helps to overcome water pollution.

Question 41.
Complete the following equation by identifying the microbes involved in bioremediation.

Answer:
A – Ideonella sakaiensis
B – Dehalococcoides species.

12th Bio Zoology Guide Microbes in Human Welfare Three Marks Questions and Answers

Question 1.
Differentiate between Prebiotics and Probiotics.
Answer:
Prebiotics :
Prebiotics are compounds in food (fibers) that induce the growth or activity of beneficial microorganisms.

Probiotics :
Probiotics are live microorganisms intended to provide health benefits when consumed, generally by improving or restoring the gut flora.

Question 2.
How yogurt is produced?
Answer:
Yogurt is produced by bacterial fermentation of milk, and lactic acid is produced as a byproduct. Microorganisms such as Streptococcus thermophilus and Lactobacillus bulgaricus coagulate the milk protein and convert the lactose in the milk to lactic acid. The flavour in yogurt is due to acetaldehyde.

Question 3.
Name the scientists who were awarded Nobel Prize in 1945 for discovering the penicillin drug & its role as an antibiotic.
Answer:
(a) Alexander Fleming (b) Earnest Chain (c) Howard Florey

Question 4.
Both Tetracycline and Streptomycin are broad-spectrum antibiotics yet functionally discriminate. How?
Answer:
Though Tetracycline & Streptomycin are broad-spectrum antibiotics, Tetracycline acts as a bacteriostatic antibiotic whereas streptomycin acts as a bactericidal antibiotic against both gram-positive and gram-negative bacteria.

Question 5.
Under which condition does a microbe gains resistance against antibiotics?
Answer:
Antibiotic resistance occurs when bacteria develop the ability to defeat the drug designed to kill or inhibit their growth. It is one of the most acute threats to public health. Antibiotic resistance is accelerated by the misuse and overuse of antibiotics, as well as poor infection prevention control. Antibiotics should be used only when prescribed by a certified health professional. When the bacteria become resistant, antibiotics cannot fight against them and the bacteria multiply. Narrow spectrum antibiotics are preferred over broad-spectrum antibiotics. They effectively and accurately target specific pathogenic organisms and are less likely to cause resistance.

Question 6.
Write a short on “Pathaneer”.
Answer:
In some parts of South India, a traditional drink called pathaneer is obtained from fermenting sap of palms and coconut trees. A common source is tapping of unopened spadices of coconut. It is a refreshing drink, which on boiling produces jaggery or palm sugar. When pathaneer is left undisturbed for few hours it gets fermented to form today with the help of naturally occurring yeast, to form a beverage that contains 4 percent alcohol. After 24 hours toddy becomes unpalatable and is used for the production of vinegar.

Question 7.
Explain in simple about the production process of industrial alcohol.
Answer:
The process of ethanol production starts by milling a feedstock followed by the addition of dilute or fungal amylase (enzyme) from Aspergillus to break down the starch into fermentable sugars. Yeast is then added to convert the sugars to ethanol which is then distilled off to obtain ethanol which is upto 96 percent in concentration. Ethanol and biodiesel represents the first generation of biofuel technology. Ethanol is often used as a fuel, mainly as a biofuel additive for gasoline

Question 8.
Write the percent of alcohol in Beer, Wine and Distilled spirits
Answer:
Beer – 3 to 5 % alcohol; Wine – 9 to 14% alcohol; Distilled spirits – 35 to 50% alcohol.

Question 9.
Complete the table.

Answer:
A – Germinatd Barley;
B- Rum;
C- Fermented sap of palm and coconut tree;
D- Saccharomyces cerevisiae.

Question 10.
Name the enzymes used as clarifier for Bottled juices.
Answer:
PeCtinase, Protease, Cellulase.

Question 11.
Write a short note on bio diesel.
Answer:
Biodiesel is a fuel made from vegetable oils, fats or greases. Biodiesel fuel can be used in diesel engines without altering the engine. Pure biodiesel is non-toxic, biodegradable and produces lower level of air pollutants than petroleum-based diesel fuel. The Government of India approved the National Policy on Biofuels in December 2009 and identified Jatropha curcas as the most suitable oilseed for biodiesel production. Pongamia species is also a suitable choice for production of biodiesel.

Question 12.
Name the act enforced by government to conserve water bodies. Also mention its objectives.
Answer:
National river conservation plan (NRCP) was enacted in 1995 to improve the water quality of the rivers, which are the major fresh water resources in our country. This important assignment taken up under the NRCP includes,

• To capture the raw sewage flowing into the river through open drains and divert them for treatment.
• Setting up sewage treatment plants for treating the diverted sewage.
• Construction of low cost sanitation toilets to prevent open defecation on river banks.

Question 13.
Give an account on MFC.
Answer:
A microbial fuel cell is a bio-electrochemical system that drives an electric current by using bacteria and mimicking bacterial interaction found in nature. Microbial fuel cells work by allowing bacteria to oxidize and reduce organic molecules. Bacterial respiration is basically one big redox reaction in which electrons are being moved around. A MFC consists of an anode and a cathode separated by a proton exchange membrane. Microbes at the anode oxidize the organic fuel generating protons which pass through the membrane to the cathode and the electrons pass through the anode to the external circuit to generate current.

Question 14.
How Bacillus thuringiensis provides insect resistance?
Answer:
During sporulation Bacillus thuringiensis produces crystal proteins called Delta-endotoxin which is encoded by cry genes. Delta-endotoxins have specific activities against the insects of the orders Lepidoptera, Diptera, Coleoptera and Hymenoptera. When the insects ingest the toxin crystals their alkaline digestive tract denatures the insoluble crystals making them soluble. The cry toxin then gets inserted into the gut cell membrance and paralyzes the digestive tract. The insect then stops eating and starves to death.

Question 15.
Give an account on Weedicides.
Answer:
Weedicides are substances, which destroy weeds without harming the useful plants. Bioweedicides are compounds and secondary metabolites derived from microbes such as fungi, bacteria or protozoa. The first bioherbicide developed in 1981 was a Mycoheribicide derived from the fungus Phytophthora palmivora. It controls the growth of strangler vine in citrus crops.

Question 16.
What are the Bio-fertilizers? Mention their role in agriculture?
Answer:
Biofertilisers are the formulation of living microorganisms that enrich the nutrient quality of the soil. They increase physicochemical properties of soils such as soil structure, texture, water holding capacity, cation exchange capacity and pH by providing several nutrients and sufficient organic matter.

Question 17.
What is mycorrhiza? How does it benefit the plants?
Answer:
A symbiotic association between a fungus and the roots of the plants is called mycorrhiza. The fungal symbiont in these associations absorbs the phosphorus from soil and transfers to the plant. Plants having such association show other benefits such as resistance to rootbome pathogens, tolerance to salinity, drought, enhances plant growth and developments. For example, many members of the genus Glomus form mycorrhiza.

Question 18.
Write a brief note on Cyanobacteria & its significances.
Answer:
Cyanobacteria (or) blue-green algae (BGA) are prokaryotic free-living organisms that can fix nitrogen. Oscillatoria, Nostoc, Anabaena, Tolypothrix are well known nitrogen-fixing cyanobacteria. Their importance is realized in the waterlogged paddy fields where Cyanobacteria multiply and fix molecular nitrogen. Cyanobacteria secrete growth-promoting substances like indole-3-acetic acid, indole-3- butyric acid, naphthalene acetic acid, amino acids, proteins, vitamins which promote plant growth and production.

Question 19.
Define bio-remediation and its types.
Answer:
The use of naturally occurring or genetically engineered microorganisms to reduce or degrade pollutants is called bioremediation. Bioremediation is less expensive and more sustainable than other remediations available. It is grouped into in-situ bioremediation (treatment of contaminated soil or water in the site) and ex-situ bioremediation (treatment of contaminated soil or water that is removed from the site and treated).

12th Bio Zoology Guide Microbes in Human Welfare Five Marks Questions and Answers

Question 1.
Explain the role of microbes in the production of enzymes & bio-active molecules?
Answer:
Microbes used for the production of chemicals like organic acids and enzymes. Examples of organic acid producers are Aspergillus niger for citric acid, Acetobacter aceti for acetic acid, Rhizopus oryzae for fumaric acid, Clostridium butyricum for butyric acid and Lactobacillus for lactic acid.

Yeast (Saccharomyces cerevisiae) and bacteria are used for the commercial production of enzymes. Lipases are used in detergent formulations and are used for removing oily stains from the laundry. Bottled juices are clarified by the use of pectinase, protease and cellulase. Rennet can also be used to separate milk into solid curds for cheese making. Streptokinase ’ produced by the bacterium Streptococcus and genetically engineered Streptococci are used as “clot buster” for removing clots from the blood vessels of patients who have undergone myocardial infarction.

Cyclosporin A, an immunosuppressant used in organ transplantation is produced from the fungus Trichoderma polysporum. It is also used for its anti-inflammatory, anti-fungal and anti-parasitic properties. Statins produced by the yeast Monascus purpureus have been used to lower blood cholesterol levels. It acts by competitively inhibiting the enzyme responsible for the synthesis of cholesterol. Recombinant human insulin has been produced predominantly using E. coli and Saccharomyces cerevisiae for therapeutic use in humans.

Question 2.
Describe the stages of the Sewage treatment process.
Answer:
Sewage treatment is usually performed in the following three stages.

Primary treatment: Primary treatment involves the physical removal of solid and particulate organic and inorganic materials from the sewage through filtration and sedimentation. Floating debris is removed by sequential filtration. Then the grit (soil and small pebbles) are removed by sedimentation. All solids that settle form the primary sludge and the supernatant forms the effluent. The effluent from the primary settling tank is taken for secondary treatment.

Secondary treatment or biological treatment: The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it. This allows vigorous growth of useful aerobic microbes into floe (masses of bacteria associated with fungal filaments to form mesh-like structures). While growing, these microbes consume the major part of the organic matter in the effluent.

This significantly reduces the BOD (Biochemical oxygen demand or Biological oxygen demand). BOD refers to the amount of oxygen that would be consumed if all the organic matter in one liter of water were oxidized by bacteria. The sewage water is treated till the BOD is reduced. The greater the BOD of the wastewater more is its polluting potential. Once the BOD of sewage water is reduced significantly, the effluent is then passed into a settling tank where the bacterial “floes” are allowed to sediment.

This sediment is called activated sludge. A small part of activated sludge is pumped back into the aeration tank to serve as the inoculum. The remaining major part of the sludge is pumped into large tanks called anaerobic sludge digesters. Here, the bacteria which grow anaerobically, digest – the bacteria and the fungi in the sludge. During this digestion, bacteria produce a mixture of gases such as methane, hydrogen sulfide and CO₂. These gases form biogas and can be used as a source of energy.

Tertiary treatment: Tertiary treatment is the final process that improves the quality of the wastewater before it is reused, recycled or released into natural water bodies. This treatment removes the remaining inorganic compounds and substances, such as nitrogen and phosphorus. UV is an ideal disinfectant for wastewater since it does not alter the water quality – except for inactivating microorganisms. UV is a chemical free process that can completely replace the existing chlorination system and also inactivates chlorine-resistant microorganisms like Cryptosporidium and Giardia.

Question 3.
How microbes are used in Gobar gas production?
Answer:
Biogas is a mixture of different gases produced by the breakdown of organic matter in the absence of oxygen. Biogas can be produced from raw materials such as agricultural wastes, manure, municipal wastes, plant material, sewage, food waste, etc., Biogas is produced under anaerobic condition, when the organic materials are converted through microbiological reactions into gas and organic fertilizer.

Biogas primarily consists of methane (63 percent), along with CO₂ and hydrogen. Methane producing bacteria are called methanogens and one such common bacterium is Methanobacterium. Biogas is devoid of smell and bums with a blue flame without smoke. The Methanogens are also present in anaerobic sludge and rumen of cattle. In rumen, these bacteria help in the breakdown of cellulose. The excreta of cattle called dung is commonly called “Gobar”. Gobar gas is generated by the anaerobic decomposition of cattle dung. It consists of methane, CO₂ with some hydrogen, nitrogen and other gases in trace amounts.

In a biogas plant, anaerobic digestion is carried out in an air tight cylindrical tank known as digester. It is made up of concrete bricks and cement or steel. Bio-wastes are collected and slurry of dung is fed into this digester. It has a side opening into which organic materials for digestion are incorporated for microbial activity.

Anaerobic digestion is accomplished in three stages: solubilization, acidogenesis and methanogenesis. The outlet is connected to a pipe to supply biogas. The slurry is drained through another outlet and is used as fertilizer. Biogas is used for cooking and lighting. The technology of biogas production was developed in India mainly due to the efforts of Indian Agricultural Research Institute (IARI) and Khadi and Village Industries Commission (KVIC).

Question 4.
Bioremediation & microbes – Discuss.
Answer:
Aerobic microbes degrade the pollutants in the presence of oxygen. They mainly degrade pesticides and hydrocarbons. Pseudomonas putida is a genetically engineered microorganism (GEM). Ananda Mohan Chakrabarty obtained patent for this recombinant bacterial strain. It is multiplasmid hydrocarbon-degrading bacterium which can digest the hydrocarbons in the oil spills.

Nitrosomonas europaea is also capable of degrading benzene and a variety of halogenated organic compounds including trichloroethylene and vinyl chloride. Ideonella sakaiensis is currently tried for recycling of PET plastics. These bacteria use PETase and MHETase enzymes to breakdown PET plastic into terephthalic acid and ethylene glycol.

Anaerobic microbes degrade the pollutants in the absence of oxygen. Dechloromonas aromatica has the ability to degrade benzene anaerobically and to oxidize toluene and xylene. Phanerochaete chrysosporium an anaerobic fungus exhibits strong potential for bioremediation ‘ of pesticides, polyaromatic hydrocarbons, dyes, trinitrotoluene, cyanides, carbon tetrachloride, etc.,

Dehalococcoides species are responsible for anaerobic bioremediation of toxic trichloroethene to nontoxic ethane. Pestalotiopsis microspora is a species of endophytic fungus capable of breaking down and digesting polyurethane. This makes the fungus a potential candidate for bioremediation projects involving large quantities of plastics.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
One can notice large holes in Swiss cheese. Which bacterium is responsible for these holes. What are these holes due to?
Answer:
Propionibacterium shermanii.
The holes are due to the formation of large amount of carbon-di-oxide during fermentation.

Question 2.
Name the resource agent of statins. How does this bioactive molecule functions in our body?
Answer:
Monascus purpureus (Yeast).
Statins acts effectively to reduce the blood cholestrol level.

Question 3.
Cyclosporin-A is an immuno supprenant. Suggest any one area where it can be used efficiently? Explan.
Answer:
Being an immuno-suppresant, cyclosporin-A can be used in organ transplant surgeries/pattent to overcome graft rejection.

Question 4.
Name the blank spaces a, b, and c of the given table:

Answer:
(a) Clostridium butyricum
(b) Fungus
(c) Cyclosporin A

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 9 Applications of Biotechnology Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 9 Applications of Biotechnology

12th Bio Zoology Guide Applications of Biotechnology Text Book Back Questions and Answers

Question 1.
The first clinical gene therapy was done for the treatment of ……………………..
(a) AIDS
(b) Cancer
(c) Cystic fibrosis
(d) SCID
Answer:
(d) SCID

Question 2.
Dolly, the sheep was obtained by a technique known as ……………………..
(a) Cloning by gene transfer
(b) Cloning without the help of gametes
(c) Cloning by tissue culture of somatic cells
(d) Cloning by nuclear transfer
Answer:
(d) Cloning by nuclear transfer

Question 3.
The genetic defect adenosine deaminase deficiency may be cured permanently by ……………………..
(a) Enzyme replacement therapy
(b) periodic infusion of genetically engineered lymphocytes having ADA cDNA
(c) administering adenosine deaminase activators
(d) introducing bone marrow cells producing ADA into embryo at an early stage of development.
Answer:
(d) introducing bone marrow cells producing ADA into embryo at an early stage of development.

Question 4.
How many amino acids are arranged in the two chains of Insulin?
(a) Chain A has 12 and Chain B has 13
(b) Chain A has 21 and Chain B has 30 amino acids
(c) Chain A has 20 and chain B has 30 amino acids
(d) Chain A has 12 and chain B has 20 amino acids
Answer:
(b) Chain A has 21 and Chain B has 30 amino acids

Question 5.
PCR proceeds in three distinct steps governed by temperature, they are in order of………………………
(a) Denaturation, Annealing, Synthesis
(b) Synthesis, Annealing, Denaturation
(c) Annealing, Synthesis, Denaturation
(d) Denaturation, Synthesis, Annealing
Answer:
(a) Denaturation, Annealing, Synthesis

Question 6.
Which one of the following statements is true regarding DNA polymerase used in PCR?
(a) It is used to ligate introduced DNA in recipient cells
(b) It serves as a selectable marker
(c) It is isolated from a Virus
(d) It remains active at a high temperature
Answer:
(d) It remains active at a high temperature

Question 7.
ELISA is mainly used for……………………..
(a) Detection of mutations
(b) Detection of pathogens
(c) Selecting animals having desired traits
(d) Selecting plants having desired traits
Answer:
(b) Detection of pathogens

Question 8.
Transgenic animals are those which have
(a) Foreign DNA in some of their cells
(b) Foreign DNA in all their cells
(c) Foreign RNA in some of their cells
(d) Foreign RNA in all their cells
Answer:
(b) Foreign DNA in all their cells

Question 9.
Recombinant Factor VIII is produced in the …………………….. cells of the Chinese Hamster
(a) Liver cells
(b) blood cells
(c) ovarian cells
(d) brain cells
Answer:
(c) ovarian cells

Question 10.
Vaccines that use components of a pathogenic organism rather than the whole organism are called ……………………..
(a) Subunit recombinant vaccines
(b) attenuated recombinant vaccines
(c) DNA vaccines
(d) conventional vaccines
Answer:
(a) Subunit recombinant vaccines

Question 11.
Mention the number of primers required in each cycle of PCR. Write the role of primers and DNA polymerase in PCR. Name the source organism of the DNA polymerase used in PCR.
Answer:

• For each cycle of PCR two primers are required.
• Primers are the small fragments of single-stranded DNA or RNA that serve as a template for initiating DNA polymerization.
• DNA polymerase is an enzyme that synthesizes DNA molecules by pairing the Deoxyribo Nucleotides leading to the formation of new strands.
• DNA polymerase used in PCR is Taq polymerase which is isolated from a thermophilic bacteria called Thermus aquatics.
• Taq polymerase will remain active ever at a very high temperature (80°C) and hence used in the PCR amplification technique.

Question 12.
How is the amplification of a gene sample of interest carried out using PCR?
Answer:
Denaturation, renaturation or primer annealing, and synthesis or primer extension, are the three steps involved in PCR. The double-stranded DNA of interest is denatured to separate into two individual strands by high temperature. This is called denaturation. Each strand is allowed to hybridize with a primer (renaturation or primer-annealing). The primer-template is used to synthesize DNA by using Taq – DNA polymerase.

During denaturation, the reaction mixture is heated to 95 °C for a short time to denature the target DNA into single strands that will act as a template for DNA synthesis. Annealing is done by the rapid cooling of the mixture, allowing the primers to bind to the sequences on each of the two strands flanking the target DNA.

During primer extension or synthesis the temperature of the mixture is increased to 75°C for a sufficient period of time to allow Taq DNA polymerase to extend each primer by copying the single-stranded template. At the end of incubation, both single template strands will be made partially double-stranded. The new strand of each double-stranded DNA extends to a variable distance downstream. These steps are repeated again and again to generate multiple forms of the desired DNA. This process is also called DNA amplification.

Question 13.
What has genetically engineered Insulin?
Answer:
The insulin synthesized by recombinant DNA technology is called genetically engineered Insulin. It was the first-ever pharmaceutical product of rDNA technology. In 1986, human insulin was marked under the trade name Humulin.

Question 14.
Explain how “Rosie” is different from a normal cow.
Answer:
Rosie was the first transgenic cow. It produced human protein-enriched milk, which contained the human alpha-lactalbumin (2.4 gm/liter). This milk was a nutritionally balanced food for infants than the normal milk of cows.

Question 15.
How was Insulin obtained before the advent of rDNA technology? What were the problems encountered?
Answer:
Conventionally, Insulin was isolated and refined from the pancreas of pigs and cows to treat diabetic patients. Though it is effective, due to minor structural changes, the animal insulin caused allergic reaction in a few patients.

Question 16.
ELISA is a technique based on the principles of antigen-antibody reactions. Can ~ this technique be used in the molecular diagnosis of a genetic disorder such as Phenylketonuria?
Answer:
Yes, ELISA test can be done to diagnose phenylketonuria. The affected person does not produce the enzyme phenylalanine hydroxylase. If specific antibodies are developed against the enzyme and ELISA is performed, the unaffected person will show positive result due to antigen and antibody reaction, whereas the affected individual produces negative results. [Note: phenylketonuria is an inherited metabolic disorder that causes the accumulation of phenylalanine (an amino acid) in body cells due to defect in the synthesizing of an enzyme phenylalanine hydroxylase]

Question 17.
Gene therapy is an attempt to correct a Genetic defect by providing a normal gene into the individual. By this the function can be restored. An alternate method would be to provide gene product known as enzyme replacement therapy, which would also restore the function. Which in your opinion is a better option? Give reasons for your answer.
Answer:
Though both Gene therapy and Enzyme replacement therapy helps to restore genetic defects, Gene therapy is much better than Enzyme replacement therapy. Because, in Gene therapy, once the defective gene is repaired using a normal gene, the affected individual gains complete recovery whereas, in Enzyme replacement therapy, the respective enzyme or protein has to be provided periodically and does not offer a permanent cure. Moreover, when compared to Gene therapy, Enzyme replacement therapy is highly expensive.

Question 18.
What are transgenic animals? Give examples.
Answer:
Transgenesis is the process of introduction of extra (foreign/exogenous) DNA into the genome of the animals to create and maintain stable heritable characters. The foreign DNA that is introduced is called the transgene and the animals that are produced by DNA manipulations are called transgenic animals or genetically engineered or genetically modified organisms. e.g. Mice, Cow

Question 19.
If a person thinks he is infected with HIV, due to unprotected sex, and goes for a blood test. Do you think a test such as ELISA will help? If so why? If not, why?
Answer:
Yes, ELISA is a highly sensitive and precise procedure and can detect antigens even in the range of a nanogram. So, it can be used to detect HIV in blood.’

Question 20.
Explain how ADA deficiency can be corrected?
Answer:
The right approach for SCID treatment would be to give the patient a functioning ADA which breaks down toxic biological products.

In some children ADA deficiency could be cured by bone marrow transplantation, where defective immune cells could be replaced with healthy immune cells from a donor. In some patients it can be treated by enzyme replacement therapy, in which functional ADA is injected into the patient.

During gene therapy the lymphocytes from the blood of the patient are removed and grown in a nutrient culture medium. A healthy and functional human gene, ADA cDNA encoding this enzyme is introduced into the lymphocytes using a retrovirus. The genetically engineered lymphocytes are subsequently returned to the patient. Since these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes. The disease could be cured permanently if the gene for ADA isolated from bone marrow cells are introduced into the cells of the early embryonic stages.

Question 21.
What are DNA vaccines?
Answer:
Genetic immunization by using DNA vaccines is a novel approach that came into being in 1990. The immune response of the body is stimulated by a DNA molecule. A DNA vaccine consists of a gene encoding an antigenic protein, inserted onto a plasmid, and then incorporated into the cells in a target animal. DNA instructs the cells to make antigenic molecules which are displayed on its surfaces. This would evoke an antibody response to the free-floating antigen secreted by the cells. The DNA vaccine cannot cause the disease as it contains only copies of a few of its genes. DNA vaccines are relatively easy and inexpensive to design and produce.

Question 22.
Differentiate between Somatic cell gene therapy and Germline gene therapy.
Answer:
Somatic Cell Gene Therapy:

1. Therapeutic genes transferred into the somatic cells.
2. Introduction of genes into bone marrow cells, blood cells, skin cells, etc.
3. It Will not be inherited in later generations.

Germ Line Gene Therapy:

1. Therapeutic genes transferred into the germ cells.
2. Genes introduced into eggs and sperms.
3. Heritable and passed on to later generations.

Question 23.
What are stem cells? Explain its role in the field of medicine.
Answer:
Stem cells are undifferentiated cells found in most of multicellular animals. These cells maintain their undifferentiated state even after undergoing numerous mitotic divisions.

Stem cell research has the potential to revolutionize the future of medicine with the ability to regenerate damaged and diseased organs. Stem cells are capable of self-renewal and exhibit ‘cellular potency’. Stem cells can differentiate into all types of cells that are derived from any of the three germ layers ectoderm, endoderm, and mesoderm.

Question 24.
One of the applications of biotechnology is ‘gene therapy” to treat a person born with a hereditary disease
(i) What does “gene therapy” mean?
(ii) Name the hereditary disease for which the first clinical gene therapy was used.
(iii) Mention the steps involved in gene therapy to treat this disease.
Answer:
(i) Gene therapy is the process in which the defective genes are replaced with normal genes leading to the expression of the proper phenotype.
(ii) SCID (Severe Combined Immuno Deficiency) disease was the first disease treated by using gene therapy.
(iii) There are two strategies involved in gene therapy namely; Gene augmentation therapy, which involves the insertion of DNA into the genome to replace the missing gene product; and Gene inhibition therapy, which involves insertion of the anti-sense gene which inhibits the expression of the dominant gene.

Question 25.
PCR is a useful tool for early diagnosis of Infectious disease. Elaborate.
Answer:
The specificity and sensitivity of PCR is useful for the diagnosis of inherited disorders (genetic diseases), viral diseases, bacterial diseases, etc., The diagnosis and treatment of a particular disease often require identifying a particular pathogen. Traditional methods of identification involve culturing these organisms from clinical specimens and performing metabolic and other tests to identify them.

The concept behind PCR-based diagnosis of infectious diseases is simple – if the pathogen is present in a clinical specimen its DNA will be present.
Its DNA has unique sequences that can be detected by PCR, often using the clinical specimen (for example, blood, stool, spinal fluid, or sputum) in the PCR mixture.

Question 26.
What are recombinant vaccines? Explain the types.
Answer:
Vaccines developed by using recombinant DNA technology are called recombinant vaccines. Subunit recombinant vaccines, attenuated recombinant vaccines, DNA vaccines are the types of recombinant vaccines.

Question 27.
Explain why cloning of Dolly, the sheep was such a major scientific breakthrough?
Answer:
The development of Dolly was a remarkable achievement in the scientific field and it demonstrates that the DNA from differentiated adult cells can also be used to develop into an entire organism.

Question 28.
Mention the advantages and disadvantages of cloning.
Answer:

• Offers benefits for clinical trials and medical research. It can help in the production of proteins and drugs in the field of medicine.
• Aids stem cell research.
• Animal cloning could help to save endangered species.
• Animal and human activists see it as a threat to biodiversity saying that this alters evolution which will have an impact on populations and the ecosystem.
• The process is tedious and very expensive.
• It can cause animals to suffer.
• Reports show that animal surrogates were manifesting adverse outcomes and cloned animals were affected with the disease and have a high mortality rate.
•  It might compromise human health through the consumption of cloned animal meat.
• Cloned animals age faster than normal animals and are less healthy than the parent organism as discovered in Dolly
• Cloning can lead to the occurrence of genetic disorders in animals.
• More than 90% of cloning attempts fail to produce viable offspring.

Question 29.
Explain how recombinant Insulin can be produced.
Answer:
Production of insulin by recombinant DNA technology started in the late 1970s. This technique involved the insertion of a human insulin gene on the plasmids of E. coli. The polypeptide chains are synthesized as a precursor called pre-pro insulin, which contains A and B segments linked by a third chain (C) and preceded by a leader sequence. The leader sequence is removed after translation and the C chain is excised, leaving the A and B polypeptide chains

Question 30.
Explain the steps involved in the production of recombinant hGH.
Answer:
Using recombinant DNA technology hGH can be produced. The gene for hGH is isolated from the human pituitary gland cells. The isolated gene is inserted into a plasmid vector and then is transferred into E. coli. The recombinant E. coli then starts producing human growth hormone. The recombinant E. coli are isolated from the culture and mass production of hGH is carried out by fermentation technology.

12th Bio Zoology Guide Applications of Biotechnology Additional Important Questions and Answers

12th Bio Zoology Guide Applications of Biotechnology One Mark Questions and Answers

Question 1.
Statement 1: Human Insulin is a polypeptide Statement 2: It is composed of 52 amino acids
(a) Statement 1 is true. Statement 2 is false.
(b) Statement 1 is false. Statement 2 is true.
(c) Both statements 1 and 2 are true.
(d) Both statements 1 and 2 are false.
Answer:
(a) Statement 1 is true. Statement 2 is false.

Question 2.
Statement 1: Rosie was the first transgenic goat.
Statement 2: Meat is enriched with human protein.
(a) Statement 1 is true. Statement 2 is false.
(b) Statement 1 is false. Statement 2 is true.
(c) Both statements 1 and 2 are true. ‘
(d) Both statements 1 and 2 are false.
Answer:
(d) Both statements 1 and 2 are false.

Question 3.
Statement 1: Recombinant Hepatitis B vaccine is a live vaccine.
Statement 2: It is obtained by cloning HB antigen gene in yeast.
(a) Statement 1 is true. Statement 2 is false.
(b) Statement 1 is false. Statement 2 is true.
(c) Both statements 1 and 2 are true.
(d) Both statements 1 and 2 are false.
Answer:
(b) Statement 1 is false. Statement 2 is true.

Question 4.
Statement 1: ADA deficiency was the first disease treated by gene therapy.
Statement 2: ADA is an autosomal recessive metabolic disorder.
(a) Statement 4 is true. Statement 2 is false.
(b) Statement 1 is false. Statement 2 is true.
(c) Both statements 1 and 2 are true.
(d) Both statements 1 and 2 are false.
Answer:
(c) Both statements 1 and 2 are true.

Question 5.
Statement 1: Attenuated recombinant vaccines are live vaccines.
Statement 2: Polio is a live vaccine.
(a) Statement 1 is true. Statement 2 is false.
(b) Statement 1 is false. Statement 2 is true.
(c) Both statements 1 and 2 are true.
(d) Both statements 1 and 2 are false.
Answer:
(c) Both statements 1 and 2 are true.

Question 6.
Assertion (A): Interferons are used to treat herpes zoster.
Reason (R): Interferons are antiviral proteins.
(a) R explains A.
(b) Both A and Rare incorrect.
(c) A is correct. R is incorrect. y
(d) A and R are correct. R does not explain A.
Answer:
(a) R explains A.

Question 7.
Assertion (A): PCR is an amplification technique used in biotechnology.
Reason (R): Using PCR multiple copies of DNA can be generated.
(a) R explains A.
(b) Both A and Rare incorrect.
(c) A is correct. R is incorrect.
(d) A and R are correct. R does not explain A.
Answer:
(a) R explains A.

