Mahesh

Students can go through AP Inter 1st Year Commerce Notes 6th Lesson Joint Stock Company – Formation will help students in revising the entire concepts quickly.

AP Inter 1st Year Commerce Notes 6th Lesson Joint Stock Company – Formation

→ Joint Stock Company is one kind of business unit.

→ It is a corporate business unit.

→ A joint stock company is commenced with a minimum of 7 members and maximum members are unlimited.

→ Joint stock company is also known as a public limited company which is governed by the Indian Companies Act, 1956.

→ The capital amount is contributed to the company by 15 members through the purchase of shares. So, it is called “Share capital”.

→ The members invest their money by purchasing the shares of the company, they are known as “Shareholders”.

→ The joint stock company is an artificial person created by law, it enjoys a separate legal entity.

→ The liability of the members is limited.

→ Company form of organization is divided into two types,

1. Private Limited Company
2. Public Limited Company

→ Company form of organization is commenced with the issue of prospects and the formation process is completed by obtaining the certificate of commencement of business.

Students can go through AP Inter 1st Year Commerce Notes 5th Lesson Partnership will help students in revising the entire concepts quickly.

AP Inter 1st Year Commerce Notes 5th Lesson Partnership

→ Partnership Business Firm is one of the business units.

→ It is a Non-corporate Business Unit.

→ Partnership firm is an outcome of an agreement between two or more persons to share profits or losses among them.

→ Partnership is established by partnership agreement among partners.

→ Partnership Agreement Registered it. is known as Partnership Deed.

→ In India partnership is formed with the rules and regulations of the Indian Partnership Act, 1932.

→ The partnership is a part of a Limited Liability Partnership.

→ The partnership was established with a minimum of 2 partners, according to the Indian Partnership Act, 1932.

→ According to section II of the Indian Companies Act 1956, the maximum limit of partners in case of a partnership for banking business is 10, and in case of other than banking business of the partnership is 20.

→ In order to overcome the limitations of sole proprietorship concerns i.e. limited capital, limited managerial ability, and extending the size of the business, the viable and feasible option is a partnership form of organisation.

→ The liability of partners is unlimited and there is no separate legal entity to this organisation.

→ Partnership form of organisation can be dissolved with the mutual consent of the partners.

→ Dissolution of partnership occurs due to the partner giving notice in writing to other partners, expiry of the term of agreement or business, insolvency of the firm, or in the event of the court order.

Students can go through AP Inter 1st Year Commerce Notes 4th Lesson Joint Hindu Family Business & Co-op Society will help students in revising the entire concepts quickly.

AP Inter 1st Year Commerce Notes 4th Lesson Joint Hindu Family Business & Co-op Society

→ Joint Hindu Family Business is also one of the kinds of business units.

→ Joint Hindu Family Business is a form of a business organisation run by the Hindu Undivided Family, wherein the family members of three successive generations own the business jointly.

→ The head of the family is known as ‘Karta’. He manages the business and family.

→ In the JHF, other members are called ‘Co-parceners’. All of them have equal ownership right over the properties-of business.

→ The membership of the JHF is acquired by virtue of birth in the same family.

→ No restriction for minors to become members of the business in JHF.

→ JHF business is governed by two laws i.e., ‘Dayabhaga’ and ‘Mitakshara’.

→ A cooperative society is formed particularly to provide services to its members and to the society in general.

→ Cooperative society enjoys perpetual succession.

→ According to the needs of the people cooperative societies are divided into different types.

→ Individuals, producers, consumers, farmers, etc. who are in need and wish to protect themselves can go for cooperatives.

Students can go through AP Inter 1st Year Commerce Notes 3rd Lesson Forms of Business Organization will help students in revising the entire concepts quickly.

AP Inter 1st Year Commerce Notes 3rd Lesson Forms of Business Organization

→ Business is one of the human economic activities. Profit is a consideration of business.

→ Business is an economic entity i.e., an artificial person.

→ Business units may be classified into two types.

1. Noncorporate units
2. Corporate units

→ Sole proprietorship concern is one of the noncorporate units.

→ Each and every business concern must have its own merits and demerits.

→ Sole proprietorship business is owned by only one person and controlled by a single individual.

→ The complete risk in sole proprietorship concern is borne by a sole trader.

→ The sole trade liability is an unlimited liability because the sole proprietorship firm has no separate legal entity.

→ The sole trader and sole proprietorship firms both were the same as per law.

→ To commencement of sole proprietorship firm legal formalities are very less.

→ In sole proprietorship concerns, decisions should be taken by only one person i.e., the sole trader.

Students can go through AP Inter 1st Year Commerce Notes 2nd Lesson Business Activities will help students in revising the entire concepts quickly.

AP Inter 1st Year Commerce Notes 2nd Lesson Business Activities

→ All business activities are economic activities, a man is engaged in, to earn his livelihood by producing and distributing goods and rendering services.

→ Business may be defined as a human activity directed towards producing or acquiring wealth through buying and selling of goods.

→ Industry refers to the production of consumer goods and capital goods, creating form utility.

→ Commerce is part of the business. It deals with buying and selling goods and services. Commerce is concerned only with the exchange of goods. It includes all those activities which are related to the transfer of goods from the production place to the consumption place.

→ Trade means the purchase and sale of goods with a profit motive. It involves the exchange of goods and services between buyers and sellers.

→ Aids to trade include transport, communication, warehousing, banking, insurance, and advertising.

Students can go through AP Inter 1st Year Commerce Notes 1st Lesson Concept of Business will help students in revising the entire concepts quickly.

AP Inter 1st Year Commerce Notes 1st Lesson Concept of Business

→ The term business refers to ‘the state of being busy”.

→ Business is one of the human economic activities. Business is an economic activity that involves the regular transfer or exchange of goods and services for earning profit.

→ Business creates utilities by producing and selling goods and services to satisfy human wants.

→ Time, place, and possession of values are created by business enterprises.

→ Every business enterprise has both economic and social objectives.

→ The obligation of any business enterprise is to protect and serve the public interest as they operate within a society.

→ The Business organisations must be responsible to different Interest groups like owners, employees, suppliers, customers, government, etc.

Students can go through Telangana & Andhra Pradesh BIEAP TS AP Inter 1st Year Commerce Notes Pdf Download in English Medium and Telugu Medium to understand and remember the concepts easily. Besides, with our AP Sr Inter 1st Year Commerce Notes students can have a complete revision of the subject effectively while focusing on the important chapters and topics.

Students can also go through AP Inter 1st Year Commerce Study Material and AP Inter 1st Year Commerce Important Questions for exam preparation.

AP Intermediate 1st Year Commerce Notes

• Chapter 1 Concept of Business Notes
• Chapter 2 Business Activities Notes
• Chapter 3 Forms of Business Organization Notes
• Chapter 4 Joint Hindu Family Business & Co-op Society Notes
• Chapter 5 Partnership Notes
• Chapter 6 Joint Stock Company – Formation Notes
• Chapter 7 Formation of a Company Notes
• Chapter 8 Sources of Business Finance-I Notes
• Chapter 9 Sources of Business Finance-II Notes
• Chapter 10 Micro, Small and Medium Enterprises (MSMEs) Notes
• Chapter 11 Multi National Corporations (MNCs) Notes
• Chapter 12 Emerging Trends in Business Notes

These TS AP Intermediate 1st Year Commerce Notes provide an extra edge and help students to boost their self-confidence before appearing for their final examinations. These Inter 1st Year Commerce Notes will enable students to study smartly and get a clear idea about each and every concept discussed in their syllabus.

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 13th Lesson Ecological Adaptation, Succession and Ecological Services Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 13th Lesson Ecological Adaptation, Succession and Ecological Services

Very Short Answer Questions

Question 1.
Climax stage is achieved quickly in secondary succession as compared to primary succession. Why?
Answer:
In secondary succession, the species that Invade depend on the condition of the soil, availability of water, the environment as also the seeds or other propagules present. Since soil is already there, the rate of succession is much faster and hence climax stage is also reached quickly than primary succession.

Question 2.
Among Bryophytes, lichens and ferns which one is a pioneer species in a xeric succession.
Answer:
Among Bryophytes, lichens and Ferns, lichens are the pioneer species in a xeric succession.

Question 3.
Give any two examples of xerarch succession.
Answer:
Crustose lichens – Rhizocarpon, Lecanora.
Foliose lichens – Parmelia, Dermetocarpon.
Mossess – Funaria

Question 4.
Name the type of land plants that can tolerate the salinities of the sea.
Answer:
Halophytes.
Ex : Rhizophora.

Question 5.
Define Heliophytes and Sciophytes. Name a plant from your locality that i$ either Heliophyte or Sciophyte.
Answer:
Plants grow in direct sunlight are called heliophytes.
Ex : Tridax, grass.

Plants grow in shady places are called Sciophytes.
Ex : Ferns, Mosses.

Question 6.
Define population and community.
Answer:
A group of similar individuals belonging to the same species found in an area is called population. An assemblage of all the populations belonging to different species occuring in an area is called community.

Question 7.
Define communities? Who classified plant communities into hydrophytes, meso- phytes and xerophytes?
Answer:
An assemblage of all the populations belonging to different species occuring in an area are called communities. “Eugene Warming” classified plant communities into Hydrophytes, mesophytes and xerophytes.

