Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.6 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6

Question 1.
Find the absolute extrema of the following functions on the given closed interval.
(i) f(x) = x² – 12x + 10; [1, 2]
(ii) f(x) = 3x⁴ – 4x³; [-1, 2].
(iii) f(x)= 6x^{\(\frac { 4 }{ 3 }\)} – 3x^{\(\frac { 1 }{ 3 }\)}; [-1, 1]
(iv) f(x) = 2 cos x + sin 2x; [0, \(\frac { π }{ 2 }\) ]
Solution:
(i) f(x) = x² – 12x + 10;
f'(x) = 2x – 12
f'(x) = 0 ⇒ 2x – 12 = 0
x = 6 ∉ (1, 2)
Now, Evaluating f(x) at the end points x = 1, 2
f(1) = 1 – 12 + 10 = -1
f(2) = 4 – 24 + 10 = -10
Absolute maximum f(1) = -1
Absolute minimum f(2) = -10

(ii) f(x) = 3x⁴ – 4x³
f'(x) = 12x³ – 12x²
f'(x) = 0 ⇒ 12x²(x – 1) = 0
⇒ x = 0 or x = 1
[Here x = 0, 1 ∈ [-1, 2]]
Now f (-1) = 4
f(0) = 0
f(1) = -1
f(2) = 16
so absolute maximum = 16 and absolute minimum = -1

Question 2.
Find the intervals of monotonicities and hence find the local extremum for the following functions:
(i) f(x) = 2x³ + 3x² – 12x
(ii) f(x) = \(\frac { x }{ x-5 }\)
(iii) f(x) = \(\frac { e^x }{ 1-e^x }\)
(iv) f(x) = \(\frac { x^3 }{ 3 }\) – log x
(v) f(x) = sin x cos x+ 5, x ∈ (0, 2π)
Solution:
(i) f(x) = 2x³ + 3x² – 12x
f'(x) = 6x² + 6x – 12
f'(x) = 0 ⇒ 6(x² + x – 2) = 0
(x + 2)(x – 1) = 0
Stationary points x = -2, 1

Now, the intervals of monotonicity are
(-∞, -2), (-2, 1) and (1, ∞)
In (-∞, -2), f'(x) > 0 ⇒ f(x) is strictly increasing.
In (-2, 1), f'(x) < 0 ⇒ f(x) is strictly decreasing.
In (1, ∞), f'(x) > 0 ⇒ f(x) is strictly increasing.
f(x) attains local maximum as f'(x) changes its sign from positive to negative when passing through x = -2.
Local maximum
f(-2) = 2 (-8) + 3 (4) – 12 (-2)
= -16 + 12 + 24 = 20
f(x) attains local minimum as f'(x) changes its sign from negative to positive when passing through x = 1.
∴ Local minimum f(1) = 2 + 3 – 12 = -7

(ii) f(x) = \(\frac { x }{ x-5 }\)
f'(x) = \(\frac { (x-5)(1)-x(1) }{ (x-5)^2 }\) = –\(\frac { 5 }{ (x-5)^2 }\)
f'(x) = 0, which is absured
But in f(x) = \(\frac { x }{ x-5 }\)
The function is defined only when x < 5 or x > 5
∴ The intervals are (-∞, 5) and (5, ∞)
In the interval (-∞, 5), f'(x) < 0
In the interval (5, ∞), f'(x) < 0
∴ f(x) is strictly decreasing in (-∞, 5) and (5, ∞)

When x = 0, f(x) becomes undefined.
∴ x = 0 is an excluded value.
∴ The intervals are (-∞, 0) ∪ (0, ∞) in – (-∞, ∞), f'(x) > 0
∴ f(x) is strictly increasing in (- ∞, ∞) and there is no extremum.

(iv) f(x)= \(\frac { x^3 }{ 3 }\) – log x
f'(x) = x² – \(\frac { 1 }{ x }\)
f'(x) = 0 ⇒ x³ – 1 = 0 ⇒ x = 1
The intervals are (0, 1) and (1, ∞).
i.e., when x > 0, the function f(x) is defined in the interval (0, 1), f'(x) < 0
∴ f(x) is strictly decreasing in (0, 1) in the interval (1, ∞), f'(x) > 0
∴f(x) is strictly increasing in (1, ∞)
f(x) attains local minimum as f'(x) changes its sign from negative to positive when passing through x = 1
∴ Local minimum
f(1) = \(\frac { 1 }{ 3 }\) – log 1 = \(\frac { 1 }{ 3 }\) – 0 = \(\frac { 1 }{ 3 }\)

(v) f(x) = sin x cos x + 5, x ∈ (0, 2π)
f'(x) = cos 2x
f'(x) = 0 ⇒ cos 2x = 0
Stationary points
x = \(\frac { π }{ 4 }\), \(\frac { 3π }{ 4 }\), \(\frac { 5π }{ 4 }\), \(\frac { π }{ 4 }\) ∈x = (0, 2π)

In the interval (0, \(\frac { π }{ 4 }\)), f'(x) > 0 ⇒ f(x) is strictly increasing.
In the interval (\(\frac { π }{ 4 }\), \(\frac { 3π }{ 4 }\)), f'(x) < 0 ⇒ f(x) is strictly decreasing.
In the interval (\(\frac { 3π }{ 4 }\), \(\frac { 5π }{ 4 }\)), f'(x) > 0 ⇒ f(x) is strictly increasing.
In the interval (\(\frac { 5π }{ 4 }\), \(\frac { 7π }{ 4 }\)), f'(x) < 0 ⇒ f(x) is strictly decreasing.
In the interval (\(\frac { 7π }{ 4 }\), 2π), f'(x) > 0 ⇒ f(x) is strictly increasing.
f'(x) changes its sign from positive to negative when passing through x = \(\frac { π }{ 4 }\) and x = \(\frac { 5π }{ 4 }\)
∴ f(x) attains local maximum at x = \(\frac { π }{ 4 }\) and \(\frac { 5π }{ 4 }\)

f'(x) changes its sign from negative to positive when passing through x = \(\frac { 3π }{ 4 }\) and x = \(\frac { 7π }{ 4 }\)
∴ f(x) attains local maximum at x = \(\frac { 3π }{ 4 }\) and x = \(\frac { 5π }{ 4 }\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Botany Guide Pdf Chapter 7 Ecosystem Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 7 Ecosystem

12th Bio Botany Guide Ecosystem Text Book Back Questions and Answers

I. Choose the most suitable answer from the given four alternatives and write the option code and the corresponding answer.

Question 1.
Which of the following is not a abiotic component of the ecosystem?
a. Bacteria
b. Humus
c. Organic compounds
d. Inorganic compounds
Answer:
d. Inorganic compounds

Question 2.
Which of the following is / are not a natural ecosystem?
a. Forest ecosystem
b. Rice field
c. Grassland ecosystem
d. Desert ecosystem
Answer:
b. Rice field

Question 3.
Pond is a type of
a. forest ecosystem
b. grassland ecosystem
c. marine ecosystem
d. fresh water ecosystem
Answer:
d. fresh water ecosystem

Question 4.
Pond ecosystem is
a. not self sufficient and self regulating
b. partially self sufficient and self regulating
c. self sufficient and not self regulating
d. self sufficient and self regulating
Answer:
d. self sufficient and self regulating

Question 5.
Profundal zone is predominated by heterotrophs in a pond ecosystem, because of
a. with effective light penetration
b. no effective light penetration
c. complete absence of light
d. a and b
Answer:
b. no effective light penetration

Question 6.
Solar energy used by green plants for photosynthesis is only
a. 2-8%
b.2-10%
c.3-10%
d.2-9%
Ans :
b. 2 – 10%

Question 7.
Which of the following ecosystem has the highest primary productivity?
a. Pond ecosystem
b. Lake ecosystem
c. Grassland ecosystem
d. Forest ecosystem
Answer:
d. Forest ecosystem

Question 8.
Ecosystem consists of
a. decomposers
b. producers
c. consumers
d. all of the above
Answer:
d. all of the above

Question 9.
Which one is in descending order of a food chain
a. Producers →  Secondary consumers → Primary consumers→  Tertiary consumers
b. Tertiary consumers →  Primary consumers →  Secondary consumers →  Producers
c. Tertiary consumers →  Secondary consumers →  Primary consumers→  Producers
d. Tertiary consumers → Producers →  Primary consumers →  Secondary consumers
Answer:
c. Tertiary consumers →  Secondary consumers →  Primary consumers→  Producers

Question 10.
Significance of food web is/are
a. it does not maintain stability in nature
b. it shows patterns of energy transfer
c. it explains species interaction
d. b and c
Answer:
d. b and c

Question 11.
The following diagram represents

a. pyramid of number in a grassland ecosystem
b. pyramid of number in a pond ecosystem
c. pyramid of number in a forest ecosystem
d. pyramid of biomass in a pond ecosystem
Answer:
c. pyramid of number in a forest ecosystem

Question 12.
Which of the following is / are not the mechanism of decomposition
a. Eluviation
b. Catabolism
c. Anabolism
d. Fragmentation
Answer:
c. Anabolism

Question 13.
Which of the following is not a sedimentary cycle
a. Nitrogen cycle
b. Phosphorous cycle
c. Sulphur cycle
d. Calcium cycle
Answer:
a. Nitrogen cycle

Question 14.
Which of the following are not regulating services of ecosystem services
i) Genetic resources
ii) Recreation and aesthetic values
iii) Invasion resistance
iv) Climatic regulation
a. i and iii
b. ii and iv
c.iandii
d. iandiv
Answer:
c. i and ii

Question 15.
The productivity of the profundal zone will be low. Why?
Answer:
The producers of the pond ecosystem depend on phytoplankton through photosynthesis. The profundal zone lies below the limnetic zone with no effective light penetration, hence the productivity rate is very low.

Question 16.
Discuss the gross primary productivity is more efficient than net primary productivity.
Grass primary productivity (GPP)
Answer:
The grass primary productivity is the total amount of food energy (or) biomass produced by autotrophs through the process of photosynthesis without respiratory loss in the plant.

Net primary productivity (NPP)

• But the net primary productivity is the proportion of energy, which remains after respiration loss from the gross primary productivity in the plant. NPP = GPP – Respiration
• So, gross primary productivity is more efficient than net primary productivity.

Question 17.
Pyramid of energy is always upright. Give reasons
Answer:
The energy pyramid represents a successive energy flow at each trophic level in an ecosystem. There is a gradual decrease in energy transfer at successive tropic levels from producers to higher levels, hence the pyramid of energy is always upright.

Question 18.
What will happen if all producers are removed from ecosystem?
Answer:

• The removal of all the producers would cause the collapse of the entire food web.
• Primary consumers (or) herbivores, which feed on producers directly would die off because they would have nothing to eat.
• All the other animals in the food web would die too, because their food supplies would have gone.
• The population of the consumers would fall as the population of the producer fell.
• Producers are maintaining C02 and oxygen level in the atmosphere.
• If the producers are removed C02 and 02 cycle will be in imbalance in the atmosphere.
• It will leads to the depletion of respiratory gas oxygen as it is a by product of photosynthesis

Question 19.
Construct the food chain with the following data. Hawk, plants, frog, snake, grasshopper.
Answer:

• Food chain with plants → grasshopper → frog → snake → hawk
• The movement of energy from producers upto top carnivores is known as food chain.
• In this food chain energy flows from producers (plants) to primary consumers (grasshopper) to secondary consumer (frog) to tertiary consumer (snake) to predator (hawk)
• It shows linear network link.

Question 20.
Name of the food chain which is generally present in all type of ecosystem. Explain and write their significance.
Answer:
Detritus food chain is generally present in all type of ecosystem.

Detritus food chain:
This type of food chain begins with dead organic matter which is an important source of energy. A large amount of organic matter is derived from dead plants, animals, and their excreta. This type of food chain is present in all ecosystems.

The transfer of energy from the dead organic matter, is transferred through a series of organisms called detritus consumers (detritivores)- small carnivores – large (top) carnivores with repeated eating and being eaten respectively. This is called the detritus food chain.

Significance:

• The detritus (dead plants, animals and their excreta) are breakdown into simple organic matter by the decomposers.
• It is an essential process for recycling and balancing the nutrient pool in an ecosystem.

Question 21.
Shape of pyramid in a particular ecosystem is always different in shape. Explain with example.
Answer:

In a forest ecosystem the pyramid of number is spindle in shape, it is because the base (T₁) of the pyramid occupies large sized trees (Producer) which are lesser in number. Herbivores (T₂) (Fruit-eating birds, elephant and deer) occupying second trophic level, are more in number than the producers. In final trophic level (T₄), .tertiary consumers (lion) are lesser in number than the secondary consumer (T₃) (fox and snake).

Question 22.
Generally human activities are against to the ecosystem, whereas you a student how will you help to protect the ecosystem?
Answer:
To protect the ecosystem, we have to practice the following in our daily life.

• Buy and use only eco-friendly products and recycle them.
• Grow more trees.
• Choose sustained farm products (vegetables, fruits, greens, etc.) > Reduce the use of natural resources.
• Recycle the waste and reduce the amount of waste you produce.
• Reduce consumption of water and electricity.
• Reduce or eliminate the use of household chemicals and pesticides.
• Maintain your cars and vehicles properly. (In order to reduce carbon emission)
• Create awareness and educate about ecosystem protection among your friends and family members and ask them to find out solution to minimize this problem.

Question 23.
Generally in summer the forest are affected by natural fire. Over a period of time it recovers itself by the process of successions. Find out the types of succession and explain.
Answer:

1. Primary succession:
The development of plant community in a barren area where no community existed before is called primary succession. The plants which colonize first in a barren area is called pioneer species or primary community or primary colonies. Generally, Primary succession takes a very long time for the occurrence in any region.
Example: Microbes, Lichen, Mosses.

2. Secondary succession:
The development of a plant community in an area where an already developed community has been destroyed by some natural disturbance (Fire, flood, human activity) is known as secondary succession.

Generally, This succession takes less time than the time taken for primary succession. Example: The forest destroyed by fire may be re-occupied by herbs over period of times.

3. Autogenic succession:
Autogenic succession occurs as a result of biotic factors. The vegetation reacts with its environment and modifies its own environment causing its own replacement by new communities. This is known as autogenic succession.

Example: In forest ecosystem, the larger trees produce broader leaves providing shade to the forest floor area. It affects the shrubs and herbs which require more light (heliophytes) but supports the shade tolerant species (sciophytes) to grow well.

4. Allogenic succession:
Allogeneic succession occurs as a result of abiotic factors. The replacement of existing community is caused by other external factors (soil erosion, leaching, etc.,) and not by existing organisms.

Example : In a forest ecosystem soil erosion and leaching alter the nutrient value of the soil leading to the change of vegetation in that area.

5. Autotrophic succession:
If the autotrophic organisms like green plants are dominant during the early stages of succession it is called autotrophic succession, this occurs in the habitat which is rich in inorganic substances. Since, green plants dominate in the beginning of this succession, there is a gradual increase in organic matter and subsequently the energy flow in the ecosystem.

6. Heterotrophic succession:
If heterotrophic organisms like bacteria, fungi, actinomycetes, and animals are dominant during the early stages of succession it is called heterotrophic succession. Such a succession takes place in organic habitats. Since heterotrophs dominate in the beginning of such succession, there will be a gradual decrease in the energy content.

Question 24.
Draw a pyramid from following details and explain in brief.
Quantities of organisms are given – Hawks – 50, plants – 1000. rabbit and mouse – 250 +250, pythons and lizard- 100 + 50 respectively.
Answer:

• A graphical representation of the amount of organic material (biomass) present at each successive trophic level in an ecosystem is called pyramid of biomass.
• In grassland and forest ecosystems there is a gradual decrease in biomass of organisms at successive trophic levels from producers to top carnivores (tertiary consumer)
• Therefore these two ecosystems show pyramids as upright

1. No.of producers → 1000
2. No.of primary consumers → 500
3. No.of secondary consumer → 150
4. No.of the tertiary consumer is lesser than Secondary consumers (50)

Question 25.
Various stages of succession are given below. From that rearrange them accordingly. Find out the type of succession and explain in detail.
Reed-swamp stage, phytoplankton stage, shrub stage, submerged plant stage, forest stage, submerged free-floating stage, marsh meadow stage.
Answer:
Reed-swamp stage, phytoplankton stage, shrub stage, submerged plant stage, forest stage, submerged free floating stage and marsh medow stage.

(1) Phytoplankton stage – It is the first stage of succession consisting of the pioneer community like blue green algae, green algae, diatoms, bacteria, etc., The colonization of these organisms enrich the amount of organic matter and nutrients of pond due to their life activities and death. This favors the development of the next serai stages.

(2) Submerged plant stage – As the result of death and decomposition of planktons, silt brought from land by rain water, lead to a loose mud formation at the bottom of the pond. Hence, the rooted submerged hydrophytes begin to appear on the new substratum.
Example: Vallisneria and Hydrilla etc. The death and decay of these plants will build up the substratum of pond to become shallow.

(3) Submerged free floating stage – During this stage, the depth of the pond will become almost 2-5 feet Hence, the rooted hydrophytic plants and with floating large leaves start colonising the pond.
Example: Rooted floating plants like Nelumbo, Nymphaea and some free floating species like Azolla, and Pistia are also present in this stage. By death and decomposition of these plants, further the pond becomes more shallow.

(4) Reed-swamp stage – It is also called an amphibious stage. During this stage, rooted floating plants are replaced by plants which can live successfully in aquatic as well as aerial environment.
Example: Typha, Phragmites, Sagittaria and Scirpus etc. At the end of this stage, water level is very much reduced, making it unsuitable for the continuous growth of amphibious plants.

(5) Marsh meadow stage – When the pond becomes swallowed due to decreasing water level, species of Cyperaceae and Poaceae colonise the area. They form a mat-like vegetation with the help of their much branched root system. This leads to an absorption and loss of large quantity of water. At the end of this stage, the soil becomes dry and the marshy vegetation disappears gradually and leads to shurb stage.

(6) Shrub stage – Here areas are invaded by terrestrial plants like shrubs (Salix and Comus) and trees (Populus and Alnus). These plants absorb large quantity of water and make the habitat dry. Further, the accumulation of humus with a rich flora of microorganisms produce minerals in the soil, ultimately favouring the arrival of new tree species in the area.

(7) Forest stage – It is the climax community of hydrosere. A variety of trees invade the area and develop any one of the diverse type of vegetation.
Example.Temperate mixed forest (Ulmus, Acer and Quercus), Tropical rain forest (Artocarpus and Cinnamomum ) and Tropical deciduous forest (Bamboo and Tectona).

12th Bio Botany Guide Ecosystem Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
The most stable and productive ecosystem seen on the earth is _____
a. Mangrove ecosystem
b. Grassland ecosystem
c. Pond ecosystem
d. Forest ecosystem
Answer:
d. Forest ecosystem

Question 2.
In an ecosystem, the energy flow is always
a. Unidirectional
b. Top to bottom
c. Chain form
d. Multi directional
Answer:
a. Unidirectional

Question 3.
Grass 2 Goat 2 Man. This is the grazing food chain choose the correct option.
a. Goat is primary producer, secondary trophic level and herbivore
b. Grass is a primary producer, herbivore, and first trophic level.
c. Goat is a primary consumer, herbivore, second trophic level
d. Goat is a primary consumer, herbivore, first trophil level
Answer:
c. Goat is a primary consumer, herbivore, second trophic level

Question 4.
Choose the incorrect statement from following
a. Pyramid of energy is always upright
b. In grass land and forest ecosystem pyramid of biomass as upright.
c. Pyramid of number shows three different shape of pyramids like upright, spindle and inverted
d. Food web is used for the construction of ecological pyramid.
Answer:
d. Food web is used for the construction of ecological pyramid.

Question 5.
Assertion : If the decomposers were removed completely from the ecosystem the functioning of ecosystem will be adversely affected.
Reason : The cycling of nutrients between abiotic and biotic components will be blocked
a. A and R are correct
b. A and R are not correct
c. A is correct R is wrong
d. R is not a correct explanation for A
Answer:
a. A and R are correct

Question 6.
Assertion : An ecological pyramid is a diagrammatic (or) graphic representation of the trophic structure and function.
Reason : Various trophic levels of a food chain are considered in ecological pyramid.
a. Both are not correct
b. R is not related to A
c. Both A and R are wrong
d. R is the correct explanation of A
Answer:
d. R is the correct explanation of A

Question 7.
Match the following
Read this statement and fill it with correct (A) and (B)

Carbon stored in fossil fuel is ……(A)……….. and carbon stored in the biosphere is …………..(B)……..
The only one correct option for the two blank is

Answer:
b. Grey carbon – Green carbon

Question 8.
Match the following and find the correct answer

a. (i) – B; (ii) – C; (iii) – D; (iv) – A
b. (i) – C; (ii) – D; (iii) – B; (iv) – A
c. (i) – B; (ii) – A; (iii) – D; (iv) – C
d. (i) – D; (ii) – C; (iii) – B; (iv) – A
Answer:
d. (i) – D; (ii) – C; (iii) – B; (iv) – A

Question 9.
Choose the incorrect pair

Answer:
d. Fragmentation – Release of inorganic nutrients from the humus.

Question 10.
Choose the correct pair

a. B and C
b. A and C
c. B, C and D
d. A, B and D
Answer:
d. A, B and D

Question 11.
The plants which colonize first in a barren …………..
area is called
a. Pioneers
b. Serai
c. Autogenic
d. Allogenic
Answer:
a. Pioneers

Question 12.
The term ‘ecosystem’ was proposed by……………..in the year 1935.
a. A.G. Hoxley
b. A.G.Tansley
c. Odum
d. Lindeman
Answer:
b. A.G. Tansley

Question 13.
The position of organisms in food chain is refers to
a. Ecosystem
b. Trophic level
c. Food chain
d. Ten percent law
Answer:
b. Trophic level

Question 14.
Temperate mixed forest, tropical rain forest tropical deciduous forest are
a. Climax communities
b. T ertiary communities
c. Primary communities
d. Secondary communities
Answer:
a. Climax communities

Question 15.
Assertion and Reason
Assertion : Photosynthetically active radiation (PAR) which is between the range of 400 – 700 nm wave length.
Reason : At night PAR is zero and during midday in the summer, PAR often reaches 2000 – 3000 millimoles/square meter/second
a. R does not explains A
b. R explaining A
c. A is correct R is wrong
d. Both A and R are wrong
Answer:
a. R does not explains A

Question 16.
Choose the correct statement:

Question 17.
Choose the correct statement
a. Only 2 – 10% of solar energy is used by green plants for photo synthesis
b. Only 56% of the solar energy is used by green plants for photosynthesis.
c. Productivity can be expressed in terms of kcal / m2 / 10 year
d. Limnology is the study about ocean.
Answer:
a. Only 2 – 10% of solar energy is used by green plants for photo synthesis

Question 18.
Blue carbon ecosystems is related to …………….
a. Carbon sequestration
b. Productivity
c. Visibility
d. Phosphorus cycle
Answer:
a) Carbon sequestration

Question 19.
Choose the correct pair related to this statement
Secondary productivity can be defined as
a. Amount of energy in the tissues of consumer (or) heterotrophs
b. Amount of biomas formation
c. Rate of energy formation
d. Rate of energy utilization
Answer:
a. Amount of energy in the tissues of consumer (or) heterotrophs

Question 20.
Which one of the following will be the shape of the pyramid. If you consider the following statement.
“No. of fruit eating birds, elephant, deer depends on large sized tree (producer) which are lesser in number and lesser number of secondary consumer (fox and snake) and final trophic level tertiary consumer lion.

Answer:
d.

Question 21.
Choose the odd one out
a. Flagship species
b. Rehabilitation
c. Maintaining biodiversity
d. Anthropogenic activities
Answer:
d. Anthropogenic activities

Question 22.
Ecological succession refers to …………………
a. Gradual, fairly changes and pH development of a given area
b. Linking of ecosystem
c. Energy transfer
d. Biotic communities
Answer:
a. Gradual, fairly changes and pH development of a given area

Question 23.
Succession occur in which area ?
Answer:
Flooded, earthquake and anthropogenic area

Question 24.
Plant succession in saline water is…………
Answer:
Halosere

Question 25.
The replacement of existing community by other factors like soil erosion, leaching etc. is …………..
Answer:
Allogenic succession.

Question 26.
The type of succession takes less time to the time taken for primary succession ?
Answer:
Secondary succession

Question 27.
The type of succession in which organisms like bacteria, fungi, actinomycetes and animals are dominant during its early stages is …………..
a. Heterotrophic
b. Allogenic
c. Autotrophic
d. Autogenic
Answer:
a. Heterotrophic

Question 28.
Match the following:

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

Question 29.
Pioneer community like blue green algae, green algae, diatoms, bacteria etc are present in ……….. stage of hydrosere.
Answer:
Phytoplankton stage

Question 30.
Rooted floating plants like Nelumbo Nymphaea and Trapa and free floating SPS wolffia and lemna are present in this stage is
Answer:
Submerged free floating stage

Question 31.
Submerged hydrophytes like char a, utricularia, vallisneria, hydrilla are present in the ………… stage of hydrosere.
Answer:
Submerged plant.

Question 32.
Mat – like vegetation with the help of much branched root system is the character of ………….. stage of hydrosere
Answer:
Marsh meadow.

Question 33.
Species of cyperaceae and poaceae like carex, juncus cyperus and eleocharis colonise in this i area is
Answer:
Marsh meadow stage

Question 34.
Salix and cornus (shrubs) and trees like populous and alnus are present in the …………… stage.
Answer:
Shrub stage

Question 35.
Say true or false
Reduce, reuse, recycle are “three R” s for waste management.
Answer:
True

Question 36.
…………… is the best example for urban eco restoration in the state of Tamilnadu.
Answer:
Urban ecosystem restoration model.

Question 37.
Initiation of plant succession on barren rock is …………………
Answer:
Lithosere

Question 38.
Succession with minimal amount of water is
a. Merosere
b. Psammosere
c. Halosere
d. Xerosere
Answer:
d. Xerosere

Question 39.
Succession in a fresh water ecosystem is
a. Merosere
b. Hydrosere
c. Lithosere
d. Halosere
Answer:
b. Hydrosere

Question 40.
Study of interaction between living and non- living components is
a. Biomass
b. Food chain
c. Food web
d. Ecosystem
Answer:
d. Ecosystem

Question 41.
Which of the following is not related to photosynthetic active radiation.
a. 400-700 nm
b. 10% is held by ozone
c. 2-10% by green plants
d. 46% reaches earth surface
Answer:
d. 46% reaches earth surface

Question 42.
Which of the following light is efficient for
photo-synthesis.
a. Blue and red
b. Blue and green
c. Blue and white
d. Blue and violet
Answer:
a. Blue and red

Question 43.
Which of the following determine the productivity of different ecosystem.
a. Fixation of radiant energy
b. Population
c. Size of ecosystem
d. Number of plants
Answer:
a. Fixation of radiant energy

Question 44.
Which is the right representation of detritus food chain?
a. Grass Earthworm → Black bird → Hawk
b. Grass Mouse → Snake Eagle
c. Fallen leaves → Earthworm → Black bird →Hawk
d. Plants → Rabbit → Snake → Eagle
Answer:
c. Fallen leaves → Earthworm → Black bird →Hawk

Question 45.
NPP of whole biosphere is estimated to be …………. about billion tons dry weight/year
a. 140
b. 170
c. 150
d.160
Answer:
b. 170

Question 46.
NPP of oceanic producers is only ……………… billiob tons/year in unit time.
a. 55
b. 45
c. 56
d. 54
Answer:
a. 55

Question 47.
During energy transfer from one trophic level to other only 10% stored at every level this is related to
a. First law of thermodynamics
b. Second law of thermodynamics
c. Ten percent law
d. Law of thermodynamics
Answer:
c. Ten percent law

Question 48.
The commonly occurring pioneer species in
xerich succession
a. Lichens
b. Mosses
c. Bryophytes
d. Pteridophytes
Answer:
a. Lichens

Question 49.
Lowest productivity is seen in
a. Ocean
b. Grass land
c. forest
d. savannah
Answer:
a. Ocean

Question 50.
The term biosphere is used for
a. Ecosystem
b. Plants and animals
c. All living organism
d. The part of the earth with life
Answer:
d. The part of the earth with life

Question 51.
Bio geo chemical cycle refers to
a. Cycling of nutrients
b. Cycling of nutrients with ecosystem
c. Water cycle
d. Cycling of chemicals substrances
Answer:
b. Cycling of nutrients with ecosystem

Question 52.
Which of the following does not contain phosphorous?
a. Phospholipids
b. DNA, RNA
c. ATP, NADP
d. Respiration
Answer:
d. Respiration

Question 53.
The ecosystem with (or) without human interference is
a. Terrestrial
b. Natural
c. Artificial
d. Lotic
Answer:
b. Natural

Question 54.
Examine the ecological pyramid given below and select the type of it represent

a. Upright pyramid of biomass
b. Upright pyramid of number
c. Inverted pyramid of biomass
d. Inverted pyramid of number
Answer:
a. Upright pyramid of biomass

Question 55.
The organisms which eat both plants and animals are called ……………….
Answer:
Omnivores

Question 56.
When sparrow eats insects and worms it is a
a. Primary consumer
b. Secondary consumer
c. Tertiary consumer
d. Carnivore
Answer:
b. Secondary consumer

Question 57.
Imagine number of fruit eating birds and insects feeding or a big tree what kind of pyramid would be ?
a. Inverted pyramid of number
b. Inverted pyramid of energy
c. Upright pyramid
d. Upright pyramid of energy
Answer:
a. Inverted pyramid of number

Question 58.
Which of the following is natural ecosystem.
a. Pond and lake
b. Rice field
c. Maize fi eld
d. Aquarium
Answer:
a. Pond and lake

Question 59.
Which one of the following food chain refers inverted pyrmid? Q23D
a. Grasses → Rats → Snake → Hawk
b. Banyan tree → Birds → Beetles → Fungi
c. Phytoplanktons → Zooplanktons → Fishes → snakes
d. Plants Rabbits → Fox → Hawk
Answer:
c. Phytoplanktons → Zooplanktons → Fishes → snakes

Question 60.
The quantity of energy present in the universe is constant it is related to
a. Second law of thermodynamics
b. First law of thermodynamics
c. Ten percent lawd. Law of thermodynamics
Answer:
b. First law of thermodynamics

II. Two Marks

Question 1.
What do you mean standing quality (or) Standing state of abiotic components ?
Answer:

• Abiotic components play vital role in an ecosystem.
• The total inorganic substances present in any ecosystem at a given time is called standing quality (or) standing state.