Question 8.
The B-chain of Insulin is composed of ………………….. amino acids
(a) 70
(b) 30
(c) 45
(d) 60
Answer:
(b) 30

Question 9.
The gene for the formation of factor VIII is located in ……………….
(a) 20th Chromosome
(b) 12th Chromosome
(c) X-chromosome
(d) Y-chromosome
Answer:
(c) X-chromosome

Question 10.
The genetic defect in the synthesis of factor VIII results in……………….
(a) Polycythemia
(b) Anaemia
(c) Thalassemia
(d) Haemophilia
Answer:
(c) Haemophilia

Question 11.
Name the scientists who discovered Interferons?
Answer:
Alick Issac and Jean Lindemann

Question 12.
Which is the first synthetic vaccine produced?
(a) Polio Vaccine
(b) Hepatitis B Vaccine
(c) BCG Vaccine
(d) MMR Vaccine
Answer:
(A) Hepatitis B Vaccine

Question 13.
Identify the incorrect statement.
(i) The first clinical gene therapy was given by French Anderson.
(ii) For a four-year-old boy with ADA deficiency.
(iii) ACD is an autosomal dominant metabolic disorder.
(iv) Where patients have non-functioning B – lymphocytes.
(a) i and iv only
(b) ii, iii and iv
(c) i, ii and iv
(d) all the above
Answer:
(b) ii, iii and iv

Question 14.
Identify the correct statement(s).
(i) Totipotency is the ability of single cell to produce a whole organism.
(ii) Pluripotency refers to ability of stem cell with apotential to differentiate into any kind of germ layers.
(iii) Unipotency refers to ability of stem cell to differentiate into one cell type.
(iv) Oligopotency refers to stem cells to differentiate into few cell types.
(a) i and iii (b) ii and iv (c) i and iv (d) all the above
Answer:
(d) all the above

Question 15.
Identify those proper sequence of ELISA testing.
(a) Coating → Blocking → Detection → Read out
(b) Detection → Read out → Coating → Blocking
(c) Read out→ Coating→ Detection → Blocking
(d) Blocking → Detection → Read out → Coating
Answer:
(a) Coating → Blocking → Detection→ Read out

Question 16.
PCR technique was developed by
(a) Eva Engvall
(b) Peter Perlmanin
(c) Kary Mullis
(d) Wilmut
Answer:
(c) Kary Mullis

Question 17.
Arrange the steps of PCR in proper sequence.
(a) Denaturation, Primer extension, Renaturation
(b) Renaturation, Denaturation, Primer extension
(c) Primer extension, Denaturation, Renaturation
(d) Denaturation, Renaturation, Primer extension
Answer:
(d) Denaturation, Renaturation, Primer extension

Question 18.
The first cloned organism was………….
(a) Goat
(b) Cow
(c) Sheep
(d) Pig
Answer:
(c) Sheep

Question 19.
The first transgenic clone of sheep was called as ……………….
(a) Rosie
(b) Dolly
(c) Sameera
(d) Joel
Answer:
(b) Dolly

Question 20.
In the cloning process of Dolly, how many embryos were implemented by Ian Wilmut and Campbell, out of which one successful Dolly was developed?
(a) 267
(b) 211
(c) 287
(d) 307
Answer:
(b) 277

Question 21.
The term Biotechnology was coined by…………..
Answer:
Karl Ereky

12th Bio Zoology Guide Applications of Biotechnology Two Marks Questions and Answers

Question 1.
How insulin controls blood sugar level?
Answer:
Insulin controls the blood sugar level by facilitating the cellular uptake and utilisation of glucose for the release of energy.

Question 2.
State the role of Somatostatin and Somatotropin in human beings.
Answer:
Both somatostatin and somatotropin are peptide hormones which helps in growth and development by increasing the uptake of amino acids and promoting protein synthesis.

Question 3.
Mention the manifestation of the disease – Haemophilia-A
Answer:
Haemophilia A is a X-linked disease which is characterized by prolonged clotting time and * internal bleeding.

Question 4.
Define Interferons.
Answer:
Interferons are proteinaceous, antiviral, species specific substances produced by mammalian cells when infected with viruses. They stimulate the cellular DNA to produce antiviral enzymes which inhibit viral replication and protect the cells.

Question 5.
Who discovered Interferons? On which basis it was classified?’
Answer:
Interferons were discovered by Alick Isaacs and Jean Lindemann in 1957. It is classified as a, P and y interferons based on the structure.

Question 6.
Name the disease that are treated by using interferons.
Answer:
Interferons are used for the treatment of various diseases like cancer, AIDS, multiple sclerosis, hepatitis C and herpes zoster.

Question 7.
Recombinant vaccines are better than conventional ones – Justify.
Answer:
The recombinant vaccines are generally of uniform quality and produce less side effects as compared to the vaccines produced by conventional methods.

Question 8.
Point out four types of recombinant vaccines.
Answer:

1. Subunit recombinant vaccines
2. Attenuated recombinant vaccines
3. Edible vaccines
4. DNA vaccines

Question 9.
What are subunit recombinant vaccines? Mention its advantages.
Answer:
Vaccines that use components of a pathogenic organism rather than the whole organism are called subunit vaccines. The advantages of these vaccines include their purity in preparation, stability and safe use.

Question 10.
Define Attenuated recombinant vaccines.
Answer:
Attenuated recombinant vaccines include genetically modified pathogenic organisms (bacteria or viruses) that are made nonpathogenic and are used as vaccines. Such vaccines are referred to as attenuated recombinant vaccines.

Question 11.
List out the benefits of recombinant vaccines.
Answer:
Vaccines produced by recombinant techniques have definite advantages like producing target proteins, long-lasting immunity and trigger immune response only against specific pathogens with less toxic effects.

Question 12.
Name the two strategies involved in gene therapy
Answer:

1. Gene augmentation therapy.
2. Gene inhibition therapy.

Question 13.
Comment on SCID.
Answer:
ADA deficiency or SCID (Severe Combined Immuno Deficiency) is an autosomal recessive metabolic disorder. It is caused by the deletion or dysfunction of the gene coding for ADA enzyme. In these patients the nonfunctioning T-Lymphocytes cannot elicit immune responses against invading pathogens.

Question 14.
Differentiate between Gene augmentation therapy and gene inhibition therapy.
Answer:
Gene augmentation therapy which involves insertion of DNA into the genome to replace the missing gene product and Gene inhibition therapy which involves insertion of the anti sense gene which inhibits the expression of the dominant gene.

Question 15.
Define the terms (a) Totipotency (b) Unipotency
Answer:
Totipotency is the ability of a single cell to divide and produce all of the differentiated cells in an organism.
Unipotency refers to the ability of the stem cells to differentiate into only one cell type.

Question 16.
What are the best sources of stem cells in mammals?
Answer:
Placenta, Umbilical cord, amniotic sac, amniotic fluid.

Question 17.
Write the names of any two molecular diagnostic techniques used for early diagnosis of diseases?\
Answer:
(a) Polymerase Chain Reaction (PCR) technique.
(b) Enzyme Linked Immuno Sorbent Assay (ELISA)

Question 18.
What does ELISA stands for? Who invented this technique?
Answer:
Enzyme-Linked ImmunoSorbent Assay (ELISA). It was invented by Eva Engvall and Peter Perlman.

Question 19.
Name the various kinds of ELISA.
Answer:
There are four kinds of ELISA namely, Direct ELISA, Indirect ELISA, sandwich ELISA and competitive ELISA.

Question 20.
Simply define the PCR technique. Also mention its inventor.
Answer:
The Polymerase Chain Reaction (PCR) is an invitro amplification technique used for synthesizing multiple identical copies (billions) of DNA of interest. The technique was developed by Kary Mullis in the year 1983.

Question 21.
Expand PCR and name the steps involved in the process.
Answer:
PCR – Polymerase Chain Reaction.
Denaturation, Renaturation or Primer annealing and Primer extension are the three steps in PCR technique.

Question 22.
For which disease does the first clinical gene therapy was done? Who accomplished it?
Answer:
The first clinical gene therapy was done for SCID. Severe Combined Immuno Deficiency disease is caused by ADA deficiency. It was done by French Anderson in 1990.

Question 23.
Define Transgenesis.
Answer:
Transgenesis is the process of introduction of foreign DNA (exogenous DNA) into the genome of the other organism to create and maintain stable heritable characters.

Question 24.
What are the Genetically Modified Organisms?
Answer:
Transgenesis is the process of introduction of extra (foreign/ exogenous) DNA into the genome of the animals to create and maintain stable heritable characters. The foreign DNA that is introduced is called the transgene and the animals that are produced by DNA manipulations are called transgenic animals or the genetically engineered or genetically modified organisms.

Question 25.
What does Biological Product refers to?
Answer:
A biological product is a substance derived from a living organism and used for the prevention or treatment of disease. These products include antitoxins, bacterial and viral vaccines, blood products and hormone extracts.

Question 26.
Define cloning. Name the first organism developed by cloning.
Answer:
Cloning is the process of producing genetically identical individuals of an organism either naturally or artificially. The first cloned organism is a sheep named Dolly.

Question 27.
Who developed Dolly? How many embryos were aborted to develop a single Dolly?
Answer:
Dolly- The first cloned organism (sheep) was developed by Ian Wilmut and Campbell. Out of 29 embryos implanted only one Dolly was developed.

Question 28.
Define Biotechnology.
Answer:
Biotechnology is defined as “any technological application that uses biological systems, living organisms or derivatives thereof, to make or modify products or processes for specific use”.

12th Bio Zoology Guide Applications of Biotechnology Three Marks Questions and Answers

Question 29.
Briefly explain the structure of insulin.
Answer:
Human insulin is synthesized by the (5 cells of Islets of Langerhans in the pancreas. It is formed of 51 amino acids which are arranged in two polypeptide chains, A and B. Polypeptide chain A has 21 amino acids while the polypeptide chain B has 30 amino acids. Both A and B chains are attached together by disulfide bonds.

Question 30.
Who was the first to discover the role of insulin against diabetes? From which organism does was insulin isolated?
Answer:
Best and Banting in 1921, isolated insulin from the pancreatic islets of a dog and demonstrated its effectiveness against diabetes.

Question 31.
How “Rosie” differs from a normal cow? Explain.
Answer:
Rosie, the first transgenic cow produced human protein-enriched milk, which contained the human alpha-lactalbumin. The protein-rich milk (2.4 gm/litre) was a nutritionally balanced food for new bom babies than the normal milk produced by the cows.

Question 32.
Point out any two microbes that play a crucial role in recombinant DNA technology.
Answer:

1. Saccharomyces cerevisiae
2. Escherichia coli

Question 33.
What are Edible vaccines?
Answer:
Edible vaccines are prepared by molecular pharming using the science of genetic engineering. Selected genes are introduced into plants and the transgenic plants are induced to manufacture the encoded protein. Edible vaccines are mucosal targeted vaccines which cause stimulation of both systemic and mucosal immune response. At present edible vaccines are produced for human and animal diseases like measles, cholera, foot and mouth disease and hepatitis.

Question 34.
How the recombinant hepatitis B vaccine is produced in the laboratory?
Answer:
Recombinant hepatitis B vaccine as a subunit vaccine is produced by cloning hepatitis B surface antigen (HbsAg) gene in the yeast, Saccharomyces cerevisiae.

Question 35.
Suggest few methods to treat SCID.
Asnwer:
SCID caused by ADA deficiency could be cured by bone marrow transplantation where defective immune cells could be replaced with healthy immune cells from donor. It can also be treated by enzyme replacement therapy in which functional ADA is injected into patient’s body where it breaks down toxic biological product.

Question 36.
How gene therapy is done to treat ADA deficiency?
Answer:
During gene therapy, the lymphocytes from the blood of the patient are removed and grown in a nutrient culture medium. A healthy and functional human gene, ADA cDNA encoding this enzyme is introduced into the lymphocytes using a retrovirus. The genetically engineered lymphocytes are subsequently returned to the patient. Since these cells are not immortal, the patient requires a periodic infusion of such genetically engineered lymphocytes. The disease could be cured permanently if the gene for ADA isolated from bone marrow cells is introduced into the cells of the early embryonic stages.

Question 37.
How does Somatic cell therapy differ from germline gene therapy?
Answer:
Somatic cell therapy involves the insertion of a fully functional and expressible gene into a target somatic cell to correct a genetic disease permanently whereas Germline gene therapy involves the introduction of DNA into germ cells which is passed on to successive generations. Gene therapy involves the isolation of a specific gene and making its copies and inserting them into target cells ’to make the desired proteins.

Question 38.
Differentiate between Pluripotency and Multipotency.
Answer:
Pluripotency: refers to a stem cell that has the potential to differentiate into any of the three germ layers-ectoderm, endoderm, and mesoderm.
Multipotency: refers to the stem cells that can differentiate into various types of cells that are related. For example, blood stem cells can differentiate into lymphocytes, monocytes, neutrophils etc.

Question 39.
Write a short note on stem cell banks.
Answer:
Stem cell banking is the extraction, processing and storage of stem cells, so that they may be used for treatment in the future, when required. Amniotic cell bank is a facility that stores stem cells derived from amniotic fluid for future use. Stem cells are stored in banks specifically for use by the individual from whom such cells have been collected and the banking costs are paid. Cord Blood Banking is the extraction of stem cells from the umbilical cord during childbirth. While the umbilical cord and cord blood are the most popular sources of stem cells, the placenta, amniotic sac and amniotic fluid are also rich sources in terms of both quantity and quality.

Question 40.
State any two uniqueness of the ELISA test.
Answer:
ELISA is highly sensitive and can detect antigen even in nanograms.
ELISA test does not require radioisotopes or radiation counting apparatus.

Question 41.
What is the ELISA test?
Answer:
ELISA – Enzyme-Linked ImmunoSorbent Assay is a biochemical procedure done to detect the presence of specific antibodies or antigens or hormones in a sample of serum, urine etc.

Question 42.
Elucidate the methodology of the ELISA test.
Answer:
During diagnosis, the sample suspected to contain the antigen is immobilized on the surface of an ELISA plate. The antibody specific to this antigen is added and allowed to react with the immobilized antigen. The anti-antibody is linked to an appropriate enzyme-like peroxidase. The unreacted anti-antibody is washed away and the substrate of the enzyme (hydrogen peroxidase) is added with certain reagents such as 4-chloronaphthol. The activity of the enzyme yields a coloured product indicating the presence of the antigen.

Question 43.
Whether PCR is applicable for RNA molecules? Explain.
Answer:
The PCR technique can also be used for amplification of RNA in which case it is referred to as reverse transcription PCR (RT-PCR). In this process, the RNA molecules (mRNA) must be converted to complementary DNA by the enzyme reverse transcriptase. The cDNA then serves as the template for PCR.

Question 44.
How PCR helps forensic personnel?
Answer:
PCR technique can also be used in the field of forensic medicine. A single molecule of DNA from bloodstains, hair, semen of an individual is adequate for amplification by PCR. The amplified DNA is used to develop a DNA fingerprint which is used as an important tool in forensic science. Thus, PCR is very useful for the identification of criminals. PCR is also used in the amplification of a specific DNA segments to be used in gene therapy.

Question 45.
Role of PCR in phylogenetics. Explain.
Answer:
The differences in the genomes of two different organisms can be studied by PCR. PCR is very important in the study of evolutions, more specifically phylogenetics. As a technique that can amplify even minute quantities of DNA from any source, like hair, mummified tissues, bones or any fossilized materials.

Question 46.
Enumerate the use of biological products.
Answer:
Antibodies are substances that react against the disease-causing antigens and these can be produced using transgenic animals as bioreactors. Monoclonal antibodies, which are used to treat cancer, heart disease and transplant rejection are produced by this technology. Natural protein adhesives are non-toxic, biodegradable and rarely trigger an immune response, hence could be used to reattach tendons and tissues, fill cavities in teeth, and repair broken bones.

Question 47.
Name the principles underlying the cloning technique.
Answer:
(a) Nuclear transfer (b) Totipotency (the ability of a cell to develop into an entire organism)

12th Bio Zoology Guide Applications of Biotechnology Five Marks Questions and Answers

Question 48.
Explain in detail about stem cell therapy.
Answer:
Stem cells are undifferentiated cells found in most multicellular animals. These cells maintain their undifferentiated state even after undergoing numerous mitotic divisions.

Stem cell research has the potential to revolutionize the future of medicine with the ability to regenerate damaged and diseased organs. Stem cells are capable of self-renewal and exhibit ‘cellular potency’. Stem cells can differentiate into all types of cells that are derived from any of the three germ layers ectoderm, endoderm arid mesoderm.

In mammals, there are two main types of stem cells – embryonic stem cells (ES cells) and ‘ adult stem cells. ES cells are pluripotent and can produce the three primary germ layers ectoderm, mesoderm and endoderm. Embryonic stem cells are multipotent stem cells that can differentiate into a number of types of cells. ES cells are isolated from the epiblast tissue of the inner cell mass of a blastocyst. When stimulated ES can develop into more than 200 cells types of the adult body. ES cells are immortal ie. they can proliferate in a sterile culture medium and maintain their undifferentiated state.

Adult stem cells are found in various tissues of children as well as adults. An adult stem cell or somatic stem cell can divide and create another cell similar to it. Most of the adult stem cells are multipotent and can act as a repair system of the body, replenishing adult tissues. The red bone marrow is a rich source of adult stem cells.
The most important and potential application of human stem cells is the generation of cells and tissues that could be used for cell-based therapies. Human stem cells could be used to test new drugs.

Question 49.
Describe the role of PCR in clinical field.
Answer:
PCR In Clinical Diagnosis: The specificity and sensitivity of PCR is useful for the diagnosis of inherited disorders (genetic diseases), viral diseases, bacterial diseases, etc., The diagnosis and treatment of a particular disease often require identifying a particular pathogen. Traditional methods of identification involve culturing these organisms from clinical specimens and performing metabolic and other tests to identify them.

The concept behind PCR-based diagnosis of infectious diseases is simple – if the pathogen is present in a clinical specimen its DNA will be present. Its DNA has unique sequences that can be detected by PCR, often using the clinical specimen (for example, blood, stool, spinal fluid, or sputum) in the PCR mixture. PCR is also employed in the prenatal diagnosis of inherited diseases by using chorionic villi samples or cells from amniocentesis. Diseases like sickle cell anemia, P-thalassemia and phenylketonuria can be detected by PCR in these samples. cDNA from PCR is a valuable tool for diagnosis and monitoring retroviral infections – eg. Tuberculosis by Mycobacterium tuberculosis.

Several virally induced cancers, like cervical cancer caused by Papillomavirus, can be detected by PCR. Sex of human beings and live stocks, embryos fertilized in-vitro can be determined by PCR by using primers and DNA probes specific for sex chromosomes. PCR technique is also used to detect sexlinked disorders in fertilized embryos.

Question 50.
Enumerate the steps involved in producing transgenic animals.
Answer:
The various steps involved in the production of transgenic organisms are,

• Identification and separation of the desired gene.
• Selection of a vector (generally a virus) or direct transmission.
• Combining the desired gene with the vector.
• Introduction of the transferred vector into cells, tissues, embryo or mature individual.
• Demonstration of integration and expression of a foreign gene in transgenic tissue or animals. Transgenic animals such as mice, rat, rabbit, pig, cow, goat, sheep and fish have been produced.

Question 51.
List out the uses of Transgenesis.
Answer:

• Transgenesis is a powerful tool to study gene expression and developmental processes in higher organisms.
• Transgenesis helps in the improvement of genetic characters in animals.Transgenic animals serve as good models for understanding human diseases which help in the investigation of new treatments for diseases. Transgenic models exist for many human diseases such as cancer, Alzheimer’s, cystic fibrosis, rheumatoid arthritis and sickle cell anemia.
• Transgenic animals are used to produce proteins which are important for medical and pharmaceutical applications.
• Transgenic mice are used for testing the safety of vaccines.
• Transgenic animals are used for testing toxicity in animals that carry genes which make them sensitive to toxic substances than non-transgenic animals exposed to toxic substances and their effects are studied.
• Transgenesis is important for improving the quality and quantity of milk, meat, eggs and wool production in addition to testing drug resistance.

Question 52.
Describe the procedure by which Dolly was developed.
Answer:
Dolly was the first mammal (Sheep) clone developed by Ian Wilmut and Campbell in 1997. Dolly, the transgenic clone was developed by the nuclear transfer technique and the phenomenon of totipotency. Totipotency refers to the potential of a cell to develop different cells, tissues, organs and finally an organism.

The mammary gland udder cells (somatic cells) from a donor sheep (ewe) were isolated and subjected to starvation for 5 days. The udder cells could not undergo normal growth cycle, entered a dormant stage and became totipotent. An ovum (egg cell) was taken from another sheep (ewe) and its nucleus was removed to form an enucleated ovum. The dormant mammary gland cell/udder cell and the enucleated ovum were fused. The outer membrane of the mammary cell was ruptured allowing the ovum to envelop the nucleus.

The fused cell was implanted into another ewe which served as a surrogate mother. Five months later dolly was bom. Dolly was the first animal to be cloned from a differentiated somatic cell taken from an adult animal without the process of fertilization.

Question 53.
What are the ethical issues about cloning.
Answer:
Biotechnology has given to the soceity cheap drugs, better fruits and vegetables, pest-resistant crops, indigenous cure to diseases and lot of controversy. This is mainly because the major part of the modem biotechnology deals with genetic manipulations. People fear that these genetic manipulations may lead to unknown consequences.

The major apprehension of recombinant DNA technology is that unique microorganism either inadvertently or deliberately for the purpose of war may be developed that could cause epidemics or environmental catastrophe. Although many are concerned about the possible risk of genetic engineering, the risks are in fact slight and the potential benefits are substantial.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
The immune system of a person is suppressed,. In ELISA test, the result is positive (i) Name the disease associated with this.
(ii) Why did he loose his immunity?
Answer:
AIDS caused by Human Immuno Virus.
In AIDS, the pathogen destroys the T-lymphocytes which forms the major immune resouce of our body.

Question 2.
Why do children cured by enzyme replacement therapy for ADA deficiency need periodic treatment? Suggest a permanent solution for this issue.
Answer:
During gene therapy the lymphocytes from the blood of the patient are removed and grown in a nutrient culture medium. A healthy and functional human gene, ADA cDNA encoding this enzyme is introduced into the lymphocytes using a retrovirus. The genetically engineered lymphocytes are subsequently returned to the patient. Since these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes. The disease could be cured permanently if the gene for ADA isolated from bone marrow cells are introduced into the cells of the early embryonic stages.

Question 3.
Saccharomyces cerevisiae, acts as a best host than Encherichia coli for the production of recombinant interferons. Yes or No? Support your answer.
Answer:
Yes. The Saccharomyces cerevisiae, is the best source of recombinant interferon than E-coli. Since E-coli does not possess the machinery for glycolysation of protein.

Question 4.
Isolation of blood to treat Haemophilia A is practically impossible. Give reason.
Answer:
a) Requirement of large quantity of blood.
b) Risk of transmission of blood related diseases like AIDS.

Question 5.
Functional Insulin differs from its pre-hormonal form. How?
Answer:
Pro-Insulin contains A and B segments linked by a C – chain and preceded by a leader sequence. Whereas the functional Insulin contains only A and B chain formed by the excision of C-chain and leaders sequence after translation.

Question 6.
Whether PCR can be done for RNA molecules? Explain.
Answer:
The PCR technique can also be used for amplification of RNA in which case it is referred to as reverse transcription PCR (RT-PCR). In this process, the RNA molecules (mRNA) must be converted to complementary DNA by the enzyme reverse transcriptase. The cDNA then serves as the template for PCR.

Question 7.
Suggest any two techniques for early diagnosis of bacterial/viral human diseases.
Answer:
PCR and ELISA

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 8 Differentials and Partial Derivatives Ex 8.1 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.1

Question 1.
Let f(x) = \(\sqrt[3] { x }\). Find the linear approximation at x = 27. Use the linear approximation to approximate \(\sqrt[3] { 27.2 }\)
Solution:
x = 27
f(x) = \(\sqrt[3] { 27 }\) = 3
We need to find the value of \(\sqrt[3] { 27.2 }\)
We know that
f(x₀ + Δx) = f(x₀) + f'(x₀) Δx

∴ Approximate value of \(\sqrt[3] { 27.2 }\) = 3.0074

Question 2.
Use the linear approximation to find approximate value of
(i) (123)^2/3
(ii) \(\sqrt[4] { 15 }\)
(iii) \(\sqrt[4] { 26 }\)
Solution:
(i) (123)^2/3
Let x₀ = 125, Δx = -2
f(x) = x^2/3, f(x₀) = 25
We know that
f(x₀ + Δx) = f(x₀) + f'(x₀) Δx

= 25 – 0.2666
(123)^2/3 = 24.7334

(ii) \(\sqrt[4] { 15 }\) = (15)^1/4
f(x) = x^1/4, f(x₀) = (16)^1/4 = 2
We know that
f(x₀ + Δx) = f(x₀) + f’(x₀) Δx

= 2 – 0.03125
(15)^1/4 = 1.96875

(iii) (26)^1/3
f(x) = x^1/3, f(x₀) = (27)^1/3 = 2, Δx = -1
We know that
f(x₀ + Δx) = f(x₀) + f’(x₀) Δx

= 3 – 0.370
(26)^1/3 = 2.963

Question 3.
Find a linear approximation for the following functions at the indicated points
(i) f(x) = x³ – 5x + 12, x₀ = 2
(ii) g(x) = \(\sqrt { x^2+9 }\), x₀ = -4
(iii) h(x) = \(\frac { x }{ x+1 }\), x₀ = 1
Solution:
(i) We know that the linear approximation
L(x) = f(x₀) + f’(x₀)(x – x₀)
f(x) = x³ – 5x + 12
f'(x) = 3x² – 5
f'(x₀) = f'(2) = 12 – 5 = 7
f(x₀) = f(2) = 8 – 10 + 12 = 10
L(x) = 10 + 7 (x – 2)
= 10 + 7x – 14
= 7x – 4

(ii) g(x) = \(\sqrt { x^2+9 }\), x₀ = -4
g(x₀) = g(14) = \(\sqrt {16+9 }\) = 5

(iii) h(x) = \(\frac { x }{ x+1 }\), x₀ = 1

Question 4.
The radius of a circular plate is measured as 12.65. cm instead of the actual length 12.5 cm find the following in calculating the area of the circular plate:
(i) Absolute error
(ii) Relative error
(iii) Percentage error
Solution:
Actual radius of the circular plate = 12.5 cm
Measured radius of the circular plate = 12.65
dr = 12.65 – 12.5
= 0.15
A = π r²
dA = 2π r dv
Change in Area
A(12.65) – A(12.5) = dA
= 2π × 12.5 × 0.15
= 3.75 π
Exact calculation of the Area changes gives
A(12.65) – A(12.5) = π(12.65)² – π(12.5)²
= 160.0225 π – 156.25 π
= 3.7725 π
Absolute error = 3.7725 π – 3.75 π
= 0.0225 π cm²
Relative error
= \(\frac { 3.7725π-3.75π }{ 3.7725π }\)
= \(\frac { 0.0225π }{ 3.7725π }\)
= 0.00596
= 0.006 cm²
Percentage error = Relative error × 100
= 0.006 × 100
= 0.6% .

Question 5.
A sphere is made of ice having radius 10 cm. Its radius decreases from 10 cm to 9.8 cm. Find approximations for the following:
(i) Change in the volume.
(ii) Change in the surface area
Solution:
(i) Given r = 10
dr = 10 – 9.8 = 0.2
Volume v = \(\frac { 4 }{ 3 }\)πr³
dv = \(\frac { 4 }{ 3 }\).3πr²dv
Change in volume
v(10) – v(9.8) = 4π(10)²(0.2)
= 80π cm³

(ii) Surface area of the sphere
S(r) = 4πr²
S'(r) = 8πr
Change in surface area at r = 10 is
= S'(r) [10 – 9.8]
= 8π (10) (0.2) = 16π cm²
∴ Surface Area decreases by 16π cm²

Question 6.
The time T, taken for a complete oscillation of a single pendulum with length l, is given by the equation T = 2π\(\sqrt{\frac { l }{ g }}\) where g is a constant. Find the approximate percentage error in the calculated value of T corresponding to an error of 2 percent in the value of l.
Solution:
Given T = 2π\(\sqrt{\frac { l }{ g }}\)
On taking log both sides, we get
log T = log 2 + log π + \(\frac { 1 }{ 2 }\) log l – \(\frac { 1 }{ 2 }\) log g
On differentiating both sides w. r. to l, we get

So, the percentage error in T is 1%

Question 7.
Show that the percentage error in the nth root of a number is approximately \(\frac { 1 }{ n }\) times the percentage error in the number.
Solution:
Let x be the number
Let y = x^1/n
log y = \(\frac { 1 }{ n }\) log x
Taking differentiate on both sides, we have

= \(\frac { 1 }{ n }\) times the percentage error in the number.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 7 Human Health and Diseases Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

12th Bio Zoology Guide Human Health and Diseases Text Book Back Questions and Answers

Question 1.
A 30 year old woman has bleedy diarrhoea for the past 14 hours, which one of the following organisms is likely to cause this illness?
(a) Streptococcus pyogens
(b) Clostridium difficile
(c) Shigella dysenteriae
(d) Salmonella enteritidis
Answer:
(c) Shigella dysenteriae

Question 2.
Exo-erythrocytic schizogony of Plasmodium takes place in ………………….
(a) RBC
(b) Leucocytes
(c) Stomach
(d) Liver
Answer:
(d) Liver

Question 3.
The sporozoites of Plasmodium vivax are formed from ………………….
(a) Gametocytes
(b) Sporoblasts
(c) Oocysts
(d) Spores
Answer:
(c) Oocysts

Question 4.
Amphetamines are stimulants of the CNS, whereas barbiturates are ………………….
(a) CNS stimulant
(b) both a and b
(c) hallucinogenic
(d) CNS depressants
Answer:
(d) CNS depressants

Question 5.
Choose the correctly match pair.
(a) Amphetamines – Stimulant
(b) LSD-Narcotic
(c) Heroin – Psychotropic
(d) Benzodiazepine – Pain killer
Answer:
(a) Amphetamines – Stimulant

Question 6.
The Athlete’s foot disease in human is caused by ………………….
(a) Bacteria
(b) Fungi
(c) Virus
(d) Protozoan
Answer:
(b) Fungi

Question 7.
Cirrhosis of liver is caused by chronic intake of ………………….
(a) Opium
(b) Alcohol
(c) Tobacco
(d) Cocaine
Answer:
(b) Alcohol

Question 8.
The sporozoite of the malarial parasite is present in …………………..
(a) saliva of infected female Anopheles mosquito.
(b) RBC of human suffering from malaria.
(c) Spleen of infected humans.
(d) Gut of female Anopheles mosquito
Answer:
(a) saliva of infected female Anopheles mosquito

Question 9.
Where do the following events in the life cycle of Plasmodium takes place?
(a) Fertilization – ………………….
(b) Development of gametocytes – ………………….
(c) Release of sporozoites – ………………….
(d) Schizogony – ………………….
Answer:
(a) Fertilization – Gut of mosquito
(b) Development of gametocytes – Human RBC’s
(c) Release of sporozoites – From Mosquito to the human blood
(d) Schizogony – Human liver cells

Question 10.
Paratope is an ………………….
(a) Antibody binding site on variable regions
(b) Antibody binding site on heavy regions
(c) Antigen binding site on variable regions
(d) Antigen binding site on heavy regions
Answer:
(c) Antigen binding site on variable regions

Question 11.
Allergy involves ………………….
(a) IgE
(b) IgG
(c) IgA
(d) IgM
Answer:
(a) IgE

Question 12.
Spread of cancerous cells to distant sites is termed as ………………….
(a) Metastasis
(b) Oncogenes
(c) Proto-oncogenes
(d) Malignant neoplasm
Answer:
(a) Metastasis

Question 13.
AIDS virus has ………………….
(a) Single stranded RNA
(b) Double stranded RNA
(c) Single stranded DNA
(d) Double stranded DNA
Answer:
(a) Single stranded RNA

Question 14.
B cells that produce and release large amounts of antibody are called ………………….
(a) Memory cells
(b) Basophils
(c) Plasma cells
(d) killer cells
Answer:
(c) Plasma cells

Question 15.
Given below are some human organs.
Identify one primary and one secondary lymphoid organ. Explain its role.
Liver, thymus, stomach, thyroid, tonsils
Answer:
Identify one primary and one secondary lymphoid organ. Explain its role.
Liver, thymus, stomach, thyroid, tonsils Primary lymphoid organ – thymus Secondary lymphoid organ – tonsils Immunocompetency of T cells is achieved in Thymus Whereas tonsils help to fight viral and bacterial infections.

