Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 15 Plant Growth and Development Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

11th Bio Botany Guide Plant Growth and Development Text Book Back Questions and Answers

Part – I.

Question 1.
Select the wrong statement from the following:
(a) Formative phase of the cells retain the capability of cell division.
(b) In elongation phase development of central vacuole takes place.
(c) In maturation phase thickening and differentiation takes place.
(d) In maturation phase, the cells grow further.
Answer:
(d) In maturation phase, the cells grow further.

Question 2.
If the diameter of the pulley is 6 inches, length of pointer is 10 inches and distance travelled by pointer is 5 inches. Calculate the actual growth in length of plant.
a) 3 inches
b) 6 inches
c) 12 inches
d) 30 inches
Answer:
options are wrong, (correct Ans = 1.5 inches)
Solution:
Step I:
Diameter of the Pulley=6 inches
Radius of the pulley \(=\frac{6}{2}\)= 3 inches
Actual growth in length= Distance travailed by pointer x Radius of the pulley Length of the pointer
=\(\frac{5 \times 3}{10}\) =1.5=1.5.
Answer:
1.5 inches

Question 3.
In unisexual plants, sex can changed by the application of
a) Ethanol
b) Cytokinins
c) ABA
d) Auxin
Answer:
c) ABA

Question 4.
Select the correctly matched one
A) Humanurine i) Auxin-B
B) Corn gram oil ii) GA₃
C) Fungs iii) Abscisic acid II
D) Herring fish sperm iv) Kinetin
E) Unripcrnaizegrains v) AuxinA
F) Young cotton boils vi) Zeatin

Answer:
b) A – v, B – i, C – ii, D – iv, E – vi, F – iii

Question 5.
Seed dormancy allows the plants to:
(a) overcome unfavorable climatic conditions
(b) develop healthy seeds
(c) reduce viability
(d) prevent deterioration of seeds
Answer:
(a) overcome unfavorable climatic conditions

Question 6.
What are the parameters used to measure growth of plants?
Answer:
Growth in plants can be measured in terms, of

• Increase in length or girth (roots and stems)
• Increase in fresh or dry weight
• Increase in area or volume (fruits and leaves)
• Increase in a number of cells produced.

Question 7.
What is plasticity?
Answer:
Plasticity refers to the environmental heterophylly seen in Butter cup plant (Ranunculus). In this aquatic plant, the leaves in the air is normal, where as the leaves submerged underwater are highly thin and hairy highly adapted to do carbon assimilation Developmental heterophlly seen in the juvenile plant leaves of cotton and corianter. Where the young leaves have a different shape from the mature leaves is not considered as plasticity.

Question 8.
Write the physiological effects of Cytokinins.
Answer:

1. Cytokinin promotes cell division in the presence of auxin (IAA).
2. Induces cell enlargement associated with IAA and gibberellins
3. Cytokinin can break the dormancy of certain light-sensitive seeds like tobacco and induces seed germination.
4. Cytokinin promotes the growth of lateral bud in the presence of apical bud.
5. Application of cytokinin delays the process of aging by nutrient mobilization. It is known as the Richmond Lang effect.
6. Cytokinin:
   • increases rate protein synthesis
   • induces the formation of inter-fascicular cambium
   • overcomes apical dominance
   • induces the formation of new leaves, chloroplast and lateral shoots.
7. Plants accumulate solutes very actively with the help of cytokinins.

Question 9.
Describe the mechanism of photoperiodic induction of flowering.
Answer:
Mechanism of photoperiodic induction of flowering.

• The physiological change on flowering due to the relative length of light and darkness (photoperiod) is called Photoperiodism.
• The photoperiod required to induce flowering is called critical day length. Eg. 12 hours in Maryland Mammoth’s Tobacco Xanthium 15.05 hours.

Photoperiodic induction:

• An appropriate photoperiod in 24 hours cycle constitutes one inductive cycle. Plants may require one or more inductive cycles for flowering.
• The phenomenon of conversion of leaf primordia into flower primordia under the influence of suitable inductive cycles is called photoperiodic induction. Example: Xanthium (SDP) -1 inductive cycle and Plantago (LDP) -25 inductive cycles.

Site of photoconductive perception:

• Leaves are the parts that receive photoperiodic stimulus (PPS), again it is only leaves that synthesize floral hormones and translocate them to the apical tip to promote flowering.
• This can be demonstrated by experiments conducted in the Cocklebur plant. Which is an SD plant. The nature of flower-producing stimulus has been elusive so far. It is believed by physiologists that a hormone is responsible for it, Chailakyan (1936) named it as Florigen It is not possible to isolate it.

Procedure	Observation	Inference
1. Take potted plant A and defoliate the plant subject it to SD – a condition	There is no induction of flowering	No leaf to receive stimulus or induction of flowering
2.  Take potted plant B – and defoliate all, except one leaf subject it to SD – condition.	There is the induction of flowering	One leaf is enough to receive stimulus or induction of flowering.
3.  Take potted plant C – and defoliate it and subject it to LD condition	There is no induction of flowering	no leaf to receive stimulus or induction of flowering
4. Take potted plant D and subject all leaves to LD but one leaf to SD	There is the induction of flowering	One leaf is enough to receive an induction in the SD condition

Question 10.
Give a brief account of programmed cell death (PCD).
Answer:
Senescence is controlled by plants’ own genetic program and the death of the plant or plants part consequent to senescence is called Programmed Cell Death. In short senescence of an individual cell is called PCD. The proteolytic enzymes involving PCD in plants are phytases and in animals are caspases. The nutrients and other substrates from senescing cells and tissues are remobilized and reallocated to other parts of the plant that survives.

The protoplasts of developing xylem vessels and tracheids die and disappear at maturity to make them functionally efficient to conduct water for transport. In aquatic plants, aerenchyma is normally formed in different parts of the plant such as roots and stems which enclose large air spaces that are created through PCD. In the development of unisexual flowers, male and female flowers are present in earlier stages, but only one of these two completes its development while the other aborts through PCD.

Part-II.

11th Bio Botany Guide Plant Growth and Development Additional Important Questions and Answers

I. Choose The Correct Answer

Question 1.
The open form of the growth occurs in:
(a) leaves and flowers
(b) stem and root
(c) leaves and stem
(d) stem and flowers
Answer:
(b) stem and root

Question 2.
An example of a De-Differentiating cell is ………………
a) Tracheary element
b) shoot apex
c) Cork cambium
d) root apex
Answer:
c) Cork cambium

Question 3.
Primary growth of the plant is due to the activity of:
(a) phloem parenchyma
(b) phloem meristem
(c) vascular cambium
(d) apical meristem
Answer:
(d) apical meristem

Question 4
Choose the natural Auxin of the following
a) Anti Auxin
b) NAA
c) 2.4.D
d) IndoleAcetic Acid (IAA)
Answer:
d) Indole Acetic Acid (IAA)

Question 5.
Thickening and differentiation of cells take place during:
(a) elongation phase
(b) formative phase
(c) maturation phase
(d) flowering phase
Answer:
(c) maturation phase

Question 6.
The hormone present in Coconut milk is
a) Gibberellins
b) Ethylene
c) Cytokinin
d) Auxin
Answer:
c) Cytokinin

Question 7.
The total growth of the plant consists of four phases in the following order.
(a) Log phase, lag phase, decelerating phase and maturation phase
(b) Log phase, lag phase, maturation phase and decelerating phase
(c) Lag phase, log phase, maturation phase and decelerating phase
(d) Lag phase, log phase, decelerating phase and maturation phase
Answer:
(d) Lag phase, log phase, decelerating phase and maturation phase

Question 8.
Which of the following Phytóhormone does not occur naturally in plants?
a) 2. 4. D
b) GibberellicAcid
c) 6. Furfuryl amino purine
d) IAA
Answer:
a) 2.4.D

Question 9.
Absence of light may lead to the yellowish color in plants and this is called:
(a) venation
(b) etiolation
(c) estivation
(d) vernation
Answer:
(b) etiolation

Question 10.
Apical dominance is caused when Auxin
a) Concentration is more than Cytokinins
b) Concentration is less than Cytokinins
c) and Cytokinin concentration are equal
d) and Cytokinin concentration are fluctuating
Answer:
a) Concentration is more than Cytokinins

Question 11.
Indicate a plant growth regulator from the following:
(a) cytocin
(b) cytokinins
(c) acetic acid
(d) methylene
Answer:
(b) cytokinins

Question 12.
Which prevents premature fall of fruit?
a) NAA
b) Ethylene
c) GA₃
d) Zeatin
Answer:
a) NAA

Question 13.
The activity of synergistic effect involves the activity of:
(a) auxin and gibberellins
(b) auxin and ethylene
(c) ABA and gibberellins
(d) none of the above
Answer:
(a) auxin and gibberellins

Question 14.
The term Auxin was coined by
a) Went
b) Darwin
c) Smith
d) Garner
Answer:
a) Went

Question 15.
The term auxin was first coined by:
(a) Charles Darwin
(b) Kogl
(c) F.W. Went
(d) Smith
Answer:
(c) F.W. Went

Question 16.
The term Gibberellin was coined by
a) Went
b) Kurosawa
c) Skoog
d) Yabuta
Answer:
d) Yabuta

Question 17.
Indicate a synthetic auxin.
(a) Indole Acetic Acid
(b) Phenyl Acetic Acid
(c) Indole Butyric Acid
(d) Naphthalene Acetic Acid
Answer:
(d) Naphthalene Acetic Acid

Question 18.
The mineral required for the synthesis of IAA is
a) Copper
b) Magnesium
c) Zinc
d) Boron
Answer:
c) Zinc

Question 19.
Auxin stimulates:
(a) transpiration
(b) respiration
(c) flowering
(d) none of the above
Answer:
(b) respiration

Question 20.
The most widely occurring Cytokinin in plants is
a) ABA
b) NAA
c) TNT
d) IPA
Answer:
d) IPA

Question 21.
Who established the structure of gibberellic acid?
(a) Brain etal
(b) Kurosawa
(c) Cross et al
(d) Yabuta and Sumiki
Answer:
(c) Cross etal

Question 22.
The term Florigen was coined by
a) Maheswari
b) Chailakyan
c) R Gane
d) Richmond Lang
Answer:
b) Chailakyan

Question 23.
Cytokinins inducing cell division was first demonstrated by:
(a) Haberlandt
(b) Charles Darwin
(c) Clarke
(d) Hubert
Answer:
(a) Haberlandt

Question 24.
Which of the following is a bioassay for Cytokinins?
a) Chlorophyll preservation test
b) Dwarf maize Assay
c) Seed germination Assay test
d) Neem cotyledon Assay
Answer:
d) Neem cotyledon Assay

Question 25.
Indicate correct statements.
(i) Genes are intracellular factors for growth.
(ii) Temperature has no role in the growth of plant.
(iii) Oxygen has a vital role in the growth of plants.
(iv) CIN ratio of soil does not affect the growth of plant.
(a) (i) and (iv)
(b) (ii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iii)
Answer:
(c) (i) and (iii)

Question 26.
Avena curvature test as a BioAssay for
a) Auxins
b) GA₃
c) Cytokinin
d) Ethylene
Answer:
a) Auxins

Question 27.
The stress phytohormones (Abscisic acid) was first isolated by:
(a) Linn et al
(b) Addicott et al
(c) Edward et al
(d) Stone and Black
Answer:
(b) Addicott et al

Question 28.
The Gibberellins have been commercially exploited for
a) increasing the size of grapefruits
b) inducing rooting in stem cuttings
c) breaking the dormancy in seeds
d) production of disease-resistant varieties
Answer:
c) breaking the dormancy in seeds

Question 29.
Pick out the correct statement from the following:
(i) Abscisic acid is found abundantly inside the chloroplast of green cells.
(ii) ABA is a powerful growth promotor.
(iii) ABA is formed from the pentose phosphate pathway.
(iv) ABA has anti-auxin and anti-gibberellin properties.
(a) (i) and (iv)
(b) (i) and (ii)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(a) (i) and (iv)

Question 30.
Biennials can be induced to flower in the first season itself by treatment with
a) Auxin
b) Kinetin
c) GA
d) ABA
Answer:
c) GA

Question 31.
Pea and barley are classified under:
(a) short-day plants
(b) short long day plants
(c) long day plants
(d) long short day plants
Answer:
(c) long day plants

Question 32.
Auxin a was isolated from human urine by
a) F.W. went
b) Charles Darwin
c) Kogl and Haugen Smith
d) Denny
Answer:
c) Kogl and Haugen smith

Question 33.
Usually, Xanthiumpensylvanicum will flower under:
(a) long day condition
(b) short long day condition
(c) photo neutral condition
(d) short-day condition
Answer:
(d) short-day condition

Question 34.
The most widely occurring Cytokinin in plants is
a) Indole Acetic Acid (LAA)
b) Indole Butyric Acid (IBA)
c) Pentenyl Adenine (IPA)
d) Naphthalene  Acetic Acid (NAA)
Answer:
c) Pentenyl Adenine (IPA)

Question 35.
Who found out the phytochrome in plants?
(a) Butler et al
(b) Michell et al
(c) Boumick et al
(d) Gamers and Allard
Answer:
(a) Butler et al

Question 36.
Scientists, those who are connected with Ethylene
(I) Denny
(II) R. Gane
(III) Kurosawa
(IV) Cocken
Options:
a) (I) (II) & (III)
b) (II) (III) & (IV)
c) (I) (II) & (IV)
d) (I) (III) & (IV)
Answer:
c) (I)(II)& (IV)

Question 37.
Pick out the wrong statement from the following:
(a) Vernalization increases the cold resistance of plants
(b) It increases the resistance of plants to fungal disease
(c) Vemalizatiqn increase the vegetative period of the plant
(d) It accelerates the plant breeding
Answer:
(c) Vemalizatiqn increase the vegetative period of the plant

Question 38.
Day-neutral plants are
a) Sugarcane & Coleus
b) Bryophyllum& Night Jasmine
c) Wheat, rice & Oats.
d) Potato, Tomato & Cotton
Answer:
d) Potato, Tomato & Cotton

Question 39.
In apple and plum, the method of breaking seed dormancy involves the process of:
(a) impaction
(b) Scarification
(c) exposing to red light
(d) Stratification
Answer:
(d) Stratification

Question 40.
Xanthium (Cocklebur) requires …………….. hours of light to induce flowering,
a) 12
b) 9
c) 15.05
d) 13.05
Answer:
c) 15.05

Question 41.
The hormone that cannot be isolated
a) IAA
b) ABA
c) NAA
d) Florigen
Answer:
d) Florigen

Question 42.
The term Photoperiodism was coined by
a) Went
b) Butler
c) Gamer
d) Skoog
Answer:
c) Garner

Question 43.
ABA acts as antagonistic to
a) Ethylene
b) Cytokinin
c) Gibberellic acid
d) IAA
Answer:
c) Gibberellic acid

Question 44.
If a short-day plant, flowering is induced by
a) Long nights
b) Photo periods less than 12 hrs
c) Photoperiods shorter than critical value and uninterrupted long night
d) Short photoperiods and interrupted long nights
Answer:
c) Photoperiods shorter than critical value and uninterrupted long night.

Question 45.
Phytochrome is
a) Reddish phytohormone
b) Bluish biliprotein pigment
c) Photoreceptor of apical bud
d) Unstable pigment molecule
Answer:
b) Bluish biliprotein pigment

Question 46.
The growth & ripening is induced by Ethylene in
a) Tropical fruits
b) Temperate fruits
c) Climacteric fruits
d) Nonclimacteric fruits
Answer:
c) Climacteric fruits

Question 47.
The bioassay of ABA was done with
a) Rice
b) Wheat
c) Maize
d) Barley
Answer:
a) Rice

Question 48.
Four types of senescence were recognized by
a) Leopold
b) Gamer
c) Addicott
d) Cocken et al
Answer:
a) Leopold

Question 49.
The final stage of senescence is
a) PCD
b) Scarification
c) Yellowing
d) Abscission
Answer:
d) Abscission

Question 50.
Match & Find out the Correct Answer

Column I	Column II
1.  Yabuta&Sumiki	a) Identified Ethylene
2.  Lethan & Miller	b) Isolated Auxin from Human urine
3.  Cockenetal	c) Isolated and identified Zeatin
4.  Kogi & Haugen Smith	d) Isolated Gibberellin in Crystal form

Answer:
b) D C A B

Question 51.
Match the following and Find the Correct Answer

I. Auxin	a) Bolting
II. ABA	b) Induces Respiration
III. Gibberellin	c) Cell division
IV. Ethylene	d) Weedicide
V. Cytokinin	e) Closure of stomata

Answer:
b) D E A B C

II. Assertion (A) & Reason (R)

Question 52.
a. Both A & R are true and ‘R’ is the correct explanation of A
b. Both A & R are true but ‘R’ is not the correct explanation of A
c. A is true but R is False
d. Both A and ‘R’ are False
Assertion (A): The shoot Apical meristems are the only source of Auxin synthesis
Reason (R): Dormancy of lateral buds over Apical buds is due to Auxin
Answer:
C. A is true but R is False

Question 53.
Assertion (A): Hormones are also called Growth regulator
Reason (R): Hormones promote or inhibit plant growth
Answer:
A. Both Assertion (A) and Reason (R) are true and Reason is the correct explanation of Assertion.

Question 54.
Assertion (A): In many land Mammoth flowering occurred at different times at different latitude
Reason (R): Many land Mammoth is a tobacco variety
Answer:
b. Both Assertion (A) and, Reason (R) are true and Reason is not the correct explanation of Assertion.

III. 2 Mark Questions

Question 1.
Define closed form of growth in plants.
Answer:
Leaves, flowers, and fruits are limited in growth or of determinate or closed-form growth.

Question 2.
Compare between Absolute and Relative growth rates
Answer:

Absolute growth	Relative growth
An increase in the total growth of two organs measured and compared per unit time is called Absolute growth rate	The growth of the given system per unit time expressed per unit initial parameter is called relative growth rate

Question 3.
Name the phases of growth in ‘S’ shaped growth curve.
Answer:

• Lag phase
• Log phase
• Decelerating phase
• Maturation phase

Question 4.
Mention the phase of growth in plants
Answer:
I. Formative phase
II. Elongation phase
III. Maturation phase

Question 5.
Distinguish between absolute growth rate and relative growth, rate.
Answer:
Absolute growth rate:
An increase in total growth of two organs measured and compared per unit time is called absolute growth rate.

Relative growth rate:
The growth of the given system per unit time expressed per unit initial parameter is called relative growth rate.

Question 6.
What is the Grand period of growth
Answer:
The total period from initial to the final stage of growth is called Grand period of growth.
When plotted against time the growth curve is ‘S’ shaped, (sigma curve) it is also known as Grand Period curie consists of 4 phases

1. Lag,
2. Log,
3. Decelerating,
4. Maturation.

Question 7.
What is meant by the dedifferentiation of plant cells?
Answer:
Differentiated cells, after multiplication again lose the ability to divide and mature to perform specific functions. This is called redifferentiation, eg: Secondary xylem and Secondary phloem.

Question 8.
Define Phytohormone.
Answer:
The chemical substances synthesized by plants and thus naturally occuring are known as Phytohormones. Eg. Auxin, Gibberellins.
Recently 2 groups – Brassinosteroids, Polyamines were also known to behave like hormones.

Question 9.
Mention any two synthetic auxins.
Answer:

• 2, 4 – Dichloro Phenoxy Acetic Acid (2, 4 – D)
• 2, 4, 5 – Trichloro Phenoxy Acetic Acid (2, 4, 5 – T)

Question 10.
State 3 characteristic features of phytohormones.
Answer:

1. They are produced in root tips and stem tips and leaves (do not have specialized cells or organs for secretion)
2. The transfer of hormones takes place through the conducting system (xylem and phloem)
3. They are required in trace quantities
4. They either promote, inhibit or modify growth.

Question 11.
Name the natural auxins present in plants.
Answer:

• Indole Acetic Acid (IAA)
• Indole Propionic Acid (IPA)
• Indole Butyric Acid (IBA)
• Phenyl Acetic Acid (PAA)

Question 12.
Give the historial significance of Agent Orange
Answer:

• Mixture of two phenoxy herbicides – 2.4. D and 2.4.5 T together known as Agent orange.
• This Agent orange, was used by USA in Vietnam war as chemical warfare weapon to defoliate forests in Vietnam.

Question 13.
Does the trimming of plants in gardens have any scientific explanation?
Answer:

• Yes, trimming of plants removes apical buds and hence apical dominance is prevented the lateral buds sprout and give a beautiful bushy appearance and aesthetic value.
• Also in tea estates, this trimming develops more lateral branches and more tea leaves thus it has commercial significance.

Question 14.
Where do you find cytokinin hormone in plants?
Answer:
The distribution of cytokinin in plants is not as wide as those of auxin and gibberellins but found mostly in roots. Cytokinins appear to be translocated through xylem.

Question 15.
What is bolting?
Answer:

• When treated with Gibberellins the rose the plants (genetic dwarf) exhibit excessive internodal growth.
• This sudden elongation of a stem followed by flowering is called bolting.

Question 16.
What is Richmond Lang effect?
Answer:
Application of Cytokinin delays the process of aging by nuitrient mobilization.

Question 17.
Why do you call Abscisic acid (ABA) as stress hormone?
Answer:
It inhibits the shoot growth and promotes growth of root system. This character protect the plants from water stress. Hence, ABA is called as stress hormone.

Question 18.
Define photoperiodism & Critical day length.
Answer:

• The physiological change on flowering due to relative length of light and darkness (Photoperiod) is called
  Photoperiodism.
• The photoperiod required to induce flowering is called critical day length Eg.
  • Mary land mammoth (tobacco variety) requires 12 hours of light.
  • Cocklebur required 15.05 hours of light.

Question 19.
Write down the importance of photoperiodism in plants.
Answer:

• The knowledge of photoperiodism plays an important role in hybridisation experiments.
• Photoperiodism is an excellent example of physiological pre-conditioning that is using an external factor to induce physiological changes in the plant.

Question 20.
What is the importance of photoperiodism?
Answer:

• The knowledge of photoperiodism an important role in hybridization experiments.
• It is an excellent example of physiological preconditioning that is using an external factor to induce physiological changes in the plant.

Question 21.
What is meant by Epigeal germination?
Answer:
During epigeal germination, cotyledons are pushed out of the soil. This happens due to the elongation of the hypocotyl.
Eg: Castor and Bean.

Question 22.
Define Vernalization.
Answer:

• It is a process by which many annuals and biennials are induced to flower when subjected to low-temperature exposure.
• T.d. Lysenko first used the term.

Question 23.
Define the term phytogerontology.
Answer:
The branch of botany which deals with ageing, abscission and senescence is called Phytogerontology.

Question 24.
Distinguish between Epigeal and Hypogeal germination.
Answer:

Epigeal	Hypogeal
Cotyledons pussed out of the soil	Cotyledons remain below the soil due to rapid elongation of epicotyls.
Happens due to the elongation of the hypocotyl Eg. Castor & Bean	Eg. Maize

Question 25.
Define seed dormancy and what are its types.
Answer:
The condition of a seed when it fails to germinate even in suitable environmental condition is called seed dormancy.
There are two types
(I) Innate dormancy (II) Imposed dormancy.

Question 26.
What is Scarification?
Answer:

• By mechanical and chemical treatments like cutting or chipping of hard tough sed coat and use of organic solvents to remove waxy or fatty compounds are called scarification.
• It is a method of breaking dormancy of the seeds.

Question 27.
Distinguish between Re differentiation and Devernalization.
Answer:

Redifferentiation	Devernalization
Differentiated cells after multiplication again lose the ability to divide and mature to perform specific functions, is called Re differentiation. Eg. Sec.Xylem & Sec.	Phloem The reversal of the effect of vernalization is called Devemalization.

Question 28.
Define Senescence.
Answer:

• Ageing or getting old is called senescence.
• It refers to all collective, progressive and deteriorative processes which ultimately lead to complete loss of organization and function (Eg. leaves turn yellow and fall off from plant).

Question 29.
What is impaction in seed dormancy.
Answer:

• In some seeds water and oxygen are unable to penetrate micropyle due to blockage by cork cells.
• These seeds are shaken vigorously to remove the plug
• The process of removing the plug or block is called impactation.

Question 30.
What is called stratification in seed dormancy?
Answer:

• The break dormancy, some plant seeds have to be exposed to well aerated, moist conditions under low temperature (0°c to 10°c) for weeks to months.
• This kind of seed dormancy breaking treatment is known as stratification.
• The stratified soil layers should be given a low-temperature treatment for a certain period so as to induce germination.
• Eg. the Seeds of Rosaceae plants Apple, Plum, Peach, etc.

Question 31.
What are the 4 types of Senescence?
Answer:
Leopold (1961) explained 4 types they are

1. Overall senescence
2. Top senescence
3. Deciduous senescence
4. Progressive senescence.

Question 32.
What is the Abscission layer or Abscission Zone?
Answer:
Abscission is marked internally at the place of petiole by a distance zone of few layers of thin-walled cells arranged transversely. This zone is called Abscission Zone, which leads to Abscission of the leaf.

Question 33.
The photoperiodic response will not be possible in a defoliated plant. Give scientific reasons.
Answer:
Yes, a defoliated plant will not respond to photoperiodic change because the hormonal substance responsible for flowering is present in the leaves of the plant.

Question 34.
What is gas chromatography?
Answer:

• It is a bioassay technique by which Ethylene can be measured.
• It helps in the detection of the exact amount of ethylene from different plant tissues like lemon and orange.

Question 35.
Give the occurrence and precursors of Gibberellins and Cytokinins.
Answer:

Character	Gibberellins	Cytokinin
Occurrence	Produced by plant parts like an embryo, roots, and young leaves near the tip. Immature seeds are rich in Gibberellins.	Formed in root apex shoot apex like Auxin. Also formed in buds & young fruits.
Precursor	Formed by 5C precursor, Iso prenoidunit called Iso Pentenyl Pyrophosphate (IPP) through a number of intermediates primary precursor – Acetate.	Derived from purine-Adenine.

IV. 3 Mark Questions

Question 1.
Explain Arithmetic growth rate and Geometric growth rate by diagrams.
Answer:

Question 2.
Explain stages in growth by drawing the sigmoid curve.
Answer:

• The total period from the initial to the final stage of growth is called the Grand period of growth.
• The graph that is drawn by taking time and rate of growth is ‘S’ shaped. It is known as a sigmoid curve.

It has 4 stages:

1. Lag phase
2. Log phase
3. Decelerating phase
4. Maturation phase

Question 3.
Mention the internal factors, that affect the growth of plants.
Answer:

• Genes are intracellular factors for growth.
• Phytohormones are intracellular factors for growth, eg: auxin, gibberellin, cytokinin.
• C/N ratio.

Question 4.
Mention the Agricultural role of Auxin.
Answer:

• Eradicate weeds: Eg. 2.4 D and 2.4.5.7
• Formation of seedless fruits: (Parthenocarpic fruits) Eg. Synthetic Auxin.
• Break dormancy.
• Induction of flowering: In pineapple NAA induce flowering
• Increase the number of female flowers: Eg. Cucurbita.

Question 5.
List out the agricultural applications of auxins.
Answer:

• It is used to eradicate weeds, eg: 2,4 – D and 2,4,5 – T.
• Synthetic auxins are used in the formation of seedless fruits (Parthenocarpic fruit).
• It is used to break the dormancy in seeds.
• Induce flowering in Pineapple by NAA & 2,4 – D.
• Increase the number of female flowers and fruits in cucurbits.

Question 6.
What are the Precursors of Gibberellins?
Answer:

• Gibberellins are chemically related to terpenoids (natural rubber, Carotenoids, and steroids) formed by 5-C precursors and an Isoprenoid unit called Iso Pentenyl Pyrophosphate (IPP) through a number of intermediates.
• The primary precursors are Acetate.

Question 7.
What are the uses of ethylene in agriculture?
Answer:

• Ethylene normally reduces flowering in plants except in Pineapple and Mango.
• It increases the number of female flowers and decreases the number of male flowers.
• Ethylene spray in the cucumber crop produces female flowers and increases the yield.

Question 8.
Explain the mechanism of Vernalization by Hypothesis of hormonal involvement.
Answer:
I. Vernalization: According to Purvis

 

 

 

 

II. Devernalization

 

 

Question 9.
What are the practical applications of Vernalization?
Answer:

• It shortens the vegetative period and induces the plant to flower earlier.
• It increases the cold resistance of the plants.
• It increases the resistance of plants to fungal disease.
• Plant breeding can be accelerated.