Question 2.
In most of the ecosystem, which one is called autotrophs why?
Answer:

• Autotrophs are organism which can manufacture the organic compounds from simple inorganic components through a process called photo synthesis.
• Eg. Green plants. They are called as producers.

Question 3.
What is standing crop?
Answer:

• The amount of living materials present in a population at a given time is known as standing crop.
• Which is expressed in terms of number (or) biomass per unit area.

Question 4.
What are the functions of an ecosystem?
Answer:

• Its functions are energy creation and sharing of energy.
• It is the way to cycling of materials between the living and nonliving component of the ecosystem.

Question 5.
Why biomass production is called productivity of an ecosystem?
Answer:

• The rate of biomass production per unit area in a unit time is called productivity.
• It can be expressed in terms of gm / m2 / yr (or) kcal / m2 / yr

Question 6.
What is biomass? How it is measured?
Answer:

• The total quantity (or) weight of organism in a give area is called biomass.
• It can be measured as fresh weight (or) dry weight (or) carbon weight of organisms.

Question 7.
Write the difference between gross primary productivity and gross secondary productivity ?
Answer:
Grass Primary Productivity :

1. The total amount of food energy (or) biomass produced in an ecosystem
2. It is done by autotrophs through the process called photosynthesis

Grass Secondary Productivity :

1. The total amount of plant material ingested minus (-) materials lost as faeces.
2. It is done by herbivores by the process called ingestion.

Question 8.
Differentiate net primary productivity from net secondary productivity ?
Answer:
Net Primary Productivity :

1. The proportion of energy which remains after respiration loss in the plant.
2. Net primary productivity is calculated in producers.

Net Secondary Productivity:

1. Stored energy (or) biomass per unit area per unit time after respiratory loss.
2. Net secondary productivity is calculated in consumers.

Question 10.
Which factors may affect the primary productivity of plants (or) producers ?
Answer:
Primary productivity of plants affected by

• Plant species of an area.
• Photosynthetic capacity
• Availability of nutrients
• Solar radiation
• Precipitation
• Soil type
• opographic factors
• Environmental factors.

Question 11.
How does NPP calculated from GPP?
Answer:

• NPP = GPP – Respiration
• Thus it is the difference between gross primary productivity and respiration is net primary productivity.

Question 12.
Why grey carbon different from brown carbon?
Answer:

• Grey carbon is stored in fossil fuel like coal, oil and biogas deposits in the lithosphere
• Brown carbon is stored in industrialized forest wood used in making commercial articles.

Question 13.
What are the sources for blue carbon and black carbon
Answer:

• Blue carbon is stored in the atmosphere.
• Black carbon is emitted from diesel engine gas and coal fired power plants.

Question 14.
What is energy flow ?
Answer:

• The transfer of energy in an ecosystem between trophic levels can be termed as energy flow.
• It is the key function in an ecosystem.

Question 15.
If crow is absent in an ecosystem what would happen?
Answer:

• The crow has omnivorous type of Nutrition.
• They eat cereals, fruit and seeds, small insects and worms.
• They eat dead & decaying animals body and keep the environment clean, so known as scavengers of the sky.
• They occupy several trophic levels in the food chain.
• When they eat some fruits they swallow the seeds which are dispersed along with the excreta.
• However unlike certain pollination and insects they are not considered vital organisms because their loss from the ecu system mav cause drastic impacts, leading to the extinction of rhe. it partk ular plant species which can’t be pollinated in the absence of the particular insect/bird.
• However when crow is absent, then its niche in the food chain will be kept vacant ecological, this will be drastic change occur some other organism mav evolve to occupy the same neche in the long run.

Question 17.
Conversion of light energy into chemical energy which law is related to this ?
Answer:

The first law of thermodynamics is related to this statement.

Question 18.
Construct the diagrammatic representation of grazing food chain.
Answer:

Question 19.
Draw the diagrammatic representation of detritus food chain.
Answer:

Question 20.
Food web is known as basic unit of ecosystem. Why ?
Answer:

• The interlocking pattern of a number of food chain form a weblike arrangement called food web.
• It is known as basic unit of ecosystem to maintain its stability in nature.
• It is called homeostasis.

Question 21.
Why ecological pyramids are called Eltonianpyramids ?
Answer:

• The concept of ecological pyramids was introduced by Charles Elton in the year 1927.
• Thus ecological pyramids are called Eltonian pyramids.

Question 22.
What is ecological pyramid ?
Answer:
Graphic representation of the trophic structure and function at successive trophic levels of an ecosystem is called ecological pyramid.

Question 23.
Why the pyramids of energy is always upright?
Answer:

• There is a gradual decrease in energy transfer at successive tropic levels from producers to the upper levels.
• Therefore, the pyramid of energy is always upright.

Question 24.
Define pyramid of energy.
Answer:
The graphical representation of energy flow of each successive trophic level in an ecosystem is called pyramids of energy.

Question 25.
Name the process which is essential for recycling and balancing the nutrient in an ecosystem?
Answer:
Decomposition is a process in which the detritus are breakdown into simple organic matter by the decomposers.

Question 26.
Write the difference between limnology and oceanography ?
Answer:
Limnology
It is the studv of biological, chemical physical and geological components of inland fresh water aquatic ecosystem (ponds, lakes etc)

Oceanography :
It is the study of biological, chemical, physica I and geological components of ocean.

Question 27.
Blue carbon ecosystems are very important to nature. Why?
Answer:

• Sea grasses and mangroves of estuarine and coastal ecosystems are the most efficient in carbon sequestration.
• Hence these ecosystems are called blue carbon ecosystem.

Question 28.
What is carbon sequestration?
Answer:
Carbon sequestration is the process of capturing and storing carbondioxide from the atmosphere in carbon sinks such as ocean and forest.

Question 29.
Odd one out and give reason.
Answer:
Provision of habitat, nutrient recycling, primary production, succession Odd one out: Succession.
Reason : Replacement of one type of plant community by other of the same place is known as succession. While others are supporting services of ecosystem services.

Question 30.
Odd one out and give reason.
Lichen, blue green algae, green algae diatoms, bacteria.
Answer:
Odd one out: Lichen
Reason : Lichen is related to primary succession. While others are related to phytoplankton stage of hydrosere.

Question 31.
Write the slogan for the safety of the environment.
Answer:
“Use Ecosystem. But don’t lost ecosystem. Make it sustainable.”

Question 32.
Define fragmentation.
Answer:
The breaking down of detritus in to smaller particles by detritivores like bacteria, fungi and earth worm is known as Fragmentation.

Question 33.
What is Humification?
Answer:
It is a process by which simplified detritus is changed into dark coloured amorphous substance called humus.

Question 34.
Define oceanography.
Answer:
It is the study of biological chemical, physical and geogical components of ocean.

Question 35.
In different food chains of different ecosystem the placement of man is not mentioned. You give placement in a suitable food chain and give reason for your answer.
Answer:

• Placement of man in the food chain is not clear, due to the various dietary choices of each human.
• Many are omnivores – consuming both plants & meat.
• They may be kept on the 3rd or even on the 4th trophic level.
• Meat eaters (cow, goat are herbivores) are a part of 3rd trophic level.
• If you were to eat Salmon, – Salmon consume other fish – and so you are in the 4th trophic

III. Three Marks

Question 1.
What is photosynthetically active radiation ?
Answer:

• The amount of light available for photosynthesis of plants is called photosynthetically active radiation.
• Which is between the range of 400 – 700 nm of wave length.
• Generally plants absorb move blue and red light for efficient photosynthesis.
• Only 2 -10% of the solar energy is used by green plants for photosynthesis.

Question 2.
Differentiate green carbon from grey carbon.
Answer:
Green Carbon :

1. It is the carbon stored in the biosphere.
2. It is done by the process of photosynthesis by green plants.

Grey Carbon :

1. It is the carbon stored in fossil fuel. (Coal, oil, biogas)
2. It is done by the process of decomposition in the lithosphere.

Question 3.
Differentiate primary productivity from secondary productivity.
Answer:

Question 4.
“The energy transformation results in the education of the free energy of the system”. What does it states ?
Answer:

• It states second law of thermodynamics.
• Usually energy transformation cannot be 100% efficient.
• As energy is transformed from one organism to another in the form of food, where as a large part of energy is dissipated as heat through respiration. (Eg.) Ten percent law.

Question 5.
What is ten percent law?
Answer:

• This law was proposed by Lindeman (1942)
• During transfer of food energy from one trophic level to other, only about 10% stored at every level and rest of them 90% is lost in respiration, decomposition in the form of heat.
• Hence the law is called ten percent law.

Question 6.
Which type of food chain is present in all ecosystem ? (or) What is detritus food chain?
Answer:

• It begins with dead organic matter.
  Which is an important source of energy
• Transfer of energy from Detritus → Detritivores → Small carnivores → Top carnivores
• This type of food chain is present in all ecosystems.

Question 7.
Complete the missing organisms of food web in grassland ecosystem.
Answer:

1. A – Rabbit
2. B – Mouse
3. C – Lizard
   

Question 8.
What are the importance of studying food web ? (or) What are significant of food web? Is there any significance in maintaining food web?
Answer:

• Food web is constructed to describe species interaction called direct interaction.
• It can be used to illustrate indirect interaction among different species.
• It can be used to reveal different patterns of energy transfer in terrestrial and aquatic ecosystems.

Question 9.
What are the different shapes of pyramids present in pyramids of number?
Answer:

• There are three different shapes of pyramids present in pyramid of number.
• Pyramid of number in grassland and pond ecosystem are always upright in shape.
• In forest ecosystem it is spindle shaped.
• In parasite ecosystem it is inverted.

Question 10.
What are the different shapes of pyramids present in pyramid of biomass?
Answer:
According to pyramid of biomass
Eco System – Shape :
Grassland and forest – Upright
Pond ecosystem – Inverted

Question 11.
Why pyramids of biomass in grass land and forest ecosystem is always upright ?
Answer:

• In grassland and forest there is a gradual decrease in biomass of organisms from producers to top carnivore.
• There fore it is always upright.

Question 12.
Why pyramids of biomass in pond ecosystem is always inverted in shape?
Answer:

• In pond ecosystem the bottom of the pyramid is occupied by the producers which are small organisms posses least biomass.
• So the value gradually increases towards the tip of the pyramid.
• Therefore pyramid of biomass is always inverted in shape.

Question 13.
Differentiate humification and mineralisation.
Answer:
Humification :

1. It is the process by which simplified detritus into dark coloured amorphous substance called humus.
2. Humus is resistant to microbial action.

Mineralisation :

1. Release of inorganic nutrients from the humus is called mineralisation.
2. Some microbes are involved in release of nutrients.

Question 14.
Define catabolism:
Answer:
The decomposers produce some extracellular enzymes in their surroundings to break down complex organic and inorganic compounds into simpler ones called catabolism.

Question 15.
What is leaching or eluviation ? (or) Which process of decomposition helpful to enrich lower layer of soil ?
Answer:
The movement of decomposed water soluble organic and inorganic compounds from the surface to the lower layer of soil by water is called eluviation (or) leaching.

Question 16.
What is bio-geo chemical cycle ?
Answer:
Circulation of nutrients within the ecosystem (or) biosphere is known as biogeo chemical cycles, (or) Cycling of materials.

Question 17.
Which cycles are called sedimentary cycles ? Why?
Answer:

• Phosphorus, sulphur, calcium are called as sedimentary cycles.
• Which are present as sediments on earth.
• Sedimentary cycles are very slow its take a long time to complete its circulation, because nutrient elements may get locked in the reservoir pool.

Question 18.
Wihat is carbon cycle?
Answer:

• The circulation of carbon between organisms and environment is known as carbon cycle.
• Cycling of carbon between organisms and atmosphere is consequence (event) of two reciprocal process of photosynthesis and respiration.

Question 19.
What is ecosystem resilience ? (or) Ecosystem robustness?
Answer:
Ecosystem is damaged by disturbances from fire, flood, predation, infection, drought, etc., removing a great amount of biomass. However, ecosystem is endowed with the ability to resist the damage and recover quickly. This ability of ecosystem is called ecosystem resilience or ecosystem robustness.

Question 20.
What are the ways to go green and save green?
Answer:

• Close the tap when not in use.
• Switch off the electrical gadgets when not in use.
• Never use plastics and replace them with biodegradable products.
• Always use ecofriendly technology and products.

Question 21.
Draw and write the 3 Rs for the safety and benefits of the environment.
Answer:

The three R’s are – reduce, reuse and recycle. It refers to the changing of one’s lifestyle for the safety and benefits of the environments (Reduce, Reuse, Recycle)

Question 22.
Which is significance of food web?
Answer:

• Food web is constructed to describe species interaction called direct interaction
• It can be used to illustrate indirect interaction among different species.
• It can be used to study bottom up or top down control of community structure.
• It can be used to reveal different patterns of energy transfer in terrestrial and aquatic ecosystems.

Question 23.
What are the types of carbon?
Answer:

1. Green carbon : carbon stored in the biosphere by the process of photosynthesis.
2. ray carbon: carbon stored in fossil fuel (Coal, oil and biogas deposits in the lithosphere).
3. Blue carbon : carbon stored in the atmosphere and oceans.
4. Brown carbon: carbon stored in industrialized forests (wood used in making commercial articles)
5. Black carbon : carbon emitted from gas, diesel engine and coal fired power plants.

IV. Five Marks

Question 1.
Define ecosystem. Describe the components of ecosystem
Answer:
The term ecosystem was proposed by A.G. Tansley (1935), who defined it as.

• The system resulting from the integration of all the living and non living factors of the environment.
  Whereas Odum (1962) defined ecosystem as
• Eco system is structural and functional unit of ecology.
• Ecosystem comprises of two major components
  i) Abiotic ii) Biotic components

i) Abiotic (non-living) components:
Edaphic Factors – Soil, air, soil water and pH of soil.
Topography factors – Latitude altitude Organic components – Carbohydrates, protein’s lipids and humic substances

Inorganic substances:
C, H, O, N and P . Abiotic components play vital role in any ecosystem hence it is important for standing quality.

ii) Biotic (living) components:
It includes all living organisms like plants, animal, fungi and bacteria.

• They form the trophic structures of any ecosystem.
• It has two components
  i) Autotrophic components ii) Heterotrophic components

i) Autotrophic components:
Autotrophs are plants (or) producers which can manufacture the organic compounds from simple inorganic components through a process called photosynthesis.

ii) Heterotrophic components:
Heterotrophs are organisms which consume the producers are called consumers. They are two types
i) Macro consumers ii) Micro consumers

• Macro consumers are herbivores, carnivores and omnivores.
• Micro consumers are called decomposers. (Eg.) Bacteria, actinomycetes and fungi
• Biotic components are essential to construct the food chain, food web and ecological pyramids.

Question 2.
What kind of solar radiation is used in photosynthesis ? (OR)
AR is not always constant because of clouds, tree shades, air, dust particles, seasons latitudes and length of the daylight availability. (OR)
What is photosynthetically active radiation ? (OR)
How it is calculated for active photosynthesis in plants ? (OR)
From the sunlight, only 2-10% of the solar energy is used by green plants for photosynthesis. Explain why?
Answer:

• The amount of light available for photosynthesis of plants is called photosynthetically active radiation.
• Which is between the range of 400 -700 nm wavelength. Which is essential for photosynthesis and plant growth.
• Generally plants absorb more blue and red light for efficient photosynthesis.
• Of the total sunlight, 34 percent that reaching the atmosphere is reflected back into the atmosphere, moreover 10% is held by ozone.
• Remaining 56% reaches the earths surface out of this 56%, only 2 – 10% of solar energy is used by green plants for photosynthesis while the remaining portion is dissipated as heat
• PAR is generally reported as millimoles / square meter / second by using silicon photo voltic detector.
• Which detect only 400 – 700 nm wavelength of light.
• PAR values range from 0 to 3000 millimoles / square meter / second.
• At night PAR is zero and during midday in summer, PAR often reaches 2000 – 3000 millimoles /square meter / second.

Question 3.
What is primary productivity ? What are its types ? What are the factors affecting primary productivity ? (or)
What is primary productivity ? How net primary productivity is calculated ? What are the factors affecting primary productivity.
Answer:
1.Primary productivity:
The chemical energy or organic matter generated by autotrophs during the process of photosynthesis and chemosynthesis is called primary productivity. It is the source of energy for all organisms, from bacteria to human.

a. Gross Primary Productivity (GPP):
The total amount of food energy or organic matter or biomass produced in an ecosystem by autotrophs through the process of photosynthesis is called gross primary pro-ductivity

b. Net Primary Productivity (NPP):
The proportion of energy which remains after respiration loss in the plant is called net primary productivity. It is also called as apparent photosynthesis. Thus the difference between GPP and respiration is known as NPP.
NPP = GPP – Respiration
NPP of whole biosphere is estimated to be
about 170 billion tons (dry weight) per year. Out of which NPP of oceanic producers is only 55 billion tons per year in unit time.

Factors affecting primary productivity:

• Factors affecting primary productivity depends upon the plant species of an area
• Their photosynthetic capacity, availability of nutrients solar radiation, precipitation, soil type and other environmental factors.
• It varies in different types of ecosystems.

Question 4.
What is secondary productivity and its types ? In what way community productivity calculated ?
Answer:
Secondary productivity:
The amount of energy stored in the tissues of heterotrophs or consumers is called secondary productivity.

a. Gross secondary productivity:
It is equivalent to the total amount of plant material is ingested by the herbivores minus the materials lost as faeces.

b. Net secondary productivity:
Storage of energy or biomass by consumers per unit area per unit time, after respiratory loss is called net secondary productivity.

Community productivity:
The rate of net synthesis of organic matter (biomass) by a group of plants per unit area per unit time is known as community productivity.

Question 5.
How does energy flow in an ecosystem, (or)
Describe about the concept of trophic level in an ecosystem, (or)
Write about various trophic level of food chain in an ecosystem.
Answer:

• A trophic level refers to the position of an organism in the food chain.
• The number of trophic levels is equal to the number of steps in the food chain.
• The green plants (producers) occupying the first trophic level (TA are called producers.
• The energy produced by the producers is utilized by the plant eaters (herbivores) they are called primary consumers and occupies the second trophic level (T₂).
• Herbivores are eaten by carnivores, which occupy the third trophic level (T₃).
  They are also called secondary consumers or primary carnivores.
• Carnivores are eaten by the other carnivores, which occupy the fourth trophic level (T₄). They are called the tertiary consumers or secondary carnivores.
• Some organisms which eat both plants and animals are called as omnivores (Crow). Such organisms may occupy more than one trophic level in the food chain.

Question 6.
How does the laws of thermodynamics explain the storage and loss of energy in an ecosystem ? (or) Write about laws of thermodynamics in an ecosystem.
Answer:
Laws of thermodynamics:
The storage and loss of energy in an ecosystem is based on two basic laws of thermodynamics.

i) First law of thermodynamics

• It states that energy can be transmitted from one system to another in various forms. Energy cannot be destroyed or created.
• As a result, the quantity of energy present in the universe is constant.
• Example: In photosynthesis, the product of starch (chemical energy) is formed by the combination of reactants (chlorophyll, H₂O, CO₂). ‘ “
  Starch is acquired from the external sources (light energy) and so there is no gain or loss in total energy. Here light energy is converted into chemical energy.
  
• Light energyv → Chemical energy

ii) Second law of thermodynamics Usually energy transformation cannot be 100% efficient. As energy is transferred from one organism to another in the form of food, a portion of it is stored as energy in living tissue, whereas a large part of energy is dissipated as heat through respiration. Example: Ten percent law Ten percent law : During transfer of food energy from one trophic level to other, only about 10% stored at every level and rest of them (90%) is lost in respiration, decomposition and in the form of heat. Hence, the law is called ten percent law.

Question 7.
What is food chain ? (or) How does movement takes place from producers to top carnivore? What are its types.
Answer:

• The movement of energy from producers upto top carnivores is known as food chain,
• Generally, there are two types of food chain, (1) Grazing food chain and (2) Detritus food chain.

1. Grazing food chain : Main source of energy for the grazing food chain is the Sun. It begins with the first link, producers (plants) The second link in the food chain is primary consumers (mouse) which get their food from producers. The third link in the food chain is secondary consumers (snake) which get their food from primary consumers. Fourth link in the food chain is tertiary consumers (eagle) which get their food from secondary consumers.

2. Detritus food chain : This type of food chain begins with dead organic matter which is an important source of energv. A large amount of organic matter is derived from the dead plants, animals and their excreta. This type of food chain is present in all ecosystems.

The transfer of energy from the dead organic matter, is transferred through a series of organisms called detritus consumers (detritivores)- small carnivores – large (top) carnivores with repeated eating and being eaten respectively. This is called the detritus food chain.

Question 8.
What is ecological pyramid ? Write about ecological pyramid of number.
Answer:
Graphic representation of trophic structure and function in successive trophic levels of an ecosystem is called ecological pyramid (or) Eltonian pyramid.

Figure 7.8 : Pyramids of numbers (individuals per unit area) in different types of ecosystems, Upright – A) Grassland ecosystem B) Pond ecosystem, Spindle shaped – C) Forest ecosystem,
Inverted – D) Parasite ecosystem

Pyramid of number:

• A graphical representation of the number of organism present at each successive trophic level in an ecosystem is called pyramid of number.
• There are different shapes, pyramid of number in grassland and pond ecosystem are always upright. Because
• There is a gradual decrease in number of organisms in each trophic level from Producer → Primary consumer then → Secondary consumer and finally → Tertiary consumer.
• Pyramid of number in forest ecosystem looks spindle shaped. Because
• In forest ecosystem, large sized tree (producer) which are lesser in number, herbivores are more in number than produces.
• In final tertiary consumers are lesser in number than the secondary consumer.
• In a parasite ecosystem the pyramid of number is always inverted because
• Its starts with single tree, there is gradual increase in the number of organisms in successive
  trophic levels from producers to tertiary consumers.
• Question 9.
  Write about the mechanism of decomposition ? What are the factors affecting decomposition, (or) Describe about an essential process for recycling and balancing the nutrient pool in an ecosystem.
  Answer:

• Decomposition is a step wise process of degradation mediated by enzymatic reactions. Detritus acts as a raw material for decomposition.
• a. Fragmentation – The breaking down of detritus into smaller particles by detritivores like bacteria, fungi and earth worm is known as fragmentation.
• b. Catabolism – The decomposers produce some extracellular enzymes in their surroundings to break down complex organic and inorganic compounds in to simpler ones. This is called catabolism
• c. Leaching or Eluviation – The movement of decomposed, water soluble organic and inorganic compounds from the surface to the lower layer of soil or the carrying away of the same by water is called leaching or eluviation.
• d. Humification – Detritus is changed into dark coloured amorphous substance called humus. It is highly resistant to microbial action, therefore decomposition is very slow.
• e. Mineralisation – The release of inorganic nutrients from the humus is called mineralisation.
• Decomposition is affected by climatic factors like temperature, soil moisture, soil pH oxygen and also the chemical quality of detritus.

Question 10.
Define pyramid of biomass. What are its types based on shape ?
Answer:

• A graphical representation of the amount of organic material (biomass) present at each successive trophic level in an ecosystem is called pyramid of biomass.
• In grassland and forest ecosystems, there is a gradual decrease in biomass of organisms at successive trophic levels from producers to top carnivores (Tertiary consumer). Therefore, these two ecosystems show pyramids as upright pyramids of biomass.
• However, in pond ecosystem, the bottom of the pyramid is occupied by the producers, which comprise very small organisms possessing the least biomass and so, the value gradually increases towards the tip of the pyramid. Therefore, the pyramid of biomass is always inverted in shape.

Figure 7.9: Pyramids of biomass (dry weight per unit area) in different types of ecosystems.
Upright – A) Grassland ecosystem Inverted – C) Pond ecosystem

Question 11.
Write about the structure of pond ecosystem with diagram.
Answer:

• Pond ecosystem is a self sustaining and self regulatory fresh water ecosystem, which shows a complex interaction between the abiotic and biotic components in it.
• A pond ecosystem consists of dissolved inorganic (CO₂, O₂, Ca, N, Phosphate) and organic substances (amino acids and humic acid) formed from the dead organic matter.
• The function of pond ecosystem is regulated by few factors like the amount of light, temperature, pH value of water and other climatic conditions.
• Biotic components : They constitute the producers, variety of consumers and decomposers (microorganisms).
• A variety of phytoplankton like Oscillatoria, Anabaena, Eudorina, Volvox and Diatoms. Filamentous algae, floating plants, rooted floating plants are the major producers of a pond ecosystem.
• zooplanktons, benthos, secondary consumers like water beetles and frogs tertiary consumers (carnivores) like duck, crane and some top carnivores which include large fish, hawk ,man, etc.
• They are also called as microconsumers. They help to recycle the nutrients in the ecosystem.
• The cycling of nutrients between abiotic and biotic components is evident in the pond ecosystem, making itself self sufficient and self regulating.

Question 12.
Write important features of a sedimentation cycle in an ecosystem.
Answer:
Sedimentary cycles are very slow. They take a long time to complete their circulation. Because during recycling, nutrient elements may get locked in the reservoir pool there by taking a very long to come out and continue circulation.

• It is a type of sedimentary cycle. Already we know that phosphorus is found in the biomolecules like DNA, RNA, ATP, NADP and phospholipid molecules of living organisms.
• Bulk quantity of phosphorus is present in rock deposits, marine sediments and guano.
  It is released from these deposits by weathering process.
• The producers absorb phosphorus in the form of phosphate ions, and then it is transferred to each trophic level of food chain through food.
• Again death of the organisms and degradation by the action of decomposers, the phosphorus is released back into the lithosphere and hydrosphere to maintain phosphorus cycle.

Question 13.
Based on the varieties of benefits obtained from ecosystem what are the various types ecosystem services ?
Answer:
The varieties of benefits obtained from the ecosystem are generally categorized into the follo wing four types

Question 14.
Discuss about the various benefits of mangrove ecosystem services, (or) What are the various benefits of mangrove ecosystem services to nature and human being?
Answer:
Mangrove ecosystem services:

• Offers habitat and act as nursery for aquatic plants and animals Provides medicine, fuel wood and timber.
• Act as bridge between sea and rivers by balancing sedimentation and soil erosion.
• Help to reduce water force during cyclones, tsunamis and high tide periods.
• Help in wind break, O₂ production, carbon sequestration and prevents salt spray from waves.

Question 15.
If we fail to protect environment, we will fail to save posterity? (or) What are the practices we have to follow to protect the ecosystem ? (or) How to protect the ecosystem ?
Answer:
To protect ecosystem, we have to practice the following in our daily life.

• Buy and use only ecofriendly products and recycle them.
• Grow more trees.
• Choose sustained farm products (vegetables, fruits, greens, etc.)
• Reduce the use of natural resources.
• Recycle the waste and reduce the amount of waste you produce.
• educe consumption of water and electricity.
• Reduce or eliminate the use of house-hold chemicals and pesticides.
• Maintain your cars and vehicles properly. (In order to reduce carbon emission)
  Create awareness and educate about ecosystem protection among your friends and family members and ask them to find out solution to minimise this problem.

Question 16.
What are the strategy of ecosystem management ?
Answer:
Strategy of ecosystem management:

• It is used to maintain biodiversity of ecosystems.
• It helps in indicating the damaged ecosystem (Some species indicate the health of the ecosystem: such species are called a flagship species).
• It is used to recognize unavoidance of ecosystem change and plan accordingly.
• It is one of the tools used for achieving sustainability of ecosystem through sustainable development programme (or projects).
• It is also helpful in identifying ecosystems which are in need of rehabilitation.

Question 17.
What are various steps involved in ecological succession ? (or) What are the characteristics of ecological succession ?
Answer:
Characteristics of ecological succession:

• It is a systematic process which causes changes in specific structure of plant community.
• It is resultant of changes of abiotic and biotic factors.
• It transforms unstable community into a stable community.
• Gradual progression in species diversity, total biomass, niche specialisation, and humus content of soil takes place.
• It progresses from simple food chain to complex food web.
• It modifies the lower and simple life form to the higher life forms.
• It creates inter-dependence of plants and animals.

Question 18.
Tabulate the differences between primary and secondary succession?
Answer:

Question 19.
Write about the different stages of hydrosere? (or) Write about the different substages of hydrosere in plant succession.
Answer:
The type of succession is hydrosere. It includes the following stages.

1. Phytoplankton stage
2. Submerged plant stage
3. Submerged free floating stage
4. Reed-swamp stage
5. Marsh rnedow stage
6. Shr ub stage
7. Forest stage

1. Phytoplankton stage:
It is the first stage of succession consisting of the pioneer community like blue green algae, green algae, diatoms, bacteria, etc., The colonization of these organisms enrich the amount of organic matter and nutrients of pond due to their life activities and death. This favours the development of the next serai stages.

2. Submerged plant stage:
As the result of death and decomposition of planktons, silt brought from land by rain water, lead to a loose mud formation at the bottom of the pond. Hence, the rooted submerged hydrophytes begin to appear on the new substratum. Example: Chara, Utricularia. The death and decay of these plants will build up the substratum of pond to become shallow. Therefore, this habitat now replaces another group of plants which are of floating type.

3. Submerged free floating stage:
During this stage, the depth of the pond will become almost 2-5 feet. Hence, the rooted hydrophytic plants and with floating large leaves start colonising the pond. Example: Rooted floating plants like Nelumbo, Nymphaea and Trapa. By death and decomposition of these plants, further the pond becomes more shallow. Due to this reason, floating plant species is gradually replaced by another species which makes new serai stage.

4. Reed-swamp stage:
It is also called an amphibious stage. During this stage, rooted floating plants are replaced by plants which can live successfully in aquatic as well as aerial environment. Example: Typha

5. Marsh meadow stage:
When the pond becomes swallowed due to decreasing water level, species of Cyperaceae and Poaceae. They form a mat-like vegetation with the help of their much branched root system. This leads to an absorption and loss of large quantity of water. At the end of this stage, the soil becomes dry and the marshy vegetation disappears gradually and leads to shurb stage.

6. Shrub stage:
As the disappearance of marshy vegetation continues, soil becomes dry. Hence, these areas are now invaded by terrestrial plants like shrubs (Salix and Cornus) and trees (Populus and Alnus). These plants absorb large quantity of water and make the habitat dry. Further, the accumulation of humus with a rich flora of microorganisms produce minerals in the soil, ultimately favouring the arrival of new tree species in the area.