Question 16.
Name and explain the type of barriers which involve macrophages.
Answer:
Phagocytic barriers involves macrophages (monocytes, neutrophils) which phagocytose and digest whole microorganisms.

Question 17.
What are interferons? Mention their role.
Answer:
Interferons are the antiviral proteins which induce antiviral state in the infected cells.

Question 18.
List out chemical alarm signals produced during inflammation.
Answer:
Serotonin, histamine and prostaglandins.

Question 19.
Explain the process of replication of retrovirus after it gains entry into the human body.
Answer:
After getting into the body of the person, the retrovirus enters into macrophages where RNA genome of the virus replicates to form viral DNA with the help of the enzyme reverse transcriptase. This viral DNA gets incorporated into the DNA of host cells and directs the infected cells to produce viral particles. The macrophages continue to produce virus and in this way acts like a HIV factory.

Question 20.
Explain the structure of immunoglobulin with a suitable diagram.
Answer:
An antibody molecule is Y-shaped structure that comprises of four polypeptide chains, two identical light chains (L) of molecular weight 25,000 Da (approximately 214 amino acids) and two identical heavy chains (H) of molecular weight 50,000 Da (approximately 450 amino acids). The polypeptide chains are linked together by di-sulphide (S-S) bonds. One light chain is attached to each heavy chain and two heavy chains are attached to each other to form a Y shaped structure. Hence, an antibody is represented by H₂L₂ The heavy chains have a flexible hinge region at their approximate middles

Each chain (L and H) has two terminal They are C – terminal (Carboxyl) and amino or N-terminal. Each chain (L and H) has two regions. They have variable (V) region at one end and a much larger constant (C) region at the other end. Antibodies responding to different antigens have very different (V) regions but their (C) regions are the same in all antibodies. In each arm of the monomer antibody, the (V) regions of the heavy and light chains combines to form an antigen – binding site shaped to ‘fit’ a specific antigenic determinant. Consequently each antibody monomer has two such antigen – binding regions. The (C) regions that forms the stem of the antibody monomer determine the antibody class and serve common functions in all antibodies.

Question 21.
What are the cells involved innate immune system?
Answer:
Cells involved in innate immunity are monocytes (macrophages), neutrophils, helper T-cells, B-cells, dendritic cells.

Question 22.
What is vaccine? What are its types?
Answer:
A vaccine is a biological preparation that provides active acquired immunity to a particular disease and resembles a disease-causing microorganism and is often made from weakened or _ attenuated or killed forms of the microbes, their toxins, or one of its surface proteins. The vaccines are classified as first, second and third-generation vaccines.

Question 23.
A person is infected by HIV. How will you diagnose for AIDS?
Answer:
A simple blood test is available that can determine whether the person has been infected with HIV. The ELISA test (Enzyme-Linked Immunosorbent Assay) detects the presence of HIV antibodies. It is a preliminary test. Western blot test is more reliable and a confirmatory test. It detects the viral core proteins. If both tests detect the presence of the antibodies, the person is considered to be HIV positive.

Question 24.
Autoimmunity is a misdirected immune response. Justify.
Answer:
Autoimmune diseases: Autoimmunity is due to an abnormal immune response in which the immune system fails to properly distinguish between self and non-self and attacks its own body. Our body produces antibodies (autoantibodies) and cytotoxic T cells that destroy our own tissues. If a disease-state results, it is referred to as an auto-immune disease. Thus, autoimmunity is a misdirected immune response.

Question 25.
List the causative agent, mode of transmission and symptoms for Diphtheria and Typhoid.
Answer:

Question 26.
A patient was hospitalized with fever and chills. Merozoites were observed in her blood. What is your diagnosis?
Answer:
The appearance of merozoites in a patient’s blood may be an indication of a malarial parasite – plasmodium.

Question 27.

1. Write the scientific name of the filarial worm that causes filariasis.
2. Write the symptoms of filariasis.
3. How is this disease transmitted?

Answer:

1. Scientific name of the filarial worm – Wuchereria bancrofti
2. Symptoms of filariasis – Inflammation of lymph nodes, obstruction of lymph vessels causes elephantiasis of limbs, genital etc.
3. Mode of disease transmission: Vector transmission (female Culex mosquito)

Question 28.
List the common withdrawal symptoms of drugs and alcohol abuse.
Answer:
The withdrawal symptoms may range from mild tremors or convulsions, severe agitation ” and fits, depressed mood, anxiety, nervousness, restlessness, irritability, insomnia, dryness of throat, etc, depending on the type of drug abuse.

Question 29.
Why do you think it is not possible to produce vaccine against ‘common cold’?
Answer:
Vaccines target specific pathogens and elicit immunity. In case of the common cold, more than 200 strains of viruses are responsible for causing the disease. Hence it is practically difficult to develop a vaccine against it.

12th Bio Zoology Guide Human Health and Diseases Additional Important Questions and Answers

12th Bio Zoology Guide Human Health and Diseases One Mark Questions and Answers

Question 1.
Pick out the disease which is caused by virus.
(a) Candidiasis
(b) Ascariasis
(c) Poliomyelitis
(d) Dysentery
Answer:
(c) Poliomyelitis

Question 2
…………….test is done to confirm typhoid.
(a) ELISA
(b) Western blot
(c) Widal test
(d) Southern blot
Answer:
(c) Widal test

Question 3.
Match the following.

Answer:
(a) a – iv, b – iii, c – i, d- ii

Question 4.
Identify the mismatched pair.

Answer:
(c) Measles – Intestine

Question 5.
Identify the wrong statement regarding polio disease.
(a) Polio is caused by a RNA virus.
(b) One of the infection site of polio is intestine.
(c) Culex mosquito acts as a vector for polio.
(d) Paralysis and respiratory failure are the symptoms
Answer:
(c) Culex mosquito acts as a vector for polio.

Question 6.
Yellow fever is a ………………… type of disease.
(a) Neurotropic
(b) Viscerotropic
(c) Pneumotropic
(d) Dermotropic
Answer:
(b) Viscerotropic

Question 7.
Which one of the following pairs is wrong.
(a) Amoebiasis – Home fly
(b) African sleeping sickness – Tsetse flies
(c) Kala-azar – Sandfly
(d) Malaria – female Anopheles mosquito
Answer:
(d) Malaria – female Anopheles mosquito

Question 8.
………………..is the malignant form of malaria.
Answer:
P. falciparum

Question 9.
Schizogony of plasmodium parasite in human liver cells returns in………………..
(a) sporozoites
(b) merozoites
(c) trophozoites
(d) schizont
Answer:
(b) merozoites

Question 10.
Incubation period of malaria is about ………….
Answer:
12 days

Question 11.
Match the following

Answer:
(a) a – iii, b – i, c – iv, d – ii

Question 12.
Assertion (A): Plasmodium vivax is a digenic parasite
Reason (R): The primary host of P. vivax is man.
(a) Both (A) and (R) are true. (R) explains (A)
(b) (A) is true (R) is false
(c) Both (A) and (R) are false
(d) (A) is false (R) is true
Answer:
(b) (A) is true (R) is false

Question 13.
Assertion (A): Dermatomycosis is a cutaneous infection.
Reason (R): Fungus belongs to the order Trichophyton,
(a) Both (A) and (R) are true. (R) explains (A)
(b) (A) is true (R) is false
(c) Both (A) and (R) are false.
(d) (A) is false (R) is true
Answer:
(a) Both (A) and (R) are true. (R) explains (A)

Question 14.
Assertion (A): Spleen is a primary lymphoid organ
Reason (R): Primary lymphoid organs trap antigen and destroy them.
(a) Both (A) and (R) are true. (R) explains (A)
(b) (A) is true (R) is false
(c) Both (A) and (R) are false.
(d) (A) is false (R) is true
Answer:
(c) Both (A) and (R) are false.

Question 15.
Assertion (A): Paratope is the antigen-binding site.
Reason (R): It is a part of antibody
(a) Both (A) and (R) are true. (R) explains (A)
(b) (A) is true (R) is false
(c) Both (A) and (R) are false.
(d) (A) is false (R) is true
Answer:
(b) (A) is true (R) is false

Question 16.
Assertion (A): HIV is a DNA virus.
Reason (R): HIV belongs to genus Lentivirus
(a) Both (A) and (R) are true. (R) explains (A)
(b) (A) is true (R) is false
(c) Both (A) and (R) are false.
(d) (A) is false (R) is true
Answer:
(d) (A) is false (R) is true

Question 17.
Secretion of HC1 in stomach is an example for ……………..
(a) Anatomical barriers
(b) Phagocytic barriers
(c) Physiological barriers
(d) Inflammatory barriers
Answer:
(c) Physiological barriers

Question 18.
Identify the incorrect statement.
(a) Antibody Mediated Immunity was elicited by T cells.
(b) It is a character of vertebrates only
(c) Immunoglobulins act against pathogens and kill them.
(d) It is also called humoral immunity
Answer:
(a) Antibody Mediated Immunity was elicited by T cells.

Question 19.
Production process of blood cells in bone marrow is called …………..
Answer:
haematopoiesis

Question 20.
Which of the following is not a feature of passive immunity?
(a) It is transient and less effective
(b) Immunological memory is present
(c) Immunity develops immediately
(d) Antibodies are obtained from outside
Answer:
(b) Immunological memory is present

Question 21.
……………. is a primary lymphoid organ of birds.
Answer:
Bursa of Fabricius

Question 22.
Match the following.

Answer:
(a) a – ii, b – i, c – iv, d – iii

Question 23.
Which is not a granulocyte?
(a) Lymphocytes
(b) Neutrophils
(c) Basophils
(d) Eosinophils
Answer:
(a) Lymphocytes

Question 24.
The L and H chains of immunoglobulin are joined by………………
(a) Hydrogen bonds
(b) disulphide bonds
(c) phosphodiester bonds
(d) ionic bond
Answer:
(b) disulphide bonds

Question 25.
………….type of Immunoglobulin is involved in allergic reactions.
Answer:
IgE

Question 26.
Identify the wrong statement.
(a) Vaccine provide passive acquired immunity
(b) It is made from attenuated or killed microbes.
(c) Vaccines teach our body how to defend from microbes.
(d) MMR is a fist generation vaccine.
Answer:
(a) Vaccine provide passive acquired immunity

Question 27.
………………developed first vaccine for small pox.
Answer:
Edward Jenner

Question 28.
The enzyme attached to RNA of HIV is ………………………..
(a) RNA polymerase
(b) reverse transcriptase
(c) primase
(d) endonuclease
Answer:
(b) reverse transcriptase

Question 29.
Infection of Ascariasis occur due to……………………..
(a) Sand fly
(b) contaminated food
(c) mosquito bite
(d) stagnant water
Answer:
(b) contaminated food

Question 30.
Which of the following statement(s) is true regarding cancer cells?
(a) Neoplasm or tumor cells show uncontrolled growth
(b) They are metastatic
(c) They lack contact inhibition
(d) They may be benign or malignant.

(a) (a) only
(b) (b) and (c)
(c) (d) only
(d) All the above
Answer:
(d) All the above

Question 31.
Study dealing with body’s defence mechanism against disease is called……………………..
(a) Pathology
(b) Immunology
(c) Microbiology
(d) Dermatology
Answer:
(b) Immunology

Question 32.
AIDS is characterized by sharp reduction in number of……………………..
(a) helper T cells
(b) killer T cells
(c) superior T cells
(d) B-cells
Answer:
(a) helper T cells

Question 33.
Plague and malaria are caused by ……………………..and……………… respectively.
(a) bacteria and virus
(b) fungi and protozoa
(c) bacteria and protozoan
(d) fungi and bacteria
Answer:
(c) bacteria and protozoan

Question 34.
A pair of fungal disease……………………..
(a) Amoebiasis, Kala-azar
(b) Candidiasis, Athlete’s foot
(c) Ascariasis, Filariasis
(d) Poliomyelitis, Amoebiasis
Answer:
(b) Candidiasis, Athlete’s foot

Question 35.
Plant source of Heroin is……………………..
(a) Poppy plants
(b) Cannabis plants
(c) Datura species
(d) Atropa species
Answer:
(a) Poppy plants

Question 36.
The test that confirms HIV positive is……………………..
(a) Western blot
(b) Northern blot
(c) Southern blot
(d) all the above
Answer:
(a) Western blot

Question 37.
Bacillary dysentery is caused due to……………………..
(t») Salmonella
(b) Shigella
(c) Clostridium
(d) Yersinia
Answer:
(b) Shigellosis

Question 38.
Cocaine is a ……………………..potent.
{a) Sedative
(b) Hallucinogen
(c) pain reliever
(d) neurotransmitter
Answer:
(b) Hallucinogen

Question 39.
Alkaloid found in tobacco is……………………..
(a) Atropine
(b) cocaine
(c) heroin
(d) nicotine
Answer:
(d) nicotine

Question 40.
……………….is a chronic memory disorder due to alcohol misuse.
(a) Cushing’s syndrome
(b) Turners’ syndrome
(c) Klinefelters’ syndrome
(d) Korsakoff syndrome
Answer:
(d) Korsakoff syndrome

12th Bio Zoology Guide Human Health and Diseases Two Marks Questions and Answers

Question 1.
What steps should be taken to maintain good health?
Answer:
Personal hygiene, regular exercise and balanced diet are very important to maintain good health.

Question 2.
Define disease.
Answer:
Disease can be defined as a disorder or malfunction of the mind or body. It involves morphological, physiological and psychological disturbances which may be due to environmental factors or pathogens or genetic anomalies or life style changes.

Question 3.
According to the WHO, what is health?
Answer:
The World Health Organization [WHO] defines health as ‘a state of complete physical, mental and social wellbeing and not merely absence of diseases’.

Question 4.
Differentiate between infectious and non-infections disease.
Answer:
Infectious disease
Disease which are transmitted form one person to another are called infectious disease, e.g. Common cold

Non-infectious disease :
Disease which do not transmitted form one person to another are called non-infectious disease, e.g. Anaemia.

Question 5.
Name any two fungal disease and helminthic disease.
Answer:
Fungal disease: Candidiasis, Athlete’s foot
Helminthic disease: Ascariasis, Filariasis

Question 6.
Mention the causative organism of the following.
(a) Tetanus (b) Bubonic plague (c) Pneumonia (d) Cholera
Answer:
(a) Tetanus : Clostridium tetani
(b) Bubonic plague : Yersinia pestis
(c) Pneumonia : Streptococcus pneumoniae
(d) Cholera : Vibrio cholerae

Question 7.
Classify viral disease based on the organ of infection.
Answer:
(a) Pneumotropic diseases
(b) Dermotropic diseases
(c) Viscerotropic diseases
(d) Neurotropic diseases

Question 8.
Write the symptoms of viral hepatitis.
Answer:
Liver damage, jaundice, nausea, yellowish eyes, fever and pain in the abdomen.

Question 9.
Name the different species of malarial parasites and also mention which is the fatal one?
Answer:
Plasmodium vivax
Plasmodium ovale
Plasmodium malariae
Plasmodium falciparum .
Plasmodium falciparum causes fatal malaria.

Question 10.
Name the causative agent and confirmatory test for Typhoid.
Answer:
Typhoid is caused by Salmonella typhi. Widal test can be done to confirm typhoid.

Question 11.
Mention the three phases in the life of plasmodium parasite with their respective host.
Answer:
(a) Schizogony occurs in man.
(b) Gamogony occurs in man.
(c) Sporogony occurs in gut of mosquito.

Question 12.
What causes shivering in malarial patient?
Answer:
After infection of RBC with sporozoite, when the RBC ruptures a toxic substance haemozoin is released which causes shivering chills and high fever followed by sweating.

Question 13.
Malaria Eradication programme launched by WHO is a failure. Why?
Answer:
In the 1950’s the World Health Organisation (WHO) introduced the Malaria eradication programme. This programme was not successful due to the resistance of Plasmodium to the drugs used to treat it and resistance of mosquito’s to DDT and other insecticides.

Question 14.
Filarial worm and Plasmodium both are digenic parasites. Man is a common host for both parasites. Name the other host.
Answer:
The other host of filarial worm is female Culex mosquito
The other host of plasmodium is female Anopheles mosquito

Question 15.
Name the causative organism of filariasis and mention the site of infection of parasite.
Answer:
Causative organisms of filariasis is Wuchereria bancrofti (filarial worm). The site of infection is lymph vessels and lymph nodes.

Question 16.
What do you mean by the term personal hygiene?
Answer:
Personal hygiene refers to maintaining one’s body clean by bathing, washing hands, trimming fingernails, wearing clean clothes and also includes attention to keeping surfaces in the home and workplace, including toilets, bathroom facilities, clean and pathogen-free.

Question 17.
Define Immunity and Susceptibility.
Answer:
The overall ability of body to fight against the disease causing pathogen is called immunity. It is also called disease resistance and the lack of immunity is known as susceptibility.

Question 18.
How skin and mucus membrane act as barriers for infections?
Answer:
Skin prevents the entry of microbes. Its acidic environment (pH 3-5) retards the growth of microbes.
Mucus membrane entraps foreign microorganisms and competes with microbes for attachment.

Question 19.
What is diapedesis?
Answer:
Tissue damage and infection induce leakage of vascular fluid, containing chemotactic signals like serotonin, histamine and prostaglandins. They influx the phagocytic cells into the affected area. This phenomenon is called diapedesis.

Question 20.
Write any two differences between CMI and AMI.
Answer:
Cell Mediated Immunity (CMI) :

• In CMI, pathogens are destroyed by cells without producing antibodies.
• It is carried out by T cells, Macrophages, NK cells .

Antibody Mediated Immunity (AMI) :

• In AMI, pathogens are destroyed by antibodies.
• It is carried out by B cells, T helper cells, APC cells.

Question 21.
Define haematopoiesis.
Answer:
The process of production of blood cells in the bone marrow is called haematopoiesis.

Question 22.
Which is the primary lymphoid organs of birds? Mention its location and role.
Answer:
Bursa of Fabricius is a primary lymphoid organ of birds. It is attached to the dorsal side of the cloaca. B lymphocytes mature in the bursa and bring about humoral immunity.

Question 23.
What are the primary lymphoid organs of mammals?
Answer:
Bone marrow and thymus gland.

Question 24.
What are Peyer’s patches?
Answer:
Peyer’s patches are oval-shaped areas of thickened tissue that are embedded in the mucus- secreting lining of the small intestine of humans and other vertebrate animals. Peyer’s patches contain a variety of immune cells, including macrophages, dendritic cells, T cells, and B cells.

Question 25.
Point out any four peripheral lymphoid organs.
Answer:
Lymph nodes, spleen, tonsils, MALT.

Question 26.
Write a brief note on GALT.
Answer:
Gut-associated lymphoid tissue (GALT) is a component of the mucosa associated lymphoid tissue (MALT) which works in the immune system to protect the body from invasion in the gut.

Question 27.
Why secondary immune response is more effective than primary immune response?
Answer:
Due to the establishment of immunological memory of the first encounter, the secondary immune response highly intense and effective.

Question 28.
Name the Agranulocytes involved in immune response.
Answer:
(a) Lymphocyte (b) Monocytes

Question 29.
Why dendritic cells are called so?
Answer:
Dendritic cells are called so because its covered with long, thin membrane extensions that resemble dendrites of nerve cells. These cells present the antigen to T-helper cells.

Question 30.
How many dendritic cells are identified? Name them.
Answer:
Four types of dendritic cells are known. They are langerhans, interstitial cells, myeloid and lymphoid cells.

Question 31.
What are Haptens?
Answer:
Haptens are substance that are non-immunogenic but can react with the products of a specific immune response.

Question 32.
Distinguish between Epitope and Paratope.
Answer:
Epitope :
Epitope is an antigenic determinant and is ‘ the active part of an antigen.

Paratope :
A paratope is the antigen-binding site and is a part of an antibody which recognizes and binds to an antigen.

Question 33.
Draw a simplified diagram of immunoglobulin.
Answer:

Question 34.
On which basis, the antibodies are classified? Name them.
Answer:
The antibodies are classified into five major categories, based on their physiological and biochemical properties. They are IgG (gamma), IgM(mu), IgA (alpha), IgD(delta) and IgE (epsilon).

Question 35.
Name any four functions of antibodies.
Answer:
The functions of immunoglobulin are agglutination, precipitation, opsonisation, neutralization.

Question 36.
Which type of bonds are developed between an antigen and antibody? Name them.
Answer:
The bonds that hold the antigen to the antibody combining site are all non-covalent in nature. These include hydrogen bonds, electrostatic bonds, Van der Waals forces and hydrophobic bonds.

Question 37.
What is antibody affinity?
Answer:
Antibody affinity is the strength of the reaction between a single antigenic determinant and a single combining site on the antibody.

Question 38.
What do you mean by third generation vaccine?
Answer:
Third generation vaccine contains the purest and the highest potency vaccines which are synthetic in generation. The latest revolution in vaccine is DNA vaccine or recombinant vaccine.

Question 39.
Define vaccination.
Answer:
Vaccination is the process of administrating a vaccine into the body or the act of introducing a vaccine into the body to produce immunity to a specific disease.

Question 40.
Define allergy and allergen.
Answer:
The exaggerated response of the immune system to certain antigens present in the environment is called allergy (allo-altered, erg-reaction). The substances to which such an immune response is produced are called allergens.

Question 41.
What is Anaphylaxis?
Answer:
Anaphylaxis is the classical immediate hypersensitivity reaction. It is a sudden, systematic, severe and immediate hypersensitivity reaction occurring as a result of rapid generalized mast-cell degranulation.

Question 42.
Expand (a) MALT (b) NACO
Answer:
(a) MALT – Mucosal-Associated Lymphoid Tissue
(b) NACO – National AIDS Control Organisation

Question 43.
How will you define Autoimmunity?
Answer:
Autoimmunity is due to an abnormal immune response in which the immune system fails to properly distinguish between self and non-self and attacks its own body.

Question 44.
What do you mean by drug abuse?
Answer:
The intake of certain drugs for a purpose other than their normal clinical use in an amount and frequency that impair one’s physical, physiological and psychological functions is called drug abuse.

Question 45.
Name any 4 natural cannabinoids.
Answer:
Marijuana, Ganja, Hashish and Charas.

Question 46.
Mention any two drugs to treat insomnia patient.
Answer:

1. methamphetamine
2. Lysergic acid diethylamide (LSD)

Question 47.
Name the antibody responsible for allergic reaction. Also mention two chemicals released during allergic response.
Answer:
IgE (epsilon) antibody is responsible for allergic reaction.
Histamine and serotonin are the chemicals released during allergic reaction.

Question 48.
Name an opioid drug and its source plant. How it is useful in medical field?
Answer:
Morphine is an opioid drug which is extracted from flowers of the poppy plant.
It is used as a strong pain killer during surgery.

Question 49.
What is withdrawal symptom? Name any two symptoms.
Answer:
If the intake of the drug or alcohol is abruptly stopped, he or she would develop withdrawal symptoms. In a sense, the body becomes confused and protests against the absence of the drug. The withdrawal symptoms are nervousness and insomnia.

Question 50.
Name the plant source of Cocaine and Heroin.
Answer:
Cocaine is obtained from Erythroxylum coca.
Heroin is obtained from poppy plant.

Question 51.
What is Liver cirrhosis?
Answer:
Alcohol interferes with the ability of the liver to break down fat. Over time fat accumulation and high levels of alcohol destroy the liver cells and a scar tissue grows in the place of dead cells. This scarring of the liver is called “Liver cirrhosis”.

Question 52.
Define Korsakoff syndrome.
Answer:
Korsakoff syndrome, a chronic memory disorder is most commonly caused by alcohol misuse.

Question 53.
What are the benefits of exercising our body?
Answer:
Participating in an exercise programme can:

• Increase self-esteem
• Boost self-confidence
• Create a sense of empowerment
• Enhance social connections and relationships

12th Bio Zoology Guide Human Health and Diseases Three Marks Questions and Answers

Question 54.
Write the name of causative agent, infection site, mode of transmission and any two symptoms of Chikungunya.
Answer:
Causative agent – Alpha virus
Infection site – Nervous system
Mode of transmission – Aedes aegypti (Mosquito)
Symptoms – Fever, headache, joint pain and swelling.

Question 55.
Draw and label the parts of Entamoeba histolytica.
Answer:

Question 56.
Give a brief account of Kala-azar.
Answer:
Kala – azar or visceral leishmaniasis is caused by Leishmania donovani, which is transmitted by the vector Phlebotomus (sand fly). Infection may occur in the endothelial cells, bone marrow, liver, lymph glands and blood vessels of the spleen. Symptoms of Kala azar are weight loss, anaemia, fever, enlargement of spleen and liver.

Question 57.
Comment on Malaria vaccine.
Answer:
Malaria vaccine is used to prevent malaria. The only approved vaccine as of 2015 is RTS,S(Mosquirix). It requires four injections and has relatively low efficacy (26-50%). Due to this low efficacy, WHO does not recommend the use of RTS,S vaccine in babies between 6 lr and 12 weeks of age.

Question 58.
What is innate immunity?
Answer:
Innate immunity is the natural phenomenon of resistance to infection which an individual possesses right from the birth. The innate defense mechanisms are non-specific in the sense that they are effective against a wide range of potentially infectious agents. It is otherwise known as non-specific immunity or natural immunity

Question 59.
Explain (a) Cell Mediated Immunity (b) Antibody Mediated Immunity
Answer:
(a) Cell-mediated immunity
When pathogens are destroyed by cells without producing antibodies, then it is known as cell mediated immune response or cell mediated immunity. This is brought about by T cells, macrophages and natural killer cells.

(b) Antibody mediated immunity or humoral immunity When pathogens are destroyed by the production of antibodies, then it is known as antibody mediated or humoral immunity. This is brought about by B cells with the aid of antigen . presenting cells and T helper cells. Antibody production is the characteristic feature of vertebrates only.

Question 60.
Secondary response is a booster response – Explain.
Answer:
The secondary immune response occurs when a person is exposed to the same antigen again. During this time, immunological memory has been established and the immune system can start producing antibodies immediately. Within hours after recognition of the antigen, a new army of plasma cells are generated. Within 2 to 3 days, the antibody concentration in the blood rises steeply to reach much higher level than primary response. This is also called as y “booster response”.

Question 61.
What are lymphoid organs? Mention its types.
Answer:
The organs involved in the origin, maturation and proliferation of lymphocytes are called lymphoid organs. Based on their functions, they are classified into primary or central lymphoid organs and secondary or peripheral lymphoid organs.

Question 62.
Classify antigens based on origin.
Answer:
On the basis of origin, antigens are classified into exogenous antigens and endogenous antigens.
The antigens which enter the host from the outside in the form of microorganisms, pollens, drugs, or pollutants are called exogenous antigens. The antigens which are formed within the individual are endogenous antigens. The best examples are blood group antigens.

Question 63.
Point out the factors that determine the binding force between antigen – antibody reactions.
Answer:
The binding force between antigen and antibody is due to three factors. They are closeness between antigen and antibody, noncovalent bonds or intermolecular forces and affinity of antibody.

Question 64.
Why it is important to study antigen – antibody interaction?
Answer:
The chief application of antigen – antibody reactions are to determine blood groups for v
transfusion, to study serological ascertainment of exposure to infectious agents, to develop immunoassays for the quantification of various substances, to detect the presence or absence v,
of protein in serum and to determine the characteristics of certain immunodeficiency diseases. ^

Question 65.
Explain opsonisation property of antibodies.
Answer:
Opsonisation or enhanced attachment is the process by which a pathogen is marked of ingestion and destruction by a phagocyte.
Opsonisation involves the binding of an opsonin i.e., antibody, to a receptor on the pathogen’s cell membrane. After opsonin binds to the membrane, phagocytes are attracted to the pathogen. So, opsonisation is a process in which pathogens are coated with a substance called an opsonin, marking the pathogen out for destruction by the immune system. This results in a much more efficient phagocytosis.

Question 66.
Give an example for (a) First generufkmvaccine (b) Second generation vaccine (c) Third generation vaccine
Answer:
(a) First generation vaccine – MMR vaccine
(b) Second generation vaccine – Hepatitis-B vaccine
(c) Third generation vaccine – DNA Vaccine

Question 67.
Name the diseases for which vaccines were developed by Louis Pasteur.
Answer:
Cholera, Anthrax and Rabies

Question 68.
How AIDS patient fail to develop immunity?
Answer:
AIDS is caused by Human Immuno Deficiency Virus (HIV). It selectively infects helper T cells. The infected helper T cells will not stimulate antibody production by B-cells resulting in loss of natural defence against viral infection.

Question 69.
Suggest few methods to prevent AIDS.
Answer:
Prevention of AIDS is the best option. Advocating safe sex and promoting regular check-up, safe blood for transfusion, use of disposable needles, use of condoms during sexual contact, prevention of drug abuse, AIDS awareness programme by NACO (National AIDS Control Organisation), NGOs (Non-Governmental Organisations) and WHO are to prevent the spreading of AIDS.

Question 70.
State immunological surveillance theory.
Answer:
The concept of immunological surveillance postulates that the primary function of the immune system is to “seek and destroy” malignant cells that arise by somatic mutation. The efficiency of the surveillance mechanism reduces either as a result of ageing or due to congenital or acquired immunodeficiencies, leads to increased incidence of cancer. Thus, if immunological surveillance is effective, cancer should not occur. The development of tumour represents a lapse in surveillance.

Question 71.
What is contact inhibition? How it is related to tumours growth?
Answer:
Normal cells show a property called contact inhibition, which inhibits uncontrolled growth. Cancer cells do not have this property. As a result, cancerous cells divide continuously giving rise to mass of tissues called tumours.