Question 8.
Hydrophytes show reduced xylem. Why?
Answer:
All submerged organs are capable of absorbing water that’s why Hydrophytes show reduced xylem.

Short Answer Type Questions

Question 1.
What are hydrophytes? Briefly discuss the different kinds of hydrophytes with example.
Answer:
Hydrophytes are the plants which grow in water or in very wet places. According to their relation to water and air hydrophytes are classified into five categories. They are

1. Free floating Hydrophytes :
They float freely on the surface of the water and have no contact with soil.
E.g. : Pistia, Eichhornia, Wolffia, Salvinia, Azolla.

2. Rooted hydrophytes with floating leaves :
They are attached to the muddy soil by roots. Their leaves have long petioles which keep them floating on the surface of water.
E.g. : Nymphaea, Nelumbo and Victoria regia.

3. Submerged suspended hydrophytes :
They are completely submerged and suspended in water, but not rooted in the mud and have no contact with air.
E.g. : Hydrilla, Ceratophyllum, Utricularia and Najas.

4. Submerged rooted hydrophytes :
These plants are completely submerged in water. They are attached to the muddy soil by roots.
E.g. : Vallisneria, Potamogeton etc.

5. Amphibious plants :
These live partly in water and partly in air.
E.g. : Sagittaria, Ranunculus, Limnophila.

Some amphibious plants grow around water bodies, with water touching them. They will survive in dry periods also.
E.g. : Typha, Cyperus etc.

Question 2.
Enumerate the morphological adaptations of hydrophytes,
Answer:

1. Roots may be absent or poorly developed. In some plants (Salvinia) submerged leaves compensate for roots.
2. Root caps are usually absent. In some amphibious plants which grow in mud, roots are well developed with root caps. In some plants root caps are replaced by root pockets.
   E.g. : Eichhornia.
3. Roots if present, are generally fibrous, reduced in length, uribranched or poorly branched.
4. Stem is long, slender and flexible.
5. Leaves are thin and either long and ribbon shaped (vallisnaria) or long and linear {potamogeton) or finely dissected (caratophyllum).
6. Floating leaves are large and flat with their upper surfaces coated with wax (Nymphaea, Nelumbium).

Question 3.
List out the anatomical adaptations of hydrophytes.
Answer:

1. Cuticle is totally absent in the submerged parts of the plants. It may be present in the form of a thin film on the surface of parts exposed to atmosphere.
2. The epidermis is composed of thin walled cells. They perform absorption and assimilation as all cells contain chloroplasts.
3. Stomata are totally absent in submerged hydrophytes as the gaseous exchange takes place by diffusion.
4. In Nymphaea, Nelumbium, the leaves are epistomatous.
5. All Hydrophytes contain aerenchyma that helps in gaseous exchange and buoyancy.
6. Mechanical tissues like Collenchyma and Sclerenchyma are poorly developed.
7. Xylem is poorly developed.
8. Secondary growth is absent

Question 4.
Write a brief account on classification of xerophytes.
Answer:
On the basis of their Drought resisting Capacity, Xerophytes are generally classified into the following three categories.

1. Ephemerals :
They are called “drought evaders” or “drought escapers”. They are annuals which complete their life cycle with in a period of 6-8 weeks. They are found in dry zones, e.g. : Tribulus, chenopodium.

2. Succulents :
These are called “drought avoiding plants”. They are fleshy due to storage of water in the form of mucilage. The stored water is sparingly utilised during dry periods.
Examples :
a) Stem succulents : Opuntia, Euphorbia.
b) Leaf succulents : Bryophyllum, Aloe, Agave etc.
c) Root succulents : Asparagus, Ceiba.

Non-succulents :
They are called true xerophytes. These are perennial plants which can withstand prolonged period of drought.
E.g. : Casuarina, Nerium, Ziziphus, Calotropis, Acacia, etc. .

Question 5.
Enumerate the morphological adaptations of xerophytes.
Answer:

1. Roots are long with extensive branching spread over wide areas.
2. Root hairs and root caps are very well developed.
3. Mostly the stem is stunted, woody, hard and covered with thick bark.
4. Stems are usually covered by hairs or waxy coatings.
5. Leaves are very much reduced to small, scale like and sometimes modified into spines to reduce the rate of transpiration.
6. Certain Xerophytes shed their leaves during the dry period.
   Ex : Capparis.

Question 6.
Give in detail the anatomical adaptations shown by xerophytes.
Answer:

1. Epidermis is covered with thick cuticle to reduce the rate of transpiration.
2. Epidermal cells may have silica crystals.
3. Epidermis may be multilayered as in leaves of Nerium.
4. Stomata are generally confined to lower epidermis of leaves called hypostomatous. They are present in pits called sunken stomata.
5. Hypodermis is parenchymatous to check evaporation of water.
   Ex : Calotropis.
6. Mechanical tissues are very well developed.
7. Vascular tissues are very well developed.

Question 7.
Define plant succession. Differentiate primary and secondary successions.
Answer:
The gradual and fairly predictable change in the species composition of a given area is called plant succession.

Differences :

Question 8.
Define ecosystem/ecological services. Explain in brief with regard to pollination.
Answer:
The processes by which the environment produces resources that we often take for granted such as cleen water, timber and habitat for fisheries and pollination of native and agricultural plants is called Ecosystem /Ecological services.

The transfer of pollengrains to fertilize the ovaries of flowers is called pollination. It is an essential part of a healthy ecosystem. Most flowering plants require pollinators to produce fruits and seeds. So pollinators play a significant role in the production of more food crops in the world. Declines of pollinator activity could have serious economic repercussions throughout the world.

The most important pollinator for Agriculture is Honeybee. Over 1,00,000 invertebrate species such as bees, moths, butterflies, beetles and flies serve as pollinators worldwide. At least 1035 species of vertebrates including birds, mammals and reptiles also pollinate many plant species. Continued declines in pollinator activity could mean rising costs for pollinator dependent fruits and vegetables and the disruption of entire ecological systems.

Question 9.
Write about the measure to be taken to sustain ecological functions.
Answer:

1. Choose products produced with methods that conserve resources, minimize waste and reduce or eliminate environmental damage.
2. Prefer products made with methods that reduce the use of pesticides and artificial fertilizers.
3. Reduce consumption and waste production.
4. Support usage of renewable energy alternatives.
5. Use public transit, cycle or walk to conserve natural resources and to reduce pollution and enjoy the health benefits.
6. Participate in developing community garden and tree plantation programmes.
7. Avoid the usage of pesticides and follow methods of natural pest control.
8. Use native plants in the garden and provide habitat for wildlife.

Question 10.
What measure do you suggest to protect the pollinators?
Answer:

1. Creating own pollinator friendly garden using a wide variety of native flowering plants.
2. Reducing the use of pesticides used in and around your home.
3. Encouraging local clubs or school groups to build artificial habiutats like butterfly gardens, bee boards and bee boxes.
4. Supporting agriculture enterprises with pollinato-friendly practices to minimize pesticide use.
5. Encouraging government agencies to take into account the full economic benefits of wild pollinators when formulating policies for agriculture and other land uses.
6. To develop techniques for cultivating native pollinator species for crot pollination.

Long Answer Type Question

Question 1.
Give an account of ecosystem services with reference to carbon fixation and oxygen release.
Answer:
Trees are essential to carbon sequestration keeping excess carbon from entering the atmosphere. The main chemical flow between forest and atmosphere is the exchange of CO₂ and O₂. Forests provide a vast bank for CO₂ and a large amount of CO₂ is deposited in its timber. It plays an essential role in maintaining a dynamic balance between CO₂ & O₂ in atmosphere. According to photosynthesis equation 180 gm of Glucose and 193 gm O₂ are produced by using 264 gm of CO₂ and 108 gm of water and 677.2 K.cal. of solar energy.

180 gm of Glucose can be transformed to 162 gm of polysaccharide inside the plant. So whenever plant produces, 162 gm of dry organic matter 264 gm of CO₂ will be fixed. Then the total amount of the dry organic matter of the reserve forests can be estimated. It provides a foundation for reckoning the total amount of CO₂ fixation by the forests in the reserve.

Natural ecosystems may have helped to stabilize climate and prevent overheating of the earth by removing more of the greenhouse gas, CO₂ from the atmosphere. Many countries have established a carbon tax system to reduce emissions of the greenhouse gases, especially to cut down CO₂ and CO in atmosphere.

Ecosystem services – oxygen release :
Trees and plankton play a big role in release of oxygen; which depends on the species of tree, its age, its health, and also on the trees surroundings. “A mature leafy tree produces as much oxygen in a season as 10 peole in hale in a year”, or A single mature tree can absorb carbon dioxide at a rate of 48 Ibs/year and release enough oxygen back into the atmosphere to support 2 human beings.

One acre of trees annually consumes the amount of CO₂ equivalent to that produced by driving an average can for 26,000 miles. That same acre of trees also produces enough oxygen for 18 people to breathe for a year.

Submerged macrophytes release O₂ and enrich dissolved O₂ in water. The plants and planktons are described as “the Lungs of the World”, taking billion of tonnes of CO₂ and exhaling billions of tones of O₂.

Micro organisms also contribute to the oxygen release in direct and indirect ways.