7. Forest stage:
It is the climax community of hydrosere. A variety of trees invade the area and develop any one of the diverse type of vegetation. Example: Temperate mixed forest 207

Question 20.
Write about the significance of plant succession.
Answer:
Significance of Plant Succession:

• Succession is a dynamic process. Hence an ecologist can access and study the serai stages of a plant community found in a particular area.
• The knowledge of ecological succession helps to understand the controlled growth of one or more species in a forest.
• Utilizing the knowledge of succession, even dams can be protected by preventing siltation
• It gives information about the techniques to be used during reforestation and afforestation.
• It helps in the maintenance of pastures.
• Plant succession helps to maintain species diversity in an ecosystem.
• Patterns of diversity during succession are influenced by resource availability and disturbance by various factors.
• Primary succession involves the colonization of habitat of an area devoid of life.
• Secondary succession involves the reestablishment of a plant community in disturbed area or habitat.
• Forests and vegetation that we come across all over the world are the result of plant succession.

Question 21.
What is carbon cycle ? Draw the diagrammatic sketch showing carbon cycle ?
Answer:
The circulation of carbon between organisms and environment is known as the carbon cycle

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.5 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5

Question 1.
Evaluate the following limits, if necessary use L’ Hôpital’s Rule:
\(\lim _{x \rightarrow 0}\) \(\frac { 1-cosx }{ x^2 }\)
Solution:
\(\lim _{x \rightarrow 0}\) \(\frac { 1-cosx }{ x^2 }\)

Question 2.
\(\lim _{x \rightarrow ∞}\) \(\frac { 2x^2-3 }{ x^2-5x+3 }\)
Solution:

Question 3.
\(\lim _{x \rightarrow ∞}\) \(\frac { x }{ log x }\)
Solution:
\(\lim _{x \rightarrow ∞}\) \(\frac { x }{ log x }\) [ \(\frac { ∞ }{ ∞ }\) indeterminate form
Applying L’ Hôpital’s Rule
\(\lim _{x \rightarrow ∞}\) \(\frac { 1 }{ \frac{1}{x} }\) = \(\lim _{x \rightarrow ∞}\) x = ∞

Question 4.
\(\lim _{x \rightarrow \frac{π}{2}}\) \(\frac { secx }{ tanx }\)
Solution:
\(\lim _{x \rightarrow \frac{π}{2}}\) \(\frac { secx }{ tanx }\) [ \(\frac { ∞ }{ ∞ }\) indeterminate form
Simplifying, we get
\(\lim _{x \rightarrow \frac{π}{2}}\) \(\frac { 1 }{ sinx }\) = \(\frac { 1 }{ sin \frac{π}{2} }\) = 1

Question 5.
\(\lim _{x \rightarrow ∞}\) e^-x√x
Solution:
\(\lim _{x \rightarrow ∞}\) e^-x√x [0 × ∞ indeterminate form
The other form is \(\lim _{x \rightarrow ∞}\) \(\frac { √x }{ e^x }\)
[0 × ∞ indeterminate form
Applying L’ Hôpital’s Rule
= \(\lim _{x \rightarrow ∞}\) \(\frac { 1 }{ 2 \sqrt{xe^x} }\)
= 0

Question 6.
\(\lim _{x \rightarrow ∞}\) (\(\frac { 1 }{ sinx }\) – \(\frac { 1 }{ x }\))
Solution:

Question 7.
\(\lim _{x \rightarrow 1}\) (\(\frac { 2 }{ x^2-1 }\) – \(\frac { x }{ x-1 }\))
Solution:

Question 8.
\(\lim _{x \rightarrow 0^+}\) x^x
Solution:
\(\lim _{x \rightarrow 0^+}\) x^x [0° indeterminate form
Let g(x) = x^x
Taking log on both sides
log g(x) = log x^x
log g(x) = x log x
\(\lim _{x \rightarrow 0^+}\) log g(x) = \(\lim _{x \rightarrow 0^+}\) x log x [0 × ∞ indeterminate form

Question 9.
\(\lim _{x \rightarrow ∞}\) (1 + \(\frac { 1 }{ x }\)) ^x
Solution:

Applying L’ Hôpital’s Rule

Exponentiating we get, \(\lim _{x \rightarrow ∞}\) g(x) = e¹ = e

Question 10.
\(\lim _{x \rightarrow \frac{π}{2}}\) (sin x) ^{tan x}
Solution:
\(\lim _{x \rightarrow \frac{π}{2}}\) (sin x)^{tan x} [1^∞ indeterminate form]
Let g(x) = (sin x) ^{tan x}
Taking log on both sides,
log g(x) = tan x log sin x
\(\lim _{x \rightarrow \frac{π}{2}}\) log g(x) = \(\lim _{x \rightarrow \frac{π}{2}}\) \(\frac { log sin x }{ cot x }\)
[ \(\frac { 0 }{ 0 }\) Indeterminate form
Applying L’ Hôpital’s Rule
= \(\lim _{x \rightarrow \frac{π}{2}}\) (\(\frac { cotx }{ -cosec^2x }\)) = -1
exponentiating, we get
\(\lim _{x \rightarrow \frac{π}{2}}\) g(x) = e^-1 = \(\frac { 1 }{ e }\)

Question 11.
\(\lim _{x \rightarrow 0^+}\) (cos x) ^{\(\frac { 1 }{ x^2 }\)}
Solution:
\(\lim _{x \rightarrow 0^+}\) (cos x) ^{\(\frac { 1 }{ x^2 }\)} [1^∞ indeterminate form
let g(x) = (cos x)^{\(\frac { 1 }{ x^2 }\)}
Taking log on both sides,

Question 12.
If an initial amount A₀ of money is invested at an interest rate r compounded n times a year, the value of the investment after t years is A = A₀(1 + \(\frac { r }{ n }\))^nt. If the interest is compounded continuously, (that is as n → ∞), show that the amount after t years is A = A₀e^rt.
Solution:

Applying L-Hospital’s Rule

Hence Proved.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Botany Guide Pdf Chapter 6 Principles of Ecology Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 6 Principles of Ecology

12th Bio Botany Guide Principles of Ecology Text Book Back Questions and Answers

I. Choose the correct evaluation

Question 1.
Arrange the correct sequence of ecological hierarchy starting from lower to higher level.
a. Individual organism → Population Landscape → Ecosystem
b. Landscape → Ecosystem → Biome → Biosphere
c. Community → Ecosystem → Landscape → Biome
d; Population → organism → Biome → Landscape
Answer:
c. Community → Ecosystem → Landscape → Biome

Question 2.
Ecology is the study of an individual species is called
i) Community ecology
ii) Autecology
iii) Species ecology
iv) Synecology
a. i only
b. ii only
c. i and iv only
d. ii and iii only
Answer:
d. ii and iii only

Question 3.
A specific place in an ecosystem, where an organism lives and performs its functions is
a. habitat
b. niche
c. landscape
d. biome
Answer:
b. niche

Question 4.
Read the given statements and select the correct option.
i) Hydrophytes possess aerenchyma to support themselves in water.
ii) Seeds of Viscum are positively photoblastic as they germinate only in presence of light.
iii) Hygroscopic water is the only soil water available to roots of plant growing in soil as it is present inside the micropores.
iv) High temperature reduces use of water and solute absorption by roots.

a. i, ii, and iii only
b. ii, iii and iv
c. ii and iii only
d. i and ii only
Answer:
b. ii, iii and iv correct answer

Question 5.
Which of the given plant produces cardiac glycosides?
a. Calotropis
c. Nepenthes
b. Acacia
d. Utricularia
Answer:
a. Calotropis

Question 6.
Read the given statements and select the correct option.
i) Loamy soil is best suited for plant growth as it contains a mixture of silt, sand and clay.
ii) The process of humification is slow in case of organic remains containing a large amount of lignin and cellulose.
iii) Capillary water is the only water available to plant roots as it is present inside the micropores.
iv) Leaves of shade plant have more total chlorophyll per reaction centre, low ratio of chi a and chi b are usually thinner leaves.
a. i, ii and iii only
b. ii, iii and iv only
c. i, ii and iv only
d. ii and iii only
Answer:
d. ii and iii only

Question 7.
Read the given statements and select the correct option.
Statement A: Cattle do not graze on weeds of Calotropis.
Statement B : Calotropis have thorns and spines, as defense against herbivores.
a. Both statements A and B are incorrect.
b. Statement A is correct but statement B is incorrect.
c. Both statements A and B are correct but statement B is not the correct explanation of statement A.
d. Both statements A and B are correct and statement B is the correct explanation of statement A.
Answer:
b. Statement A is correct but statement

Question 8.
In soil water available for plants is
a. gravitational water
b. chemically bound water
c. capillary water
d. hygroscopic water
Answer:
c. capillary water

Question 9.
Read the following statements and fill up the blanks with correct option.
i) Total soil water content in soil is called ……………………………..
ii) Soil water not available to plants is called ……………………………
iii) Soil water available to plants is called ……………………..

Answer:
a) i – Holard, ii-Echard, iii-Chresard

Question 10.
Column I represent the size of the soil particles and Column II represents type of soil components. Which of the following is correct match for the Column I and Column II

Answer:
c) I (iii), II (ii), III (i), IV (iv)

Question 11.
The plant of this group are adapted to live partly in water and partly above substratum and free from water
a. Xerophytes
b. Mesophytes
c. Hydrophytes
d. Halophytes
Answer:
d. Halophytes

Question 12.
Identify the A, B, C and D in the given table

Answer:
a) A (+) B – Parasitism, C (-), D – Amensalism

Question 13.
Ophrys an orchid resembling the female of an insect so as to able to get pollinated is due to phenomenon of
a. Myrmecophily
b. Ecological equivalents
c. Mimicry
d. None of these
Answer:
c. Mimicry

Question 14.
A free living nitrogen fixing cyanobacterium which can also form symbiotic association with the water fern Azolla
a. Nostoc
b. Anabaena
c. chlorella
d. Rhizobium
Answer:
b. Anabaena

Question 15.
Pedogenesis refers to
a. Fossils
b. Water
c. Population
d. Soil
Answer:
d. Soil

Question 16.
Mycorrhiza promotes plant growth by
a. Serving as a plant growth regulators
b. Absorbing inorganic ions from soil
c. Helping the plant in utilizing atmospheric nitrogen
d. Protecting the plant from infection
Answer:
b. Absorbing inorganic ions from soil

Question 17.
Which of the following plant has a non-succulent xerophytic and thick leathery leaves with waxy coating
a. Bryophyllum
b. Ruscus
c. Nerium
d. Calotropis
Answer:
c. Nerium

Question 18.
In a fresh water environment like pond, rooted autotrophs are
a. Nymphaea and typha
b. ‘ CeratophyllumandUtricularia
c. Wolffia and pistia
d. Azolla and lemna
Answer:
a. Nymphaea and typha

Question 19.
Match the following and choose the correct combination from the options given below:

Answer:
d (I) iv, (II) iii, (III) ii, (IV) v, (V) i

Question 20.
Strong, sharp spines that get attached to animal’s feet are found in the fruits of
a. Ar gemone
b. Ecballium
c. Heritier
d. Crossandra
Answer:
a. Argemone

Question 21.
Sticky glands of Boerhaavia and Cleome support
a. Anemochory
b. Zoochory
c. Autochory
d. Hydrochory
Answer:
b. Zoochory

Question 22.
Define ecology.
Answer:
Ecology is the study of the reciprocal relationship between living organisms and their environment.

Question 23.
What is the ecological hierarchy? Name the levels of ecological hierarchy.
Answer:
Ecological hierarchy is the interaction of organisms with their environment results in the establishment of a grouping of organisms. Which is called ecological hierarchy (or) ecological levels of organisation.

Question 24.
What are ecological equivalents? Give one example.
Answer:
Taxonomically different species occupying similar habitats (Niches) in different geographical regions are called Ecological equivalents.
E.g: Certain species of epiphytic orchids of Western Ghats of India differ from the epiphytic orchids of South America. But they are epiphytes.

Question 25.
Distinguish habitat and niche
Answer:
Habitat

1. A specific phsical space occupied bv an organism (species)
2. The same habitat may be shared by many organisms (species)
3. Habitat specificity is exhibited by organisms.

Niche

1. A functional space occupied by an organism in the same ecosystem
2. A single niche is occupied by a single species
3. Organisms may change their niche with time and season.

Question 26.
Why are some organisms called eurythermal and some others stenohaline?
Answer:

1. Eurythermal: Organisms which can tolerate a wide range of temperature fluctuations.
   Example: Zostera.
2. Stenothermal: Organisms that can tolerate only a small range of temperature variations.
   Example: Mango.

Question 27.
‘Green algae are not likely to be found in the deepest strata of the ocean’. Give at least one reason.
Answer:

• The deepest strata of the ocean is dark and sufficient light is not available for the photosynthesis of green algae.
• Algae need brackish water for its growth. Which is also not available in the deepest strata of the ocean.

Question 28.
What is Phytoremediation?
Answer:
Phytoremediation refers to the using of living green plants to overcome soil or water contamination.
E.g: Growing Eichhomia in cadmium enriched soil reduces the level of cadmium.

Question 29.
What is the Albedo effect and write their effects?
Answer:

• The albedo effect is due to greenhouse effect.
• Aerosols (suspension of fine solid (or) liquid particles in gas) with small particles is reflecting the solar radiation entering the atmosphere is known as Albedo effect.
• It reduces the temperature, photosynthesis and respiration
• The sulphur compounds present in the aerosol are responsible for acid rain due to acidification of rain water and destroy the ozone.

Question 30.
The organic horizon is generally absent from agricultural soils because tilling, e.g., ploughing, buries organic matter. Why is an organic horizon generally absent in desert soils?
Answer:
Organic horizon is generally absent in deserts because of low content of organic matter due to scarcity of plant and animal remains or excreta.

Question 31.
Soil formation can be initiated by biological organisms. Explain how?
Answer:

• Soil formation is initiated by the weathering process.
• Biological weathering takes place when organisms like bacteria, fungi, lichens and plants helps in the breakdown of rocks through the production of acids and certain chemical substances.

Question 32.
Sandy soil is not suitable for cultivation. Explain why?
Answer:
Sandy soil has a high porosity leading to decreased water retention hence unfit for cultivation.

Question 33.
Describe the mutual relationship between the fig and wasp and comment on the phenomenon that operates in this relationship.
Answer:

• Mutualism interaction exist between fig tree and wasp
• In fig tree there is a tight one to one relationship with a pollinator species of wasp and no other species.
• The wasp pollinates the fig while finding egg lav ing sites and in turn, the fig offers the wasp developing seeds, as food for the developing larvae.

Question 34.
Lichen is considered as a good example of obligate mutualism. Explain.
Answer:

• It is an interaction between two species of organisms in which both are benefitted from the obligate association.
• Lichens is a mutual association of algae and a fungus.
• The alga is usually green alga (or) blue green alga. The fungus is an ascomycete (or) basidiomycete.
• It is believed that alga contributes organic food from photosynthesis and the fungus is able to absorb water and mineral salts.
• The fungus can also conserver water and this enables lichens to grow in extremely dry conditions where no other plants can exist.

Question 35.
What is mutualism? Mention any two examples where the organisms involved are commercially exploited in modern agriculture.
Answer:

• Mutualism is an interaction between two species of organisms in which both are benefitted from their association.
• Eg: 1 – Water Fern (Azolla) and Nitrogen fixing Cyanobacterium (Anabaena)
• Eg: 2 – Roots of terrestrial plants and fungal hyphae – Mycorrhiza.

Question 36.
List any two adaptive features evolved in parasites enabling them to live successfully on their host?
Answer:
Holoparasites:
The organisms which are dependent upon the host plants for their entire nutrition are called Holoparasites. They are also called total parasites.

Examples:

• Cuscuta is a total stem parasite of the host plant Acacia, Duranta, and manv other plants. Cuscuta even gets flower inducing hormone from its host plant.
• Balanophora, orobanche, and refflesia are the total root parasites found on higher plants.

Hemiparasites:
The organisms which derive only water and minerals from their host plant while synthesizing their own food by photosynthesis are called Hemiparasites. They are also called partial parasites.
Examples:
Viscum and Loranthus are partial stem parasites.

Question 37.
Mention any two significant roles predation plays in nature.
Answer:
Predation:

• It is an interaction between two species, one of which captures, kills, and eats up the other.
• The species which kills is called a predator and the species which is killed is called prey.
• The predator is benefitted while the prey is harmed.

Examples:
A number of plants like Drosera (Sundew Plant), Nepenthes (Pitcher Plant), Diaonaea (Venus flytrap), Utricularia (Bladderwort), and Sarracenia are predators

• which consume insects and other small animals for their food as a source of nitrogen.
• They are also called insectivorous plants.
• Many herbivores are predators. Cattles, Camels, Goats, etc., frequently browse on the tender shoots of herbs, shrubs, and trees.
• Generally, annuals suffer more than perennials.
• Grazing and browsing may cause remarkable changes in vegetation.
• Nearly 25 percent of all insects are known as phytophagous (feeds on plant sap and other parts of the plant)

Many defense mechanisms are envoloved to avoid their predations by plants.
Calotropis produces highly poisonous cardiac glycosides.
Tobacco: Produces nicotine.
Coffee: coffee plants produce coffeine.

Question 38.
How does an orchid ophrys ensure its pollination by bees?
Answer:
The plant, Ophrys an orchid, the flower looks like a female insect to attract the male insect to get pollinated by the male insect and it is otherwise called ‘floral mimicry.’

Question 39.
Water is very essential for life. Write any three features for plants which enable them to survive in water scarce environment.
Answer:

• Xerophytes are the plants which are living in dry (or) xeric condition are known as xerophytes.

Adaptations of xerophytes:

• Root system is well developed and is greater than that of shoot system.
• In some xerophytes all the internodes in the stem are modified into a fleshy leaf structure called phvlloclades (Opuntia)
• In some the petiole is modified into a fleshy leaf like structure called phyllode (Acacia melanoxylon).

Question 40.
Why do submerged plants receive weak illumination than exposed floating plants in a lake?
Answer:

• Submerged plants which receive weak illumination because
• Submerged plants are completely immersed in water and not in contact with the atmosphere (or) surface of the water.
• The floating hydrophytes float freely (or) float their leaves and flowers on the surface of water do not allow light to pass inside the lake.
• So submerged plants receive weak illumination than exposed floating plants.

Question 41.
What is vivipary? Name a plant group which exhibits vivipary.
Answer:

•  Vivipary is the special type of seed germination
• During germination, the seed is till attached to the parent plant and nourished by it.
• Vivipary generally occurs in mangrove plants.
• The mangrove plants are medium sized trees which grow in salty marshes of sea coasts. (Eg.) Rhizophora, Sonneratia, Avicennia.
• The seeds of this plant cannot germinate on the marshy habitat because of the excessive salt concentration and lack of oxygen.
• The radicle of the plant elongates considerably and projects out of the fruit.
• Then dark like seedling breaks off from the parent plant.
• Then radicle immediately forms new roots and establishes the seedling as a new plant.

Question 42.
What is thermal stratification? Mention their types.
Answer:
Thermal Stratification is usually found in aquatic habitats. The change in the temperature profile with increasing depth in a water body is called thermal stratification. There are three kinds of thermal stratifications.

1. Epilimnion – The upper layer of warmer water.
2. Metalimnion – The middle layer with a zone of a gradual decrease in temperature.
3. Hypolimnion – The bottom layer of colder water.

Question 43.
How is rhytidome act as the structural defence by plants against fire?
Answer:

• Rhytidome is the structural defense by plants against fire
• The outer bark of trees which extends to the last formed periderm is called Rhytidome.
• It is composed of multiple layers of suberized periderm, cortical and phloem tissues.
• It protects the stem against fire, water loss, invasion of insects and prevents infections by microorganisms.

Question 44.
What is myrmecophily?
Answer:
Sometimes, ants take their shelter on some trees such as Mango, Litchi, Jamun, and Acacia, etc. These ants act as bodyguards of the plants against any disturbing agent and the plants, in turn, provide food and shelter to these ants. This phenomenon is known as Myrmecophily. Example: Acacia and its ants.

Question 45.
What is a seed ball?
Answer:

• It is a method of human aided seed dispersal r Seed ball is an ancient Japanese technique of encasing
• seeds in a mixture of clay and soil humus (also in cow dung) and scattering them on suitable ground, not planting of trees manually.
• This method is suitable for barren and degraded lands for tree regeneration and vegetation before the monsoon period where the suitable dispersal agents become rare.

Question 46.
How is anemochory differ from zoochory?
Answer:

Question 47.
What is co evolution?
Answer:
The interaction between organisms, when continues for generations, involves reciprocal changes in genetic and morphological characters of both organisms. This type of evolution is called Co-evolution. It is a kind of co-adaptation and mutual change among interactive species.
Examples:

• Corolla length and probosci’s length of butterflies and moths (Habenaria and Moth).
• Bird’s beak shape and flower shape and size.

Question 48.
Explain Raunkiaer classification of the world’s vegetation based on temperature.
Answer:
Based on the temperature prevailing in an area, Raunkiaer classified the world’s vegetation into the following four types.

They are megatherms, mesotherms, microtherms, and hekistotherms.
In thermal springs and deep sea hydrothermal vents where average temperature exceed 100°c.

Based on the range of thermal tolerance, organisms are divided into two types.

• Eurythermal: Organisms which can tolerate a wide range of temperature fluctuations.
  (Eg.) Zostera (A marine Angiosperm) and Artemisia tridentata.
• Stenothermal: Organisms which can tolerate only a small range of temperature variations.
  (Eg.) Mango and Palm (Terrestrial Angiosperms).

Mango plants do not and cannot grow in temperate countries like Canada and Germany.

Thermal stratifications:
It is usually found in aquatic habitat.
The change in the temperature profile with increasing depth in a water body is called thermal stratification. There are three kinds of thermal stratification

1. Epilimniotn The upper layer of warmer water
2. Metalimnion The middle layer with a zone of gradual decrease in temperature.
3. Hypolimnion The bottom layer of colder water.

Temperature based zonation:
Variations are latitude and altitude do affect the temperature of the vegetation on the earth’s surface.
Latitude: Latitude is an angle which ranges from 0° at the equator to 90° at the poles.
Altitude: How High a place is located above the sea level is called the altitude of the place.

Question 49.
List out the effects of fire on plants.
Answer:
Effects of fire:

• A fire has a direct lethal effect on plants
• Burning scars are suitable places for the entry of parasitic fungi and insects
• It brings out the alteration of light, rainfall, nutrient cycle, the fertility of the soil, pH, soil flora, and fauna
• Some fungi which grow in the soil of burnt areas called pyrophilous. (Eg.) Pyronema confluens.

Question 50.
What is the soil profile? Explain the characters of different soil horizons.
Answer:
Soil Profile:

• Soil is commonly stratified into horizons at different depth. These layers differ in their physical, chemical and biological properties.
• This succession of super-imposed horizons is called a soil profile.

Question 51.
Give an account of various types of parasitism with examples.
Answer:
(a) Parasitism: It is an interaction between two different species in which the smaller partner (parasite) obtains food from the larger partner (host or plant). So the parasitic species is benefitted while the host species is harmed. Based on the host-parasite relationship, parasitism is classified into two types they are holoparasite and hemiparasite.

(b) Holoparasites: The organisms which are dependent upon the host plants for their entire nutrition are called Holoparasites. They are also called total parasites.
Examples:

• Cuscuta is a total stem parasite of the host plant Acacia, Duranta and many other plants. Cuscuta even gets flower inducing hormone from its host plant.
• Balanophora, Orobanche and Rafflesia are the total root parasites found on higher plants.

(c) Hemiparasites: The organisms which derive only water and minerals from their host plant while synthesizing their own food by photosynthesis are called Hemiparasites. They are also called partial parasites.
Examples:

• Viscum and Loranthus are partial stem parasites.
• Santalum (Sandal Wood) is a partial root parasite.
  The parasitic plants produce the haustorial roots inside the host plant to absorb nutrients from the vascular tissues of host plants.

Question 52.
Explain different types of hydrophytes with examples.
Answer:
The plants which are living in water or wet places are called hydrophytes. According to their relation to water and air, they are sub-divided into the following categories:

• Free floating hydrophytes: These plants float freely on the surface of the water. They remain in contact with water and air, but not with soil. Examples: Eichhornia, Pistia and Wolffia (smallest flowering plant).
• Rooted floating hydrophytes : In these plants, the roots are fixed in mud, but their leaves and flowers are floating on the surface of water. These plants are in contact with soil, water and air. Examples: Nelumbo, Nymphaea, Potomogeton, and Marsilea.
• Submerged floating hydrophytes: These plants are completely submerged in water and not in contact with soil and air. Examples: Ceratophyllum and Utricularia.
• Rooted- submerged hydrophytes : These plants are completely submerged in water and rooted in soil and not in contact with air. Examples: Hydrilla, Vallisneria, and Isoetes.

Amphibious hydrophytes (Rooted emergent hydrophytes): These plants are adapted to both aquatic and terrestrial modes of life. They grow in shallow water. Examples: Ranunculus, Typha and Sagittaria.
Hygrophytes: The plants which can grow in moist damp and shady places are called hygrophytes. (Eg.) Habenaria (Orchid), Mosses (Bryophytes), etc.

Question 53.
Enumerate the anatomical adaptations of xerophytes.
Answer:
Anatomical adaptations:

• Presence of multilayered epidermis with heavy cuticle to prevent water loss due to transpiration.
• The hypodermis is well developed with sclerenchymatous tissues.
• Sunken shaped stomata are present only in the lower epidermis with hairs in the sunken pits.
• Scotoactive type of stomata found in succulent plants.
• Vascular bundles are well developed with several layered bundle sheath.
• Mesophyll is well differentiated into palisade and spongy parenchyma.
• In succulents the stem possesses a water storage region.

Question 54.
List out any five morphological adaptations of halophytes.
Answer:
Morphological adaptations:

• The temperate halophytes are herbaceous but the tropical halophytes are mostly bushy
• In addition to the normal roots, many stilt roots are developed
• A special type of negatively geotropic roots called pneumatophores with pneumathodes to get sufficient aeration are also present. They are called breathing roots.
  Example: Avicennia

• Presence of thick cuticle on the aerial parts of the plant body
• Leaves are thick, entire, succulent and glossy. Some species are aphyllous (without leaves).
• Vivipary mode of seed germination is found in halophytes

Question 55.
What are the advantages of seed dispersal?
Answer:
Advantages of seed dispersal:

1. Seeds escape from mortality near the parent plants due to predation by animals or getting diseases and also avoiding competition.
2. Dispersal also gives a chance to occupy favourable sites for growth.
3. It is an important process in the movement of plant genes, particularly this is the only method available for self-fertilized flowers and maternally transmitted genes in outcrossing plants.
4. Seed dispersal by animals helps in conservation of many species even in human-altered ecosystems.
5. Understanding of fruits and seed dispersal acts as a key for proper functioning and establishment of many ecosystems from deserts to evergreen forests and also for the maintenance of biodiversity conservation and restoration of ecosystems.

Question 56.
Describe dispersal of fruit and seeds by animals.
Answer:
Birds and mammals, including human beings, play an efficient and important role in the dispersal of fruit and seeds. They have the following devices,

i. Hooked fruit: The surface of the fruit or seeds have hooks,(Xanthium), barbs (Andropogon), spines (Aristida) by means of which they adhere to the body of animals or clothes of human beings and get dispersed.

ii. Sticky fruits and seeds:

• Some fruits have sticky glandular hairs by which they adhere to the fur of grazing animals. Example: Boerhaavia and Cleome.
• Some fruits have a viscid layer which adheres to the beak of the bird which eats them and when they rub them on to the branch of the tree, they disperse and germinate. Example: Cordia and Alangium

iii. Fleshy fruits: Some fleshy fruits with conspicuous colours are dispersed by human beings to distant places after consumption. Example: Mango and Diplocyclos

12th Bio Botany Guide Principles of Ecology Additional Important Questions and Answers

I. Match the following

Question 1.

a. A – (4), B – (1), C – (3), D – (2)
b. A – (3), B – (1), C – (2), D – (4)
c. A – (2), B – (1), C – (4), D – (3)
d. A – (2), B – (4), C – (1), D – (3)
Answer:
d. A – (2), B – (4), C – (1), D – (3)

Question 2.

a. A – (4), B – (3), C – (2), D – (1)
b. A – (3), B – (1), C – (4), D – (2)
c. A – (1), B – (2), C – (3), D – (4)
d. A – (2), B – (3), C – (4), D – (1)
Answer:
b. A – (3), B – (1), C – (4), D – (2)

Question 3.

a. A – (1), B – (2), C – (3), D – (4)
b. A – (2), B – (3), C – (1), D – (4)
c. A – (3), B – (4), C – (1), D – (2)
d. A – (2), B – (3), C – (1), D – (4):
Answer:
c. A – (3), B – (4), C – (1), D – (2)

Question 4.

a. -(1), B – (2), C – (3), D – (4)
b. A -(2), B – (3), C – (1), D – (4)
c. A – (4), B – (3), C – (2), D – (1)
d. A – (3), B – (4), C – (1), D – (2)
Answer:
d. A – (3), B – (4), C – (1), D – (2)

Question 5.

. A – (ii), B – (iv), C – (i), D – (iii) b. A – (ii), B – (iii), C – (iv), D – (i)
c. A – (iii), B – (iv), C – (i), D – (ii) d. A – (iv), B – (iii), C – (ii), D – (i)

Question 6.

Anwer:
a. A – (ii), B – (iv), C – (i), D – (iii)

Question 7.

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

Question 8.
Match the following
Read the statement and fill it with correct (A) and (B)
In Halophytes …………….. shape sclereids and …………………. heavily thickened spicules that provide mechanical strength to the cortex are present in the stem.

Answer:
c) A – Star- B – H-shape

9. Match the following
Read the statement and fill it with correct (A), (B) and (C)
It is the structural defence by plants against fire. The outer bark of trees which extends to the last formed ………… A……………. is called Rhytidome. It is composed of multiple layers of suberised …………..B……………… cortical and ……….C………….. tissues.