Question 72.
Differentiate between normal cells and cancer cells.
Answer:
Normal Cells :

1. Small uniformly shaped nuclei Relatively large cytoplasmic volume
2. Conformity in cell size and shape Cells arranged into discrete tissue
3. May possess differentiated cell structures Normal presentation of cell surface markers
4. Lower levels of dividing cells Cell tissues clearly demarcated

Cancer Cells :

1. Large, variable shaped nuclei Relatively small cytoplasmic volume
2. Variation in cell size and shape Disorganised arrangement of cells
3. Loss of normal specialised features Elevated expression of certain cell markers
4. Large number of dividing cells Poorly defined tumor boundaries

Question 73.
Write a note on Heroin.
Answer:
Heroin (smack) is chemically diacetyl morphine, which is white, odourless and bitter crystalline compound. It is obtained by acetylation of morphine, which is extracted from flowers of the poppy plant.

Question 74.
“Smoking and Tobacco chewing is injurious to health” – Comment on the statement.
Answer:
Tohacco is smoked, chewed and used as snuff. It increases the carbon monoxide content of blood and reduces the concentration of haem bound oxygen, thus causing oxygen deficiency in the body. Tobacco contains nicotine, carbon monoxide and tars, which cause problems in the heart, lung and nervous system. Adrenal glands are stimulated by nicotine to release adrenaline and nor adrenaline which increases blood pressure and heart beat.

Question 75.
Point out the symptoms of mental depression.
Answer:
Signs and symptoms of mental depression:

• Loss of self confidence and self esteem
• Anxiety
• Not being able to enjoy things that are usually pleasurable or interesting.

12th Bio Zoology Guide Human Health and Diseases Five Marks Questions and Answers

Question 76.
Name any five viral diseases, their causative agents, infection site, mode of transmission and their symptoms.
Answer:

Question 77.
Describe the life cycle of plasmodium parasite.
Answer:
Plasmodium vivax is a digenic parasite, involving two hosts, man as the secondary host and female Anopheles mosquito as the primary host. The life cycle of Plasmodium involves three phases namely schizogony, gamogony and sporogony.

The parasite first enters the human blood stream through the bite of an infected female Anopheles mosquito. As it feeds, the mosquito injects the saliva containing the sporozoites. ‘The sporozoite within the blood stream immediately enters the hepatic cells of the liver. Further in the liver they undergo multiple asexual fission (schizogony) and produce merozoites. After being released from liver cells, the merozoites penetrate the RBC’s.
Inside the RBC, the merozoite begins to develop as unicellular trophozoites.

The trophozoite grows in size and a central vacuole develops pushing them to one side of cytoplasm and becomes the signet ring stage. The trophozoite nucleus then divides asexually to produce the schizont. The large schizont shows yellowish – brown pigmented granules called Schuffners granules. The schizont divides and produces mononucleated merozoites.

Eventually the erythrocyte lyses, releasing the merozoites and haemozoin toxin into the blood stream to infect other erythrocytes. Lysis of red blood cells results in cycles of fever and other symptoms. This erythrocytic stage is cyclic and repeats itself approximately every 48 to 72 hours or longer depending on the species of Plasmodium involved. The sudden release of merozoites triggers an attack on the RBCs.

Occasionally, merozoites differentiate into macrogametocytes and microgametocytes. When these are ingested by a mosquitoe, they develop into male and female gametes respectively. In the mosquito’s gut, the infected erythrocytes lyse and male and female gametes fertilize to form a diploid zygote called ookinete. The ookinete migrates to the mosquito’s gut wall and develop into an oocyte.

The oocyte undergoes meiosis by a process called sporogony to form sporozoites. These sporozoites migrate to the salivary glands of the mosquito. The cycle is now completed and when the mosquito bites another human host, the sporozoites are injected and the cycle begins a new.

The pathological changes caused by malaria, affects not only the erythrocytes but also the spleen and other visceral organs. Incubation period of malaria is about 12 days. The early symptoms of malaria are headache, nausea and muscular pain. The classic symptoms first develop with the synchronized release of merozoites, haemozoin toxin and erythrocyte debris into the blood stream resulting in malarial paroxysms – shivering chills, high fever followed by sweating. Fever and chills are caused partly by malarial toxins that induce macrophages to release tumour necrosis factor (TNF-a) and interleukin.

Question 78.
Give an account of helminthic disease.
Answer:
Helminthes are mostly endoparasitic in the gut and blood of human beings and cause diseases called helminthiasis. The two most prevalent helminthic diseases are Ascariasis and Filariasis. Ascaris is a monogenic parasite and exhibits sexual dimorphism. Ascariasis is a disease caused by the intestinal endoparasite Ascaris lumbricoides commonly called the round worms.

It is transmitted through ingestion of embryonated eggs through contaminated food and water. Children playing in contaminated soils are also prone to have a chance of transfer of eggs from hand to mouth. The symptoms of the disease are abdominal pain, vomiting, headache, anaemia, irritability and diarrhoea. A heavy infection can cause nutritional deficiency and – severe abdominal pain and causes stunted growth in children.

It may also cause enteritis, hepatitis and bronchitis. Filariasis is caused by Wuchereria bancrofti, commonly called filarial worm. It is found in the lymph vessels and lymph nodes of man. Wuchereria bancrofti is sexually dimorphic, viviparous and digenic. The life cycle is completed in two hosts, man and the female Culex mosquitoe. The female filarial worm gives rise to juveniles called microfilariae larvae. In the lymph glands, the juveniles develop into adults. The accumulation of the worms block the lymphatic system resulting in inflammation of the lymph nodes. In some cases, the obstruction of lymph vessels causes elephantiasis or filariasis of the limbs, scrotum and mammary glands.

Question 79.
Tabulate the various types of innate immunity and their action mechanism.
Answer:

Question 80.
Point out the differences between active and passive immunity.
Answer:
Active Immunity :

1. Active immunity is produced actively by host’s immune system.
2. It is produced due to contact with pathogen or by its antigen.
3. It is durable and effective in protection.
4. Immunological memory is present.
5. Booster effect on subsequent dose is possible.
6. Immunity is effective only after a short period.

Passive Immunity :

1. Passive immunity is received passively and there is no active host participation.
2. It is produced due to antibodies obtained from outside.
3. It is transient and less effective.
4. No memory.
5. Subsequent dose is less effective.
6. Immunity develops immediately.

Question 81.
How primary immune response differ from secondary immune response?
Answer:

Question 82.
Explain the structure and role of thymus in primary lymphoid organ.
Answer:
Thymus is most active during the neonatal and pre-adolescent periods. The thymus is a flat and bilobed organ located behind the stemun, above the heart. Each lobe of the thymus contains numerous lobules, separated from each other by connective tissue called septa. Each lobule is differentiate into an outer cortex and an inner medulla.

Within each lobule, the devoloping T cells called thymocytes are arranged into outer cortex and inner medulla. The inner medulla contains immature thymocytes and outer cortex has matured thymocytes. One of its main secretions is the hormone thymosin. It stimulates the T cell to become mature and immunocompetent. By the early teens, the thymus begins to atrophy and is replaced by adipose tissue.

Question 83.
Describe the structure of lymph node with a labelled diagram.
Answer:
Lymph node has three zones. They are the cortex, paracortex and medulla. The outer most layer of the lymph node is called cortex, which consists of B-lymphocytes, macrophages, and follicular dendritic cells. The paracortex zone is beneath the cortex, which is richly populated by T lymphocytes and interdigitating dendritic cells. The inner most zone is called the medulla which is sparsely populated by lymphocytes, but many of them are plasma cells, which actively secrete antibody molecules. As the lymph enters, it slowly percolates through the cortex, paracortex and medulla, giving sufficient chance for the phagocytic cells and dendritic cells to trap the antigen brought by the lymph.

The lymph leaving a node carries enriched antibodies secreted by the medullary plasma cells against the antigens that enter the lymph node. Sometimes visible swelling of lymph nodes occurs due to active immune response and increased concentration of lymphocytes. Thus swollen lymph nodes may signal an infection. There are several groups of lymph nodes. The most frequently enlarged lymph nodes are found in the neck, under the chin, in the armpits and in the groin.

Question 84.
Explain the types of cells involved in immune system.
Answer:
Lymphocytes: About 20-30% of the white blood cells are lymphocytes. They have a large nucleus filling most of the cell, surrounded by a little cytoplasm. The two main types of lymphocytes are B and T lymphocytes. Both these are produced in the bone marrow. B lymphocytes (B cells) stay in the bone marrow until they are mature. Then they circulate around the body. Some remain in the blood, while others accumulate in the lymph nodes and spleen. T lymphocytes leave the bone marrow and mature in the thymus gland. Once mature, T cells also accumulate in the same areas of the body as B cells.

Lymphocytes have receptor proteins on their surface. When receptors on a B cell bind with an antigen, the B cell becomes activated and divides rapidly to produce plasma cells. The plasma cells produce antibodies. Some B cells do not produce antibodies but become memory cells. These cells are responsible for secondary immune response. T lymphocytes do not produce antibodies. They recognize antigen-presenting cells and destroy them. The two important types of T cells are Helper T cells and Killer T cells.

Helper T cells release a chemical called cytokine which activates B cells. Killer cells more around the body and destroy cells which are damaged or infected. Apart from these cells neutrophils and monocytes destroy foreign cells by phagocytosis. Monocytes when they mature into large cells, they are called macrophages which performs phagocytosis on any foreign organism.

Dendritic cells are called so because its covered with long, thin membrane extensions that resemble dendrites of nerve cells. These cells present the antigen to T-helper cells. Four types of dendritic cells are known. They are langerhans, interstitial cells, myeloid and lymphoid cells.

Question 85.
Write in detail about various types of antigen-antibody reactions.
Answer:
Different types of antigen and antibody reactions:
The reaction between soluble antigen and antibody leads to visible precipitate formation, which is Called precipitin reaction. Antibodies that bring about precipitate formation on reacting with antigens are called as precipitins.

Whenever a particulate antigen interacts with its antibody, it would result in clumping or agglutination of the particulate antigen, which is called agglutination reaction. The antibody involved in bringing about agglutination reaction is called agglutinin.

Opsonisation or enhanced attachment is the process by which a pathogen is marked of ingestion and destruction by a phagocyte. Opsonisation involves the binding of an opsonin i.e., antibody, to a receptor on the pathogen’s cell membrane. After opsonin binds to the membrane, phagocytes are attracted to the pathogen. So, opsonisation is a process in which pathogens are coated with a substance called an opsonin, marking the pathogen out for destruction by the immune system. This results in a much more efficient phagocytosis.

The neutralization reactions are the reactions of antigen-antibody that involve the elimination of harmful effects of bacterial exotoxins or a virus by specific antibodies. These neutralizing substances i.e. antibodies are known as antitoxins. This specific antibody is produced by a host cell in response to a bacterial exotoxin or corresponding toxoid (inactivated toxin).

Question 86.
Describe the structure of HIV with a diagram.
Answer:
The human immunodeficiency virus belongs to the genus Lentivirus. When observed under the electron microscope, HIV is seen as a spherical virus, 100-120 nm in diameter, containing a dense core surrounded by a lipoprotein envelope.

The envelope has glycoprotein (gp) spikes termed gp 41 and gp 120. At the core, there are two large single stranded RNA. Attached to the RNA are molecules of reverse transcriptase.

It also contains enzymes like protease and ribonuclease. The core is covered by a capsid made of proteins. This is followed by another layer of matrix proteins as shown above.

Question 87.
Suggest some of the ways to prevent drug and alcohol abuse.
Answer:
1. Effectively dealing with peer pressure:
The biggest reason for teens to start on drugs is due to their friends / peer groups imposing pressure on them. Hence, it is important to have a better group of friends to avoid such harmful drugs and alcohol.

2. Seeking help from parents and peers:
Help from parents and peer group should be sought immediately so that they can be guided appropriately. Help may even be sought from close and trusted friends. Getting proper advice to sort out their problems would help the young to vent their feelings of anxiety and guilt.

3. Education and counselling:
Education and counselling create positive attitude to deal with many problems and to accept disappointments in life.

4. Looking for danger signs:
Teachers and parents need to look for sign that indicate tendency to go in for addiction.

5. Seeking professional and medical assistance:
Assistance is available in the form of highly qualified psychologists, psychiatrists and de¬addiction and rehabilitation programmes to help individuals to overcome their problems.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
Identify the mismatched pair and give reason.
(a) Plague – Yersinia pestis
(b) Filariasis – Wuchereria bancrofti
(c) Dermatomycosis – Trypanosoma gambiense
(d) Common cold – Rhinovirus
Answer:
Dermatomycosis – Trypanosoma gambiense is a mismatched pair. Dermatomycosis is a cutaneous infection caused by fungus.

Question 2.
In which form does the malarial parasite enter the human body through mosquito? What is the target site of the parasite immediately after entering the host body?
Answer:
Sporozoites.
After entering the bloodstream of the host it finally reaches the liver cell.

Question 3.
A boy of ten years had suffered from chicken-pox. His grandmother consoled him that he is not expected to have it for the rest of his life. Whether his grandmother is right? If so how it happens?
Answer:
The infection produces not only antibodies but also generates memory in the lymphocytes. These cells recognize the pathogen on the subsequent attacks and generate antibodies to neutralize the pathogen, so his grandmother is right and the boy will not get the disease for the rest of his life.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 12 Environmental Issues Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 12 Environmental Issues

12th Bio Zoology Guide Environmental Issues Text Book Back Questions and Answers

Question 1.
Right to Clean Water is a fundamental right, under the Indian Constitution
(a) Article 12
(b) Article 21
(c) Article 31
(d) Article 41
Answer:
(c) Article 31

Question 2.
With which of the following, the Agenda 21’ of Rio Summit, 1992 is related to?
(a) Sustainable development
(b) Combating the consequences of population
(c) Mitigation norms of Green House Gases (GHG_s) emission
(d) Technology transfer mechanism to developing countries for ‘clean energy’ production
Answer:
(a) Sustainable development

Question 3.
Which among the following awards instituted by the Government of India for individuals or communities from rural areas that have shown extraordinary courage and dedication in protecting Wildlife?
(a) Indira Gandhi Paryavaran Puraskar
(b) Medini Puraskar Yojana
(c) Amrita Devi Bishnoi Award
(d) Pitambar Pant National Award
Answer:
(c) Amrita Devi Bishnoi Award

Question 4.
The ‘thickness’ of Stratospheric Ozone layer is measured in/on:
(a) Sieverts units
(b) Dobson units
(c) Melson units
(d) Beaufort Scale
Answer:
(b) Dobson units

Question 5.
Which among the following is the most abundant Green-House-Gas (GHG) in the Earth’s atmosphere?
(a) Carbon dioxide
(b) Water Vapour
(c) Sulphur Dioxide
(d) Tropospheric Ozone
Answer:
(a) Carbon dioxide

Question 6.
As per 2017 statistics, the highest per capita emitter of Carbon dioxide in the world is
(a) USA
(b) China
(c) Qatar
(d) Saudi Arabia
Answer:
(b) China

Question 7.
The use of microorganism metabolism to remove pollutants such as oil spills in the water bodies is known as ……………..
(a) Biomagnification
(b) Bioremediation
(c) Biomethanation
(d) Bioreduction
Answer:
(b) Bioremediation

Question 8.
The Ozone Day is observed every year on September 16 as on this day in 1987 the …………………. was signed for launching efforts to arrest the depletion of the fragile ozone layer in the stratosphere that prevents the harmful ultra-violet rays of the Sun from reaching the Earth. Fill the correct word in blank.
(a) Montreal Protocol
(b) Geneva Protocol
(c) Kyoto Protocol
(d) Nagoya Protocol
Answer:
(a) Montreal Protocol

Question 9.
Which among the following always decreases in a Food chain across tropic levels?
(a) Number
(b) Accumulated chemicals
(c) Energy
(d) Force
Answer:
(c) Energy

Question 10.
In the E-waste generated by the Mobile Phones, which among the following metal is most abundant?
(a) Copper
(b) Silver
(c) Palladium
(d) Gold
Answer:
(a) Copper

Question 11.
The Hydrochlorofluorocarbons (HCFC_s) are the compounds which have the following molecules:
(a) Hydrogen
(b) Carbon
(c) Chlorine
(d) Fluorine
Answer:
(c) Chlorine

Question 12.
SMOG is derived from:
(a) Smoke
(b) Fog
(c) Both A and B
(d) Only A
Answer:
(c) Both A and B

Question 13.
Excess of fluoride in drinking water causes:
(a) Lung disease
(b) Intestinal infection
(c) Fluorosis
(d) None of the above
Answers:
(c) Fluorosis

Question 14.
Expand (i) CFC (ii) AQI (iii) PAN
Answer:
(i) CFC: Chlorofluorocarbons
(ii) AQI: Air Quality Index
(iii) PAN: Peroxyacetyl nitrate

Question 15.
What is SMOG and how it is harmful for us?
Answer:
Smog is a type of air pollution caused by tiny particles in the air. The word comes from a mixture of the words smoke and fog. Today, smog generally refers to photochemical smog, which is created when sunlight reacts with nitrogen oxides and volatile organic compounds found in fossil fuel emissions from automobiles, factories, and power plants. These reactions create ground-level ozone and particulate matter, reducing visibility. Smog can make breathing more difficult, especially for people with asthma. Smog also affects plants and animals. It damages crops as well as causes health problems in pets, farm animals and human beings. Smog has also been known to cause corrosive damage to buildings and vehicles.

Question 16.
List all the wastes that you generate, at home, school or during your trips to other places. Could you very easily reduce the generation of these wastes? Which would be difficult or rather impossible to reduce?

Yes, we can reduce the generation of daily wastes in our life. For example, using jute/cloth bags instead of plastic bags, using the plastic drum as flower pots, etc.
E-wastes and plastic wastes are generally difficult to biodegrade.

Question 17.
Discuss the causes and effects of global warming. What measures need to be taken to control global warming?
Answer:

1. Phase down or ban the use of CFCs (CFC free refrigerants).
2. Minimizing the use of chemicals such as halons and halocarbons.
3. Creating awareness about ozone-depleting agents.
4. UV rays may penetrate deep into the skin and can lead to premature skin aging and wrinkling of the skin; suppression of the immune system, skin cancer (melanoma), and chronic effects leading to eye damage.
5. DNA damage can result from free radicals and reactive oxygen and. photons can damage the DNA itself.

Question 18.
What would Earth be like without the greenhouse effect?
Answer:
The greenhouse effect is vital for the sustenance of life. Greenhouse gases like CO₂, water vapour etc absorb some of the reflected sun’s radiation and radiate back it to the Earth surface, thus maintaining the Earth’s warm condition. Without this effect, life on Earth would be difficult or rather impossible for existence or become hostile to most living organisms.

Question 19.
Write notes on the following:
a. Eutrophication
b. Algal Bloom
Answer:
a. Eutrophication
Eutrophication refers to the nutrient enrichment in water bodies leading to a lack of oxygen and will end up in the death of aquatic organisms.
b. Algal Bloom
Algal Bloom is excess growth of algae due to abundant excess nutrients imparting distinct color to water.

Question 20.
What effect can fertilizer runoff have on an aquatic ecosystem?
Answer:
When run-off from land containing nutrients’ reaches water bodies like lakes, it results in a dense growth of plant life. This phenomenon is called Eutrophication.

Question 21.
How can we control eutrophication?
Answer:
Eutrophication can be controlled by reducing the use of fertilizers in agricultural lands, checking the runoff from fields, planting vegetations along the stream beds thereby the nutrients will be uptaken by plants.

Question 22.
Why does an ozone hole form over Antarctica?
Answer:
Ozone hole in Antarctica is due to special atmospheric and chemical conditions that existed there and nowhere else in the globe.

Question 23.
Mention the causes of enhanced ultraviolet radiation.
Answer:
UV rays may penetrate deep into the skin and can lead to premature skin aging and wrinkling of skin; suppression of the immune system, skin cancer (melanoma) and chronic effects leading to eye damage. DNA damage can result from free radicals and reactive oxygen and photons can damage the DNA itself.

Question 24.
Discuss the role of women in the protection and conservation of forests.
Answer:
Women and communities have played a crucial role in environmental conservation programmes.

Example 1: Chipko movement was started in 1974 in India. In this women from the village hugged the trees and prevented the felling of trees by contractors.

Example 2: Amrita Devi, from Khejarli village of Jodhpur district, Rajasthan sacrificed her life to maintain Bishoni Dharma. Here the King of Jodhpur ordered his men to get the wood of Khejri trees by cutting them to construct his new palace. On hearing this Amrita Devi and many others had hugged the trees to save them from cutting. But her effort went on vain. Amrita Devi and others were killed by King’s men. This resistance by the people of the village forced the King to give up the idea of cutting trees.

Question 25.
Discuss the role of an individual to reduce environmental pollution.
Answer:

• Decrease waste generation
• Use efficient transportation
• Reduce energy consumption
• Planting trees

Question 26.
How does recycling help to reduce pollution?
Answer:
Recycling prevents the emission of many greenhouse gases which are the major cause of global warming. It also reduces water pollutants and saves energy.

Question 27.
What is the primary purpose of the Kyoto Protocol?
Answer:
The primary purpose of Kyoto Protocol is to reduce the onset of global warming by reducing the concentration of greenhouse gases in the atmosphere.

Question 28.
In what way Peyang conserves the forest?
Answer:
The ‘Forest man of India’, Jadav Payeng who created 1,360 acres of dense and defiant forest was born in Arunasapori (a river island on the Brahmaputra). He had just completed his Class X exams in1979 when he started-to sow the seeds and shoots on the eroded island covered with sand and silt. Thirty-six years later he had converted the once unproductive land into a forest.

Payeng’s forest is now home to five Royal Bengal tigers, over a hundred deer, wild boar, vultures, and several species of birds. For his remarkable initiative, the Jawaharlal Nehru University invited Payeng in 2012 on Earth Day and honoured him with the title of the ‘Forest Man of India’.

Question 29.
Discuss briefly the following:
(a) Catalytic converter
(b) Greenhouse gases
(c) Ecosan
Answer:
(a) A catalytic converter is a device that converts toxic gases and pollutants into less toxic pollutants by catalyzing redox reactions.

(b) Gases that trap the heat within the atmosphere are called Greenhouse gases. E.g: CO₂ and CO

(c) About 150 liters of wastewater at an average is generated by an Indian individual daily, and a large amount of it is generated from toilets. Ecological sanitation (EcoSan) is a sustainable system for handling human excreta by using dry composting toilets. EcoSan toilets not only reduce wastewater generation but also generate natural fertilizer from recycled human excreta, which forms an excellent substitute for chemical fertilizers. This method is based on the principle of recovery and recycling of nutrients from excreta to create a valuable supply for agriculture.

Question 30.
What are some solutions to toxic dumping in our oceans?
Answer:
Reducing energy usage, Eat sustainable seafood, use reusable plastic products, properly dispose of hazardous materials. Avoid littering on the beach and buying ocean-friendly products.

Question 31.
Describe how deforestation might contribute to global warming.
Answer:
Deforestation refers to the cutting down of trees. Trees are one of the major sources of CO₂ uptake. CO₂ is a major greenhouse gas. If there is large-scale destruction of forests, the level of CO₂ will be increased leading to global warming.

Question 32.
How does forest conservation help to reduce air pollution?
Forests decrease the CO₂ level and increase the oxygen level of the atmosphere. Particulate matters in the air will settle down on leaves. Thus they help to reduce water pollution.

12th Bio Zoology Guide Environmental Issues Additional Important Questions and Answers

12th Bio Zoology Guide Environmental Issues One Mark Questions and Answers

Question 1.
The gaseous envelope which surrounds the Earth is called…………………
(a) Stratosphere
(b) Atmosphere
(c) Troposphere
(d) Ozonosphere
Answer:
(b) Atmosphere

Question 2.
…………….. are the major causes of CO pollution in large cities and towns.
(a) Fossil fuels
(b) Ocean
(c) Deforestation
(d) Automobiles
Answer:
(d) Automobiles

Question 3
…………… and ………… are the major cause of acid rain
(a) Sulphur dioxide and Hydrogen peroxide
(b) Hydrogen peroxide and Sulphuric acid
(c) Hydrochloride and sulphur dioxide
(d) Sulphur dioxide and Nitrogen oxide
Answer:
(d) Sulphur dioxide and Nitrogen oxide.

Question 4.
What is the name of the app published by the Central Pollution Control Board that provides updates on AQI
(a) Hamear
(b) Jhoan
(c) Sameer
(d) Industan
Answer:
(c) Sameer

Question 5.
PAN stands for ……………..
(a) Peroxy acetic nitrogen
(b) Perchloro acetate
(c) Peroxyacetyl nitrate
(d) Peractyl nitroxide
Answer:
(d) Peroxyacetyl nitrate

Question 6.
Average human consumption of oxygen per day is ……………….
(a) 280 L
(b) 550 L
(c) 620L
(d) 730L
Answer:
(b) 550 L

Question 7.
Name the famous flagship programme of Central Government launched to achieve pollution abatement and rejuvenation of River Ganga.
Answer:
Namami Gange

Question 8.
The intensity of noise is measured in…………..
(a) Dobson
(b) Hertz
(c) Decibel
(d) Frequency
Answer:
(c) Decibel

Question 9.
According to noise pollution rules 2000, the permissible level of noise in the commercial area is………….. during the day and …………. during nitght.
(a) 55 db, 65 db
(b) 65 db, 55 db
(c) 70 db, 60 db
(d) 75 db, 65 db
Answer:
(b) 65 db, 55 db

Question 10.
Which is not a physical method of wastewater treatment.
(i) Floatation
(ii) Filtration
(iii) Phydro remediation
(iv) Oxidation
(a) i and iii
(b) ii and iv
(c) i and ii
(d) iii and iv
Answer:
(d) iii and iv

Question 11.
Match List I with List II

Answer:
a – iv, b – i, c – ii, d – iii

Question 12.
Which is not a method of disposal of radioactive waste.
(a) Dilute and dispense
(b) Delay and decay
(c) Recycle and reuse
(d) Limit generation
Answer:
(e) Recycle and reuse

Question 13.
E – Wastes are basically…………………..
(a) Poly iodinated biphenyl based compounds
(b) Polychlorinated biphenyl based compounds
(c) Polyhydroxy biphenyl based compounds
(d) Poly acetyl biphenyl based compounds
Answer:
(c) Polyhydroxy biphenyl based compounds

Question 14.
Identify the correct statement indicating 4Rs of treating water.
(a) Regenerate, Reduce, Reuse and Recycle
(b) Refuse, Reduce, Rejenuvate, and Reuse
(c) Redeem, Refuse, Rejenuvate and Reduce
(d) Refuse, Reduce, Rescue and Recycle.
Answer:
(d) Refuse, Reduce, Rescue and Recycle.

Question 15.
UN conference on Sustainable development in 2012 was held at ……………….
(a) Ruanda
(b) Rio de Janeiro
(c) Geneva
(d) Stockholm
Answer:
(b) Rio de Janeiro

Question 16.
The molecular formula for ozone is………………
(a) O₂
(b) O₄
(c) O₃
(d) O₇
Answer:
(c) O₃

Question 17.
World Ozone Day was observed on……………..
(a) September 16^th
(b) October 12^th
(e) December 1^st
(d) August 18^th
Answer:
(a) September 16th

Question 18.
Who is the leader of Chipko Movement?
Answer:
Sunderlal Bahuguna

Question 19.
I am called the Forest Man of India? Who am I?
Answer:
Jadav Payeng

Question 20.
Identify the incorrect statement.
(i) EcoSan toilets is a sustainable way of handling human excreta by using dry composting toilets
(ii) It reduces wastewater generation
(iii) It is based on the recovery and recycling of nutrients from excreta
(iv) EcoSan toilets are used in several parts of India and Srilanka.
(a) i and ii only
(b) iii and iv only
(c) all the above
(d) none of the above
Answer:
(d) none of the above

Question 21.
What is the name of the action plan for sustainable development framed at the Rio conference in 1992?
(a) Action 21
(b) Declaration 21
(c) Protocol 21
(d) Agenda 21
Answer:
(d) Agenda 21

Question 22.
Eutrophication is a result of…………..
(a) Agricultural and sewage waste
(b) Vehicle emission
(c) Pesticides
(d) Industrial effluent
Answer:
(c) Pesticides

Question 23.
BOD stands for………………
(a) Biological Oxidation Demand
(b) Biotic Oxygen Deficient
(c) Biological Oxygen Deficit
(d) Biochemical Oxidation Deficit
Answer:
(a) Biological Oxidation Demand

Question 24.
The stratosphere is mainly depleted by…………..
(a) Excess CO
(b) CFC’_s
(c) Ozone
(cl) Excess CO₂
Answer:
(b) CFC’_s

Question 25.
Treatment of sewage involves
(a) Floatation, Filtration and Sedimentation of suspended particles
(b) Aerating it for bacterial action
(c) Removal of nitrates and phosphates
(d) All of the above
Answer:
(d) All of the above

Question 26.
Assertion (A): The ozone layer protects the UV rays entering the Earth.
Reason (R): UV rays may cause melanoma
(a) A is right R is wrong
(b) A is the wrong R is right
(c) Both A and R are correct
(d) R explains A
Answer:
(d) R explains A

Question 27.
Assertion (A): Evolution of Greenhouse gases leads to Global warming
Reason (R): The energy released by the greenhouse gases move away from the atmospheric surface
(a) A is right R is wrong
(b) A is wrong R is right
(c) Both A and R are correct
(d) R explains A
Answer:
(a) A is right R is wrong

Question 28.
Statement (1): Incomplete combination of fossil fuels releases CO
Statement (2): CO is a GHG
(a) Statement 1 is true, statement 2 is false
(b) Statement 1 is false, statement 2 is true
(c) Both Statements 1 and 2 are true
(d) Both statements 1 and 2 are false
Answer:
(c) Both statements 1 and 2 are true.

Question 29.
Statement (1): The intensity of noise is measured in the dobson (dB) unit.
Statement (2): Noise provides immense bliss.
(a) Statement 1 is true, statement 2 is false
(b) Statement 1 is false, statement 2 is true
(c) Both Statements 1 and 2 are true
(d) Both statements 1 and 2 are false
Answer:
(d) Both statements 1 and 2 are false.

Question 30.
The threshold of noise level is ………………
Answer:
120db

Question 31
Chemicals used in agriculture as fertilizer and pesticides are generally called as
Answer:
Agrochemicals

12th Bio Zoology Guide Environmental Issues Two MarkS Questions and Answers

Question 1.
Define pollution.
Answer:
Pollution is any undesirable change in the physical, chemical and biological characteristics of the environment due to natural causes and human activities.

Question 2.
What is air pollution?
Answer:
The alterations or changes in the composition of the Earth’s atmosphere by natural or human activities (anthropogenic factors) are referred as Air Pollution.

Question 3.
Classify air pollutants.
Answer:
Air pollutants can be

• discharge of dusts or particulate matter (PM: 2.5 and 10)
• discharge of gases (SO_x, NO_x, CO and CO₂)

Question 4.
What are particulate matters? Give examples.
Answer:
Particulate matters are tiny particles of solid matter suspended in a gas or liquid. Combustion of fossil fuels, fly ash produced in thermal power plants, forest fires, asbestos mining units and cement factories are the main sources of particulate matter pollution.