Question 10.
What is meant by the viability of seeds?
Answer:

• Viable means the living condition of the seed
• The shelf life of the seed after which it cannot germinate is known as the viable period
• It varies from plant to plant

Name of the plant	Viability
1. Oxalis seeds	Few days
2.  Lotus seeds	More than 1000 years
3.  Judean Dale palm (Methuselah)	More than 2000 years

Question 11.
Differentiate between climacteric and Non-climacteric fruits.
Answer:

Climacteric fruits	Non-Climacteric fruits
1.There is a sharp rise in respiration rate near the end of the development of fruit. 2. The ripening on demand can be induced in these fruits by exposing them to normal air conditioning. 3. Epthan secrete Ethylene continuously about 1 ppm of ethylene Eg. Lemon, Apples, Banana, Mango	These fruits cannot be ripened by exposure to ethylene so-known as non-climacteric fruits. Eg. Grapes, Watermelon orange.

Question 12.
Differentiate between scarification & Stratification in breaking seed dormacy
Answer:

Scarification	Stratification
Mechanical and chemical treatment either by cutting, chipping or use of organic solvents to remove waxy or fatty compounds is called scarification.	Rosaceous plants (Apple, Plum Peach, and Cherry) will not germinate until they have been exposed to well derated, moist conditions under low temperature (1°c to 10°c) for weeks to months and this treatment is known as stratification.

Question 13.
Mention the factors causing dormancy of seeds.
Answer:

• Hard, tough seed coat causes barrier effect as impermeability of water, gas and restriction of the expansion of embryo prevents seed germination.
• Many species of seeds produce imperfectly developed embryos called rudimentary embryos which promotes dormancy.
• Lack of specific light requirement leads to seed dormancy.
• A range of temperatures either higher or lower cause dormancy.
• The presence of inhibitors like phenolic compounds which inhibits seed germination cause dormancy.

Question 14.
What are the factors that affect senescence?
Answer:

Name of the factor	Effect of senescence
ABA & Ethylene	Accelerates
Auxin & Cytokinin Nitrogen deficiency	reduces increases
Nitrogen supply	retards
High temperature in vernalized seeds	Accelerates
Low temperature	Retards
Water stress	Accumulation of ABA leading to senescence

Question 15.
What are the morphological and Anatomical changes due to Abscission?
Answer:

• Abscission Zone: formed at the base of petiole
• Greenish grey in colour by rows of 2 to 15 cells thick primary wall and middle lamella
• The dissolution of by pectinase & Cellulase
• Formation tyloses – that block conduction of vessels
• Degradation of chlorophyll – Colour of leaves changes and leaves fall off.
• After Abscission – Suberization of outer layer of cells by the development of periderm.

Question 16.
Write down the significance of Abscission.
Answer:
1. Abscission separates dead parts of the plant like old leaves and ripe fruits.
2. Helps in dispersal of fruits and continuing the life cycle.
3. Abscission of leaves (in deciduous plants) helps in water conservation during summer.
4. Helps in vegetative propagation (Shedding of gemmae or plantlets) Eg. Bryophyta.

V. 5 Mark Questions

Question 1.
Describe an experiment to measure the growth of a plant or By lever Auxanometer measure the rate of growth in stem tip.
Answer:
Experiment:
1. Arc auxanometer:
The increase in the length of the stem tip can easily by measured by an arc auxanometer. If consists of a small pulley to the axis of which is attached a long pointer sliding over a graduated arc. A thread one end of which is tied to the stem tip and another and to a weight passes over the pulley tightly. As soon as the stem tip increases in length, the pulley moves and the pointer slide over the graduated arc (Refer Figure) The reading is taken. The acutal increase in the lengthwm stem is then calculated by knowing the length of the pointer and the radius of the pulley. If the radius of the pulley is 4 inches and the length of pointer 20 inches the actual growth is measured as follows:

Actual growth in length\(=\frac{\text { Distance travelled by the pointer radius of the }}{\text { Length of the pointer }}\)
For example, actual growth in length \(=\frac{10 \times 4 \text { inches }}{20 \text { inches }}\) = 2 inches

Question 2.
Explain the physiological effect of Auxin? Add a note on its agricultural applications.
Answer:
Cell elongation:
Promotes cell elongation in stem & Coleoptile

Root growth:
At extremely low concentration – promote root growth, at high concentrations it inhibits elongation of roots, but induce more lateral roots

Apical dominance:
Suppression of growth of lateral buds – by apical bud is known as Apical dominance
Prevents Abscission

Secondary growth:
Promotion of cell division in the cambium, responsible for secondary growth this property is exploited in tissue culture. (Callus foundation)

Respiration Stimulates respiration Induces Vascular differentiation.

Question 3.
Give the Agricultural application of Auxin.
Answer:
Weedicide
2.4.D & 2.4.5. T – Weedicides to remove weeds

Induce parthenocarpy
Synthetic auxins used to induce parthenocarpy (formation of seedless fruits).

Break dormancy
Used to break seed dormancy.

Induce flowering in Pineapple Eg. NAA & 2.4.D

Induce female flowers (numbers)
Eg. Cucumber.

Question 4.
Explain physiological effects of Gibberellins
Answer:

• Induction of cell division & cell elongation – Extraordinary stem elongation.
• Reversal of dwarfism & Bolting – Rosette (genetic dwarfism) plants when treated with Gibberellins exhibit excessive enter nodal growth – This sudden elongation of a stem followed by flowering is called Bolting.
• Breaks dormancy – in Potato tubers.
• Biennials flower in the 1 st year – Instead of cold exposure, if biennials treated with Gibberellins flower in the 1st year itself.

Question 5.
Write an essay on the role of ethylene on plant physiology and agriculture.
Answer:
Almost all plant tissues produce ethylene gas in minute quantities.
1. Discovery:
In 1924, Denny found that ethylene stimulates the ripening of lemons. In 1934, R. Gane found that ripe bananas contain abundant ethylene. In 1935, Cocken et al., identified ethylene as a natural plant hormone.

2. Occurrence:
Maximum synthesis occurs during climacteric ripening of fruits and tissues undergoing senescence. It is formed in almost all plant parts like roots, leaves, flowers, fruits and seeds.

3. Transport in plants:
Ethylene can easily diffuse inside the plant through intercellular spaces.

4. Precursor:
It is a derivative of amino acid methionine, linolenic acid and fumaric acid.

5. Bioassay (Gas Chromatography):
Ethylene can be measured by gas chromatography. This technique helps in the detection of exact amount of ethylene from different plant tissues like lemon and orange.

6. Physiological Effects:

• Ethylene stimulates respiration and ripening in fruits.
• It stimulates radial growth in sterft and roof and inhibits linear growth.
• It breaks the dormancy of buds, seeds and storage organs.
• It stimulates the formation of an abscission zone in leaves, flowers and fruits. This makes the leaves to shed prematurely.
• Inhibition of stem elongation (shortening the internode).
• In low concentration, ethylene helps in root initiation.
• Growth of lateral roots and root hairs. This increases the absorption surface of the plant roots.
• The growth of fruits is stimulated by ethylene in some plants. It is more marked in climacteric fruits.
• Ethylene causes epinasty.

7. Agricultural role:

• Ethylene normally reduces flowering in plants except in Pineapple and Mango.
• It increases the number of female flowers and decreases the number of male flowers.
• Ethylene spray in cucumber crops produces female flowers and increases the yield.

Question 6.
Explain the physiological Effects of Cytokinins.
Answer:

• With IAA – Promotes cell division With IAA & GA – Induces cell enlargement
• Breaks dormancy of light-sensitive seeds (tobacco) induces seed germination.
• Promotes growth of lateral
• buds even in the presence of apical bud.
• Delays the process of aging by nutrient mobilization known as Richmond Lang effect.
• Induces rate of protein synthesis.
• Induces the formation of interfascicular cambium Overcomes apical dominance
• Induces the formation of new leaves chloroplast and lateral shoots.
• Induces Accumulation of solutes.

Question 7.
Write down the physiological effects of Ethylene
Answer:

• Stimulates respiration and thereby ripening of fruits
• Stimulates radial growth in stem and root and inhibits linear growth.
• breaks dormancy of
  1. buds
  2. Seeds
  3. Storage Organs
• Stimulates abscission 2 one formation in
  1. leaves
  2. flowers
  3. fruits (so leaves shed prematurely)
• Prevents stem elongation by preventing internodal growth
• Root growth in low concentration
• Stimulates growth of lateral roots and root hairs and increase the absorptive surface
• Ripening of fruits – Increases ripening in climacteric fruits (Mango, banana) etc.
• It causes epinasty

Question 8.
Describe the methods of breaking the dormancy of seeds in plants.
Answer:
The dormancy of seeds can be broken by different methods. These are:
1. Scarification:
Mechanical and chemical treatments like cutting or chipping of hard tough seed coat and use of organic solvents to remove waxy or fatty compounds are called Scarification.

2. impaction:
in some seeds, water and oxygen are unable to penetrate micropyle due to blockage by cork cells. These seeds are shaken vigorously to remove the plug which is called Impaction.

3. Stratification:
Seeds of rosaceous plants (Apple, Plum, Peach, and Cherry) will not germinate until they have been exposed to well aerated, moist conditions under low temperature (0°C to 10°C) for weeks to months. Such treatment is called Stratification.

4. Alternating temperatures: Germination of some seeds is strongly promoted by alternating daily temperatures. An alternation of low and high temperature improves the germination of seeds.

5. Light:
The dormancy of photoelastic seeds can be broken by exposing them to red light.

Question 9.
Define photoperiodism – Classify plants based on photoperiodism
Answer:
a. The physiological change on flowering due to the relative length of light and darkness is called photoperiodism.

• Gamer and Allard (1920) coined the term
• They studied photoperiodism in Biloxi variety of soybean (Glycine max) and Many land mammoth varieties of tobacco.

b. Depending on photoperiodic responses plants are classified into several types.
1. L.D. Plants (Long Day) The photoperiod required to induce flowering is called critical day length depending on critical day length if it is long (more than 12 hours) and with short nights. Eg. Pea Barley and Oats
Short LD Plants: These are Long day plants but need short day length during the early period of growth for flowering Eg. Wheat, Rye

2. SD Plants: Plants requiring short critical day length for flowering or a long night.
Eg. Tobacco, Cocklebur, Soya, Rice, and Chrysanthemum.
Long SD Plants: Actually SD plants but need long days during the early period of growth for flowering Eg. Some SPS of Bryophyllum & Night Jasmine.

3. Intermediate day plants:
These require a photoperiod between a long day and a short day for flowering Eg. Sugarcane and coleus.

4. Day Neutral plants:
There are a number of plants which can flower in all possible photoperiods, known as photo neutral or hiterterminate plants. Eg. Potato, Rhododendron, Tomato & Cotton.

Question 10.
Describe the role of phytochrome in inducing Flowering
Answer:
Definition:
It is a bluish biliprotein responsible for the perception of light in the photophysiological process, existing in two different forms is mainly involved in flower induction, (i.e) Pr and PFr.

• Butler et al(1959) named the pigment.
• It exists in two interconvertible forms

Pr	PFr
1. red light absorbing form 2. Absorbs red lgiht of wavelength 660 nm 3. Biologically inactive form & stable found in the diffused state in cytoplasm 4. Promotes flowering in SD plants and inhibits flowering LD plants.	1. Far-red light absorbing form 2. Absorbs far-red light of wavelength 730 nm 3. Biologically active and it is unstable Associated with a hydrophobic area of the membrane system 4. Promotes flowering in LD plants and inhibit flowering in SD plants.

Mechanism:

Other functions:
Play a role in seed germination and changes in membrane conformation.

Question 11.
Write an Essay on Vernalization
Answer:
Definition:

• Many biennials and perennials are induced to flower by low-temperature exposure (O°c to 5°c) This process is called Vernalization.
• T.D. Lysenko – Coined the term.

Mechanism of Vernalization:
2 theories explain the mechanism of vernalization.

1. Hypothesis of Phasic development (T.D. Lysenko),
The development of the annual plant has 2 phases.

1. Thermostate-Vegetatine stage requiring low temperature and suitable moisture.
2. Photo stage -high temperature needs to synthesize florigen.

2. Hypothesis of hormonal involvement (Purvis 1961)

Vernalization has several steps

The technique of Vernalization:

• Seeds soaked in water
• Allowed to germinate at 10°C to 12°C
• Transferred to low temperature for few days to 30 days (3°C to 5°C).
• Germinated seeds after the low temp, treatment are allowed to dry & then sown.
• Quickened flowering than untreated (control seedling)

Devernalization:
The reversal of the effect of vernalization is called Devemalization.

• Practical Applications:
• Vernalization shortens the vegetative period and induces the plant to flower earlier
• It increases cold resistance
• It increases fungal resistance
• It accelerates Plant Breeding.

Question 12.
Define Senescence and give its types
Answer:
Definition:
Getting old or Ageing is call d senescence in plants.
It refers to all collective, progressive, and deteriorative processes which ultimately lead to complete loss of organization and function.
Types – 4 types (Leopold -1961)

1. Overall senescence: When the entire plant gets affected and dies – Eg. Annuals – Wheat & Soybeans, Perennials – Agave & Bamboo
2. Top senescence: Occur in aerial parts only Eg. Parrennials – Banana and Gladiolus
3. Deciduous senescence: Occur only in leaves Eg. Decidual plants – Elm and Maple
4. Progressive Senescence: Occur in Annuals occur in old leaves first followed by new leaves than stem and finally root system.

Question 13.
Explain the physiology of senescence and Factors affecting senescence Physiology of Senescence :
Answer:

• Change in the structure of cells
• Vacuoles act like lysosome-secrete hydrolytic enzymes.
• Reducation in photosynthetic rate (due to loss of chlorophyll & accumulation of anthocyanin)
• The decrease in Starch content,  Protein content
• Decrease in …………. r RNA level due to increased activity of enzyme RNA ase
• Degeneration of DNA – by increased activity of enzyme DNA ase

Factors affecting senescence :

Name of the factor	Effect of senescence
ABA & Ethylene	Accelerates
Auxin & Cytokinin Nitrogen deficiency	reduces increases
Nitrogen supply	retards
High temperature in vernalized seeds	Accelerates
Low temperature	Retards
Water stress	Accumulation of ABA leading to senescence

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 11 Transport in Plants Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants

11th Bio Botany Guide Transport in Plants Text Book Back Questions and Answers

Part – I

Question 1.
In a fully turgid cell:
(a) DPD = 10 atm; OP = 5 atm; TP = 10 atm
(b) DPD = 0 atm; OP =10 atm; TP = 10 atm
(c) DPD = 0 atm; OP = 5 atm; TP = 10 atm
(d) DPD = 20 atm; OP = 20 atm; TP = 10 atm
Answer:
(b) DPD = 0 atm; OP =10 atm; TP = 10 atm

Question 2.
Which among the following is correct?
i) apoplast is fastest and operate in nonliving part
ii) Transmembrane route includes vacuole
in) Symplast interconnect the nearby cell through plasma desmata
iv) Symplast and the transmembrane route is in the living part of the cell
a) i and ii
b) ii and iii
c) iii and iv
d) i, ii, iii, iv
Answer:
d) i, ii, iii, iv

Question 3.
What type of transpiration is possible in the xerophyte Opuntia?
(a) Stomatal
(b) Lenticular
(c) Cuticular
(d) All the above
Answer:
(b) Lenticular

Question 4.
Stomata of a plant open due to
a) Influx of K⁺
b) Effrilx of K⁺
c) Influx of Cl^–
d) Influx of OH^–
Answer:
a) Influx of K⁺

Question 5.
Munch hypothesis is based on:
(a) translocation of food due to TP gradient and imbibition force
(b) ranslocation of food due to TP
(c) translocation of food due to imbibition force
(d) None of the above
Answer:
(b) ranslocation of food due to TP

Question 6.
If the concentration of salt in the soil is too high and the plants may wilt even if the field is thoroughy irrigated. Explain
Answer:
High salt concentration results in high be osmotic potential of the soil solution, so the plant has to use more energy to absorb water. Under extreme salinity conditions, plants may be unable to absorb water and will wilt even if the surrounding soil is thoroughly irrigated. This is also referred to as the osmotic or water deficit effect of salinity.

Question 7.
How phosphorylase enzyme open the stomata in starch sugar interconversion theory?
Answer:

• The discovery of enzyme phosphorylase in guard cells by Hanes (1940) greatly supports the starch-sugar interconversion theory.
• The enzyme phosphorylase hydrolyses starch into sugar and high PH followed and the opening takes place during the night.

Day:

1.
2. Photosynthesis occur
3. pH – increased
4. Movement of water from
5. subsidiary cells to guard cells
6. Guard cells become turgid
7. Opening of stomata

Night:

1.

2. No Photosynthesis
3. pH – lowered
4. Movement of water from guard cells
5. Guard cells become flaccid
6. Closure of stomata

Question 8.
List out the non-photosynthetic parts of a plant that need a supply of sucrose?
Answer:
The non-photosynthetic parts of a plant that need a supply of sucrose:

1. Roots
2. Tubers
3. Developing fruits and
4. Immature leaves.

Question 9.
What are the parameters which control water potential?
Answer:
1. Slatyer and Taylor (1960) introduced the concept of water potential.
Definition – water potential is the potential energy of water in a system – compared to pure water when temperature and pressure are kept constant.

2. It is also a measure of how freely water molecules can move in a particular environment or system. Water potential is denoted by the Greek symbol  Ψ (psi) and measured in Pascal (Pa). At standard temperature, the water potential of pure water is zero

3. Addition of solute to pure water decreases the kinetic energy thereby decreasing the water potential, from zero to negative.

4. So, Comparatively a solution always has low water potential than pure water. In a group of cells with different water potential, a water potential gradient is generated.

5. Water will move from higher water potential to lower water potential.
When potential ( Ψ) can be determined by. Solute concentration or Solute potential ( Ψ_s) Pressure potential ( Ψ_p)
By correlating two factors, water potential is written as (Ψw=Ψ_s)+Ψ_p

a) Solute potential (Ψ_s) or Osmotic potential

• Denotes the effect of dissolved solute on water potential.
• In pure water, the addition of solute reduces its free energy and lowers the water potential value from zero to negative.
• Thus the value of solute potential is always negative. In a solution at standard atmospheric pressure, water potential is always equal to solute potential (Ψw = Ψ_s ).

b) Pressure Potential (Ψ_p)

• Pressure potential is a mechanical force working against the effect of solute potential.
• Increased pressure potential will increase water potential and water enters cells and cells become turgid.
• This positive hydrostatic pressure within the cell is called Turgor, pressure likewise, withdrawal of water from the cell decreases the water potential and the cell becomes flaccid.

Question 10.
An artificial cell made of selectively permeable membrane immersed in a beaker (in the figure) Read the values and answer the following questions?

a) Draw an arrow to indicate the direction of water movement
b) Is the solution outside the cell isotonic, hypotonic or hypertonic?
c) Is the cell isotonic, hypotonic, or hypertonic?
d) Will the cell become more flaccid, more turgid or stay in original size?
e) With reference to artifical cell state, the process is endomosis or exomosis? Give reasons
Answer:
a)

(b) Outside solution in hypotonic.
(c) The cell is hypertonic.
(d) The cell becomes more turgid.
(e) The process is endo – osmosis because the solvent (water) moves inside the cell.

Reason: Endomosis is defined as the osmotic entry of solvent into a cell when it is placed in pure water/Hypotonic solution. The solution in the beaker outside the cell is pure water. ( Ψw = 0), and water enters into the artificial cell which is placed inside the beaker of pure water, (i.e) from hypotonic to hypertonic solution.

Part II 

11th Bio Botany Guide Transport in Plants Additional Important Questions and Answers

I – Choose The Correct Answers

Question 1.
In plants, cell to cell transport is aided by:
(a) diffusion alone
(b) osmosis alone
(c) imbibition alone
(d) all the three above
Answer:
(d) all the three above

Question 2.
The smell from a lightened incense stick or mosquito coil or open perfume bottle in a closed room is due to
a) Osmosis
b) Facilitated diffusion
c) Simple diffusion
d) imbibition
Answer:
c. Simple diffusion

Question 3.
Which of the following statements are correct?
(i) Cell membranes allow water and non-polar molecules to permeate by simple diffusion.
(ii) Polar molecules like amino acids can also diffuse through the membrane.
(iii) Smaller molecules diffuse faster than larger molecules.
(iv) Larger molecules diffuse faster than smaller molecules.

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

Question 4.
Solute potential is also known as
a) Water potential
b) Pressure potential
c) Osmotic potential
d) Maic potential
Answer:
c. Osmotic potential

Question 5.
The swelling of dry seeds is due to a phenomenon called:
(a) osmosis
(b) transpiration
(c) imbibition
(d) none of the above
Answer:
(c) imbibition

Question 6.
Cell A has an osmotic potential of -20 bars and a pressure potential of +6 bars. What will be its water potential?
a) -14 bars
b) +14 bars
c) -20 bars
d) +20 bars
Answer:
a. -14 bars

Question 7.
The OP and TP of two pairs of cells A – B, and X-Y are under
a) Cell A: OP=-I0atm, TP=4atm
b) Cell B : OP = l0atm, TP = 6atm
c) Cell X: Op =-l0atm, TP = 4atm
d) CeIlY: OP = -Katm, TP = 4atm
The net movement of water shall be from
a) A toB and X to Y
b) A to B and Y toX
c) B to A and X to Y
d) B to A and Y to X
Answer:
d. B to A and Y to X

Question 8.
Water potential is influenced by which of the two factors among the given four
I) Concentration
II) Pressure
III) Temperature
IV) gravity
a) I & II
b) II & III
c) III & IV
d) I & IV
Answer:
a) I & II

Question 9.
………………………. is equal to TP and is positive except plasmolysed cell and in xylem vessel where it is negative
a) Water potential
b) Pressure potential
c) Solute potential
d) Hydrostatic potential
Answer:
b. Pressure potential

Question 10.
Diffusion Pressure Deficit (DPD) was termed by Meyer in:
(a) 1928
(b) 1828
(c) 1936
(d) 1938
Answer:
(d) 1938

Question 11.
Imbibants present in plants are generally
a) Hydrothermic
b) Hydrostatic
c) Hydrophilic
d) Hydrophobic
Answer:
c. Hydrophilic

Question 12.
Kramer (1949) recognised two distinct mechanisms, which independently operate in the absorption of water in plants are:
(a) osmosis and diffusion
(b) imbibition and diffusion
(c) diffusion and absorption
(d) active absorption and passive absorption
Answer:
(d) active absorption and passive absorption

Question 13.
Root pressure is totally apsent in Gymnosperms because
a) Trachea absent
b) Tracheids absent
c) Trees are tall
d) Trees are comparatively short
Answer:
a. Trachea absent

Question 14.
When respiratory inhibitors like KCN, chloroform are applied:
(a) there is a decrease in the rate of respiration and an increase in the rate of absorption of water.
(b) there is an increase in the rate of respiration and a decrease in the rate of absorption of water.
(c) there is a decrease in the rate of respiration and also a decrease in the rate of absorption of water.
(d) there is an increase in the rate of respiration and also in the rate of absorption of water.
Answer:
(c) there is a decrease in the rate of respiration and also a decrease in the rate of absorption of water.

Question 15.
Find the DPD in a flaccid cell if its OP is 10
a) 20
b) 30
c) 10
d) 40
Answer:
c.10

Question 16.
Pulsation theory was proposed by:
(a) Strasburger
(b) Godsey
(c) J.C. Bose
(d) C.V. Raman
Answer:
(c) J.C. Bose

Question 17.
When a cell is kept in 0.5m solution of sucrose it’s volume does not alter. If the same cell is placed in 0.5M solution of sodium chloride, the volume of the cell
a) Increase
b) Decrease
c) cell will be pIasrnoysed
d) Will does not show any change
Answer:
d. Will does not show any change

Question 18.
Indicate the correct statements:
(i) Root pressure is absent in gymnosperms.
(ii) Root pressure is totally absent in angiosperms.
(iii) There is a relationship between the ascent of sap and root pressure.
(iv) There is no relationship between the ascent of sap and root pressure.

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

Match The Following & Find Out The Correct Order

Question 19.
I) Water potential – A) Turgor pressure
II) Solute potential – B) Osmotic potential + Pressure potential
III) Matric potential – C) Osmotic potential
IV) Pressure potential – D) Imbibition pressure

Answer:
b) B C D A

Question 20.
I) Leaves – A) Antitransport
II) Seed – B) Transpiration
III) Roots – C) Negative osmotic potential
IV) Aspirin – D) Imbibition
V) Plasmolyced cell – E. Absorption

Answer:
a) C B D E A

Question 21.
I) Transport of substance from a region of lower concentration to a region of higher concentration is with the expenditure of energy – A. Antiport
II) The movement of two types of molecules across the membrane in opposite direction – B. Symport The movement of a molecule across III) a membrane independent of other molecules – C. Active port
IV) The movement of two types of molecules across the membrane in the same direction – D. Uniport

Answer:
c) C D A B

Question 22.
I) Passive transport – A) Uphill transport
II) Active transport – B) Short distance transport
HI) Cell to cell transport – C) Long-distance transport
IV) Ascent of sap – D) Downhill transport

Answer:
b) D A B C

Question 23.
The length and breadth of stomata is:
(a) about 10 – 30μ and 2 – 10μ respectively
(b) about 10 – 14μ and 3 – 10μ respectively
(c) about 10 – 40μ and 3 – 10μ respectively
(d) about 5 – 30μ and 5 – 10μ respectively
Answer:
(c) about 10 – 40μ and 3 – 10μ respectively

Question 24.
A membrane that permits the solvent and not the solute to pass through it is termed is
a) Permeable,
b) impermeable
c) semipermeable
d) differentially permeable
Answer:
c. Semi permeable

Question 25.
Who did observe that stomata open in light and close in the night:
(a) Unger
(b) Sachs
(c) Boehm
(d) Von Mohl
Answer:
(d) Von Mohl

Question 26.
The phosphorylase enzyme in guard cells supports the starch-sugar interconversion theory. The above reaction is:
(a) oxidation reaction
(b) hydrolyses reaction
(c) reduction reaction
(d) none of the above
Answer:
(b) hydrolyses reaction

Question 27.
If a cell kept in a solution of unknown concentration gets deplasmolysed the solution is
a) hypotonic
b) hypertonic
c) isotonic
d) detonic
Answer:
a. hypotonic

Question 28.
A cell placed in a strong salt solution will shrink because
a) the cytoplasm will decompose
b) mineral salts will break the cell wall
c) salt will leave the cell
d) water will leave by exosmosis
Answer:
d. water will leave by exosmosis

Question 29.
Phenyl Mercuric Acetate (PMA), when applied as a foliar spray to plants:
(a) induces partial stomatal closure for two weeks.
(b) induces partial stomatal opening for two weeks.
(c) induces partial stomatal closure for four weeks.
(d) induces stomatal closure permanently
Answer:
(a) induces partial stomatal closure for two weeks.