Answer:
c) A-Periderm, B – Periderm, C – Phloem

II. Pick out the correct pair.

A) Oxylophytes – Plants living ice surface
B) Hollard – Total soil water content
C) Chresard – Water not available to plants
D) Echard – Water available to plants
Answer:
B) Hollard – Total soil water content

III. Choose the incorrect statements and select the option.

Question 1.
A) Some fungi which grow in soil of burnt areas called pyrophilous
B) Pyronema confluens is the indicator of fire.
C) Lagoons salinity will be more than 100%
D) Low temperature with high humidity helps the plants to become disease-free
a. None of these
b. A alone
c. A, B and C
d. D alone
Answer:
d. D alone

Question 2.
A) Applied ecology help us to manage and conserve the natural resource.
B) Niche – The term was coined by the naturalist Roswell Hill Johnson.
C) Halophytes are the plants which lives in saltwater.
D) Metalimnion is a zone of a gradual decrease in temperature.
a. A and B
b. A, B, and C
c. C and D
d. C alone
Answer:
d. C alone

Question 3.
Choose the incorrect one with respect to type of soil particles.

Answer:

Question 4.
A) R – Horizon consists of parent bedrock.
B) C – Horizon is partially weathered horizon.
C) B – Horizon is rich in minerals.
D) A – Horizon is often rich in humus and minerals.
a. A, B and C
b. B, C, D
c. None of these
d. C, A, and B
Answer:
c. None of these

Question 5.
a. Hollard is total soil water content
b. Echard is water not available to plant
c. The visible part of the light is made up of wavelengths from 400 nm to 700 nm
d. July 06 is Van Mahostav day
Answer:
d. July 06 is Van Mahostav day

VI. Choose the incorrect pair

Question 1.

a. A – 3), B – 4), C – 2), D – 1)
c. A – 2), B – 3), C – 4), D – 1)
b. A – 4), B – 2), C – 3), D – 1)
d. A – 1), B – 2), C – 3), D – 4)
Answer:
a. A – 3), B – 4), C – 2), D – 1)

Question 2.
Choose the incorrect pair which is related to mimicry.
A) Ophrys an orchid is an example of floral mimicry.
B) Carausius morosus – Stick insect is an example of protective mimicry.
C) Phyllium frondosum – Leaf insect another example of protective mimicry.
D) Ants take their shelter on Mango, Litchi
a. A, Band C b. B, C and D c. C alone d. None of these
Answer:
d. None of these

V. Assertion and Reason

Question 1.
Assertion(A): Hypolimniotic layer of water is always cold.
Reason (R): The water holds the temperature of the soil at the bottom of pond
a. (A) correct; (R) wrong
b. Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c. Both (A) and (R) are correct; (R) is the correct explanation of (A)
d. Both (A) and (R) are wrong.
Answer:
a. (A) correct; (R) wrong

Question 2.
Assertion: Non-succulent plants are called drought resistant plants.
Reason: They have many adaptation to resist dry conditions.
a. A is correct R is wrong
b. A is the wrong R is correct
c. Both are wrong
d. A is correct R is the correct explanation of A
Answer:
d. A is correct R is the correct explanation of A

VI. odd man out

Question 1.
Choose the odd man out:
Hornbills,
Slitz size of pollina of Apocynaceae,
Birds of scrub jungles,
Leg size of insect camel’s foot climber
Answer:
Leg size of insect camel’s foot climber

Question 2.
Choose the odd man out:
Isoetes
Hydrilla,
Potamogeton
Ranunculus Mukia.
Answer:
Ranunculus Mukia

Question 3.
Choose the odd man out:
Argemone,
Mollugo,
Tribulus
Bryophyllum
Answer:
Bryophyllum

VII. Pictorial Questions

Question 1.
Observe the diagram and choose their part from the following option.

a. Petiole, phyllode
b. Stem, leaf
c. Spines, leaves
d. Stem, Scale leaves
Answer:
d. Stem, Scale leaves

Question 2.
Observe the diagram and write the name of this plant.

Answer:
Ceratophyllum

Question 3.
Observe the diagram and choose the correct vegetation option for A, B, C and D

a. Tundra, ice, treeline, desert
b. Rain, coniferous, tropical, grass land
c. Grass, conifer, tropical, deciduous
d. Coniferous, deciduous, grassland
Answer:
d. Coniferous, deciduous, grassland

Question 4.
Draw the diagram showing altitudinal zonation of vegetation.

VIII. True or False

Question 1.
Read the sentences and select the correct options stating which ones are true (T) and which ones are false (F).
A) Juglone which inhibits the growth of seedlings of Apple, Tomato and Alfalfa around it.
B) Penicillin which inhibits the growth of staphylococcus bacteria.
C) Amensalism is not antibiosis.
D) Trichoderma inhibit the growth of Aspergillus.

Answer:
C) T, T, F, T

IX. Fill in the blanks Answers

1. Roots and hulls of Black Walnut Juglans nigra secretes an alkaloid……………………………………….
Answer:
Juglone

2. The plants which behave as xerophytes at summer and behave as mesophvtes during rainv season is……………………………….
Answer:
tropophytes

3. The ……………………………….wave length of spectrum is less strongly absorbed by plants.
Answer:
green (500 – 600 nm)

4. The rate of photosynthesis is maximum at blue (400 – 500 nm) and……………………………….
Answer:
red (600 – 700 nm)

5. ……………………………….is well known factor needed for the physiological process of plants.
Answer:
Light

6. In climatology diurnal cycle is the basic form of climatic pattern in every ………………………………..
Answer:
24 hrs.

7. The altitudinal limit of normal tree growth is about ………………………………..
Answer:
3000 to 4000

8. ……………………………….are organisms, which derive onlv water and minerals from their host plant for synthesizing their own food.
Answer:
Hemiparasites

9. ……………………………….is the smallest free floating hvdrophvtes.
Answer:
Wolffia

10. Scotoactive tvpe of stomata found in ……………………………….plants.
Answer:
succulent

11. ……………………………….are plants which grow perched on other plants.
Answer:
Epiphytes

12. The plants which are living in moderate conditions (neither too wet nor too drv) are known as……………………………….
Answer:
Mesophytes

13. ……………………………….is the world forest dav.
Answer:
March – 21

14. Earth dav falls on……………………………….
Answer:
April – 22

15. International ozone dav is celebrated on……………………………….
Answer:
September -16

X. Choose the correct answer

Question 1.
Which one of the following is not related to mutualism.
a. Anabaena present in coralloid root of cycas.
b. Wasps present in fruits of fig.
c. Tillandsia grows on the bark of oak and pine trees.
d. Cyanobacterium (Nostoc) found in the thalloid body of Anthoceros.
Answer:
c. Tillandsia grows on the bark of oak and pine trees.

Question 2.
Dionaea, Bladder wort and sarracenia are
a. Epiphytes
b. Commensals
c. Predators
d. Parasites
Answer:
c. Predators

Question 3.
Tobacco produces nicotine, coffee plants produce caffeine, cinchona plant produce quinine it is meant for…………….
a. Predators
b. Defence mechanism
c. Proto cooperation
d. Holoparasites
Answer:
b. Defence mechanism

Question 4.
Which of the following is not related to parasitism.
a. Acacia, Duranta – cuscuta
b. Balanophora,Orabanche and Refflesia – higher plants.
c. Pitcher plant with insect
d. Viscum and Loranthus – Stem parasite
Answer:
c. Pitcher plant with insect

Question 5.
Which plant pods explodes with a loud noise like cracker?
a. Boerhaavia
b.Cleome
c. Bauhinia vahlii
d.Ecballium elatrium
Answer:
d. Ecballium elatrium

Question 6.
Casuarina, Nerium, Ziziphus and Acacia examples for
a.Truexerophytes
b. Succulents
c. Ephemerals
d. Phyllode
Answer:
a. True xerophytes

Question 7.
Root pockets are present in
a. Eichhornia
b. Nelumbo
c. Potamogeton
d. Ceratophyllum
Answer:
a. Eichhornia

Question 8.
Stenophagic means
a. The organism can survive by taking wide range of food.
b. The organism can survive by taking narrow range of food.
c. The organism can live in water with wide range of salinity
d. The organism can live in water with narrow range of salinity.
Answer:
a. The organism can survive by taking wide range of food.

Question 9.
In euphorbia, acacia, ziziphus and capparis are modified into spines.
a. Stipules
b. Scales
c. Leaves
d. Bud
Answer:
a. Stipules

Question 10.
Latitude, altitude, direction of mountain, steepness of mountain etc are factors.
a. Topographic
b. Ecotone
c. Altitude
d. Edge effect
Answer:
a. Topographic

Question 11.
The roots of orchids which contain special type of spongy tissue called
a. Xylem
b. Phloem
c. Parenchyma
d. V elamen
Answer:
d. Velamen

Question 12.
Thorns of Bougainvillea, spines of opuntia and latex of cacti protect them from
a. Drought
b. Parasites
c. Predators
d. Insects
Answer:
c. Predators

Question 13.
Which seeds showing highest longevity in plant kingdom.
a. Lotus
b. Hydrilla
c. Nymphaea
d. Marsilea
Answer:
a. Lotus

Question 14.
plants are completing life cycle within a short period.
a. Lotus
b. Hydrilla
c. Nymphaea
d. Marsilea
Answer:
a. Lotus

XI. Two Marks

Question 1.
What is biotope and ecotope?
Answer:
The environment of any community is called biotope.
The habitat and niche of any organism is called ecotope.

Question 2.
What is biome?
Answer:
Biome is a large naturally occurring community of flora and fauna occupying in a major habitat. (Eg.) forest, fundra

Question 3.
Difference between evergreen forest and sclerophyllous forest?
Answer:
Evergreen forest
It is found where heavy rainfall occurs throughout the year.

Sclerophyllous forest
It is found where heavy rainfall occurs during winter and low rainfall during summer.

Give reason for the following

• Species of the grass lands of western ghats of India differ from the grass species of temperate grass lands of steppe in North America. But they are all ecologically primary producers.
• Taxonomically different species occupying similar habitats (Niche) in different geographical regions are called ecological equivalents.

Question 4.
What are the various latitudinal zonation of vegetation ?
Answer:
Tropical rain forest, grassland (or) desert, deciduous forest, coniferous forest, treeline (or) tundra, snow (or) ice.

Question 5.
What are the various altitudinal zonation of vegetation?
Answer:
Tropical rain forest, grassland (or) desert, deciduous forest, coniferous forest, tundra snow are the various vegetation of altitudinal zonation.

Question 6.
Differentiate between euryhaline and stenohaline.
Answer:
Euryhaline Stenohaline

1. Organisms which can live in water with wide range of salinity.
2. (Eg.) Marine algae and marine angiosperms

stenohaline.

1. Organisms which can withstand only small range of salinity.
2. (Eg.) Plants of estuaries

Question 7.
What is albedo effect ?
Answer:

• Aerosols with small particles is reflecting the solar radiation entering the atmosphere called albedo effect.
• It reduces the temperature limits, photo synthesis and respiration.

Question 8.
Define : Indicators of fire (or) Why pteris and pyronema are called as fire indicators?
Answer:

• Some pteris are well adapted to grow in burnt and highly disturbed area
• Pteris (fern) and pyronema (fungus) indicates the burnt up and fire disturbed areas.
• So they are called indicators of fire.

Question 9.
What is edge effect?
Answer:

• Some species are found in the ecotone areas border between forest and grassland due to the effect of the environment of the two habitats. This is called the edge effect.
• (Eg.) Owl in the ecotone area between forest and grassland.

Question 10.
What is the ecotone area?
Answer:
The transition zone between two ecosystems. (Eg.) The border between forest and grassland.

Question 11.
Why do valleys are rich in vegetation compare to the steepness of the mountain?
Answer:

• The steepness of the mountain (or) hill allows the rain to run off.
• Asa result the loss of water causes water deficit and quick erosion of the top soil resulting in poor vegetation.
• On the other hand the plains and valley are rich in vegetation due to the slow drain of surface water and retention of water in the soil.

Question 12.
Draw the picture and mark the A and B.
Write the name of the plant which is present on the host trunk.
Answer:

Question 13.
Write the examples for the following type of holo parasites (or) total parasites.
a) Total stem parasite:………………
b) Total root parasite on higher plants :……………., …………………, ……………..
Answer:
a) Cuscuta
b) Balanophora, Orobanche and Reflesia.

Question 14.
Write the examples for the following type of hemi parasites (or) partial parasites.
a) Partial stem parasite :……………… , ………………
b) Partial root parasite:………………
Answer:
a) Viscum and Loranthus
b) Santalum (sandal wood)

Question 15.
“The utricularia (Bladderwort) competes with tiny fishes for small crustaceans and insects” What type of interaction exist in the above examples.
Answer:

• Inter-specific competition is exist in the above examples.
• It is an interaction between individuals of different species for common need.

Question 16.
What is intra specific competition ?
Answer:

• It is an interaction between individuals same species.
• It is very severe because all the members of species have similar requirement of food habitat, pollination etc.

Question 17.
What is the competition?
Answer:
It is an interaction between two organisms (or) species in which both the organisms (or) species are harmed.

Question 18.
What is junglone?
Answer:
Roots and hulls of black walnut junglone nigra secretes an alkaloid junglone.

• Which inhibits the growth of seedlings of apple,tomato and alfalfa around it.
• It is an example for amensalism type of interspecific interaction.

Question 19.
What are trichophyllous plants? Give example.
Answer:
In xerophytes, the leaves and stem are covered with hairs are called trichophyllous plants. (Eg.) cucurbits, (melothria and mukia)

Question 20.
Orchids, money plant and lianas are epiphytes. Why? (or) What are epiphytes give an example.
Answer:

• Many orchids ferns, lianas, money plant usnea (lichen) are some of examples of epiphytes.
• These plants which are found on other plants and growing without harming them are called epiphytes.

Question 21.
What are tropophytes?
Answer:
Tropophytes are plants which behave as xerophytes at summer and behave as mesophytes (or) hydrophytes during rainy season.

Question 22.
Seeds of maple gyrocarpus, dipterocarpus and terminalia exhibit which type of adaptations for dispersal of fruits and seeds?
Answer:
Seeds (or) whole fruits are flattened to form a wing for dispersal by wind.

Question 23.
Guess !! Who am I………….? I am dispersed by ant and I have caruncle.
Answer:

• Caruncle is a structure found in micropylar region of euphorbiaceae seeds, that attract ants. Which feed the caruncle to their larvae.
• Then ants leave the seed to their waste disposal area. Where the seeds germinate.
• This type of seed dispersal called myrmecophily.

Question 24.
What are xerophytes ? What are its types ?
Answer:
Xerophytes are plants, which are living in dry (or) xeric conditions are known as xerophytes. They are a) Physical dryness, b) Physiological dryness

Question 25.
In some habitats, water is sufficiently present but plants are unable to absorb it. Why? How do you call it?
Answer:

• It is called physiological dryness.
• In these habitats, water is sufficiently present but plants are unable to absorb it because of the absence of capillary spaces.
• (Eg.) Plants in salty and acidic soil.

Question 26.
Define pedology
Answer:
The study of soils is called pedology

Question 27.
What is palaeoclimatology? Give an example.
Answer:

• Helps to reconstruct past climates of our planet and flora, fauna and ecosystem in which they lived.
• Example : Air bubbles trapped in ice for tens of thousands of years with fossilized pollen, coral, plant and animal debris.

Question 28.
What is sclerophyllous forests?
Answer:
Found where heavy rainfall occurs during winter and low rainfall during summer.

Question 29.
Define Anemometer?
Answer:

• Air in motion is called wind.
• It is also a vital ecdogical factor.
• The atmospheric air contains a number of gases, particles and other constituents.
• Anemometer is the instrument used to measure the speed of wind.

Question 30.
Which is Nitrogen fixation?
Answer:
Rhizoblum (Bacterium) forms nodules in the roots of leguminous plants and lives symbiotically.
The Rhizobium obrains food from leguminous plant and in turn fixes atmospheric nitrogen in to nitrate, making it available to host plants.

XII. Three Marks

Question 1.
Is there any limit of tree growth related to altitude ? (or) What is treeline or timberline?
Answer:

• Timberline (or) tree line is an imaginary line in a mountain (or) higher areas of land that marks the level above which trees do not
  grow.
• The altitudinal limit of normal tree growth is about 3000 to 4000 m.

Question 2.
How does Rhytidome protect the plant against forest fire?
Answer:

• It is the structural defense by plant against.
• The outer bark of trees which extends to the last formed periderm is called Rhytidome.
• It is composed of multiple layers of suberized periderm cortical and phloem tissue. It protects the stem against fire, water loss, invasion of insects and prevents infections by microorganism

Question 3.
Which branch of ecology help us to manage and conserve natural resources, particularly ecosystem? (or) What is applied ecology (or) environmental technology?
Answer:
Applied ecology (or) environmental technology helps us to manage and conserve natural resource particularly ecosystems, forest and wild life conservative management.

Environmental management involves bio-diversity, conservation, ecosystem restoration, habitat management, invasive SPS management protected areas management and also help us plan landscapes and environmental impact designing for the futuristic ecology.

Question 4.
What are ecological factors (or) environmental factors?
Answer:

• Many organisms co-exist in an environment.
• The environment includes physical, chemical biological component.
• When a component surrounding an organism affects the life of an organism called factor.
• These factors may be biotic and abiotic.

Question 5.
What are the environmental factors affecting a plant life?
Answer:

• Climatic factors : Sunlight, precipitation wind, carbon dioxide and water vapour.
• Biotic factors : Birds, insects, man, grazing animals, rodents, plant pathogens and epiphytes.
• Edaphic factors : Soil slope, soil water, physical nature of soil, minerals, soil air.

Question 6.
What are the various effects of light upon a green plant ?
Answer:
Various effects of light on green plants are photosynthesis, opening and closing of sto¬mata movements, germination of seeds, flowering tuber formation, runner production stem and leaf formation.

Question 7.
What is phytoremediation ? Given Example.
Answer:
Some plants are used to remove cadmium from contaminated soil is known as phyto remediation. tM.-fTiVM
(Eg.) Rice and eichhornia tolerate cadmium by binding it to their proteins.
Soybean and tomato manage to tolerate cadmium by storing into few group of cells.

Question 8.
Which soil is ideal for cultivation ? (or) Why loamy soil is ideal for cultivation ?
Answer:

• Loamy soil is ideal for cultivation. It consist of 70% sand and 30% clay (or) silt.
• It ensures good retention and proper drainage of water.
• The porosity of soil provides adequate aeration and allows the penetration of roots.

Question 9.
What is soil profile (or) super-imposed horizons?
Answer:

• Soil is commonly stratified into horizons at different depth.
• These layers differ in their physical, chemical and biological properties.
• This succession of super-imposed horizons is called soil profile.

Question 9.
Differentiate species ecology from community ecology.
Answer:
Species Ecology :
It is the study of a group of individual of particular species, (population)

Community Ecology :
It is the study of several species that are living together as a community (made up of several populations)

Question 10.
Why do vegetation at different altitude are varies ?
Answer:
Height above the sea level forms the altitude.
At high altitudes

• The velocity of wind remains high.
• Temperature and air pressure-decreases.
• While humidity and intensity of light increases.

Due to these factors vegetation at different altitudes varies showing distinct zonation.

Question 11.
Why do different types of vegetation occur from equator to poles (or) latitudes?
Answer:

• Latitudes represent distance from the equator.
• Temperature values are maximum at the equator and decrease gradually towards poles.
• So different types of vegetation occur in latitude.

Question 12.
What is topography? WTiat are its factors?
Answer:

• The surface features of earth are called topography.
• Its factors include, latitude, altitude, direction of mountain, steepness of mountain.

Question 13.
Spanish Moss-Tillandsia grows on the bark of oak and pine trees. Which type of interactions exist in the above examples ?
Answer:

• They were epiphytes showing commensalism type of positive interaction.
• In which one is benefitted and other is neither benefitted nor harmed.
• The species that derives benefit is called the commensal, while the other species is called the host.

Question 14.
What is velamen?
Answer:

• The epiphytic higher plant gets its nutrients and water from the atmosphere with the help of their hygroscopic roots.
• These roots contain special type of spongy tissue called velamen.

Question 15.
What is proto cooperation?
Answer:
An interaction between organisms of different species in which both organisms benefit but neither is dependent on the relationship.
(Eg.) Soil bacteria, fungi and plants growing in the soil.

Question 16.
Draw the picture and mark A, B and C

Answer:
A – Tendril, B – Pitcher, C – Lamina

Question 17.
Draw the diagram mark A, B and C.
Answer:
A – Haustoria, B – Host, C – Parasite

a) Name the parasite with example.
Answer:
Holoparasite Cuscuta.

Question 18.
What are holo parasites ? (or) total parasites.
Answer:
The organisms which are dependent upon the host plants for their entire nutrition are called holo parasites (or) total parasites. (Eg.) Cuscuta.

Question 19.
Trichoderma (fungus) inhibits the growth of fungus aspergillus. Why ? (or) What is amensalism? (or) WTiat is antibiosis?
Answer:

• It is an inter specific interaction in which one species is inhibited while the other species is neither benefitted nor harmed.
• The inhibition is achieved by secretion of chemicals called allelopathic substances.

Question 20.
Phyllium frondosum, carausius morosus exhibit what type of interactions ? (or) WTiat is mimicry?
Answer:

• It exhibits mimicry of inter specific interactions (or) mimicry of co-evolutionary dynamics.
• Mimicry is a phenomenon in which living organism modifies its form appearance structure (or) behaviour and looks like another living organism as a self defence and increases the chance of their survival.

Question 21.
What is kairomones ? (or) Which type of defence induced by the predator to the progeny of wild radish?
Answer:
kairomone is a chemical substance emitted by pieris rapae caterpillar (butterfly) exposed to wild radish gets the capacity to transmit defence induced by predator (butterfly) to progeny of wild radish.

Question 22.
Name the phenomenon which is exhibited by acacia and acacia ants ? What does it says?
Answer:

• Sometimes, ants take their shelter on some trees such as Mango, Litchi, Jamun, Acacia etc.
• These ants act as body guards of the plants against any disturbing agent and the plants in turn provide food and shelter to these ants.
• This phenomenon is known as Myrmecophily. Example: Acacia and acacia ants.

Question 23.
What is hygrophytes ? Why do they call so ?
Answer:
The plants which can grow in moist damp and shady places are called hygrophytes.
(Eg.) Habenaria (orchid),
Mosses (Bryophytes)

Question 24.
How do mangroves work to protect us from natural disaster with example.
Answer:

• Mangroves protect vulnerable coastal areas from wave action by holding the soil together and prevent coastal erosion.
• Out of three districts of Tamil Nadu (Nagapattinam, Thanjavur and Thiruvarur), Muthupet (Thiruvarur district) was less damaged by Gaja Cyclone (Nov-2018) due to the presence of mangrove forest.

Question 25.
Some plants are called drought evaders. Why?
Answer:

• Ephemerals are called drought evaders (or) drought escapers because
• These plants complete their life cycle within a short period, (single season)
• (Eg.) Argemone, Mollugo, Tribulus, Tephrosia.

Question 26.
What is the source of energy in deep sea?
Answer:

• In deep sea (>500 m) the environment is dark and its inhabitants are not aware of the existence of celestial source of energy called sun.
• Dead sea organisms use chemical energy rather than energy from sunlight.
• Chemosynthesis is a process, special bacteria use this process to produce energy without using sunlight.

Question 27.
Write any four ecologically important days.
Answer:

• March 21 – World Forest Day
• May 22 – World Biodiversity Day
• June 05 – World Environment Day
• July 07 – Van Mohot Stav day.

Question 28.
What is phylloclades (or) fleshy leaf? Give an example
Answer:
In some xerophytes all the internode in the stem are modified in to a fleshy leaf structure is called phylloclades.
Ex.→ opuntia

Question 29.
What is breathing roots? Give an example.
Answer:
A special type of negatively geotropic roots called pneumatophores with pneumathodes to get sufficient aeration are also present.
They are called breathing roots.
Example : Avicennia

Question 30.
Define cladode? Give an example
Answer:
In some of the others single or occasionally two internodes modified into fleshy green structure called cladode Example: Asparagus

Question 31.
Define phyllode?
Answer:
In some the petiole is modified into a fleshy leaf like structure called phyllode (Acacia melonoxylon)

XIII. Five Marks

Question 1.
What are the various zonation based on climatic factor temperature and its effects ?
Answer:
Temperature based zonation:

• Variations in latitude and altitude do affect the temperature and the vegetation on the earth surface.
• Timber line / Tree line : It is an imaginary line in a mountain or higher areas of land that marks the level above which trees do not grow. The altitudinal limit of normal tree growth is about 3000 to 4000m.

Effects of temperature:

• The following physiological processes are influenced by temperature:
• Temperature affects the enzymatic action of all the bio-chemical reactions in a plant body.
• It influences CO2 and O2 solubility in the biological systems. Increases respiration and stimulates growth of seedlings.
• Low temperature with high humidity can spread diseases to plants
• The varying temperature with moisture determines the distribution of the vegetation types.

Question 2.
What are the important climatic effects of wind on plants ?
Answer:
Effects of wind:

• Wind is an important factor for the formation of rain
• Causes wave formation in lakes and ocean, which promotes aeration of water
• Strong wind causes soil erosion and reduces soil fertility
• Increases the rate of transpiration
• Helps in pollination in anemophilous plants
• It also helps in dispersal of many fruits, seeds, spores, etc.
• Strong wind may cause up-rooting of big trees
• Unidirectional wind stimulates the development of flag forms in trees.

Question 3.
Write about important edaphic factors which affect vegetation of plants ?
Answer:
The important edaphic factors which affect vegetation are as follows:
1. Soil moisture:
Plants absorbs rain water and moisture directly from the air.

2. Soil water:
Soil water is more important than any other ecological factors affecting the distribution of plants. Rain is the main source of soil water. Capillary water held between pore spaces of soil particles and angles between them is the most important form of water available to the plants.

3. Soil reactions:
Soil may be acidic or alkaline or neutral in their reaction. pH value of the soil solution determines the availability of plant nutrients. The best pH range of the soil for cultivation of crop plants is 5.5 to 6.8.

4. Soil nutrients:
Soil fertility and productivity is the ability of soil to provide all essential plant nutrients such as minerals and organic nutrients in the form of ions.

5. Soil temperature:
Soil temperature of an area plays an important role in determining the geographical distribution of plants. Low temperature reduces use of water and solute absorption by roots.

6. Soil temperature:
The spaces left between soil particles are called pore spaces which contains oxygen and carbon-di-oxide.

7. Soil organisms:
Many organisms existing in the soil like bacteria, fungi, algae, protozoans, nematodes, insects, earthworms, etc. are called soil organisms.

Question 4.
What is topography? What are the various topographic factors involves (or) influence on the climate of any area ?
Answer:
The surface features of earth are called topography. Its factors include, latitude, altitude, direction of mountain, steepness of mountain.

a. Latitudes and altitudes:

• Latitudes represent distance from the equator. Temperature values are maximum at the equator and decrease gradually towards poles.
• So different types of vegetation occur in latitude.
• Height above the sea level forms the altitude. At high altitudes
• The velocity of wind remains high. Temperature and are pressure – decrease While humidity and intensely of light increases.
• Due to these factors vegetation at difference altitudes varies showing distinct zonation.

b. Direction of mountain:
North and south faces of mountain (or) hill possess different types of flora and fauna because thev differ in their humidity, rainfall light intensity, light duration and temperature regions.

c. Steepness of the mountain:

• The steepness of the mountain (or) hill allows the rain to run off.
• As a result the loss of water causes water deficit and quick erosion of the top soil resulting in poor vegetation.
• On the other hand the plains and valley are rich in vegetation due to the slow drain of surface water and retention of water in the soil.

Question 5.
What are various biotic factors which is exist between the organism ?
Answer:
The interactions among living organisms such as plants and animals are called biotic factors.

Positive interactions:
When both (or) one the participating species are benefited. (Eg.) Mutualism and commen¬salism.

Mutualism:

• It is an interaction between two species of organisms in which both are benefitted from the obligate association.
• Lichens is a mutual association of an algae and a fungus.
• The alga is usually green alga (or) blue green alga. The fungus is an ascomycete (or) basidiomycetes.
• It is believed that alga contributes organic food from photosynthesis and the fungus is able to absorb water and mineral salts.
• The fungus can also conserve water and this enables lichens to grow in extreme dry conditions where no other plants can exist.

Nitrogen fixation:

• Rhizobium (Bacterium) forms nodules in the roots of leguminous plants and lives symbiotically.
• The Rhizobium obtains food from leguminous plant and in turn fixes atmospheric nitrogen into nitrate, making it available to host plants.
• Example: Water fern (Azolla) and Nitrogen fixing cyanobacterium (Anabaena)
• Anabaena present in coralloid roots of cycas

Commensalism:
Many orchids ferns, lianas, money plant usnea (lichen) are some of the examples of epiphytes.
These plants which are found on other plant and growing without harming them are called epiphytes.

Negative interactions:

a. Predation:
It is an interaction between two species, one of which captures, kills and eats up the other. The species which kills is called a predator and the species which is killed is called a prey. The predator is benefitted while the prey is harmed.

Examples:
A number of plants like Drosera (Sun dew Plant), Nepenthes (Pitcher Plant), Dionaea (Venus fly trap), Utricularia (Bladder wort) and Sarracenia are predators which consume insects and other small animals for their food as a source of nitrogen. They are also called as insectivorous plants.

Many herbivores are predators. Cattles, Camels, Goats etc., frequently browse on the tender shoots of herbs, shrubs and trees. Generally annuals suffer more than the perennials. Grazing and browsing may cause remarkable changes in vegetation. Nearly 25 percent of all insects are known as phytophagous(feeds on plant sap and other parts of plant)

Question 6.
Write the morphological adaptations of xerophytes.
Answer:
Morphological Adaptations:
In root:

• Root system is well developed and is greater than that of shoot system.
• Root hairs and root caps are also well developed.
• In Xerophytic plants with the leaves and stem are covered with hairs are called tricho- phyllous plants.
• Eg: Cucubits (Melothria andMukia)

In stem:

• Stems are mostly hard and woody. They may be aerial or underground.
• The stems and leaves are covered with wax coating or covered with dense hairs.
• In some xerophytes all the internodes in the stem are modified into a fleshy leaf structure called phylloclades (Opuntia)
• In some of the others single or occasionally two internodes modified into fleshy green structure called cladode (Asparagus)
• In some the petiole is modified into a fleshy leaf like structure called phyllode (Acacia melanoxylon)

a) A succulent xerophyte: Phylloclade – opuntia
b) Non succulent: Perennial – Capparis
c) Cladode of Asparagus
d) Phyllode – Acacia
aquatic and terrestrial modes of life. They grow in shallow water.
Examples: Ranunculus, Typha and Sagittaria.