Question 5.
Name the secondary pollutant in photochemical smog. Mention its adverse effect.
Answer:
Peroxyacetyl nitrate (PAN) is a secondary pollutant present in photochemical smog. It is thermally unstable and decomposes into phenoxyethanol radicals and nitrogen dioxide gas causing eye irritation.

Question 6.
How ozone hole is developed?
Answer:
Ozone depletion: Thinning of the stratospheric ozone layer is known as ozone depletion. Such depletion causes the ‘ozone hole’, resulting in the poor screening of the harmful UV rays and an increase in incidences of skin cancer. Some of the common agents that deplete ozone are CFC_s.

Question 7.
List out the greenhouse gases. (Any four)
Answer:
a) CO₂
b) CFC_s
c) Methane
d) Nitrous oxide

Question 8.
What is Acid rain?
Answer:
Acid rain is a form of precipitation that contains acidic components, such as sulfuric acid or nitric acid. It damages trees, crops and harms marine animals (coral reefs), and induces corrosion.

Question 9.
Point out any two non-point sources of water pollution.
Answer:

• Agriculture chemical runoff.
• Dumping of plastics in water bodies.

Question 10.
How will you define noise pollution?
Answer:
Sound that is unwanted and undesirable or can disrupt one’s quality of life is called noise. When there is a lot of ‘noise’ in the environment, it is termed Noise Pollution.

Question 11.
‘Expand USEPA and MOEFCC.
Answer:
USEPA: the United States Environmental Protection Agency.
MOEFCC: Ministry of Environment, Forest, and Climate Change.

Question 12.
What are agrochemicals?
Answer:
Chemicals which are used in agriculture for the growth of plants and pest control are called agrochemicals or agrichemicals.

Question 13.
Which organism is affected by colony collapse syndrome? What will be the impact?
Answer:
Colony collapse syndrome in Honey bees due to pesticides/herbicides can lead to the destruction of hives and lower agricultural productivity.

Question 14.
Define the term eutrophication.
Answer:
When run-off from land containing nutrients reaches water bodies like lakes, it results in a dense growth of plant life. This phenomenon is called Eutrophication.

Question 15.
What is meant by accelerated eutrophication?
Answer:
Pollutants from anthropogenic activities like effluents from the industries and homes can radically accelerate the aging process. This phenomenon is known as Cultural or Accelerated Eutrophication.

Question 16.
List out the chemical methods of wastewater treatment.
Answer:
Chemical methods of wastewater treatment include:

• Generation of insoluble solids.
• Produce an insoluble gas.
• Produce biologically degradable substances from a non-biodegradable substance.
• Oxidize or reduce to produce a non-objectionable substance.

Question 17.
Expand RZWT and DEWATS.
Answer:
RZWT: Root Zone Wastewater Treatment
DEWATS: Decentralized Wastewater Treatment System

Question 18.
What are the medical wastes?
Answer:
Any kind of waste that contains infectious material generated by hospitals, laboratories, medical research centers, Pharmaceutical companies, and Veterinary clinics is called medical wastes.

Question 19.
What does 4R stand for?
Answer:
‘4R’ refers to Refuse, Reduce, Reuse, and Recycle methods to manage plastic waste.

Question 20.
Define deforestation.
Answer:
Deforestation is the destruction of forests in order to clear the land and make it available for other uses.

12th Bio Zoology Guide Environmental Issues Three Marks Questions and Answers

Question 21.
Classify degradable pollutants.
Answer:
Based on the time taken to breakdown into their ingredients, degradable pollutants are classified as rapidly degradable (non-persistent) and slowly degradable (persistent).

a) Rapidly degradable or non-persistent pollutants: These can be broken down by natural processes. Domestic sewage and vegetable waste are examples of such pollutants.

b) Slowly degradable or persistent pollutants: These are pollutants that remain in the environment for many years in an unchanged condition and take decades or longer to degrade, as in the case of DDT.

Question 22.
What are the main sources of air pollution?
Answer:
The main sources of air pollution are:

• Transport sources – cars, buses, airplanes, trucks, and trains.
• Stationary sources – power plants, incinerators, oil refineries, industrial facilities, and factories.
• Area sources – agricultural – wood/stubble burning and fireplaces.
• Natural sources – wind-blown dust, wildfires, and volcanoes.

Question 23.
Define AQI. What will be the air quality if AQI is between 0-50?
Answer:
Air Quality Index (AQI) is a number used by government agencies to communicate to the public how polluted the air is at a given time. If AQI is between 0-50 then the quality of air is good.

Question 24.
Explain the three main sources of water pollution.
Answer:
There are three main types of sources: point sources, non-point sources, leaks, and spills. Point sources: Discharge of pollutants at specific locations through pipelines or sewers into the water body. Factory effluents, sewage, underground mines, oil wells, oil tankers, and agriculture are common point sources.

Non-point sources: Sources that cannot be traced to a single site of discharge like acid rain, dumping of the plastics in water bodies, and agricultural chemical runoff are common examples.

Leaks and Spills: This occurs mostly due to shipping collisions, offshore oil rigs, oil leakages, and discharges into the sea.

Question 25.
What will be the impact of water pollution on the ecosystem?
Answer:
Destruction of ecosystems: Ecosystems, especially aquatic systems, can be severely affected or destroyed by water pollution. Water pollutants affect existing niches and habitats and the survival of organisms. Soil fertility is affected and the system becomes uninhabitable.

Question 26.
Suggest a few ways to control water pollution.
Answer:

• Regulate or control pollutant(s) discharge at the point of generation.
• Wastewater can be pretreated by scientific methods before discharge to municipal treatment sources.
• Setting up of Sewage Treatment Plants (STP) and Effluent Treatment Plants (ETP).
• Regulate or restrict the use of synthetic fertilizers and pesticides.
• Public awareness and peoples’ involvement is essential.

Question 27.
Write a note on the ‘Namami Ganga’ programme.
Answer:
Namami Gange (National Mission for Clean Ganga) Programme is an Integrated Conservation Mission approved as the ‘Flagship Programme’ of the Union Government in June 2014 with a budget outlay of 20,000 crores to accomplish the twin objectives of effective abatement of pollution, conservation, and rejuvenation of River Ganga.

Question 28.
List out the sources of noise pollution.
Answer:
Vehicle engines, air horns, audio-video systems, trains, low flying aircraft, factory machines, sirens, motors, drillers and crushers, compressor machines, crackers, explosives, and modem supersonic transports are the common sources of noise pollution.

Question 29.
State the role of Government in controlled Noise pollution.
Answer:
Article 48-A and Article 51-A of the Constitution of India, Noise Pollution (Regulation and Control) Rules 2000, and Tamil Nadu State Environment Policy 2017 are some of the legal relief from noise pollution.

According to Noise Pollution (Regulation and Control) Rules, 2000, the permissible limit of noise in areas categorized as commercial is 65 decibels (dB) during the day and 55 dB during the night.

Question 30.
Comment on Bio-magnification of DDT.
Answer:
When non-degradable substances enter the food chain, they do not get metabolized or broken down or expelled and instead get transferred up the trophic levels of the food chain. During this process, they show an increase in concentration which is referred to as biomagnification. This results in increased toxicity and may even be lethal. This phenomenon is well established for mercury and DDT.

Question 31.
Explain the biological methods of wastewater treatment.
Answer:
Biological methods of Wastewater treatment:

• Bioremediation of wastewater includes aerobic treatment (oxidation ponds and aeration lagoons) and anaerobic treatment (anaerobic bioreactors and anaerobic lagoons).
• Phytoremediation of wastewater includes constructed wetlands, Root Zone Wastewater Treatment (RZWT), and Decentralized Waste Water Treatment System (DEWATS).

Question 32.
Write a note on organic farming.
Answer:
Organic farming is a method of a farming system which primarily aims at cultivating the land and raising crops in such a way, so as to keep the soil alive and in good health by use of organic wastes (crop, animal and farm wastes, aquatic wastes) and other biological materials along with beneficial microbes (biofertilizers) to release nutrients to crops for increased sustainable production in an eco-friendly pollution-free environment.

Question 33.
How radioactive wastes are generated?
Answer:
Radioactive wastes are generated during various operations of the nuclear power plant. Radioactive waste can be in gas, liquid, or solid form, and its level of radioactivity can vary. The waste can remain radioactive for a few hours or several months or even hundreds of thousands of years. Depending on the level and nature of radioactivity, radioactive wastes can be classified as Exempt Waste, Low and Intermediate Level Waste, and High-Level Waste.

Question 34.
What are geological repositories?
Answer:
Geological Repositories – A deep geological repository is a nuclear waste repository excavated deep within a stable geologic environment. It is suited to provide a high level of long-term isolation and containment without future maintenance. In India at Tarapur and Kalpakkam, a wet storage facility of Spent Fuel is the main mode of storage.

Question 35.
Give an account of E-waste.
Answer:
Electronic waste or e-waste describes discarded electrical electronic devices as well as any refuse created by discarded electronic devices and components and substances involved in their manufacture or use. Their disposal is a growing problem because electronic equipment frequently contains hazardous substances.

In a personal computer, for example, there may be lead (Pb) in the cathode ray tube (CRT) and soldering compound, mercury (Hg) in switches and housing, and cobalt (Co) in steel components, among other equally toxic substances. E-wastes are basically PCB (Polychlorinated biphenyl) based, which are non-degradable.

Question 36.
What will be the impact of global warming?
Answer:
Large-scale global warming will have a significant impact on people and nature. As global average temperatures rise, precipitation patterns could be, affected. Extreme wet and dry conditions can be expected (flooding and desertification). Coastal areas shall become more vulnerable to storm surges as sea level rises. Plant and animal species will migrate or disappear in response to climate change.

Global warming can directly affect the flora and fauna. This could also result in a shortage of food and even lead to a food crisis, and affect the health of the people and organisms.

Question 37.
Explain the process of Ozone formation in the atmosphere.
Answer:
Ozone is found in the layer of the atmosphere called the Stratosphere. It acts as a protective covering that absorbs ultraviolet (UV) radiation from the Sun. The ozone molecule (O₃) consists of three oxygen atoms. It is formed when atmospheric oxygen (O₂) on exposure to solar radiation breaks into two oxygen atoms; each atom then joins up with a single oxygen atom. The ozone molecule is unstable. It soon decays again to form molecular oxygen. This cycle is a continuous process in the upper reaches of the stratosphere.

Question 38.
How Ozone depletion can be controlled?
Answer:
Ozone layer depletion can be controlled by

•  Phase down or ban the use of CFCs (CFC free refrigerants).
• Minimizing the use of chemicals such as halons and halocarbons.
• Creating awareness about ozone-depleting agents.

Question 39.
What will be the effect of Ozone depletion on Earth?
Answer:
UV rays may penetrate deep into the skin and can lead to premature skin aging and wrinkling of the skin; suppression of the immune system, skin cancer (melanoma), and chronic effects leading to eye damage. DNA damage can result from free radicals and reactive oxygen and photons can damage the DNA itself.

Question 40.
Write a brief note on the Chipko movement.
Answer:
The Bishnois, who are known conservators of their forest, were an inspiration to many people’s participatory movements for Environmental protection in India. The Chipko movement resisted the destruction of forests of India in the 1970s. Sunderlal Bahuguna was the leader of this movement. People in the movement hugged the trees and prevented the felling of trees by contractors.

Question 41.
What do you mean by Ecosan Toilets?
Answer:
About 150 liters of wastewater at an average is generated by an Indian individual daily, and a large amount of it is generated from toilets. Ecological sanitation (EcoSan) is a sustainable system for handling human excreta by using dry composting toilets. clean toilets not only reduce wastewater generation but also generate natural fertilizer from recycled human excreta, which forms an excellent substitute for chemical fertilizers. This method is based on the principle of recovery and recycling of nutrients from excreta to create a valuable supply for agriculture. ‘EcoSan’ toilets are being used in several parts of India and Sri Lanka.

12th Bio Zoology Guide Environmental Issues Five Marks Questions and Answers

Question 42.
List out the effects of air pollution.
Answer:

• Affects all organisms as they depend on the atmosphere for respiration.
• Causes irritation in the throat, nose, lungs, and eyes. It causes breathing problems and aggravates existing health conditions such as emphysema and asthma.
• Contaminated air reduces the body’s defense mechanism and decreases the body’s capacity to fight other infections in the respiratory system.
• Frequent exposure to polluted air increases the risk of cardiovascular diseases. Breathing air that is filled with fine particulate matter can induce hardening of the arteries, triggering cardiac arrhythmia, or even a heart attack.’
• People who exercise outdoors can sometimes be susceptible to the adverse effects of air pollution because it involves deeper and faster breathing. Hence it is advisable to walk or jog in the mornings in places with ample tree cover.
• Gas leaks can be lethal or affect the quality of air in the affected area.
• CO in the atmosphere interferes with O₂ transport since hemoglobin has a greater affinity for carbon monoxide. At low concentration, it causes headaches and blurred vision. In higher concentration, it can lead to coma and death.

Question 43.
How air pollution can be controlled?
Answer:
Certain measures help to remove pollutants, reduce their presence or prevent their entry into the atmosphere.

• Trees are the best remedy for urban particulate and gaseous pollution.
• Forests act as carbon sinks and lungs of the planet.
• Catalytic converters in vehicles help to reduce polluting gases drastically.
• Diesel exhaust filters in automobiles cut particulates.
• Electrostatic precipitators reduce the release of industrial pollutants.
• Cost-effective air pollution treatment systems like indoor plants and high-performance biofilters can improve indoor air quality.

Question 44.
Point out the effects of Noise pollution.
Answer:

• According to the USEPA (the United States Environmental Protection Agency), there are direct links between noise and health. Heart disease, high blood pressure, stress-related illness, sleep disruption, hearing loss (deafness), and productivity loss are the problems related to noise pollution.
• Increased stress and tension, nervousness, irritability, anxiety, depression, and panic attacks.
• Peptic ulcer, severe headache, and memory loss.
• Marine animals are affected by noise pollution from offshore activities and port activities.
• Firecrackers frighten animals. Birds are often affected by increased air traffic.

Question 45.
How agrochemicals affect the ecosystem?
Answer:

• May kill beneficial bacteria and soil organisms.
• Can cause eutrophication in water bodies.
• Fect aquatic animals and their productivity.
• Pesticide containing water, even in trace quantities is unfit for human consumption.
• Particles (aerosols) and residues of these chemicals cause air pollution.
• Inhalation of contaminated air can cause respiratory problems.
• Consumption can lead to poisoning, side effects, and after-effects.
• Chemicals can cause skin rashes and irritation of the eyes.
• Many of these chemicals are reported to be carcinogenic.
• They can trigger hormonal disorders and neurotoxicity.
• Beneficial insects and animals can be affected.

Question 46.
Categorize the solid waste and its sources.
Answer:

Question 47.
How radioactive waste is managed? Suggest a few methods of disposal of radioactive ‘ waste.
Answer:
Radioactive waste management involves the treatment, storage, and disposal of liquid, airborne, and solid effluents from the nuclear industry.
Methods of disposal of radioactive wastes are:

• Limit generation – Limiting the generation of waste is the first and most important consideration in managing radioactive wastes.
• Dilute and disperse – For wastes having low radioactivity, dilution, and dispersion are adopted.
• Delay and decay – Delay and decay is frequently an important strategy because much of the radioactivity in nuclear reactors and accelerators are very short-lived.

Concentrate and confine process – Concentrating and containing is the objective of treatment activities for longer-lived radioactivity. The waste is contained in corrosion-resistant containers and transported to disposal sites. Leaching of heavy metals and radionuclides from these sites is a problem of growing concern.

Higher Order Thinking Skills (HOTs) Questions
Question 1.
Complete the following:
(a) Smoke + ______ = Smog
Answer:
Fog

Question 2.
What is the unit of measuring ozone thickness?
Answer:
Dobson Unit

Question 3.
List any four adverse effects of noise.
Answer:
High blood pressure Stress-related ailments Sleep disruption Hearing impairment

Question 4.
Name the scientist who discovered the ozone layer
Answer
Charles Fabry and Henri Buisson

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.10 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.10

Choose the most suitable answer from the given four alternatives:

Question 1.
The volume of a sphere is increasing in volume at the rate of 3π cm³/ sec. The rate of change of its radius when radius is \(\frac { 1 }{ 2 }\) cm
(a) 3 cm/s
(b) 2 cm/s
(c) 1 cm/s
(d) \(\frac { 1 }{ 2 }\) cm/s
Solution:
(a) 3 cm/s
Hint:
Volume V = \(\frac { 4 }{ 3 }\) πr³
Given \(\frac { dV }{ dt }\) = 3π cm³/sec
Differentiating w.r.t. ‘t’ r = \(\frac { 1 }{ 2 }\) cm

Rate of change of radius is 3 cm/sec.

Question 2.
A balloon rises straight up at 10 m/s. An observer is 40 m away from the spot where the balloon left the ground. Find the rate of change of the balloon’s angle of elevation in radian per second when the balloon is 30 metres above the ground.
(a) \(\frac { 3 }{ 25 }\) radan/sec
(b) \(\frac { 4 }{ 25 }\) radian/sec
(c) \(\frac { 1 }{ 5 }\) radian/sec
(d) \(\frac { 1 }{ 3 }\) radian/sec
Solution:
(b) \(\frac { 4 }{ 25 }\) radian/sec
Hint:

Question 3.
The position of a particle moving along a horizontal line of any time t is given by s(t) = 3t² – 2t – 8. The time at which the particle is at rest is
(a) t = 0
(b) t = \(\frac { 1 }{ 3 }\)
(c) t = 1
(d) t = 3
Solution:
(b) t = \(\frac { 1 }{ 3 }\)
Hint:
s(t) = 3t² – 2t – 8
Velocity V = \(\frac { ds }{ dt }\) = 6t – 2
When the particle comes to rest,
velocity V = 0
6t – 2 = 0
t = \(\frac { 1 }{ 3 }\)

Question 4.
A stone is thrown, up vertically. The height reaches at time t seconds is given by x = 80t – 16t². The stone reaches the maximum! height in time t seconds is given by
(a) 2
(b) 2.5
(c) 3
(d) 3.5
Solution:
(b) 2.5
Hint:
x = 80t – 16t²
Velocity V = \(\frac { ds }{ dt }\) = 80 – 32t
When it reaches the maximum height
V = 0 ⇒ 80 – 32t = 0
t = \(\frac { 80 }{ 32 }\) = 2.5

Question 5.
Find the point on the curve 6y = x³ + 2 at which y-coordinate changes 8 times as fast as x-coordinate is
(a) (4, 11)
(b) (4, -11)
(c) (-4, 11)
(d) (-4, -11)
Solution:
(a) (4, 11)
Hint:

x = ±4
When x = 4, y = 11
∴ Point on the curve is (4, 11).

Question 6.
The abscissa of the point on the curve f(x) = \(\sqrt { 8-2x }\) at which the slope of the tangent is -0.25?
(a) -8
(b) -4
(c) -2
(d) 0
Solution:
(b) -4
Hint:
f(x) = \(\sqrt { 8-2x }\)
f'(x) = –\(\frac { 2 }{ 2\sqrt { 8-2x } }\) = –\(\frac { 1 }{ \sqrt { 8-2x } }\)
Slope of the tangent is – 0.25
ie„ f'(x) = -0.25
–\(\frac { 1 }{ \sqrt { 8-2x } }\) = -0.25 = \(\frac { -1 }{ 4 }\)
\(\sqrt { 8-2x }\) = 4
8 – 2x = 16
x = –\(\frac { 8 }{ 2 }\) = -4
Abscissa x = -4

Question 7.
The slope of the line normal to the curve f(x) = 2 cos 4x at x = \(\frac { π }{ 12 }\) is
(a) -4√3
(b) -4
(c) –\(\frac { √3 }{ 12 }\)
(d) 4√3
Solution:
(c) –\(\frac { √3 }{ 12 }\)
Hint:
f(x) = 2 cos 4x
f'(x) = -8 sin 4x
Slope of the normal at x = \(\frac { π }{ 12 }\) is

Question 8.
The tangent to the curve y² – xy + 9 = 0 is vertical when
(a) y = 0
(b) y = ±√3
(c) –\(\frac { 1 }{ 2 }\)
(d) y = ±3
Solution:
(d) y = ±3
Hint:
y² – xy + 9 = 0 ……… (1)
2y\(\frac { dy }{ dx }\) – (x\(\frac { dy }{ dx }\) + y) = 0
\(\frac { dy }{ dx }\) (2y – x) = y
\(\frac { dy }{ dx }\) = \(\frac { y }{ 2y-x }\)
When the tangent is vertical \(\frac { dy }{ dx }\) = ∞
i.e., \(\frac { y }{ 2y-x }\) = \(\frac { 0 }{ 1 }\)
⇒ 2y – x = 0
2y = x
sub in (1)
y² – 2y² + 9 = 0
⇒ y² = 9
y = ±3

Question 9.
The angle between y² = x and x² = y at the origin is
(a) tan^-1 \(\frac { 3 }{ 4 }\)
(b) tan^-1 (\(\frac { 4 }{ 3 }\))
(c) \(\frac { π }{ 2 }\)
(d) \(\frac { π }{ 4 }\)
Solution:
(c) \(\frac { π }{ 2 }\)
Hint:
y² = x and x² = y are the standard forms of parabolas for which y-axis and x-axis are the two tangents respectively.
Angle between x-axis and y-axis is \(\frac { π }{ 2 }\)

Question 10.
The value of the limit \(\lim _{x \rightarrow 0}\) (cot x – \(\frac { 1 }{ x }\)) is
(a) 0
(b) 1
(c) 2
(d) ∞
Solution:
(d) ∞
Hint:

Question 11.
The function sin⁴ x + cos⁴ x is increasing in the interval
(a) [ \(\frac { 5π }{ 8 }\), \(\frac { 3π }{ 4 }\) ]
(b) [ \(\frac { π }{ 2 }\), \(\frac { 5π }{ 8 }\) ]
(c) [ \(\frac { π }{ 4 }\), \(\frac { π }{ 2 }\) ]
(d) [ 0, \(\frac { π }{ 4 }\) ]
Solution:
(c) [ \(\frac { π }{ 4 }\), \(\frac { π }{ 2 }\) ]
Hint:
f(x) = sin⁴x + cos⁴x
f'(x) = 4 sin³ x cos x – 4 cos³ x sin x
f'(x) = 0 ⇒ 4 sin x cos x (sin²x – cos²x) = 0
sin x = 0; cos x = 0; sin² – cos² x = 0
x = 0; x = \(\frac { π }{ 2 }\); sin² x = cos² x
x = \(\frac { π }{ 4 }\)

In [0, \(\frac { π }{ 2 }\) ], f'(x) = -ve ⇒ f(x) is decreasing
In [ \(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\) ], f'(x) = +ve ⇒ f(x) is increasing

Question 12.
The number is given by Rolle’s theorem for the function x³ – 3x², x ∈ [0, 3] is
(a) 1
(b) √2
(c) \(\frac { 3 }{ 2 }\)
(d) 2
Solution:
(d) 2
Hint:
f(x) = x³ – 3x²
f'(x) = 3x² – 6x
f'(x) = 0
⇒ 3x (x – 2) = 0
x = 0, 2

Question 13.
The number given by the Mean value theorem for the function \(\frac { 1 }{ x }\), x ∈ [1, 9] is
(a) 2
(b) 2.5
(c) 3
(d) 3.5
Solution:
(c) 3
Hint:

Question 14.
The minimum value of the function |3 – x| + 9 is
(a) 0
(b) 3
(c) 6
(d) 9
Solution:
(d) 9
Hint:
f(x) = |3 – x | + 9
Minimum value of |3 – x | = 0
Minimum value of |3 – x| + 9 = 0 + 9 = 9 and No Maximum value.

Question 15.
The maximum slope of the tangent to the curve y = e^x sin x, x ∈ [0, 2π] is at
(a) x = \(\frac { π }{ 4 }\)
(b) x = \(\frac { π }{ 2 }\)
(c) x = π
(d) x = \(\frac { 3π }{ 2 }\)
Solution:
(b) x = \(\frac { π }{ 2 }\)
Hint:
y = e^x sin x, x ∈ [0, 2π] dy
Slope ‘S’ = \(\frac { dy }{ dx }\) = e^x cos x + e^x sin x
S = e^x (cos x + sin x)
\(\frac { dS }{ dx }\) = e^x (-sin x + cos x) + (cos x + sin x)e^x
= e^x (2 cos x)
For maximum or minimum,
\(\frac { dS }{ dx }\) = 0 ⇒ 2e^x cos x = 0
e^x = 0 is not possible
∴ cos x = 0
x = \(\frac { π }{ 2 }\)

Question 16.
The maximum value of the function x² e^-2x, x > 0 is
(a) \(\frac { 1 }{ e }\)
(b) \(\frac { 1 }{ 2e }\)
(c) \(\frac { 1 }{ e^2 }\)
(d) \(\frac { 4 }{ e^4 }\)
Solution:
(c) \(\frac { 1 }{ e^2 }\)
Hint:
Let f(x) = x²e^-2x, x > 0
f'(x) = -2x² e^-2x + e^-2x (2x)
f'(x) = 0 ⇒ -2xe^-2x(x – 1) = 0
x = 0 and x = 1
f(x) attains maximum at x = 1 as f”(x) < 0
when x = 1
∴ Maximum value f(1) = (1)² e^-2 = \(\frac { 1 }{ e^2 }\)

Question 17.
One of the closest points on the curve x² – y² = 4 to the point (6, 0) is
(a) (2, 0)
(b) (√5, 1)
(c) (3, √5)
(d) (\(\sqrt { 13 }\), -√3)
Solution:
(c) (3, √5)
Hint:
x² – y² = 4
y² = x² – 4
y = ±\(\sqrt { x^2-4 }\)
Any point on the curve is (x, ± \(\sqrt { x^2-4 }\))
Distance between (6, 0) and (x, ± \(\sqrt { x^2-4 }\)) is \(\sqrt { (x-6)^2+x^2-4 }\)
Substituting all’ the given options, we get minimum distance.
∴ Required point is (3, √5)

Question 18.
The maximum value of the product of two positive numbers’, when their sum of the squares is 200, is
(a) 100
(b) 25√7
(c) 28
(d) 24\(\sqrt { 14 }\)
Solution:
(a) 100
Hint:
Given x² + y² = 200
y² = 200 – x²
y = \(\sqrt { 200-x^2 }\)
Product P = xy = x\(\sqrt { 200-x^2 }\)
\(\frac { dP }{ dx }\) = \(\frac { x(-2x) }{ 2\sqrt{200-x^2} }\) + \(\sqrt{200-x^2}\) …. (1)
\(\frac { dP }{ dx }\) = 0 ⇒ -2x² + 200 = 0
x² = 100
x = 10
∴ y = \(\sqrt{200-100}\) = 10
∴ Maximum product is P = xy
= (10) (10)
= 100

Question 19.
The curve y = ax⁴ + bx² with ab > 0
(a) has no horizontal tangent
(b) is concave up
(c) is concave down
(d) has no points of inflection
Solution:
(d) has no points of inflection
Hint:
y = ax⁴ + bx²
\(\frac { dy }{ dx }\) = 4ax³ + 2bx
\(\frac { d^2y }{ dx^2 }\) = 12ax² + 2b
\(\frac { d^2y }{ dx^2 }\) = 0 ⇒ 12ax² + 2b = 0
x² = –\(\frac { b }{ 6a }\)
x is unreal.
Hence no points of inflection.

Question 20.
The point of inflection of the curve y = (x – 1)³ is
(a) (0, 0)
(b) (0, 1)
(c) (1, 0)
(d) (1, 1)
Solution:
(c) (1, 0)
Hint:
y = (x – 1)³
\(\frac { dy }{ dx }\) = 3(x – 1)²
\(\frac { d^2y }{ dx^2 }\) = 6(x – 1)
\(\frac { d^2y }{ dx^2 }\) = 0 ⇒ 6(x – 1) = 0
x = 1
y = f(x)
\(\frac { dy }{ dx }\) = f'(x)
\(\frac { d^2y }{ dx^2 }\) = f”(x)
In (-∞, 1), f”(x) < 0, curve is concave down In (1, ∞), f”(x) > 0, curve is concave up
f”(x) changes its sign when passing through x = 1
when x = 1, y = 0
∴ (1, 0) is the point of inflection.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 6 Evolution Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 6 Evolution

12th Bio Zoology Guide Evolution Text Book Back Questions and Answers

Question 1.
The first life on Earth originated …………………..
(a) in air
(b) on land
(c) in water
(d) on mountain
Answer:
(c) in water

Question 2.
Who published the book “Origin of species by Natural Selection” in 1859?
(a) Charles Darwin
(b) Lamarck
(c) Weismann
(d) Hugo de Vries
Answer:
(a) Charles Darwin

Question 3.
Which of the following was the contribution of Hugo de Vries?
(a) Theory of mutation
(b) Theory of natural Selection
(c) Theory of inheritance of acquired characters
(d) Germplasm theory
Answer:
(a) Theory of mutation

Question 4.
The wings of birds and butterflies is an example of………………………………..
(a) Adaptive radiation
(b) convergent evolution
(c) divergent evolution
(d) variation
Answer:
(b) Convergent evolutions

Question 5.
The phenomenon of “Industrial Melanism” demonstrates……………………….
(a) Natural selection
(b) induced mutation
(c) reproductive isolation
(d) geographical isolation
Answer:
(a) Natural selections

Question 6.
Darwin’s finches are an excellent example of……………………….
(a) connecting links
(b) seasonal migration
(c) adaptive radiation
(d) parasitism
Answer:
(c) Adaptive radiations

Question 7.
Who proposed the Germplasm theory?
(a) Darwin
(b) August Weismann
(c) Lamarck
(d) Alfred Wallace
Answer:
(b) August Weismann

Question 8.
The age of fossils can be determined by
(a) electron microscope
(b) weighing the fossils
(c) carbon dating
(d) analysis of bones
Answer:
(c) Carbon dating

Question 9.
Fossils are generally found in
(a) igneous rocks
(b) metamorphic rocks
(c) volcanic rocks
(d) sedimentary rocks
Answer:
(d) Sedimentary rocks

Question 10.
Evolutionary history of an organism is called……………………….
(a) ancestry
(b) ontogeny
(c) phylogeny
Answer:
(c) Phylogeny

Question 11.
The golden age of reptiles was……………………….
(a) Mesozoic era
(b) Cenozoic era
(c) Paleozoic era
(d) Proterozoic era
Answer:
(a) Mesozoic era

Question 12.
Which period was called “Age of fishes”?
(a) Permian
(b) Triassic
(c) Devonian
(d) Ordovician
Answer:
(c) Devonian

Question 13.
Modem man belongs to which period?
(a) Quaternary
(b) Cretaceous
(c) Silurian
(d) Cambrian
Answer:
(a) Quaternary

Question 14.
The Neanderthal man had the brain capacity of………………………..
(a) 650 – 800cc
(b) 1200cc
(c) 900cc
(d) 1400cc
Answer:
(d) 1400cc

Question 15.
List out the major gases seems to fie found in the primitive Earth.
Answer:
CO₂, NH₃, UV and Water vapour

Question 16.
Explain the three major categories in which fossilization occur.
Answer:
(i) Actual remains is the most common method of fossilization. When marine animals die, their hard parts such as bones and shells, etc. are covered with sediments and are protected from further deterioration. They get preserved as such as they are preserved in vast ocean the salinity in them prevents decay. The sediments become hardened to form definite layers or strata. For example, Woolly Mammoth that lived 22 thousand years ago were preserved in the frozen coast of Siberia as such. Several human beings and animals living in die ancient city of Pompeii were preserved intact by volcanic ash which gushed out from Mount Vesuvius.