Ex : Cyanobacteria releases O₂ in a direct way. The other supporting services include Nutrient cycling through decomposition of fallen Logs in forests, soil formation by bacteria and lichens.

Intext Questions

Question 1.
Categorise the following plants into hydrophytes halophytes, mesophytes and xerophytes and give reasons.
a) Salvinia b) Opuntia c) Rhizophora d) Mangifera
Answer:
a) Salvinia is a hydrophyte, It grows on the surface of water.
b) Opuntia is a xerophyte, grows in xeric (dry) areas.
c) Rhizophora is a Halophyte which tolerates the salinities of the sea.
d) Mangifera is a mesophyte, grows in habitats where water is neither scarce nor not abundant.

Question 2.
In a pond, we see plants which are free-floating ; rooted-submerged; rooted emergent; rooted with floating leaves, write the type of plants against each of them.
Answer:

Question 3.
Undertake the following a part of learning process.
a) Identify and assess ecological services found in your area.
b) Think of measures or means to sustain such ecological services.
c) Observe the type of plants or crops grown in your area.
d) Enumerate ecological servies of your area.
e) find out the ecological goods of natural forests commonly used in your area.
f) Observe the biotic agents of pollination for ornamental flowring plants and or agricultural crops in your locality.
Answer:
a) Ecological services :

1. Purification of air and water.
2. Detoxification and decomposition of water.

b) Measures to sustain Ecological services are

1. Reduce consumption and waste production.
2. Avoid the usage of pesticides.

c) Crops grown in our area are
a) Paddy b) Maize c) Black gram d) green gram e) Crotalaria (fodder) g) vegetables.

d) Ecological services :

1. Purification of air and water.
2. Mitigation of floods and droughts
3. Decomposition of wastes

e) Ecological goods :
a) Clean air b) Fresh water c) Food d) fibre e) Timber f) Medicines.

f) Biotic agents of pollination
a) Insects b) Birds c) Animals (Bats, Snails, Snakes).

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 12th Lesson Histology and Anatomy of Flowering Plants Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 12th Lesson Histology and Anatomy of Flowering Plants

Very Short Answer Questions

Question 1.
The transverse section of a plant material shows the following anatomical features.
a) The vascular bundles are conjoint, scattered, and surrounded by a sclerenchymatous bundle sheath.
b) Phloem parenchyma is absent. What will you identify it as?
Answer:
Monocot Stem.

Question 2.
Why are the xylem and phloem called complex tissues?
Answer:
Xylem and phloem are permanent tissues having more than one type of cell that work together. So-called complex tissues.

Question 3.
How is the study of plant anatomy useful to us?
Answer:
First of all the study of plant Anatomy helps us understand the way a plant functions, carrying out its routine activities like transpiration, photosynthesis, growth and repair. Second, it helps botanists and agriculture scientists to understand the disease and cure for the plants. Anatomy is way to understand the larger system of Ecology on this planet.

Question 4.
Protoxylem is the first formed xylem. If the protoxylem lies rodialy next to pholem what kind of arrangement of xylem would you call it? Where do you find it?
Answer:
Radial vascular Bundle. They are found in Roots.

Question 5.
What is the function of phloem parenchyma?
Answer:
Phloem parenchyma stores food material and other substances like resins, latex and mucilage.

Question 6.
a) What is present on the surface of the leaves which helps the plant to prevent loss of water but is absent in roots?
b) What is the epidermal cell modification in plants which prevents water loss?
Answer:
(a) Cuticle
(b) Trichomes.

Question 7.
Which part of the plant would show the following?
(a) Radical vascular bundle (b) Polyarch xylem (c)Well developed pith (d) Exarch xylem
Answer:
a) Root
b) Monocot Root
c) Monocot Root
d) Roots

Question 8.
What, are the cells that make the leaves curl in plants during water stress? Give an example.
Answer:
Bulliform cells. Ex : Monocot leaf (grass).

Question 9.
What constitutes the Vascular combial ring?
Answer:
Intrafascicular cambium + Interfascicular cambium constitutes cambial ring.

Question 10.
Give one basic functional difference between phellogen and phelloderm.
Answer:
Phellogen is also called cork cambium which appears in cortex and produces cork and phelloderm. The cells in the phellogen are thin walled, rectangular.

Phelloderm :
The cells which are formed towards inside from phellogen constitutes phelloderm or secondary cortex. The cells are parenchymatous.

Question 11.
If one debarks a tree, what parts of the plant are removed?
Answer:
Periderm and secondary phloem.

Short Answer Type Questions

Question 1.
State the location and function of different types of meristems.
Answer:
Based on the position, meristems are classified into three types.
A) Apical Meristems :
The meristems that are present at the tip of the root and at the tip of the stem or branches are called Apical Meristems. They help in the linear growth of the plant body.

B) Intercalary Meristems :
The Meristems that are present in between mature tissues are known as Intercalary Meristems. They occur in grasses. They also contribute to the formation of the primary plant body.

C) Lateral Meristems :
The meristems that occur in the mature regions of roots and shoots of many plants are called lateral meristems. They help in increase in thickness of the plant organs (Root, Stem). Ex : Vascular cambium, cork cambium.

Question 2.
Cut a transverse section of young stem of a plant from your garden and observe it under the microscope. How would you ascertain whether it is a monocot stem or a dicot stem? Give Reasons.
Answer:
The transverse section of a young stem which was observed under Microscope shows some characters to indicate that it is either Monocot or Dicot. They are ;

With the above differences. We can observe the stem, whether it belongs to Dicot or Monocot stem.

Question 3.
What is periderm? How does periderm formation take place in the dicot stems?
Answer:
Phellogen, phellem and phelloderm are collectively known as “Periderm”.

Periderm :
Due to the formation secondary vascular tissues inside the stele, a pressure is cortex on the epidermis causing it to rupture. In the mean while, a secondary protective layer is formed called ‘periderm’ from the cortex. The secondary growth in the cortex begins with the appearence of a meristematic layer of cells from the middle part of the cortex.

This is called “Phellogen or cork cambium”. The cells of the phellogen divide periclinally and cuts of new cells on either side. The cells produced towards outside are called cork cells or phellem and the cells produced towards inside are called “secondary cortex cells or phelloderm”. The phellogen (cork cambium), phellum (cork) and phelloderm (secondary cortex) together constitute “periderm”.

Question 4.
A transverse section of the trunk of a tree, shows concentric rings which are known as annual rings. How are these rings formed? What is the significance of these rings?
Answer:
Annual rings :
In temperate regions and cold regions the activity of the cambium is influenced by the seasonal variations. During the favourable season, i.e., in spring, when more leaves and flowers are formed, the plant requires large amounts of water and mineral salts. Hence the wood formed in this period shows more number of xylem vessels with wider lumens.

This is known as spring wood or early wood. The colour of this wood is light. During the unfavourable season i.e., in autumn, the plants are less active and do not require more water and mineral salts. Hence the wood produced in this period shows less number of xylem vessely with narrow lumens. This is known as autumn wood or late wood. It is dark coloured. In this way two types of secondary xylem (wood) are produced in one year. They appear in the form of dark and light coloured circles alternately in a mature tree trunk. These are called annual rings or growth rings or seasonal rings.

By counting the number of annual rings, the approximate age of trees can be estimated. This branch of science is known as “dendrochronology or growth ring analysis”. For Ex : The age of sequoid dendron, presently growing in America, is estimated to be about 3500 years. In tropical countries like India, annual rings do not appear clearly, as the seasonal variations are not sharp. Hence these are called growth marks.

Question 5.
What is the difference between lenticels and stomata?
Answer:

Question 6.
Write the precise function of
(a) Sieve tube (b) Interfasicular cambium (c) Collenchyma (d) Sclerenchyma.
Answer:
a) Sieve tube :
They are the more advanced type of conducting cells are and found in phloem of Angiosperms. They are elongated cells, arranged end to end and functioning to conduct food materials through out the plant. They are a nucleate living cells.

b) Interfasicular cambium :
The parenchymatous cells present between vascular Bundles, become meristamatic and form a cambium called Interfascicular cambium.

c) Collenchyma :

1. It is a living mechanical tissue cellwall is composed of cellulose, hemicellulose and pectin.
2. Collenchyma cells with chloroplasts perform photosynthesis.
3. They provide mechanical support to the growing parts of the plants such as young stem and petiole of a leaf.

d) Sclerenchyma :
It is Dead mechanical tissue. Cell wall is made up of lignin. Intercellular spaces are absent.

1. Fibres are useful in textile and jute industries. ,
2. Fibres give mechanical support to the plant parts.
3. Sclerids give mechanical support to the plant parts.

Question 7.
The stomatal pore is guarded by two kidney shaped guard cells. Name the epidermal cells surrounding the guard cells. How does a guard cell differ from an epidermal cell? Use a diagram to illustrate your answer.
Answer:
Stomata are structures present in the epidermis of leaves. Each stomata is composed of two bean shaped cells known as guard cells. The outer walls of guard cells are thin and the inner walls are thick. The guard cells possess chloroplasts and regulate the opening and closing of stomata. Sometimes a few epidermal cells, become specialized in their shape and size are called subsidiary cells. Differences between Guard cell and Epidermal cells.