Question 30.
The osmotic pressure of cell sap is maximum in
a) Hydrophytes
b) Halophytes
c) Xerophytes
d) Mesophytes
Answer:
b. Halophytes

Question 31.
Say true or false and on that basis choose the right answer.
I) In facilitated diffusion, molecules move across the cell membrane with the help of special proteins, with the expenditure of energy
II) Porin is a larger transport protein, facilitates smaller molecules to pass through.
III) Aquaporins are recognized to transport urea, CO₂, NH3 metalloid & ROS
IV) The carrier proteins structure does not get modified due to its association with the molecules

Answer:
d. False True True False

Question 32.
I) Hypertonic is a strong solution (low solvent/high solute/ low Ψ )
II) Hypotonic is a weak solution (high solvent/low or zero solutes/ high Ψ)
III) Hypertonic is the weak solution (high solvent/low or zero solutes/high Ψ)
IV) Hypotonic is a strong solution (low solvent / high solute/low Ψ)

Answer:
b. True True False False

Question 33.
From sieve elements sucrose is translocated into sink organs such as root, tubers etc and this process is termed as:
(a) Xylem unloading
(b) Xylem uploading
(c) Phloem unloading
(d) Phloem uploading
Answer:
(c) Phloem unloading

Question 34.
The value of pure water is zero in which three aspects of the given options
I) Osmotic pressure
II) Osmotic potential
III) Water potential
IV) Pressure potential
a) I, II, & III
b) II, III & IV
c) I, Ill & IV
d) I, II & IV
Answer:
a. I, II & III

Question 35.
Gases such as oxygen and carbon dioxide cross the cell membrane by
a) Passive diffusion through the lipid bilayer
b) Primary active transport
c) Specific gas transport proteins
d) Secondary active transport
Answer:

Question 36.
Hydathodes are generally present in plants that grow in:
(a) dry places
(b) moist and shady places
(c) sunny places
(d) deserts
Answer:
(b) moist and shady places

Question 37.
Why sugars are transported in the form of su-crose in phloem?
a) It is inactive and highly soluble
b) It is active
c) It yields high ATP
d) It is lighter in weight.
Answer:
a. It is inactive and highly soluble

Question 38.
Unloading of pholem at sink includes
a) Passive transport
b) diffusio
c) Osmosis
d) Active transport
Answer:
d. Active transport

Question 39.
The liquid coming out of the hydathode of grasses is:
(a) pure water
(b) not pure water
(c) a solution containing a number of dissolved substances
(d) saltwater
Answer:
(c) a solution containing a number of dissolved substances

Question 40.
In a flaccid cell
a) DPD = OP
b) DPD = TP
c) TP = OP
d) OP = O
Answer:
a. DPD = OP

Question 41.
The pathway of water movement involving living part of a cell is
a) Apoplast pathway
b) symplast pathway
c) Transmembrane pathway
d) Lateral conduction
Answer:
b. Symplast pathway

Question 42.
The ascent of sap is
a) Upward movement of water in plants
b) downward movement of water in plants
c) upward and downward movement of water plants
d) None of the above
Answer:
a. upward movement of the water plants

Question 43.
High tensile strength of water is due to
a) Adhesion only
b) cohesion only
c) Both (a) and (b)
d) None of these
Answer:
c. Both (a) and (b)

Question 44.
Maximum transpiration occur in
a) Mesophytes
b) Xerophytes
c) Hydrophytes
d) Epiphytes
Answer:
a. Mesophytes

Question 45.
Supply ends in transport of solutes are
a) green leaves
b) root and stem
c) xylem and phloem
d) Hormones and enzymes
Answer:
c. Xylem and phloem

Question 46.
For guttation in plants, the process responsible is
a) Root pressure
b) Atmospheric pressure
c) Imbibition
d) None of these
Answer:
a. Root pressure

Question 47.
Which of the following theories for Ascent of sap was proposed by famous Indian scientist. J.C. Bose.
a) Transpiration pull theory
b) Pulsation theory
c) Root pressure theory
d) Atmospheric pressure theory
Answer:
b. Pulsation theory

Question 48.
Which of the following plant material is an efficient water imbibant?
a) Lignin
b) Pectin
c) Cellulose
d) Agar
Answer:
d. Agar

Question 49.
Which of the following helps in the Ascent of sap?
a) Root pressure
b) Transpiration
c) Capillarity
d) All the above
Answer:
d. All the above

Question 50.
In a girdled plant which of the following dies first?
a) Shoot
b) root
c) Both die simultaneously
d) None – the plant survives
Answer:
b. root

Question 51.
Assertion:-A Imbibition is also diffusion
Reason -R The movement of water in the above process is along a concentration gradient.
a) Both A and Rare true and R is correct explanation of A
b) Both A and R are true but R is not the correct explanation of A
c) A true but R false
d) Both A and Rare false
Answer:
a) Both A and R are True and R is correct explanation of A

Question 52.
Assertion: – A In rooted plant, the transport of water and minerals in xylem is essentially multi-directional
Reason – R Organic compound and nuitrient undergoes undirectional transport only
Answer:
d) Both A and R are false

Question 53.
Assertion: – A The adsorption of water by solid particles of an adsorbant with out forming a solution is known as imbibition
Reason: – R The liquid which is imbided is known as imbibate
Answer:
b) Both A and R are true but R is not the correct explanation of A

Question 54.
Assertion: – A In phloem loading, food is transported to the sink
Reason – R Food is transported from source to sink ‘
Answer:
d) Both Assertion ‘A’ and Reason ‘R’ are false

Question 55.
Assertion – A: Xylem a principal water conducting ’
Reason -R: It has been recognised by girdling or ringing experiments
Answer:
a) Both A and R are True R is the correct explanation of A

Question 56.
Assertion: – A In phloem, sugar are translocated in non reducing form
Reason – R Non reducing sugars are most reactive sugars
Answer:
c) Assertion is true but Reason is false

Question 57.
Assertion: AIn ringing experiment a narrow continuous band of tissues external to the phloem is removed
Reason: R Ringing experiment proves that phloem is involved in water transport ’
Answer:
d) Both A and R are false

II. Two Mark Questions

Question 1.
What is the need for the transport of materials in plants?
Answer:
Water absorbed from roots must travel up to leaves by xylem for food preparation by photosynthesis. Likewise, food prepared from leaves has to travel to all parts of the plant including roots.

Question 2.
What is osmosis
Answer:
It is a special type of diffusion almost same like simple diffusion but has a selectively permeable membrane is here, through which osmosis occur.
(OR)
It is the movement of water molecules from a place of its higher concentration, to the place of its lower concentration through a semipermeable membrane.

Question 3.
Define the term diffusion.
Answer:
The net movement of molecules from a region of their higher concentration to a region of their lower concentration along a concentration gradient until an equilibrium is attained.

Question 4.
The touch me plant closes its leaves at the touch – Explain.
Answer:

• In the ‘Touch me not’ plant the touching act as stimulus, and it closes the leaves.
• When we touch the plant, at that time the stem releases some chemicals, which force water to move out of the cell leading to the loss of Turgor pressure and the leaves droop down However after sometime they become normal.

Question 5.
What is meant by Porin?
Answer:
Porin is a large transporter protein found in the outer membrane of plastids, mitochondria and bacteria which facilitates smaller molecules to pass through the membrane.

Question 6.
Define water potential
Answer:

• The potential energy of water in a system compared to pure water when both temperature and pressure are ketp same.
• It is a measure of how freely water molecules can move in a given environment
• Water potential of pure water is = 0

Question 7.
Define Diffusion Pressure Deficit.
Answer:

• Termed by Meyer (1938)
• The difference between the Diffusion pressure of the solution and its solvent at a particular temperature and atmospheric pressure of the solution and its solvent at a particular temperature and atmospheric pressure is called DPD.

Question 8.
Differentiate between short distance and Long Distance Transport.
Answer:

SDT	LDT
1. Cell to cell Transport Involve few cells ni lateral direction	Transport with in the network of xylem and phloem
2. Connecting link to xylem bind phloem from root hairs to leaf tissues	Direct vertical – main Transport
3. Eg. Diffusion, Osmosis etc	4. Eg. Ascent of sap & Translocation of solutes.

Question 9.
Differentiate between Passive & Active Transport
Answer:

PT	AT
1. Down hill Transport (Phyical)	Up hill Transport (Biological)
2. Occur According to concentration gradient	Occur against concentration gradient
3. No expenditure of energy	There is expenditure of energy obtained from Respiration
4. Eg. Diffusion – Facilitated Diffusion osmosis etc.	Eg. Na+ K+ ATP are pump.

Question 10.
Give two examples of the phenomenon of Imbibition.
Answer:
two examples for the phenomenon of Imbibition:

1. The swelling of dry seeds.
2. The swelling of wooden windows, tables, doors due to high humidity during the rainy season.

Question 11.
Explain carbonic Acid Exchange theory.
Answer:

• Soil solution act as a medium of ion-exchange
• The CO₂ released by roots combine with water to form carbonic acid (H₂CO₃)
• Carbonic acid dissociates into H⁺ + HCO₃ in the soil solution.
• H⁺ ions exchange with cations adsorbed on clay particles and cations from micelles get released int c.

Question 12.
Give Answer in a sentence or two Distinguish between (i) Exomosis & Endomosis (ii) Apoplast & Symplast (iii) Cohesion & Adhesion (v) Influx & Efflux
Answer:

I) Exomosis	Endomosis
The osmotic outflow of water, when cell placed in hypertonic solution	Osmotic inflow into the cell when placed in hypotonic solution or water
Eg. Preservation of Jam, Jellies, pickles	Eg. Swelling of Dry grapes placed in water

II) Apoplast	Symplast
System of adjacent cell walls – continuous throughout except at the asparian strips of endodermis in the roots	System of interconnected protoplasts of neighbouring cells in plants

III) Cohesion	Adhesion
Attraction between molecules of a similar kind	The attraction between molecules of different kind

IV) Influx	Efflux
The entry of ion into the cell is known as Influx	The exit of ion from the cell into outside is known as Efflux
It can be active or passive	It can be active or passive.

Question 13.
What is meant by osmotic pressure?
Answer:
When a solution and its solvent (pure water) are separated by a semipermeable membrane, the pressure is developed in the solution, due to the presence of dissolved solutes. This is called osmotic pressure (OP).

Question 14.
Define Root Pressure.
Answer:

• Stephen Hales – coined the term
• Stoking (1956) Defined the term.
• A pressure developing in the tracheary elements of the xylem as a result of metabolic activities of the root.

Question 15.
Define the term osmosis.
Answer:
Osmosis (Latin: Osmos – impulse, urge) is a special type of diffusion. It represents the movement of water or solvent molecules through a selectively permeable membrane from the place of its higher concentration (high water potential) to the place of its lower concentration (low water potential).

Question 16.
Why plants transport sugars as sucrose and not as starch or Monosaccharide (Glucose & Fructose)
Answer:

Name	Type	Properties
1.Starch	Polysaccharide (non reducing sugar)	Insoluble in water cann’t be transport
2. Glucose & Fructose	Monosaccharides (reducing sugar)	Soluble in water but less efficient in energy storage & reactive
3. Sucrose	Disaccharide(non reducing sugar)	Soluble in water, even at high concentration, low viscosity, more efficient in energy storage no reducing ends make it inert than glucose & Fructose.

Question 17.
What are the three types of plasmolysis?
Answer:
Three types of plasmolysis occur in plants:

1. Incipient plasmolysis
2. Evident plasmolysis
3. Final plasmolysis.

Question 18.
Identify the diagram and Neatly label the parts
Answer:

The given diagram is the structure of Hydathode
A-Guard cell
B-Epithem
C-Tracheids

Question 19.
Identify the Diagram & Label the parts.
Answer:

The given diagram explain Reverse osmosis
A – Pressure
B – Pure water
C – Saltwater
D – Membrane

Question 20.
Differentiate between Ascent of sap and Translocation of solute.
Answer:

Ascent of sap	Translocation of solute
The upward transport of water along with dissolved minerals from roots to the aerial parts is called as Ascent of sap.	The transport of food from the site of synthesis to the site of utilization or from source to sink is known as Translocation of organic solutes (a dissolved substance)
Occur through Xylem	Occur through Phloem

Question 21.
Give any two objections to starch-sugar interconversion theory.
Answer:
Two objections to starch – sugar interconversion theory:

1. In monocots, the guard cell does not have starch.
2. There is no evidence to show the presence of sugar at a time when starch disappears and stomata open.

Question 22.
Differentiate between cuticular and Lenticular Transpiration.
Answer:

Cuticular Transpiration	Lenticular Transpiration
Loss of water through cuticle is known as cuticular Transpiration	Some pores
It is only about 5 to 10% of the total Transpiration	Present on the woody surface of stem (bark) are known as Lenticels
The thicker the cuticle, the lesser will be the Transpiration. Eg. xerophytes	The loss of water from the lenticel is Lenticular Transpiration – It is only about 0.1 % of the total.

Question 23.
Mention any two uses of anti – transpirants.
Answer:
Two uses of anti – transpirants:

1. Anti – transpirants reduce the enormous loss of water by transpiration in crop plants.
2. Useful for seedling transplantations in nurseries.

Question 24.
Give notes an Aquaporin.
Answer:

• Water pore – Aquaporin in KBC was discovered by Peter Agre (Nobel Prize for chemistry – 2003)
• Water channel protein is present in PM.
• Regulate the massive amount of water transport across PM
• 30 types of Aquaporins are known from maize

They also transporter

• glycerol
• urea
• CO2
• NH
• metalloids & Reactive oxygen species (ROS)

Function:

• They increase the permeability of the membrane of water
• They confer drought and salt, stress tolerance.

Question 25.
Define the term Ion – Exchange.
Answer:
Ions of external soil solution are exchanged with the same charged (anion for anion or cation for cation) ions of the root cells.

Question 26.
A. Differentiate between Cohesion and Adhesion and
B. Add a note on their significance.
Answer:
A.

Cohesion	Adhesion
The strong mutual attraction between water molecules is called cohesion or cohesive force.	The Attraction between a water molecule and the wall of the xylem element is called adhesion.

B. The cohesive and Adhesive forces work together to form an unbroken continuous water column in xylem.
The magnitude of cohesive force is much high (350 atm) and is more than enough to ascent sap in the tallest
trees.

III. 3 Mark Questions

Question 1.
Compare and Contrast Diffusion & Osmosis.
Answer:

Diffusion	Osmosis
1. The net movement of molecules from a region of their higher concentration to a region of their lower concentration along a concentration gradient until an equilibrium is attained	It is a special type of diffusion – There is movement of water or solvent molecules through a selectively permeable membrane from a place of its higher concentration to its lower concentration until an equilibrium is attained.
2. it is independent of the living system	It is also independent of the living system
3. Passive process	Passive process
4. Obvious in solids gases & liquids Only in liquid molecules Eg. diffusion of sugar in water	Eg. Dry grapes, when kept in water swells, & becomes turgid.

Question 2.
Differentiate between osmotic pressure it and osmotic potential
Answer:

Osmotic pressure	Osmotic potential
1. The hydrostatic pressure developed in a solution. due to the presence of dissolved solutes when it is separated from a pure solvent by a semi-permeable membrane.	The ratio between the number of solvent particles and the number of solute particles in a solution or (lowering of free energy of water in a system due to the presence of solute particles
2. develops only in a confined system.	develops in confined or an open system
3. The value is positive, though it is numerically equal to osmotic potential	The value is negative though it is numerically opposite to osmotic pressure.

Question 3.
Do you have an R.O. Purifier ¡n your house? Explain the principle behind it.
Answer:

• Yes / No – R.O. is working on the principle of osmosis. but in the reverse direction.
• In regular osmosis water moves from its higher concentration to its lower concentration through the selectively permeable membrane but here water moves from lower concentration to higher concentration through selectively permeable membrane.
• Since against concentration gradient, there is the expenditure of energy, to apply pressure, to force water in a reverse direction.
• Eg- Desalination plants to purify seawater also work like R-O-Purifiers Movement of Water in house hold usage.

Question 4.
Difference between various plasmolysis types
Answer:

Question 5.
Define Antitranspirant.
Answer:
Antitranspirant is any material applied to plants to retard or reduce the rate of transpiration – without disturbing the process of gaseous exchange, for respiration and photosynthesis.
Eg. Colourless plastics silicone oil and low viscosity waxes.

Question 6.
What are the inducers of stomatal closure.
Answer:

• Natural antitranspirants usually induce stomatal closure
  Eg. CO2 – inhibits photorespiration – thereby induces stomata! closure
• Some chemicals, when applied as a foliar spray can induce stomatal closure for 2 – 3 weeks.
  Eg. (PMA) Phenyl Mercuric Acetate & (ABA) Abscisic Acid.

Question 7.
Fill in the blanks in the tabulations given below

The Study	Year	Scientist associated with it
1. The concept of water potential	1960	………………………………….
2. Active and Passive absorptions	1949	…………………………………
3. Pulsation theory	1923	…………………………………….

Answer:
1) Slatyer & Taylor 2) Kramer 3) J.C. Bose

Question 8.

Nature of membrane	Definition	Example
1. Impermeable	1. ……………………………..	suberized. cutinizedcell walls
2. ……………………………….	Allow diffusion of solvent molecules, do not allow the passage of solute molecules	Parched paper
3. Selectively permeable	biomembranes allow some solutes to pass in addition to solvent molecules	3. ………………………………….

Answer:
1) Inhibit the movement of both solvent and solute molecules
2) Semipermeable
3) Tonoplast & plasmalemma

Question 9.
Give the flow chart to cell to cell transport in plants.
Answer:

Question 10.
Explain the capillary theory of Boehm (1809).
Answer:
Capillary theory: Boehm (1809) suggested that the xylem vessels work like a capillary tube. This capillarity of the vessels under normal atmospheric pressure is responsible for the ascent of sap. This theory was rejected because the magnitude of the capillary force can raise water level only up to a certain height. Further, the xylem vessels are broader than the tracheid which actually conducts more water and against the capillary theory.

Question 11.
Explain Phloem loading?
Answer:
Definition:
The products of Photosynthesis from the Mesophyll of leaves to sieve elements of phloem is known as phloem loading. (Just like the cement sack manufactured in a factory being loaded in a vehicle to be transported the respective site)
It involves 3 steps.
Step I:

1. The chloroplast has photosynthate in the form of starch or Trlose phosphate
2. It is transported to the cytoplasm, where it is converted into Sucrose.

Question 12.
Explain the theory of photosynthesis in guard cells observed by Von Mohl with its demerits.
Answer:
Von Mohl (1856) observed that stomata open in light and close in the night. According to him, chloroplasts present in the guard cells photosynthesize in the presence of light resulting in the production of carbohydrate (Sugar) which increases osmotic pressure in guard cells. It leads to the entry of water from other cells and the stomatal aperture opens. The above process vice versa in the night leads to the closure of stomata.

Demerits:

1. The chloroplast of guard cells is poorly developed and incapable of performing photosynthesis.
2. The guard cells already possess much amount of stored sugars.

IV. 5 Mark Questions

Question 1.
Explain ‘routes’ of Water Absorption in the roots.
Answer:

• Introduction
• Root hair & other epidermal cells – By imbibition absorb water from soil –
• By osmosis moves radically & centripetally – across
  • cortex
  • Endodermis
  • Pencycle & Xylem

There are 3 Routes

• Apoplast
• Symplast
• Transmembrane route

I. Apoplast ( GK – Apo – Away) Everything external to PM
1. Cell walls
2. Extra Cellular Space
3. Interior of dead cells (vessel elements Tracheids)
Movement is continuous exclusively through the cell wall or nonliving part of the plant without crossing any membrane.

II. Symplast (GK – Sym = within)
Entire mass of cytosol of all the living cells in a plant + plasmo desmata + inter connecting cytoplasmic channel.
In the movement water has to cross PM, to enter cytoplasm of outer root cell; then move within adjoining
cytoplasm through plasmodesmata around the vacuoles without the necessity to cross more membrane it reaches xylem.

III. Trans – Membrane Route

• Water enters a cell on one side and exits from the other side.
• It crusses 2 membranes for each cell (also through to no plast).

Question 2.
Draw & Explain the structure of Stomata.
Answer:
1. Definition:
The epidermis of leaves and green stems possess many small pores called – Stomata

2. Length & Breadth
The length – 10- 40μ The Breadth – 3 – 10μ
Number Mature leaves contain 50- 500 stomata / mm²

3. Structure
a. Guard Cells – A pair of Kidney shaped cells (semilunar) surrounded a small opening called stoma
b. Subsidiary Cells – Guard cells attached to surrounding epidermal cells known as subsidiary cells or accessory cells.

• The inner wall of guard cell is thicker
• The stoma open into an interior substomatal cavity.

Question 3.
Explain osmosis by Potato osmoscope Experiment.
Answer:
Aim : To demonstrate osmosis by Potato osmoscope
Apparatus used: Potato tuber, beaker containing water, sugar solution and pin.

Definition:
Diffusion of water or solvent from the region of higher water potential to a region of lower water potential
is known as osmosis.

Procedure:
Take a peeled potato tuber and make a cavity inside with the help of a knife fill the cavity with concentrated sugar solution and mark the initial level.
Place this set up in a beaker containing pure water After 10 minutes observe the sugar solution level and record your observation.

Observation:
There is rise in the level of the solution. in the cavity of the tuber due to osmosis
Inference: Osmosis has occured, through the potato osmoscope

Question 4.
Measure Transpiration with Ganong’s Photometer
Answer:
Aim: To measure the rate of Transpiration with Ganong’s Potometer
Apparatus needed: Ganong’s Potometer, a twig, beaker, water, split rubber cork, and vaseline.

Procedure:

• Ganong’s Potometer is a horizontal graduated tube which is bent in opposite directions at the ends.
• A reservoir is fixed to the horizontal tube hear the wider end Reservoir has stop cock to regulate water flow.
• A twig is fixed to the wider arm through the split cork. The apparatus is filled with water with water from reservoir.
• The apparatus is made air tight by applying vaseline.
• The other bent end of the horizontal tube is dipped into a beaker containing coloured water.

An air bubble is introduced into the graduated tube at the narrow end. Keep the apparatus in bright sunIght
and observe

Observation:
As the twig transpires, the air bubble move towards the twig.
This loss is compensated by water ohsorption from the beaker.

inference:
By the experiment we can study the rate of Transpiration and rate of transpiration is equal to the rate of water absorption.

Question 5.
Explain Mechanism of Translocation by Munch Mass flow Hypothesis
Answer:
Munch – Proposed it in 1930 Crafts – elaborated it in 1938
Definition: Organic substances (solute) move from a region of high osmotic pressure (mesophyll) to
region of low OP along TP gradient.

Example – Physical system :
Chamber ‘A’ & chamber ‘B’ made up of semi permeable membrane connected by a tube ‘T’
A – Contain highly concentrated sugar solution (hypertonic)
B – Contain dilute sugar solution (hypotonic)
A – draws water from the reservoir by Endosmosis – TP of chamber ‘A’ increased

• Continuous entry of water in to A – TP increased
• Flow of solute from chamber A to B thro TP gradient.
• The movement continues till both Aand B attain isotonic condition (equilibrium)
  (However if new sugar solution added to A system will start to run again)
  Example (Biological system)
• Chamber A (Source) – (Equivalent to) – Mesophyll cells of leaves (High concentration of soluble food)
• Chamber B (Sink) – (Equivalent to) – Cells of stem & Roots (Consumption end)
• TubeT – (Analogous to) – Sieve tube to phloem

Steps :
1. Xylem (Reservoir) – Movement of water (Endomosis) – Mesophyll cells (TP increase)
2. Mesophyll cells (High TP) Source – enmass movement of  organic solutes through Phloem by TP Gradient  – Cells of stem & Root (low TP) (Sink)

Question 6.
Explain the theory of K⁺ transport – or Explain the mechanism of stomatal movement
Answer:
Introduction:
Levit (1974) – Proposed it
Raschke (1975) – Elaborated it
Steps:

This process of exchange of ions is called Actie ion exchange ( consume ATP) or Energy

• Increased K⁺ ions in the Guard cells – balanced by CP ions
• Increase in solute concentration (Hypertonic) Decrease in water potential
• Water enters into Guard cells from subsidiary cells
• Wall pressure increase Turgor pressure, Turgid guard cells – fall apart & opens the stoma

• Exit of H⁺
• Intake of K⁺
• Exit of K⁺
• Loss of H2O
• Uptake of H2O⁺
• Turgidity of Guard Cells
• Accumulation of CO2 – Lowering of pH
• Opening of Stoma.
• Activation of ABA
• Closure of Stoma.

Question 7.
Explain Cytochrome Pump Theory (or) Explain Carrier concept of Active Absorption, through cytochrome Pump theory.
Answer:
Lunde gardth & Burstom (1933)- Proposed the Cytochrome

Pump theory:

• There is correlation between Respiration & Anion absorption.
• when a plant is transferred from water to salt solution, the rate of respiration increases – known as Anion respiration – or salt respiration

The Assumptions of Cytochrome pump theory:

• The mechanism of anion and cation absorption is different.
• Anion – absorption – through cytochrome pump or chain by Active process
• An oxygen gradient is responsible for oxidation at outer surface of the membrane and reduction at the inner surface.

Explanation:

• On the inner surface, the enzyme dehydrogenase Produces protons (W) and electrons (e)

• Anions are picked up by oxidized cytochrome oxidase and transferred to the other members of the chain.
• Theory assumes the passive movement of cations (C⁺) along the electrical gradient created by the accumulation of anions (A^–) at the inner surface of the membrane.

Defects :

• • Cations also induce respiration
  • to fail to explain the selective uptake of ions
  • It explains absorption of anions only.

Question 8.
Explain the opening and closing of stomata by a starch – sugar – Interconversion theory.
Answer:
i) Lloyd (1908)
According to him, turgidity of Guard cell is due to interconversion of starch → sugar

• Day time:
  Guard cells have sugar → so turgid → opening of stomata
• Nighttime:
  Guard cells have starch → so loose turgidity (become flaccid) → closure of stomata

ii) Sayre (1920)
According to him, the pH of Guard cell determine opening and closing of stomata

• Day time: Guard cells have high pH →so turgid → opening of stomata
• Nighttime: Guard cells have low pH → become flaccid → closure of stomata to be elaborate
• Day time: Utilisation of CO₂. in photosynthesis → Starch into sugar → high pH → high Turgor pressure→Opening of Stomata
• Night Time: No photosynthesis, so the accumulation of CO₂ → sugar to starch → low pH → decrease in TP → closure of stomata

iii) Hanes (1940)
According to Hanes – Enzyme phosphorylase is responsible for starch sugar conversion in the guard cells.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 14 Classes and Objects Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 14 Classes and Objects

11th Computer Science Guide Classes and Objects Text Book Questions and Answers

Book Evaluation

Part I

Choose The Correct Answer
Question 1.
The variables declared inside the class are known as data members and the functions are known as
a) data functions
b) inline functions
c) member functions
d) attributes
Answer:
c) member functions

Question 2.
Which of the following statements about member functions are True or False?
i) A member function can call another member function directly with using the dot operator.
ii) Member function can access the private data of the class.
a) i-True, ii-True
b) i-False, ii-True
c) i-True, ii-False
d) i-False, ii-False
Answer:
b) i-False, ii-True

Question 3.
A member function can call another member function directly, without using the dot operator called as
a) sub function
b) sub member
c) nesting of member function
d) sibling of member function
Answer:
c) nesting of member function

Question 4.
The member function defined within the class behave like
a) inline functions
b) Non inline function
c) Outline function
d) Data function
Answer:
a) inline functions

Question 5.
Which of the following access specifier protects data from inadvertent modifications?
a) Private
b) Protected
c) Public
d) Global
Answer:
a) Private

Question 6.
class x
{
inty;
public:
x(int z)
{
y=z;
}
} x1[4];
intmain( )
{
x x2(10);
return 0;
}
How many objects are created for the above program?
a) 10
b) 14
c) 5
d) 2
Answer:
c) 5

Question 7.
State whether the following statements about the constructor are True or False.
i) constructors should be declared in the private section.
ii) constructors are invoked automatically when the objects are created.
a) True, True
b) True, False
c) False, True
d) False, False
Answer:
c) False, True

Question 8.
Which of the following constructor is executed for the following prototype ?
add display (add &); // add is a class name
a) Default constructor
b) Parameterized constructor
c) Copy constructor
d) Non Parameterized constructor
Answer:
c) Copy constructor

Question 9.
What happens when a class with parameterized constructors and having no default constructor is used in a program and we create an object that needs a zero- argument constructor?
a) Compile-time error
b) Domain error
c) Runtime error
d) Runtime exception
Answer:
a) Compile-time error

Question 10.
Which of the following create a temporary instance?
a) Implicit call to the constructor
b) Explicit call to the constructor
c) Implicit call to the destructor
d) Explicit call to the destructor
Answer:
b) Explicit call to the constructor

Part – II

Very Short Answers

Question 1.
What are called members?
Answer:
The class comprises members. Members are classified as Data Members and Member functions. Data members are the data variables that represent the features or properties of a class. Member functions are the functions that perform specific tasks in a class.

Question 2.
Differentiate structure and class though both are user-defined data types.
Answer:
The only difference between structure and class is the members of the structure are by default public whereas it is private in class.

Question 3.
What is the difference between the class and object in terms of oop?
Answer:
Object:

• Object is an instance of a class.
• Object is a real-world entity such as pen, laptop, mobile, chair, etc.
• Object allocates memory when it is created.

Class:

• Class is a blueprint or template from which objects are created.
• Class is a group of similar objects.
• Class doesn’t allocate memory when it is created.

Question 4.
Why it is considered a good practice to define a constructor though a compiler can automatically generate a constructor?
Answer:
A user-defined constructor is the best method of initialise array of objects and normal objects.

Question 5.
Write down the importance of the destructor.
Answer:
The purpose of the destructor is to free the resources that the object may have acquired during its lifetime. A destructor function removes the memory of an object which was allocated by the constructor at the time of creating an object.