Question 7.
Write anatomical and physiological adaptation of hydrophytes ?
Answer:
Anatomical adaptations:

• Cuticle is either completely absent or if present it is thin and poorly developed
• Single layer of epidermis is present
• Cortex is well developed with aerenchyma
• Vascular tissues are poorly developed. In emergent forms vascular elements are well developed.
• Mechanical tissues are generally absent except in some emergent forms. Pith cells are sclerenchymatous.

Physiological adaptations of hydrophytes:

• Hydrophytes have the ability to withstand anaerobic conditions.
• They possess special aerating organs.

Question 8.
What are epiphytes ? What are its morphologic adaptations?
Answer:
Epiphytes:
Epiphytes are plants which grow perched on other plants (Supporting plants). They use the supporting plants only as shelter and not for water or food supply.
Orchids, Lianas, Hanging Mosses and Money plant.

Morphological adaptations:

• Root system is extensively developed. These roots may be of two types. They are Clinging roots and Aerial roots.
• Clinging roots fix the epiphytes firmly on the surface of the supporting objects.
• Aerial roots are green coloured roots which may hang downwardly and absorb moisture from the atmosphere with the help of a spongy tissue called velamen.
• Stem of some epiphytes are succulent and develop pseudo bulb or tuber.
• Generally the leaves are lesser in number and may be fleshy and leathery Myrmecophily is a common occurrence in the epiphytic vegetation to prevent the predators.

Question 9.
Write about anatomical and physiological adaptations of epiphytes ?
Answer:
Anatomical adaptations:

• Multilayered epidermis is present. Inner to the velamen tissue, the peculiar exodermis layer is present.
• Presence of thick cuticle and sunken stomata greatly reduces transpiration.
• Succulent epiphytes contain well developed parenchymatous cells to store water.

Physiological adaptations:
Special absorption processes of water by velamen tissue.

Question 10.
Write about anatomical and physiological adaptations of halophytes ?
Answer:
Anatomical adaptations:

• Epidermal cells of stem is heavy cutinized, almost squarish and are filled with oil and tannins.
• Star’ shaped sclereids and ‘H’ shaped heavy thickened spicules that provide mechanical strength to cortex are present in the stem.
• The leaves may be dorsiventral or isobilateral with salt secreting glands. Physiological adaptations:
• High osmotic pressure exists in some plants
• Seeds germinate in the fruits of mother plant itself (Vivipary).

Question 11.
What are the adaptations found in hy drochory plants with example ?
Answer:
Dispersal of seeds and fruits by water usually occurs in those plants which grow in or near water bodies. Adaptation of hy drochory are

• Obconical receptacle with prominent air spaces. Example: Neliimbo.
• Presence of fibrous mesocarp and light pericarp. Example: Coconut.
• Seeds are light, small, provided with aril which encloses air. Example: Nymphaea.
• The fruit may be inflated. Examples:

Heritiera littoralis.

• Seeds by themselves would not float may be carried by water current.
  Example: Coconut.

Question 12.
Write about the various adaptation of wind dispersal plant (or) anemochory plants.
Answer:
The individual seeds or the whole fruit may be modified to help for the dispersal by wind. Wind dispersal of fruits and seeds is quite common in tall trees.

• Minute seeds: Seeds are minute, very small, light and with inflated covering. Example: Orchids.
  Wings : Seeds or whole fruits are flattened to form a wing. Examples: Maple, Gyrocarpus, Dipterocarpus and Terminalia
• Feathery Appendages : Seeds or fruits may have feathery appendages which greatly increase their buoyancy to disperse to high altitudes. Examples: Vernonia and Asclepias.
• Censor mechanisms : The fruits of many plants open in such a way that the seeds can escape only when the fruit is violently shaken by a strong wind. Examples: Aristolochia and Poppy.

Question 13.
What is autochory? Write about its explosive mechanism for dispersal of fruits and seeds.
Answer:

• Some fruits burst suddenly with a force enabling to throw seeds to a little distance away from the plant. Autochory shows the following adaptations.
• Mere touch of some plants causes the ripened
  fruit to explode suddenly and seeds are thrown out with great force. Example: Impatiens (Balsam), Hura.
• Some fruits when they come in contact with water particularly after a shower of rain, burst suddenly with a noise and scatter the seeds. Examples: Ruellia and Crossandra.
• Certain long pods explode with a loud noise like cracker, scattering the seeds in all directions. Example: Bauhinia vahlii (Camel’s foot climber)
• As the fruit matures, tissues around seeds are converted into a mucilaginous fluid, due to which a high turgor pressure develops inside the fruit which leads to the dispersal of seeds.
• Example: Ecballium elaterium (Squirting cucumber) Gyrocarpus and Dipterocarpus.

Question 14.
Explain the role of wind as a vital Ecological
Answer:

• Air in motion is called wind. It is also vitral ecological factor.
• The atmospheric air contains a number of gases, particles and other constituents.
• The composition of gases in atmosphere is as follows
• Nitrogen – 78%, Oxygen – 21%., Carbon-di¬oxide – 0.03%., Argon and other gases – 0.93%.
• The other components of wind are water vapour, gaseous pollutants, dust, smoke particles, microorganisms pollen grains, spores etc.
• Anemometer is the instrument used to measure the speed of wind.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4

Question 1.
Write the Maclaurin series expansion of thef following functions:
(i) e^x
(ii) sin x
(iii) cos x
(iv) log (1 – x); – 1 ≤ x ≤ 1
(v) tan^-1 (x); -1 ≤ x ≤ 1
(vi) cos² x
Solution:
(i) Let f(x) = e^x
f(x) = e^x f'(0) = e° = 1
f(x) = e^x f'(0) = e° = 1
f”(x) = e^x f”(0) = e° = 1
Maclaurin ‘s expansion is

(ii) Let f(x) = sin x
f(x) = sin x; f(0) = 0
f'(x) = cos x; f'(0) = 1
f”(x) = -sin x; f”(0) = 0
f”‘(x) = -cos x; f”'(0) = -1
f^IV(x) = sin x; f^IV(0) = 0
f^V(x) = cos x; f^V(0) = 1
f^VI(x) = -sin x; f^VI(0) = 0
f^VII(x) = -cos x; f^VII(0) = -1
Maclaurin ‘s expansion is

(iii) Let f(x) = cos x
f(x) = cos x ; f(0) = 1
f'(x) = -sin x ; f'(0) = 0
f”(x) = -cos x ; f”(0) = -1
f”'(x) = sin x ; f”'(0) = 0
f^IV(x) = cos x ; f^IV(0) = 1
f^V(x) = -sin x ; f^V(0) = 0
f^VI(x) = -cos x ; f^VI(0) = -1
Maclaurin ‘s expansion is

(iv) log (1 – x); – 1 ≤ x ≤ 1

Maclaurin ‘s expansion is

(v) tan^-1 (x); -1 ≤ x ≤ 1
f(x) = tan^-1 x ; f(0) = 0
f'(x) = \(\frac { 1 }{ 1+x^2 }\) f'(0) = 1
= 1 – x² + x⁴ – x⁶ + …..
f”(x) = -2x + 4x³ – 6x⁵ + ….. f”(0) = 0
f”'(x) = -2 + 12x² – 30x⁴ + ….. f”(0) = -2
f^IV(x) = 24x – 120x³ + …… f^IV(0) = 0
f^V(x) = 24 – 360x² + ….. f^V(0) = 24 .
f^VI(x) = -720x + ….. f^VI(0) = 0
f^VII(x) = -720 + … f^VII(0) = -720
Maclaurin ‘s expansion is

(vi) Let f(x) = cos² x
f(x) = cos² x ; f(0) = 1
f'(x) = -2 cos x sin x ; f'(0) = 0
= -sin 2 x
f”(x) = -2 cos 2x ; f”(0) = -2
f”‘(x) = 4 sin 2x ; f”‘(0) = 0
f^IV(x) = 8 cos 2x ; f^IV( 0) = 8
f^V(x) = -16 sin 2x ; f^V(0) = 0
f^VI(x) = -32 cos 2x ; f^VI(0) = -32
Maclaurin’s expansion is

Question 2.
Write down the Taylor series expansion, of the function log x about x = 1 upto three non-zero terms for x > 0.
Solution:
Let f(x) = log x
Taylor series of f(x) is

Question 3.
Expand sin x ascending powers x – \(\frac { π }{ 4 }\) upto three non-zero terms.
Solution:
Let f(x) = sin x

Taylor series of f(x) is

Question 4.
Expand the polynomial f(x) = x² – 3x + 2 in power of x – 1.
Solution:
Let f(x) = x² – 3x + 2
f(x) = x² – 3x + 2 ; f(1) = 0
f'(x) = 2x – 3 ; f'(1) = -1
f”(x) = 2 ; f”(1) = 2
Taylor series of f(x) is
f(x) = \(\sum_{n=0}^{n=\infty}\) a_n (x – 1)ⁿ, where a_n = \(\frac { f^{(n)} (1)}{ n! }\)
∴ The required expansion is
x² – 3x + 2 = 0 – \(\frac { 1(x-1) }{ 1! }\) + \(\frac { 2(x-1)^2 }{ 2! }\)
= -(x – 1) + (x – 1)²

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Botany Guide Pdf Chapter 9 Plant Breeding Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 9 Plant Breeding

12th Bio Botany Guide Plant Breeding Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
Assertion : Genetic variation provides the raw material for selection.
Reason : Genetic variations are differences in genotypes of the individuals.
a) Assertion is right and reason is wrong.
b) Assertion is wrong and reason is right.
c) Both reason and assertion is right.
d) Both reason and assertion is wrong.
Answer:
b) Assertion is wrong and reason is right.

Question 2.
While studying the history of domestication of various cultivated plants ………………. were recognized earlier.
a) Centres of origin
b) Centres of domestication
c) Centres of hybrid
d) Centres of variation
Answer:
a) Centres of origin

Question 3.
Pick out the odd pair …………..
a) Mass selection – Morphological characters
b) Purline selection – Repeated self pollination
c) Clonal selection — Sexually propagated
d) Natural selection – Involves nature
Answer:
c) Clonal selection – Sexually propagated

Question 4.
Match Column I with Column II

Answer:
b) i -III, ii-I, iii-IV, iv-II

Question 5.
The quickest method of plant breeding is
a) Introduction
b) Selection
c) Hybridization
d) Mutation breeding
Answer:
b) Selection

Question 6.
Desired improved variety of economically useful crops are raised by
a) Natural selection
b) hybridization
c) mutation
d) biofertilisers
Answer:
b) hybridization

Question 7.
Plants having similar genotypes produced by plant breeding are called
a) clone
b) haploid
c) autopolyploid
d) genome
Answer:
a) clone

Question 8.
Importing better varieties and plants from outside and acclimatising them to local environment is called
a) cloning
b) heterosis
c) selection
d) introduction
Answer:
d) Introduction

Question 9.
Dwarfing gene of wheat is
a) pal 1
b) Atomita 1
c) Norin 10
d) pelita 2
Answer:
c) Norin 10

Question 10.
Crosses between the plants of the same variety are called
a) interspecific
b) inter varietal
c) intra varietal
d) inter generic
Answer:
c) Intra varietal

Question 11.
Progeny obtained as a result of repeat self pollination a cross pollinated crop to called
a) pure line
b) pedigree line
c) inbreed line
d) heterosis
Answer:
b) Pure line

Question 12.
Jaya and Ratna are the semi dwarf varieties of
a) wheat
b) rice
c) cowpea
d) mustard
Answer:
b) Rice

Question 13.
Which one of the following are the species that are crossed to give sugarcane varieties with high sugar, high yield, thick stems and ability to grow in the sugarcane belt of North India?
a) Saccharum robustum and Saccharum officinarum
b) Saccharum barberi and Saccharum officinarum
c) Saccharum sinense and Saccharum officinarum
d) Saccharum barberi and Saccharum robustum
Answer:
b)Saccharum barberi and Saccharum officinarum

Question 14.
Match column I (crop) with column II (Corresponding disease resistant variety) and select the correct option from the given codes.

Answer:
b) I-(ii), II-(i), IH-(iii), IV-(iv)

Question 15.
A Wheat variety, Atlas 66 which has been used as a donor for improving cultivated wheat, which is rich in
a) iron
b) carbohydrates
c) proteins
d) vitamins
Answer:
c) proteins

Question 16.
Which one of the following crop varieties correct matches with its resistance to a disease

Answer:
a) Pusa Komal – Bacterial blight

Question 17.
Which of the following is incorrectly paired?
a) Wheat – Himgiri
b) Milch breed – Sahiwal
c) Rice – Ratna
d) Pusa Komal – Brassica
Answer:
d) Pusa Komal – Brassica

Question 18.
Match list I with list II

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

Question 19.
Differentiate primary introduction from secondary introduction
Answer:

Question 20.
How are microbial innoculants used to increase the soil fertility?
Answer:

Biofertilizers or microbial innoculants are defined as preparations containing living cells or latent cells of efficient strains of microorganisms that help crop plants uptake of nutrients by their interactions in the rhizosphere when applied through seed or soil.

They are efficient in fixing nitrogen, solubilising phosphate and decomposing cellulose. They are designed to improve the soil fertility, plant growth, and also the number and biological activity of beneficial microorganisms in the soil. They are ecofriendly organic agro inputs and are more efficient and cost effective than chemical fertilizers.

Question 21.
What are the different types of hybridization?
Answer:
Types of hybridization:

(i) Intravarietal hybridization:
The cross between the plants of same variety. such crosses are useful only in the self. pollinated crops.

(ii) Intervarietal hybridization:
The cross between the plants belonging to two different varieties of the same species and is also known as intraspecific hybridization.

(iii) Interspecific hybridization:
The cross between the plants belonging to different species belonging to the same genus is also called intragenic hybridization.
Example:
Gossypium hirsutum
Gossypium arboreum

(v) Intergeneric hybridization:

• The crosses are made between the plants belonging to two different genera.
• The disadvantages are hybrid sterility time consuming and expensive procedure.
  Example : Raphanobrassica x Triticale

Question 22.
Explain the best suited type followed by plant breeders at present?
Answer:
Mutation breeding represents a new method of conventional breeding procedures as they have the advantage of improving the defect without losing an agronomic and quality character in agriculture and crop improvement. Mutation means the sudden heritable changes in the genotype or phenotype of an organism. Gene mutations are of considerable importance in plant breeding as they provide essential inputs for evolution as well as for recombination and selection. It is the only method for improving seedless crops.

Question 23.
Write a note on heterosis.
Answer:

• The superiority of the FI hybrid in performance over its parents is called heterosis or hybird vigour.
• G.H. Shull was the first scientist to use the term heterosis in 1912.
• Heterosis are of the following types.
• Euheterosis, Mutational Euheterosis, Balanced Euheterosis and Pseudoheterosis

(i) Euheterosis:
This is the true heterrosis which is inherited and is further classified as.

(a) Mutational Euheteosis:
Simplest type of euheterosis and results from the sheltering or eliminating of the deleterious unfavourable often lethal, recessive, mutant genes by their adaptively superior dominant alleles in cross pollinated crops.

(b) Balanced Euheterosis :
well balanced gene combinations which is more adaptive to environmental conditions and agricultural usefulness.

(ii) Psuedohetrosis:
Also termed as luxuriance progeny possess superiority over parents in vegetative growth but not in yield and adaptation usually sterile or poorly fertile.

Question 24.
List out the new breeding techniques involved in developing new traits in plant breeding.
Answer:
New Breeding Techniques (NBT) are a collection of methods that could increase and accelerate the development of new traits in plant breeding. These techniques often involve genome editing, to modify DNA at specific locations within the plants to produce new traits in crop plants. The various methods of achieving these changes in traits include the following.

• Cutting and modifying the genome during the repair process by tools like CRISPR /Cas.
• Genome editing to introduce changes in a few base pairs using a technique called Oligonucleotide-directed mutagenesis (ODM).
• Transferring a gene from an identical or closely related species (cisgenesis).
• Organizing processes that alter gene activity without altering the DNA itself (epigenetic methods)

12th Bio Botany Guide Plant Breeding Additional Important Questions and Answers

I. Match the Following

Question 1.

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

Question 2.

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

Question 3.

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

Question 4.

A) a – i, b – ii, c – iv, d – iii
B) a – ii, b – iii, c – iv, d – i
c) a – iii, b – iv, c – i, d – iii
D) a – iv, b – i, c – iii, d – ii
Answer:
B) a – ii, b – iii, c – iv, d – i

II. Choose the Odd man Out

Question 1.
a) Rhizoblum
b) Azolla
c) Trichoderma
d) Arbuscular mycorhizae
Answer:
c) Trichoderma

Question 2.
a) Anabaena
b) Amanita
c) Nostoc
d) Azospirillum
Answer:
b) Amanita

Question 3.
a) Natural selection
b) Mass selection
c) Purelinc selection
d) Clonal selection
Answer:
a) Natural selection

Question 4.
a) Cesium
b) Nitromethyl
c) Urea
d) X-ray
Answer:
d) X-ray

III. Choose the incorrect Pair

Question 1.

Answer:
d) South America – Onion

Question 2.

Answer:
C) Dr. K. Ramiah – Wheat breeder

Question 3.

Answer:
a) N2 Fixing Bacteria – Bacillus

Question 4.

Answer:
c) Green Manuring – Sonora – 63

IV. Choose the incorrect Pair

Question 5.

Answer:
b) Blast resistant Rice -1940

Question 6.

Answer:
a) Green Manuring – Tephrosia purpurea

Question 7

Answer:
c) M.S. Swaminathan – TN 1 – hybrid rice

Question 8.

Answer:
d) Cauliflower – Pusa shubhra

V. Assertion and Reason

Question 1.
Assertion : Growing of green manure crops and use of these crops as manure is called Green Manuring.
Reason : It helps to increase the nitrogen in the soil.
a) A & R are true.
b) A & R are Wrong.
c) A is true and R is Wrong.
d) A is wrong and R is true.
Answer:
a) A & R are true.

Question 2.
Assertion : Dr. M.S. Swaminathan is called “Father of green revolution in India.
Reason : He strived hard for conservation of traditional rice varities.
a) Both A & R is True.
b) Both A & R is Wrong.
c) A is True and R is Wrong.
d) A is wrong and R is True.
Answer:
c) A is True and R is Wrong

Question 3.
Assertion(A): A variety formed by pure line selection method shows more homozggosity with respect to all genes.
Reason(R): The pure line plants are produced by asexual method of vegetative propagation method.
a) (A) is correct; (R) is wrong
b) (A) is wrong; (R) is correct
c) (A) is correct; (R) does not explain
d) (A) is correct; (R) explain (A)
Answer:
a) (A) is correct; (R) is wrong

Question 4.
Assertion : Earliest record of Agriculture is found in the fertile Crescent region in and around River Nile
Reason : Approximately 12000, years ago Fertile crescent region is in and around. Tigris and Euphrates river.
a) Both A & R are True.
b) Both A & R are Wrong.
c) A is True and R is Wrong.
d) A is Wrong and R is True.
Answer:
d) A is Wrong and R is True

VI. Choose the Correct Statements

Question 1.
a) Vavilov Studied 247 Cultivated plants
b) Zhukovsky divided the whole world into 12 mega gene centres.
c) Valvilo intially propose 12 main geographic centres of origin
d) Harlan put forward the concept of mega gene centre for the origin of cultivated plants
Answer:
b) Zhukovsky divided the whole world into 12 mega gene centres

Question 2.
a) Foxtail mullet was domesticated by India
b) Wheat & Pea war domesticated by Ethiopia.
c) Castor and Coffee was domesticated by central East
d) Tomato & Pineapple was domesticated by south America
Answer:
d) Tomato & Pineapple was domesticated by south America

Question 3.
a) Seaweeds has more than 70 minerals vitamins and enzymes
b) Trichoderma is a parasitic fungi.
c) Rhizobium is a pathogenic bacteria
d) Azolla is submerged water fern
Answer:
a) Seaweeds has more than 70 minerals vitamins and enzymes

Question 4.
a) Hirngiri is a wheat variety that is resistant to Bacterial blight disease.
b) A Variety of CowPea, Pusakomal is resistent to Hill bunt disease.
c) Pusa shubra is a cauliflower variety that is resistant to Black rot disease
d) Pusa swarnim is a variety of Brassica that is resistent to TMV disease
Answer:
c) Pusa shubra is a cauliflower variety that is resistant to Black disease

VII. Choose the incorrect Statements

Question 1.
a) Mr. Jayaraman was a disciple of Dr. Nammalvar
b) Normal E. Borlaug was awarded Noble prize for peace in 1970.
c) M.S. Swaminathan produced the first semidwarf fertiliser responsive hybrid variety of rice TN 1..
d) Green revolution the term was Coined by Muller
Answer:
d) Green revolution the term was Coined by Muller

Question 2.
The disadvantage of pureline selection is
a) It is difficult to distinguish between hereditary variation from environmental variation.
b) New genotypes are never created so they are less stable to environmental fluctuations
c) The genotype is unchanged for a long period of time.
d) The plants show more heterozygosity.
Answer:
c) The genotype is unchanged for a long period of time.

Question 3.
The possible changes in the plant species due to domestication are
a) Adaptation to a greater diversity of environments.
b) Uniform flowering and fruiting.
c) Drop in Yield
d) Change in breeding system
Answer:
c) Drop in yield

Question 4.
a) Rhizobiurn is best suited for the wheat fields.
b) Azolla that fixes the atmospheric nitrogen along with blue green algae.
c) Arbuscular mycorrhizae also assures water availability
cl) Sea weed liquid fertilizer improves resistance of plants to frost and disease.
Answer:
a) Rhizobium is best suited for the wheat fields

VIII. Choose the correct answer.

Question 1.
The domesticated crop of Mesoamerica is …………………..
a) Tomato
b) Pine apple
c) Sweet Potato
d) Rubber
Answer:
a) Tomato

Question 2.
The domesticated plant of the chiloe centre.
a) Maize
b)Potato
c) Tobacco
d) Olive
Answer:
a) Maize

Question 3.
The domesticated plant of the Near East is______
a) Rye
b) Rice
c) hemp
d) Cotton
Answer:
a) Rye

Question 4.
Name the rice variety with saline tolerance and pest resistance.
a) Wild-type rice
b) Atomita – 2
c) Dwarf rice variety
d) Golden rice
Answer:
b) Atomita -2

Question 5.
Biofertilizers could be also called as …………………
a) Viral inoculants
b) Myco inoculants
c) Protozoan inoculants
d) Bacterial Inoculants
Answer:
d) Bacterial Inoculants

Question 6.
Beauveria species act as a parasite on …………… species.
a) Mammals
b) Aves
c) Arthropod
d) Amphibians
Answer:
c) Arthropod

Question 7.
Damping off of ……………. is caused by Rhizoctonia solani.
a) Tomato
b) Potato
c) Millet
d) Maize
Answer:
a) Tomato

Question 8.
Pongamia pinnata is an important plant species useful for ………………. manure.
a) Organic
b) Potassium rich
c) Green leaf
d) Calcium rich
Answer:
c) Green leaf

Question 9.
Match the following and find the correct answer
(i) Rhizobium – (A) Water ferm
(ii) Trichoderma – (B) Green manuring
(iii) Azolla – (C) Symbiotic bacterium
(iv) Crotolaria – (D) Free living fungus
a) (i) B; (ii) C; (iii) D; (iv) A
b) (i) C; (ii) D; (iii) B; (iv) A
c) (i) C; (ii) D; (iii) A; (iv) B
d) (i) B; (ii) D; (iii) C; (iv) A
Answer:
c) (i) C; (ii) D; (iii) A; (iv) B

Question 10.
More Vigorous hybrid corn was developed in _______
a) 1926
b) 1943
c) 1950
d) 1936
Answer:
a) 1926

Question 11.
In 10,000 BC Domestication of ……………….. was done.
a) Maize
b) Paddy
c) Wheat
d) Sugar cane
Answer:
c) Wheat

Question 12.
The newly introduced plant was carefully examined by the process called …………….
a) PCR Method
b) Dots Method
c) Cisgenesid
d) Quarantine
Answer:
d) Quarantine

Question 13.
National Bureau of plant Genetic Resources is located at ……………….
a) Rangpuri
b) Andhra
c) Pune
d) Bihar
Answer:
a) Rangpuri

Question 14.
Hereditary Variation cannot be distinguished from environmental variation in ………………
a) Pureline selection
b) Clonal Selection
c) Mass Selection
d) Hybridization
Answer:
c) Mass Selection

Question 15.
Johannsen in 1903 coined the word.
a) Clonal Selection
b) Pure line
c) Mass Selection
d) Heterosis
Answer:
b) Pure line

Question 16.
Which one of the following is a biopesticide?
a) Azolla
b) Rhizobium
c) Beauveria
d) Hevea
Answer:
c) Beauveria

Question 17.
…………….. is the best suited measure for maintaining hybrid vigour.
a) Asexual reproduction
b) Vegetative Propagation
c) Grafting
d) Cutting
Answer:
b) Vegetative Propagation

Question 18.
…………. is also termed as luxuriance.
a) Euheterosis
b) Heterosis
c) Mutational heterosis
d) Pseudo heterosis
Answer:
d) Pseudo heterosis

Question 19.
Muller and Stadler coined the term ………………….
a) Mutation Breeding
b) Modern Breeding
c) Plant Breeding
d) Poly ploidy
Answer:
a) Mutation Breeding

Question 20.
Mutation Breeding is the only method of improving …………… crops.
a) Multi seeded
b) single seeded
C) seedless
d) Nutritional
Answer:
c) seedless

Question 21.
Bose Research institute at Calcutta is the first …………. in India.
a) Botanical garden
b) planatorium
c) Zoological park
d) Gamma garden
Answer:
d) Gamma garden

Question 22.
Which one of the following selection method takes longer time in bringing about desired variation?
a) clonal selection
b) Mass selection
c) pureline selection
d) Natural selection
Answer:
d) Natural selection

Question 23.
(A) was originally grown in North India
(B) was orginally grown in South India
a) (A) Saccharum officinarum (B) Saccharum bareri
b) (A) Saccharu bareri (B) Saccharum officinarum
c) (A) Saccharum coarctatum (B) Saccharum alopecuroidum
d) (A) Saccharum alopecuroidum (B) Saccharu coarctatum
Answer:
b) (A) Saccharu bareri (B) Saccharum officinarum

Question 24.
Green revolution is the ………………. Agricultural revolution.
a) Third
b) second
c) third
d) fourth
Answer:
a) Third

Question 25.
…………….. received the national award for best genome saviour. .
a) Nel jeyaraman
b) C.T.Patel
c) Dr.B.P.Pal
d) N.G.P.Rao
Answer:
a) Nel jeyaraman

Question 26.
……………. is the breeding of crops for improved nutritional quality
a) Plant Breeding
b) Heterosis
c) Mutation
d) Biofortification
Answer:
d) Biofortification

Question 27.
High aspartic acid maize leads to resistance to ………………….
a) aphids
b) jassids
c) maize stem borer
d) leaf roller
Answer:
c) maize stem borer

Question 28.
Which one of the following is not a free living N2 fixing organism?
a) Anabaena azollae
b)Azotobacter
c) Clostridium
d) Nostoc
Answer:
c) Clostridium

IX. Fill in the blanks.

1. De candolle in his ……………. studied 247 caltivated plants.
Answer:
Origen of cultivated plants

2. Bamboo eas demesticated by ……………….
Answer:
China

3. He was an eminet Sorghum breeder,devoloped World’s first hybrid of Sorghum CSH-1.
Answer:
N.G.P.Rao.

4. ……………… is used as a biofertilizer for wetland rice cultivation
Answer:
Azolla.

5. Vavilov in the year ………………. converted 8 main geographic centres of origin to 12
Answer:
1935

6. Harlan says that the centre of crop plants means the places of ……………… origin of the crop plants.
Answer:
Agricultural

7. The ……………… was domesticate only in the Chiloe centre
Answer:
Potato

8. ………………. could be also called as Bioinoculants
Answer:
Biofertilizers

9. C.T.Patel devoloped World’s first …………….. hybrid
Answer:
Cotton

10. Choudhary ram dhan made …………….. as punjab granary of India.
Answer:
Wheat

11. Azdla is used as biofertiliser for wetland rice cultivation and is known to contribute ……………./ha/crop.
Answer:
40-60 kg

12. …………….. is Associated with Phycomycetous fungi and angiosperm roots.
Answer:
Arbuscular Mycorhizzoe

13. ……………… contains Auxin,Cytokinin and Gibberellins
Answer:
Seaweed liquid fertiliser

14. ……………… species are free living fungi that are common in soil and root ecosystem
Answer:
Trichoderma

15. Damping of tomato is caused by ……………..
Answer:
Rhizoctonia Solani

16. It is one of the most important green manure crops
Answer:
Tephrosia Purpurea

17. The double helix structure of DNA was identified by ……………. and …………….
Answer:
James Watson,Francis Crick

18. By 2050 we will …………….. need more food to feed the rapid growing population.
Answer:
50%

19. Crop domestication started early during ……………….
Answer:
10,000 Bc

20. ………………. Corn was developed using targetted breeding.
Answer:
Waxy

21. Rice variety of ……………. introduced from Philippines
Answer:
IR8

22. NBPGR is located in Chennai at …………………..
Answer:
Meenembakkan

23. Introduced plants get adapted to the new environment is called as ……………….
Answer:
Acclimatization

24. Tea varieties collected from China and North East India initially grown in Botanical garden of ……………….
Answer:
Kolkatta
Selection

25. ______ is the oldest and basic method of plant breeding
Answer:
Preliminary

26. In clonal selection …………… yield trial takes place during 3rd year.
Answer:
Intrageneric

27. Interspecific hybridization is also called as _____
Answer:
hybridization

28. Green revolution scheme began in ……………. in 1940’s.
Answer:
Mexico

29. In 2005 …………….. organized a first ever traditional paddy seed festival in his farm as an individual.
Answer:
Nel jayaraman

30. …………… is a cleaving protein.
Answer:
Cas9

X. Two Marks

Question 1.
What is Economic Botany?
Answer:
It is the study of the relationship between people and economically important plants.