(ii) Petrifaction – When animals die the original portion of their body may be replaced molecule for a molecule by minerals and the original substance being lost through disintegration. This method of fossilization is called petrifaction. The principal minerals involved in this type of fossilization are iron pyrites, silica, calcium carbonate, and bicarbonates of calcium and magnesium.

(iii) Natural moulds and casts – Even after disintegration, the body of an animal might leave indelible impression on the soft mud which later becomes hardened into stones. Such impressions are called moulds. The cavities of the moulds may get filled up by hard minerals and get fossilized, which are called casts. Hardened faecal matter termed as coprolites, occur as tiny pellets. Analysis of the coprolites enables us to understand the nature of diet, the prehistoric animals thrived.

Question 17.
Differentiate between divergent evolution and convergent evolution with one example for each.
Answer:
Divergent Evolution :
Divergent evolution is a result of homology. Erg: The wings of bird and the forelimbs of human both are homologous structures modified according to functions. In birds, it is used for flight and in humans used for writing and other purposes.

Convergent Evolution :
Convergent evolution is a result of analogy, E.g: Root modification in sweet potato, and stem modification in potato are analogous structures both performing same function i.e., storage,

Question 18.
How does Hardy-Weinberg’s expression (p² + 2pq + q² = 1) explain that genetic equilibrium is maintained in a population? List any four factors that can disturb the genetic equilibrium.
Answer:
The allele frequencies in a population are stable and are constant from generation to generation in the absence of gene flow, genetic drift, mutation, recombination and natural selection. If a population is in a state of Hardy Weinberg equilibrium, the frequencies of alleles and genotypes or sets of alleles in that population will remain same over generations. Evolution is a change in the allele frequencies in a population over time. Hence population in Hardy Weinberg is not evolving.

Suppose we have a large population of beetles, (infinitely large) and appear in two colours ’ dark grey (black) and light grey, and their colour is determined by ‘A’ gene. ‘AA’ and ‘Aa’ beetles are dark grey and ‘aa’ beetles are light grey. In a population let’s say that ‘ A’ allele has frequency (p) of 0.3 and ‘a’ allele has a frequency (q ) of 0.7. Then p+q= 1.

If a population is in Hardy Weinberg equilibrium, the genotype’frequencycan be estimated by Hardy Weinberg equation.

(p + q)² = p² + 2pq + q²
p² = frequency of AA
2pq= frequency of Aa
q² = frequency of aa
p = 0.3, q = 0.7 then,
p² = (0.3)2 = 0.09 = 9 %AA
2pq = 2(0.3) (0.7) = 0.42 = 42 % Aa
q² = (0.7)2 0.49 = 49 % aa’

Hence the beetle population appears to be in Hardy- Weinberg equilibrium. When the beetles in Hardy- Weinberg equilibrium reproduce the allele and genotype frequency in the next generation would be: Let’s assume that the frequency of ‘A’ and ‘a’ allele in the pool of gametes that make the next generation would be the same, then there would be no variation in the progeny. The genotype frequencies of the parent appears in the next generation.
(i.e. 9% AA, 42% Aa and 49% aa).

If we assume that the beetles mate randomly (selection of male gamete and female gamete in the pool of gametes), the probability of getting the offspring genotype depends on the genotype of the combining parental gametes.

Question 19.
Explain how mutations, natural selection and genetic drift affect Hardy Weinberg equilibrium.
Answer:
Natural selection occurs when one allele (or combination of alleles of differences) makes an organism more or less fit to survive and reproduce in a given environment. If an allele reduces fitness, its frequencies tend to drop from one generation to the next.

The evolutionary path of a given gene (i.e) how its allele’s change in frequency in the population across generation, may result from several evolutionary mechanisms acting at once. For example, one gene’s allele frequencies might be modified by both gene flow and genetic drift, for another gene, mutation may produce a new allele, that is favoured by natural selection.

Genetic drift / Sewall Wright Effect is a mechanism of evolution in which allele frequencies of a population change over generation due to chance (sampling error). Genetic drift occurs in all population sizes, but its effects are strong in a small population. It may result in a loss of some alleles (including beneficial ones) and fixation of other alleles. Genetic drift can have major effects, when the population is reduced in size by natural disaster due to bottle neck effect or when a small group of population splits from the main population to form a new colony due to founder’s effect.

Although mutation is the original source of all genetic variation, mutation rate for most organisms is low. Hence new mutations on allele frequencies from one generation to the next is usually not large.

Question 20.
How did Darwin explain fitness of organisms?
Answer:
Organisms struggle for food, space and mate. As these become a limiting factor, competition exists among the members of the population. Darwin denoted struggle for existence in three ways Intra specific struggle between the same species for food, space and mate.
Interspecific struggle with different species for food and space.

Struggle with the environment to cope with the climatic variations, flood, earthquakes and drought, etc.
According to Darwin, nature is the most powerful selective force. He compared origin of species by natural selection to a small isolated group. Darwin believed that the struggle for existence resulted in the survival of the fittest. Such organisms become better adapted to the changing environment.

Question 21.
Mention the main objections to Darwinism.
Answer:
Some objections raised against Darwinism were Darwin failed to explain the mechanism of variation.

1. Darwinism explains the survival of the fittest but not the arrival of the fittest.
2. He focused on small fluctuating variations that are mostly non-heritable.
3. He did not distinguish between somatic and germinal variations.
4. He could not explain the occurrence of vestigial organs, overspecialization of some organs like large tusks in extinct mammoths and oversized antlers in the extinct Irish deer, etc.

Question 22.
Taking the example of Peppered moth, explain the action of natural selection. What do you call the above phenomenon?
Answer:
Natural selection can be explained clearly through industrial melanism. Industrial melanism is a classical case of Natural selection exhibited by the peppered moth, Bistort betularia. These were available in two colours, white and black. Before industrialization peppered moth both white and black coloured were common in England. Pre-industrialization witnessed white colpured background of the wall of the buildings hence the white coloured moths escaped from their predators.

Post industrialization, the tree trunks became dark due to smoke and soot let out from the industries. The black moths camouflaged on the dark bark of the trees and the white moths were easily identified by their predators. Hence the dark coloured moth population was selected and their number increased when compared to the white moths. Nature offered positive selection pressure to the black coloured moths. The above proof shows that in a population, organisms that can adapt will survive and produce more progenies resulting in increase in population through natural selection.

Question 23.
Darwin’s finches and Australian marsupials are suitable examples of adaptive radiation – Justify the statement.
Answer:
Darwin’s finches are the birds whose common ancestor arrived on the Galapagos about 2 million years ago. During that time, Darwin’s finches have evolved into 14 recognized species differing in body size, beak shape and feeding behavior. Changes in the size and form of the beak have enabled different species to utilize different food resources such as insects, seeds, nectar from cactus flowers and blood from iguanas, all driven by Natural selection. Genetic variation in the ALX1 gene in the DNA of Darwin finches is associated with variation in the beak shape. Mild mutation in the ALX1 gene leads to phenotypic change in the shape of the beak of the Darwin finches.

Marsupials in Australia and placental mammals in North America are two subclasses of mammals they have adapted in similar way to a particular food resource, locomotory skill or climate. They were separated from the common ancestor more than 100 million years ago and each lineage continued to evolve independently.

Despite temporal and geographical separation, marsupials in Australia and placental mammals in North America have produced varieties of species living in similar habitats with similar ways of life. Their overall resemblance in shape, locomotory mode, feeding and foraging are superimposed upon different modes of reproduction. This feature reflects their distinctive evolutionary relationships.

Over 200 species of marsupials live in Australia along with many fewer species of placental mammals. The marsupials have undergone adaptive radiation to occupy the diverse habitats in Australia, just as the placental mammals have radiated across North America.

Question 24.
Who disproved Lamarck’s Theory of acquired characters? How?
Answer:
Lamarck’s “Theory of Acquired characters” was disproved by August Weismann who conducted experiments on mice for twenty generations by cutting their tails and breeding them. All mice bom were with tail. Weismann proved that change in the somatoplasm will not be transferred to the next generation but changes in the germplasm will be inherited.

Question 25.
How does the Mutation Theory of De Vries differ from Lamarck and Darwin’s view on the origin of new species?
Answer:
According to de Vries, sudden and large variations were responsible for the origin of new species, whereas Lamarck and Darwin believed in gradual accumulation of all variations as the causative factors in the origin of new species.

Question 26.
Explain stabilizing, directional, and disruptive selection with examples.
Answer:
i. Stabilising selection (centripetal selection): This type of selection operates in a stable environment as shown in fig. The organisms with average phenotypes survive whereas the extreme individuals from both ends are eliminated. There is no speciation but the phenotypic stability is maintained within the population over a generation. For example, measurements of sparrows that survived the storm clustered around the mean, and the sparrows that failed to survive the storm clustered around the extremes of the variation showing stabilizing selection.

ii. Directional Selection: The environment” which undergoes gradual change is subjected to directional selection, as shown in fig. This type of selection removes the individuals from one end towards the other end of phenotypic distribution. For example, size differences between male and female sparrows. Both male and female look alike externally but differ in body weight. Females show directional selection in relation to body weight.

iii. Disruptive selection: (centrifugal selection) When homogenous environment changes into heterogenous environment this type of selection is operational as shown in fig. The organisms of both the extreme phenotypes are selected, whereas individuals with average phenotype are eliminated. This results in splitting of the population into subpopulation/species. This _ is a rare form of selection but leads to formation of two or more different species. It is also ( called adaptive radiation. (E.g:) Darwin’s finches beak size in relation to seed size inhabiting Galapagos islands. Group selection and sexual selection are other types of selection. The two major group selections are Altrusim and Kin selection.

Question 27.
Rearrange the descent in human evolution.
Answer:
Australopithecus → Homo erectus → Homo sapiens → Ramapithecus → Homo habilis
Ramapithecus → Australopithecus → Homo habilis → Homo erectus → Homo sapiens

Question 28.
Differentiate between the eating habit and brain size of Australopithecus and Ramapithecus.
Answer:

Question 29.
How does the Neanderthal man differ from the modern man in appearance?
Answer:
Neanderthal man differ from the modem human in having semierect posture, flat cranium, sloping forehead, thin large orbits, heavy brow ridges, protruding jaws and no chin.

Question 30.
Mention any three similarities found common in Neanderthal man and Homo sapiens.
Answer:
Common characters showed by Neanderthal man and Homo sapiens are:

• Usage of Fire
• Burying of dead bodies
• Protecting themselves from predators

Question 31.
According to Darwin, organic evolution is due to………………….
(а) Intraspecific competition
(b) Interspecific competition
(c) Competition within closely related species. ‘
(d) Reduced feeding efficiency in one species due to the presence of interfering species.
Answer:
(d) Reduced feeding efficiency in one species due to the presence of interfering species.

Question 32.
A population will not exist in Hardy – Weinberg equilibrium if……………….
(a) Individuals mate selectively
(b) There are no mutations
(c) There is no migration
(d) The population is large
Answer:
(a) Individuals mate selectively

12th Bio Zoology Guide Evolution Additional Important Questions and Answers

12th Bio Zoology Guide Evolution One Mark Questions and Answers

Question 1.
Identify the incorrect statement in concern with Neanderthals.
(a) Neanderthal human were found in Germany.
(b) They possessed flat cranium.
(c) They used to bury their dead.
(d) Their brain size is of 650 – 800 cc.
Answer:
(d) Their brain size is of 650 – 800 cc

Question 2.
Which of the following statement does not satisfy Hardy Weinberg principle?
(a) A population undergoing random mating
(b) Small-sized population
(c) Population where there is no mutation or gene flow
(d) Absence of natural selection
Answer:
(b) Small-sized population

Question 3.
Match column I with column II

(a) a – iii b – ii c – iv d – i
(b) a – iv b – iii c – i d – ii .
(c) a – iii b – iv c – i d – ii
(d) a – ii b – iii c – i d – iv
Answer:
(a) a – iii b-ii c-iv d-i

Question 4.
Placental mammals develop during……..
(a) Eocene
(b) Oligocene
(c) Pliocene
(d) Paleocene
Answer:
(d) Paleocene

Question 5.
Identify the correct sequence from oldest to youngest
(а) Cambrian → Permian → Devonian → Silurian → Ordovician
(б) Permian → Silurian → Devonian → Ordovician → Cambrian
(c) Permian → Devonian → Silurian → Cambrian → Ordovician
(d) Cambrian → Ordovician → Silurian → Devonian → Permian
Answer:
(d) Cambrian → Ordovician → Silurian Devonian → Permian

Question 6.
Match the scientists with their terminologies used

(a) a – iii b-iv c-ii d- i
(b) a – ii b-iv, c -i d- Hi
(c) a – Hi b – i c- iv d- ii
(d) a-i b-iv c-iii d-ii
Answer:
(b) a – ii b-iv c-i d – iii

Question 7.
Anatomical structures that have similar functions but not similar structures are called
(a) Homologous structures
(b) Vestigial structures
(c) Analogous structures
(d) Generalized structures
Answer:
(c) Analogous structures

Question 8.
Who propounded the theory of recapitulation?
(a) Ernst Von Haeckel
(b) Charles Darwin
(c) Thomas Huxley
(d) Oparin
Answer:
(c) Ernst Von Haeckel

Question 9.
Mammal in human male is…………….
(a) Atavistic organ
(b) Rudimentary Organ
(c) Vestigial organ
(d) Homologous structure
Answer:
(c) Vestigial organ

Question 10.
Which of the following is/are not examples of analogous structure
(a) Wings of Birds and Bats
(b) Wings of Birds and Insects
(c) Thom of Bougainvillea and Tendril of cururbita
(d) Flippers of Penguins and Dolphins
(i) a, b, c (ii) a and c (iii) b and d (iv) All the above
Answer:
(ii) a and c

Question 11.
identify the mismatched pairs
(a) Thom of Bougainvillea and Tenrdril of cucurbita – Analogy
(b) Forelimbs of whale and cat – Analogy
(c) Octopus eye & Mammalian eye – Homology
(d) Root of sweet potato & stem of potato – Homology
Answer:
(a) Thorn of Bougainvillea & Terdril of crucurbita – Analog

Question 12.
Witnesses for evolution are found in ……………….
(a) Rocks
(b) Ocean beds
(c) Fossils
(d) Desert
Answer:
(c) Fossils

Question 13.
Assertion (A): Oparin used the term coacervates
Reason (R): Coacervates are colloidal particles in an aqueous environment
(a) Both A and Rare incorrect
(b) Both A and R are correct
(c) Both A and R are correct. R explains A.
(d) A is correct R is incorrect
Answer:
(c) Both A and R are correct. R explains A.

Question 14.
According to the theory of spontaneous generation, life originated from…………….
(a) Cosmic particles
(b) Non-living materials
(c) Coacervates
(d) Sea
Answer:
(b) Non-living materials

Question 15.
Assertion (A): Hardy – Weinberg principle states that allelic frequency of a population remains constant
Reason (R) : Constancy is maintained through natural selection and mutation
(a) A is true R is false
(b) A is false R is true
(c) Both A and R are true
(d) R explains A
Answer:
(a) A is true R is false

Question 16.
Calculate the allelic frequency of Aa. frequency of 0.7
(a) 0.67
(b) 0.42
(c) 0.36
(d) 0.59
Answer:
(b) 0.42

Question 17.
Match the following Evolutionary Human

(d) a – iv b-i c – ii d – iii
ib) a – ii b – iv c – iii d – i
(c) a – ii b – iii c – iv d – i
{d) a – iii b-i c – ii d – iv
Answer:
(a) a-iv b-i c-ii d-iii

Question 18.
Genetic drift leads to………………
(a) Mutation
(b) Bottleneck effect
(c) Immigration
(d) Isolation
Answer:
(b) Bottleneck effect

Question 19.
Atavism refers to…………….
(a) Inheritance of trial by mother
(b) Inheritance of trial by father
(c) Criss-cross inheritance
(d) Inheritance of characters not shown by parents
Answer:
(d) Inheritance of characters not shown by parents

12th Bio Zoology Guide Evolution Two Marks Questions and Answers

Question 1.
State the theory of spontaneous generation.
Answer:
According to the theory of spontaneous generation or Abiogenesis, living organisms originated from non-living materials and occurred through stepwise chemical and molecular evolution over millions of years. Thomas Huxley coined the term abiogenesis.

Question 2.
List the four eras of geological time scale.
Answer:

1. Precambrian era
2. Paleozoic era
3. Mesozoic era
4. Cenozoic era

Question 3.
Which periods of the Paleozoic era are referred to as

• Age of fishes
• Invertebrates

Answer:

• Age of fishes – Devonian period
• Age of invertebrates – Cambrian period

Question 4.
Point out the epochs of Carboniferous period.
Answer:

•  Pennsylvanian
• Mississippian

Question 5.
Compare relative dating with absolute dating.
Answer:
Relative dating is used to determine a fossil by comparing it to similar rocks and fossils of known age. Absolute dating is used to determine the precise age of a fossil by using radiometric dating to measure the decay of isotopes.

Question 6.
Wing of a cockroach and the wing of parrot. What do you infer from this statement with reference to evolution?
Answer:
Both the wings of cockroach and bird are different in structure but similar in their function. Thus, they are the analogous structure that brings about convergent evolution.

Question 7.
Name the scientists who propounded the following theories.

1. Mutation theory
2. Chemical theory of evolution

Answer:

1. Mutation theory was propounded by Hugo de Vries.
2. The chemical theory of evolution was propounded by Oparin and Haldane

Question 8.
Define fossilization and mention its types.
Answer:
‘ Fossilization is the process by which plant and animal remains are preserved in sedimentary rocks. It is of three major types,

• Actual remains
• Petrifaction
• Natural moulds and casts.

Question 9.
Name the principle minerals involved in petrifaction.
Answer:
Iron pyrites, silica, calcium carbonate and bicarbonates of calcium and magnesium.

Question 10.
What is meant by petrifaction?
Answer:
When animals die the original portion of their body may be replaced molecule for molecule by minerals and the original substance being lost through disintegration. This method of fossilization is called petrifaction. The principle minerals involved in this type fossilization are iron pyrites, silica, calcium carbonate and bicarbonates of calcium and magnesium.

Question 11.
Define analogous organ with an example.
Answer:
Organisms having different structural patterns but similar function are termed as analogous structures. For example, the wings of birds and insects are different structurally but perform the same function of flight that brings about convergent evolution.

Question 12.
Mention any four organs homologous to human hand.
Answer:
Flippers of whale, wings of bat, wings of bird and forelimb of horse.

Question 13.
Thorn of Bougainvillea and tendrils of Pisum sativum represent homology. How?
Answer:
The thorn of Bougainvillea and the tendrils of Curcurbita and Pisum sativum represent homology. The thorn in former is used as a defence mechanism from grazing animals and the tendrils of latter is used as a support for climbing.

Question 14.
Which type of evolution is brought out by homologous structures and analogous structures?
Answer:
Homologous structures brings about divergent evolution. Analogous structures brings about convergent evolution.

Question 15.
What are vestigial organs? Give example.
Answer:
Structures that are of no use to the possessor, and are not necessary for their existence are called vestigial organs. Vestigial organs may be considered as remnants of structures which were well developed and functional in the ancestors, but disappeared in course of evolution due to their non-utilization. E.g: Human appendix.

Question 16.
Human appendix is a vestige. Give reason.
Answer:
Human appendix is the remnant of caecum which is functional in the digestive tract of
herbivorous animals like rabbit. Cellulose digestion takes place in the caecum of these .
animals. Due to change in the diet containing less cellulose, caecum in human became functionless and is reduced to a vermiform appendix, which is vestigial.

Question 17.
What are connecting link? Give example.
Answer:
The organisms which possess the characters of two different groups (transitional stage) are called connecting links. Example Peripatus (connecting link between Annelida and Arthropoda) Archaeopteryx (connecting link between Reptiles and Aves).

Question 18.
Name one fossilised connecting link between reptiles and Aves also one living connecting link between Annelida and Arthropoda.
Answer:
Archaeopteryx – connecting link between Reptiles and Aves.
Peripatus – Connecting link between Annelida and Arthropoda.

Question 19.
Why it is considered as a connecting link?
Answer:
Peripatus is a worm that shown the characters of both Annelidia and Arthropoda. Hence it is a connecting link between Annelida and Arthropoda. ‘

Question 20.
Atavistic organs – comment.
Answer:
Sudden appearance of vestigial organs in highly evolved organisms is called atavistic organs. Example, presence of tail in human baby is an atavistic organ.

Question 21.
Define Ontogeny and Phytogeny.
Answer:
Ontogeny refers to the life history of an individual.
Phytogeny refers to the evolutionary history of a race. ‘

Question 22.
Who proposed the theory of recapitulation? State the theory.
Answer:
Ernst Von Haeckel proposed the theory of recapitulation, which states that life history of an individual briefly repeats the evolutionary history of the race. ,

Question 23.
Name few Neo – Lamarckists.
Answer:
Cope, Osborn, Packard and Spencer.

Question 24.
Who proposed the theory of acquired characters? Also mention the scientist who disproved it.
Answer:
The theory of acquired characters was proposed by Jean Baptise de Lamarck and it was disproved by August Weismann.

Question 25.
Point out the basic principles of Darwin’s theory of evolution.
Answer:
Over production, struggle for existence, Universal occurence of variation, Survival of fittest and Natural selection.

Question 26.
Name any four Neo – Darwinists.
Answer:
Gregor Mendel, August Weismann, Russel Wallace and Heinrich.

Question 27.
Enumerate the salient features of mutation theory.
Answer:

• Mutations or discontinuous variation are transmitted to other generations.
• In naturally breeding populations, mutations occur from time to time.
• There are no intermediate forms, as they are fully fledged.
• They are strictly subjected to natural selection.

Question 28.
Who proposed Mutation theory? Name the organism on which the experiment was carried out.
Answer:
Mutation theory was put forth by Hugo de Vries. Based on the experiments in Oenothera lamarckiana (The evening primrose plant).

Question 29.
What are the basic factors of modern synthetic theory that leads to evolution?
Answer:
Gene mutation, Chromosomal mutation, Genetic recombination, Natural selection and Reproductive isolation.

Question 30.
Name the scientists who supported modern synthetic theory.
Answer:
Sewell Wright, Dobzhansky, Huxley and Simpson.

Question 31.
Define point mutation.
Answer:
Gene mutation refers to the changes in the structure of the gene. It is also called gene / point mutation. It alters the phenotype of an organism and produces variations in their offsprings.

Question 32.
Point out the factors that alters allelic frequency of a population.
Answer:
Natural selection, Genetic drift, Mutation and Geneflow

Question 33.
Mention any two differences between Homo habilis and Homo erectus
Answer:
Homo habilis :

1. The brain capacity was between 650-800 cc.
2. They were probably vegetarians.

Homo erectus :

1. The brain capacity was around 900 cc.
2. They probably ate meat

Question 34.
Write a brief note on Homo sapiens with respect to evolution.
Answer:
Homo sapiens or modem human arose in Africa some 25,000 years ago and moved to other continents and developed into distinct races. They had a brain capacity of 1300 – 1600 cc. “fhey started cultivating crops and domesticating animals.

Question 35.
Define evolution.
Answer:
The term evolution describes heritable changes in one or more characteristics of a population of species from one generation to the other.

12th Bio Zoology Guide Evolution Three Marks Questions and Answers

Question 36.
Write a short note on Big Bang theory.
Answer:
Big bang theory explains the origin of universe as a singular huge explosion in physical terms. The primitive Earth had no proper atmosphere, but consisted of ammonia, methane, hydrogen and water vapour. The climate of the Earth was extremely high. UV rays from the Sun split up water molecules into hydrogen and oxygen. Gradually the temperature cooled and the water vapour condensed to form rain. Rain water filled all the depressions to form water bodies. Ammonia and methane in the atmosphere combined with oxygen to form carbon dioxide and other gases.

Question 37.
Theory of chemical evolution states that organisms have evolved from inorganic substances. If so, what was the atmospheric condition that favoured evolution?
Answer:
The atmosphere was devoid of O₂, and with high level of CO₂ NH₀3 and UV radiations.

Question 38.
Name the periods of Mesozoic era. Also mention the flora and fauna dominates during that periods.
Answer:
Mesozoic era is divided into three periods namely Triassic, Jurassic and Cretaceous.
Dominating Fauna : Reptiles and Dinosaurs Dominating Flora : Conifers, Ferns and Ginkgon.

Question 39.
Which era is referred as Age of Mammals? What are the periods of that era? And also mention the fauna during the periods.
Answer:
Cenozoic era is called as Age of Mammals.
Tertiary and Quaternary are the two periods of Cenozoic era.
Tertiary periods marks the abundance of mammalian fauna. Quaternary period marks the beginning of human social life.

Question 40.
Write a short note on Cenozoic era.
Answer:
Cenozoic era (Age of mammals) is subdivided into two periods namely Tertiary and Quaternary. Tertiary period is characterized by abundant mammalian fauna. This period is subdivided into five epochs namely, Paleocene (placental mammals, Eocene (Monotremes except duck billed Platypus and Echidna, hoofed mammals and carnivores), Oligocene (higher placental mammals appeared), Miocene (origin of first man like apes) and Pliocene (origin of man from man like apes). Quaternary period witnesses decline of mammals and beginning of human social life.

Question 41.
Name the gaseous mixture used in Urey – Miller’s experiment. Which type of physical force is applied to generate amino acids?
Answer:
Ammonia, Methane, Hydrogen, Water vapour are the gaseous mixture allowed to circulate over electric discharge from a tungsten electrode.

Question 42.
Which is the most common methods of fossilization? Explain how it occurs.
Answer:
Actual remains – The original hard parts such as bones, teeth or shells are preserved as such in the Earth’s atmosphere. This is the most commpn method of fossilization. When marine animals die, their hard parts such as bones and shells, etc., are covered with sediments and are protected from further deterioration. They get preserved as such as they are preserved in vast ocean; the salinity in them prevents decay. The sediments become hardened to form definite layers or strata. For example, Woolly Mammoth that lived 22 thousand years ago were preserved in the frozen coast of Siberia as such. Several human beings and animals living in the ancient city of Pompeii were preserved intact by volcanic ash which gushed out from Mount Vesuvius. ,

Question 43.
What are coprolites? Mention its role in phytogeny.
Answer:
Coprolites are the hardened faecal matters occurs as small pieces. Analysing the coprolites helps to understand the nature of diet of pre-historic animals.

Question 44.
What are moulds and casts?
Answer:
Even after disintegration, the body of an animal might leave indelible impression on the soft mud which later becomes hardened into stones. Such impressions are called moulds. The cavities of the moulds may get filled up by hard minerals and get fossilized, which are called casts.

Question 45.
How will you compute the age of fossil?
Answer:
The age of fossils can be determined using two methods namely, relative dating and absolute dating. Relative dating is used to determine a fossil by comparing it to similar rocks and fossils of known age. Absolute dating is used to determine the precise age of a fossil by using radiometric dating to measure the decay of isotopes.

Question 46.
“Ontogeny recapitulates phylogeny” – comment on the statement with example.
Answer:
The embryonic stages of a higher animal resemble the adult stage of its ancestors. Appearance of pharyngeal gill slits, yolk sac and the appearances of tail in human embryos are some of the examples.

Question 47.
Biogenetic law is not universal – justify.
Answer:
The biogenetic law is not universal and it is now thought that animals do not recapitulate the adult stage of any ancestors. The human embryo recapitulates the embryonic history and not the adult history of the organisms.

Question 48.
How macro molecules like DNA and RNA play their crucial role in evolutionary history?
Answer:
Molecular evolution is the process of change in the sequence composition of molecules such as DNA, RNA and proteins across generations. It uses principles of evolutionary biology and population genetics to explain patterns in the changes of molecules.

One of the most useful advancement in the development of molecular biology is proteins and other molecules that control life processes are conserved among species. A slight change that occurs over time in these conserved molecules (DNA, RNA and protein) are often called molecular clocks. Molecules that have been used to study evolution are cytochrome c (respiratory pathway) and rRNA (protein synthesis).

Question 49.
Explain the principles of Lamarckian theory.
Answer:

i. The theory of use and disuse – Organs that are used often will increase in size and those that are not used will degenerate. Neck in giraffe is an example of use and absence of limbs in snakes is an example for disuse theory.
ii. The theory of inheritance of acquired characters – Characters that are developed during ’ the life time of an organism are called acquired characters and these are then inherited.

Question 50.
Write a note on Mutation theory.
Answer:
Hugo de Vries put forth the Mutation theory. Mutations are sudden random changes that occur in an organism that is not heritable. De Vries carried out his experiments in the Evening Primrose plant {Oenothera lamarckiana) and observed variations in them due to mutation. According to de Vries, sudden and large variations were responsible for the origin of new species whereas Lamarck and Darwin believed in gradual accumulation of all variations as the causative factors in the origin of new species.

Question 51.
What do you mean by “adaptive radiation”? Give example.
Answer:
The evolutionary process which produces new species diverged from a single ancestral form becomes adapted to newly invaded habitats is called adaptive radiation. Adaptive radiations are best exemplified in closely related groups that have evolved in relatively short time. Darwin’s finches and Australian marsupials are best examples for adaptive radiation.

Question 52.
Darwins finches are the classical examples studied for adaptive radation. Explain.
Answer:
Darwin’s finches are the birds whose common ancestor arrived on the Galapagos about 2million year ago. During that time, Darwin’s finches have evolved into 14 recognized species differing in body size, beak shape and feeding behavior. Changes in the size and form of the beak have enabled different species to utilize different food resources such as insects, seeds and nectar from cactus flowers and blood from iguanas, all driven by Natural selection. Genetic variation in the ALX1 gene in the DNA of Darwin finches is associated with variation in the beak shape. Mild mutation in the ALX1 gene leads to phenotypic change in the shape of the beak of the Darwin finches.

Question 53.
What is micro evolution?
Answer:
Microevolution (evolution on a small scale) refers to the changes in allele frequencies within a population. Allele frequencies in a population may change due to four fundamental forces of evolution such as natural selection, genetic drift, mutation and gene flow.

Question 54.
Name the major types of Natural Selection.
Answer:

1. Stabilising Selection
2. Directional Selection
3. Disruptive Selection

Question 55.
What do you mean by gene flow?
Answer:
Movement of genes through gametes or movement of individuals in (immigration) and out (emigration) of a population is referred to as gene flow. Organisms and gametes that enter the population may have new alleles or may bring in existing alleles but in different proportions than those already in the population. Gene flow can be a strong agent of evolution.