Part – III

Short Answers

Question 1.
Rewrite the following program after removing the syntax errors if any and underline the errors:
#include<iostream>
#include<stdio.h>
classmystud
{ intstudid =1001;
char name[20];
public
mystud( )
{ }
void register ( ) {cin>>stdid;gets(name);
}
void display ( )
{ cout<<studid<<“: “<<name<<endl;}
}
int main( )
{ mystud MS;
register.MS( );
MS.display( );
}
Answer:
MODIFIED PROGRAM:
#include<iostream>
#include<stdio.h>
class mystud
{
int studid;
char name[20];
public:
mystud( )
{
studid=1001;
}
void register ( )
{
cin>>stdid;
gets(name);
}
void display ( )
{
cout<<studid<<“: “<<name<<endl;
}
};
int main( )
{
mystud MS;
MS.reqister( );
MS.display( );
}

Question 2.
Write with an example of how will you dynamically initialize objects?
Answer:
Dynamic initialization of Objects:
When the initial values are provided during runtime then it is called dynamic initialization.
Program to illustrate dynamic initialization
#include
using namespace std;
class X
{
int n;
float avg;
public:
X(int p,float q)
{
n=p;
avg=q;
}
void disp( )
{
cout<<“\n Roll numbe:-” <<n;
cout<<“\nAverage :-“<<avg;
}
};
int main( )
{
int a ; float b;
cout<<“\nEnter the Roll Number”;
cin>>a; .
cout<<“\nEnter the Average”;
cin>>b;
X x(a,b); // dynamic initialization
x.disp( );
return 0;
}
Output
Enter the Roll Number 1201
Enter the Average 98.6
Roll number:- 1201
Average :- 98.6

Question 3.
What are advantages of declaring constructors and destructor under public access ability?
Answer:

When constructor and destructor are declared under public:

1. we can initialize the object while declaring it.
2. we can explicitly call the constructor.
3. we can overload constructors and therefore use multiple constructors to initialize objects automatically.
4. we can destroy the objects at the end of class scope automatically (free unused memory).

However, some C++ compiler and Dev C++ do not allow to declare constructor and destructor under private section. So it is better to declare constructor and destructor under public section only.

Question 4.
Given the following C++ code, answer the questions (i) & (ii).
Answer:
class TestMeOut
{
public:
~TestMeOut( ) //Function 1
{
cout<<“Leaving the examination hall”<<endl;
}
TestMeOut( ) //Function 2
{
cout<<“Appearing for examination'<<endl;
}
void MyWork( ) //Function 3
{
cout<<“Attempting Questions//<<endl;
}
};
i) In Object-Oriented Programming, what is Function 1 referred to as and when does it get invoked/called?
Function 1 is called a destructor. It will be automatically invoked when the object goes out of scope (ie. at the end of a program).

ii) In Object-Oriented Programming, what is Function 2 referred to as, and when does it get invoked/called?
Function2 is called a constructor. It will be automatically invoked when an object comes into scope.(ie. at the time of object creation).

Question 5.
Write the output of the following C++ program code:
Answer:
#include<iostream>
using namespace std;
class Calci
{
char Grade; .
int Bonus;
public:
Calci( )
{
Grade=’E’;
Bonus=0;
}//ascii value of A=65
void Down(int G)
{
Grade-=G;
}
void Up(int G)
{
Grade+=G;
Bonus++;
}
void Show( )
{
cout<<Grade<<“#”<<Bonus<<endl;
}
};
int main( )
{
Calci c;
c.Down(3);
c.Show( );
c.Up(7);
c.Show( );
c.Down(2);
c.Show( );
return 0;
}
Output

Part – IV

Explain In Detail

Question 1.
Explain nested class with example.
Answer:
When one class becomes a member of another class then it is called Nested class and the relationship is called containership. When a class is declared within another class, the inner class is called a Nested class (i.e. the inner class) and the outer class is known as the Enclosing class. The nested class can be defined in private as well as in the public section of the Enclosing class.

Classes can be nested in two ways:

1. By defining a class within another class
2. By declaring an object of a class as a member to another class
3. By defining a class within another class

C++ program to illustrate the nested class
#include<iostream>
using namespace std;
class enclose
{
private:
int x;
class nest
{
private :
int y;
public:
int z;
void prn( )
{
y=3;z=2;
cout<<“\n The product of”
< <y< <‘*'< <z<<“= “< <y*z< <“\n”;
}
}; //inner class definition over
nest m1;
public:
nest n2;
void square( )
{
n2.prn( ); //inner class member function is called by its object
x=2;
n2.z=4;
cout<<“\n The product of” <<n2.z<<‘*'<<n2.z<<“=”n2.z*n2.z<<“/n”;
cout<<“\n The product of” <<x<<‘*'<<x<<“= “<<x*x;
}
}; //outer class definition over
int main( )
{
enclose e;
e.square( ); //outer class member function is called
}
Output
The product of 3*2=6
The product of 4*4=16
The product of 2*2=4

In the above program the inner class nest is defined inside the outer class enclose, nest is accessed by enclose by creating an object of nest

Question 2.
Mention the differences between constructor and destructor.
Answer:

CONSTRUCTOR	DESTRUCTOR
The name of the constructor must be same as that of the class.	The destructor has the same name as that of the class prefixed by the tilde character
A constructor can have parameter list.	The destructor cannot have arguments.
The constructor function can be overloaded.	Destructors cannot be overloaded i.e., there can be only one destructor in a class.
Constructor cannot be inherited but a derived class can call the base class constructor.	Destructor cannot be inherited.
The constructor is executed automatically when the object is created.	The destructor is executed automatically when the control reaches the end of class.
Allocated memory space for the object.	Destroy the object.

Question 3.
Define a class RESORT with the following description in C++ :
Answer:
Private members:
Rno // Data member to storeroom number
Name //Data member to store user name
Charges //Data member to store per day charge
Days //Data member to store the number of days
Compute ( ) // A function to calculate total amount as Days * Charges and if the
//total amount exceeds 11000 then total amount is 1.02 * Days *Charges

Public member:
getinfo( ) // Function to Read the information like name , room no, charges and days
dispinfo ( ) // Function to display all entered details and total amount calculated
//using COMPUTE function
PROGRAM
using namespace std;
#include<iostream>
class RESORT
{
private:
int Rno,Days,Charges;
char Rname[20];
int compute( )
{
if (Days * Charges >11000)
return (Days * Charges * 1.02);
else
return(Days * Charges);
}
public:
getinfo( )
{
cout<<“\nEnter customer name : “;
cin>>Rname;
cout<<‘nEnter charges per day : “;
cin>>Charges;
cout< <‘nEnter Number of days : “;
cin>>Days;
cout<<‘n Enter Room Number : “;
cin>>Rno;
}
dispinfo( )
{
cout<<‘nRoom Number :
“<<Rno;
cout<<‘nCustomer name :
“<<Rname;
cout<<‘nCharges per day :
“<<Charges;
cout<<‘nNumber of days :
“<<Days;
cout<<‘nTotal Amount :
“<<compute( );
}
int main( )
{
RESORT Obj;
Obj,getinfo( );
Obj.dispinfo( );
Output

Question 4.
WrIte the output of the following:
Answer:
#include<iostream>
#indude<stdio.h>
using namespace std;
class sub
{
int day, subno;
public :
sub(int,int); // prototype
void printsub( )
{
cout<<” subject number: “<<subno;
cout<<” Days : ” <<day;
}
};
sub::sub(int d=150,int sn=12)
{
cout<<endl<<“Constructing the object
“<<endl;
day=d;
sub no=sn;
}
class stud ‘
{
int rno;
float marks; public:
stud( )
{
cout<< “Constructing the object of
students “<<endl;
rno=0;
marks=0.0;
}
void getval( )
{
cout<<“Enter the roll number and the marks secured”; cin>>rno>>marks;
}
void printdet( )
{
cout<<“Roll no : “<<rno<<“Marks : “<<marks<<endl; .
}
};
class admission
{
sub obj;
stud objone;
float fees; ,
public :
admission ( )
{
cout<< “Constructing the object of admission “<<endl;
fees=0.0;
}
void printdet( )
{
objone.printdet( );
obj.printsub();
cout<<“fees : “<<fees<<endl;
}
};
int main( )
{
system (“cls”);
admission adm;
cout<<endl<< ?’Back in main ()”;
return 0;
}
Output

Question 5.
Write the output of the following.
Answer:
#indude<iostream>
#include<stdio,h>
using namespace std;
class P
{
public:
P( )
{
cout<< “\nConstructor of class P }
~P( )
{
cout< < “\nDestructor of class P
}
};
class Q
{
public:
Q( )
{
cout< <“\nConstructor of class Q “;}
~ Q( )
{
cout<< “\nDestructor of class Q
}
};
class R
{
P obj1, obj2;
Q obj3;
public:
R( )
{
cout<< “\nConstructor of class R “;}
~R( )
{
cout<< “\nDestructor of class R
}
};
int main ( )
{
R oR; :
Q oq;
Pop;
return 0;
Output

11th Computer Science Guide Classes and Objects Additional Questions and Answers

Choose The Correct Answer (1 Mark)

Question 1.
The most important feature of C++ is ………………..
(a) object
(b) class
(c) public
(d) All the above
Answer:
(b) class

Question 2.
How many features are commonly present in OOP languages?
a) 3
b) 2
c) 4
d) 5
Answer:
c) 4

Question 3.
Calling a member function of an object is also known as ……………….. to object.
(a) call function
(b) call by value
(c) call by reference
(d) sending message
Answer:
(d) sending message

Question 4.
……………… is a way to bind the data and its associated functions together,
a) Class
b) Array
c) Structure
d) All the above
Answer:
a) Class

Question 5.
When one class become a member of another class, the relationship is called ………………..
(a) containership
(b) partnership
(c) friendship
(d) all the above
Answer:
(a) containership

Question 6.
The body of the class is defined inside the ………………. brackets.
a) Angle < >
b) Square [ ]
c) Curly { }
d) None of these
Answer:
c) Curly { }

Question 7.
……………….. can be defined either in private or in the public section of a class.
(a) Object
(b) Data type
(c) Memory
(d) constructor
Answer:
(d) constructor

Question 8.
The members of the structure are by default ………………..
a) Private
b) Public
c) Protected
d) None of these
Answer:
b) Public

Question 9.
There are ……………….. ways to create an object using the parameterized constructor.
(a) 3
(b) 2
(c) 1
(d) 4
Answer:
(c) 1

Question 10.
The class body contains ………………….
a) Data members
b) Member functions
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 11.
The class body has………………… access specifiers.
a) Three
b) Four
c) Two
d) Five
Answer:
a) Three

Question 12.
The class body has…………….. access specifiers.
a) Private
b) Public
c) Protected
d) All the above
Answer:
d) All the above

Question 13.
…………………. is a visibility label.
a) Private
b) Public
c) Protected
d) All the above
Answer:
d) All the above

Question 14.
……………….. allows preventing the functions of a program to access directly the internal representation of a class type.
a) Data Hiding
b) Data Capturing
c) Data Processing
d) None of these
Answer:
a) Data Hiding

Question 15.
The access restriction to the class members is specified by ……………. section within the class
body.
a) Private
b) Public
c) Protected
d) All the above
Answer:
d) All the above

Question 16.
A …………………. member is accessible from where outside the class but within a program.
a) Private
b) Public
c) Protected
d) All the above
Answer:
b) Public

Question 17.
We can set and get the value of public data members using ………………… function.
a) Member
b) Nonmember
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 18.
A …………….. member cannot be accessed from outside the class.
a) Private
b) Public
c) Protected
d) All the above
Answer:
a) Private

Question 19.
Only the class member functions can access ……………… members.
a) Private
b) Public
c) Protected
d) All the above
Answer:
a) Private

Question 20.
……………….. members can be accessed in child classes.
a) Private
b) Public
c) Protected
d) All the above
Answer:
c) Protected

Question 21.
If all members of the class are defined as …………….. then the class become frozen.
a) Private
b) Public
c) Protected
d) All the above
Answer:
a) Private

Question 22.
If all members of the class are defined as ………………….. then the object of the class can not access anything from the class,
a) Private
b) Public
c) Protected
d) All the above
Answer:
a) Private

Question 23.
………………. are the data variables that represent the features or properties of a class.
a) Data members
b) Member functions
c) Both A and B
d) None of these
Answer:
a) Data members

Question 24.
………………… are the functions that perform specific tasks in a class.
a) Data members
b) Member functions
c) Both A and B
d) None of these
Answer:
b) Member functions

Question 25.
Member functions are called as …………………..
a) Methods
b) Attributes
c) Properties
d) None of these
Answer:
a) Methods

Question 26.
Data members are also called as ……………….
a) Methods
b) Attributes
c) Properties
d) None of these
Answer:
b) Attributes

Question 27.
Classes contain a special member function called as ……………………
a) Constructors
b) Destructors
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 28.
The member functions of a class can be defined in ……………….. ways.
a) Two
b) Three
c) Four
d) Five
Answer:
a) Two

Question 29.
The member functions of a class can be defined in ………………. way.
a) Inside the class definition
b) Outside the class definition
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 30.
When a member function is defined Inside a class, it behaves like ………………. functions.
a) Inline
b) General
c) Local
d) None of these
Answer:
a) Inline

Question 31.
If a function is inline, the compiler places a copy of the code of that function at each point where the function is called at ………………….
a) Run Time
b) Compile time
c) Both A and B
d) None of these
Answer:
b) Compile time

Question 32.
When Member function defined outside the class, and then it is be called as ………………….
member function.
a) Outline
b) Non-inline
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 33.
When Member function defined outside the class using …………….. operator.
a) Scope resolution
b) Membership
c) Reference
d) Conditional
Answer:
a) Scope resolution

Question 34.
The class variables are called ………………….
a) Object
b) Attributes
c) Procedures
d) None of these
Answer:
a) Object

Question 35.
Objects are also called as …………………. of class.
a) Instant
b) Instance
c) Attributes
d) None of these
Answer:
b) Instance

Question 36.
Objects can be created in ………………… methods.
a) Three
b) Four
c) Two
d) Five
Answer:
c) Two

Question 37.
Objects can be created as ……………….
a) Global object
b) Local object
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 38.
…………….. objects can be used by any function in the program.
a) Global object
b) Local object
c) Either A or B
d) None of these
Answer:
a) Global object

Question 39.
If an object is declared outside all the function bodies or by placing their names immediately after the closing brace of the class declaration then it is called as ……………….
a) Global object
b) Local object
c) Either A or B
d) None of these
Answer:
a) Global object

Question 40.
If an object is declared within a function then it is called ……………….
a) Global object
b) Local object
c) Either A or B
d) None of these
Answer:
b) Local object

Question 41.
……………….. object can not be accessed from outside the function.
a) Global
b) Local
c) Either A or B
d) None of these
Answer:
b) Local

Question 42.
No separate space is allocated for …………………. when the objects are created.
a) Member functions
b) Data members
c) Both A and B
d) None of these
Answer:
a) Member functions

Question 43.
Memory space required for the ……………….
a) Member functions
b) Data members
c) Both A and B
d) None of these
Answer:
b) Data members

Question 44.
The members of a class are referenced (accessed) by using the object of the class followed by the ………………. operator.
a) Scope resolution
b) Conditional
c) Dot (membership)
d) None of these
Answer:
c) Dot (membership)

Question 45.
Calling a member function of an object is also known as ……………….
a) Sending a message to object
b) Communication with the object
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 46.
An array which contains the class type of element is called …………………
a) Array of objects
b) Structure Objects
c) Block of objects
d) None of these
Answer:
a) Array of objects

Question 47.
The ………………… of the outline member function given in a class specification, instructs the compiler about its visibility mode.
a) Name
b) Prototype
C) Data type
d) None of these
Answer:
b) Prototype

Question 48.
A member function can call another member function of the same class directly without using the dot operator is called ………………………
a) Nesting of the member function
b) Invariant Members
c) Variant Members
d) None of these
Answer:
a) Nesting of the member function

Question 49.
A member function can call another member function of the same class for that you do not
need a(n) …………………..
a) Member function
b) Data Member
c) Object
d) None of these
Answer:
c) Object

Question 50.
A member function can access ………………. functions.
a) Public
b) Private
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 51.
…………………. operator will reveal the hidden file scope(global) variable.
a) Membership
b) Conditional
c) Scope resolution
d) All the above
Answer:
c) Scope resolution

Question 52.
When an object is passed by ……………… the function creates its own copy of the object and works on it.
a) Value
b) Reference
c) Either A or B
d) None of these
Answer:
a) Value

Question 53.
When an object is passed by …………….. changes made to the object inside the function do not affect the original object.
a) Value
b) Reference
c) Either A or B
d) None of these
Answer:
a) Value

Question 54.
When an object is passed by ……………….. its memory address is passed to the function so the called function works directly on the original object used in the function call.
a) Value
b) Reference
c) Either A or B
d) None of these
Answer:
b) Reference

Question 55.
When an object is passed by ………………………. any changes made to the object inside the function definition are reflected in the original object.
a) Value
b) Reference
c) Either A or B
d) None of these
Answer:
b) Reference

Question 56.
Member Functions can ………………….
a) Receive object as an argument
b) Return an object
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 57.
When one class become a member of another class then it is called ……………. class.
a) Nested
b) Inline
c) External
d) Global
Answer:
a) Nested

Question 58.
When one class becomes a member of another class then the relationship is called ………………….
a) Containership
b) Nesting
c) Parent-Child
d) None of these
Answer:
a) Containership

Question 59.
Classes can be nested in ……………….. ways.
a) Three
b) Two
c) Four
d) Five
Answer:
b) Two
Question 60.
Classes can be nested in ………………..way.
a) By defining a class within another class
b) By declaring an object of a class as a member to another class
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 61.
When a class is declared within another class, the inner class is called a Nested class (ie the inner class) and the outer class is known as …………………… class.
a) Enclosing
b) Abstract
c) Transit
d) None of these
Answer:
a) Enclosing

Question 62.
The nested class can be defined in ………………….. section of the Enclosing class.
a) Private
b) Public
c) Either Private or Public
d) None of these
Answer:
c) Either Private or Public

Question 63.
Whenever an object of a class is declared as a member of another class it is known as a _____ class.
a) Abstract
b) Container
c) Literal
d) None of these
Answer:
b) Container

Question 64.
Instantiating object is done using ……………….
a) Constructor
b) Destructor
c) Data abstraction
d) Data hiding
Answer:
a) Constructor

Question 65.
A(n) ………………..in C++ can be initialized during the time of their declaration.
a) Array
b) Structure
c) Array or Structure
d) None of these
Answer:
c) Array or Structure

Question 66.
Member function of a class can access all the members irrespective of their associated …………………..
a) Access specifier
b) Data type
c) Return type
d) Argument
Answer:
a) Access specifier

Question 67.
When an instance of a class comes into scope, a special function called the ……………. gets executed.
a) Constructor
b) Destructor
c) Data abstraction
d) Data hiding
Answer:
a) Constructor

Question 68.
The constructor function name has the same name as the …………….. name.
a) Object
b) Class
c) Data member
d) None of these
Answer:
b) Class

Question 69.
The constructors return ………………
a) int
b) char
c) float
d) nothing
Answer:
d) nothing

Question 70.
…………………. are not associated with any data type
a) Constructor
b) Data member
c) Data abstraction
d) Member functions
Answer:
a) Constructor

Question 71.
……………… can be defined either inside class definition or outside the Class definition.
a) Constructor
b) Destructor
c) Data abstraction
d) Member functions
Answer:
a) Constructor

Question 72.
A constructor can be defined in ……………… section of a class.
a) Private
b) Public
c) Either Private or Public
d) None of these
Answer:
c) Either Private or Public

Question 73.
If a constructor is defined in ………………… section of a class, then only its object Can be created in any function.
a) Private
b) Public
c) Either Private or Public
d) None of these
Answer:
b) Public

Question 74.
The main function of the constructor is ……………………….
a) To allocate memory space to the object
b) To initialize the data member of the class object
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 75.
A constructor that accepts no parameter is called …………………… constructor.
a) Null
b) Default
c) Empty
d) None of these
Answer:
b) Default

Question 76.
Identify the correct statement from following with respect to constructor.
a) If a class does not contain an explicit constructor (user defined constructor) the compiler automatically generate a default constructor implicitly as an inline public member.
b) In the absence of user defined constructor the compiler automatically provides the default constructor. It simply allocates memory for the object.
c) Parameterized constructor is achieved by passing parameters to the function.
d) All the above
Answer:
d) All the above

Question 77.
A constructor which can take arguments is called ……………….. constructor.
a) Parmeterized
b) Default
c) Empty
d) None of these
Answer:
a) Parmeterized

Question 78.
……………….. type of constructor helps to create objects with different initial values.
a) Parmeterized
b) Default
c) Empty
d) None of these
Answer:
a) Parmeterized

Question 79.
Declaring a constructor with arguments hides the ……………………
a) Data members
b) Compiler generated constructor
c) Member functions
d) None of these
Answer:
b) Compiler generated constructor

Question 80.
………………… Constructor is used to creating an array of objects.
a) Default
b) Parameterized
b) Overloaded
d) None of these
Answer:
a) Default

Question 81.
There are ………………. ways to create an object using the parameterized constructor.
a) Three
b) Two
c) Four
d) Five
Answer:
b) Two

Question 82.
……………… is a way to create an object using the parameterized constructor,
a) Implicit call
b) Explicit call
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 83.
In …………….. method, the parameterized constructor is invoked automatically
whenever an object is created.
a) Implicit call
b) Explicit call
c) Either A or B
d) None of these
Answer:
a) Implicit call

Question 84.
In the………………….. method, the name of the constructor is explicitly given to invoking the parameterized constructor.
a) Implicit call
b) Explicit call
c) Either A or B
d) None of these
Answer:
b) Explicit call

Question 85.
………………… method is the most suitable method as it creates a temporary object
a) Implicit call
b) Explicit call
c) Either A or B
d) None of these
Answer:
b) Explicit call

Question 86.
The chance of data loss will not arise in ………………….. method.
a) Implicit call
b) Explicit call
c) Either A or B
d) None of these
Answer:
b) Explicit call

Question 87.
A ……………….. object lives in memory as long as it is being used in an expression.
a) Temporary
b) Nested
c) Inline
d) None of these
Answer:
a) Temporary

Question 88.
A constructor having a reference to an already existing object of its own class is called ………………….. constructor.
a) Reference
b) Value
c) Copy
d) Move
Answer:
c) Copy

Question 89.
A copy constructor is called ……………..
a) When an object is passed as a parameter to any of the member functions
b) When a member function returns an object
c) When an object is passed by reference to an instance of its own class
d) All the above
Answer:
d) All the above

Question 90.
The constructors are executed in the …………………. of the object declared.
a) Order
b) Reverse order
c) Either A or B
d) None of these
Answer:
a) Order

Question 91.
When the initial values are provided during runtime then it is called ………………….. initialization.
a) Static
b) Dynamic
c) Run time
d) None of these
Answer:
b) Dynamic

Question 92.
…………………. constructor can have parameter list.
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
a) Constructor

Question 93.
No return type can be specified for ………………….
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 94.
The ……………….. function can be overloaded.
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
a) Constructor

Question 95.
The compiler generates a ………………… in the absence of a user-defined.
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
a) Constructor

Question 96.
Compiler generated constructor is ………………….. member function.
a) private
b) protected
c) public
d) None of these
Answer:
c) public

Question 97.
The ………………. is executed automatically,
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 98.
The ……………. is executed automatically when the object is created.
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
a) Constructor

Question 99.
When a class object goes out of scope, a special function called the ………………. gets executed.
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
b) Destructor

Question 100.
The destructor has the same name as the class tag but prefixed with a …………………
a) ~ (tilde)
b) #
c) @
d) None of these
Answer:
a) ~ (tilde)

Question 101.
A …………………… is a special member function that is called when the lifetime of an object ends.
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
b) Destructor

Question 102.
The ………………… cannot have arguments,
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
b) Destructor

Question 103.
There can be ……………… destructor in a class.
a) Two
b) Three
c) Only one
d) Four
Answer:
c) Only one

Question 104.
_____ cannot be inherited.
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
c) Both A and B

Very Short Answer (2 Marks)

Question 1.
Define methods of a class and write its types.
Answer:
The class comprises members. Member functions are called methods. The member functions of a class can be defined in two ways.

1. Inside the class definition
2. Outside the class definition

Question 2.
Why classes are needed?
Answer:
Classes are needed to represent real-world entities that not only have data type properties but also have associated operations. It is used to create user-defined data types.

Question 3.
What is called as nesting of member functions?
Answer:
Only the public members of a class can be accessed by the object of that class, using the dot operator. However, a member function can call another member function of the same class directly without using the dot operator. This is called as nesting of member functions.

Question 4.
What are the visibility labels of a class body?
Answer:
The class body has three visibility labels viz., private, public, and protected. The Visibility labels are also called as access specifiers.

Question 5.
What is a parameterized constructor?
Answer:
A constructor which can take arguments is called a parameterized constructor. This type of constructor helps to create objects with different initial values. This is achieved by passing parameters to the function.

Question 6.
What happened if all the members of a class are defined as private?
Answer:
If all members of the class are defined as private, then the class becomes frozen.
The object of the class can not access anything from the class.

Question 7.
Write about objects.
Answer:
A class specification just defines the properties of a class. To make use of a class specified, the variables of that class type have to be declared. The class variables are called objects. Objects are also called an instance of the class.

For example:
student s;
In the above statement ‘s’ is an instance of the class student.

Question 8.
How many ways objects can be created for a class? Give its types.
Answer:
Objects can be created in two methods:

1. Global object
2. Local object

Question 9.
What do you mean by an array of objects?
Answer:
An array which contains the class type of element is called an array of objects. It is declared and defined in the same way as any other type of array.

Example:
class stock
{
int itemno;
float price; public:
}s[5];
Here s[5] is an array of objects.

Question 10.
What is a nested member function?
Answer:
A member function can cal! another member function of the same class directly without using the dot operator. This is called as nesting of member functions.

Question 11.
How many ways objects can be passed to function argument?
Answer:
Objects can also be passed in both ways

• Pass By Value
• Pass By Reference

Question 12.
What is a container class?
Answer:
Whenever an object of a class is declared as a member of another class it is known as a container class. In the container-ship, the object of one class is declared in another class.

Question 13.
What is the need for a constructor in a class?
Answer:
Instantiating object is done using constructor. An array or a structure in C++ can be initialized during the time of their declaration using constructor. The constructor function initializes the class object.

Question 14.
What are the functions of a constructor?
Answer:
The main functions of the constructor are:

• To allocate memory space to the object and
• To initialize the data member of the class object.

Question 15.
What is a default constructor?
Answer:
Default constructor:
A constructor that accepts no parameter is called the default constructor.
For example in the class Data program Data::Data( ) is the default constructor.
Using this constructor objects are created similar to the way the variables of other data types are created.

Example:
int num; //ordinary variable declaration
Data d1; // object declaration
If a class does not contain an explicit constructor the compiler automatically generates a default constructor implicitly as an inline public member.

Question 16.
What is the significance of default constructor?
Answer:
Default constructors are very useful to crate objects without having specific initial value. It is also used to create array of objects.

Question 17.
How many ways a constructor can be invoked?
Answer:
There are two ways to create an object using parameterized constructor:

1. Implicit call
2. Explicit call

Question 18.
What is copy constructor?
Answer:
A constructor having a reference to an already existing object of its own class is called copy constructor.
In other words Copy Constructor is a type of constructor which is used to create a copy of an already existing object of a class type.

Question 19.
What is the order of constructor invocation?
Answer:
The constructors are executed in the order of the object declared. (If it is in same statement left to right)

For example:
Test t1;
Test t2; // the order of constructor execution is first for t1 and then for t2.
Consider the following example
Sample s1,s2,s3 ; //The order of construction is s1 then s2 and finally s3

Question 20.
What do you mean by dynamic initialization of Object?
Answer:
When the initial values are provided during runtime then it is called dynamic initialization.

Question 21.
Write a note on the destructor.
Answer:

• When a class object goes out of scope, a special function called the destructor gets executed.
• The destructor has the same name as the class tag but prefixed with a ~(tilde).
• The destructor function also returns nothing and it does not associate with any data type

Question 22.
What is the need for a destructor in a class?
Answer:
The purpose of the destructor is to free the resources that the object may have acquired during its lifetime. A destructor function removes the memory of an object which was allocated by the constructor at the time of creating an object

Question 23.
Define destructor.
Answer:
A destructor is a special member function that is called when the lifetime of an object ends and destroys the object constructed by the constructor. Normally it is declared under the public visibility of a class.