Question 2.
When did people started practising Agriculture?
Answer:

• Archeological evidence for earliest record of agriculture is found in the fertile crescent region in and around Tigris and Euphrates river valleys, approximate about 12,000 years
  ago

Question 3.
What was the contribution by De Candolle towards agriculture?
Answer:

• He studied 247 cultivated plant species.
• He attempted to solve the mystery about the anscestral form, region of domestication and history.

Question 4.
How many geographic centres were proposed by vavilov?
Answer:

• Initially he proposed eight main geographic centres of origin.
• Later by dividing few centres into two or three centres and added a new centre USA.
• Thus making the 8 centres of Origin into 12.

Question 5.
Define Biofertilizers.
Answer:

• It is defined as preparations containing living cells of efficient strains of micro organisms that help in crop yield.

Question 6.
What is component of seaweed liquid fertilizer?
Answer:

• It contains Cytokinin, Gibberellins, and auxin a part from macro & micro nutrients.
• It has Alginates, Carbohydrate, 70 types of minerals, Vitaming and enzymes.

Question 7.
Trichoderma has been recognised as bio¬control agent – Give Reasons.
Answer:

• It control plant diseases
• Ability to enhance root growth.
• Increases crop productivity.
• Provides resistance to abiotic stress.
• Helps in uptake and use of nutrients.

Question 8.
What is plant introduction?
Answer:

• Introduction of genotypes from a place where it is normally grown to a new place.
• eg. IR & Rice from Philippines.

Question 9.
Define Acclimatization?
Answer:

• The adjustment or adaptation of the introduced plant in the changed environment is called acclimatization.

Question 10.
What is Quarantine?
Answer:

• All the introduced Crop must be free from presence of weeds, insects and disease causing Organisms.
• It has to be carefully examined by the process called quarantine.
• A strict isolation imposed to prevent the spread of disease.

Question 11.
What is Natural Selection?
Answer:

• This is a rule in the nature.
• It results in evolution reflected in the Darwinian’s principle survival of the fittest.
• It takes longer time to bring about desired Variation.

Question 12.
What is Arificial Selection?
Answer:

• It is a human involved process.
• Having better crop from a mixed population.
• The individuals differ in character.

Question 13.
Name the three types of Artificial Selection.
Answer:
a) Mass Selection,
b) Pureline Selection,
c) Clonal Selection

Question 14.
What is Emasculation
Answer:

• It is a process of removal of anthers to prevent self pollination before the opening of a flower.

Question 15.
What are the various types of Mutagens?
Answer:

• Physical Mutagen – UV short wave, X-ray, Alpha, Beta and Gamma Waves.
• Chemical Mutagen – Cesium, Ethylmethane sulfonate Nitromethyl and Urea.

Question 16.
What is Gamma Garden?
Answer:

• It is a form of mutation breeding.
• The Radioactive sources are cobalt – 60 and Caesium -137
• The first Gamma Garden in India is Bose Research institute at Calcutta.

Question 17.
What are the benefits of polyploidy?
Answer:

• It often exhibit increase hybrid vigour.
• Increase the tolerance to both biotic and abiotic stresses

Question 18.
How Polyploidy can be induced? List out the products achieved through polyploidy.
Answer:

• Polyploidy can be induced by the use of cochicine to double the chromosome number.
• Seedless Tomato, Apple, Watermelon and orange.

Question 19.
Define Green Revolution.
Answer:

• It is the cumulative result of a series of research, development, innovation and technology transfer initiatives in Agriculture.

Question 20.
Write about the contribution of Dr. M.S.Swaminathan in mutation breeding.
Answer:

• He is the pioneer mutation breeder.
• He has produce sharbati Sonora is the amber grain coloured variety of wheat.
• He is responsible for green revolution in India.

Question 21.
What are the objectives considered in Breeding to improve the nutritional quality of plants?
Answer:
To improve protein, oil, vitamin, Micronutrient and mineral content and quality

Question 22.
Tabulate certain crops and the pest resistant Variety?
Answer:

Question 23.
Expand the following.
Answer:

• CRISPR – Clustered Regularly Interspaced short Palindromic Repeats
• ODM – Oligonucleotide – Directed Mutagenesis.

Question 24.
Which method of artificial vegetative reproduction is good in plants. Give reason for your answer.
Answer:
Tissue culture is the modern artifical reproductive technique.

Reasons:

•  large scale production of clones (genitically uniform population) from the callus tissue – developed from the chosen explant cells (invitro)
• Both crop and tree species useful in forestry can be produced with desirable characters in large numbers with in a short span of time.

Question 25.
What are the advantages of rice field?
Answer:

• Azolla is a free floating water fern that fixes atmospheric nitrogen in association with nitrogen fixing blue green algae. Anabaena azolla.
• It is used as a bio-fertilizer for wetland rice cultivation and is known to contribute 40-60 kh/hal/crop.

Question 26.
Write the any three names of plant species useful in green manure?
Answer:

• Cassia fistula
• Sesbania grandiflora
• Azadirachta indica

Question 27.
What is Bio-pesticides with example
Answer:
Bio-pesticides are biologically based agents used for the control of plant pests.
Example: Trichoderma.

XI. Differentiate Type Questions

Question 1.
Differentiate Biofertilizers and Green manuring
Answer:

Question 2.
Differentiate Pureline selection and clonal selection
Answer:

Question 3.
Differentiate Mutation Breeding and Polyploid Breeding
Answer:

Question 4.
Differentiate Intervarietal and Interspecific Hybridization.
Answer:

XII. Three marks

Question 1.
What are the fields involved in Economic Botany?
Answer:

• Agronomy, Anthropology, Archaeology, Chemistry, trade and commerce.

Question 2.
Define Domestication.
Answer:

• It is the process of bringing a plant species under the control of humans and gradually changing it through careful selection.

Question 3.
What is Organic Agriculture?
Answer:

• It is as alternative agricultural system.
• It is due to rapidly changing farming practices.
• It is a production system that sustains the health of the soils, ecosystems and people.
• It is based on ecological processes biodiversity and cycles adapted to local conditions

Question 4.
Write about the role played by Beauveria as a Bio-pesticide
Answer:

• It is an entomo-pathogenic fungi
• It acts as a parasite on various arthropod that causes white muscardine disease.
• It also controls datnping off of tomato caused by Rhizoctonia Solani.

Question 5.
Differentiate Green manuring and Green leaf manuring.
Green Manuring

• Growing of green manure crops and use these crops directly in the field by ploughing.
• It increases the nitrogen in the soil.
• It helps to improve the physical property of the soil.
  eg. Crotalaria juncea.

Green Leaf Manuring

• Application of green leaves twigs, shrubs, plants growing in wasteland and field bunds.
• The important plant species useful for green leaf manure are – cassia fistula. Delonix regia.

Question 6.
Define plant breeding
Answer:

• Plant breeding is the science of improvement of crop varities with higher yield better quality, resistance to diseases and shorter durations which are suitable to particular environment.

Question 7.
What is the plan of thrid Agricultural Revolution
Answer:
The Green revolution or thrid Agricultural Revolution is the intensive plan of 1960’s to increase crop yield in developing countries by introducing the high yielding, resistant varieties, increased irrigation facilities, fertilizer application and better agricultural management.

Question 8.
Draw a flow chart showing the steps involved in plant Breeding.
Answer:
The main steps in plant breeding are given below

Question 9.
Write about NBPG R
Answer:

• It is responsible for introduction and maintence of germplasm of various agricultural and horticultural station in our country.
• It is also responsible for maintenance of plant materials of botanical and medicinal interest.
• It is located at Rangpuri/New Delhi with four regional station at Amristar, Kolkata, Mumbai and Chennai.

Question 10.
Draw the Schematic difference between Mass & Pureline selection.
Answer:

Question 11.
What is hybridization?
Answer:

• Hybridization is the method of producing new crop verities by crossing of plants that are genetically different.
• It offers improvement in crop by combining the desirable character of two or more varities.
• The first natural hybridization was observed by Cotton Mather in maize.

Question 12.
A type of Heterosis is called as Pseudoheterosis – Why?
Answer:

• It is called as pseudoheterosis because the progeny is superior over parents by vegetative growth.
• They are usually sterile.
• It is also called as Luxuriance.

Question 13.
Tabulate certain crop variety and the disease to which they are resistant.
Answer:

Question 14.
What is Norm 10?

• It is dwarfing genes with high photosynthetic rate and increases the respiratory activity.
• Gonjiro Inazuka was scientist, who selected the Semi- dwarf wheat variety – Norm 1o.
• This variety helped one billion lives from hunger and starvation.

Question 15.
How can we develop resistance to Insect pests in plants?

• Insect resistance in host crop plants may be due to morphological, biochemical 0r physiological characteristics.
• Hairy leaves in plants – Jassids in cotton
  – Cereal leaf bettle in wheat.
• Solids Stem – Stem sawfly.
  Smooth leaves and nectar less cotton – Bol worms.
  High aspartic acid, low nitrogen and sugar content – Maize stem borers.

Question 16.
What are the plant breeding tools used to improve the crop varieties?
Answer:

• Genetic Engineering, plant tissue culture, protoplasmic fusion, Molecular marking and DNA finger printing.

Question 17.
Define mutation breeding?
Answer:
Mutation means the sudden heritable changes in the genotype or phenotype of an organism.

Question 18.
What is Bio-fortification?
Answer:
Breeding crops with higher levels of vitamins and minerals or higher protein and healthier fats is the most practical means to improve public health.

Question 19.
Comment on sugercane.
Answer:

• Saccharum bareri was originally grown in north India.
• It is poor sugar content and yield.
• Tropical canes grown in south India saccharurn officinarum had thicker stems and higher sugar content.
• It did not grow well in North India
• These two species were successfully crossed to get sugar cane varieties combining the desirable qualities of high yield thick stems high sugar and ability to grows in the sugar cane areas of North India.

XIII. Five Marks

Question 1.
What are the possible changes in the plant due to domestication?
Answer:

• Adaption to the environmental alteration.
• Adaption to wider geographical range.
• Uniformity in flowering and fruiting.
• Increased size of fruits and seeds.
• Change in breeding system.
• Increase in yield.
• Increased resistance for disease and pest.
• Developing seedless parthenocarpic fruit.
• Enhancing the taste and nutritional composition.

Question 2.
Enumerate the History of Agriculture:
Answer:

• 1807 – Alexander Yon Humboldt considered that original & source of most and their origin is an importable secret.
• 1868 – Darwin’s theory proposed that natural selection and hybridization led to the origin of cultivate plants.
• 1883 – De Candolle studied 247 cultivated plants species and was able to find the ancestral form, region of domestication and history.
• 1887-1943 – Vavilov studied about the diverse forms of plants based on various criteria like morphology,cytology etc., He proposed 8 geographic centres and later developed it to 12 centres.
• 1968 – Zhukovsky put forward the concept of mega gene and divided the whole world into 12 mega gene centres.
• 1971 – Harlen believed that agriculture originated independently in three different areas.
  – There were non centres are the area were the crop has been shifted.

Question 3.
Tabulate the vavilov’s centres of crop origin and crop domesticated.
Answer:
Vavilov’s centres of crop origin and crops domesticated.

Question 4.
Write an essay on Indian plant Breeders.
Answer:

• Dr.M.S.Swaminathan- He is pioneer in mutation breeder.
• Sir.T.S.Venkatraman- An eminent sugarcane breeder.
• Dr.B.P.Pal-Developed superior disease resistant varieties of wheat
• Dr.K.Ramiah- Eminent Rice Breeder.
  N.G.P.Rao-An eminent sorghum breeder, developed world’s first hybrid of sorghum (CSH-1).
• C.T.Patel-Developed world’s first cotton hybrid.
• Choudhary Ram Dhan – wheat breeder,developed C-591 wheat which made Punjab as wheat granary of india.

Question 5.
Differentiate Rhizobium from Azolla as bio-fertilizers.
Answer:

• Bio-fertilizers could also be called as microbial cultures or bacteria) fertilizers.
• They are efficient in fixing nitrogen improve soil fertility, eco-friendly and cost effective.

Rhizobium:

• It resides in the root nodules of leguminous plants.
• It fixes the atmospheric Nitrogen.
• It increases yield of paddy by 15-40%

Azolla:

• Free floating water fern, with blue green algae fixes the Nitrogrn.
• It increases the yield of rice.
• It decomposes quickly.

Arbuscular Mycorrhizae:

• They can dissolve the phosphates found in the soil.
• It provides strength to resist disease, germ and unfavourable weather.

Seaweed Liquid Fertilizer:

• It contains cytokinin, gibberellins and Auxin
• Most of it are made from kelp ( Brown Algae).
• The alginates in it react with metals in soil and retain moisture for a long time.
• They have more than 70 minerals vitamin and enzymes.
• Seeds soaked in seaweed germinate rapidly.

Question 6.
Tabulate the classification of Biofertilizers.
Answer:

Question 7.
Discuss about Biopesticides in detail.
Answer:

• They are biologically based agents used for the control of plant pests.
• They are ecofriendly , Non – toxic and cheaper the chemical pesticide.

Trichoderma:

• They are free living fungi in the soil.
• They control plant disease.
• It has the ability to enhance root growth development.
• Increases the crop productivity.
• It helps in resisting Abiotic stress.
• It increases the uptake and use of nutrients.

Beauveria:

• It is an entomo – pathogenic fungus
• It parasitse arthropods that cause white muscardine and controls damping off of tomato

Question 8.
Enumerate the objectives of plant Breeding.
Answer:

• To increase yield, Vigour and fertility of the crop.
• To increase tolerance to environmental condition, Salinity, Temperature and drought.
• To prevent premature falling of buds fruits etc.,
• To improve the maturation of both the male and female gametes at the same time.
• To develop resistance to pathogens and pests.
• To develop photosensitive and thermos – Sensitive Varieties.

Question 9.
Discuss about the types of selection.
Answer:

• Selection is the oldest and basic method of plant breeding.
• There are two main types of selection – Natural and Artificial.

Natural Selection:

• It occurs naturally.
• It takes longer time in bringing about desired variation.
• It reflected the Darwinian principle.

Artificial Selection:

• It is a human involved process.
• Producing better crop from a mixed population.

a) Mass Selection :

• Large number of plants of similar phenotype are selected and crossed to get a new variety.
• After repeated selection for five to six years it is distributed to the farmers.

b) Pureline Selection:

• Plants obtained as a result of self pollination from a single homozygous individual.
• The progeny shows homozy gosity with respect to all genes.
• New Genotypes are never created,

c) Clonal Seection:

• The progenies that are asexually propagated resembles the parent genetically.
• The progeny is multiplied to form clone.
• The genotype of a clone remains unchanged for a long period of time.

Question 10.
Describe the steps involved in Hybri-dization.
Answer:
a) Selection of Parents :

• Male and female plants of desired characters are selected.
• It should be tested for their homozygosity

b) Emasculation:

• It is removal of anther before blooming.
• It avoids self pollination

c) Bagging:
The stigma is protected against any undesirable pollen grains, by covering with a bag.

d) Crossing:
Transfer of pollen grains from selected male flower to the stigma of the emasculated female flower.

e) Harvesting Seeds and Raising Plants :

• Due to fertilization seeds form.
• These seeds are grown into a new generation.

Question 11.
Discuss about the types of heterosis.
Answer:
a) Euheterosis:

• It is the true heterosis.
• It is inherited

b) Mutational Euheterosis :

• It is the simplest form.
• Removal of harmful, recessive, mutant genes by superior dominant allele in cross pollinated crops.

c) Balanced Euheterosis:

• Well balanced gene combination.
• More adaptive to environment and agricultural usefulness.

d) Psuedo heterosis :

• Also known as luxuriance.
• Progeny possess superiority over parents in regetative Growth.
• But not in yield & adaptation.
• Usually sterile or poorly fertile.

Question 12.
Describe polyploid Breeding.
Answer:

• The plants which posses more than two sets of chromosomes are called Polyploids.
• It is the major force in the evolution of both wild and cultivated plants.
• Polyploid of exhibit hybrid vigour.
• Increases tolerance to biotic and abiotic stresses.
• Polyploidy results in reduced fertility and producing seedless varieties.
• If chromosomes is doubled by itself it is autopolyploidy.
• Triploid condition in Sugarbeets result in Vigour.
• Colchicine used to double the chromosomes. Eg. Triticale and Raphanobrassica

Question 13,
Biofortification is the breeding crops with
higher levels of nutrients. Justify it..
Answer:
Breeding of improved nutritional quality like
– Protein content and quality.
– Oil content and quality
– Vitamin Content
– Micro & Macro nutrient content.

• In 2000 – Maize hybrid had twice the nutrient value than the parents.
• Wheat – Atlass 66 having high protein content.
• Iron fortified rice can be developed.
• Vitamin A enriched vegetable.
• Vitamin C enriched vegetables and iron and calcium crops also developed.

Question 14.
Enumerate the New Breeding Techniques.
Answer:

• It is a collection of methods that could increase the development of new traits in plant breeding.
• It often involve genome editing.
• Cutting and modifying the genome during the repair process by tools like CRISPR.
• Genome editing to introduce changes in few base pairs using a technique called ODM.
• Transferring a gene from an identical or closely related species (Cisgenesis)
• Organising process that alter gene activity without altering the DNA itself.

Question 15.
Ramu and Somu are farmers. Ramu cultivated the crops by self fertilization method. Somu cultivated the crops from mixed population.
(i) Who will get new variety?
(ii) Write the advantage and disadvantages of their selection.
Answer:
(i) Somu will get the new variety. Because he had selected the mixed population method.

(ii) Advantages of self fertilization method:
The repeated self pollination from a single homozygous individual produces a variety that shows more homozygosity with respect to all genes.

(iii) Dis advantages:

• The major disadvantage of this type is that it never creates new genotypes.
• The plants produced are also less adaptible and less stable to the environmental fluctuations.

Advantages of Mixed population method
In this method, a large number of plants of similar phenotype (or) morphological characters are selected and their seeds are mixed together to constitute a newer variety’.

Disadvantages:
The disadvantage of mixed population method is that it is difficult to distinguish the hereditary variation from environmental variation .

Question 16.
Mention the benefits of seed treatment?
Answer:

• Prevents spread of plant disease
• Protects seed from seedling blights
• Improves germination
• Improves germination
• provides protection from storage insects
• controls soil insects.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.3 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.3

Question 1.
Explain why Rolle’s theorem is not applicable to the following functions in the respective intervals.
(i) f(x) = |\(\frac { 1 }{ x }\)|, x ∈ [-1, 1]
(ii) f(x) = tan x, x ∈ [0, π]
(iii) f(x) = x – 2 log x, x ∈ [2, 7]
Solution:
(i) f(x) = |\(\frac { 1 }{ x }\)|, x ∈ [-1, 1]
f(-1) = 1
f(1) = 1
⇒ f(-1) = f(1) = 1
But f(x) is not differentiable at x = 0
∴ Rolle’s theorem is not applicable.

(ii) f (x) = tan x
f(x) is not continuous at x = \(\frac{\pi}{2}\).
So Rolle’s Theorem is not applicable.

(iii) f(x) = x – 2 log x
f(x) = x – 2 log x
f(2) = 2 – 2 log 2 = 2 – log 4
f(7) = 7 – 2 log 7 = 7 – log 49
f(2) ≠ f(7)
So Rolle’s theorem is not applicable.

Question 2.
Using the Rolle’s theorem, determine the values of x at which the tangent is parallel to the x-axis for the following functions:
(i) f(x) = x² – x, x ∈ [0, 1]
(ii) f(x) = \(\frac { x^2-2x }{ x+2 }\), x ∈ [-1, 6]
(iii) f(x) = √x – \(\frac { x }{ 3 }\), x ∈ [0, 9]
Solution:
(i) f(x) = x² – x, x ∈ [0, 1]
f(0) = 0, f(1) = 0
⇒ f(0) = f(1) = 0
f(x) is continuous on [0, 1]
f(x) is differentiable on (0, 1)
Now, f'(x) = 2x – 1
Since, the tangent is parallel to x-axis then
f'(x) = 0 ⇒ 2x – 1 = 0
x = \(\frac { x }{ 3 }\) ∈ (0, 1)

(ii) f(x) = \(\frac { x^2-2x }{ x+2 }\), x ∈ [-1, 6]
f(-1) = \(\frac { 1+2 }{ -1+2 }\) = 3
f(6) = \(\frac { 36-12 }{ 8 }\) = \(\frac { 24 }{ 8 }\) = 3
⇒ f (-1) = 3 = f(6)
f(x) is continuous on [- 1, 6]
f(x) is differentiable on (- 1, 6)
Now, f'(x)

Since the tangent is parallel to the x-axis.
f'(x) = 0
⇒ x² + 4x – 4 = 0
⇒ x = –\(\frac { 4±\sqrt{16+16} }{ 2 }\)
x = –\(\frac { 4±4√2 }{ 2 }\) = -2 ± 2√2
x = -2 ± 2√2 ∈ (-1, 6)

(iii) f(x) = √x – \(\frac { x }{ 3 }\), x ∈ [0, 9]
f(0) = 0, f(9) = √9 – \(\frac { 9 }{ 3 }\) = 3 – 3 = 0
⇒ f(0) = 0 = f(9)
f(x) is continuous on [0, 9]
f(x) is differentiable on (0, 9)
Now f'(x) = \(\frac { 1 }{ 2√x }\) – \(\frac { 1 }{ 3 }\)
Since, the tangent is parallel to x-axis.
f'(x) = 0

Question 3.
Explain why Lagrange’s mean value theorem is not applicable to the following functions in the respective intervals:
(i) f(x) = \(\frac { 1 }{ 2√x }\), x ∈ [-1, 2]
(ii) f(x) = |3x + 1|, x ∈ [-1, 3]
Solution:
(i) f(x) = \(\frac { 1 }{ 2√x }\), x ∈ [-1, 2]
f(0) = undefined
∴ f(x) is not continuous at x = 0
Hence, Lagrange’s mean value theorem is not applicable.

(ii) f(x) =|3x + 1|, x ∈ [-1, 3]
The function is not differentiable at x = \(\frac{-1}{3}\).
So Lagrange’s mean value theorem is not applicable in the given interval.

Question 4.
Using the Lagrange’s mean value theorem determine the values of x at which the tangent is parallel to the secant line at the end points of the given interval:
(i) f(x) = x³ – 3x + 2, x ∈ [-2, 2]
(ii) f(x) = (x – 2) (x – 7), x ∈ [3, 11]
Solution:
f(x) = x³ – 3x + 2, x ∈ [-2, 2]
f(x) is continuous in [- 2, 2]
f(x) is differentiable in (- 2, 2)
f(-2) = (-2)³ – 3 (-2) + 2 = – 8 + 6 + 2 = 0
f(2) = (2)³ -3(2) + 2 = 8 – 6 + 2 = 4
∴ f(x) is defined in the given interval.
Given that tangent is parallel to the secant line of the curve between x = -2 and x = 2.

(ii) f(x) = (x – 2)(x – 7), x ∈ [3, 11]
f(x) is continuous in [3, 11]
f(x) is differentiable in (3, 11)
f(3) = (3 – 2) (3 – 7) = (1) (-4) = -4
f(11) = (11 – 2) (11 – 7) = (9) (4) = 36
∴ f(x) is defined in the given interval.
Given that the tangent is parallel to the secant line ofthe curve between x = 3 and x = 11.
∴ f'(c) = \(\frac { f(b)-f(a) }{ b-a }\)
2c – 9 = \(\frac { 36+4 }{ 11-3 }\) where f'(x) = 2x – 9
2x – 9 = \(\frac { 40 }{ 8 }\) = 5
2c = 14 ⇒ c = 7 ∈ (3, 11)
∴ x = 7.

Question 5.
Show that the value in the conclusion of the mean value theorem for
(i) f(x) = \(\frac { 1 }{ x }\) on a closed interval of positive numbers [a, b] is \(\sqrt { ab }\)
(ii) f(x) = Ax² + Bx + C on any interval [a, b] is \(\frac { a+b }{ 2 }\)
Solution:

(ii) f(x) = Ax² + Bx + C, x ∈ [a, b]
f'(x) = 2Ax + B
By Mean Value Theorem,

Question 6.
A race car driver is racing at 20th km. If his speed never exceeds 150 km/hr, what is the maximum distance he can cover in the next two hours?
Solution:
Let a = 0, b = 2 and the interval is [0, 2] and f(0) = 20 (given)
We need to find f(2)
By Lagrange’s Mean Value Theorem,
f(b) – f(a) ≤ f'(c) [b – a]
f(b) – 20 ≤ 150(2 – 0)
f(b) ≤ 300 + 20
f(b) ≤ 320
∴ Maximum distance f(2) = 320 km.

Question 7.
Suppose that for a function f(x), f'(x) ≤ 1 for all 1 ≤ x ≤ 4. Show that f(4) – f(1) ≤ 3.
Solution:
Given: For f(x), f'(x) ≤ 1 for all 1 ≤ x ≤ 4
∴ a = 1, b = 4.
By Lagrange’s Mean Value Theorem,
f(b) – f(a) ≤ f'(c) (b – a)
f(4) – f(1) ≤ 1(4 – 1)
f(4) – f(1) ≤ 3
Hence Proved.

Question 8.
Does there exist a differentiable function f(x) such that f(0) = -1, f(2) = 4 and f (x) ≤ 2 for all x. Justify your answer.
Solution:
Given:For f(x), f'(x) ≤ 2, f(0) = -1, f(2) = 4
∴ a = 0, b = 2
By Lagrange’s Mean Value Theorem,
f(b) – f(a) ≤ f'(c)(b – a)
f(2) – f(0) ≤ f'(c) (2 – 0)
\(\frac { 4+1 }{ 2 }\) ≤ f'(c) ⇒ \(\frac { 5 }{ 2 }\) ≤ f'(c) ≤ 2 (given)
f(x) cannot be a differentiable function in (0, 2) as f'(x) cannot be 2.5.

Question 9.
Show that there lies a point on the curve f(x) = x(x + 3)e^-π/2, -3 ≤ x ≤ 0 where tangent drawn is parallel to the x-axis.
Solution:
f(x) = x(x + 3)e^-π/2, -3 ≤ x ≤ 0
f(x) is continuous in [-3, 0]
f(x) is differentiable in (- 3, 0)
f(-3) = -3 (-3 + 3)e^-π/2 = 0
f(0) = 0
⇒ f(-3) = f(0) = 0
Since the tangent is parallel to x-axis.
f'(c) = 0
e^-π/2 (2c + 3) = 0 where f'(x) = e^-π/2 (2x + 3)
2c + 3 = 0
c = –\(\frac { 3 }{ 2 }\) ∈ (-3, 0)
∴ The point lies on the curve.

Question 10.
Using Mean Value Theorem prove that for, a > 0, b > 0, |e^-a – e^-b| < |a – b|
Solution:
Let f(x) = e^-x
f'(x) = e^-x
By Lagrange’s Mean Value Theorem,

Hence Proved.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Botany Guide Pdf Chapter 5 Plant Tissue Culture Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 5 Plant Tissue Culture

12th Bio Botany Guide Plant Tissue Culture Text Book Back Questions and Answers

I. Choose the correct answer from the given option:

Question 1.
Totipotency refers to.
a) capacity to generate genetically identical plants.
b) capacity to generate a whole plant from any plant cell/explant.
c) capacity to generate hybrid protoplasts.
d) recovery of healthy plants from diseased plants.
Answer:
b) capacity to generate a whole plant from any plant cell / explant

Question 2.
Micro propagation involves
a) vegetative multiplication of plants by using micro – organisms.
b) vegetative multiplication of plants by using small explants.
c) vegetative multiplication of plants by using microspores.
d) Non – vegetative multiplication of plants by using microspores and megaspores.
Ans:
b) vegetative multiplication of plants by using small explants.

Question 3.
Match the following

Answer:
c) 1-B, 2-A, 3-D, 4-C

Question 4.
The time duration for sterilization process by using autoclave is _____ minutes and the temperature is
a) 10 to 30 minutes and 1250 C
b) 15 to 30 minutes and 1210 C
c) 15 to 20 minutes and 1250 C
d) 10 to 20 minutes and 1210 C
Answer:
b) 15 to 30 minutes and 1210 C

Question 5.
Which of the following statement is correct.
a) Agar is not extracted from marine algae such as seaweeds.
b) Callus undergoes differentiation and produces somatic embryoids.
c) Surface sterilization of explants is done by using mercuric bromide
d) PH of the culture medium is 5.0 to 6.0
Answer:
b) Callus undergoes differentiation and produces somatic embryoids.

Question 6.
Select the incorrect statement from given statement
a) A tonic used for cardiac arrest is obtained from Digitalis purpuria
b) Medicine used to treat Rheumatic pain is extracted from Capsicum annum
c) An anti malarial drug is isolated from Cinchona officinalis.
d) Anti – carcinogenic property is not seen in Catharanthus roseus.
Answer:
d) Anti – carcinogenic property is not seen in Catharanthus roseus

Question 7.
Virus free plants are developed from
a) Organ culture
b) Meristem culture
c) Protoplast culture
d) Cell suspension culture
Answer:
b) Meristem culture

Question 8.
The prevention of large scale loss of biological interity.
a) Biopatent
b) Bioethics
c) Biosafety
d) Biofuel
Answer:
c) Biosafety

Question 9.
Cryopreservation means it is a process to preserve plant cells, tissues or organs
a) at very low temperature by using ether.
b) at very high temperature by using liquid nitrogen
c) at very low temperature of -196 by using liquid nitrogen
d) at very low temperature by using liquid nitrogen
Answer:
c) at very low temperature of -196 by using liquid nitrogen

Question 10.
Solidifying agent used in plant tissue culture is
a) Nicotinic acid
b) Cobaltous chloride
c) EDTA
d) Agar
Answer:
d) Agar

Question 11.
What is the name of the process given below? Write its 4 types.
Answer:

These are the basic steps in plat Tissue culture technology
The process is plant tissue culture. Based on the explants, plant tissue culture is classified as:

1. Organ culture
2. Meristem culture
3. Protoplast culture
4. Cell culture

Question 12.
How will you avoid the growth of microbes in the nutrient medium during the culture process? What are the techniques used to remove the microbes?
Answer:
The microbial growth in the culture medium can be overcome by autoclaving the medium at Plant Tissue Culture II 121°C (15 psi) for 15 to 30 minutes.

Chemical sterilization using chemicals, sterilizing using UV radiation. Alcoholic sterilization using ethanol, autoclaving and filtration, etc., are the various techniques used to remove microbes.