Question 8.
Point out the differences in the anatomy of leaf of peepal (Ficus religiosa) and Maize (Zea mays). Draw the diagrams and label the differences.
Answer:

Dicot leaf	Monocot leaf
1. Stomata are more in the lower epidermis than in the upper epidermis.	1. Stomata are equally distributed on both the sides.
2. Bulliform cells are absent.	2. Bulliform cells are present in the upper epidermis.
3. Mesophyll is differentiated into palisade and spongy tissue.	3. Mesophyll is undifferentiated.
4. Bundle sheath extensions are generally parenchymatous.	4. Bundle sheath extensions are generally sclerenchymatous.
5. They are dark green on the upper surface, less green on the lower surface. (Dorsiventral)	5. They are in same colour on both the surfaces. (Isobilateral)

Question 9.
Cork cambium forms tissues that form the cork. Do you agree with this statement? Explain.
Answer:
Yes, cork cambium or phellogen is formed during secondary growth of Dicot stem. Cork cambium or phellogen cuts off on both sides. The cells produced towards outerside differentiates into cork or phellem and the inner cells differentiates into secondary cortex or phelloderm. The cork cells are impervious to water due to deposition of suberin in the cell wall.

Question 10.
Name the three basic tissue systems in the flowering plants. Give the tissue names under each system.
Answer:
In flowering plants, there are three tissue systems are present namely.

1. Epidermal tissue system
2. Ground tissue system
3. Vascular tissue system.

1) Epidermal tissue system :
It consists of epidermis, cuticle, stomata, unicellular hairs and Multicellular trichomes.

2) Ground tissue system :
It consists of simple tissues like parenchyma, collenchyma and sclerenchyma. These cells are present in cortex, pericycle, pith, Medullary rays, Hypodermis, Endodermis layers. In leaves, the ground tissue consist of thin walled chloroplast containing cells called Mesophyll.

3) Vascular tissue system :
It consists of complex tissues, the xylem and the phloem.

Long Answer Type Questions

Question 1.
Explain the process of secondary growth in the stems of woody Angiosperms with the help of schematic diagrams. What is its significance?
Answer:
Formation of cambium ring :
In the primary structure of dicto stem, the stele shows vascular bundles in the form of a ring. Each vascular bundle consists of cambium in between the xylem and phloem. This is called Inter fascicular cambium. In between the vascular bundles, there are medullary rays. From the cells of medullary rays intrafascicular cambium is formed. The Inter fascicular and intrafascicular cambia fuse to form a continuous cambial ring called “vascular cambium”.

Activity of the vascular cambial ring :
The cells of vascular cambium divide repeatedly by periclinal method and produce new cells on both the sides. The cells which are produced outside develop into secondary phloem and those produced to the inner side develop into secondary xylem (wood). Generally more secondary xylem is produced than the secondary phloem. The secondary xylem consists of xylem vessels, tracheids, xylem fibres and xylem parenchyma. The secondary phloem consists of sieve tubes, companion cells, phloem fibres and phloem parenchyma.

In the cambium two types of initiating cells are found. They are 1. Fusiform initials and 2. Ray initials. The fusiform initials give rise to the secondary xylem and the secondary phloem. The ray initials produce phloem rays (bast rays) to the outside and xylem rays (wood rays) to the inside. They are helpful in lateral conduction and storage. They are called secondary medullary rays.

Annual rings :
In temperate regions and cold regions the activity of the Cambium is influenced by the seasonal variations. During the favourable season, i.e., in spring, when more leaves and flowers are formed, the plant requires large amounts of water and mineral salts. Hence the wood formed in this period shows more number of xylem vessels with wider lumens. This is known as spring wood or early wood. The colour .of this wood is light during the unfavourable season ie., in autumn, the plants are less active and do not requirfe more water and mineral salts.

Hence the wood produced in the period shows less number of xylem vessels with narrow lumens. This is knwon as Autumn wood or late wood. It is dark coloured. In this way two types of secondary xylem (wood) are produced in one year. They appear in the form of dark and light coloured circles alternately in a mature tree trunk. These are called Annual rings or growth rings or seasonal rings.

By counting the number of annual rings, the approximate age of trees can be estimated. This branch of science is known as “dendrochronology or growth ring analysisFor example the age of sequoid dendron, presently growing in America, is estimated to be about 3500 years, in tropical countries like India, annual rings do not appear clearly, as the seasonal variations are not sharp. Hence these are called growth marks.

Heart wood and sapwood :
With the increase in the age of the tree, the wood undergoes a number of hysical and chemical changes. The older wood gradually loses water and stores food substances and becomes infilterated with various organic compounds such as oils, gums, resins, tannins, colouring agents and aromatic substances. Hence the older xylem present in the centre appears dark in colour. This is called heart wood or duramen.

It is very hard highly durable. Heart wood cannot conduct water and salts because of the growth of tyloses in the lumens of xylem vessels. The heart wood gives mechanical strength to the tree.

The newly formed secondary xylem is found in the peripheral part of the tree trunk. This is called sapwood or alburnum. It is light in colour and is active in conducting water, mineral salts and storage of food materials. As time passes on, the sap wood gradually changes into heart wood. Hence the sap wood remains uniformly thick.

Periderm :
As the secondary xylem and secondary phloem are formed inside the stele, a pressure is exerted on the epidermis, causing its rupture. Mean while a secondary protective layer formed from the middle or inner part of the cortex become meristematic and acts as phellogen or cork cambium. These cells divide periclinally and cuts of new cells towards outside called cork or phellem and towards inside called secondary cortex or phelloderm. The phellogen, phellem and phelloderm together constitute periderm.

At certain regions, the phellogen cuts off closely arranged parenchymatous cells on the outer side instead of cork cells, called complementary cells. These cells soon rupture the epidermis forming a lens-shaped oepnings called lenticels. They permit the exchange of gases between the outer atmosphere and the internal tissues.

Question 2.
Draw illustrations to bring out the anatomical differences between
a) Monocot root and Dicot root
b) Monocot stem and Dicot stem.
Answer:
a) Monocot root and Dicot root

b) Monocot stem and Dicot stem.

Question 3.
What are simple tissues? Describe various types of simple tissues.
Answer:
Tissues which are made up of only one type of cells are called simple tissues. They are of three types. They are parenchyma, Collenchyma and Sclerenchyma.

1) Parenchyma :
It is living tissue. It occupies the major part of the plant body and the cells are isodiametric. They may be spherical, oval, polygonal or elongated in shape. Their walls are thin and made up of cellulose. They have small intercellular spaces. It performs photosynthesis, storage, secretion, Healing of wounds secretion, Buoyancy by storing air etc.

2) Collenchyma :
It is simple living mechanical tissue, occurs below the epidermis of Dicot plants. It consists of cells which are thickened at the corners due to deposition of cellulose, hemicellulose and pectin. The cells are oval or spherical or polygonal in shape and contain chloroplasts. Intercellular spaces are absent. They may be angular or lacunar or Lemellar type. They help in assimilation and provide mechanical support to young stems and petiole of a leaf.

3) Sclerenchyma :
It is a dead Mechanical tissue, consists of long, narrow cells with thick and lignified cell walls having pits. They are usually dead cells and without protoplasts. Based on the form, structure, it may be either fibres or sclereids. The fibres are thick walled, elongated and pointed cells, gives mechanical support and also used in Jute Industries. The sclereids are spherical, oval or cylindrical, highly thickened dead cells, found in the fruit walls of nuts, pulp of fruits like guava and sapota, seed coats of legumes and leaves of tea. They also provide Mechanical support to organs.

Question 4.
What are complex tissues? Describe various types of complex tissues
Answer:
Tissues which are made up of more than one type of cells and work together as a unit are called complex tissues. They are of two types namely xylem and phloem.

Xylem :
It is a conducting tissue, helps in conduction of water and minerals form roots to stem and leaves. It is composed of four elements namely tracheids, vessels, xylem fibres and xylem parenchyma.

Tracheids are elongated or tube like cells with thick and lignified walls and tapered ends. These are dead and are without protoplasm. Vessels are long cylindrical tube like structures, made of many cells with lignified walls and a large central cavity. Vessels are interconnected through perforations in their walls. They are, also dead cells. Both Tracheids and vessels are the main water transporting channels. Xylem fibres are long, with thick walls and narrow lumens, gives mechanial strength, xylem parenchyma cells are living and thin walled cells, made up of cellulose. They store food materials in the form of starch of fat.

Phloem :
It is a complex tissue, helps in conduction of food materials. It is composed of sieve tube elements, companion cells, phloem parenchyma and phloem fibres. Sieve tube elements are long tube like structures and are associated with companion cells. Their end walls are performted in a sieve like manner to form the sieve plates. A mature sieve element shows a peripheral cytoplasm and a large vacuole but lacks a nucleus. The companion cells are specialised parenchymatous cells which are connected to sieve tube by pit fields the companion cells help in maintaining the pressure gradient in the sieve tubes.

Phloem parenchyma is made up of cylindrical cells which have dense cytoplasm and nucleus. The cell wall is composed of cellulose and has pits. It stores food material and other substances like resins, Latex and Mucilage. Phloem fibres are long elongated unbranched cells with pointed apices. The cell wall is thick and is made up of lignin. Phloem fibres of jute flax and hemp are used commerically.