Short Answers (3 Marks)

Question 1.
Explain the local object with an example.
Answer:
If an object is declared within a function then it is called a local object.
It cannot be accessed from outside the function.
# include
# include
using namespace std
class add  //Global class
{
int a,b; public:
int sum; void
getdata()
{
a = 5; b = 10; sum
= a + b;
}
} a1;
add a2;
int main()
{
add a3;
a1.getdata();  //global object
a2.getdata();  //global object
a3.getdata();
cout << a1 .sum;  //Local object for a global class
cout << a2.sum;
cout << a3.sum;
return 0;   //public data member accessed from outside the class
}
Output:
151515

Question 2.
Write about private, protected, and public members of a class.
Answer:
The Public Members:
A public member is accessible from anywhere outside the class but within a program.

The Private Members:
A private member cannot be accessed from outside the class. Only the class member functions can access private members. By default, all the members of a class would be private.

The Protected Members:
A protected member is very similar to a private member but they can be accessed in child classes which are called derived classes (inherited classes).

Question 3.
What is a constructor?
Answer:
The definition of a class only creates a new user-defined data type. The instances of the class type should be instantiated (created and initialized). Instantiating objects is done using the constructor. An array or a structure in C++ can be initialized during the time of their declaration.

The initialization of a class type object at the time of declaration similar to a structure or an array is not possible because the class members have their associated access specifiers (private or protected or public). Therefore Classes include special member functions called constructors. The constructor function initializes the class object.

Question 4.
Explain memory allocation of objects.
Answer:
Memory allocation of objects:
All the objects belonging to that class use the same member function, no separate space is allocated for member functions when the objects are created.
Memory space required for the member variables are only allocated separately for each object because the member variables will hold different data values for different objects.
Memory for Objects for p1 and p2 is illustrated:

Question 5.
Explain the default constructor with an example.
Answer:
A constructor that accepts no parameter is called the default constructor. For example in the class data program Data::Data() is the default constructor. Using this constructor objects are created similar to the way the variables of other data types are created.

Example:
int num; //ordinary variable declaration
Data d1; // object declaration

Question 6.
How will you refer members of the class? Give its syntax and an example.
Answer:
The members of a class are referenced (accessed) by using the object of the class followed by the dot (membership) operator and the name of the member.

The general syntax for calling the member function is:
Object_name.function_name (actual parameter); For example consider the following illustration:

Question 7.
Explain the different methods of passing an object to the function argument.
Answer:
Pass By Value:
When an object is passed by value the function creates its own copy of the object and works on it. Therefore any changes made to the object inside the function do not affect the original object.

Pass By Reference:
When an object is passed by reference, its memory address is passed to the function so the called function works directly on the original object used in the function call. So any changes made to the object inside the function definition are reflected in the original object.

Question 8.
Write about constructor.
Answer:
When an instance of a class comes into scope, a special function called the constructor gets executed. The constructor function name has the same name as the class name. The constructors return nothing. They are not associated with any data type. It can be defined either inside class definition or outside the class definition.

Question 9.
What is a parameterized constructor?
Answer:
Parameterized Constructors:
A constructor which can take arguments is called a parameterized constructor. This type of constructor helps to create objects with different initial values. This is achieved by passing parameters to the function.

Example:
class simple
{
private:
int a,b;
public:
simple(int m, int n)
{
a= m ;
b= n;
cout< < “\n Parameterized Constructor of class-simple
}
};

Question 10.
What do you mean by the implicit and explicit call of a constructor?
Answer:
Implicit call:
In this method, the parameterized constructor is invoked automatically whenever an object is created.
For example, simple s1(10,20); in this for creating the object si parameterized constructor is automatically invoked

Explicit call:
In this method, the name of the constructor is explicitly given to invoking the parameterized constructor so that the object can be created and initialized.

For example:
simple s1=simple(10,20); //explicit call

Question 11.
When copy constructor Is executed? Give examples.
Answer:
A copy constructor is called

• When an object is passed as a parameter to any of the member functions
  Example: void simple: :putdata(simple x);
• When a member function returns an object
  Example: simple get data( ) { }
• When an object is passed by reference to an instance of its own class
  For example: simple1, s2(s1); // s2(s1) calls copy constructor

Explain in Detail (5 Marks)

Question 1.
Explain how to define class members?
Answer:
Definition of class members:
Class comprises of members. Members are classified as Data Members and Member functions.

• Data members are the data variables that represent the features or properties of a class.
• Member functions are the functions that perform specific tasks in a class.
• Member functions are called methods, and data members are also called attributes.

Example:

Defining methods of a class:
Without defining the methods (functions), class definition will become incomplete. The member functions of a class can be defined in two ways.

• Inside the class definition
• Outside the class definition

Inside the class definition:
When a member function is defined inside a class, it behaves like inline functions. These are called Inline member functions.

Outside the class definition:
When Member function defined outside the class just like normal function definition (Function definitions you are familiar with) then it is being called as an outline member function or non-inline member function. Scope resolution operator (::) is used for this purpose.

The syntax for defining the outline member function is:
return_type class_name :: function name (parameter list)
{
function definition

Class using Inline and Outline member function:
# include<iostream>
using namespace std;
class Box
{
// no access specifier mentioned
double width;
public:
double length;
//inline member function definition
void printWidth( )
{
cout<<“\n The width of the box is…”<<width;
}
//prototype of the function
void setWidth(double w);
};
// outline member function definition
void Box :: setWidth(double w)
{
width=w;
}
int main( )
{
// object for class Box
Box b;
// Use member function to set the width.
b.setWidth(10.0);
//Use member function to print the width.
b.printWidth( );
return 0;
Output
The width of the box is… 10

Question 2.
What are the ways to create an object using the parameterized constructor with an example?
Answer:
There are two ways to create an object using the parameterized constructor:
1. Implicit call: In this method, the parameterized constructor is invoked automatically whenever an object is created. For example, simple s1( 10,20); in this, for creating the object s1 parameterized constructor is automatically invoked.

2. Explicit call: In this method, the name of the constructor is explicitly given to invoking the parameterized constructor so that the object can be created and initialized.

#include
using namespace std;
class simple
{
private:
int a, b;
public:
simple(int m,int n)
{
a = m;
b = n;
cout << “\n Constructor of class – simple invoked for implicit and explicit call” << endl;
}
void putdata()
{
cout << “\n The two integers are…” << a << ‘\t’ << b << endl;
cout << “\n The sum of the variables” << a << “+” << b << “=” << a + b;
}
};
int main()
{
simple s1(10,20); //implicit call
simple s2 = simple(30,45); //explicit call
cout << “\n\t\tObject 1\n”;
s1.putdata();
s2.putdata();
return 0;
}
Output:
Constructor of class – simple invoked for the implicit and explicit call
Constructor of class-simple invoked for the implicit and explicit call

Object 1
The two integers are… 10 20
The sum of the variables 10 + 20 = 30

Object 2
The two integers are… 30 45.
The sum of the variables 30 + 45 = 75

Question 3.
What are the characteristics of a destructor?
Answer:
Characteristics of destructors:

• The destructor has the same name as that of the class prefixed by the tilde character
• The destructor cannot have arguments.
• It has no return type.
• Destructors cannot be overloaded i.e., there can be only one destructor in a class.
• In the absence of a user-defined destructor, it is generated by the compiler.
• The destructor is executed automatically when the control reaches the end of the class scope to destroy the object.
• They cannot be inherited.

Evaluate Yourself

Question 1.
Define a class in general and in C++’s context.
Answer:
Classes represent real-world entities that not only have data type properties but also have associated operations.
In C++ class is a way to bind the data and its associated functions together. It is a user-defined data type.

Question 2.
What is the purpose of a class specifier?
Answer:
Data hiding is one of the important features of Object Oriented Programming which allows preventing the functions of a program to access directly the internal representation of a class type.
The access restriction to the class members is specified by class specifies like public, private, and protected sections within the class body.

Question 3.
Compare a structure and a class in C++ context.
Answer:
The only difference between structure and class is the members of structure are by default public where as it is private in class.

Question 4.
Compare private and public access specifier.
Answer:
Public members:
A public member is accessible from anywhere outside the class but within a program. We can set and get the value of public data members even without using any member function.

Private members:
A private member cannot be accessed from outside the class. Only the class member functions can access private members. By default all the members of a class would be private.

Question 5.
What is a non-inline member function? Write its syntax.
Answer:
When Member function defined outside the class just like normal function definition (Function definitions you are familiar with) then it is being called as an outline member function or non-inline member function. Scope resolution operator (::) is used for this purpose.
The syntax for defining the outline member function is:
Syntax:
return_type class_name :: function_name (parameter list)
{
function definition
}

Activity – 1
State the reason for the invalidity of the following code fragment.

(i)	(ii)
class count { int first int second; public: int first };	class item { int prd; }; int prdno;

Answer:

• Data member first is duplicated and it is defined with two scopes( both private and public). It is invalid.
• Object name prefix with the only class name. No data type allowed in between class name and object name.

Activity – 2
class area
{
int s;
public:
void calc( );
};
Write an outline function definition for calc( ); which finds the area of a square
Answer:
int area :: calc( ) .
{
return(s * s);
}

Activity – 3
Identify the error in the following code fragment
class A
{
float x;
void init( )
{
Aa1;
X1.5=1; .
}
};
void main( )
{
A1.init( );
}
Answer:
Error:
Local object can not be accessed from outside the function. Al is the local object, so it can not be accessed in main( );

Activity – 4
What is the size of the objects s1, s2?
class sum
{
int n1,n2;
public:
void add( )
{
int n3=10;n1=n2=10;
}
} s1,s2;
Answer:
The size of the object SI and S2 is 8 bytes each in Dev C++, In Turbo C++ 4 bytes each.
Program to test the memory requirement:
class sum
{
int n1,n2;
public:
void add( )
{
int n3-10;
n1=n2=10;
}
} s1,s2;
using namespace std;
#include<iostream>
int main( )
{
cout<<sizeof(s1)<< “”<<sizeof(s2);
}
Output

Activity – 5
i) Write member function called display with no return.
class objects.
ii) Try the output of the above coding with the necessary modifications.
PROGRAM
#indude<iostream>
using namespace std;
class compute
{
int n1, n2;
public :
void init (int a, int b)
{
n1 = a;
n2 = b;
}
int n;
int add ( )
return (n1+n2);;
{
int prd ( )
{
return (n1*n2);
}
};
compute c1, c2;
void display(compute &objl,compute &obj2)
{
c1.init(12,15);
c2.init(8,4);
objlm = obj1.add( );
obj2.n = obj2.add( );
cout<<“\n Sum of object-1 “<<obj1.n;
cout<<“\n Sum of object-2 “<<obj2.n;
cout<<“\n Sum of the two objects are”<<obj1.
n+obj2.n;
c1.init(5,4);
c2.init(2,5);
obj1.n = obj1.prd( );
obj2.n = obj2.prd( );
cout<<“\n Product of object-1 “<<objl.n;
cout<<“\n Product of object-2 “<<obj2.n;
cout<<“\n Product of the two objects are “<<objl.n*obj2.n;
}
int main( )
{
display(c1,c2);
return 0;
}
output

Activity – 6
#include<iostream>
using namespace std;
class Sample
{
int i,j;
public :
int k;
Sample( )
{
i=j=k=0;//constructor defined inside the class
}
};
int main( )
{
Sample s1;
return 0;
}
Output
In the above program justify your reason for no output.
Answer:
Constructor alone is defined without output statement. When the above program is executed, the constructor executed. But no output on the screen because of missing cout

Hands-On Practice

Question 1.
Define a class employee with the following specification.
Answer:
private members of class Employee
empno- integer
ename – 20 characters
basic-float
netpay, hra, da, – float
calculate ( ) – A function to find the basic+hra+da with float return type

public member functions of class employee
havedata( ) – A function to accept values for empno, ename, basic, hra,
da and call calculate( ) to compute netpay
dispdata( ) – A function to display all the data members on the screen
PROGRAM
using namespace std;
#include<iostream>
#include<iomanip>
class Employee
{
private :
int empno;
char ename[20];
float basic,hra,da,netpay;
float calculate( )
{
return (basic+hra+da);
}
public:
void have data( )
{
cout<<setw(35)<<“Enter Employee number :”;
cin>>empno;
cout<<setw(35)<<“Enter Employee name :”;
cin>>ename;
cout<<setw(35)<<“Enter Basic pay :”;
cin>>basic; ,
cout<<setw(35)<<“Enter House Rent Allowance (HRA):”;
cin>>hra; .
cout<<setw(35)<<“Enter Dearness Allowance (DA):”;
cin>>da;
netpay = calculate( );
}
void dispdata( )
{
cout<<“\nEMPLOYEE DETAILS\n\n”;
cout<<setw(35)<<“Employee number :”<<empno<<endl;
cout<<setw(35)<<“Employee name :”<<ename<<endl;
cout<<setw(35)<<“Basic pay :”<<basic<<endl;
cout<<setw(35)<<“House Rent Allowance (HRA) :”<<hra<<endl; cout<<setw(35)<<“Dearness Allowance (DA) :”<<da<<endl;
cout<<setw(35)< <“Netpay :”<<netpay<<endl;
}
};
int main( )
{
Employee e;
e.havedata( );
e.dispdata( );
}
Output

Question 2.
Define a class MATH with the following specifications.
Answer:
private members:
num1, num2, result – float
init( ) function to initialize num1, num2 and result to zero .

protected members:
add( ) function to add num1 and num2 and store the sum in result
diff( ) function to subtract num1 from num2 and store the difference in the result

public members:
getdata( ) function to accept values for num1 and num2
menu( ) function to display menu
1. Add…
2. Subtract…
invoke add() when the choice is 1 and invoke prod when the choice is 2 and also display the result.
PROGRAM
using namespace std;
#include<iostream>
#include<iomanip>
class MATH
{
private:
float num1,num2,result;
init( )
{
num1=0;
num2=0;
result=0;
}
protected: void add( )
{
result = num1+num2;
}
void diff( )
{
result = num1 – num2;
}
public:
getdata( )
{
cout<<“\nEntertwo numbers “;
cin>>num1>>num2;
} ‘ menu( )
{
int choice;
cout<<“\n1.Add …”;
cout<<“\n2.Subtract …….”;
cout<<“\nEnter your choice :”; cin>>choice;
switch(choice)
{
case 1: getdata( );
add( );
cout<<“\nAdded value is”<<result;
break;
case 2: getdata( );
diff( );
cout<<“\nSubtracted value is “<<result;
break;
default: cout<<“\End”;
}
}
};
int main( )
{
MATH m;
m. menu( );
}
Output

Question 3.
Create a class called Item with the following specifications.
Answer:
private members:
code, quantity- Integer data type
price – Float data type
getdata( )-function to accept values for all , data members with no return

public members:
taxt – float
dispdata( ) member function to display code,quantity,price and tax .The tax is calculated as if the quantity is more than 100 tax is 2500 otherwise 1000.
PROGRAM
using namespace std;
#include<iostream>
#include<iomanip>
class Item
{
private:
int code,quantity;
float price;
void getdata( )
{
cout<<“\nEnter product code “; cin>>code;
cout<<“\nEnter quantity “; cin>>quantity;
cout<<“\nEnter price “; cin>> price;
}
public:
float tax;
void display( )
{
getdata( );
if(quantity>100)
tax = 2500;
else
tax = 1000;
cout<<endl<<setw(25)<< “Product code : “<<code<<endl<<endl;
cout<<setw(25)<<“Quantity : ” <<quantity<<endl<<endl;
cout<<setw(25)<<“Unit price :” <<price<<endl<<endl;
cout<<setw(25)<<“Total Amount: ” < cout<<setw(25)<<“Net Bill amount : ”
<<quantity* price+tax<<endl<<endl;
}
};
int main( )
{
Item i; i.display( );
}
Output

Question 4.
Write the definition of a class FRAME in C++ with the following description.
Answer:
Private members:
FramelD – Integer data type
Height, Width, Amount – Float data type
SetAmount( ) -Member function to calculate and assign amount as 10*Height*Width

Public members:
GetDetail( ) Afunction to allow user to entervalues of FramelD, Height, Width. This function should also call SetAmount() to calculate the amount.
ShowDetail( ) A function to display the values of all data members.
PROGRAM
using namespace std;
#include<iostream>
#include<iomanip>
class FRAME
{
private:
int FrameId;
float Height, Width, Amount;
void SetAmount()
{
Amount = 10 * Height * Width;
}
public:
void Getdetails( )
{
cout<<“\nEnter Frame Id : “; cin>> FrameId;
cout<<“\nEnter Frame Height: cin>> Height;
cout<<“\nEnter Frame Width : “; cin>>Width;
SetAmount( );
}
void ShowDetaiis( )
{
cout<<endl<setw(25)<<“Frame Id :” <<FrameId<<endl<<endl; cout<<setw(25)<<“Frame Height:” <<Height<<endl<endl;
cout<<setw(25)<<“Frame Width :”<<Width<<endk<endl;
cout<<setw(25)<<“Total Amount:”
<<Amount< <endl< <endl;
}
int main( )
{
FRAME F;
F.Getdetails( );
F.ShowDetails( );
}
Output

Question 5.
Define a class RESORT in C++ with the following description:
Answer:
Private Members:
Rno //Data member to store Room No
RName //Data member to store customer name
Charges //Data member to store per day charges Days //Data member to store a number of days of stay
COMPUTE( ) //A function to calculate and return Amount as
//Days*Chagres and if the value of Days*Charges is more than 5000 then as 1.02*Days*Charges

Public Members:
Getinfo( ) //A function to enter the content Rno, Name, Charges //and Days Displayinfo( ) //A function to display Rno, RName, Charges, Days and
// Amount (Amount to displayed by calling function COMPUTE())
PROGRAM
using namespace std;
#include<iostream>
class RESORT
{
private:
int Rno,Days,Charges;
char Rname[20];
int compute( )
{
if (Days * Charges >5000)
return (Days * Charges * 1.02);
else
return(Days * Charges);
}
public:
getinfo( )
{
cout<<“\nEnter customer name :”;
cin>>Rname;
cout<<“\nEnter charges per day :”;
cin>>Charges;
cout<<“\nEnter Number of days :”;
cin>>Days;
cout<<“\nEnter Room Number :”;
cin>>Rno;
}
dispinfo( )
{
cout<<“\nRoom Number : “<<Rno;
cout<<“\nCustomer name :
“<<Rname;
cout«”\nCharges per day :
“<<Charges;
cout<<“\nNumber of days :
“<<Days;
cout<<“\nTotal Amount :
“<<compute( );
}
};
int main( )
{
RESORT Obj;
Obj.getinfo( );
Obj.dispinfo( );
}
Output

Question 7.
Answer:
struct pno
{
int pin;
float balance;
}
Create a BankAccount class with the following specifications

protected members
pno_obj //array of 10 elements
init(pin) // to accept the pin number and initialize it and initialize
// the balance amount is 0

public members
deposit(pin, amount):
Increment the account balance by accepting the amount and pin. Check the pin number for matching. If it matches increment the balance and display the balance else display an appropriate message withdraw(self, pin, amount):
Decrement the account balance by accepting the amount and pin. Check the pin number for matching and balance is greater than 1000 and amount is less than the balance. If it matches withdraw the amount and display the balance else display an appropriate message
PROGRAM
using namespace std;
#include<iostream>
#include<iomanip>
struct pno
{
int pin;
float balance;
};
class BankAccount
{
public:
pno pno_obj[10];
void deposit(int pn,float amt)
{
for(int i=0;i<10;i++)
if(pno_obj[i].pin == pn)
{
pno_obj[i].balance = pno_obj[i].balance + amt;
cout<<“\nTransaction successful!
cout<<“\nBalance amount in your account is”<< pno_obj[i].balance; break;
}
void withdraw(int self,int pn,float amt)
{
for(int i=0;i<10;i++)
{
if(pno_obj[i].pin== pn)
{
if (pno_obj[i].balance>1000 && amt < pno_obj[i].balance)
{
pno_obj[i].balance=pno_obj[i]. balance- amt;
cout<<“\nTransaction successful”;
cout<<“\nBalance amount in your account is “<< pno_obj[i].balance; break;
}
} }
}
};
int main( )
{
int pin_no, tamt;
BankAccount b;
// initialization of objects with pin and balance amount as 0
for(int i=0;i<10;i++)
{
b.pno_obj[i].pin=i+l;
}
int choice;
while(choice !=3)
{
cout <<“\n1. Deposit”;
cout <<“\n2.Withdrawal”;
cout<<“\n3.Exit”;
cout<<“\nEnter your choice “; cin >>choice;
switch(choice)
{
case 1:
cout<<“\nEnter PIN
cin>>pin_no;
cout<<“\nEnter Deposit amount”;
cin>>tamt;
b.deposit(pin_no,tamt); break;

case 2:
cout<<“\nEnter PIN cin>>pin_no;
cout<<“\nEnter Withdrawal amount”;
cin>>tamt;
cout<<“\nEnter 1 for Self 2 for Others :”;
int type;
cin>>type;
b.withdraw(type,pin_no,tamt); break;
default: cout<<“\nTransaction completed”;
}
}
}
Output

Question 8.
Define a class Hotel in C++ with the following description:
Answer:
Private Members:
Rno //Data member to store Room No
Name //Data member t store customer name
Charges //Data member to store per day charges
Days //Data member to store number of days of stay
Calculate() //A function to calculate and return Amount as
//Days*Chagres and if the value of Days*Charges is more than 12000 then as 1.2*Days*Charges

Public Members:
Hotel( ) //to initialize the class members
Getinfo( ) //A function to enter the content Rno, Name, Charges //and Days
Showinfo( ) //A function to display Rno, RName, Charges, Days and
//Amount (Amount to displayed by calling function CALCULATE( ))
PROGRAM
using namespace std;
#include<iostream>
#include<string.h>
class Hotel
private:
int Rno,Days,Charges;
char Name[20];
1 int Calculate( )
{
if (Days * Charges >12000)
return (Days * Charges * 1.02);
else
return(Days * Charges);
}
public:
Hotel( )
{
Rno=0;
Days=0;
Charges=0;
strcpy(Name,””);
}
void Getinfo( )
{
cout<<“nEnter customer name :”;
cin>>Name;
cout<<“nEnter charges per day : “;
cin>>Charges;
cout<<“\nEnter Number of days :”;
cin>>Days;
cout<<“\nEnter Room Number :”;
cin>>Rno;
}
void Showinfo( )
{
cout<<“\nRoom Number: “<<Rno;
cout<<“\nCustomer name : “<<Name;
cout<<“\nCharges per day : “<<Charges;
cout<<“\nNumber of days : “<<Days;
cout<<“\nTota! Amount: “<<Calculate( );
}
};
int main( )
{
Hotel obj;
obj.Getinfo( );
obj.Showinfo( )
}
Output

Question 9.
Define a class Exam in C++ with the following description:
Answer:
Private Members:
Rollno – Integer data type
Cname – 25 characters
Mark – Integer data type

public:
Exam(int,char[],int) //to initialize the object ~Exam() // display message “Result will be intimated shortly”
void Display( ) // to display all the details if the mark is above 60 other wise display “Result Withheld”
PROGRAM
using namespace std;
#include<iostream>
#include<string.h>
class Exam
{
private:
int Rollno,Mark;
char Cname[25];
public:
Exam(int r,char n[25],int m)
{
Rollno = r;
Mark = m;
strcpy(Cname,n);
}
~Exam( )
{
cout<<“\n\nResult will be intimated shortly”;
}
void Display( )
{
if (Mark>60)
{
cout<<“\n\nRoll Number : “<<Rollno;
cout<<“\nCandidate name : “<<Cname;
cout<<“\nMark :”<<Mark;
}
else
{
cout<<“\n\nRoll Number : “<<Rollno;
cout<<“\nCandidate name : “<<Cname;
cout<<“\nResult Withheld”;
}
}
};
int main( )
{
Exam obj 1(1011,”SURYA”,78),obj2( 1012,”JOHN”,44);
objl.Display( );
obj2. Display( );
}
Output

Question 10.
Define a class Student in C++ with the following specification:
Answer:
Private Members:
A data member Rno(Registration Number) type long
A data member Cname of type string A data member Agg_marks (Aggregate Marks) of type float
A data member Grade of type char
A member function setGrade () to find the grade as per the aggregate marks obtained by the student. Equivalent aggregate marks range and the respective grade as shown below.
Aggregate Marks -Grade
>=90 – A
Less than 90 and >=75 – B
Less than 75 and >=50 – C
Less than 50 – D
Public members:
A constructor to assign default values to data members:
A copy constructor to store the value in another object
Rno=0, Cname=”N.A” Agg_marks=0,0
A function Getdata ( ) to allow users to enter values for Rno.Cname, Aggjnarks and call functionsetGrade ( ) to find the grade.
A function dispResult( ) to allow user to view the content of all the data members.
A destructor to display the message “END”
PROGRAM .
using namespace std;
#include<iostream>
#include<string.h>
#include<iomanip>
class Student
{
private:
long Rno;
char Cname[25],Grade;
float Agg_marks;
void Setgrade()
{
if (Ag g_ma rks >=90)
Grade = ‘A’;
else if(Agg_marks>=75)
Grade = ‘B’;
else if(Agg_marks>=50)
Grade = ‘C’;.
else
Grade = ‘D’;
}
public:
Student( )
{
Rno = 0;
Agg_marks = 0;
strcpy(Cname,””);
Grade=”;
}
Student(Student &s) .
{
Rno = s.Rno;
Agg_marks = s.Agg_marks;
strcpy(Cname,s.Cname);
Grade=s.Grade;

~Student( )
{
cout<<“\nEND”; >
void Getdata( )
{
cout<<“\nEnter Register Number”;
cin>>Rno;
cout<<“\nEnter Candidate Name
cin>>Cname;
cout<<“\nEnter Aggrigate Mark”;
cin>>Agg_marks;
Setgrade( );
}
void dispResult( )
{
cout<<setw(30)<<“Candidate Register Number
“<<Rno<<endl<<endl;
cout< <setw(30)<< “Candidate Name : “<<Cname<<endk<endI;
cout<<setw(30)<<“Aggrigate Mark : “<<Agg_marks«endk<endl;
cout<<setw(30)<<“Grade :
“<<Grade<<endk<endl;
}
};
int main( )
{
Student s1;
s1.Getdata( );
Student s2(s1);
COut<<“\nFIRST CANDIDATE DETAIL \n\n”; s1.dispResult( );
cout<<“\nSECOND CANDIDATE DETAIL \n\n”;
s2.dispResult( );
}
Output

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 1 The Living World Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 1 The Living World

11th Bio Zoology Guide Living World Text Book Back Questions and Answers

Part I

Question 1.
A living organism is differentiated from a non-living structure based on
a. Reproduction
b. Growth
c. Metabolism
d. All the above
Answer:
d. All the above

Question 2.
A group of organisms having similar traits of a rank is
a. Species
b. Taxon
c. Genus
d. Family
Answer:
b. Taxon

Question 3.
Every unit of classification regardless of its rank is
a. Taxon
b. Variety
c. Species
d. Strain
Answer:
a. Taxon

Question 4.
Which of the following is not present in the same rank?
a. Primata
b. Orthoptera
c. Diptera
d. Insecta
Answer:
a. Primata

Question 5.
What taxonomic aid gives comprehensive information about a taxon?
a. Taxonomic Key
b. Herbarium
c. Flora
d. Monograph
Answer:
a. Taxonomic Key

Question 6.
Who coined the term biodiversity?
a. Walter Rosen
b. AG Tansley
c. Aristotle
d. AP de Candole
Answer:
a. Walter Rosen

Question 7.
Cladogram considers the following characters
a. Physiological and Biochemical
b. Evolutionary and Phylogenetic
c. Taxonomic and systematic
d. None of the above
Answer:
b. Evolutionary and Phylogenetic

Question 8.
The molecular taxonomic tool consists of
a. DNA and RNA
b. Mitochondria and Endoplasmic reticulum
c. Cell wall and Membrane proteins
d. All the above
Answer:
a. DNA and RNA

Question 10.
Differentiate between probiotics and pathogenic bacteria
Answer:

Probiotics	Pathogenic bacteria
1. It converts the milk into curd (Eg.) Lactobacillus	It causes diseases in plants and animals
2. It decomposes debris.	Tomato – bacterial species
3. By the action of fermentation vinegar is produced. (Eg.) Acetobacter	Anthrax, Tuberculosis Pneumonia Tetanus,

Question 11.
Why mule is sterile?
Answer:
Mule gets one set of chromosomes (32) from male parent, horse and one set of chromosomes (31) from female parent, donkey. These two sets of chromosomes do not match with each other and cannot produce gametes by meiosis. Hence mule is sterile in nature.

Question 12.
What is the role of Charles Darwin in relation to the concept of species?
Answer:
Charles Darwin’s book on Origin of Species explains the evolutionary connections of species by the process of natural selection.