Question 13.
Write the various steps involved in cell suspension culture
Answer:
Definition: The culture of single cells or small aggregate of cells invitro in liquid medium is called cells suspension culture.

Preparation steps:

Production of Secondary Metabolites:

• Alkaloids, flavonoids, terpenoids, phenolic compounds, and recombinant proteins.
• secondary metabolites are chemical compounds that are not required by the plant for normal growth and development.
• The process of production of secondary metabolites can be scaled up and automated using bio-reactors for commercial production.
• Many strategies such as biotransformation, elicitation, and immobilization have been used to make cell suspension cultures more efficient in the production of secondary metabolites.

Question 14.
What do you mean Embryoids? Write its application.
Answer:
Somatic embryogenesis is the formation of embryos from the callus tissue directly and these embryos are called Embryoids or from the in vitro cells directly form pre-embryonic cells which differentiate into embryoids.
Applications:

1. Somatic embryogenesis provides potential plantlets which after the hardening period can establish into plants.
2. Somatic embryoids can be used for the production of synthetic seeds.
3. Somatic embryogenesis is now reported in many plants such as Allium sativum, Hordeum vulgare, Oryza sativa, Zea mays and this is possible in any plant.

Question 15.
Give examples of micropropagation performed in plants.
Answer:
Micropropagations are performed in many plants.
Examples:

1. Pineapple
2. banana
3. strawberry
4. Potato, etc

Question 16.
Explain the basic concepts involved in plant tissue culture.
Answer:
Basic concepts of plant tissue culture are totipotency, differentiation, differentiation, and redifferentiation.

1. Totipotency: The property of live plant cells that they have the genetic potential when cultured in a nutrient medium to give rise to a complete individual plant.

2. Differentiation: The process of biochemical and structural changes by which cells become specialized in form and function.

3. Redifferentiation: The further differentiation of already differentiated cell into another type of cell. For example, when the component cells of callus have the ability to form a whole plant in a nutrient medium, the phenomenon is called redifferentiation.

4. Dedifferentiation: The phenomenon of the reversion of mature cells to the meristematic state leading to the formation of callus is called dedifferentiation. These two phenomena of redifferentiation and dedifferentiation are the inherent capacities of living plant cells or tissue. This is described as totipotency.

Question 17.
Based on the material used, how will you classify culture technology? Explain it.
Answer:
Based on the explants some other plant tissue culture types are:
1. The culture of embryos

• anthers
• ovaries
• roots
• shoots etc
  

2. Meristem culture
The culture of any plant meristematic tissue on culture media.

3. Protoplast culture

• Protoplasts (cells without a cell wall, but plasma membrane) are used to regenerate whole plants from single-cell protoplasts of 2 different plants fused into hybrids – later by PTC – develop into many plantlets.
• This process of formation of somatic hybrids into somatic hybridization.

4. Cell culture

• The formation of cell suspension from the callus
• The cells are separated from the callus tissue and used for cell suspension culture

Question 18.
Give an account on Cryopreservation. The parts such as,
Answer:
Cryopreservation, also known as Cryo-conservation, is a process by which protoplasts, cells, tissues, organelles, organs, extracellular matrix, enzymes or any other biological materials are subjected to preservation by cooling to a very low-temperature of-196°C using liquid nitrogen. At this extremely low temperature, any enzymatic or chemical activity of the biological material will be totally stopped and this leads to the preservation of material in dormant status.

Later these materials can be activated by bringing to room temperature slowly for any experimental work. Protective agents like dimethyl sulphoxide, glycerol, or sucrose are added before the cryopreservation process. These protective agents are called cryoprotectants since they protect the cells, or tissues from the stress of freezing temperature.

Question 19.
What do you know about Germplasm conservation? Describe it. Definition
Answer:
Living genetic resources such as pollen, seeds, or plant tissue materials are preserved in living conditions for future use for many hybridization crop improvement research works. Eg. Pollen banks, Seedbanks

Purpose

• To maintain viability and Fertility for future use
• Gene bank, DNA bank of elite plants are maintained to keep
  • biological diversity
  • food security

Question 20.
Write the protocol for artificial seed preparation
Answer:
Later these seeds are grown in vitro medium and converted into plantlets. These plantlets require a hardening period (either greenhouse or hardening chamber) and then shifted to normal environmental conditions.

12th Bio Botany Guide Plant Tissue Culture Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
Invitro means.
a) In a test tube.
b) inside the body
c) inside the cell
d) in a laboratory
Answer:
a) In a test tube

Question 2.
The concept of Totipotency was proposed by.
a) Hildbrandt
b) Haberlandt
c) Chilton
d) Takebe et-al
Answer:
b) Haberlandt

Question 3.
The scientist developed root cultures, used Knop’s solution along with 3 vitamins is.
a) Murashige & Skoog
b) P.R. White
c) Kanta et-al
d) E.C. Steward
Answer:
b) P.R. White

Question 4.
Virus-free Dahlia and Potato plants are produced by.
a) Morel
b) Martin
c) Morel & Martin
d) E.C steward
Answer:
c) Morel & Martin

Question 5.
The Indian scientists developed in vitro production of haploid embryos from
a) ovule of Nicotiana
b) anthers of Datura
c) gametes of Dahlia
d) Zygote of Carrot
Answer:
b) anthers of Datura

Question 6.
Melchers & Co workers produced
a) Somatic hybrid of Nicotiana species
b) Intergeneric hybrid between potato & tomato
c) Interspecific hybrid of Nicotiana glauca and Nicotiana longs dorffii
d) test tube fertilization in flowering plants
Answer:
b) Intergeneric hybrid between potato & tomato

Question 7.
The growth hormones added in MS – medium are
a) Auxin & Gibberellins
b) IAA & Kinetin
c) Gaibberelline & cytokinin
d) Auxin & ABA
Answer:
b) IAA & Kinetin

Question 8.
Somatic embryogenes is not applied in
a) Oryza sativa
b) Hordeum vulgare
c) Ficus bengaliensis
d) Avena sativa
Answer:
c) Ficus bengaliensis

Question 9.
Which one of the following is a correct set?
a) Vincristine Cinchona officinalis Anti carcinogen
b) Capsacin catharanthus roseus – Antimalarial
c) Digoxin Digitalis purpuria Cardiac tonic
d) Codeine Capsicum annum Analgesic
Answer:
c) Digoxin Digitalis purpuria Cardiac tonic

Question 10.
Germ plasm conservation does not include
a) DNA bank
b) Seed bank
c) SWISS bank
d) pollen bank
Answer:
c) SWISS bank

Question 11.
This is not of the strategies used to make cell suspension
a) biotrans formation
b) elicitation
c) immobilization
d) filtration
Answer:
d) filtration

Question 12.
Choose the odd man out with regard to protoplasmic fusion
a) somatic hybridization
b) Protoplasmic fusion
c) Embryoids
d) Polyethylene Glycol
Answer:
c) Embryoids

Question 13.
This is not a technique in PTC?
a) organ culture
b) Meristem culture
c) Cell culture
d) M.S. culture
Ans:Answer:d) M.S. culture

Question 14.
Which one of the following is the correct steps in the direct embryogenesis?

Answer:
b

Question 15.
Protoplasts are transferred to sucrose solution to
a) retain osmotic pressure
b)retain viability
c) restore solubility
d) sterilize the protoplast
Answer:
b) retain viability

Question 16.
Plants those can’not be subjected to hybridization technique can be raised by?
a) somatic embryogenesis
b) PTC
c) somatic hybridization
d) meristem culture
Answer:
c) somatic hybridization

Question 17.
Indole alkaloids used as bio medicine is got from
a) phyllanthus amaras
b) Acalypha indica
c) Catharanthus roseue
d) Avena sativa
Answer:
c) Catharanthus roseus

Question 18.
Virus is free in
a) cell culture
d) cambial culture
b) protoplasm culture
c) Apical meristem culture
Answer:
c) Apical meristem culture

Question 19.
From the following secondary metabolites which one is used as cardioc tonic
a) capsaicin
b) Quinine
c) Codeine
d) Digoxin
Answer:
d) Digoxin

II. Match the following

Question 20.

A) a-2, b-1, c-5, d-3, e-4
B) a-1, b-2, c-3, d-4, e-5
C) a-5, b-4, c-3, d-2, e-1
D) a-4, b-3, c-2, d-1, e-5
Answer:
a) a-2, b-1, c-5, d-3, e-4

Question 21.

A) a -1, b – 2, c-3, d-4, e-5
B) a – 4, b-3, c-5, d-1, e-2
C) a – 5, b – 4, c-3, d-2, e-1
D) a – 3, b-1 c-2, d-5, e-4 .
Answer:
b) a – 4, b-3, c-5, d-1, e-2

Question 22.

A) a-4, b-3, c-2, d-1
B) a-2, b-4, c-1, d-3
C) a-3, b-1, c-4, d-2
D) a-1, b-2, c-3, d-4
Answer:
C) a-3, b-1, c-4, d-2

III. Choose the incorrect Statement

Question 23.
a) The plant material used in tissue culture should be surface sterilized
b) Callus is a mass of unorganized growth of plant cells or tissues in invivo culture
c) The fusion product of protoplasts without a nucleus of different cells is called cybrid
d) Bioreactors are used for the production of secondary metabolites in a commercial way
Answer:
b) Callus is a mass of unorganized growth of plant cells or tissues in invivo culture

Question 24.
Which one of the following statements is true regarding IPR?
a) The discoverer has the full rights on his / her property
b) IPR – includes only the process of the product, not trade secrets.
c) IPR is not protected by laws formed by the country.
d) The discoverer can use his discovery for his own company but can not sell it to others.
Answer:
a) The discoverer has the full rights on his/her property.

IV. Choose the correct Statement

Question 25.
a) The HGP was founded in 2010 as an integral part of ELSI
b) GEAC is an apex body under the UNO
c) GMOs-GEMs & Trans genie plants approval are not coming under the scanning of GEAC
d) The release of genetically engineered organisms and products into the environment need at least three levels of field trials such as BRL -1, BRLII & BRL III
Answer:
d) The release of genetically engineered organisms and products into the environment need at least three levels of field trials such as BRL -1, BRL II & BRL III

Question 26.
a) ‘Takepe’ regenerated tobacco plants from isolated mesophyll protoplasts.
b) Morel & Martin formulated Bioethics.
c) The photoperiod needs for Tissue culture is 12-18 hours of light.
d) The PH medium for Tissue culture should be below 5
Answer:
a) Takepe’ regenerated tobacco plants from isolated mesophyll protoplasts

V. In each of the following questions, two statements are given – one as Assertion (A) and the other one is Reason (R) Mark the correct answer as

Question 27.
Assertion: High yielding plants can be raised in large number by Micropropagation.
Reason: Micropropagation maintain high standards of homogeneity
a) If both ‘A’ and ‘R’ are true and ‘R’ is the correct explanation of A
b) It both A’ and ‘R’ are true but ‘R’ is not the correct explanation of A
c) It A is true but ‘R’ is false d) If both A & R are false
Answer:
a) If both ‘A’ and ‘R’ are true and ‘R’ is the correct explanation of A

Question 28.
Assertion: A major advantage of tissue culture is protoplast fusion.
Reason: It produces a genetically uniform population.
a) If both ‘A’ and ‘R1 are true and ‘R’ is the correct explanation of A
b) It both A’ and ‘R’ are true but ‘R’ is not the correct explanation of A
c) It A is true but ‘R’ is false
d) If both A & R are false
Answer:
c) It A is true but ‘R’ is false

Question 29.
Assertion(A): The explants are sterilized by mercuric chloride
Reason(R): Sterilization prevents the growth of other microorganisms in the Culture medium
a) (A) correct; (R) wrong
b) (A) wrong: (R) correct
c) Both (A) and (R) are correct; but (R) is not the explanation to (A)
d) Both (A) and (R) are correct; (R) is the explanation of (A)
Answer:
b) (A) wrong: (R) correct

VI. Two Marks

Question 1.
What are the contributions of Haberlandt to PTC?
Answer:

• He did the in-vitro culture of plant cells
• He used Knop’s salt solution as a culture medium
• He only proposed the concept – Totipotency

Question 2.
What is the special contribution of Murashige and Skoog?
Answer:

• They formulated a tissue culture medium
• A landmark in PTC, because it is the most frequently medium for all kinds of tissue culture work.

Question 3.
Who developed first interspecific somatic hybrid?
Answer:
Carlson & co-worker obtained protoplast fusion between Nicotiana glauca & Nicotiana longdorffii, and developed the first interspecific somatic hybrid in 1971

Question 4.
Define Totipotency?
Answer:

• The inherent genetic potential of any living plant cell, when cultured in the nutrient medium can develop into a complete individual plant.
• One of the basic concepts exploited in tissue culture.

Question 5.
What are the components of Knop’s solution?
Answer:
I. It contains various salts dissolved in Sucrose solution

• Calcium Chloride: 3.0 gm
• Potassium Nitrate: 1.0 gm
• Magnesium Sulphate: 1.0 gm
• Dibasic Potassium Phosphate: 1.0 gm

II. Sucrose: 50 gm(optimal)
III. Deionized Water: 1000ml

Question 6.
Distinguish between Redifferentiation and Dedifferentiation.
Answer:
Redifferentiation :
The ability of callus tissue to develop into shoot & root (embryoid)

Dedifferentiation :
Reversion of mature tissue into meristematic state leading to the formation of callus.

Question 7.
Notes on PEG.
Answer:

• PEG is Poly Ethylene Glycol.
• It is the fusogenic agent that facilitates the fusion of 2 different protoplasts coming together in somatic hybridization to produce cybrid.

Question 8.
What is Agar?
Answer:

• Agar is a mucilaginous polysaccharide obtained from marine algae (seaweeds)
• Gelladium, Gracilaria, Gellidiella.
• The Agar is a solidifying agent used in culture media preparation.

Question 9.
Notes on Autoclave.
Answer:

• An autoclave is a device used to do wet steam sterilization.
• Autoclaving at 15 psi (121°C) for 15-30 minutes.
• Glassware, forceps, scalpels, and all accessories are subjected to autoclaving for

Question 10.
What are the minor nutrients added in MS medium?
Answer:

• Sodium molybdate
• Cupric sulphate
• Cobaltous chloride.

Question 11.
Why do we subject plantlets to hardening?
Answer:
Hardening slowly steadily helps the plantlets from the conditions of readymade medium, light & temperature of the laboratory, to which they were used, to the conditions of light, temperature & soil in the natural environment.

Question 12.
What is cybrid?
Answer:
The fusion product of a protoplast without a nucleus of different cells is called a cybrid.

Question 13.
What are the various components of MS- Medium?
Answer:

• Macronutrients, Micronutrients, Minor nutrients
• Iron stock
• Vitamins.
• Growth Hormone all in specific measurement & along with these solidifying agent- Agar is also added.

Question 14.
How to remove the cell wall of a plant cell.
Answer:
The chosen leaf tissue is immersed in the following solutions.

• 0.5% macrozyme. 2% onozuka cellulose enzyme dissolved in 13% sorbitol or mannitol kept at pH 5.4 at 25°c incubated during the night.
• After a gentle teasing of the cells, the protoplasts are obtained.
• Then they are transferred to 20% sucrose solution to retain viability.
• Finally by centrifuging the protoplasts are isolated.

Question 15.
What is organogenesis?
Answer:

• The morphological changes in the callus leading to the formation of the shoot, root, and then plantlets. The plantlets formation has 2 steps
• Root formation is known as Rhizogenesis
• Shoot formation is known as Caulogenesis.

Question 16.
Distinguish between callus & clone
Answer:
Callus :
It is the mass of unorganized growth of plant cells or tissues in in-vitro -culture medium.

Clone :
The clone develops from callus – which gets differentiated into many plantlets known as clones (i.e) genetically uniform population.

Question 17.
What is meant by hardening?
Answer:

• Hardening is the gradual exposure of invitro developed plantlets in humid chambers in diffused light – or transferred to – greenhouse setup.
• This enables them to get acclimatized to grow under normal field conditions.

Question 18.
How are the syn seeds produced?
Answer:

• Somatic embryoids – can be used in the production of syn seeds.
• They are nothing but somatic embryoids encapsulated in Agarose gel or calcium alginate/sodium
  alginate.

Question 19.
Give the tabulation of a few secondary metabolites their plant sources.
Answer:

Question 20.
Give the IPR – aspects in India
Answer:

Question 21.
Expand the following.
PTC, HEPA, RCGM -, GE AC, ELSI, GMO
Answer:
PTA – Plant Tissue Culture
HEPA – High-Efficiency Particulate Air
RCGM – Review Committee on Genetic Manipulation
GEAC – Genetic Engineering Approval Committee
ELSI – Ethical Legal and Social Implications
GMO – Genetically Modified Organism
GEM – Genetically Engineered Micro Organism

Question 22.
Name the cryoprotectants used in Cryopreservation
Answer:

• Dimethyl sulphoxide, glycerol, or sucrose are added before cryopreservation process.
• They protect the cells and tissues from the stress of freezing temperature, So known as Cryo protectants.

Question 23.
How is ELSI research funded?
Answer:
A percentage of the HGP – budget at the National Institute of Health & the V S Department of Energy was devoted to ELSI – research.

Question 24.
What is Biosafety?
Answer:
Biosafety is the prevention of large – scale loss of biological integrity, focusing both on ecology and human health.

Question 25.
Differentiate of Organ culture and meristem culture
Answer:
Organ culture :
The culture of embryos anthers, ovaries, roots, shoots

Meristem culture :
The culture of plant meristematic tissue on culture media
Give the tabulation of few secondary metaboltes a their plant sources.

Question 26.
What is somatic Embryogenesis?
Answer:
Somatic embryogenesis is the formation of embryos from the callus tissue directly and these embryos are called Embryoids or from the pre-embryonic cells which differentiate into embryoids.

VII. Three Marks

Question 1.
Give the name of few culture media used in PTC & their nature.
Answer:

• M.S. Nutrient Medium (Muroshige & Skoog -1992)
  It has carbon sources, suitable vitamins & hormones
• B5 – Medium (Gamborg.et.al 1968)
• White Medium (White 1943)
• Nitsch’s Medium (Nitsch & Nitsch 1969)
  The medium may be solid or semisolid or liquid – For solidification, a gelling agent such as agar is added.

Question 2.
Explain the Induction of Callus.
Answer:
Steps
I) Inoculation: Sterile segment of leaf, stem, tuber or root or (explant) is transferred to the sterile nutrient medium (MS – medium – + Auxins)

II) Incubation: The inoculated medium + auxins are incubated at 25 °C ± 2°C in an alternate light & dark period of 12 hours.

III) Induction of Callus:
The cell division occurs & the upper surface of the explant develop into a callus.
Callus – is a mass of unorganized growth of plant cells/tissue in-vitro – culture medium

Question 3.
Write the flow chart of plant Regeneration pathway.
Answer:
Plant Regeneration Pathway
From the explants, plants can be regenerated by somatic embryogenesis or organogenesis.

Question 4.
What are the application of somatic embrogenesis
Answer:

• It provides potential → after hardening becomes plantlets
• Used for production of synthetic seeds
• Eg. Allium sativum, Hordeum Vulgare, Oryza – sativa, Zee mays etc.,

Question 5.
Distinguish between Somaclonal Variations & Gametoclonal variations (Invitro Condition)
Answer:
Somaclonal Variations :
Variation found in somatic parts such as

• Leaf, stem
• root, tuber
• propagule etc

Gametoclonal variations:
Variations found in plants regenerated in vitro by gametes & gametophytes

Question 6.
Why there is a need to produce Virus-free plants?
Answer:

• Chemicals can be used to control fungal and bacterial mycoplasma pathogens but not viruses generally.
• Viral pathogens also cause great economic loss to the crops.
• Shoot meristem culture – help to produce virus-free plants because shoot meristem is free of viruses.

Question 7.
What are the Advantages of Artificial seeds?
Answer:

• Number/ time / cost – Millions of seeds produced / at any time / cheaper cost.
• Method – Easy method to produce genetically engineered plants.
• Quality – Seeds with desirable traits are produced.
• Storage – can be stored for long time use by Cryopreservation method.
• Nature of plants – Plants – Produced are identical
• Period of dormancy – greatly reduced
• Growth & Lifespan – grow faster, plants have a shorter life span

Question 8.
What are the applications of plant tissue culture?
Answer:

• Somatic hybridization → Improve hybrids produced
• Somatic embryoids → develop into syn – seeds help to conserve biodiversity
• Meristem & Shoot tip culture → production of Disease Resistant Varieties
• Production of plants → Stress resistant → herbicide tolerant → Drought tolerant
• Micropropagation → Large number of plantlets produced in a short time & throughout the year of both
• crop plants & true species – Used in Forestry

Question 9.
Write down the protocol for the micropropagation of banana.
Answer:

Question 10.
Write down the protocol for virus-free meristem tip culture.
Answer:

Question 11.
Which is the best conventional method to introduce disease resistance capacity into a plant? Explain.
Answer:

• Plant tissue culture is the conventional method which is also known as micropropagation.
• In this method, we take the meristematic tissue of the plant, referred to as explant is cultured over the given conditions of temperature and humidity, which makes the plant disease resistant.

Question 12.
What are the 3 parts of a patent? Explain them.
Answer:
It has 3 parts

1. The grant
2. The specifications
3. The claim

The grant

• It is a signed document (actually agreement) that grants patent rights to the inventor.
• It is filled at the patent office, (not published)

The Specifications

• • It is a narrative describing the invention & how it was carried out.
• Specifications & their claims are published from the patent office.

The Claim
The scope of the invention to be protected by the patent, preventing others from practicing it.

Question 13.
Write down the – general steps in patenting
Answer:

Question 14.
What is IPR? Explain the various aspects of if.
Answer:

• It is a category of properly include products created through one’s knowledge, research & creativity.
• It includes v Copyrights v Patents & v Trademarks
• It also includes v trade secrets v publicity rights v moral rights v rights against unfair competitions
• It also includes – designs & geographical indications

Other Various aspects :
The above-mentioned property of the discovery should not be exploited by others without legal permission or by getting proper authorization.
Rights – must be protected by the enforcement of laws framed by a country.

Question 15.
What are the future prospects of Biotechnology?
Answer:

• It will bring in a great revolution like the computer revolution.
• It will lead to new scientific – revolutions that would change the lives & future of people.
• Major challenges will be met and major changes incomprehensible in many aspects of modern life.

Question 16.
What is the function of GEAC?
Answer:

• It regulates -manufacturing, use, import, export, and storage of hazardous microbes or genetically modified organisms (GMOs) and cells in the country.
• It approves – activities involving large-scale use of hazardous microbes and recombinants in research & Industrial production.
• It is responsible – for approval of proposals relating to the release of GEO and products into the environment including experimental field trials (Biosafety Research Level – trial – I and II are known as BRL – I and BRL – II)

Question 17.
Write short notes on Ethical issues in Genomic Research?
Answer:

• Privacy and fairness in the use of genetic information, including the potential for genetic discrimination in employment and insurance.
• The integration of new genetic technologies such as genetic testing, into the practice of clinical medicine.
• Ethical issues surrounding the design and conduct of genetic research with people, including the process of informed consent.

Question 18.
Which is Laboratory Facilities for PTC?
Answer:
Washing facility for glassware and ovens for drying glassware.
Medium preparation room with autoclave, electronic balance, and PH meter.

Culture facility:
Growing the plant inoculated into culture tubes at 22-280C with the illumination of light 2400 lux, with a photoperiod of 8 -16 hours and relative humidity of about 60%

VIII. Five Marks.

Question 1.
Give the milestones in PTC – (Any 5 only)
Answer:

• Haberlandt (1902) – In-vitro culture of plant cells – (using knop’s salt solution + glucose & peptone)
• He proposed the Totipotency concept.
• P.R.White (1934) – In Knop’s solution + 3 vitamins (Pyridine, thiamine & nicotinic acid → developed root culture)
• F.C.Steward (1948) – used coconut water → produced cell proliferates from carrot explants.
• Morel & Martin (1952, 55) – Produced virus-free plants by shoot meristem culture →  Eg. Dahlia, Potato.
• Murashige & Skoog (1962) – Most frequently used culture medium for all kinds of tissue culture work.
• Guha & Maheswari (1964) – developed in-vitro production of haploid embryos from another of Datura.
• Vasil & Hildbrandt (1965) – developed a tobacco plant by micropropagation.

Question 2.
List down the culture conditions PTC.
Answer:
PH :

• PH of medium – should lie between 5.6 to 6 – Temperature
• Incubation of culture normally at temperature 25°C ± 2°C for optimal growth.

Humidity & Light Intensity

• 50-60% relative humidity
• 16-hours of photo period by the illumination of cool white fluorescent tubes of approximately 1000 lux

Aeration :

• Provided by shaking of flasks or tubes of liquid culture of Automatic shaker
• Aeration of the medium bypassing with filter-sterilized air.

Question 3.
What are the needed Lab – facilities for PTC?
Answer:
Washing & drying facility (oven) for the glassware
Medium preparation room with

• autoclave
• electronic balance
• PH meter etc., Maintain aseptic condition in,

a) Laminar air flow bench a positive pressure ventilation, unit

• (High-Efficiency Particulate Air (HEPA) filter to maintain the aseptic condition.
• Culture facility
• growing the ex-plant – inoculate into culture tube at 22 – 28°C with the illumination of light 2000 lux with 8-6 hours photoperiod, the relative humidity of about 60%

Question 4.
Explain various steps in Protoplast culture.
Answer:
Protoplasts are cells without a cell wall but with a cell membrane or plasma membrane.
1. Isolation of protoplast

2. Fusion of protoplast (Agglutination & Fusion)
Protoplast (A) + Protoplast (B) – fused in to one in the presence of Fusogenic agent PEG in 25 – 30% concentration (Poly Ethylene Glycol) with Ca++ ions.

3. Culture of protoplast:
Protoplast viability is tested with Fluorescein diacetate – before culture.
MS – Medium – used – (with some modifications) droplet, plating or Micro drop array technique.

a. Incubation: done in continuous light (1000 – 2000 lux) at 25°C.
The cell wall formation occurs within (24-48 hrs).
The first division of new cells occurs between 2-7 days of culture.

4. Selection of somatic hybrid cells:
The fusion product of protoplasts without a nucleus of different cells – (cybrid)
Cybrid is also known as Somatic hybrid the process is known as somatic hybridization

Question 5.
What is meant by biosafety? Explain.
Answer:

• It deals with the application of knowledge, techniques & equipment with strict guidelines in biological laboratories & related industries,
• to prevent large scale loss of
  • biological integrity
  • ecology
  • human health aspects
• to minimize human error and technical flaws & failures which contribute to unnecessary.
• exposures & disposal of – pathogenic microbes & hazardous chemicals, to regularise, risk management assessment and to set in best safeguard measures as per need.

Question 6.
Expand ELSI & What is meant by Bioethics.
Answer:

• ELSI – represents Ethical legal and social Implications.
• Advancements in biotechnology such as,
  • In Agriculture – Transgenic plants
  • In the pharmaceutical Industry – genotherapy
  • Advancements of medicine etc.,
• The biotechnological applications have raised controversies, hurting social beliefs, raising legal
  issues certain ecological principles & moral values.
• So it is high time to regularise legally the modern biotechnological applications & manipulation as Bioethics, for the welfare of humanity & other plant & animal communities of our world.

Question 7.
Write about Potential risks and consideration for safety aspects.
Answer:

• Pathogenicity – of living organisms & viruses natural or genetically modified to infect i) humans, ii) animals, iii) plants causing diseases
• Toxicity of allergy – associated with microbial production.
• Antibiotic-Resistant Microbes – increasing in number day by day.
• Disposal problem – regard to spent microbial biomass & purification of effluents.
• Safety aspects – regard to – i) contamination, ii) infection, iii) mutant strains
• regard to industrial use of microorganisms containing invitro recombinants.

Question 8.
List down organizations implementing Bio-safety guidelines.
Answer:
IBSCs – Institutional Bio-Safety Committees monitor the research activity at the institutional level.
RCGM – The Review Committee on Genetic manipulation, functioning in the Department of Biotechnology (DBT) monitors the risky research activities in the laboratories.
GEAC – Genetic Engineering Approval Committee
– (Ministry of Environment and Forest)
– has the power to use GMO at a commercial level and open field trials of transgenic

• crops
• industrial product
• health care products

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.2 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.2

Question 1.
Find the slope of the tangent to the following curves at the respective given points.
(i) y = x⁴ + 2x² – x at x = 1
(ii) x = a cos³ t, y = b sin³ t at t = \(\frac { π }{ 2 }\)
Solution:
(i) y = x⁴ + 2x² – x
Differentiating w.r.t. ‘x’
\(\frac { dy }{ dx }\) = 4x³ + 4x – 1
Slope of the tangent (\(\frac { dy }{ dx }\))_(x=1)
= 4(1)³ + 4(1) – 1
= 4 + 4 – 1 = 7
(ii) x = a cos³ t, y = b sin³ t
Differenriating w.r.t. ‘t’
\(\frac { dx }{ dt }\) = – 3a cos² t sin t
\(\frac { dy }{ dt }\) = 3b sin² t sin t

Question 2.
Find the point on the curve y = x² – 5x + 4 at which the tangent is parallel to the line 3x + y = 7.
Solution:
y = x² – 5x + 4
Differentiating w.r.t. ‘x’
Slope of the tangent \(\frac { dy }{ dx }\) = 2x – 5
Given line 3x + y = 7
Slope of the line = –\(\frac { 3 }{ 1 }\) = -3
Since the tangent is parallel to the line, their slopes are equal.
∴ \(\frac { dy }{ dx }\) = -3
⇒ 2x – 5 = -3
2x = 2
x = 1
When x = 1, y = (1)² – 5 (1) + 4 = 0
∴ Point on the curve is (1, 0).

Question 3.
Find the points on curve y = x³ – 6x² + x + 3 where the normal is parallel to the line x + y = 1729.
Solution:
y = x³ – 6x² + x+ 3
Differentiating w.r.t. ‘x’
Slope of the tangent \(\frac { dy }{ dx }\) = 3x² – 12x + 1
Slope of the normal = \(\frac { 1 }{ 3x^2 – 12x + 1 }\)
Given line is x + y = 1729
Slope of the line is – 1
Since the normal is parallel to the line, their slopes are equal.
\(\frac { 1 }{ 3x^2 – 12x + 1 }\) = -1
3x² – 12x + 1 = 1
3x² – 12x =0
3x(x – 4) = 0
x = 0, 4
When x = 0, y = (0)³ – 6(0)² + 0 + 3 = 3
When x = 4, y = (4)³ – 6(4)² + 4 + 3
= 64 – 96 + 4 + 3 = -25
∴ The points on the curve are (0, 3) and (4, -25).