Question 5.
Describe the internal structure of dorsiventral leaf with the help of labelled diagram.
Answer:
Transverse section of a dorsiventral leaf (dicot leaf) shows 3 important parts. They are

1. Epidermis,
2. Mesophyll and
3. Vascular byndles.

1. Epidermis :
Epidermis is present on the both the upper surface (adaxial) and the lower surface Cabaxial) of the leaf. The epidermis present on the adaxial surface is called upper epidermis and on the abaxial surface is called lower epidermis. The epidermis is made up of one row of barrel shaped cells, which are arranged compactly without intercellular spaces. The cells are filled with vacuolated and nucleated protoplast. On outerside of the epidermis a waxy layer called Cuticle is present.

Stomata are present, more on the lower surface than on the upper surface. Each stoma is surrounded by two kidney shaped guard cells. They are chlorophyllous and regulate the opening and closing of stomata. Epidermis shows multicellular uniseriate hairs. The cells of leaf hairs are filled with water. They protect the inner tissues by absorbing the heat and prevents evaporation of water from the leaf surface. The stomata help in the gaseous exchange and also promote transpiration.

2. Mesophyll :
The ground tissue that extends between the upper and lower epidermal layers is called the mesophyll. It is composed of thin walled parenchyma with chloroplasts. It is chiefly concerned with the synthesis of carbohydrates. In dicot leaves mesophyll is differentiated into two parts namely,
i) Palisade parenchyma and
ii) Spongy parenchyma.

i) Palisade parenchyma :
Part of the mesophyll found beneath the upper epidermis is called ‘palisade tissue’. It shows elongated, columnar cells arranged in 1-3 vertical rows. Narrow intercellular spaces are present between the cells. In these cells, large numbers of chloroplasts are found nearer to the cell wall. Palisade tissue is primarily concerned with the manufacture of carbohydrates in the presence of sunlight.

ii) Spongy parenchyma :
Part of the mesophyll found towards the lower epidermis is called spongy tissue. It shows 3-5 rows of irregular shaped cells that are arranged loosely with large intercellular spaces. Some intercellular spaces present in the vicinity of the stomata are very large, forming air chambers (air cavities). In thise cells, number of chloroplasts is less. That is why the upper surface of leaf is dark green and the lower surface is light green in colour. Spongy tissue has a primary role in gaseous exchange, apart from the synthesis of food materials.

3. Vascular bundles :
Vascular bundles are extended in the mesophyll in the form of veins. They help in supplying water, mineral salts and food materials all over the leaf surface. Veins also provide mechanical strength to the leaf.

The vascular bundles are conjoint, collateral and closed. The xylem is present on the upper side and phloem on the lower side. Cambium is absent between them. Xylem shows vessles, tracheids, parenchyma and fibres. Phloem shows sieve tubes companion cells and phloem parenchyma.

Each vascular bundle is surrounded by a layer of specialised mesophyll cells that are arranged closely and compactly without intercellular spaces. This layer is called bundle sheath or border parenchyma. The bundle sheath cells divide and grow towards the upper and lower epidermal layers. These are called bundle sheath extensions. They help in the conduction of food materials form the mesophyll to the vascular bundles.

Question 6.
Describe the internal structure of an isobilateral leaf with the help of labelled diagram.
Answer:
The internal structure of a monocot leaf (isobilateral leaf) shows 3 main parts, namely 1. Epidermis, 2. Mesophyll and 3. Vascular bundles.

1) Epidermis :
Epidermis is present on both the upper surface (adaxial) and the lower surface (obaxial) of the leaf. The epidermis is made up of barrel shaped cells which are arranged compactly without intercellular spaces. The cells are filled with vacuolated cytoplasm and possess a single nucleus but chloroplasts are absent. Epidermal hairs are absent. Epidermis is externaly covered by a waxy layer called cuticle. The number of stomata on both the sides is almost equal. In some monocots like grasses some cells of upper epidermis are enlarged and specialised, called Bulliform cells or Motor cells. They are thin walled and are filled with water. They help in rolling and unrolling of the leaf. The epidermis gives protection to the inner tissues, regulates the transpiration and helps in gaseous exchange.

2) Mesophyll :
It is present between the upper and lower epidermal layers. It is made up of several layers of columnar cells or spongy cells, that are loosely arranged showing intercellular spaces. They contain chloroplasts. The mesophyll is the cheief photosynthetic tissue of the leaf.

3) Vascular bundles :
Numerous vascular bundles are present in the mesophyll in the form of veins. The vascular bundles are conjoint, collateral and closed. Xylem is present towards upper side and phloem towards lower side.

Each vascular bundle is enclosed by a layer of specialised mesophyll cells called border parenchyma or bundle sheath. Sometimes, bundle sheath is composed of dead sclerenchymatous tissue. The bundle sheath cells divide and grow towards both the sides of the vascular bundle. They are called bundle sheath extensions. They help in the conduction of materials from the mesophyll to the vascular bundle. They also give mechanical strength to the leaf.

Question 7.
Distinguish between the following :
a) Exarch and endarch condition of protoxylem.
b) Stele and vascular bundle.
c) Protoxylem and metaxylem.
d) Interfasicular cambium and Intrafasicular cambium.
e) Open and closed vascular bundles.
f) Stem hair and root hair.
g) Heat wood and sap wood.
h) Spring wood and Autumn wood.
Answer:
a)

b)

c)

d)

e)

f)

g)

h)

Question 8.
What is stomatal apparatus? Describe the structure of stomata with a labelled diagrams.
Answer:
The stomatal aparture, Guard cells and the subsidiary cells together called “Stomatal apparatus”.

Structure of Stomata :
Stomata are the structures present in the epidermis of leaves. Stomata regulate the process of transpiration and gaseous exchange. Each stoma is composed of two bean-shaped Guard cells. In grasses, the guard cells are dumbbell shaped. The outer walls of guard cells are thin and inner walls are highly thickened.

The guard cells possess chloroplasts and regulate the opening and closing of stomata. Some times, a few epidermal cells become specialised in their shape and size and are known as subsidiary cells. The stomatal aperture, guard cells and the subsidiary cells are together called stomatal apparatus.

Question 9.
Describe the T.S of a dicot stem.
Answer:
The structure of young dicot stem can be clearly understood by observing the transverse section of stem of Helianthus annus (sunflower). It shows three major zones, namely epidermis, cortex and stele.

1. Epidermis :
It is the outer most layer of rectangular or tubular cells arranged compactly without any intercellular spaces. On outer surface of epidermis, a waxy layer called Cuticle is found. The cuticle is chemically composed of a substance cutin. The cell walls of epidermis also show the presence of cutin. Stomata are present in the epidermis. Multicellular trichomes develop on the epidermis. The cuticle and the trichomes check the evaporation of water and protect the stem from high temperature.

The epidermal layer gives protection to the inner tissues and also prevents the evaporation of water from the plant body. Through stomata, the epidermis allows the exchange of gases and promotes transpiration. Trichomes prevent entry of pathogens.

2. Cortex :
The part extending between the epidermis and the stele is known as cortex. The cortex is smaller than the stele. It shows three sub-zones, namely, i) Hypodermis ii) General Cortex and iii) Endodermis.

i) Hypodermis :
This is the outermost part of cortex and composed of 3-6 rows of collenchy matous cells. It is found beneath the epidermis and helps in providing tensile strength (elasticity) to the stem. The cells are arranged compactly without intercellular spaces and show excessively thickened corners. The cells are filled with active vacuolated cytoplasm possessing chloroplasts. Thus, the hypodermis also helps in the assimilation of food materials. It also gives mechanical strength.

ii) General Cortex :
It is found beneath the hypodermal layer and is made up of 5 – 10 rows of thin walled, living parenchyma cells with or without intercellular spaces. These cells may be isodiametric or oval or spherical. The outer layers of cells contain chloroplasts and in the inner layers leucoplasts are found. The general cortex is primarily concerned with the assimilation and storage of food materials.

iii) Endodermis :
The inner most layer of cortex is called endodermis. The cells are barrel shaped, compactly arranged without intercellular spaces. The endodermis cells contain vacuolated protoplasts and show starch grains. So it is also known as ‘Starch Sheath’.

3) Stele :
The central conducting cylinder is called the ‘Stele’. It occupies a major part of the stem. It Is composed of 4 parts.
i) Pericycle ii) Vascular bundles iii) Pith or Medulla and iv) Medullary rays.
i) Pericycle:
It is present in the form of a discontinuous ring and is made up of 3-5 rows of the thick walled, dead, lignified cells which gives mechanical strength to the stele. It appears as semilunar patches of sclerenchyma above the vascular bundles with intervening masses of parenchyma.

ii) Vascular bundles :
About 15-20 vascular bundles are arranged in the form of a ring. This arrangement is called eustele. Each vascular bundle is wedge or top shaped. In the vascular bundels xylem and phloem are arranged on the same radius. A meristematic layer of cells called cambium is present in between the xylem and phloem. So they are called conjoint, collateral, open vascular bundles. Xylem is at the lower side and phloem at the upper side of the vascular bundle.