Question 13.
Why elephants and other wild animals are entering the human living areas?
Answer:

• For the construction of houses, dams, and factories forests are destroyed. The area surface of forests is also getting reduced.
• As the bull elephant is hunted for their tusks the cow elephant during breeding season enters in to the dwelling area of people.

Question 14.
What is the difference between a Zoo and a wildlife sanctuary?

Zoo	Wildlife Sanctuary
1. They have formed artificially.	It’s a place of nature.
2. Animals are in houses within the enclosure.	Animals roam freely in their natural surrounding.
3. They are formed for the purpose of free time enjoyment of people.	They are not formed for the purpose of enjoyment.

Question 15.
Can we use recent molecular tools to identify and classify organisms?
Answer:
The recent molecular taxonomical tools can be used to identify and classify the organism. The following molecular techniques and approaches are used in molecular tools.

1. DNA barcoding – Short genetic marker in an organism’s DNA to identify whether it belongs to a particular species.
2. DNA hybridization – Measures the degree of genetic similarity between pools of DNA sequences.
3. DNA fingerprinting – to identify an individual from a sample of DNA by looking at unique patterns in their DNA.
4. Restriction Fragment Length Polymorphism (RFLP) Analysis- the difference in homologous DNA sequences can be detected by the presence of fragments of different lengths after digestion of DNA samples.
5. Polymerase chain reaction (PCR) sequencing- to amplify a specific gene or portion of a gene.

Question 16.
Explain the role of Latin and Greek names in Biology.
Answer:
Aristotle (384 to 322 BC) was the first to classify all animals in his Historia Animalium in Latin. He classified the living organisms into plants and animals. Animals were classified as walking (terrestrial), flying (birds), and swimming (aquatic) based on their locomotion.

He classified the animals with red blood cells as Enaima and those without red blood cells as Anaima. Though his method of classification had limitations, his contribution to biology was remarkable. Theophrastus did his research on the classification of plants. He was known as the Father of Botany.

Part II 

11th Bio Zoology Guide The Living World Additional Important Questions and Answers

Question 1.
The smallest taxon among the following is ………. (PMT-94)
(a) class
(b) order
(c) species
(d) genus
Answer:
(c) species

Question 2.
Aristotle has classified organisms based on the following category of locomotion.
a. Walking & bore dwellers
b. Flying & arboreal
c. Swimmers & aquatic
d. All the above.
Answer:
c. Swimmers & aquatic
d. All the above.

Question 3.
Species is
(a) not related to evolution
(b) specific class of evolution
(c) specific unit of evolution
(d) fertile specific unit in the evolutionary history of a race
Answer:
(d) fertile specific unit in the evolutionary history of a race

Question 4.
Whose researchers confirm that species is a fundamental unit of classification.
a. John Ray
b. R.H. Whittaker
c. CarlWoese
d. Cavalier-Smith
Answer:
a. John Ray

Question 5.
A group of plants or animals with similar traits of any rank is kept under ………. (PMT-96)
(a) species
(b) genus
(c) order
(d) taxon
Answer:
(d) taxon

Question 6.
Who has developed binomial nomenclature.
a. Carolus Linnaeus
b. Augustin
c. Aristotle
d. Ernst Haeckel
Answer:
a. Carolous Linnaeus

Question 7.
New systematic and the concept of life was given by ……….. (BHU-98)
(a) Huxley
(b) Odom
(c) Elton
(d) Linnaeus
Answer:
(c) Elton

Question 8.
The three domains classification is based on the difference in the gene.
a. 60s rRNA
b. 70s rRNA
c. l6s rRNA
d. m RNA
Answer:
c. l6s rRNA

Question 9.
Which of the following will form a new species? (PMT-98)
(a) inter breeding
(b) variations
(c) differential reproduction
(d) none of the above
Answer:
(b) variations

Question 10.
Find out the correct sequence by matching.
A. Augustin Pyramus de Candole – Father of Botany
B. Aristotle – Father of Modern Taxonomy
C. Carolus Linnaeus – Father of Taxonomy
D. Theophrastus – Introduces Taxonomy
Answer:
D. Theophrastus – Introduces Taxonomy

Question 11.
Crosses between animals – Match.
A. Male Horse + Female Donkey – Tigon
B. Male Donkey + Female Horse – Tiger
C. Male Lion + Female Tiger – Mule
D. Male Tiger + Female Lion – Hinny
a) A-II, B -1, C – IV, D – III
b) A-IV, B -1, C – II, D – III
c) A-I, B-II, C-III, D-IV
d) A-IV, B-I, C-II, D-III
Answer:
A-II, B -I, C – IV, D – III

Question 12.
In classification, the category below the level of family is ……….. (CET-98)
(a) class
(b) species
(c) phylum
(d) genus
Answer:
(d) genus

Question 13.
Find out the wrong pair
a. Peacock – Pavocristatus
b. Tiger – Pantheratigeris
c. Man -Homosapiens
d. Domestic crow – Salcopopsindica
Answer:
d. Domestic crow – Salcopopsindica

Question 14.
Find the correct match.
1. John ray -a. Five kingdom concept
2. Linnaeus -b. Cladogram
3. Ernest Haeckel -c. Binomial nomenclature
4. R.H. Whittaker – d. Methodus Plantarum
a. 1 -d,2-c,3-b,4-a
b. 1-a,2-b,3-c,4-d
c. 1 – c, 2 – a, 3 – b, 4 – d
d. 1 – d, 2 – c, 3 – a, 4 – b
Answer:
a. 1 -d,2-c,3-b,4-a

Question 15.
Which of the following taxons cover a greater number of organisms?(PMT-2001)
(a) order
(b) family
(c) genus
(d) phylum
Answer:
(d) phylum

Question 16.
Which class of protozoa is totally parasitic? (BHU 1994)
(a) Sporozoa
(b) Mastigophora
(c) Ciliate
(d) Sarcodina
Answer:
(a) Sporozoa

Question 17.
Reproduction in Paramecium is controlled by ………. (BHU 1999).
(a) flagella
(b) cell wall
(c) micronucleus
(d) macronucleus
Answer:
(c) micronucleus

Question 18.
In the life cycle of Plasmodium, exflagellation occurs in ……….. (BHU 2007)
(a) sporozoites
(b) microgametes
(c) macrogametes
(d) signet ring
Answer:
(b) microgametes

Question 19.
Excretion in Amoeba occurs through ………… (DPMT 1997)
(a) lobopodia
(b) plasma membrane
(c) uroid portion
(d) contractile vacuole
Answer:
(d) contractile vacuole

Question 20.
Method of dispersal in Amoeba is ……….. (DPMT 1995)
(a) locomotion
(b) encystment
(c) sporulation
(d) binary fission
Answer:
(b) encystment

(2 marks)

II. Very Short Questions

Question 1.
Define ecosystem.
Answer:
The ecosystem is defined as a community of living organisms (plants and animals), non-living things (minerals, climate, soil, sunlight, and water), and their interrelationships, e.g. Forest and grassland.

Question 2.
What are the unique characteristic features of living organisms?
Answer:

• Cellular organization
• Nutrition
• Respiration
• Metabolism
• Movement
• Reproduction
• Excretion
• Homeostasis

Question 3.
The mating between different species produces sterile offsprings.
Answer:
The maternal and paternal chromosomes of the offsprings produced by the mating between different species are not identical and hence gametes are not produced by meiotic division.

Question 4.
What is the need for classification?
Answer:
The basic need for classification are:

• To identify and differentiate closely related species.
• To know the variation among the species.
• To understand the evolution of the species.
• To create a phylogenetic tree among the different groups.
• To conveniently study living organisms.

Question 5.
Why are molecular tools used now to study taxonomy?
Answer:
Molecular tools are accurate and authentic. Hence they are used to study taxonomy.

Question 6.
What are the features of systematics?
Answer:

• Identifying, describing, naming, arranging, preserving, and documenting the organisms.
• Investigating the evolutionary history of the species, their adaptations to the environment, and the interrelationship among species.

Question 7.
What is the phylogenetic tree?
Answer:
It’s a method of representing evolutionary relationships with the help of a tree diagram known as a cladogram.

Question 8.
What are the limitations of Aristotle’s classification?
Answer:
Many organisms were not fitting into his classification. Frogs have lungs and they are amphibians while their larva, the tadpole is aquatic and respires through gills. It is difficult to classify frogs according to his method. All flying organisms such as birds, bats, flying insects were grouped together. Ostrich, emu, and penguin are flightless birds and hence they cannot be classified by his method.

Question 9.
What are the three domains of life indicate?
Answer:
This system emphasizes the separation of prokaryotes into two domains.

Question 10.
What is numerical taxonomy?
Answer:
The evaluation of resemblances and differences of organisms through statistical methods followed by computer analysis to establish the numerical degree of relationship among them is known as numerical taxonomy.

Question 11.
What is the seven taxonomic hierarchy?
Answer:

1. Kingdom
2. Phyla
3. Class
4. Order
5. Family
6. Genus
7. Species

Question 12.
Define species?
Answer:
It is a group of animals having similar morphological features and is reproductively isolated to produce fertile offspring.

Question 13.
What is cladistic classification?
Answer:
Cladistic classification is the method of classifying organisms based on genetic differences among all species in a phylogenetic tree.

Question 14.
Define order?
Answer:
Order is an assemblage of one or more related families which show few common features. (Eg) Family Candiae and Felidae are placed in the order Carnivora.

Question 15.
Define class.
Answer:
Class includes one or more related orders with some common characters.

Question 16.
Define Phylum.
Answer:
The group of classes with similar distinctive characteristics constitute phylum.

Question 17.
Define animal kingdom.
Answer:
All living animals belonging to various phyla are included in the kingdom.

Question 18.
What are the features that we have to keep in mind in naming them scientifically?
Answer:

• Morphology
• Genetic information
• Habitat
• Feeding pattern
• Adaptations
• Evolutions

Question 19.
Define species.
Answer:
A species is a group of organisms that have similar morphological and physiological characters which can interbreed to produce fertile offsprings.

Question 20.
What are taxonomical keys?
Answer:
Keys are based on a comparative analysis of the similarities and dissimilarities of organisms. There are separate keys for different taxonomic categories.

Question 21.
Distinguish between species and genus.
Answer:
Species:

• A species is a group of interbreeding populations having similar characters.
• It is the basic unit of classification e.g. Felis Domestica, genus species.

Genus:

• Genus is a group of related species.
• It is the second level in classification, e.g., Felis margarita genus species.

Question 22.
Define Zoological parks.
Answer:
These are places where wild animals are kept in protected environments under human care.
It enables us to study their food habits and behaviour.

Question 23.
What is binomial system of nomenclature?
Answer:
The system of naming the organism with two names, generic name, and specific (species) name is known as the binomial system of nomenclature, e.g. Pavo cristatus – Indian pea fowl.

Question 24.
What is printed taxonomical tools?
Answer:

• Identification cards
• Description
• Field guides
• Manuals

Question 25.
What is the phylogenetic tree?
Answer:
It is the inferred evolutionary relationships upon similarities and differences in their physical or genetic characters.

Question 26.
What is trinomial system of nomenclature?
Answer:
The system of naming the organism with three names, generic name, specific name (species), and sub-species name is known as the trinomial system of nomenclature, e.g. Corvus splendens -Indian house crow.

Question 27.
What are shared characters?
Answer:
A shared character is one that two lineages have in common.

Question 28.
What are derived characters?
Answer:
Derived character is one that evolved in the lineage leading up to a clade.

Question 29.
Vandaloor Zoological park.
Answer:

• It is situated in the South-Western Part of Chennai.
• It spreads over an area of 1500 acres.
• It is one of the largest zoological parks in India.
• The Zoo houses 2553 species of both flora and fauna.

(3 marks)

III. Short Questions

Question 1.
Define ecosystem.
Answer:
The ecosystem is defined as a community of living organisms (plants and animals), non-living things (minerals, climate, soil, sunlight, and water), and their interrelationships, e.g. Forest and grassland.

Question 2.
On which criteria the systematic classification is done?
Answer:

• Evolutionary history.
• Environmental adaptations.
• Environmental relationship.
• The interrelationship between species.

Question 3.
Give an account of Aristotle’s classification?
Answer:

•  In his book ‘History of Animals,’ he classifies plants and animals into two categories.
• Based on locomotion walking, flying, swimming,
• He classifies the organisms on the basis of blood.
• He classifies the animals into two as ‘Enaima’ with blood and those without blood as’ Anaima’

Question 4.
Who has developed the five-kingdom classification?
Answer:

1. R.H. Whittaker proposed the five-kingdom classification.
2. It is based on cell structure.
3. Mode of nutrition.
4. Mode of reproduction.
5. Phylogenetic relationships.

The kingdoms are

• Monera
• Protista
• Fungi
• Plantae
• Animalia

Question 5.
What are the special features of frogs that are identified in Western Gauts?
Answer:

• This frog has shiny purple skin.
• There is a light blue ring around the eyes.
• It has a pointy big nose.
• It’s Zoological name Nasikabatrachus Bhupathi.

Question 6.
What are the rules to be followed in the nomenclature of organisms?
Answer:
The scientific name should be italicized in printed form and the generic name and specific name should be underlined separately if it is handwritten.

• The first alphabet of the generic name should be of uppercase.
• The specific name (species) should be in lower case letters.
• The name or abbreviated name of the scientist who first published the scientific name may be written after the specific (species) name along with the year of publication, e.g. Felis Leo Linn., 1958.
• If the specific (species) name is framed after any person’s name, the name of the species shall end with i, ii or ae. e.g. Ground – dwelling lizard Cyrtodactylus varadgiri.

Question 7.
Name some Automated species identification tools or cyber tools.
Answer:
ALIS: Automated Leafhopper Identification System.
DAISY: Digital Automated Identification System.
ABIS: Automatic Bee Identification System.
SPIDA: Species Identified Automatically (spiders, wasp, bee wing characters).
Draw wing: Honey bee wing identification.

 (5 marks)

V. Essay Questions

Question 1.
List the defects of Aristotle’s classification.
Answer:

• Aristotle’s classification system had limitations and many organisms were not fitting into his classification.
• The tadpoles of frogs are born in water and have gills but when they metamorphosed into adult frogs they have lungs and can live both in water and on land. There is no answer to this question.
• Based on locomotion birds bats and flying insects were grouped either just by observing one single characteristic feature the flying ability.
• On the contrary to the above-said example, the ostrich emu and penguin are all birds but cannot fly. He did not classify them as birds.

Question 2.
What is special about the Domain Archaea?
Answer:

• This domain includes single-celled organisms the prokaryotes.
• They have the ability to grow in extreme conditions like volcano vents hot springs and polar ice caps hence are called extremophiles.
• They are capable of synthesizing their food without sunlight and oxygen by utilizing hydrogen sulphide and other chemicals from the volcanic vents.
• Some of them produced methane.
• Few live in salty environments and called Halophiles.
• Some thrive in acidic environments and are called thormoacidophiles.

Question 3.
What is special about the domain bacteria?
Answer:

1. Bacterias are prokaryotic.
2. They do not have a definite nucleus and do not have histones.
3. They have circular DNA.
4. They do not possess membrane-bound organelles except for 70s ribosomes.
5. Their cell wall contains peptidoglycans.
6. Many are decomposers. Some are photo-synthesizers and few cause diseases.
7. There are beneficial probiotic bacteria. (Eg.) Cyanobacteria produce oxygen.

Question 4.
What is the significance of cladistic classification?
Answer:
Cladistic classification takes into account ancestral characters (traits commons for the entire group) and derived characters (traits whose structure and function differ from the ancestral characters). The accumulation of derived characters resulted in the formation of new subspecies.

Question 5.
What are the basic roles to be followed in naming the animals?
Answer:

• The scientific name should be italicized in printed form and if handwritten it should be underlined separately.
• The generic name’s first alphabet should be in uppercase.
• The specific name should be in lower case.
• The scientific names of any two organisms are not similar.
• The name of the scientist who first publishes the scientific name may be written after the species name along with the year of publication.
• (Eg.) Lion – Felis Leo Linn. 1758 (or) Felis Leo L. 1758

Question 6.
What are the rules to be followed in the nomenclature of organisms?
Answer:
The scientific name should be italicized in printed form and the generic name and specific name should be underlined separately if it is handwritten.

• The first alphabet of the generic name should be of uppercase.
• The specific name (species) should be in lower case letters.
• The name or abbreviated name of the scientist who first published the scientific name may be written after the specific (species) name along with the year of publication, e.g. Felis Leo Linn., 1958.
• If the specific (species) name is framed after any person’s name, the name of the species shall end with i, ii, or ae. e.g. Ground – dwelling lizard Cyrtodactylus varadgirii.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 12 Trends in Economic Zoology Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter Chapter 12 Trends in Economic Zoology

11th Bio Zoology Guide Trends in Economic Zoology Text Book Back Questions and Answers

Part I

Question 1.
Which one of the following is not related to vermiculture?
a) Maintains soil fertility
b) Break down of inorganic matter
c) Gives porosity, aeration and moister holding capacity
d) Degradation of non biodegradable solid waste

a) a and b is correct
b) c and d is correct
c) b and d is not correct
d) a and c is not correct
Answer:
c) b and d is not correct

Question 2.
Which one of the following is not an endemic species of earthworm?
a) Perionyx
b) Lampito
c) Eudrillus
d) Octochaetona
Answer:
c) Eudrillus

Question 3.
Match the following
1) Bombyxmori-
a) Champa – I) Muga
2) Antheraeaassamensis
b) Mulberry – II) Eri
3) Antheraeamylitta-
c) Arjun – III) Tassar
4) Attacus ricini –
d) Castor – IV) Mulberry

Select the correct one
a) 1-b-IV
b) 2-a-I
c) 3-c -III
d) 4-d-II
Answer:
c) 3-c -III

Question 4.
Silk is obtained from
a) Laccifer lacca
b) Nosema bombycis
c) Attacus ricird
d) Attacus mylitta
Answer:
c) Attacus ricird

Question 5.
Assertion: Nuptial flight is a unique flight taken by the queen bee followed by serveral drones.
Reason: The queen bee produces a chemical substance called pheromone. The drones in that area are attracted to the pheromone and then mating takes place.
a) Assertion and reason is correct but not related
b) Assertion and reason is incorrect but related
c) Assertion and reason is correct but related
d) Assertion and reason is incorrect but not related
Answer:
c) Assertion and reason is correct but related

Question 6.
Rearing of honey bee is called
a) Sericulture
b) Lac culture
c) Vermiculture
d) Apiculture
Answer:
d) Apiculture

Question 7.
Which of the statement regarding Lac insect
a) A microscopic, resinous crawling scale insect
b) Inserts its proboscis into plant tissue suck juices and grows
c) Secretes lac from the hind end of body.
d) The male lac insect is responsible for large scale production of lac.
Answer:
d) The male lac insect is responsible for large scale production of lac.

Question 8.
Aquaponics is a technique which is
a) A combination of aquaculture and fish culture
b) A combination of aquaculture and hydroponics
c) A combination of vermiculture and hydroponics
d) A combination of aquaculture and prawn culture
Answer:
b) A combination of aquaculture and hydroponics

Question 9.
Prawn belongs to the class
a) Crustacea
b) Annelida
c) Coelenterata
d) Echinodermata
Answer:
a) Crustacea

Question 10.
Pearl oyster belongs to the class
a) Gastropoda
b) Cephalopoda
c) Scaphapoda
d) Pelecypoda
Answer:
d) Pelecypoda

Question 11.
Inland fisheries are
a) deep sea fishing
b) capturing fishes from sea coast
c) Raising and capturing fishes in fresh water
d) oil extraction from fish
Answer:
c) Raising and capturing fishes in fresh water

Question 12.
Induced breeding technique is used in
a) Marine fishery
b) Capture fishery
c) Culture fishery
d) Inland fishery
Answer:
d) Inland fishery

Question 13.
Isinglass is used in
a) Preparation
b) Clearing of wines
c) Distillation of wines
d) Preservation of wines
Answer:
b) Clearing of wines

Question 14.
Animal husbandry is the science of rearing, feeding and caring, breeding and disease control of animals. It ensures supply of proper nutrition to our growing population through activities like increased production and improvement of animal products like milk, eggs, meat, honey etc.
a) Poultry production depends upon the photoperiod. Discuss
b) Polyculture of fishes is of great importance.
Answer:
(a) Light is an important aspect of poultry production. Light stimulates the secretion of FSH and LH. A wavelength between 400 and 700 nm is required. The decrease in the photoperiod will affect egg production.

(b) A few selected fishes belonging to different species are stocked together in proper proportion in a pond. This mixed farming is termed composite fish farming or polyculture. It is of great importance because

1. All available riches are fully utilized
2. Compatible species do not harm each other
3. There is no competition among different species

Question 15.
Assertion: The best quality of pearl is known as lingha pearl and obtained from marine oysters.
Reason: Nacre is secreted continuously by the epithelial layer of the mantle and deposited around the foreign particle
a) Assertion is true Reason is false
b) Assertion and Reason is false
c) Asserion is false but Reason is true
d) Assertion and Reason are true
Answer:
d) Assertion and Reason are true

Question 16.
Choose the correctly matched pair
a) Egg layers-Brahma
c) Dual purpose-White Plymouth rock
b) Broiler types -Leghorn
d) Ornamental breeds -Silkie
Answer:
d) Ornamental breeds -Silkie

Question 17.
Write the advantages of vermicomposting.
Answer:

1. Vermicomposting provides excellent organic manure for sustainable agro-practices.
2. Marketing of vermicompost can provide a supplementary income.
3. Vermicompost is rich in essential plant nutrients.
4. It improves soil structure, texture, aeration, and water holding capacity and prevents soil erosion.
5. It is rich in nutrients and an eco-friendly amendment to soil for farming and terrace gardening,
6. It enhances seed germination and ensures good plant growth.

Question 18.
Name the three castes in a honey bee colony.
Answer:

1. Queen bee
2. Drone
3. Worker bee

Question 19.
Name the following

1. The largest bee in the colony:
2. The kind of flight which the new virgin queen takes along with the drones out of the hive:

Answer:

1. The queen.
2. Nuptial flight.

Question 20.
What are the main duties of a worker bee?
Answer:
Each worker has to perform different types of work in her lifetime. During the first half of her life, she becomes a nurse bee attending to indoor duties such as secretion of royal jelly, prepares bee- bread to feed the larvae, feeds the queen, takes care of the queen and drones, secretes beeswax, builds combs, cleans and fans the beehive. Then she becomes a soldier and guards the beehive. In the second half of her life lasting for three weeks, she searches and gathers the pollen, nectar, propolis, and water.

Question 21.
What happens to the drones after the mating flight?
Answer:
They die after copulation.

Question 22.
Give the economic importance of silkworm?
Answer:

1. The rearing of silkworm on a commercial scale is called sericulture.
2. It is an agro-based industry comprising of
   • Cultivation of food plants for the silkworms.
   • Rearing of silkworms.
   • Reeling and spinning of silk.
3. Silk fibers are utilized in preparing silk clothes.
4. Silk is used in industries and for military purposes.
5. Silk is used in the manufacture of fishing fibres, parachutes, cartridge bags, insulation coils for telephone, wireless receivers, tyres of racing cars, filter fibres, in medical dressings and suture materials.

Question 23.
What are the nutritive values of fishes?
Answer:
Economic importance of fish:-
Fishes form a rich source of protein food and provide a good staple food to tide over the nutritional needs of man. Fish species such as sardines, mackerel, tuna, herrings have high amino acids concentration particularly histidine which is responsible for the meaty flavor of the flesh. It is rich in fat such as omega 3 fatty acids. Minerals such as calcium, magnesium, phosphorus, potassium, iron, manganese, iodine and copper.

Some of the fish by-products are: Fish oil is the most important fish by-product. It is derived from fish liver and from the fish body.

Fish liver oil is derived from the liver which is rich in vitamin A and D, whereas fish body oil has high content of iodine, not suitable for human consumption, but is used in the manufacture of laundry soaps, paints and cosmetics. Fish meal is prepared from fish waste after extracting oil from the fish.

The dried wastes are used to prepare food for pig, poultry and cattle. The wastes obtained during the preparation of fish meal are widely used as manure.

Isinglass is high-grade collagen produced from dried air bladder or swim bladder of certain fishes viz. catfish and carps. The processed bladder which is dissolved in hot water forms gelatin having adhesive property. It is primarily used for clarification of wine, beer and vinegar.

Question 24.
Give the economic importance of prawn fishery?
Answer:
The flesh of prawns is palatable and rich in glycogen, a protein with low-fat content.

Question 25.
Give the economic importance of lac insect.
Answer:
Economic importance of Lac:

1. Lac is largely used as a sealing wax and adhesive for optical instruments.
2. It is used in the electric industry, as it is a good insulator.
3. It is used in preparations of shoe and leather polishes and as a protective coating of wood.
4. It is used in laminating paper board, photographs, engraved materials and plastic moulded articles.
5. Used as a filling material for gold ornaments.

Question 26.
List any three common uses of shellac.
Answer:

1. Shellac with denatured alcohol is used to remove dust on the walls.
2. Coating of metals with shellac prevents rusting.
3. Shellac coating on citrus fruits increases their shelf life.

Question 27.
Name any two trees on which lac insect grows?
Answer:

1. Acacia catechu
2. Acacia nilotica

Question 28.
What is seed lac?
Answer:
The lac after grinding the dust particles are removed. The resultant lac is called ‘seed lac’.

Question 29.
Define cross-breeding?
Answer:
Breeding between a superior male of one breed with a superior female of another breed is known as cross-breeding.

Question 30.
What are the advantages of artificial insemination?
Answer:
Advantages of artificial insemination:

1. It increases the rate of conception
2. It avoids genital diseases
3. Semen can be collected from injured bulls which have desirable traits
4. Superior animals located apart can be bred successfully.

Question 31.
Discuss the various techniques adopted in cattle breeding?
Answer:
There are two methods of animal breeding, namely inbreeding and outbreeding.
1. Inbreeding:
Breeding between animals of the same breed for 4-6 generations is called inbreeding.

2. Outbreeding:
The breeding between unrelated animals is called outbreeding. It is done in three ways;

• Outcrossing: It is the breeding between unrelated animals of the same breed but having no common ancestry. The offspring of such a cross is called an outcross.
• Crossbreeding: Breeding between a superior male of one breed with a superior female of another breed. The cross-bred progeny has superior traits (hybrid vigour or heterosis).
• Interspecific hybridization: In this method of breeding mating is between male and females of two different species.

Question 32.
Mention the advantages of MOET.
Answer:
Multiple Ovulation Embryo Transfer Technology (MOET) is a method of propagation of animals with desirable traits. This technology is used to produce high milk-yielding females and high-quality meat-yielding bulls in a short time.

Question 33.
Write the peculiar characters of the duck.
Answer:
The peculiarity of ducks:
The ducks body is fully covered with oily feathers. They have a layer of fat under their skin which prevents it from getting wet. They lay eggs at night or in the morning. The ducks feed on rice bran, kitchen wastes, waste fish and snails.