Question 4.
Find the points on the curve y² – 4xy = x² + 5 for which the tangent is horizontal.
Solution:
y² – 4xy = x² + 5 ………… (1)
Differentiating w.r.t. ‘x’

When the tangent is horizontal(Parallel to X-axis) then slope of the tangent is zero.
\(\frac { dy }{ dx }\) = 0 ⇒ \(\frac { x+2y }{ y-2x }\) = 0
⇒ x + 2y = 0
x = -2y
Substituting in (1)
y² – 4 (-2y) y = (-2y)² + 5
y² + 8y² = 4y² + 5
5y² = 5 ⇒ y² = 1
y = ±1
When y = 1, x = -2
When y = – 1, x = 2
∴ The points on the curve are (- 2, 1) and (2, -1).

Question 5.
Find the tangent and normal to the following curves at the given points on the curve.
(i) y = x² – x⁴ at (1, 0)
(ii) y = x⁴ + 2e^x at (0, 2)
(iii) y = x sin x at (\(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\))
(iv) x = cos t, y = 2 sin² t at t = \(\frac { π }{ 3 }\)
Solution:
(i) y = x² – x⁴ at (1, 0)
Differentiating w.r.t. ‘x’
\(\frac { dx }{ dy }\) = 2x – 4x³
Slope of the tangent ‘m’ = (\(\frac { dx }{ dy }\))_{(1, 0)}
= 2 (1) – 4 (1)³ = -2
Slope of the normal –\(\frac { 1 }{ m }\) = \(\frac { -1 }{ -2 }\) = \(\frac { 1 }{ 2 }\)
Equation of tangent is
y – y₁ = m (x – x₁)
y – 0 = – 2 (x – 1)
y = -2x + 2
2x + y – 2 = 0
Equation of Normal is
y – y₁ = –\(\frac { 1 }{ m }\)(x – x₁)
y – 0 = \(\frac { 1 }{ 2 }\)(x – 1)
2y = x- 1
x – 2y – 1 = 0

(ii) y = x⁴ + 2e^x at (0, 2)
Differentiating w.r.t. ‘x’
\(\frac { dy }{ dx }\) = 4x³ + 2e^x
Slope of the tangent ‘m’
(\(\frac { dy }{ dx }\))_{(0, 2)} = 4(0)³ + 2e⁰ = 2
Slope of the Normal –\(\frac { 1 }{ m }\) =-\(\frac { 1 }{ 2 }\)
Equation of tangent is
y – y₁ = m(x – x₁)
⇒ y – 2 = 2(x – 0)
⇒ y – 2 = 2x
⇒ 2x – y + 2 = 0
Equation of Normal is
y – y₁ = –\(\frac { 1 }{ m }\) (x – x₁)
y – 2 = –\(\frac { 1 }{ 2 }\)(x – 0)
2y – 4 = -x
x + 2y – 4 = 0

(iii) y = x sin x at (\(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\))
Differentiating w.r.t. ‘x’
\(\frac { dy }{ dx }\) = x cos x + sin x
Slope of the tangent ‘m’ = (\(\frac { dy }{ dx }\))_{(π/2, π/2)}
= \(\frac { π }{ 2 }\) cos \(\frac { π }{ 2 }\) + sin \(\frac { π }{ 2 }\) = 1
Slope of the Normal –\(\frac { 1 }{ m }\) = -1
Equation of tangent is
y – y₁ = m(x – x₁)
⇒ y – \(\frac { π }{ 2 }\) = 1 (x – \(\frac { π }{ 2 }\))
⇒ x – y = 0
Equation of Normal is
y – y₁ = –\(\frac { 1 }{ m }\)(x – x₁)
⇒ y – \(\frac { π }{ 2 }\) = -1(x – \(\frac { π }{ 2 }\))
⇒ y – \(\frac { π }{ 2 }\) = -x + \(\frac { π }{ 2 }\)
⇒ x + y – π = 0

(iv) x = cos t, y = 2 sin² t at t = \(\frac { π }{ 2 }\)
at t = \(\frac { π }{ 3 }\), x = cos \(\frac { π }{ 3 }\) = \(\frac { 1 }{ 2 }\)
at t = \(\frac { π }{ 3 }\), y = 2 sin² \(\frac { π }{ 3 }\) = 2(\(\frac { 3 }{ 4}\)) = \(\frac { 3 }{ 2 }\)
Point is (\(\frac { 1 }{ 2 }\), \(\frac { 3 }{ 2 }\))
Now x = cos t y = 2 sin² t
Differentiating w.r.t. ‘t’,
\(\frac { dx }{ dt }\) = -sin t; \(\frac { dy }{ dt }\) = 4 sin t cos t
Slope of the tangent

Slope of the Normal –\(\frac { 1 }{ m }\) = \(\frac { 1 }{ 2 }\)
Equation of tangent is
y – y₁ = m(x – x₁)
⇒ y – \(\frac { 3 }{ 2 }\) = -2(x – \(\frac { 1 }{ 2 }\))
⇒ 2y – 3 = – 4x + 2
⇒ 4x + 2y – 5 = 0
Equation of Normal is
y – y₁ = –\(\frac { 1 }{ m }\)(x – x₁)
⇒ y – \(\frac { 3 }{ 2 }\) = \(\frac { 1 }{ 2 }\)(x – \(\frac { 1 }{ 2 }\))
⇒ 2 (2y – 3) = 2x – 1
⇒ 4y – 6 = 2x – 1
⇒ 2x – 4y + 5 = 0

Question 6.
Find the equations of the tangents to the curve y = 1 + x³ for which the tangent is orthogonal with the line x + 12y = 12.
Solution:
Curve is y = 1 + x³
Differentiating w.r.t ‘x’,
Slope of the tangent ‘m’ = \(\frac { dy }{ dx }\) = 3x²
Given line is x + 12y = 12
Slope of the line is –\(\frac { 1 }{ 12 }\)
Since the tangent is orthogonal with the line, the slope of the tangent is 12.
∴ \(\frac { dy }{ dx }\) = 12
i.e 3x² = 12
x² = 4
x = ±2
When x = 2, y = 1 + 8 = 9 ⇒ point is (2, 9)
When x = -2, y = 1 – 8 = -7 ⇒ point is (-2, -7)
Equation of tangent with slope 12 and at the j point (2, 9) is
y – 9 = 12 (x – 2)
y – 9 = 12x – 24
12x – y – 15 = 0
Equation of tangent with slope 12 and at the point (-2, -7) is
y + 7 = 12 (x + 2)
y + 7 = 12x + 24
12x – y + 17 = 0

Question 7.
Find the equations of the tangents to the curve y = –\(\frac { x+1 }{ x-1 }\) which are parallel to the line x + 2y = 6.
Solution:

Given line is x + 2y = 6
Slope of the line = –\(\frac { 1 }{ 2 }\)
Since the tangent is parallel to the line, then the slope of the tangent is –\(\frac { 1 }{ 2 }\)
∴ \(\frac { dy }{ dx }\) = \(\frac { 2 }{ (x-1)^2 }\) = –\(\frac { 1 }{ 2 }\)
(x – 1)² = 4
x – 1 = ±2
x = -1, 3
When x = – 1, y = 0 ⇒ point is (-1, 0)
When x = 3, y = 2 ⇒ point is (3, 2)
Equation of tangent with slope –\(\frac { 1 }{ 2 }\) and at the point (-1, 0) is
y – o = –\(\frac { 1 }{ 2 }\)(x + 1)
2y = -x – 1 ⇒ x + 2y + 1 = 0
Equation of tangent with slope –\(\frac { 1 }{ 2 }\) and at the point (3, 2) is 2
y – 2 = –\(\frac { 1 }{ 2 }\) (x – 3)
2y – 4 = -x + 3
x + 2y – 7 = 0.

Question 8.
Find the equation of tangent and normal to the curve given by x – 7 cos t andy = 2 sin t, t ∈ R at any point on the curve.
Solution:
x = 7 cos t and y = 2 sin t, t ∈ R
Differentiating w.r.t. ‘t’,
\(\frac { dx }{ dt }\) = -7 sin t and \(\frac { dy }{ dt }\) = 2 cos t
Slope of the tangent ‘m’
\(\frac { dy }{ dx }\) = \(\frac{\frac { dy }{ dt }}{\frac{ dx }{ dt }}\) = \(\frac { 2 cot t }{ -7 sin t }\)
Any point on the curve is (7 Cos t, 2 sin t)
Equation of tangent is y – y₁ = m (x – x₁)
y – 2 sint = –\(\frac { 2 cot t }{ 7 sin t }\) (x – 7 cos t)
7y sin t – 14 sin² t = -2x cos t + 14 cos² t
2x cos t + 7 y sin t – 14 (sin² t + cos² t) = 0
2x cos t + 7y sin t – 14 = 0
Now slope of normal is –\(\frac { 1 }{ 3 }\) = \(\frac { 7 sin t }{ 2 cos t }\)
Equation of normal is y – y₁ = –\(\frac { 1 }{ m }\)(x – x₁)
y – 2 sin t = \(\frac { 7 sin t }{ 2 cos t }\) (x – 7 cos t)
2y cos t – 4 sin t cos t = 7x sin t – 49 sin t cos t 7x sin t – 2y cos t – 45 sin t cos t = 0

Question 9.
Find the angle between the rectangular hyperbola xy = 2 and the parabola x² + 4y = 0
Solution:
Given curves are xy = 2 ……… (1)
x² + 4y = 0 ………. (2)
Now solving (1) and (2)
Substituting (1) in (2)
⇒ x² + 4(2/x) = 0
x³ + 8 = 0
x³ = -8
x = -2
Substituting in (1) ⇒ y = \(\frac { 2 }{ -2 }\) = -1
∴ Point of intersection of (1) and (2) is (-2, -1)
xy = 2 ⇒ y = \(\frac { 2 }{ x }\) ……….. (1)
Differentiating w.r.t. ‘x’

The angle between the curves

Question 10.
Show that the two curves x² – y² = r² and xy = c² where c, r are constants, cut orthogonally.
Solution:
Given curves are x² – y² = r² ……….. (1)
xy = c² …….. (2)
Let (x₁, y₁) be the point of intersection of the given curves.
(1) ⇒ x² – y² = r²
Differentiating w.r.t ‘x’,
2x – 2y \(\frac { dx }{ dy }\) = 0
\(\frac { dx }{ dy }\) = \(\frac { x }{ y }\)
now (\(\frac { dx }{ dy }\))(_x1,y1) = m₁ = \(\frac { x_1 }{ y_1 }\)
(2) ⇒ xy = c²
Differentiating w.r.t ‘x’,

Hence, the given curves cut orthogonally.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Botany Guide Pdf Chapter 4 Principles and Processes of Biotechnology Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 4 Principles and Processes of Biotechnology

12th Bio Botany Guide Principles and Processes of Biotechnology Text Book Back Questions and Answers

I. Choose the correct answer from the given option:

Question 1.
Restriction enzymes are.
a) Not always required in genetic engineering
b) Essential tools in genetic engineering.
c) Nucleases that cleave DNA at specific sites,
d) both b and c.
Answer:
d) both b and c

Question 2.
Plasmids are
a) circular protein molecules
b) required by bacteria.
c) tiny bacteria.
d) confer resistance to antibiotics.
Answer:
d) confer resistance to antibiotics

Question 3.
EcoRI cleaves DNA at.
a) AGGGTT
b) GTATATC.
c) GAATTC
d) TATAGC.
Answer:
c) GAATTC

Question 4.
Genetic engineering is
a) making artificial genes
b) hybridization of DNA of one organism to that of the others
c) production of alcohol by using micro organisms
d) making artificial limbs, diagnostic instruments such as ECG, EEC, etc.,
Answer:
b) hybridization of DNA of one organism to that of the others

Question 5.
Consider the following statements:
I) Recombinant DNA tecimology is popularly known as genetic engineering is a stream of biotechnology which deals with manipulation of genetic materials by man invitro.
II) pBR322 is the first artificial cloning vector developed in 1977 by Boliver and Rodriguez from E.coli plasmid.
III) Restriction enzymes belongs to a class of enzymes called nucleases.
Choose the correct option regarding above statements
a) I & II.
b) I & III.
c) II & III
d) I, II, & III
Answer:
d) I, II & III

Question 6.
The process of recombinant DNA technology has the following steps
I) Amplication of the gene
II) Insertion of recombinant DNA into host cells
III) Cutting of DNA at specific location using restriction enzyme
IV) Isolation of genetic material(DNA)
Pick out the correct sequence of step for recombinant DNA technology,
a) II, III, IV, I
b) IV, II, III, I
c) I, II, III, IV
d) IV, III, I, II
Answer:
d) IV, III, I, II

Question 7.
Which one of the following palindromic base sequence in DNA can be easily cut about the middle by some particular restriction enzymes?
a) 5’ CGTTCG 3′ 3′ ATCGTA 5!
b) 5’ GATATG3′ 3′ CTACTA 5’
c) 5′ GAAHC 3′ 3′ CTTAAG 5′
d) 5′ CACGTA 3′ 3′ CTCAGT 5′
Answer:
c) 5′ GAAHC 3′ 3′ CTTAAG 5′

Question 8.
pBR 322, BR stands for
a) Plasmid Bacterial Recombination
b) Plasmid Bacterial Replication
c) Plasmid Boliver and Rodriguez
d) Plasmid Baltimore and Rodriguez
Answer:
c) Plasmid Boliver and Rodriguez

Question 9.
Which of the following one is used as a Biosensors?
a) Electrophoresis
b) Bioreactors
c) Vectors
d) Electroporation
Answer:
Correct Answer: enzymes

Question 10.
Match the following

Answer:
b) 1-c, 2-d, 3-a, 4-b

Question 11.
In which techniques Ethidium Bromide is used?
a) Southern Blotting Techniques
b) Western Blotting Techniques.
c) Polymerase Chain Reaction.
d) Agrose Gel Electroporosis.
Answer:
d) Agrose Gel Electroporosis

Question 12.
Assertion : Agrobacterium tumifaciens is popular in genetic engineering because this bacterium is associated with the root nodules of all cereals and pulse crops.
Reason : A gene incorporated in the bacterial chromosomal genome gets automatically transferred to the cross with which bacterium is associated.
a) Both assertion and reason are true. But reason is correct explanation of assertion.
b) Both assertion and reason are true. But reason is not correct explanation of assertion.
c) Assertion is true but reason is false.
d) Assertion is false but reason is true.
e) Both assertion and reason are false.
Answer:
e) Both assertion and reason are false

Question 13.
Which one of the following is not correct statement.
a) Ti plasmid causes the bunchy top disease.
b) Multiple cloning site is known as Polylinker.
c) Non viral method tranfection of Nucleic acid in cell
d) Polyactic acid is a kind of biodegradable and bioactive thermoplastic
Answer:
a) Ti plasmid causes the bunchy top disease

Question 14.
An analysis of chromosomal DNA using the southern hybridisation technique does not use
a) Electrophoresis .
b) Blotting
c) Autoradiography
d) Polymerase Chain Reaction
Answer:
d) Polymerase Chain Reaction

Question 15.
An antibiotic gene in a vector usually helps in the selection of.
a) Competent cells.
b) Transformed cells
c) Recombinant cells
d) None of the above
Answer:
b) Transformed cells

Question 16.
Some of the characteristics of Bt cotton are
a) Long fibre and resistant to aphids
b) Medium yield, long fibre and resistant to beetle pests
c) high yield and production of toxic protein crystals which kill dipteran pests
d) High yield and resistance to ball worms.
Answer:
d) High yield and resistance to bollworms

Question 17.
How do you use biotechnology in modern practice?
Answer:
Today biotechnology is a billion-dollar business around the world, applies biotechnological tools for their product improvement.

• Pharmaceutical companies.
• Breweries.
• Agro Industries & others.
• Modern biotechnology – include all methods, rDNA technology, cell fusion technology, etc.,
• Major focus of Biotechnology (see the tabulation)

Question 18.
What are the materials used to grow microorganisms like Spirulina?
Answer:
Spirulina can be grown easily on materials like waste water from potato processing plants (containing starch), straw, molasses, animal manure and even sewage, to produce large quantities.

Question 19.
You are working in a biotechnology lab with a bacterium namely E.Coli. How will you cut the nucleotide sequence? Explain it.
Answer:

• The exact kind of cleavage produced by a v restriction enzyme is important in the • design of a gene cloning experiment.
• Some cleave both strands of DNA through the centre resulting in blunt or flush end known as symmetric cuts.
• Some restriction enzymes cut the strand of DNA, a little away from the centre of palindrome sites, between the same two bases on the opposite strands, protruding and recessed ends known as sticky or cohesive end, cuts known as asymmetric cut or
  staggered cuts.
• It is necessary that the vector and the source DNA are cut with the same restriction enzyme, so that the resultant DNA fragments have the same sticky ends facilitating the action of DNA ligase to join them.

Question 20.
What are the enzymes you can used to cut terminal end and internal phosphodiester bond of nucleotide sequence?
Answer:
Restriction exonuclease are the restriction enzyme used to cut nucleotides from the terminal end of DNA. Whereas, restriction endonucleases cut the internal phosphodiester bond with DNA molecule.

Question 21.
Name the chemicals used in gene transfer.
Answer:
Director vector less Gene transfer is possible through several mediators

Chemical mediated gene transfer:
Certain chemicals like Poly Ethylene Glycol(PEG) and Dextran Sulphate.
These chemicals induce the uptake of DNA into plant protoplasts.

Question 22.
What do you know about the word pBR332?
Answer:
pBR332 – It is a reconstructed plasmid and most widely used as cloning vector.

• It contains 4361 base pairs.
• P denotes Plasmid .
• B&R – The names of Boliver and Rodriguez, the scientists developed this plasmid.
• 322 – The number of plasmid developed from their lab.
• It contains ampR & tet^R – 2 different antibiotic resistant genes & the recognition sites for several restriction enzymes (Hindlll, ECoRI, Bam H-I, Sal I, Pvu II, Pst I, Cla I) ori & antibiotic resistance genes.
• Rop – Codes for the proteins involved in the replication of the plasmid.

Question 23.
Mention the application of Biotechnology.
Answer:
Introduction: Most important applied interdisciplinary sciences of the 21 st century
It has promise for the benefits of Human Being.

Production of secondary metabolites – Biofertilizers, Biopesticides & Enzymes
Biomass Energy, Biofuel, Biorernediation phvtoremediation for environmental biotechnology.

Question 24.
What is the restriction enzymes? Mention their type with a role in Biotechnology.
Answer:
Restriction enzymes are the enzymes of bacterial origin which cleaves DNA into fragments at or near specific recognition sites within DNA molecules. This principle is used in biotechnology to cut and insert the desired gene (gene of interest) thereby generating an rDNA with desirable characters.

a) Exonucleases – remove nucleotides one at a time from the end of DNA.
Eg: Bal 31, Exonuclease III

b) Endonucleases – break the internal phosphodiester bonds with in a DNA.
Eg: Hind II, EcoRI, Pvul, Bam HI, Taql.

Three classes of Restriction endonuclease

• Type 1, II & III – which differ slightly by their mode of action
• Type II – preferred in rDNA technology as they cut DNA with in a specific sequence consisting of 4 – 8 bp.
• Hind II – cut DNA at a point of specific sequence of 6 base pairs (recognition
  sequence).
• From 200 strains 900 restriction enzymes isolated from over 230 strains of bacteria with different recognition sequences.
• Restriction endonucleases are named by a standard procedure.
• The first letter of the enzymes indicates the genus name, followed by the first two letters of the species, then comes the strain of the organism and finally a roman numeral indicating the order of discovery.
• For example ECORI is from Escherichia (E) coli (co) strain Ry 13 (R) and first endonuclease (I) to be discovered .
• It contains 2 different antibiotic resistance genes and recognition site for several restriction enzymes.
  This sequence is referred to as a restriction site and is generally – palindromic which means that the sequence in both DNA strands at this site read same in 5′ – 3′ direction and in the 3′ – 5′ direction.

Question 25.
Are there any possibilities to transfer a suitable desirable gene to host plant without vector? Justify your answer.
Answer:
Yes, it is possible to transfer a suitable desired gene to a host plant using certain chemicals, microinjection method, electroporation or by biolistics.

a. Chemical mediated gene transfer:
Chemicals Poly Ethylene Glycol (PEG) & Dextran sulphate – induce DNA uptake into plant protoplasts.

b. Microinjection:
With a fine-tipped glass needle, DNA is directly injected into the nucleus.
The protoplasts are immobilized on solid support (agarose on a microscopic slide)

c. Electroporation method of gene transferJjjJU Protoplasts, cells or tissues subjected to a pulse of high voltage electric power to make transient pores in the plasma membrane, through which uptake of foreign DNA occurs.

d. Liposome – mediated methods of gene transfer
The gene or DNA is transferred in an encapsulated form from Liposome ( the artificial phospholipid vesicles) into the vacuole of plant cells.

e. Biolistics:
The DNA particle with gold or tungsten particle (1.3 gm) coating are bombarded into the target tissue by gene gun or microprojectile gun/shotgun The bombarded cells/tissues are cultured to regenerate plants from transformed cells.

Question 26.
How will you identify vectors?
Answer:

Question 27.
Compare the various types of Blotting techniques.
Answer:

Question 28.
Write the advantages of herbicide-tolerant crops.
Answer:

Question 29.
Write the advantages and disadvantages of Bt cotton.
Answer:

Question 30.
What is bioremediation? Give some examples of bioremediation.
Answer:
Bioremediation:
It is defined as the use of microorganisms or plants to clean up the environmental pollution. It is an approach used to treat wastes including wastewater, industrial waste, and solid waste. The bioremediation process is applied to the removal of oil, petrochemical residues, pesticides, or heavy metals from soil or groundwater.

In many cases, bioremediation is less expensive and more sustainable than other physical and chemical methods of remediation. The bioremediation process is a cheaper and eco-friendly approach and can deal with lower concentrations of contaminants more effectively. The strategies for bioremediation in soil and water can be as follows:

1. Use of indigenous microbial population as indicator species for the bioremediation process.
2. Bioremediation with the addition of adapted or designed microbial inoculants.
3. Use of plants for bioremediation – green technology.

Question 31.
Write the benefits and risks of Genetically Modified Foods.
Answer:

12th Bio Botany Guide Principles and Processes of Biotechnology Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
Which one of the following is a secondary metabolite? IWriiWJ
a. Ethanol
b. Acetic acid
c. Citric acid
d. Toxic pigments
Answer:
d) Toxic pigments

Question 2.
Bio-Technology was coined by.
a. Weisner
b. Karl Prantl
c. Sanger & Gilbert
d. KarlEreky
Answer:
d) Karl Ereky

Question 3.
Traditional Bio-Technology is also known as.
a. Fermentation Biology
b. Kitchen Technology
c. Hybridization Biology
d. Transgenic Biology
Answer:
b) Kitchen Technology

Question 4.
The study of Drugs or medicines used in medical treatment is known as.
a. Pharmaceuticals
b. Biomedical Engineering
c. Chemical Engineering
d. Tissue Engineering
Answer:
a) Pharmaceuticals

Question 5.
Which of the following can be Bio Technological products?
a. Antibiotics
b. Vaccines
c. Enzymes
d. All the above.
Answer:
d) All the above

Question 6.
Multiplication of Alien DNA in organisms required
a. ROP
b. ORI
c. Stop codon
d. TATA box
Answer:
b) ORI

Question 7.
Tools of Biotechnology is used for effluent treatment, water cycling is known as
a. Process Engineering
b. Production Engineering
c. Mechanical Engineering
d. Microbial Engineering
Answer:
a) Process Engineering

Question 8.
The scientist who use a first viral vaccine to inoculate a child from smallpox is
a. Louis Pasteur
b. Edward Jenner
c. Sanger and Gilbert.
d. Arber and Nathans
Answer:
b) Edward Jenner

Question 9.
The enzyme used for making artificial sweeteners is
a. Lactose
b. Galactose
c. Invertase
d. Reductase
Answer:
c) Invertase

Question 10.
Development of Artificial gene functioning within living cells was done by
a. H.G.Khorana
b. Ian Wilmet
c. Sir Robert
d. G. Edwards
Answer:
a) H.G.Khorana

Question 11.
rDNA is also known as
a. Hybrid DNA-RNA
b. Recombinant of vector DNA and desired genes
c. Chimeric DNA
d. Bothb&c
Answer:
d) Both b & c

Question 12.
Plasmids are
a. ss DNA
b. ds DNA(linear)
c.rDNA
d. Vector DNA
Answer:
d) Vector DNA

Question 13.
pBR322 is most extensively studied
a. Foreign gene
b. r DNA
c. done
d. Plasmid DNA of Ecoli.
Answer:
d) Plasmid DNA of Ecoli

Question 14.
Restriction enzymes recognize specific
a. Palindromic region,
b. Exons
c. Introns
d. None of these
Answer:
a) Palindromic region

Question 15.
Restriction enzymes of Ecoli are
a. Hind III
b. Bam III
c. EcoRI I & EcoRI II
d. All of these.
Answer:
c) EcoRI I & EcoRI II

Question 16.
The best cloning organism for biotechnology is
a. Agrobacterium
b. Pseudomonas
c. Lambda phage
d. E. Coli
Answer:
d) E.Coli

Question 17.
The ability to form tumours is found in the plasmids of
a. E.coli
b. Pseudomonas
c. Agrobacterium tumefaciens
d. Pneumococcus
Answer:
c) Agrobacterium tumefaciens

Question 18.
Engineered bacterium carries
a. Plasmids
b. rDNA
c. c DNA
d. ssDNA
Answer:
b) rDNA

Question 19.
Electrophoresis and southern blotting techniques are used in
a. DNA fingerprinting
b. Gene Synthesis
c. gene cloning
d. All of these.
Answer:
a) DNA fingerprinting

Question 20.
In biosesnsors Green Fluorescent protein is used which is isolated from A and spliced
a. A Chlamydomonas – B Ecoli
b. A Gelidium – B Bacillus subtilis
c. A Aequorea victoria – B Arabidopsis thaliana
d. A Asoaragus – B Accacia melanoxylon
Answer:
c. A Aequorea victoria – B Arabidopsis thaliana

Question 21.
Bacteria protects themselves from viral attack by producing
a. Exonuclease
b. Endonuclease
c.DNAligase
d. Gy ase
Answer:
b) Endonuclease

Question 22.
Molecular scissor is
a. Urease
b. Helicase
c. Peptidase
d. Restriction Endonuclease
Answer:
d) Restriction Endonuclease

Question 23.
Which one of the following is used in transfer of foreign DNA to crop plants?
a. Penicillum Expansum
b. TrichodermaHarzianum
c. Meloidogyne Incognita
d. Agrobacterium tumefaciens
Answer:
d) Agrobacterium tumefaciens

Question 24.
E coli is the mostly used organism for gene cloning, because
a. It is easy to handle
b. It is growing easily under optimal condition
c. It is the safe organism d. All the above.
Answer:
d) All the above.

Question 25.
Which one of the following palindromic base sequences in DNA can be easily cut at about the middle by some particular restriction enzyme?
a. 5′ CGTTCG3’3′ ATGGTA 5′
b. 5′ GATATG 3′ 3′ CTACTA 5′
c. 5′ GAATTC 3′ 3’CTTAAG 5′
d. 5′ CACGTA3’3′ CTCAGT 5′
Answer:
c. 5′ GAATTC 3′ 3’CTTAAG 5′

Question 26.
Biolistics (gene gun) is suitable for
a. Constructing recombinant DNA by joining with vectors
b. DNA finger printing
c. Disease resistant genes
d. Transformation of plant cells
Answer:
d) Transformation of plant cells

Question 27.
For transformation micro particles coated with DNA to be bombarded with gene gun are made up of
a. Silver or Platinum
b. Platinum or Zinc
c. Silicon or Platinum
d. Gold or Tungsten
Answer:
d) Gold or Tungsten

Question 28.
Rising of dough is due to
a. Multiplication of Yeast
b. Production of CO₂
c. Emulsification
d. Hydrolysis of wheat flour starch in to sugar
Answer:
b) Production of CO₂

Question 29.
All the process after the fermentation process is known as
a. upstream process
b. downstream process
c. forward process
d. backward process
Answer:
b) downstream process

Question 30.
For making GMO, the three basic steps that are required are
a. Identification of DNA with desirable gene
b. Introduction of identified DNA into the host
c. Maintenance of introduced DNA in to the host and transfer of DNS to its progeny
d. All the above
Answer:
d) All the above

Question 31.
Zymology is the study of
a. Fermentation & its practical use
b. Name of Bioreactors
c. Upstream pro^ss
d. Downstream process
Answer:
a) Fermentation & its practical use

Question 32.
ECORI – R stands for
a. Genus
b. Species
c. Strains
d. Group
Answer:
a) Genus

Question 33.
Which is suitable for transferring an alien DNA into a plant cell?
a. CaCl₂
b. Biolistics or gene gun method
c. Micro infection
d. Heat shock
Answer:
b) Biolistics or gene gun method

Question 34.
The group of degradable biopolymers are
a. CrylAc and DMH-11
b. PHAsandPHB
c. GFPandPGA
d. DMH and HT
Answer:
b) PHAs and PHB

Question 35.
Genetically engineered human insulin is
a. Haematiri
b. Pro insulin
c. Hybridin
d. Humulin
Answer:
d) Humulin

Question 36.
Probiotics are
a. Food Allergens
b. safe antibiotics
c, Carcinogenic microbes
d. Live microbial food supplements
Answer:
d) Live microbial food supplement

Question 37.
Bt Brinjal is produced by using A and is having resistance against B.
a. A Ecoli – B Virus
b. A Virus – B Bacteria
c. A Agrobacterium – B Bacillus
d. A Agrobacterium – B Lepidopteron
Answer:
d. A Agrobacterium – B Lepidopteron

Question 38.
PCR refers to
a. A common laboratory technique of making millions of copies of a particular region of DNA
b. A biotechnological procedure of replicating DNA strands
c. Hybridization of DNA molecules in to several fragments
d. It is a test for tracing genetic defects.
Answer:
a) A common laboratory technique of making millions of copies of a particular region of DNA