Xylem consists of vessels and xylem parenchyma. There may befewtracheidsand xylem fibres. The metaxylem is towards the pericycle and protoxylem towards the pith. This is called endarch xylem. Phloem consists of sieve tubes companion cells and phloem parenchyma. Xylem and phloem are vascular tissues which conduct water, mineral salts and organic solutes respectively.

iii) Medulla :
It is the central part of the stele and filled with thin walled parenchymatous cells, showing intercellular spaces. It is well-developed, extensive and occupies a large part of the stele. The chief function of the medulla is to store food materials.

iv) Medullary rays :
The cells of the medulla extend to the periphery in between the vascular bundles. These cells are horizontal rows of thin walled, living and elongate radially, forming primary medullary rays.

The medullary rays connect the stele with the cortex and are helpful in lateral conduction.

Question 10.
Describe the T.S of a Monocot stem.
Answer:
The structure of monocot stem can be understood well by observing the T.S of stem of Zea mays. It shows 4 distinct parts., namely

1. Epidermis,
2. Hypodermis,
3. Ground tissue and
4. Vascular bundles.

Epidermis :
It is the outermost layer composed of rectangular or tubular living cells arranged closely and compactly without intercellular spaces. The cells contain vacuolated protoplasts with a single nucleus but chloroplasts are absent. A waxy layer called ‘cuticle’ is deposited on the external surface of the epidermis. Cuticle prevents the evaporation of water from the plant body. Trichomes are absent. Numerous stomata are found in the epidermis through which exchange of gases occurs.

Epidermis gives protection to the inner tissues, helps in the exchange of gases and also prevents the evaporation of water.

Hypodermis :
A distinct cortex is absent in Monocot stems. However a thick walled hypodermis is found beneath the epidermis. The cells of hypodermis are sclerenchymatous and are arranged compactly in 3 – 4 rows, without any intercellular spaces. It gives mechanical strength to the stem.

Ground tissue :
A major part of the stem is formed by an extensive soft, parencymatous tissue called the ground tissue. The peripheral layer consists of smaller cells while the inner layers show bigger cells. The cells of peripheral layers are chlorenchymatous and are concerned with assimilation of food

Vascular bundles :
Numerous vascular bundles are found scattered irregularly in the ground tissue. This kind of arrangement is called ‘atactostele1. It is considered an on advanced character. The inner vascular bundles are bigger in size and far apart from one another. The outer vascular bundles are smaller and are close to one another and found In one or two circles.

Each vascular bundle is oval in shape and shows xylem and phloem together on the same radius. There is no cambium between xylem and phloem. Hence the vascular bundles are called conjoint, collateral and closed. Xylem is at the lower side and phloem at the upper side of the vascular bundle. Each vascular bundle is enclosed by a sheath of sclenrenchymatous fibres. Hence it is called fibro vascular bundle.

Xylem consists of tracheids, vessels, fibres and parenchyma. Xylem vessels are few in number (4) and are arranged in “Y” shape. One or two protoxylem cells are crushed forming lysigeneous cavity called protoxylem lacunae which store water phloem consists of sieve tubes and companion cells. Phloem parenchyma is absent Medulla, Medullary rays and pericycle are also absent.

Question 11.
Describe the internal structure of a Dicot Root.
Answer:
The transverse section of primary dicto root can be divided into 3 zones. They are : Epidermis, Cortex and Stele.

1) Epidermis :
Epidermis is the outermost layer made up of thin walled, non-cutinised, rectangular living cells. The epidermal cells are arranged compactly without intercellular spaces. The cuticle and stomata are absent. Some epidermal cells produce tubular extensions called root hairs. Due to presence of root hairs the root epidermis is called is as epiblema or rhizodermis or piliferous layer.

The cells that give rise to root hairs are comparatively smaller than the other cells and are called trichoblasts. Root hairs help in the absorption of capillary water. The epidermis gives protection to the inner tissues and plays major role in the absorption.

2) Cortex :
The tissue extended between epidermis and stele is called ‘cortex’. Generally in roots materials. The inner layers of cells are concerned with storage of food materials. In monocot stem endodermis is absent. the cortex is bigger than the stele. Cortex can be differentiated into three parts :

1. Exodermis,
2. General cortex and
3. Endodermis.

i) Exodermis :
It is the outermost layer of cortex and composed of two to three rows of thick walled suberised cells. When the epidermal layer is removed, the exodermis acts as the protective layer. It also prevents the exit of water from the cortex. It can be observed in mature part of the root.

ii) General cortex :
It is present beneath the exodermis and is composed of several rows of thin walled, living parenchyma cells. The cells are round or oval in shape and loosely arranged showing intercellular spaces. They contain leucoplasts which store the food materials. The general cortex helps in the lateral conduction of water from the epidermis to the xylem vessels present in the stele.

iii) Endodermis :
It is the innermost layer of cortex and is made up of a single row of barrel shaped cells. The cells are compactly arranged without having any intercellular spaces. The radial and transverse walls of the endodermal cells show casparian strips that are formed by the deposition of lignin and suberin which prevent the movement of water. Therefore the endodermis acts as a barrier between the cortex and the stele.

In endodermis, some cells situated opposite to the protoxylem elements are thin walled without casparian bands. These cells are called passage cells. They help in the translocation of water and mineral salts from the cortex into the stele.

3) Stele :
The central conducting cylinder is known as ‘stele’. It is smaller than the cortex. The stele is comprised of three parts, viz., pericycle, Vascular bundles and Medulla.

i) Pericycle :
The layer of cells surrounding the stele is known as ‘pericycle’. It is usually uniseriate and composed of thin walled, rectangular, living cells which show active cell division. The pericycle gives rise to lateral roots. Some cells of the pericycle can dedifferentiate into secondary cambium which results in the secondary growth of the root.

ii) Vascular bundles :
Standards of primary xylem and phloem are found alternately on separate radii. These are called separate or radial vascular bundles’. Xylem is exarch showing protoxylem elements towards the pericycle and metaxylem elements towards the medulla. The number of vascular bundles is identified in relation to the number of xylem groups. Usually in dicot roots, four xylem bundles alternating with four phloem bundles are found. This is called ‘tetrach condition’.

There is no cambium between the vascular tissues. The ground tissue that extends between the xylem and phloem strands is called conjunctive tissue. It is usually parenchymatous. It helps in the storage of food materials. It porduces secondary cambium during secondary growth.

Medulla or Pith :
In roots, the development of xylem is centripetal and produces metaxylem towards the inner side. Sometimes, as in dicot root these metaxylem elements come closer from all sides and replace the medulla. Hence in dicot roots the medulla is very small or may be completely absent. When present, it is parenchymatous and helps in the storage of food and water.

Question 12.
Describe the internal structure of a Monocot Root.
Answer:
The internal structure of Monocot root shows 3 zones. They are :

1. Epidermis
2. Cortex and
3. Stele.

1) Epidermis :
It is the outermost layer formed by thin walled, rectangular cells, which are compactly arranged without intercellular spaces. Cuticle and stomata are absent. Some epidermal cells (trichoblasts) produce tubular extensions called root hairs. They absorb capillary water from the soil. The epidermis of root is also known as rhizodermis or epiblema or piliferous layer.

2) Cortex :
It is a wide and extensive tissue present between the epidermis and stele. It is bigger than the stele. It can be divided into three sub-zones. They are :
a) Exodermis
b) General cortex and
c) Endodermis.

a) Exodermis :
It is the outer part of the cortex and composed of one to two rows of thick walled, dead, suberised calls. In mature roots, when the outer epidermis is removed, the exodermis acts as a protective layer. It helps in preventing the exit of water from the root tissues.

b) General Cortex :
It is formed below the exodermis layer. It is composed of several rows of thin walled living cells that are arranged loosely showing intercellular spaces. The cells of cortex help in the storage of food materials and lateral conduction of water from the epidermis to the stele.

c) Endodermis :
The innermost layer of cortex and is composed of single layer of barrel shaped cells that are arranged compactly without intercellular spaces. The radical and transverse walls are wrapped by ligno-suberised bands called casparian bands.

Some cells situated opposite to the protoxylem cells are thin walled and without casparian bands. These are known as passage cells which help in the entry of water from the cortex into the stele.

3) Stele :
The central conducting cylinder. It is very prominent and bigger in size. The stele shows Pericycle, Vascular bundles and Medulla.

i) Pericycle :
The layer of cells found beneath the endodermis is known as pericycle. The cells are thin walled, parenchymatous, rectangular and compact without intercellular spaces. The cells are meristematic and divide actively producing lateral roots. In old and mature roots, the pericycle is sclerenchymatous and gives mechanical strength.

ii) Vascular bundles :
Bundles of xylem and phloem are found separately on different radii, one alternating with the other, at the peripheral boundary of the stele. These are known as radial’ or separate vascular bundles. The xylem is exarch and polyarch. More than six xylem bundles.

The ground tissue formed between the xylem and phloem stands is known as ‘conjunctive tissue’. It is usually parenchymatous. It helps in storage of food materials and provides mechanical strength.

iii) Medulla or Pith :
The wide central part of the stele is called ‘medulla or pith’. It is made up of thin walled parenchyma which primary helps in the storage of food. In some monocot roots, the medulla is composed of thick walled lignified dead cells and helps in giving mechanical strength.

Intext Questions

Question 1.
Name the various kinds of cell layers which constitute the bark.
Answer:
Periderm and secondary phloem.