Part – II

11th Bio Zoology Guide Trends in Economic Zoology Additional Important Questions and Answers

I. Choose The Best Options

Question 1.
During the process of vermiculture which of the following does not happen?
(a) Decomposition of organic food waste
(b) Supplying nutrients to the soil
(c) Use of earthworms in the process
(d) Synthesis of organic substances
Answer:
(d) Synthesis of organic substances

Question 2.
Who has proposed the term vermi tech?
a) Sultan Ismail
b) Darwin
c) Linnaeus
d) Albert William
Answer:
a) Sultan Ismail

Question 3.
The technology of composting and bioremediation of soils and other activities by application of earthworm is called? …………………
(a) Vermitech
(b) Vermicast
(c) Vermicompost
(d) Vermiculture
Answer:
(a) Vermitech

Question 4.
Which of the following is the earthworm of the native soil?
a) Periyonx Eisenia
b) Eudrilus eugeniae lampito mauritii
c) Periyonyx excavatus, octohaetona
d) all the above
Answer:
c) Periyonyx excavates, octohaetona

Question 5.
Which of the following is the exotic species used for vermicomposting?
(a) Eudrilus eugeniae
(b) Lampito mauritii
(c) Periyonyx excavatus
(d) Octochaetona serrata
Answer:
(a) Eudrilus eugeniae

Question 6.
Who has known the utility of silk even before 3000 years?
a) Europe
b) African’s
c) China
d) India
Answer:
c) China

Question 7.
Attacus ricini produces ……………. silk?
(a) Muga
(b) Tassar
(c) Mulberry
(d) Eri
Answer:
(d) Eri

Question 8.
Match and find the correct sequences.
I. Bombyx mori – a) Erisilk
II. Antheraea assamensis – b) Tisaarsilk
III. Antheraea mylitta – c) Mugasilk
IV. Attacus ricini – d) Mulberry silk

Answer:
a) I – d, II – c, III – b, IV – a

Question 9.
Match the silk moth with their respective leaves.
I. Bombyx mori – a) Castor
II. Antheraea assamensis – b) Arjun
III. Antheraea mylitta – c) Champa
IV. Attacus ricini – d) Mulberry


Answer:
a) I -d, II – c, III – b, IV – a

Question 10.
Muscardine is a- disease of silkworms caused by a …………………..
(a) Bacterium
(b) Virus
(c) Protozoan
(d) Fungus
Answer:
(d) Fungus

Question 11.
Name the branches that involve agriculture and industry.
a) Vermiculture
b) Pisciculture
c) Sericulture
d) Poultry
Answer:
c) Sericulture

Question 12.
Life span of Bombyxmori
a) 2-3 days
b) 2 – 4 days
c) 2-5 days
d) 4 – 7 days
Answer:
a) 2 – 3 days

Question 13.
What is the number of eggs laid by Bombyx mori once?
a) 500 – 800
b) 400 – 500
c) 400 – 450
d) 600-700
Answer:
b) 400 – 500

Question 14.
What is the length of matured silkworm?
a) 7.5 cm
b) 7cm
c) 8.5 cm
d) 8 cm
Answer:
a) 7.5 cm

Question 15.
What would be the length of silk fibre in a cocoon?
a) 1000-1100 m
b) 1000-1200 m
c) 1000-1500m
d) 1000 – 1300m
Answer:
b) 1000 – 1200m

Question 16.
What is meant by mariculture?
a) Rearing of mulberry plants
b) Rearing of castor plants
c) Vermiculture
d) Apiculture
Answer:
a) Rearing of mulberry plants

Question 17.
What is meant by voltinism?
a) No of broods raised
b) Cocoon
c) Spinneret
d) None of the above
Answer:
a) No of broods raised

Question 18.
What is the ideal period for mulberry culture?
a) January, February, and November, December
b) March, April and June, July
c) June, July, and Nov, December
d) October, November and December, January
Answer:
c) June, July, and November, December

Question 19.
Find the odd one out.
a) Healthy silk moths are allowed to mate for 4 hours
b) Female lays about 400 eggs in 24 hours
c) The small larvae hatch between 7-10 days after incubation
d) The optimum temperature for rearing silk moth is 25°C – 30°C
Answer:
d) The optimum temperature for rearing silk moth is 25°C – 30°C

Question 20.
Silkmoth matured in about days
a) 40 days
b) 45 days
c) 50 days
d) 35 days
Answer:
b) 4 5 days

Question 21.
The cocoon is soaked in hot water at a temperature for minutes.
a) 95°C – 97°C -10 -15 minute
b) 95°C – 97°C – 5 -10 minutes
c) 90°C – 95°C – 7 -10 minutes
d) 100°C – 105°C – 2 – 5 minutes
Answer:
a) 95°C – 97°C -10 -15 minute

Question 22.
The areas where a lot of beehives can be placed.
a) Honey culture
b) Apiculture
c) Apiaries
d) all the above
Answer:
c) apiaries

Question 23.
The nectar got from the flowers is converted into honey by the enzyme.
a) Invertase
b) Zymase
c) Lipase
d) None of the above
Answer:
a) Invertase

Question 24.
Find whether the following is true or false and find the correct sequence.
I) There is only one queen bee in a colony
II) Worker bees are around 5000 – 10000
III) The queen bee in its lifetime in 2 – 4 years lays 20 lakhs eggs.
IV) When the queen loses its fertility another worker bee becomes a queen by taking royal Jelly.
a) I – true, II – False, III – False, IV – true
b) I – true, II – False, III – true, IV – True
c) I – false, II – true, III – true, IV – true
d) I – false, II – true, III – false, IV – false
Answer:
a) I – true, II – false, III -false, IV – true.

Question 25.
How many days are needed for the worker bee to becoming mature done?
a) 15 days
b) 21 days
c) 20 days
d) 18 days
Answer:
b) 21 days

Question 26.
What is the life span of worker bees?
a) 6 weeks
b) 5 weeks
c) 7 weeks
d) 3 weeks
Answer:
a) 6 weeks

Question 27.
What are drones?
a) a male bee from unfertilised egg
b) a female bee from fertilised egg
c) a queen bee from fertilised egg
d) a worker bee after eating royal jelly becomes a queen
Answer:
a) a male bee from unfertilised egg

Question 28.
Find the wrong pair?
a) Drone cell – Comb queen bee
b) King of honeycomb – Drone
c) Swarming – Worker bee flying with the queen bee
d) Male bee – After mating it will die
Answer:
a) Drone cell – Comb queen bee

Question 29.
Read the following statement and find the correct sequence.
I) The honeycomb is built from the abdominal secretion of a worker bee
II) The chamber of the honeycomb are hexagonal in shape
III) The young stages of honey bees accommodate the lower and central cells of the hive
IV) In Apis dorsata the brood of hive cells are similar in size and shape
a) I – false, II – false, III – true, IV – true
b) I – true, II – true, III – false, IV – false
c) I – true, II – true, III – true, IV – true
d) I – false, II – true, III – true, IV – false
Answer:
c) I – true, II – true, III – true, IV – true

Question 30.
Assertion A: The pure wax is white in colour.
Reason: B The pure wax contains carotenoids pigments.
a) Assertion A true Reason B true
b) Assertion A false, Reason B false
c) Assertion A true Reason B false
d) Assertion A false, Reason B true
Answer:
c) Assertion A true Reason B false.

Question 31.
Find the correct sequence in the following.
I) Bee wax is secreted by the abdomen of the worker bees.
II) The resinous chemical substance present in the wax is called propolis.
III) The pure wax is white in colour.
IV) The yellow of the wax is due to the carotenoid pigments.
a) I – true, II – true, III – true, IV – false
b) I – false, II – false, III – true, IV – true
c) I – true, II – false, III – true, IV – true
d) I – true, II – true, III – true, IV – true
Answer:
d) I – true, II – true, III – true, IV – true

Question 32.
Find whether the statement regarding Lac insect is true or false
I) Tachardia lacca produces lac.
II) The Lac insect is a parasite on host plant karanagalli, karuvelai
III) The female lac insects are small
IV) The female lac insects after fertilisation lays about 200 to 500 eggs
a) I – true, II – false, III – false, IV – true
b) I – true, II – true, III – true, IV – false
c) I – true, II – true, III – false, IV – true
d) I – true, II – false, III – true, IV – false
Answer:
c) I – true, II – true, III – false, IV – true

Question 33.
What is the ideal salinity for fish culture in brackish water?
a) 0.5-30ppt
b) 0.1-25ppt
c) 0.1-30ppt
d) 1-10ppt
Answer:
a) 0.5 – 30 ppt

Question 34.
Which one of the following is the correct pair?
a) Exotic breed – Cyprinus carpiols
b) Apiculture – Reeling
c) Seri culture – Propolis
d) Milch breed – Malvi
Answer:
a) Exotic breed – Cyprinus Carpio

Question 35.
Name the exotic fishes of India.
a) Catlacatla
b) Lapeorohita
c) Cirrhina mirgala
d) Cyprinus carbeo
Answer:
d) Cyprinus carbeo

Question 36.
Name the vitamin found in fish oil?
a) A and E
b) A and D
c) A and C
d) A and K
Answer:
b) A and D

Question 37.
Ising glass is received from the part of the fish
a) Dried gills
b) Dried stomach
c) Dried air sacs
d) Dried liver
Answer:
c) Dried air sacs

Question 38.
What is the nutritive value of crustaceans?
a) Rich in glycogen protein with low-fat content
b) Rich in protein low in glycogen
c) Rich in glycogen low in fat content
d) Rich in protein and fat
Answer:
a) Rich in glycogen protein with low-fat content

Question 39.
These are freshwater prawns.
a) Penaeus indicus
b) Macrobrachium rosenbergi
c) Penaeus monodon
d) Metapenaeus
Answer:
b) Macrobrochium rosenbergi

Question 40.
What is the optimum temperature and PH for prawn culture in a hatching tank
a) 24°C – 30°C PH – 7 – 8
b) 20°C – 22°C PH -10 -12
c) 20°C – 23°C PH – 5 – 6
d) 19°C – 20°C PH – 9 -10
Answer:
a) 24°C – 30°C – PH – 7 – 8

Question 41.
Find the correct statement regarding the preparations of prawn form.
I) For algal growth, and for the subsequent stocking of prawns, it is essential to drain off the water and sundry the bottom.
II) Lime should be applied to absorb excess C02 and to supply calcium which is required for moulting.
III) Fertilizers like rice, bran, poultry, and cattle dung are used to increase the fertility of the soil.
IV) Preservation of prawns is done by peeling and deveining or by cooking and peeling
a) I – true, II – true, III – false, IV – false
b) I – false, II – true, III – true, IV – true
c) I – true, II – true, III – true, IV – true
d) I – true, II – false, III – true, IV – false
Answer:
c) I – true, II – true, III -true, IV – true

Question 42.
Where is pearl cultured in India for the first time?
a) Thoothukudi 1973
b) Visahapattinum 1974
c) Rameshwaram 1975
d) Kulchall973
Answer:
a) Thoothukudi 1973

Question 43.
Where are pearl oysters seen?
a) Kanyakumari coastal region a Bay of Kutch
b) Rameshwaram coastal area and Bay of Mannar
c) Nagapatinum coastal area and Thoothukudi
d) Visahapatinum coastal area and Chennai
Answer:
a) Kanyakumari coastal region a Bay of Kutch

Question 44.
The pearl oysters belong to the ‘L’ genus produce Quality pearls.
a) Gastropoda
b) Pinctada
c) Pelecypoda
d) Cephalopoda
Answer:
b) Pinctada

Question 45.
……………………….. ulturing pearl in freshwater.
a) Lamellidens
b) Mytilus
c) Loligo
d) Dentalium
Answer:
a) Lamellidens

Question 46.
The marine oysters are composed of
a) Calcium carbonate
b) Sodium carbonate
c) Potassium carbonate
d) Magnesium carbonate
Answer:
a) Calcium carbonate

Question 47.
The process of killing the silkworm cocoons is called
a) Reeling
b) Stifling
c) Spinning
d) rearing
Answer:
b) Stifling

Question 48.
Assertion A: The pearl oysters got from the sea are valuable
Reason B: The pearl oysters got from freshwater is not valuable
a) Assertion A true, B false
b) Assertion A and B reason all true
c) Assertion A false, Reason B true
d) Assertion A and Bare false
Answer:
b) Assertion A and B reason all true

Question 49.
What is the name of the breed that produces mule?
a) Outbreeding
b) Crossbreeding
c) Interspecific hybridization
d) Outbreeding
Answer:
c) Interspecific hybridization

Question 50.
Assertion X: 6 – 8 eggs can be produced by induction in an artificial method
Reason Y: The embryos at 8 – 32 celled stages are transferred to a surrogate mother.
a) Assertion x false y true
b) Assertion X true, Y false
c) Assertion X false, Y false
d) Assertion X and Y are true
Answer:
d) Assertion X and Y are true

Question 51.
Which of the cattle breed yields more milk than what they eat?
a) Vechur
b) Kankeyem
c) Gir
d) Ongole
Answer:
a) Vechur

Question 52.
Which of the following is not a milch breed?
a) Sindhi
b) Malvi
c) Jersey
d) Gir
Answer:
b) Malvi

Question 53.
Which of the following belongs to America?
a) Silkie
b) White Plymouth rock
c) Chittagong
d) Aseel
Answer:
b) White Plymouth rock

Question 54.
Name the poultry which is noted for its pugnacity?
a) Leghorn
b) Silkie
c) Brahma
d) Aseel
Answer:
d) Aseel

Question 55.
Find the odd one out
a) Leghorn – Italy
b) Chittagong – good egg yielder
c) White Plymouth rock – American breed
d) Aseel-ornamental breed
Answer:
d) Aseel – ornamental breed

Question 56.
Which among the following is a wild duck?
a) Syhlet
b) Muscori
c) Pekin
d) Mallard
Answer:
d) Mallard

Question 57.
Find the wrong statement about Duck.
a) The body is covered with oily feathers.
b) The fat layer beneath their skin prevents it from getting wet
c) They lay eggs at mid-day
d) The ducks feed on rice bran kitchen wastes and snails
Answer:
c) They lay eggs at midday

(2 marks)

II. Very Short Answer

Question 1.
What is Economic Zoology?
Answer:
Economic zoology is a branch of Science that deals with economically useful animals. It involves the study of the application of animals for human welfare.

Question 2.
How are animals classified on the basis of their economic importance?
Answer:

• Animals for food and food products.
• Economically beneficial animals.
• Animals of aesthetic importance.
• Animals for scientific research.

Question 3.
Name the endemic earthworms of India.
Answer:

1. Periy only x excavatus
2. Lampito Mauritius
3. Octo chactona serrata

Question 4.
Name the exotic species of an earthworm?
Answer:

• Eiseniafetida,
• Eudrilus eugeniae

Question 5.
Why are earthworms called ‘friends of farmers’?
Answer:
Earthworms play a vital role in maintaining soil fertility. Hence, they are called ‘friends of farmers’.

Question 6.
What decides the economic success of the industries?
Answer:

• It depends on the animals and their products.
• It depends on the proper production management and development of the next generation of farm animals.

Question 7.
What is meant by vermiculture?
Answer:
It is the process of using earthworm to decompose organic food waste into a nutrient-rich material capable of supplying necessary nutrients which help to sustain plant growth.

Question 8.
What is meant by vermicast?
Answer:
The organic matter of soil is decomposed by earthworm and becomes nutritious rich manure for plant growth.

Question 9.
What is meant by vermicompost?
Answer:
Vermicompost is composed of vermicast which contains nutrients plant growth promoters and organic matters.

Question 10.
What is meant by vermitech?
Answer:
The application of earthworm in the technology of composting and bioremediation of soil and other activities is called vermitech.

Question 11.
What are the pests of earthworms?
Answer:
Ants, springtails, centipedes, slugs, mites, certain beetle larvae, birds, rats, snakes, mice, toads and other insects or animals which feed on worms.

Question 12.
What are the special characters of vermicompost?
Answer:

• Aeration
• draining of water
• retains water.

Question 13.
What is meant by vermiwash?
Answer:

• Vermi wash is a liquid collected after the passage of water through a column of vermibed.
• It is useful as a foliar spray to enhance plant growth and field.

Question 14.
What is sericulture?
Answer:
Sericulture is an agro-based industry which denotes the commercial production of silk through silkworm rearing.

Question 15.
What is meant by sericulture?
Answer:
Sericulture denotes the commercial production of silk through silkworm rearing.

Question 16.
What is meant by mariculture?
Answer:
The cultivation of mulberry is called as Moriculture.

Question 17.
Which is the suitable period for mulberry cultivation?
Answer:
June, July, November, and December

Question 18.
What are the stages involved in the rearing process of silkworms?
Answer:
Disinfection of rearing house Incubation of eggs Brushing, young larval rearing Late age larval rearing.

Question 19.
What is meant by stifling?
Answer:
The process of killing the cocoons is called stifling,

Question 20.
What is meant by reeling?
Answer:
The process of removing the threads from the killed cocoon is called reeling.

Question 21.
What is cooking?
Answer:
The process of soaking cocoons in hot water (95° – 97°) for 10 – 15 minutes to soften the gum that binds the silk threads together is called cooking.

Question 22.
What are Uzi files?
Answer:
These files which can attack silkworm.

Question 23.
What is meant by cooking?
Answer:
The cocoons are soaked in hot water at 95° – 97°c for 10 -15 minutes to soften the gum that binds, the silk threads together are called cooking.

Question 24.
What is spun silk?
Answer:
The silk produced from silk wastes used for producing spun silk.

Question 25.
What are apiaries?
Answer:
They are areas where a lot of beehives can be placed.

Question 26.
Name the bees used for Apiculture.
Answer:

• Apis dorsata – Rock bee
• Apis florea – Little bee
• Apis indica – Indian bee
• Apismellifera – European bee
• Apis adamsoni – African bee

Question 27.
Name the types of beehives which are in practice in India?
Answer:

• Langstroth
• Newton

Question 28.
What is meant by swarming?
Answer:
The mass emergence of larvae of Lacinsect from the egg in search of a host plant is called swarming.

Question 29.
What is meant by nuptial flight?
Answer:
During the breeding season in winter, a unique flight taken by the queen bee followed by , several drones is called nuptial flight.

Question 30.
Which is called as king of the colony? Why is called so?
Answer:
The king of colony is drone hence the sole duty of the drone is to fertilize the vergin queen.

Question 31.
Name the sugar component present in the honey?
Answer:
levulose, dextrose maltose.

Question 32.
What are the uses of honey?
Answer:

• It is used as an antiseptic
• Laxative
• Sedative
• It is used in the preparation of cakes bread and biscuits.

Question 33.
What is meant by Lac culture?
Answer:
The culture of lac insect using techniques for the procurement of Lac on large scale is known as Lac culture.

Question 34.
Name the host plants on which Lac insects live?
Answer:
Karanagalli – Acacia catechu Karuvelai – Acacia nilotica Kumbadiri – Schleichera oleosa.

Question 35.
What is meant by hyper parasitism?
Answer:

• Hyper parasitism is a condition in which a secondary parasite develops within a previously existing parasite.
• The caterpillars feed upon lac insects showing hyper – parasitism.

Question 36.
What is meant by inoculation?
Answer:
The process of introducing lac insects to the host plant is called inoculation.

Question 37.
What is meant by Harvesting?
Answer:
The collection of Lac from the host plant is known as harvesting.

Question 38.
What is meant by ‘Arilac’?
Answer:
The Immature lac insects produce a lac which is called ‘arilac’.

Question 39.
What is meant by sticklac?
Answer:
Lac cut from the host plant is called ‘stick lac’.

Question 40.
What is seed Lac?
Answer:
The Lac scraped collected grounded and the dust particles are removed to produce a lac called seedlac.

Question 41.
What is meant by Shellac?
Answer:
The seedlac is sundried and then melted to produce “Shellac”.

Question 42.
What is meant by aqua ponies?
Answer:
Aquaponics is a technique which is a combination of aquaculture and growing plants in non – soil media and nutrient-laden water.

Question 43.
What are the organisms farming through aquaculture?
Answer:
Fish mollusks crustaceans and aquatic plants are farming through aquaculture.

Question 44.
How aquaculture is classified on the basis of it source?
Answer:

• Freshwater aquaculture
• Brackish water aquaculture
• Marine water aquaculture.

Question 45.
What is meant by Brackish water culture?
Answer:
Culturing of animals in the water having salinity range 0.5 – 30ppt are called brackish water cultures.

Question 46.
Name the fishes cultured through brackish water culture?
Answer:
Milkfish, Sea bass, Grey mullet Kari meen.

Question 47.
What is meant by metahaline culture?
Answer:
Culturing of animals in the salinity ranges from 36 – 40o% is called metahaline culture (eg) Artemia salina.

Question 48.
What are the organisms rearing through aquaculture?
Answer:

1. Molluscs
2. Aquatic plants
3. Crustaceans

Question 49.
Give notes on fishes of brackish water?
Answer:
Brackish water fishes spend most of its life in river mouths backwaters mangrove swamps and coastal lagoons.

Question 50.
What is meant by mariculture?
Answer:
Culturing of animals in the water salinity ranges from 35- 35% is called mariculture.

Question 51.
Give short notes on Artemia salina?
Answer:
It is a metahaline organism. It lives in high saline waters because of its high osmoregulatory capacity.

Question 52.
What is meant by composite fish culture?
Answer:
Few selected fishes belonging to different species are stocked together in proper proportion in the pond is called composite culture.

Question 53.
What are the organisms cultured through composite fish culture?
Answer:

• Catla Catla
• Labeo rohita
• Cirrhina mirgala.

Question 54.
What is meant by exotic fish culture?
Answer:
The fishes imported in to a country for fish culture are called exotic fishes and such fish culture is known as exotic fish culture.

Question 55.
Name the exotic fishes cultured in India?
Answer:

• Cyprinus carpio
• Oreochromis mossambicus.

Question 56.
What is meant by fish meal?
Answer:
Fish meal is prepared from fish waste after extracting oil from the fish.

Question 57.
What is meant by Ising glass?
Answer:

• Ising glass is high-grade collagen produced from dried air bladder.
• It is used for clarification of wine beer and vinegar.

Question 58.
What is meant by outbreeding?
Answer:
It is the breeding of unrelated animals. They do not have common ancestry for 4-6 generations.

Question 59.
What is the use of outbreeding?
Answer:
Outbreeding helps to produce hybrids with superior qualities and helps to create new breeds. New and favourable genes can be introduced into a population through outbreeding.

Question 60.
What is meant by outcrossing?
Answer:
It is the breeding between unrelated animals of the same breed but having no common ancertry.

Question 61.
What is meant by artificial insemination?
Answer:
Artificial insemination is a technique in which the semen collected from the male is injected into the reproductive tract of the selected female.

Question 62.
What is meant by the recovery period?
Answer:
Nucleated oysters are attached with floating rafts suspended into a depth of 2 to 3 metres for about 6 to 7 days is called the recovery period.

Question 63.
What is the composition of pearl?
Answer:

• Water-2-4%
• Calcium carbonate-90%
• Organic matter-3.5-5.9%
• residue-0.1-0.8%

Question 64.
Name the species of prawn?
Answer:
Penaues indicus, Penaeus monodon, Metapenaeus dobsoni, Macrobrachium rosenbergil.

Question 65.
Where are pearl oysters cultured?
Answer:
Kanyakumari Bay of Kutch.

Question 66.
What is meant by Linga pearl or best quality pearl?
Answer:
The pearl oysters in habit the ridges of rock-forming extensive pearl banks. These pearl beds produce the best quality of pearl or Linga pearl.

Question 67.
What are the types of cultivable fish?
Answer:

• Indigenous or native freshwater fish. (cg) Catla, Labeo
• Saltwater fishes acclimatized for freshwater.
• Exotic fishes or imported from other countries. (eg) Common Carps.

Question 68.
What is Hapa?
Answer:
Hatching hapas are rectangular through shaped tanks made up of mosquito net cloth supported by bampoo poles and fixed in the river.

Question 69.
What are the advantages of composite fish farming?
Answer:

•  All available niches are fully utilized.
• Compatible species do not harm each other.
• No competition among different species.

Question 70.
What is meant by animal husbandry?
Answer:
Animal husbandry is the practise of breeding and raising livestock cattle like cows, buffaloes goats and birds that are useful to human beings.

Question 71.
What is meant by breed?
Answer:
A group of animals related by descent and with similar characters like general appearance features size etc as said to belong to a breed.

Question 72.
How can we classify cattle on the basis of their utility?
Answer:

1. Milch breeds
2. Draught purpose breeds
3. Dual-purpose breeds.

Question 73.
What are characteristic features of healthy cattle?
Answer:

• They appear bright
• Alert
• Active
• Shiny coat.

Question 74.
What are the characteristic features of unhealthy cattle?
Answer:

• They appear dull
• Restless
• Change posture frequently with a drop in milk yield.

Question 75.
Name the important cattle disease?
Answer:

• Rinder pest
• Cowpox
• Anthrax
• hemorrhagic fever.

Question 76.
What is meant by poultry?
Answer:
It is the rearing and propagation of chicken ducks turkeys geese quail and guinea fowls.

Question 77.
What are the types of chicken breeds based on their utility?
Answer:

• They are egg layers
• Broiler type
• Dual type
• Games and
• ornamental types.

Question 78.
Name the native ducks.
Answer:

• Indian Runner
• Syhletmeta.

Question 79.
Name the exotic ducks.
Answer:

• Muscari
• Pekin
• Aylesbury
• Campbell

Question 80.
What are the three stages of sericulture?
Answer:

1. Cultivation of food plants for the silkworm.
2. rearing of silkworm.
3. reeling and spinning of silk.

Question 81.
What is meant by apiculture?
Answer:
Care and management of honey bees on a commercial scale for the production of honey is called apiculture.

Question 82.
What is the importance of aquaponics?
Answer:
It maintains the ecosystem balance by recycling the waste and excretory products produced by the fish.

Question 83.
What is meant by aquaponics?
Answer:
Aquaponics is a technique which is a combination of aquaculture and growing plants in non-soil media and nutrient-laden water.

Question 84.
Based on the water resources where are aquatic organisms cultured? Give examples.
Answer:

1. Freshwater culture
2. Brackish water culture, Marine culture, Cultured organism, Fishes, Prawns, Crabs and oysters.

Question 85.
What are the objectives of animal breeding?
Answer:

• To improve growth rate
• enhancing the production of milk meat egg etc.
• Increasing the quality of the animal products
• Improved resistance to disease.
• Increased reproductive rate.

3 marks

III. Short answers

Question 1.
What is meant by vermiculture?
Answer:
It is the process of using earthworm to decompose organic food waste into a nutrient-rich material capable of supplying necessary nutrients which help to sustain plant growth.

Question 2.
What is meant by worm casting?
Answer:
The worm castings are pure worm waste and nutrient-rich organic soil and composed of castings, bits of bedding, and other organic matter.

Question 3.
What are the characteristics of cultivable fishes?
Answer:
Characteristics of cultivable fishes.
The special characteristic features of cultivable fishes are:

1. Fishes should have a high growth rate in a short period of culture.
2. They should accept a supplementary diet.
3. They should be hardy enough to resist some common diseases and infections of parasites.
4. Fishes proposed for polyculture should be able to live together without interfering or attacking other fishes.
5. They should have high conversion efficiency so that they can effectively utilize the food.

Question 4.
What are the three main components of sericulture?
Answer:

1. Cultivation of food plants for the silkworms.
2. Rearing of silkworms.
3. Reeling and spinning of silk.

Question 5.
Name the predators of earthworms?
Answer:

• Ants
• centipedes
• Slugs
• Birds
• Rats
• Snakes

Question 6.
Name the honey bees used in apiculture.
Answer:

• Apis dorsate,
• Apis florea
• Apis indica
• Apis mellifera
• Apisadamsoni

Question 7.
What are the external factors affecting fish culture?
Answer:
External factors affecting fish culture. The factors that affect fish culture are temperature, light rain, water, flood, water current, turbidity of the water, pH hardness, salinity and dissolved O₂. Light and temperature also play an important role in fish breeding.

Question 8.
What are the uses of silk?
Answer:
Uses of Silk:-
1. Silk fibers are utilized in preparing silk clothes. Silk fibers are now combined with other natural or synthetic fibers to manufacture clothes like Teri-Silk, Cot-Silk etc. Silk is dyed and printed to prepare ornamented fabrics. They are generally made from Eri-silk or spun silk.

2. Silk is used in industries and for military purposes.

3. It is used in the manufacture of fishing fibers, parachutes, cartridge bags, insulation coils for telephone, wireless receivers, tyres of racing cars, filter fibres, medical dressings, and suture materials.

Question 9.
Describe the structure of Beehive?
Answer:

• The house of honey bees is a beehive.
• The hive consists of hexagonal cells made up of wax secreted by the abdomen of a worker bee.

• These hives are found hanging vertically from the rocks. building or branches of trees.
• The young stages of the honey bee accommodate the lower and central cells of the hive.
• In the rock beehives, there are separate cells for queens, workers, and drones.

Question 10.
What are the types of Prawn fishery?
Answer:
Types of prawn fishery

1. Shallow water prawn fishery – located on the west coast restricted to shallow waters.
2. Estuaries and backwaters or saline lake prawn fishery – The area of production of prawns are the backwaters seen along the Western coast, Ennur, Pulicat, Chilka lake, and Estuaries of Ganga and Brahmaputra rivers.
3. Freshwater prawn fishery – Prawns are caught from the rivers and lakes throughout India.
4. Marine prawn fishery – Most of the marine prawns are caught along the Indian coast belonging to the family Penaeidae.

Question 11.
What is meant by inbreeding? What are its effects?
Answer:
Breeding between animals of the same breed for 4 – 6 generations is called inbreeding.
Effects:

1. It increases homozygosity.
2. Exposes the harmful recessive genes.
3. Reduces fertility.
4. It produces inbreeding depression.

Question 12.
How animals are classified based on their economic importance.
Answer:

• Animals for food and food products.
• Economically beneficial animals.
• Animals of aesthetic importance.
• Animals for scientific research.

Question 13.
Give an account of earthworm based on their habitat.
Answer:
First group:

• The humus formers dwell on the surface and feed on organic matter.
• They are darker in colour.
• These worms are used for vermicomposting.

Second group:
The humus feeders are burrowing worms that are useful in making the soil porous and mixing and distributing humus throughout the soil.