Question 56.
Give an account on Genetic drift. Mention its impact over a population.
Answer:
Genetic drift is a mechanism of evolution in which allele frequencies of a population change over generation due to chance (sampling error). Genetic drift occurs in all population sizes, but its effects are strong in a small population. It may result in a loss of some alleles (including beneficial ones) and fixation of other alleles. Genetic drift can have major effects, when the population is reduced in size by natural disaster due to bottle neck effect or when a small group of population splits from the main population to form a new colony due to founder’s effect.

Question 57.
State Hardy – Weinberg equilibrium.
Answer:
The allele frequencies in a population are stable and are constant from generation to generation in the absence of gene flow, genetic drift, mutation, recombination and natural selection.

Question 58.
Write in brief about the characters of Australian ape man.
Answer:
Australopithecus lived in East African grasslands about 5 mya and was called the Australian ape man. He was about 1.5 meters tall with bipedal locomotion, omnivorous, semi erect, and lived in caves. Low forehead, brow ridges over the eyes, protruding face, lack of chin, low brain capacity of about 350 – 450 cc, human like dentition, lumbar curve in the vertebral column were his distinguishing features.

Question 59.
Who is Cro-Magnon?
Answer:
Cro-Magnon was one of the most talked forms of modem human found from the rocks of Cro-Magnon, France and is considered as the ancestor of modem Europeans. They were not only adapted to various environmental conditions, but were also known for their cave paintings, figures on floors and walls.

12th Bio Zoology Guide Evolution Five Marks Questions and Answers

Question 60.
Explain Oparin – Haldane hypothesis on evolution.
Answer:
According to the theory of chemical evolution primitive organisms in the primordial environment of the Earth evolved spontaneously from inorganic substances and physical forces such as lightning, UV radiations, volcanic activities, etc. Oparin (1924) suggested that the organic compounds could have undergone a series of reactions leading to more complex molecules.

He proposed that the molecules formed colloidal aggregates or ‘coacervates’ in an aqueous environment. The coacervates were able to absorb and assimilate organic compounds from the environment. Haldane (1929) proposed that the primordial sea served as a vast chemical laboratory powered by solar energy. The atmosphere was oxygen free and the combination of CO₂, NH₃ and UV radiations gave rise to organic compounds.

The sea became a ‘hot’ dilute soup containing large populations of organic monomers and polymers. They envisaged that groups of monomers and polymers acquired lipid membranes and further developed into the first living cell. Haldane coined the term prebiotic soup and this became the powerful symbol of the Oparin-Haldane view on the origin of life (1924-1929). Oparin and Haldane independently suggested that if the primitive atmosphere was reducing – and if there was appropriate supply of energy such as lightning or UV light then a wide range of organic compounds can be synthesized.

Question 61.
How Urey – Miller’s experiment supports the origin of life?
Answer:
Urey and Miller (1953), paved way for understanding the possible
synthesis of organic compounds that led to the appearance of living organisms is depicted in the Fjgure In their experiment, a mixture of gases was allowed to circulate over electric discharge from an tungsten electrode. A small flask was kept boiling and the steam emanating from it was made to mix with the mixture of

gases (ammonia, methane and hydrogen) in the large chamber that was connected to the boiling Fig. 6.1 Diagrammatic representation of Urey-Miller’s water. The steam condensed to form water which ran down the ‘U’ tube. Experiment was conducted continuously for a week and the liquid was analysed. Glycine, alanine, beta alanine and aspartic acid were identified. Thus Miller’s experiments had an insight as to the possibility of abiogenetic synthesis of large amount of variety of organic compounds in nature from a mixture of sample gases in which the only source of carbon was methane. Later in similar experiments, formation of all types of amino acids, and nitrogen bases were noticed.

Question 62.
Give a detailed account of Modern Synthetic Theory.
Answer:
Sewell Wright, Fisher, Mayer, Huxley, Dobzhansky, Simpson and Haeckel explained Natural Selection in the light of Post-Darwinian discoveries. According to this theory ‘ gene mutations, chromosomal mutations, genetic recombinations, natural selection and reproductive isolation are the five basic factors involved in the process of organic evolution.

• Gene mutation refers to the changes in the structure of the gene. It is also called gene/ point mutation. It alters the phenotype of an organism and produces variations in their offsprings.
• Chromosomal mutation refers to the changes in the structure of chromosomes due to deletion, addition, duplication, inversion or translocation. This too alters the phenotype of an organism and produces variations in their offspring.
• Genetic recombination is due to crossing over of genes during meiosis. This brings about genetic variations in the individuals of the same species and leads to heritable variations.
• Natural selection does not produce any genetic variations but once such variations occur » it favours some genetic changes while rejecting others (driving force of evolution).
• Reproductive isolation helps in preventing interbreeding between related organisms.

Higher Order Thinking Skills (HO’ts) Questions

Question 1.
Name the connecting link for the following groups of organisms.
(a) Annelida and Arthropoda
(b) Reptiles and Aves
(c) Pisces and Amphibians
(d) Reptiles and Mammals
Answer:
(a) Peripatus
(b) Archeopteryx
(c) Lung fish
(d) Platypus

Question 2.
Point out any four condition under which Hardy Weinberg’s equilibrium is not attained.
Answer:

•  Selected mating
• Flow of genes (either by immigration or emmigration)
• Occurance of mutation
• Definite population size

Question 3.
Why are analogous structures a result of convergent evolution?
Answer:
Analogous structures are not anatomically similar though they perform same function.

Question 4.
Organs which are of no use to the organism is called as vestige. Name any four vestigal organs that can be noticed in your body.
Answer:
Wisdom teeth, Mammae in male, Body hair and Coccyx.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 5 Molecular Genetics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

12th Bio Zoology Guide Molecular Genetics Text Book Back Questions and Answers

Question 1.
Hershey and Chase experiment with bacteriophage showed that
(a) Protein gets into the bacterial cells
(b) DNA is the genetic material
(c) DNA contains radioactive sulphur
(d) Viruses undergo transformation
Answer:
(b) DNA is the genetic material

Question 2.
DNA and RNA are similar with respect to
(a) Thymine as a nitrogen base
(b) A single-stranded helix shape
(c) Nucleotide containing sugars, nitrogen bases and phosphates
(d) The same sequence of nucleotides for the amino acid phenyl alanine
Answer:
(c) Nucleotide containing sugars, nitrogen bases and phosphates

Question 3.
A mRNA molecule is produced by
(a) Replication
(b) Transcription
(c) Duplication
(d) Translation
Answer:
(b) Transcription

Question 4.
The total number of nitrogenous bases in human genome is estimated to be about
(a) 3.5 million
(b) 35000
(c) 35 million
(d) 3.1 billion
Answer:
(d) 3.1 billion

Question 5.
E. coli cell grown on 15N medium are transferred to 14N medium and allowed to grow for two generations. DNA extracted from these cells is ultracentrifuged in a cesium chloride density gradient. What density distribution of DNA would you expect in this experiment?
(a) One high and one low density band
(b) One intermediate density band
(c) One high and one intermediate density band
(d) One low and one intermediate density band
Answer:
(d) One low and one intermediate density band

Question 6.
What is the basis for the difference in the synthesis of the leading and lagging strand of DNA molecules?
(a) Origin of replication occurs only at the 5’ end of the molecules
(b) DNA ligase works only in the 3’ → 5’ direction
(c) DNA polymerase can join new nucleotides only to the 3′ ends of the growing stand
(d) Helicases and single-strand binding proteins that work at the 5’ end
Answer:
(d) DNA polymerase can join new nucleotides only to the 3’ end of the growing stand

Question 7.
Which of the following is the correct sequence of event with reference to the central dogma?
(a) Transcription, Translation, Replication
(b) Transcription, Replication, Translation
(c) Duplication, Translation, Transcription
(d) Replication, Transcription, Translation
Answer:
(d) Replication, Transcription, Translation

Question 8.
Which of the following statements about DNA replication is not correct?
(a) Unwinding of DNA molecule occurs as hydrogen bonds break
(b) Replication occurs as each base is paired with another exactly like it
(c) Process is known as semiconservative replication because one old strand is conserved in the new molecule
(d) Complementary base pairs are held together with hydrogen bonds
Answer:
(b) Replication occurs as each base is paired with another exactly like it

Question 9.
which of the following statements is not true about DNA replication in eukaryotes?
(a) Replication begins at a single origin of replication.
(b) Replication is bidirectional from the origins.
(c) Replication occurs at about 1 million base pairs per minute.
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.
Answer:
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time

Question 10.
The first codon to be deciphered was ………………. which codes for ……………..
(a) AAA, proline
(b) GGG, alanine
(c) UUU, Phenylalanine
(d) TTT, arginine
Answer:
(c) UUU, Phenylalanine

Question 11.
Meselson and Stahl’s experiment proved
(a) Transduction
(b) Transformation
(c) DNA is the genetic material
(d) Semi-conservative nature of DNA replication
Answer:
(d) Semi-conservative nature of DNA replication

Question 12.
Ribosomes are composed of two subunits; the smaller subunit of a ribosome has a binding site for……………..and the larger subunit has two binding sites for two…………………
Answer:
mRNA, tRNA

Question 13.
Anoperonisa:
(a) Protein that suppresses gene expression
(b) Protein that accelerates gene expression
(c) Cluster of structural genes with related function
(d) Gene that switched other genes on or off
Answer:
(d) Cluster of structural genes with related function

Question 14.
When lactose is present in the culture medium:
(a) Transcription of lacy, lac z, lac a genes occurs
(b) Repressor is unable to bind to the operator
(c) Repressor is able to bind to the operator
(d) Both (a) and (b) are correct
Answer:
(d) Both (a) and (b) are correct

Question 15.
Give reasons: Genetic code is ‘universal’.
Answer:
The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial, and chloroplast genomes. However, similarities are more common than differences.

Question 16.
Name the parts marked ‘A’ and ‘B’ in the given transcription unit:

Answer:
A – Promoter site B – Structural gene

Question 17.
Differentiate – Leading strand and lagging strand
Answer:

Question 18.
Differentiate – Template strand and coding strand.
Answer:

1. Template Strand: During replication, DNA strands having the polarity 3’ → 5’ act as the template strand.
2. Coding Strand: During replication, a DNA strands having the polarity 5’ → 3’ act as a coding strand.

Question 19.
Mention any two ways in which single nucleotide polymorphism (SNPs) identified in human genome can bring revolutionary change in biological and medical science.
Answer:
Scientists have identified about 1.4 million locations, where single-base DNA differences (SNPs – Single nucleotide polymorphism – pronounced as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease-associated sequences and tracing human history.

Question 20.
State any three goals of the human genome project.
Answer:

• Identify all the genes (approximately 30000) in human DNA.
• Determine the sequence of the three billion chemical base pairs that make up the human DNA.
• To store this information in databases.

Question 21.
In E.coli, three enzymes 0- galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.
Answer:
In the absence of lactose, the repressor protein binds to the operator and prevents the transcription of structural gene by RNA polymerase, hence the enzymes are not produced. However, there will always be a minimal level of lac operon expression even in absence of lactose.

Question 22.
Distinguish between structural gene, regulatory gene and operator gene.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls the transcriptional, activity of the structural gene.

1. The structural gene codes for proteins, rRNA, and tRNA required by the cell.
2. Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
3. The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Question 23.
A low level of expression of lac operon occurs at all the time in E-coli. Justify the statement.
Answer:
One of the enzyme synthesized by lac operon is permease which is involved in the transport of lactose into the cells. If the lac operon gets inactivated, permease is not synthesized hence lactose cannot enter the cell. Lactose acts as a inducer, binding to the repressor protein and switch on the operator to initiate gene expression.

Question 24.
Why the human genome project is called a megaproject?
Answer:
The international human genome project was launched in the year 1990. It was a mega project and took 13 years to complete. The human genome is about 25 times larger than the genome of any organism sequenced to date and is the first vertebrate genome to be completed. Human genome is said to have approximately 3 >109 bp. HGP was closely associated with the rapid development of a new area in biology called bioinformatics.

Question 25.
From their examination of the structure of DNA, What did Watson and Crick infer about the probable mechanism of DNA replication, coding capability and mutation?
Answer:
Inference of Watson and Crick on DNA replication.
They concluded that each of the DNA strand in a helix act as template during DNA replication leading to formation of new daughter DNA molecules, which are complementary to parental strand, (i.e., Semi-conservative method of replication)

Inference on coding capability
During transcription, the genetic information in the DNA strand is coded to mRNA as complementary bases, (except for uracil in place of thymine in RNA)

Inference on mutation
Any changes in the nucleotide sequence of DNA leads to corresponding alteration in aminoacid sequence of specific protein thus confirming the validity of genetic code.

Question 26.
Why tRNA is called an adapter molecule?
Answer:
The transfer RNA, (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called an adapter molecule. This term was postulated by Francis Crick.

Question 27.
What are the three structural differences between RNA and DNA?
Answer:
DNA :

1. Sugar is deoxyribose sugar. .
2. Double-stranded structure.
3. Nitrogen bases are Adenine, Guanine, Cytosine and Thymine.

RNA :

1. Sugar is ribose sugar
2. Single-stranded molecule.
3. Nitrogen bases are Adenine, Guanine, Cytosine and Uracil.

Question 28.
Name the anticodon required to recognize the following codons: AAU, CGA, UAU, and GCA.
Answer:
UUA, GCU, AUA and CGU.

Question 29.
a) Identify the figure given below
b) Redraw the structure as a replicating fork and label the parts
c) Write the source of energy for this replication and name the enzyme involved in this process.
d) Mention the differences in the synthesis of protein, based on the polarity of the two template strands.
Answer:
a) Replication fork
b)

c) Deoxy nucleotide, triphosphate acts as a energy source for replication. DNA polymerase is used for replication
d) mRNA contacting information for protein synthesis will developed from DNA strand having polariy 5 ’ → 3 ’

Question 30.
If the coding sequence in a transcription unit is written as follows:
5’ TGCATGCATGCATGCATGCATGCATGC 3’
Write down the sequence of mRNA.
Answer:
mRNA sequence is 3’ACGUACGUACGUUCGUACGUACGUACG5’

Question 31.
How is the two-stage process of protein synthesis advantageous?
Answer:
The split gene feature of eukaryotic genes is almost entirely absent in prokaryotes. Originally each exon may have coded for a single polypeptide chain with a specific function. Since exon arrangement and intron removal are flexible, the exon coding for these polypeptide subunits act as domains combining in various ways to form new genes. Single genes can produce different functional proteins by arranging their exons in several different ways through alternate splicing patterns, a mechanism known to play an important role in generating both protein and functional diversity in animals.

Introns would have arose before or after the evolution of eukaryotic gene. If introns arose late how did they enter eukaryotic gene? Introns are mobile DNA sequences that can splice themselves out of, as well as into, specific ‘target sites’ acting like mobile transposon-like elements (that mediate transfer of genes between organisms – Horizontal Gene Transfer – HGT). HGT occurs between lineages of prokaryotic cells, or from prokaryotic to eukaryotic cells and between eukaryotic cells. HGT is now hypothesized to have played a major role in the evolution of life on Earth.

Question 32.
Why did Hershey and Chase use radioactively labelled phosphorous and sulphur only? Would they have got the same result if they use radiolabelled carbon and nitrogen?
Answer:
Generally proteins contain sulphur but not phosphorous and nucleic acid (DNA) contains , phosphorous but not sulphur. Hence Hershey – Chase used radioactive isotopes of sulphur (35S) and phosphorus (32P) to keep separate track of viral protein and nucleic acid in culture medium. The expected result cannot be achieved, if radioactive carbon and nitrogen is used, since these molecules are present in both DNA and proteins.

Question 33.
Explain the formation of a nucleosome.
Answer:
Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere.

The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix. The histone octameres are in close contact and DNA is coiled on the outside of nucleosome.

Question 34.
It is established that RNA is the first genetic material. Justify giving reasons.
Answer:
Three molecular biologists in the early 1980’s (Leslie Orgel, Francis Brick and Carl Woese) independently proposed the ‘RNA world’ as the first stage in the evolution of life, a stage when RNA catalysed all molecules necessary for survival and replication. The term ‘RNA world’ first used by Walter Gilbert in 1986, hypothesizes RNA as the first genetic material on Earth. There is now enough evidence to suggest that essential life processes (such as metabolism, translation and splicing etc.,) evolved around RNA.

RNA has the ability to act as both genetic material and catalyst. There are several biochemical reactions in living systems that are catalysed by RNA. This catalytic RNA is known as ribozyme. But, RNA being a catalyst was reactive and hence unstable. This led to evolution of a more stable form of DNA, with certain chemical modifications. Since DNA is a double-stranded molecule having complementary strand, it has resisted changes by evolving a process of repair. Some RNA molecules function as gene regulators by binding to DNA and affect gene expression. Some viruses use RNA as the genetic material. Andrew Fire and Craig Mellow (recipients of Nobel Prize in 2006) were of the opinion that RNA is an active ingredient in the chemistry of life.

12th Bio Zoology Guide Molecular Genetics Additional Important Questions and Answers

12th Bio Zoology Guide Molecular Genetics One Mark Questions and Answers

Question 1.
The term gene was coined by……………
Answer:
Wilhelm Johannsen

Question 2.
Whose experiment finally provided convincing evidence that DNA is the genetic material?
(a) Griffith experiment
(b) Avery, Macleod and McCarty’s experiment
(c) Hershey-Chase experiment
(d) Urey-Miller’s experiment
Answer:
(c) Hershey-Chase experiment

Question 3.
In Hershey – Chase experiment, the DNA of T₂ phase was made radioactive by using
(a) ³²P
(b) ³⁵S
(c) ³⁵P
(d) ³⁵S
Answer:
(a) ³²P

Question 4.
A nucleoside is composed of………………………………
(a) Sugar and Phosphate
(b) Nitrogen base and Phosphate
(c) Sugar and Nitrogen base
(d) Sugar, Phosphate and Nitrogenous base
Answer:
(c) Sugar and Nitrogen base

Question 5.
Identify the incorrect statement
(a) a base is a substance that accepts H+ ion
(b) Both DNA and RNA have four bases
(c) Purines have single carbon-nitrogen ring
(d) Thymine is unique for DNA
Answer:
(c) Purines have single carbon-nitrogen ring

Question 6.
Watson and Crick proposed their double helical DNA model based on the X-ray diffraction analysis of………………
(a) Erwin Chargaff
(b) Meselson and Stahl
(c) Wilkins and Franklin
(d) Griffith
Answer:
(c) Wilkins and Franklin

Question 7.
The term ‘RNA world’ was first used by………………
Answer:
Walter Gilbert

Question 8.
The distance between two consecutive base pairs in DNA is …………………
(a) 0.34 nm
(b) 3.4 nm
(c) 0.034 nm
(d) 34 nm
Answer:
(a) 0.34 nm

Question 9.
If the length of E. coli DNA is 1.36 mm, the number of base pairs is
(a) 0.36 x 10⁶m
(b) 4 x 10⁶m
(c) 0.34 x 10^-9nm
(d) 4 x 10^-9m
Answer:
(b) 4 x 10⁶m

Question 10.
Identify the proper sequence in the organisation of the eukaryotic chromosome.
(a) Nucleosome – Solenoid – Chromatid
(b) Chromatid – Nucleosome – Solenoid
(c) Solenoid – chromatin – DNA
(d) Nucleosome – solenoid – genophore
Answer:
(a) Nucleosome – Solenoid – Chromatid

Question 11.
Assertion (A) : Genophore is noticed in prokaryotes.
Reason (R) : Bacteria possess circular DNA without chromatin organisation.
(a) Both A and R are correct
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(c) R explains A

Question 12.
Assertion (A): Heterochromatin is transcriptionally active.
Reason (R): Tightly packed chromatin which stains dark.
(a) Both A and R are correct , (b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct .
(d) A is incorrect R is correct
Answer:
(d) A is incorrect R is correct

Question 13.
Assertion (A): the semi-conservative model was proposed by Hershey and Chase.
Reason (R) : The daughter DNA contains only new strands.
(a) Both A and R are incorrect
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(a) Both A and R are incorrect

Question 14.
Komberg enzyme is called as ……………..
Answer:
DNA polymerase I

Question 15.
Replication of DNA occurs at phase of cell cycle.
(a) M
(b) S
(c) G₁
(d) G₂
Answer:
(b) S

Question 16.
Semi-conservative model of replication was proved by
(a) Hershey and Chase
(b) Griffith
(c) Meselson and Stahl
(d) Macleod and McCarty
Answer:
(c) Meselson and Stahl

Question 17.
How many types of DNA polymerases does a eukaryotic cell possess?
(a) two
(b) three
(c) four
(d) five
Answer:
(d) Five

Question 18.
Identify the incorrect statement
(a) Replication occurs at ori – site of DNA
(b) Deoxy nucleotide triphosphate acts as a substrate
(c) Unwinding of DNA strand is carried out by topoisomerase
(d) DNA polymerase catalyses the polymerization at 3-OH
Answer:
(c) Unwinding of DNA strand is carried out by topoisomerase

Question 19.
The discontinuously synthesized fragments of lagging strand are called………………
Answer:
Okazaki fragments

Question 20.
Retroviruses possess ………………..as genetic material.
Answer:
RNA

Question 21.
Which is NOT a part of transcription unit?
(a) Promoter
(b) Operator
(c) Structural gene
(d) Terminator
Answer:
(b) Operator

Question 22.
Goldberg – Hogness box of eukaryotes is equivalent to of prokaryotes.
Answer:
Pribnow box

Question 23.
Okazaki fragments are joined by the enzyme during DNA replication.
Answer:
DNA ligase

Question 24.
Match the following:

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

Question 25.
The RNA polymerase of prokaryotes binds with ………………….factor to initiate polymerization.
(a) rho
(b) theta
(c) sigma
(d) psi
Answer:
(c) sigma A

Question 26.
Precursor
(a) Capung
(b) Tailing
(c) Splicing
(d) Transcribing
Answer:
(c) Splicing

Question 27.
Which of the following feature is absent in prokaryotes?
(a) Prokaryotes possess three major types of RNAs
(b) Structural genes are polycistronic
(c) Initiation process of transcription requires ‘P’ factor
(d) Split gene feature
Answer:
(d) Split gene feature

Question 28.
Which of the following sequence has completely translated?
(i) AGA, UUU, UGU, AGU, UAG
(ii) AUG, UUU, AGA, UAC, UAA
(iii) AAA, UUU, UUG, UGU, UGA
(iv) AUG,AAU,AAC,UAU,UAG
(a) i and ii
(b) ii only
(c) i and iii
(d) ii and iv
Answer:
(d) ii and iv

Question 29.
Capping of mRNA occurs using
(a) Poly A residues
(b) Methyl guanosine triphosphate
(c) Deoxy ribonucleotide triphosphate
(d) Ribonucleotide triphosphate
Answer:
(b) Methyl guanosine triphosphate

Question 30.
One of the aspect is not a feature of genetic code?
(a) Specific
(b) Degenerate
(c) Universal
(d) Ambiguous
Answer:
(d) Ambiguous

Question 31.
Which of the triplet codon is not a code of proline?
(i) CCU (ii) CAU (iii) CCG (zv) CAA
(a) i only
(b) ii and iv
(c) iii only
(d) all the above
Answer:
(b) ii and iv

Question 32.
Coding sequences found in split genes are called.
(a) Operons
(b) Introns
(c) Exons
(d) Cistron
Answer:
(c) Exons

Question 33.
Which of the following mRNA yields 6 aminoacids after translation?
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU
(ii) UGA AGA UAG GAG CAU CCC UAC UAU GAU
(iii) GUC UGC UGG GCU GAU UAA AGG AGC AUU
(iv) AUG UAC CAU UGC UGA UGC AGG AGC CCG
Answer:
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU

Question 34.
The transcription termination factor associated with RNA polymerase in prokaryotes is
(a) ∑
(b) σ
(c) ρ
(d) ∑
Answer:
(c) p

Question 35.
In a DNA double-strand, if guanine is 30%, what will be the percentage of thymine?
(a) 100%
(b) 20%
(c) 10%
(d) 70%
Answer:
(b) 20%

Question 36.
Identify the triplet pairs that code for Tyrosine
(a) UUU,UUC
(b) UAU, UAU
(c) UGC, UGU
(d) CAU, CAC
Answer:
(b) UAU, UAU

Question 37.
Match the following:

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

Question 38.
AUG code is for……………….
(a) Arginine
(b) Tyrosine
(c) Tryptophan
(d) Methionine
Answer:
(d) Methionine

Question 39.
The sequence of bases in coding strand of DNA is GAGTTAGCAGGC, then the sequence of codons in primary transcript is ……………
(d) CUCAUACGCCCG
(b) CUCAAUCGUCCG
(c) UCAGAUCUGCGC
(d) UUCAAUCGUGCG
Answer:
(b) CUCAAUCGUCCG

Question 40.
The promoter region of eukaryote is …………..
(a) TATAA
(b) AUGUT
(c) UUUGA
(d) AAAAU
Answer:
(a) TATAA

Question 41.
Match the following:

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

Question 42.
……………….number of codons, codes for cystine.
Answer:
Two

Question 43.
In sickle cell anaemia, the ……………….codon of p – globin gene is modified.
(a) Eighth
(b) Seventh
(c) Sixth
(d) Ninth
Answer:
(c) Sixth

Question 44.
Pick out the incorrect statement.
(a) tRNA acts as a adapter molecule
(b) Stop codons do not have tRNA’s
(c) Addition of amino acid leads to hydrolysis of tRNA {d) tRNA has four major loops
Answer:
(c) Addition of amino acid leads to hydrolysis of tRNA

Question 45.
Which of the following antibiotic inhibits the interaction between tRNA and mRNA?
(a) Neomycin
(b) Streptomycin
(c) Tetracycline
(d) Chloramphenicol
Answer:
(a) Neomycin

Question 47.
The cluster of genes with related function is called …………….
(a) Cistron
(b) Operon
(c) Muton
(d) Recon
Answer:
(b) Operon

Question 48.
Repressor protein of Lac operon binds to ……………of operon
(a) Promoter region
(b) Operator region
(c) terminator region
(d) inducer region
Answer:
(b) Operator region

Question 49.
Lac Z gene codes for ………………..
(a) Permease
(b) transacetylase
(c) P-galactosidase
(d) Aminoacyl transferase
Answer:
(c) p-galactosidase

Question 50.
Lac operon model was proposed by ………………
Answer:
Jacob and Monod

Question 51.
Approximate count of base pair in human genome is …………….
Answer:
3 x 10⁹ bp

Question 52.
Automated DNA sequences are developed by.
Answer:
Frederick Sanger

Question 53.
Which of the chromosome has a higher gene density?
(a) Chromosome 20
(b) Chromosome 19
(c) Chromosome 13
(d) Chromosome Y
Answer:
(b) Chromosome 19

Question 54.
Number of genes located in chromosome Y is
(a) 2968
(b) 213
(c) 2869
(d) 231
Answer:
(d) 231

Question 55.
How many structural genes are located in the lac operon of E.Coli?
(a) 4
(b) 3
(c) 2
(d) 1
Answer:
(b) 3

Question 56.
DNA fingerprinting technique was developed by
(d) Jacob and Monod
(b) Alec Jeffreys
(c) Frederick Sanger
Answer:
(b) Alec Jeffreys

Question 57.
In DNA fingerprinting, separation of DNA fragments is done by………
(a) Centrifugation
(b) Electrophoresis
(c) X-ray diffraction
(d) denaturation
Answer:
(b) Electrophoresis

Question 58.
SNP stands for
(a) Single nucleotide Polymorphism
(b) Single Nucleoside Polypeptide
(c) Single nucleotide Polymorphism
(d) Single nucleotide polymer
Answer:
(a) Single nucleotide Polymorphism

Question 59.
Specific sequences of mRNA that are not translated are……………..
Answer:
Untranslated Regions (UTR)

Question 60.
Non-coding or intervening DNA sequence is called ……………….
Answer:
Intron

Question 61.
………………is the monomer of DNA.
Answer:
Nucleotide

Question 62.
Which one of the following is wrongly matched?
(a) Transcription – Copying information from DNA to RNA
(b) Translation – Decoding information from mRNA to protein
(c) Replication – Making of DNA copies
(d) Splicing – Joining of exons with introns
Answer:
(d) ‘ Splicing – Joining of exons with introns

12th Bio Zoology Guide Molecular Genetics Two Marks Questions and Answers

Question 1.
Who proposed One gene – One enzyme hypothesis? Define it.
Answer:
George Beadle and Edward Tatum proposed the One gene – One enzyme hypothesis which states that one gene controls the production of one enzyme.

Question 2.
Differentiate nucleoside from nucleotide.
Answer:
Nucleoside :
Nucleoside subunit is composed of nitrogenous bases linked to a pentose sugar molecule.

Nucleotide :
Nucleotide subunit is composed of nitrogenous bases, a pentose sugar and a phosphate group.

Question 3.
State the key differences between DNA and RNA.
Answer:
DNA

1. DNA is made of deoxyribose sugar.
2. Nitrogenous bases of DNA are Adenine, Guanine, Cytosine and Thymine.

RNA:

1. RNA is made of ribose sugar.
2. Nitrogenous bases of RNA are Adenine,Guanine, Cytosine and Uracil.

Question 4.
Point out the nitrogenous bases of RNA.
Answer:
Adenine, Guanine, Cytosine and Uracil.

Question 5.
What makes the DNA and RNA acidic molecules?
Answer:
The phosphate functional group (P04) gives DNA and RNA the property of an acid at physiological pH, hence the name nucleic acid.

Question 6.
Which type of bond is formed

1. between a purine and pyrimidine base?
2. between the pentose sugar and adjacent nucleotide?

Answer:

1. Purine and pyrimidine bases are linked by hydrogen bonds.
2. Pentose sugar is linked to adjacent nucleotide by phosphodiester bonds.

Question 7.
DNA acts as genetic material for majority of living organisms and not the RNA. Give reasons to support the statement.
Answer:

1. RNA was reactive and hence highly unstable.
2. Some RNA molecules acts as gene regulators by binding to DNA and affect gene expression.
3. Uracil of RNA is less stable than thymine of DNA.

Question 8.
Name any two viruses whose genetic material is RNA.
Answer:

1. Tobacco Mosaic Virus (TMV)
2. Bacteriophage (B)

Question 9.
What are the properties that a molecule must possess to act as genetic material?
Answer:

• Self-replication
• Information storage
• Stability
• Variation through mutation

Question 10.
How many base pairs are present in one complete turn of DNA helix? What is the distance between two consecutive base pairs?
Answer:
There are ten base pairs in each turn with a distance of 0.34 x 10‘9 m between two adjacent base pairs.

Question 11.
What is a genophore?
Answer:
In prokaryotes such as E. coli though they do not have defined nucleus, the DNA is not scattered throughout the cell. DNA (being negatively charged) is held with some proteins (that have positive charges) in a region called the nucleoid. The DNA as a nucleoid is organized into large loops held by protein. DNA of prokaryotes is almost circular and lacks chromatin organization, hence termed genophore.

Question 12.
What is nucleosome? How many base pairs are there in a typical nucleosome?
Answer:
The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix.