Question 2.
Every 50 years, for 200 years, a nail was drilled into a tree, to the same depth and at exactly 1m above the soil surface (assuming the ground level has not changed). What will be the pattern of the four nails on the tree? Do you know the reason for your answer? If yes, give the reason?
Answer:
The pattern of four nails depends on the seasonal variations and growth of the tree.

Question 3.
Why is wood made of xylem and not of phloem?
Answer:
The walls of the xylem cells are thickened with lignin, this strengthens the walls and also makes them waterproof. Xylem also contributes greatly to the mechanical strength of the plant. Hence wood is mostly made up of secondary xylem.

Question 4.
A student estimated the age of a tree to be about 300 years. How did he anatomically estimate the age of this tree?
Answer:
By counting the number of annual rings.

Question 5.
Assume that you have removed the duramen part of a tree. Will the tree survive or die?
Answer:
The tree survives even if the duramen part of trunk is removed because of the presence of functional wood called sap wood.

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 11th Lesson Cell Cycle and Cell Division Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 11th Lesson Cell Cycle and Cell Division

Very Short Answer Questions

Question 1.
Between a Prokaryote and a eukaryote, which cell has a shorter cell division time?
Answer:
Prokaryotic cell.

Question 2.
Among Prokaryotes and eukaryotes, which one has a shorter duration of a cell cycle?
Answer:
Prokaryotic cell.

Question 3.
Which of the phases of the cell cycle is of the longest duration?
Answer:
Interphase.

Question 4.
Which tissue of animals and plants exhibits Meiosis?
Answer:
Meiosis occurs in diploid cells.

Question 5.
Given that the average duplication time of E.coli is 20 minutes. How much time will two E.coli cells take to become 32 cells?
Answer:
80 minutes.

Question 6.
Which part of the human body should one use to demonstrate stages in Mitosis?
Answer:
Cells lining the Gut and skin cells.

Question 7.
What attributes does a chromatid require to be classified as a chromosome?
Answer:
Each chromosome at Metaphase split and the two daughter chromatids now referred to as chromosoms.

Question 8.
Which of the four chromatids of a bivalent at prophase-I of Meiosis can involve in cross-over?
Answer:
Crossing over occurs between non-sisters chromatids of the homologous chromosomes.

Question 9.
If a tissue has at a given time 1024 cells. How many cycles of Mitosis had the original parental single cell undergone?
Answer:
10 Mitotic divisions.

Question 10.
An anther has 1200 pollen grains. How many pollen mother cells must have been there to produce them?
Answer:
300.

Question 11.
At what stage of cell cycle does DNA synthesis occur?
Answer:
‘S’ phase (synthesis phase).

Question 12.
It is said that one cycle of cell division in human cells, (Eukaryotic cells) take 24 hours. Which phase of the cycle, do you think occupies the maximum part of cell cycle?
Answer:
Interphase.

Question 13.
It is observed that heart cells do not exhibit cell-divisjon. Such cells do not divide further and exit phase to enter an inactive stage called of cell cycle. Fill in the blanks.

Question 14.
Identify the sub stages of Prophase n I in Meiosis in which synapse and desynapse are formed?
Answer:
Synthesis occurs in ‘zygotene’ a sub phase of Prophase I of Meiosis -I
Desynapse occurs in ‘diakinesis’ a sub phase of prophase I of Meiosis -I
G₁ Phase. Quiescent stage (G₀).

Question 15.
Name the stage of Meiosis in which actual reduction in chromosome number occurs.
Answer:
Anaphase – II

Question 16.
Mitochondria and plastids have their own DNA (genetic material). What is their fate during nuclear division like Mitosis?
Answer:
Mitochondria and plastids get distributed between the two daughter cells.

Question 17.
A cell has 32 chromosomes. It undergoes mitotic division. What will be the chromosome number during Metaphase? What would be the DNA content (c) during anaphase?
Answer:
32 chromosomes. The DNA content during anaphase gets doubled.

Question 18.
The following events occur during the various phases of the cell cycle. Fill the blanks with suitable answer against each.
a) Disintegration of Nuclear Membrane
b) Appearance of Nucleolus
c) Division of centromere
d) Replication of DNA
Answer:
a) Prophase
b) Telophase
c) Anaphase
d) S-Phase

Question 19.
While examining the Mitotic stage in a tissue, one finds some cells with 16 chromosomes and some with 32 chromosomes. What possible reasons could you give to this difference in chromosome number ? Do you think cells with 16 chromosomes could have arisen from cells with 32 chromosomes or vice versa?
Answer:
In cells with 16 chromosomes Mitosis is over, in cells with 32 chromosomes, Mitosis is not started. No. Cells with 16 chromosomes have not arisen from the cells with 32 chromosomes.

Question 20.
Two key events take place during S phase in animal cells. DNA replication and duplication of Centriole. In which parts of the cell do these events occur?
Answer:
Nucleus, Cytoplasam.

Question 21.
Name a cell that is found arrested in diplotene stage for months and years. Comment in two or three sentences how it complete cell cycle?
Answer:
Oocytes of some vertebrates.

Short Answer Type Questions

Question 1.
In which phase of meiosis are the following formed? Choose the answers from hint points given below,
a) synaptonermal complex
b) Recombination nodules
c) Appearance / activation of a Enzyme recombinase
d) Termination of chliasmata
e) Interkinesis
f) Formation of dyad of cells.
Hints :
1) Zygotene
2) Pachytene
3) Pachytene
4) Diakinesis
5) After Telophase I / before Meiosis II
6) /Telophase I / After Meiosis I
Answer:
a) Synaptonemal complex = Zygotene
b) Recombination nucleus = Pachytene
c) Appearance / activation of = Pachytene Enzyme recombinase
d) Termination of chaismata = Diakinesis
e) Interkinesis = The stage between the two meiotic divisions (After telophase I / before meiosis II)
f) Formation of dyad of cells = Telophase I / After meiosis I.

Question 2.
Mitosis results in producing two cells which are similar to each other. What would be the consequence if each of the following irregularities occurs during Mitosis?
a) Nuclear membrane fails to disintegrate
b) Duplication of DNA does not occurs.
c) Centromeres do not divide
d) Cytokinesis does not occur.
Answer:
a) Nuclear membrane fails to disintegrate : The nuclear divisions takes place.
b) Duplication of DNA does not occurs : Of two daughter cells, one cannot get DNA.
c) Centromers do not divide : Chromosomes are not distributed to daughter cells.
d) Cytokinesis does not occur : Multinucleate condition arises leading to the formation of syneytium (liquid endoplasm of coconut).

Question 2.
Describe the events of prophase – I
Answer:
Meiosis I is longer phase and consists of 5 sub phases namely Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

a) Leptotene :
The nucleus increases in size by absorbing water from the cytoplasm. The chromatin material organises into a constant number of chromosomes. The chromosomes are long, slender and show bead like structures called chromomeres.

b) Zygotene :
The chromosomes become shorter and thicker. They approach each other and form pairs. This homologous pair is called bivalent and the process of pairing is called synapsis. It is accompanied by the formation of Synaptonemal complex. The synapsis occurs at proterminal point or procentric or random means.

c) Pachytene :
Bivalent chromosomes now clearly appear as tetrads. This stage is characterised by the presence of recombination modules, the sites of which crossing over occurs between the non-sister chromatids of the homologous chromosomes. Crossing – over is the exchange of genetic material between the two homologous chromosomes. It is also an enzyme mediated process by ‘recombinase’ crossing over leads to recombination of genetic material on the two chromosomes.

d) Diplotene :
The homologous chromosomes of a bivalent begin to separate from each other except at the sites of cross overs to dissolution of synaptonemal complex. The ‘X’ shaped structures are called chaismata.

e) Diakinesis :
It is marked by terminalisation of chaismata. In this phase, the chromosomes are fully condensed and the meiotic spindle is assembled to prepare the homologous chromosomes for separation. By the end of this phase, the nuclear membrane breaks down the nucleolus disappears.

The chromatids undergo condensation, contraction and thickening.

Question 4.
Mention the significance of Meiosis.
Answer:

1. It maintains the same chromosome number in the sexually reproducing organisms.
2. It restricts the multiplication of chromosome number and maintains the stability of the species.
3. Maternal and paternal genes get exchanged during crossing over. It results in variations among the offspring.
4. All the four chromatids of homologous pair of chromosomes seggregate and go over separately to four different daughter cells this leads to variation in the daughter cells genetically.
5. Variations play an important role in the process of evolution.

Question 5.
Which division is necessary to maintain constant chromosome number in all body cell of multicellular organism and why?
Answer:
Mitosis is necessary to maintain constant number of chromosomes in all body cells in multicellular organisms because mitosis results in the production of diploid daughter cells with genetic employment usually identical to that of the parent cell. The growth of multicellular organisms is due to mitosis. Cell growth results in distributing the ratio between the nucleus and cytoplasam.

It therefore essential for the cell division to restore the nucleo-cytoplasamic ratio. A very significant contribution of mitosis is cell repair. Mitotic divisions in the meristametic tissues, the apical and the lateral Meristems, results in a continuous growth of plants throughout their life.