Question 14.
What are the uses of honey wax?
Answer:

• It is used for making candles.
• It is used for making waterproofing materials.
• It is used for making home appliances polishes for leather.
• It is used in pharmaceutical industries.

Question 15.
What is meant by Hyper parasitism?
Answer:
A condition in which a secondary parasite develops within a previously existing parasite.

Question 16.
What are the economic importance of Lac?
Answer:

• It is used as sealing wax.
• It is used as a good insulator.
• It is used in the preparation of shoes and leather polishes.
• It is a protective coating of wood.
• It is used in the preparation of plastic moulded articles.
• It is used as a filling material for gold ornament.

Question 17.
Name the breeds of cattle?
Answer:

• Milch breeds – Sindhi Jersey
• Draught breeds – KangeyamMalvi
• Dual-purpose breeds – Ongole Hariana

Question 18.
What is meant by Milch breed?
Answer:
They are high milk yielders with extended lactation. (eg) Sindhi, brown swiss.

Question 19.
What is Dual purpose?
Answer:
Cows are meant for yielding more milk and bullocks are used for better drought purpose, (eg) Ongole Hariana

Question 20.
What is draught purpose breeds?
Answer:
Bullocks are good for draught purposes (eg) Malvi.

Question 21.
What are the characteristic features of healthy cattle?
Answer:

• A healthy animal eats drinks and sleeps well regularly.
• Healthy cattle appear bright.
• Alert and active in their movement with a shiny coat.

Question 22.
What are the external factors affecting fish culture?
Answer:

1. Temperature
2. Light
3. Rain
4. Flood
5. Water current
6. Turbidity of water
7. pH

Question 23.
What is Aquaponics?
Answer:
Aquaponics is a technique which is a combination of aquaculture (growing fish)’ and hydroponics (growing plants in non-soil media and nutrient-laden water).

Question 24.
Name the species of prawn.
Answer:

• Penaeus indicus,
• Penaeus monodon,
• Metapenaeus dobson,
• Macrobrachium rosenbergii.

Question 25.
What are the three types of aquaculture on the basis of the source?
Answer:
On the basis of source, aquaculture can be classified into three categories. They are

1. Freshwater aquaculture
2. Brackish water aquaculture
3. Marine water aquaculture.

Question 26.
What are the benefits of poultry farming?
Answer:

• It does not require high capital for the construction and maintenance of poultry farming.
• It does not require a big space.
• It ensures a high return on investment.
• It provides fresh and nutritious food and has a huge global demand.
• It provides employment opportunities for the people.

Question 27.
What are the uses of dairy products?
Answer:

• Milk is a rich source of vitamin A, B₂, B₁
• It is a complete food for infants.
• Dairy products such as yoghurt cheese butter ice cream condensed milk, milk powder are produced from milk.

Question 28.
What is the importance of meat?
Answer:

• Meat is rich in protein.
• It also contains minerals like iron zinc vitamins and selenium.

Question 29.
What is Mariculture?
Answer:
Culturing of animals in the water salinity ranges from 30 – 35% is called Mariculture. Some fishes like Chanos sp, Mugil cephalus are cultured here.

Question 30.
What are poultry diseases?
Answer:
Ranikhet, Coccidiosis, Fowl pox.

5 Marks

IV. Give Detailed Answers

Question 1.
How are vermicompost produced?
Answer:
Types of Movements:

• Vermicompost is the compost produced by the action of earthworm in association with all other organisms in the compost unit.
• Vermicompost bed may be selected on upland or an elevated level.
• We have to construct a cement pit of 3 x 2 x 1m size over the ground surface using bricks.
• The vermibed should not be exposed to direct sunlight.
• The first layer of vermibed contains gravel at about 5 cm in height followed by coarses and to a thickness of 3.5 cm which will facilitate the draniage of excess water.
• The unit can now be loaded with digested biomass or animal dung.
• Earthworms such as periyonyx excavatus Eisenia fetida or Eudrilus eugeniae are introduced on the top.
• Earthworms release their castings on the surface.
• Vermiwash is a liquid collected after the passage of water through a column of vermibed. It is useful as a foliar spray to enhance plant growth.

Question 2.
Describe the life cycle of bombyx mon?
Answer:

• This moth is unisexual.
• Just after emergence male moth copulates with females for 2-3 hours.
• Just after copulation female starts laying about 400-500 eggs.
• The eggs after ten days of incubation hatch into larva called a caterpillar.
• The newly hatched caterpillar is about 3mm in length and is pale yellowish-white colour.
• The mandibulate type of mouthparts adapted to feed easily on the mulberry leaves.

• After 1st, 2nd, 3rd and 4th moulting caterpillars get transformed into 2nd, 3rd, 4th and 5th instars respectively.
• It develops salivary glands, stops feeding and undergoes pupation.
• The cafter pillars stop feeding and moves towards the comer among the leaves and secrete a sticky fluid through their silk gland.
• The secreted fluid comes out through spinnerret which hardens on exposure to air and is wrapped around the body of cater pillar in the forms of a covering called a cocoon.
• The length of continous thread secreted a caterpillar for the formation of cocoon is about 1000 -1200 metres.
• The pupal period lasts for 10-12 days and the pupae cut through the cocoon and emerge into adult moth.
• In larvel stages the larvae moults for 3 or 4 times or 5 times and become a matured moth.

Question 3.
What are the uses of vermicompost?
Answer:

• Vermicompost is rich in essential plant nutrients.
• It improves soil structure texture aeration and water holding capacity and prevents soil erosion.
• Vermicompost is rich in nutrients and an eco-friendly amendment soil for farming and terrace gardening.
• It enhances seed germination and ensures good plant growth.

Question 4.
What are the steps of insertion of the nucleus into the oyster?
Answer:
Following steps are taken for the insertion of the nucleus:

a. Fitness of oyster for operation:
The selected oysters for the insertion of the nucleus should be healthy and strong enough to overcome the stress during operation.

b. Preparation of graft tissues:
The piece of tissue which is inserted inside the mantle is called as ‘GRAFT’ tissue. The outer edges of these graft squares must be known because nacre secreting cells are found only on the outer surface of the mantle so it is essential to keep the outer surface in contact with the inserted nucleus.

c. Preparation of the nucleus:
Any small particle may function as a nucleus to initiate the pearl formation but it is reported that the calcareous nucleus is the best because the deposition of nacre was found to be more on the calcareous nucleus.

d. Insertion of the nucleus:
For the insertion of the nucleus, oysters are fixed in a desk clamp in the position of the right valve facing upward. Mantle folds are smoothly touched to expose the foot and the main body mass, followed by an incision into the epithelium of the foot and a slender channel into the main mass of one graft tissue which functions as a bed for the nucleus.

e. Post-operation care:
Nucleated oysters are placed into cages and suspended into seawater and attached with floating rafts to a depth of 2 to 3 meters for about 6 to 7 days to recover from the shocks due to operation. This period of 6 to 7 days is known as the ‘Recovery period’. About 3000 to 3600 nucleated oysters are kept in different cages suspended in seawater at 2 to 3 meters depth for 3 to 6 years and undisturbed except at the time of clearing and inspection.

f. Harvesting of pearl:
Pearls are harvested in the month of December to February which may slightly vary according to climatic conditions. After the completion of 3 years of the insertion of the nucleus, pearl oysters are harvested from the sea and the pearls are taken out from the shell.

g. Clearing of pearls:
After taking out the pearls from the oyster’s shell they are washed properly, cleared with the soap solution.

Question 5.
What are the uses of silk?
Answer:

• Silk fibres are utilized in preparing silk cloths.
• Silk fibres are combined with natural or synthetic fibres to manufacture Teri – silk and cot silk.
• Silk is dyed and printed to prepare ornamental fabrics.
• Silk is used in industries and for military purposes.
• It is used in the manufacturing of fishing fibres, parachutes insulation coils for telephones.
• They are used in the preparation of tyres for racing cars, filter fibres in medical dressings and as suture materials.

Question 6.
Name the pests and diseases of the silkworm.
Answer:
1. Predators:
They feed on silkworms.

• Birds
• ants
• crows
• kites
• rats.

2. Diseases:

• Pebrin – It is a disease caused by protozoa Nosema bombycis.
• Flacherie – It occurs in the mature larvae caused by bacteria like streptococcus and staphylococcus.
• Grasserie – It is aviral disease caused by Bombyx mori nuclear polyhedrosis virus.
• White Muscardine – It is caused by the fungus Beauveria bassiana.

Question 7.
Describe the social organization of the honey bee.
Answer:

• In the honey bee, a highly organized division of labour is found.
• There are three types of honey bees seen in the colony.
• There are one queen 10000 – 30000 workers and a few hundred drones in a colony.

Queen bee:

• It feeds on royal Jelly.
• It is only function is to lay eggs throughout its life span.
• During breeding season in winter, a unique flight takes place by the queen bee followed by several drones is called nuptial flight.
• After mating the queen bee lays about 15 lakh eggs in two to four years.

Worker bee:

• These are sterile females.
• Worker bee lives in a worker well and it takes, about 21 days to develop from the egg to adult, and its life span is about six weeks.

Works:

• It secretes royal Jelly.
• Prepares bee – bread to feed the larvae.
• Takes care of the queen and drones.
• Secretes bee wax.
• In the last three weeks, she searches and gathers the pollen nectar.

Drones:

• The drone develops from an unfertilized egg.
• It lives in a chamber called a drone cell.
• The only duty of the drone is to fertilize the queen and it is called the king of the colony.

Question 8.
Describe the structure of the Langstroth beehive?
Answer:

• Stand – It is a basal part of the hive and is adjusted to make a slope for rainwater to drain.
• Bottom board – It is situated above the stand. It has two gates one gate is an entrance the other acts as an exit.
• Brood chamber – It is an important part of the hive. It is provided with 5 to 10 frames arranged one above the other through which the workers can easily pass.
• The frame is composed of a wax – sheet which is held in a vertical position up by a couple of wires.
• It is strong that can be used repeatedly.
• Super – It is a chamber without cover and base. It is provided with many frames to provide additional space for expansion of the hive.
• Inner cover – It is a wooden piece used for covering the super with many holes for proper ventilation.
• Top cover – It is meant for protecting the colonies from rains. It is covered with a sheet which is plain and sloping.

Question 9.
What are the methods of animal breeding?
Answer:
Methods of Animal breeding:
There are two methods of animal breeding, namely inbreeding and outbreeding:

1. Inbreeding:
Breeding between animals of the same breed for 4-6 generations is called inbreeding. Inbreeding increases homozygosity and exposes the harmful recessive genes.

Continuous inbreeding reduces fertility and even productivity, resulting in “inbreeding depression”. This can be avoided by breeding selected animals of the breeding population and they should be mated with superior animals of the same breed but unrelated to the breeding population. It helps to restore fertility and yield.

2. Outbreeding:
The breeding between unrelated animals is called outbreeding. Individuals produced do not have common ancestors for 4-6 generations.

Outbreeding helps to produce new and favourable traits, to produce hybrids, with superior qualities, and helps to create new breeds. New and favourable genes can be introduced into a population through outbreeding.

Question 10.
Describe the process of rearing silkworms?
Answer:

• A typical rearing house 6 m x 4m x 3.5m is constructed on an elevated place under shade to accommodate 100 days.
• The windows and ventilators should be covered with a nylon net to restrict the entry of uziflies and other insects.
• The selected healthy silk moths are allowed to mate for 4 hours.
• The female moth is kept in a dark plastic bed it lays about 400 eggs in 24 hours.
• The small larvae hatch between 7 – 10 days.
• The larvae are kept in trays at a temperature of about 20°C-25°C.
• As the larvae grow they are transferred to fresh leaves on clean trays.
• Their maturity is achieved in about 45 days.
• At this stage, the salivary glands start secreting silk to spin cocoons.

Question 11.
What is meant by aquaponics? Describe its methods?
Answer:
1. Deepwater culture – Aquaponics is a technique which is a combination of aquaculture and growing plants in non – soil media.

• It is a raft-based method.

• In this method, the raft floats in water plants are kept in the holes of the raft, and the root floats in water. Fast-growing plants are cultivated.

2. Media-based method – It involves growing plants in inert planting media likely pellets or shales.
This method is applicable for home and hobby scale systems. A large number of fruiting plants leafy green plants herbs can be cultivated.

3. Nutrient film technique – It involves the passage of nutrient-rich water through a narrow trough or pvc pipe.
Plants are kept in the holes of the pipe to allow the roots to be in free contact with the water stream.

4. Aquavertica or vertical aquaponics – Plants are stacked in top of each other in tower systems.
This method is suitable for growing leafy green, strawberries and crops that do not need supporting solid substratum to grow.

Question 12.
What are the groups of cattle?
Answer:
(I) Dairy breeds or Milch breeds:
They are high milk yielders with extended lactation. Eg., Sindhi, Gir, Sahiwal, Jersy, Brown Swiss, Holstein cattle.

(II) Draught purpose breeds:
Bullocks are good for draught purposes. Eg. Kangayam, Malvi.

(III) Dual Purpose breeds:
Cows are meant for yielding more milk and bullocks are used for better drought purposes Eg. Ongole, Hariana.

Question 13.
What is the Natural breeding of fishes?
Answer:
Natural breeding (Bundh breeding): These are special types of ponds where natural riverine conditions or any natural water resources are managed for the breeding of culturable fishes. These bundhs are constructed in large low-lying areas that can accommodate large quantity of rainwater. The shallow area of such bundhs is used as a spawning ground.

Question 14.
Write a note on milk products?
Answer:
Milk products:
Milk is produced by dairy animals which is an emulsion of fat and lactose. Milk also contains enzymes which are destroyed during pasteurization.

Milk is a rich source of vitamin A, B„ Bp and deficient in Vitamin C. Due to its high nutrition value, it serves as a . complete food for infants. Dairy products such as yoghurt, cheese, butter, ice cream, condensed milk, curd, and milk powder processed from milk make dairy, a highly farming attraction.

Question 15.
Why carps have proved to be best suited for culture in India?
Answer:

• Feed on zooplankton and phytoplanktons decaying weeds debris and other aquatic plants.
• They can survive in turbid water with slightly higher temperatures.
• Can tolerate O₂ variations in water.
• Can be transported from one place to another easily.
• They are highly nutritive and palatable.

Question 16.
Give an account of induced breeding.
Answer:

• To improve the quality of fish seed by the artificial method of fertilization is developed.
• The gonadotrophin (FSH+ LH) secreted by the pituitary gland influences the maturation of gonads and spawning in fishes.
• The pituitary gland is removed from a healthy mature fish.
• The pituitary extract is prepared by homogenizing in 0.3% saline and centrifuged for 15 minutes at 8000rpm.
• The supernatant is injected at the base of the caudal fin.
• Male and Female fishes start to releasing gametes and are fertilized.

Question 17.
Write a short note on Game breeds?
Answer:
(I) Game breeds:
Since ancient times, special breed of roosters have been used for the sport of cockfighting.

(II) Aseel:
This breed is white or black in colour. The hens are not good egg layers but are good in the incubation of eggs. It is found in all states of India. Aseel is noted for its pugnacity, high stamina, and majestic gait and dogged fighting qualities. Although poor in productivity, this breed is well-known for its meat qualities.

Question 18.
Give an account of the culture of freshwater prawns?
Answer:

• Macrobrachium rosenbergii is seen in rivers fields and low – saline estuaries.
• For fertilization, one pair of prawn are kept in a separate tank.
• After mating the eggs are laid.
• Temperature 24°C – 30°C and PH 7-8 should be maintained in the hatching tank.
• The hatched larvae are supplied with artificial feed.
• Young ones 5cm length can be reared in fresh or slightly brackish water ponds and harvesting of prawns can be done twice a year.

Question 19.
What are the benefits of Poultry farming?
Answer:
The benefits of Poultry farming are:

1. It does not require high capital for the construction and maintenance of poultry farming.
2. It does not require a big space.
3. It ensures the high return of investment within a very short period of time.
4. It provides fresh and nutritious food and has a huge global demand.
5. It provides employment opportunities for the people.

Question 20.
What are the types of cultivable fish Types of cultivable fishes?
Answer:
Types of Cultivable fishes:

• Indigenous / Native freshwater fishes, (eg) Major Carps Catla Labeo Clarias.
• Saltwater fishes acclimatized for freshwater. (eg)Chanos, Mullet
• Exotic fishes. Imported from other countries, (eg) Common Carps.

Characteristic features of Carps:

• Feed on Zooplanktons and phytoplanktons decaying weeds debris and other aquatic plants.
• They can survive in turbid water with slightly higher temperatures.
• Can tolerate 0₂ variations in water.
• Can be transported from one place to another.
• They are highly nutritive and palatable easily

Question 21.
Give an account of the management of fish farms?
Answer:
To culture fish, one should have an idea about different stages of fish culture such as topographic situation quality source physical-chemical and biological factors of water.
Stages of fish farming:
Breeding Pond:
The first step in fish culture is the breeding of fishes therefore for proper breeding special types of ponds are prepared called breeding ponds. These ponds are prepared near the rivers of natural water resources.
a) Natural breeding:
These bundhs are constructed in large low-lying areas that can accommodate a large quantity of rainwater.
b) Induced breeding:

• This involves the removal of ova and sperm from female and male by artificial mechanical process and the eggs are fertilized.
• The fertilized eggs are removed from the spawning place and kept in hatching hapas. Fish seeds: The fish seed is collected from breeding ponds and transferred to the hatching pits.

3. Hatching Pit:
The fertilized eggs are kept in hatching pits. The hatching pits should be nearer to the breeding grounds should be smaller in size with good quality water.
There are two types of hatching pits:

1. Hatcheries: These are small-sized ponds in which unfertilized eggs are transferred and hatching happens.
2. Hatching hapas: These are rectangular trough-shaped tanks made up of mosquito net cloth supported by bamboo poles and fixed in the river.

4. Nursery Pond:
The newly hatched fries are transported from the hatching happa to nursery ponds where they grow into fingerlings.

5. Rearing Pond:
This pond is long and narrow and this pond should be free from toxicants and predators. The fingerlings are transferred to the stocking ponds.

6. Stocking Ponds:

• This pond should be devoid of weeds and predatory fishes.
• Proper organic manuring should be done to the production with cow dung.

7. Harvesting:

• Well-grown fishes are taken out for marketing is harvesting.
• Small-sized fishes are again released into the stocking ponds for further growth.
• The harvested fishes are preserved and then marketed.

Question 22.
a) What is meant by composite fish farming?
b) What is the significance?
c) What are the fishes cultured through composite fish farming.
Answer:
a. Composite fish farming: Few selected fishes belonging to different species are stocked together in proper proportion in a pond is called composite fish farming.

b. Merits :

• All available niches are fully utilized.
• Compatible species do not harm each other.
• No competition among different species is found.

c. Fishes that are cultured :

• Catla Catla
• Labeo rohita
• Cirrhinus mrigala

Question 23.
Give an account of the economic importance of fish.
Answer:
1. Fishes form a rich source of protein food the sardines, mackerel tuna herrings have high amino acid concentrations.
2. These fishes have histidine and omega fatty acids.
3. Minerals such as calcium magnesium phosphorus potassium and copper and present.
a) Fish Oil:

• It is an important fish product it is derived from fish liver and from the fish body.
• This is rich in vitamin A and O and iodine.
• It is used in the manufacture of laundry soap paints and cosmetics.

b) Fish Meal:

• It is prepared from fish waste after extracting oil from the fish.
• These dried wastes are used to prepare food for pigs, poultry, and cattle.
• The wastes obtained during the preparation of fish meals are used as manure.

c) Ising glass:

• It is high-grade collagen produced from the dried air bladder of catfish and carps.
• It is used for clarification of wine beer and vinegar.

Question 24.
Give notes on
a) Pearl Culture
b) Formation of pearl.
Answer:
a) Pearl Culture:

• Pearl is a white shining globular concretion found within the shell of an oyster.
• In India, it was cultured for the first time in 1973 at Thoothukudi.
• High-quality pearls are obtained from Genus Pinctada that can be cultured in the salinity range of 30 ppt in racks raft and long-line method.
• The pearl oysters inhabit the ridges of rocks or dead coral forming the best quality pearl.

b) Pearl Formation:

• When a foreign particle accidentally enters into the space between the mantle and shell of the oysters it adheres to the mantle.
• The mantle epithelium encloses its likes sac and starts to secrete concentric layers of nacre around it as a defensive mechanism.
• The repeated layers of calcium carbonate make the hard and glossy pearl that are separated and graded.

Question 25.
Give an account of artificial pearl culture?
Answer:

• Oysters are caught by a special type of cages.
• This cage is dipped into a sand cement mixture providing a rough surface to the cages and are suspended at a depth of 6 meters.

2. Rearing of oysters:

• The collected oysters are placed into the culture cages for a period of 10 – 20 days.
• These cages are protected from enemies like octopus, Eel, and devil fishes.

3. Insertion of the nucleus:
a) Fitness of oysters for operation:
The selected oysters for the insertion of the nucleus should be healthy and strong enough to overcome the stress during operation.

b) Preparation of graft tissues:

• The piece of tissue which is inserted inside the mantle is called “graft” tissue.
• It is essential to keep the outer surface in contact with the inserted nucleus as nacre secreting cells are found only on the outer surface.

c) Preparation of nucleus:

• Any small particle may function as a nucleus to initiate pearl formation.
• If it is calcareous the deposition of nacre was found to be more on the calcareous nucleus.

d) Insertion of the nucleus:
For the insertion of the nucleus, oysters are fixed in a desk clamp and mantle folds are smoothly touched to expose the foot followed by an incision into the epithelium of the foot and the nucleus is inserted.

e) Post Operation Care:

• Nucleated oysters are placed into cages and suspended into 2-3 metres of depth for 6-7 days.
• These periods are known as the “Recovery Period”.
• These oysters are kept for 3-6 years undisturbed.

f) Harvesting of Pearl:
After the completion of 3 years, pearl oysters are harvested. It is usually done from December to February.

g) Clearing of Pearls:
After taking out the pearls from the oyster’s shell. They are washed properly cleared with the soap solution.

Question 26.
What is inbreeding? What are its merits and demerits?
Answer:
Breeding between animals of the same breed for 4-6 generations is called inbreeding.
Merits:

• It increases homozygosity.
• It exposes the harmful recessive genes. Demerits:

1. Continuous inbreeding reduces fertility
2. And results in inbreeding depression. Avoiding inbreeding depression:
This can be avoided by breeding selected animals be mated with superior animals of the same breed but unrelated to the breeding population.

Question 27.
What is Dairying and Dairy Operation?
Answer:
Dairying is the production and marketing of milk and its products. The dairy operation consists of proper maintenance of cattle, the collection, and processing of milk, and its by-products.

Question 28.
What are the types of outbreeding?
Answer:
1. Out Crossing:
It is the breeding between unrelated animals of the same breed but having no common ancestry.
Out Cross: This method is suitable for breeding animals that are below average in productivity.

2. Crossbreeding:

• Breeding between a superior male of one breed with a superior female of another breed.
• The cross-breed progeny has superior traits. This is called hybrid vigour or heterosis.

3. Interspecific hybridization:

• Mating between male and female of two different species.
• The progeny obtained from such crosses are different from their parents.
• Mule is produced by the crosses between a male donkey and a female horse.

Question 29.
Give an account of Multiple Ovulation embryo transfer technology? MOET.
Answer:
MOET:

• This method is applied when the success rate of crossing is low even after artificial insemination.
• In this method, the follicle-stimulating hormone is administered to cows for inducing follicular maturation and superovulation.
• Instead of one egg per cycle, 6-8 eggs can be produced.
• The eggs are carefully recovered non-surgically from the genetic mother and fertilized artificially.
• The embryos at the 8 – 32 celled stage are recovered and transferred to a surrogate mother.
• This technology can be applied to cattle sheep and buffaloes.
• It can produce high milk-yielding females and high-quality meat-yielding bulls in a short time.

Question 30.
a) What is meant by dairying?
b) Classify them on the basis of their utility.
c) Give an account of dairy products.
Answer
a) Dairying:

• It is the production and marketing of milk and its products.
• The dairy operation consists of maintenance of cattle the collection, processing of the milk, and its by-products.
• These are 26 well-defined breeds of cattle and 6 breeds of buffaloes in India.

b) Classification of Cattle

1. Dairy breeds: They are high milk yielder with extended lactation, (eg) Sindhi, Jersey Sahiwal
2. Drought breed: Bullocks are good for draught purposes, (eg) Kangeyam Malvi.
3. Dual-purpose breeds: Cows are meant for yielding more milk and bullocks are used for better drought purposes (eg) Ongole Hariana.

c) Uses of dairy product:
1. Milk: Milk is a rich source of vitamin A, B2, B and deficient in Vitamin C.
Dairy products such as yoghurt cheese butter ice cream condensed milk powder are produced from milk.
2. Meat: Meat is rich in protein and also contains minerals like iron zinc vitamins and selenium.
3. Land Management: Grazing of live stocks sometimes used as a way to control weeds.
4. Manure: Manure can be spread on agriculture fields to increase crop yields.

Question 31.
Describe the types of Chicken breeds?
Answer:
1. Egg layers: These are farmed mainly for the production of eggs.
Leghorn:

• It is originated in Italy. They are small compact with a single comb.
• They mature early and begin to lay eggs at the age of 5 or 6 months.

Chittagong:

• They are good egg layers and are delicious.
• They are found in West Bengal.

2. Selection of eggs for hatching:
Eggs should be fertile medium-sized dark brown shelled and freshly laid eggs are preferred for rearing.

3. Incubation and hatching:

• The maintenance of newly laid eggs in optimum condition till hatching is called incubation.
• The fully developed chick emerges out of egg after an incubation period of 21 – 22 days.
• There are two types of incubation namely natural incubation and artificial incubation.
• In the Natural incubation method, only a limited number of eggs can be incubated by a mother hen.
• In artificial incubation, more eggs can be incubated.

4. Brooding:

• Caring and management of young chicks for 4 – 6 weeks immediately after hatching is called brooding.
• They are natural and artificial brooding. The housing of Poultry:
• To protect the poultry from the sun.
• Rain and predators it is necessary to provide housing to poultry.
• A poultry house should be moisture-proof rat-proof and it should be easily cleanable and durable.

5. Poultry feeding:
The diet of chicks should contain an adequate amount of water carbohydrates proteins fats vitamins and minerals.

Poultry Products:

• The main products of poultry farmings are eggs and meat.
• The primary aim of poultry farming is to obtain eggs.

C. Poultry by-products:

• The feathers: They are used for making pillows and quilts.
• Droppings: The droppings are rich in nitrogen potash and phosphates.
• The by-products of poultry are used as good sources of nutrients for meat-producing animals and poultry.
• These by-products supply proteins fats Vitamins and minerals.
  Poultry diseases: Ranikhet coccidiosis Fowl Pox is a common poultry disease.

Merits:

1. They can adapt themselves to all types of environmental conditions.
2. And for the breed for feed efficiency.
3. Growth rate and resistance to diseases.
4. They are exhibited in poultry shows they are calm friendly and can be maintained as pets.

Question 32.
What are the types of poultry farming? What are the stages involved in rearing? What are the poultry by-products?
Answer:
Types:

1. Organic method
2. Yarding method
3. Battery cage method
4. Furnished cage method.

Stages involved in rearing:
1. Selection of the best layer.

• An active intelligent-looking bird with a bright comb not obese should be selected.
• They are golden or light yellow coloured.

2. Broiler type:

• White Plymouthrock is a fast-growing breed and soft quality meat.
• They have white plumage throughout the body. This is an American breed. It is a fast-growing breed.

3. Dual Purpose breed:
It is known for its massive body having heavy bones. Well feathered and Peacomb is one of the important breed characters.

4. Game breeds:
Aseel:

• The hens are not good egg layers but are good in the incubation of eggs.
• Aseel is noted for its pugnacity high stamina majestic gait.
• This breed is well known for its meat qualities.

5. Ornamental breeds:
Silkie:

• They are reared as pets and for egg production and meat.
• The chicken has a fluffy plumage which is said to feel like silk and satin.
• It has black skin and bones blue ear lobes and five toes on each foot.

Question 33.
Write a note on milk products?
Answer:
Milk products:
Milk is produced by dairy animals which is an emulsion of fat and lactose. Milk also contains enzymes which are destroyed during pasteurization.

Milk is a rich source of vitamin A, B„ Bp and deficient in Vitamin C. Due to its high nutrition value, it serves as a complete food for infants. Dairy products such as yoghurt, cheese, butter, ice cream, condensed milk, curd, and milk powder processed from milk make dairy, a highly farming attraction.