Question 13.
Expand and define NHC
Answer:
NHC : Non-histone Chromosomal protein.
In eukaryotes, apart from histone proteins, additional set of proteins are required for packing of chromatin at higher level and are referred as non – histone chromosomal proteins.

Question 14.
Differentiate between Heterochromatin and Euchromatin.
Answer:
Heterochromatin:

1. Region of nucleus where the chromatin are loosely packed and stains light are called Heterochromatin.
2. Transcriptionally inactive.

Euchromatin:

1. Region of the nucleus where the chromatin are tightly packed and stains dark are called Euchromatin.
2. Transcriptionally active.

Question 15.
Which is the widely accepted model of DNA replication? Who has proved it?
Answer:
Semi-conservative replication model. It was proved by Meselson and Stahl in 1958.

Question 16.
Name the chemical substance which is called by the name

1. Kornberg Enzyme
2. Ochoa’s enzyme

Answer:

1. DNA polymerase I is also known as Komberg enzyme.
2. Polynucleotide phosphorylase is also known as Ochoa’s enzyme.

Question 17.
Name the various types of prokaryotic DNA polymerase. State their role in replication process.
Answer:

Question 18.
What is the function of Deoxy nucleotide triphosphate in replication?
Answer:
Deoxy nucleotide triphosphate acts as substrate and also provides energy for polymerization reaction.

Question 19.
Given below are some events of eukaryotic replication. Name the enzymes involved in the process.

1. Unwinding of DNA
2. Joining of Okazaki fragments
3. Addition of nucleotides to a new strand
4. Correcting the repair

Answer:

1. Helicase
2. DNA ligase
3. DNA polymerase
4. Nuclease

Question 20.
Differentiate leading strand from lagging strand
Answer:
Leading strand :

1. Leading strand has the polarity 3’ →‘ 5’.
2. Replication is continuous.

Lagging strand :

1. Lagging strand has the polarity 5’ →‘ 3’
2. Replication is discontinuous.

Question 21.
What are Okazaki fragments?
Answer:
The discontinuously synthesized fragments of the lagging strand are called the Okazaki fragments are joined by the enzyme DNA ligase.

Question 22.
What is a replication fork?
Answer:
At the point of origin of replication, the helicases and topoisomerases (DNA gyrase) unwind and pull apart the strands, forming a Y-Shaped structure called the replication fork. There are two replication forks at each origin.

Question 23.
Apart from DNA polymerase, name any other four enzymes which were involved in DNA replication of eukaryotic cell.
Answer:
DNA ligase, Topoisomerase (DNA gyrase), Helicase and Nuclease.

Question 24.
Who proposed the central dogma? Write its concept.
Answer:

Question 25.
Define transcription and name the enzyme involved in this process.
Answer:
The process of copying genetic information from one strand of DNA into RNA is termed transcription. This process takes place in presence of DNA dependent RNA polymerase.

Question 26.
What is TATA box? State its function.
In eukaryotes, the promoter has AT-rich regions called TATA box or Goldberg-Hogness box. It acts as a binding site for RNA polymerase.

Question 27.
Structural gene of eukaryotes differ from prokaryotes. How?
In eukaryotes, the structural gene is monocistronic coding for only one protein whereas in prokaryotes the structural gene is polycistronic coding for many proteins.

Question 28.
What are the two major components of prokaryotic RNA polymerase? How do they act?
Answer:
Bacterial (prokaryotic) RNA polymerase consists of two major components, the core enzyme and the sigma subunit. The core enzyme (P1, p, and a) is responsible for RNA synthesis ” whereas a sigma subunit is responsible for recognition of the promoter.

Question 29.
Distinguish between exons and introns.
Answer:
Exons:
Expressed sequences (Coding sequences) of a eukaryotic gene

Introns:
Intervening sequences (non-coding sequences) of an eukaryotic gene

Question 30.
Define splicing.
Answer:
The process of removing introns from hnRNA is called splicing.

Question 31.
What is capping and tailing?
Answer:
In capping an unusual nucleotide, methyl guanosine triphosphate is added at the 5’ end of hnRNA, whereas adenylate residues (200-300) (Poly A) are added at the 3’ end in tailing.

Question 32.
If a double-stranded DNA has 20% of cytosine, calculate the percentage of adenine in DNA.
Answer:
Cytdsine = 20, hence Guanine = 20
As per ChargafFs rule (A+T) = (G+C) =100
Percent of Thymine + Adenine = 20 + 20 = 100
(T + A) = (20 + 20) = 100
(T + A) = 100 – (20 + 20)
T + A = 100 – 40
T + A = 60
Therefore the percent of Adenine will be 60/2 = 30%.

Question 33.
Mention the dual functions of AUG.
Answer:
AUG has dual functions. It acts as a initiator codon and also codes for the amino acid methionine.

Question 34.
How many codons are involved in termination of translation. Name them.
Answer:
Three codons terminate translation process. They are UAA, UAG and UGA.

Question 35.
Degeneracy of codon – comment.
Answer:
A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.

Question 36.
Point out the exceptional categories to universality of genetic code.
Answer:
Exceptions to the universal nature of genetic code are noticed in prokaryotic mitochondrial and chloroplast genomes.

Question 37.
What are non-sense codons?
Answer:
UGA, UAA, and UAG are the non-sense codons, which terminate translation.

Question 38.
Name the triplet codons that code for

1. Tyrosine
2. Histidine

Answer:

1. Tyrosine – UAU, UAC
2. Histidine – CAU, CAC

Question 39.
Why hnRNA has to undergo splicing?
Answer:
Since hnRNA contains both coding sequences (exons) and non-coding sequences (introns) it has to undergo splicing to remove introns.

Question 40.
State the role of following codons in the translation process

1. AUG
2. UAA

Answer:

1. AUG is the initiator codon and also codes for methionine.
2. UAA is a terminator codon.

Question 41.
Given below is mRNA sequence. Mention the aminoacids sequence that is formed after its translation.
3’AUGAAAGAUGGGUAA5’
Answer:
Methionine – Lysine – Aspartic acid – Glycine

Question 42.
Name the four codons that codes valine.
Answer:
GUU, GUC, GUA and GUG.

Question 43.
The base sequence in one of the DNA strand is TAGC ATGAT. Mention the base sequence in its complementary strand.
Answer:
The complementary strand has ATCGTACTA.

Question 44.
Why t-RNA is called as adapter molecule?
Answer:
The transfer RNA (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called as adapter molecule.

Question 45.
What do you mean by charging of tRNA? Name the enzyme involved in this process.
Answer:
The process of addition of amino acid to tRNA is known as aminoacylation or charging and the resultant product is called aminoacyl- tRNA (charged tRNA). Aminoacylation is catalyzed by an enzyme aminoacyl – tRNA synthetase.

Question 46.
What is UTR’s?
Answer:
mRNA also have some additional sequences that are not translated and are referred to as Untranslated Regions (UTR). UTRs are present at both 5’ end (before start codon) and at 3’ end (after stop codon).

Question 47.
What is the S – D sequence?
Answer:
The 5’ end of the mRNA of prokaryotes has a special sequence which precedes the initial AUG start codon of mRNA. This ribosome binding site is called the Shine – Dalgarno sequence or S-D sequence. This sequences base-pairs with a region of the 16Sr RNA of the small ribosomal subunit facilitating initiation.

Question 48.
Define translation unit.
Answer:
A translation unit in mRNA is the sequence of RNA that is flanked by the start codon on 5’ end and stop codon on 3’ end and codes of polypeptide.

Question 49.
Mention the inhibitory role of tetracycline and streptomycin in bacterial translation.
Answer:
Tetracycline inhibits binding between aminoacyl tRNA and mRNA.
Streptomycin inhibits initiation of translation and causes misreading.

Question 50.
At what stage, does the gene expression is regulated?
Answer:
Gene expression can be controlled or regulated at transcriptional or translational levels.

Question 51.
What is a operon? Give example.
Answer:
The cluster of genes with related functions is called operon.
E.g: lac operon in E.coli.

Question 52.
Considering the lac operon of E.coli, name the products of the following genes.

1. a) i gene
2. (b) lac Z gene
3. (c) lac Y gene
4. (d) lac a gene

Answer:

1. i gene – repressor protein
2. lac Z gene – fS-galactosidase
3. Lac Y gene – Permease
4. lac a gene – transacetylase

Question 53.
Expand

1. ETS
2. YAC.

Answer:

1. ETS: Expressed Sequence Tags
2. YAC: Yeast Artificial Chromosomes

Question 54.
Name the human chromosome that has

1. most number of genes
2. least number of genes

Answer:

1. Chromosome 1 has the maximum number of genes (2968 genes)
2. Chromosome Y has least genes (231 genes)

Question 55.
What are SNPs? Mention its uses.
Answer:
SNPs : Single nucleotide polymorphism. It helps to find chromosomal locations for disease-associated sequences and tracing human history.

Question 56.
Mention any four areas where DNA fingerprinting can be used.
Answer:

• Forensic analysis
• Pedigree analysis
• Conservation of wildlife
• Anthropological studies

12th Bio Zoology Guide Molecular Genetics Three Marks Questions and Answers

Question 57.
Classify nucleic acid-based on sugar molecules.
Answer:
There are two types of nucleic acids depending on the type of pentose sugar. Those containing deoxyribose sugar are called Deoxyribo Nucleic Acid (DNA) and those with ribose sugar are known as Ribonucleic Acid (RNA). The only difference between these two sugars is that there is one oxygen atom less in deoxyribose.

Question 58.
Both purines and pyrimidines are nitrogen bases yet they differ. How?
Answer:
Both purines and pyrimidines are nitrogen bases. The purine bases Adenine and Guanine have double carbon-nitrogen ring, whereas cytosine and thymine bases have single carbon-nitrogen ring.

Question 59.
How 5’ of DNA differ from its 3’?
Answer:
The 5’ of DNA refers to the carbon in the sugar to which phosphate (P04V) functional group is attached. The 3’ of DNA refers to the carbon in the sugar to which a hydroxyl (OH) group is attached.

Question 60.
State Chargaff’s rule.
Answer:
According to Erwin Chargaff,
a) Adenine pairs with Thymine with two hydrogen bonds.
b) Guanine pairs with Cytosine with three hydrogen bonds.

Question 61.
Chemically DNA is more stable than RNA – Justify.
Answer:
In DNA, the two strands being complementary, if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2’ OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

Question 62.
Write in simple about semi-conservative mode of DNA replication.
Answer:
Semi-conservative replication was proposed by Watson and Crick in 1953. This mechanism of replication is based on the DNA model. They suggested that the two polynucleotide strands of DNA molecule unwind and start separating at one end. During this process, covalent hydrogen bonds are broken. The separated single strand then acts as template for the synthesis of a new strand. Subsequently, each daughter double helix carries one polynucleotide strand from the parent molecule that acts as a template and the other strand is newly synthesised and complementary to the parent strand.

Question 63.
Draw a simplified diagram of nucleosome and label it.
Answer:

Question 64.
What is a primer?
Answer:
A primer is a short stretch of RNA. It initiates the formation of new strand. The primer produces 3’-OH end on the sequence of ribonucleotides, to which deoxyribonucleotides are added to form a new strand.

Question 65.
Both strands of DNA are not copied during transcription. Give reason.
Answer:
Both the strands of DNA are not copied during transcription for two reasons.

1. If both the strands act as a template, they would code for RNA with different sequences. This in turn would code for proteins with different amino acid sequences. This would result in one segment of DNA coding for two different proteins, hence complicate the genetic information transfer machinery.

2. If two RNA molecules were produced simultaneously, double-stranded RNA complementary to each other would be formed. This would prevent RNA from being translated into proteins.

Question 66.
What do you mean by a template strand and coding strand?
Answer:
DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→5’ acts as a template, and is called the template strand. The other strand which has the polarity 5’→3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called coding strand.

Question 67.
Name the factors that are responsible for initiation and termination of transcription in prokaryotes.
Answer:
Sigma factor is responsible for initiation of transcription.
Rho factor is responsible for termination of transcription.

Question 68.
Name the major RNA types of prokaryotes and mention their role.
Answer:
In prokaryotes, there are three major types of RNAs: mRNA, tRNA, and rRNA. All three \ RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA
brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role during translation.

Question 69.
Define genetic code.
Answer:
P The order of base pairs along DNA molecule controls the kind and order of amino acids found in the proteins of an organism. This specific order of base pairs is called genetic code.

Question 70.
Explain Wobble hypothesis.
Answer:
Wobble Hypothesis is proposed by Crick (1966) which states that tRNA anticodon has the t. ability to wobble at its 5’ end by pairing with even non-complementary baSe of mRNA codon.’ According to this hypothesis, in codon-anticodon pairing the third base may not be complementary. The third base of the codon is called wobble base and this position is called wobble position. The actual base pairing occurs at first two positions only. The importance of Wobbling hypothesis is that it reduces the number of tRNAs required for polypeptide synthesis and it overcomes the effect of code degeneracy.

Question 71.
Explain the nature of eukaryotic ribosomes.
Answer:
The ribosomes of eukaryotes (80 S) are larger, consisting of 60 S and 40 S subunits. ‘S’ denotes the sedimentation efficient which is expressed as Svedberg unit (S). The larger subunit in eukaryotes consist of a 23 S RNA and 5Sr RNA molecule and 31 ribosomal proteins. The smaller eukaryotic subunit consist of 18Sr RNA component and about 33 proteins.

Question 72.
Expand and define ORF.
Answer:
Any sequence of DNA or RNA, beginning with a start codon and which can be translated into a protein is known as an Open Reading Frame (ORF).

Question 73.
What are the components of the initiation complex of prokaryotic translation?
Answer:
Initiation of translation in E. coli begins with the formation of an initiation complex, consisting of the 30S subunits of the ribosome, a messenger RNA and the charged N-formyl methionine tRNA (f^met -t RNA f^met), three proteinaceous initiation factors (IF 1, IF2, IF3), GTP (Guanine Tri Phosphate) and Mg ²⁺.

Question 74.
Explain the components of operon.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls transcriptional activity of the structural gene.

• The structural gene codes for proteins, rRNA and tRNA required by the cell.
• Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
• The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

12th Bio Zoology Guide Molecular Genetics Five Marks Questions and Answers

Question 75.
Describe the Hershey and chase experiment. What is concluded by their experiment?
Answer:
Alfred Hershey and Martha Chase (1952) conducted experiments on bacteriophages that infect bacteria. Phage T₂ is a virus that infects the bacterium Escherichia coli. When phages (virus) are added to bacteria, they adsorb to the outer surface, some material enters the bacterium, and then later each bacterium lyses to release a large number of progeny phage. Hershey and Chase wanted to observe whether it was DNA or protein that entered the bacteria. All nucleic acids contain phosphorus, and contain sulphur (in the amino acid cysteine and methionine).

Hershey and Chase designed an experiment using radioactive _ isotopes of Sulphur (³⁵S) and phosphorus (³²P) to keep separate track of the viral protein and nucleic acids during the infection process. The phages were allowed to infect bacteria in culture medium which containing the radioactive isotopes ³⁵S or ³²P. The bacteriophage that grew in the presence of ³⁵S had labelled proteins and bacteriophages grown in the presence of ³²P had labelled DNA. The differential labelling thus enabled them to identity DNA and proteins of the phage. Hershey and Chase mixed the labelled phages with unlabeled E. coli and allowed bacteriophages to attack and inject their genetic material.

Soon after infection (before lysis of bacteria), the bacterial cells were gently agitated in a blender to loosen the adhering phase particles. It was observed that only ³²P was found associated with bacterial cells and ³⁵S was in the surrounding medium and not in the bacterial cells. When phage progeny was studied for radioactivity, it was found that it carried only ³²P and not ³⁵S. These results clearly indicate that only DNA and not protein coat entered the bacterial cells. Hershey and Chase thus conclusively proved that it was DNA, not protein, which carries the hereditary information from virus to bacteria.

Question 76.
Explain the properties of DNA that makes it an ideal genetic material.
Answer:
Self Replication: It should be able to replicate. According to the rule of base pairing and complementarity, both nucleic acids (DNA and RNA) have the ability to direct duplications. Proteins fail to fulfill this criteria.

Stability: It should he stable structurally and chemically. The genetic material should be stable enough not to change with different stages of life cycle, age or with change in physiology of the organism. Stability as one of property of genetic material was clearly evident in Griffith’s transforming principle. Heat which killed the bacteria did not destroy some of the properties of genetic material. In DNA the two strands being complementary, if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2’ OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

Information storage: It should be able to express itself in the form of ‘Mendelian characters’. RNA can directly code for protein synthesis and can easily express the characters. DNA, however depends on RNA for synthesis of proteins. Both DNA and RNA can act as a genetic material, but DNA being more stable stores the genetic information and RNA transfers the genetic information.

Variation through mutation: It should be able to mutate. Both DNA and RNA are able to mutate. RNA being unstable, mutates at a faster rate. Thus viruses having RNA genome with shorter life span can mutate and evolve faster. The above discussion indicates that both RNA and DNA can function as a genetic material. DNA is more stable, and is preferred for storage of genetic information.

Question 77.
How the DNA is packed in a eukaryotic cell? ft
Answer:
In eukaryotes, organization is more complex. Chromatin is formed by a series of repeating units called nucleosomes. Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere. The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical * nucleosome contains 200 bp of DNA helix.

The histone octameres are in close contact and DNA is coiled on the outside of nucleosome. Neighbouring nucleosomes are connected by linker DNA (HI) that is exposed to enzymes. The DNA makes two complete turns around the histone octameres and the two turns are sealed off by an HI molecule. Chromatin lacking HI has a beads-on-a-string appearance in which DNA inters and leaves the nucleosomes at random places.

HI of one nucleosome can interact with 33l of the neighbouring nucleosomes resulting in the further folding of the fibre. The chrof&atin fiber in interphase nuclei and mitotic chromosomes have a diameter that vary between 200-300 nm and represents inactive chromatin. 30 nm fibre arises from the folding of nucfeosbme, chains into a solenoid structure having six nucleosomes per turn. This structure is stabilized by interaction between different HI molecules. DNA is a solenoid and packed about,%)_folds.

The hierarchical nature of chromosome structure is illustrated. Additional set of^pteins are required for packing of chromatin at higher level and are referred to as non-histone chromosomal proteins (NHC). In*,a typical nucleus, some regions of chromatin are Ibosely packed (lightly stained) and are referred to as euchromatin. The chromatin that is,-tightly packed (stained darkly) is called heterochromatin. Euchromatin is transcriptionally active and heterochromatin is transcriptionally inactive.

Question 78.
Meselson and Stahl’s experiment proved the semi-coflBptervation mode of DNA replication. Explain.
Answer:
The mode of DNA replication was determined in 1958 by Meselson and Stahl. They designed an experiment to distinguish between semi-conservative, conservative and dispersive replications. In their experiment, they grew two cultures of E.coli for many generations in separate media. The ‘heavy’ culture was grown in a medium in which the nitrogen source (NH₄Cl) contained the heavy isotope ₁₅N and the ‘ light’ culture was grown in a medium in which the nitrogen source contained light isotope ₁₄N for many generations.

At the end of growth, they observed that the bacterial DNA in the heavy culture contained only ₁₅N and in the light culture only ₁₅N. The heavy DNA could be distinguished from light DNA (₁₅N from ₁₄N) with a technique called Cesium Chloride (CsCl) density gradient centrifugation. In this process, heavy and light DNA extracted from cells in thtytwo cultures settled into two distinct and separate bands (hybrid DNA). ;

The heavy culture (₁₅N) was then transferred into a medium that had only NH₄Cl, and took samples at various definite time intervals (20 minutes duration). After the first replication, they extracted DNA and subjected it to density gradient centrifugation. The DNA settled into a band that was intermediate in position between the previously determined heavy and light bands. After the second replication (40 minutes duration), they again extracted DNA samples, and this time found the DNA settling into two bands, one at the light band position and one at intermediate position. These results confirm Watson and Crick’s semi-conservative replication hypothesis.

Question 79.
Give a detailed account of a transcription unit.
Answer:
A transcriptional unit in DNA is defined by three regions, a promoter, the structural gene and a terminator. The promoter is located towards the 5 ’ end. It is a DNA sequence that provides binding site for RNA polymerase. The presence of promoter in a transcription unit, defines the template and coding strands. The terminator region located towards the 3’ end of the coding strand contains a DNA sequence that causes the RNA polymerase to stop transcribing.

In eukaryotes the promoter has AT-rich regions called TATA box (Goldberg- Hogness box) ‘ and in prokaryotes this region is called Pribnow box. Besides promoter, eukaryotes also require an enhancer.

The two strands of the DNA in the structural gene of a transcription unit have opposite polarity. DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→5’ acts as a template, and is called the template strand. The other strand which has the polarity 5 ’→3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called the coding strand (Fig. 5.7).

The structural gene may be monocistronic (eukaryotes) or polycistronic (prokaryotes). In eukaryotes, each mRNA carries only a single gene and encodes information for only a single protein and is called monocistronic mRNA. In prokaryotes, clusters of related genes, known as operon, often found next to each other on the chromosome are transcribed together to give a single mRNA and hence are polycistronic.

Question 80.
Explain the transcription process in prokaryotes with the needed diagram.
Answer:
In prokaryotes, there are three major types of RNAs: mRNA, tRNA, and rRNA.
All three RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play a structural and catalytic role during translation. There is a single DNA-dependent RNA polymerase that catalyses the transcription of all types of RNA. It binds to the promoter and initiates transcription (Initiation).

The polymerases binding sites are called promoters. It uses nucleoside triphosphate as substrate and polymerases in a template depended fashion following the rule of complementarity. After the initiation of transcription, the polymerase continues to elongate the RNA, adding one nucleotide after another to the growing RNA chain. Only a short stretch of RNA remains bound to the enzyme, when the polymerase reaches a terminator at the end of a gene, the nascent RNA falls off, so also the RNA polymerase.

The RNA polymerase is only capable of catalyzing the process of elongation. The RNA polymerase associates transiently with initiation factor sigma (a) and termination factor rho (p) to initiate and terminate the transcription, respectively. Association of RNA with these factors instructs the RNA polymerase either to initiate or terminate the process of transcription.

In bacteria, since the mRNA does not require any processing to become active and also since transcription and translation take place simultaneously in the same compartment (since there is no separation of cytosol and nucleus in bacteria), many times the translation can begin much before the mRNA is fully transcribed. This is because the genetic material is not ‘ separated from other cell organelles by a nuclear membrane consequently; transcription and translation can be coupled in bacteria.

Question 81.
Write the salient features of genetic code.
Answer:
The salient features of genetic code are as follows:

• The genetic codon is a triplet code and 61 codons code for amino acids and 3 codons do not code for any amino acid and function as stop codon (Termination).
• The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino
• acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial, and chloroplast genomes. However, similarities are more common than differences.
• A non-overlapping codon means that the same letter is not used for two different codons. For instance, the nucleotide sequence GUTJ and GUC represents only two codons.
• It is comma less, which means that the message would be read directly from one end to the other i.e., no punctuation are needed between two codes.
• A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.
• Non-ambiguous code means that one codon will code for one amino acid.
• The code is always read in a fixed direction i.e. from 5’ → 3’ direction called polarity.
• AUG has dual functions. It acts as an initiator codon and also codes for the amino acid methionine.
• UAA, UAG (tyrosine), and UGA (tryptophan) codons are designated as termination (stop) codons and also are known as “non-sense” codons.

Question 82.
Mutations on genetic code affect the phenotype. Describe with an example.
Answer:
The simplest type of mutation at the molecular level is a change in nucleotide that substitutes one base for another. Such changes are known as base substitutions which may occur spontaneously or due to the action of mutagens. A well studied example is sickle cell anaemia in humans which results from a point mutation of an allele of (3-haemoglobin gene (PHb).

A haemoglobin molecule consists of four polypeptide chains of two types, two a chains and two β-chains. Each chain has a heme group on its surface. The heme groups are involved in the binding of oxygen. The jruman blood disease, sickle cell anaemia is due to abnormal haemoglobin. This abnormality in haemoglobin is due to a single base substitution at the sixth codon of the beta globingene from GAG to GTG in β -chain of haemoglobin. It results in a change of amino acid gluconic acid to valine at the 6th position of the β -chain. This is the classical example of point mutation that results in the change of amino acids residue glutamic acid to valine. The mutant haemoglobin undergoes polymerisation under oxygen tension causing the change in the shape of the RBC from biconcave to a sickle shaped structure.

Question 83.
Explain the mechanism of AteArperon of the E-coli.
Answer:
The Lac (Lactose) operon: The metabolism of lactose in E.coli requires three enzymes – permease, β-galactosidase (P-gat) and transacetylase. The enzyme permease is needed for entry of lactose into the cell, Pjgglactosidase brings about hydrolysis of lactose to glucose and galactose, while transacetylqSSgtransfers acetyl group from acetyl Co A to P-galactosidase. The lac operon consists of one-regulator gene (T gene refers to inhibitor) promoter sites (p), and operator site (o). Besides these, it has three structural genes namely lac z, y and lac a. The lac ‘z’ gene codes for P-gaiaqtttsidase, lac ‘y’ gene codes for permease and ‘a’ gene codes for transacetylase.

Jacob and Monod proposed the classical model of Lac operon to explain gene expression and regulation in E.coli. In lac a polycistronic structural gene is regulated by a common promoter and regulatory genfc When the cell is using its normal energy source as glucose, the ‘i’ gene transcribes a repressor mRNA and after its translation, a repressor protein is produced. It binds to the operator region of the operon and prevents translation, as a result, β-galactosidase is not produced. In the absence of preferred carbon source such as glucose, if lactose is available as an energy source for the bacteria then lactose enters the cell as a result of permease enzyme. Lactose acts as an inducer and interacts with the repressor to inactivate it. The repressor protein binds to the operator of the operon and prevents RNA polymerase from transcribing the operon. In the presence of inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase to bind to the promotor site and transcribe the operon to produce lac mRNA which enables formation of all the required enzymes needed for lactose metabolism. This regulation of lac operon by the repressor is an example of negative control of transcription initiation.

Question 84.
What are the objectives of the Human Genome project?
Answer:

• The main goals of the Human Genome Project are as follows:
•  Identify all the genes (approximately 30000) in human DNA.
• Determine the sequence of the three billion chemical base pairs that make up the human DNA.
• To store this information in databases.
• Improve tools for data analysis.
• Transfer related technologies to other sectors, such as industries.
• Address the ethical, legal and social issues (ELSI) that may arise from the project.

Question 85.
Write the salient features of the Human Genome Project.
Answer:

• Although human genome contains 3 billion nucleotide bases, the DNA sequences that encode proteins make up only about 5% of the genome.
• An average gene consists of 3000 bases, the largest known human gene being dystrophin with 2.4 million bases.
• The function of 50% of the genome is derived from transposable elements such as LINE Molecular Genetics 315 and ALU sequence.
• Genes are distributed over 24 chromosomes. Chromosome 19 has the highest gene density. Chromosomes 13 and the Y chromosomes have the lowest gene densities.
• The chromosomal organization of human genes shows diversity.
• There may be 35000-40000 genes in the genome and almost 99.9 nucleotide bases are exactly the same in all people.
• Functions for over 50 percent of the discovered genes are unknown.
• Less than 2 percent of the genome codes for proteins.
• Repeated sequences make up very large portion of the human genome. Repetitive sequences have no direct coding functions but they shed light on chromosome structure, dynamics and evolution (genetic diversity).
• Chromosome 1 has 2968 genes, whereas chromosome ’Y’ has 231 genes.
• Scientists have identified about 1.4 million locations, where single-base DNA differences (SNPs – Single nucleotide polymorphism – pronounce as ‘snips’) occur in humans.
• Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease-associated sequences and tracing human history.

Question 86.
Describe the principle involved in the DNA fingerprinting technique.
Answer:
The DNA fingerprinting technique was first developed by Alec Jeffreys in 1985. The DNA of a person and fingerprints are unique. There are 23 pairs of human chromosomes with 1.5 million pairs of genes. It is a well known fact that genes are segments of DNA which differ in the sequence of their nucleotides. Not all segments of DNA code for proteins, some DNA .segments have a regulatory function, while others are intervening sequences (introns) and still others are repeated DNA sequences.

In DNA fingerprinting, short repetitive nucleotide sequences are specific for a person. These nucleotide sequences are called variable number tandem repeats (VNTR). The VNTRs of two persons generally show variations and are useful as genetic markers. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation. The bulk DNA forms a major peak and the other small peaks are referred to as satellite DNA.

Depending on base composition (A: T rich or G : C rich), length of segment and number of repetitive units, the satellite DNA is classified into many subcategories such as micro-satellites and mini¬satellites, etc. These sequences do not code for any proteins, but they form a large portion of the human genome. These sequences show a high degree of polymorphism and form the basis of DNA fingerprinting. DNA isolated from blood, hair, skin cells, or other genetic evidences left at the scene of a crime can be compared through VNTR patterns, with the DNA of a criminal suspect to determine guilt or innocence. VNTR patterns are also useful in establishing the identity of a homicide victim, either from DNA found as evidence or from the body itself.

Question 87.
Draw a flow chart depicting the steps of the DNA fingerprinting technique.
Answer:

Higher Order Thinking Skills (HOTs) Questions

Question 1.
An mRNA strand has a series of triplet codons of which the first three codons are given below
(a) AUG (b) UUU (c) UGC
(i) Name the amino acid encoded by these triplet codons.
(ii) Mention the DNA sequence from which these triplet codons would have transcribed?
Answer:
(i) AUG codes for Methionine UUU codes for Phenylalanine UGC codes for Cysteine
(ii) TAC sequence of DNA is transcribed to AUG
AAA sequence of DNA is transcribed to UUU
ACG sequence of DNA is transcribed to UGC

Question 2.
Given below are the structures of tRNA molecules which are involved in translation process. In one tRNA, codon is mentioned but not the amino acid. In another tRNA molecule, amino acid is named and not the codon. Complete the figure by mentioning the respective amino acids and codons.
Answer:

Question 3.
A DNA fragment possesses 32 adenine bases and 32 cytosine bases. How many total number of nucleotides does that DNA fragment contains? Explain.
Answer:
128 nucleotides. Adenine always pair Thymine base.
If there are 32 adenine bases then there must be 32 Thymine bases. Similarly cytosine pairs with guanine. If cytosine bases are 32 in number the guanine bases will be equal to cytosine. So it makes a total of 128 nucleotides.

Question 4.
Following is a DNA sequence representing a part of the gene TAC TCG CCC TAT UAA CCC AAA ACC TCT using this derive A.

1. The RNA transcript
2. The spliced mRNA (consider all the codons with two Aderine bases are introns)
3. The total number of amino acids coded by the mRNA

Answer:

1. RNA transcript: AUG UGC GGG AUA GGG UUU UGG AGA
2. Spliced mRNA: AUG UGC GGG GGG UUU UGG
3. 6 amino acids are coded by mRNA

Question 5.
Complete the molecular processes by naming them

1. DNA → DNA
2. mRNA → Protein
3. RNA transcript → mRNA

Answer:

1. Replication
2. Translation
3. Splicing