Question 6.
Though redundantly described as a resting phase. Interphase does not really involve rest. Comment.
Answer:
The interphase also called phase of non-apparent division through called the resting phase is the time during which the cell is preparing for division by involving both cell growth and DNA replication. The interphase is divided into three further phases. They are :

i) ‘G₁‘ phase :
It corresponds to the internal between mitosis and initiation of DNA replication. In this the cell is metobolically active and continuously grows.

ii) ‘S’ phase :
Synthesis phase marks the period during which DNA synthesis or replication takes place. The amount of DNA per cell doubles. However, there is no increase in the number of chromosomes.

iii) ‘G₂‘ phase :
Proteins are synthesized.

Question 7.
Comment on the statement – Meiosis enables the conservation of specific chromosome number of each species even though the process per se. results in reduction of chromosome number.
Answer:
Meiosis is the mechanism by which conservation of specific chromosome number in each species is achieved across generations in sexually reproducing organisms, even though the process per sec periodically results in the reproduction of chromosome number by half. It also increases the genetic variability in the population of organisms from one generation to the next. Variations are very important for the process of evolution.

Question 8.
How does cytokinesis in plant cells differ from that in animal cells?
Answer:
In animal cells, cytokineis is achieved by the appearence of a furrow in the plasma membrane. The furrow gradually deepens and ultimately joins in the centre dividing the cell cytoplasam into two plant cells, are enclosed by a relatively inextensible cell wall. In plant cells, wall fromation starts in the centre of the cell and grows outward to meet the exiting lateral walls. The formation of the new cell wall begins with the formation of cell plate that represents the middle lamella between the walls of two adjacent cells.

Long Answer Type Questions

Question 1.
Discuss on the statement – Telophase is reverse of prophase.
Answer:
Telophase is the final stage of Mitosis in which the chromosomes that have reached their respective poles decondenses and loose their individuality. Chromosomes cluster at opposite spindle poles and their identity is lost as discrete elements. Nuclear membrane assimilates around the chromosome clusters. Nucleolus, Golgi complex and endoplasmic reticulum reform where as in prophase, chromosomal material condenses and organises to form compact chromosomes. Nuclear membrane, Nucleolus, Golgi complexes. Endoplasmic reticulum disappears. Thats why Telophase is the reverse phase to prophase.

Question 2.
What are the various stages of meiotic prophase – I? Enumerate the chromosomal events during each stage?
Answer:
Meiosis I is longer phase and consists of 5 sub phases namely Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

a) Leptotene :
The nucleus increases in size by absorbing water from the cytoplasm. The chromatin material organises into a constant number of chromosomes. The chromosomes are long, slender and show bead like structures called chromomeres.

b) Zygotene :
The chromosomes become shorter and thicker. They approach each other and form pairs. This homologous pair is called bivalent and the process of pairing is called synapsis. It is accompanied by the formation of Synaptonemal complex. The synapsis occurs at proterminal point or procentric or random means.

c) Pachytene :
Bivalent chromosomes now clearly appear as tetrads. This stage is characterised by the presence of recombination modules, the sites of which crossing over occurs between the non-sister chromatids of the homologous chromosomes. Crossing – over is the exchange of genetic material between the two homologous chromosomes. It is also an enzyme mediated process by ‘recombinase’ crossing over leads to recombination of genetic material on the two chromosomes.

d) Diplotene :
The homologous chromosomes of a bivalent begin to separate from each other except at the sites of cross overs to dissolution of synaptonemal complex. The ‘X’ shaped structures are called chaismata. The chromatids undergo condensation, contraction and thickening.

e) Diakinesis :
It is marked by terminalisation of chaismata. In this phase, the chromosomes are fully condensed and the meiotic spindle is assembled to prepare the homologous chromosomes for separation. By the end of this phase, the nuclear membrane breaks down the nucleolus disappears.

Question 3.
Differentiate between the events of mitosis and meiosis.
Answer:

Question 4.
Write brief note on the following :
a) Synaptonemal complex b) Metaphase plate
Answer:
a) Synaptonemal complex :
The homologous chromosomes approach each other and form pairs called Bivalents and the process is called synapsis. The synapsis occurs at the both ends and progresses towards their centromeres called proterminal or the synapsis starts from their centromeres and the pairing progresses towards the end of the chromosomes called procentric or the pairing occurs at various points of homologous chromosomes called Randon synapsis. The paired homologous chromosomes are joined by a thick protein containing frame work called synaptonemal complex (Sc).Sc stabilizes the pairing of the homologous chromosomes and facilitates crossing over and recombination.

b) Metaphase plate :
Metaphase chromosome is made up of two sister chromatids’which were held together by the centromere. Small disc shaped structures at the surface of of the centromeres are called Kinetochores. These serve as the sites of attachment of spindle fibres to the chromosomes that are moved into centre of the cell. Hence all chromosomes lie at the equator with one chromatid of each chromosome connected by its kinetochore to spindle fibres from one pole and its sister chromatid connected by its kinetochore to spindle fibres from the opposite pole. It is called ‘Metaphase Plate’.

Question 5.
Write briefly the significance of mitosis and meiosis in multicellular organism.
Answer:
Significance of Mitosis :

1. Growth in organisms is caused by Mitosis.
2. The daughter cells formed by Mitosis are identical with the mother cell in characters. Hence it it important in conserving the genetic diversity of the organisms.
3. In unicellular organisms, Mitosis help in reproduction.
4. The old dead and decaying cells of body are replaced with the help of Mitosis.
5. It is useful in regeneration of lost parts and for grafting in vegetative reproduction.

Significance of Meiosis :

1. Meiosis helps in the maintenance of a constant chromosome number from one generation to the next.
2. Due to crossing over, genetic recombinations are caused which help in the origin of new species and lead to evolution.
3. It helps in the formation of gametes and is thus useful in sexual reproduction.

Intext Questions

Question 1.
Name a stain commonly used to colour chromosome.
Answer:
Giemsa strain is used.

Question 2.
Name the patholqgical condition when uncontrolled cell division occurs.
Answer:
Cancer.

Question 3.
An organism has two parts of chromosomes (i.e., chromosome number = 4) Diagra-matically represent the chromosomal arrangement during different phases of Meiosis II.
Answer:
An organism has two pairs of chromosomes (i.e., chromosome number = 4) Diagrammatically represent the chromosomal arrangement during phases of meiosis – II.

Question 4.
Meiosis has events that lead to both gene recombinations as well as mendelian recombinations. Discuss.
Answer:
An organism often many phenotypic fruits in its body; are determined by at least pair of alleles (or) gener During the events of meiosis – I (crossing over) recombination events occur to pass different combination of chromosomes and consequently different combination of characters in both daughter cells.

The random assortment of the genes is due not only to crossing over but also to the random distribution of the chromosomes in first and second division. How ever, since this separation is a random process, the resulting cells will contain different chromosomal combinations even in the abscence of crossing over.

Question 5.
Both unicellular and multicellular organisms undergo mitosis. What are the differences if any observed between the two processes?
Answer:

1. In unicellulars it is referred as binary fission and in multicellulars it is referred as mitosis.
2. Mitosis allows unicellular organisms to reproduce and create more (identical) organisms.
3. In unicellular only one cell undergoes mitosis whereas in multicellular all cells undergo mitosis.

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 10th Lesson Biomolecules Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 10th Lesson Biomolecules

Very Short Answer Questions

Question 1.
Medicines are either man-made (i.e., synthetic) or obtained from living organisms like plants, bacteria, animals, etc. and hence the latter are called natural products. Sometimes natural products are chemically altered by man to reduce toxicity or side effects. Write against each of the following whether they were initially obtained as a natural product or as a synthetic chemical.
a) Pencillin_ b) Sulfonamide_ c) Vitamin C_ d) Growth Hormone_
Answer:
a) Penicillin – Natural
b) Sulfonamide – Synthetic
c) Vitamin C – Natural
d) Growth Hormone – Natural.

Question 2.
Select an appropriate chemical bond among ester bond, glycosidic bond, peptide bond and hydrogen bond and write against each of the following.
a) Polysaccharide b) Protein c) Fat d) Water
Answer:
a) Polysacharide – Glycosidic Bond
b) Protein – Peptide Bond
c) Fat – Ester Bond
d) Water – Hydrogen Bond.

Question 3.
Give one example for each of aminoacids, sugars, nucleotides and fatty acids.
Answer:
Aminoacids – Glycine, Alanine
Sugars – Cellulose [Glucose, Ribose)
Nucleotides – “Adenylic acid”
Fatty acids – Lecithin.

Question 4.
Explain the Zwitterionic form of an amino acid.
Answer:
At a specific PH the amino acid carries both the +ve and -ve charges in equal number and exists as dipolar ion. It is also called as zwitterionic form. At his point, the net charge on it is zero.

Question 5.
What constituents of DNA are linked by glylosidic bond?
Answer:
Individual Monosacharides are linked by glycosidic bond.

Question 6.
Glycine and Alanine are different with respect to one substituent on the a cabron. What are the other common substituent groups?
Answer:
Hydrogen, Carboxyl group, amino group and a variable group designated as R group.

Question 7.
Starch, Cellulose, Glycogen, Chitin are polysaccharides found among the following. Choose the one appropriate and write against each.
a) Cotton fibre_ b) Exoskeleton of cockroach_ c) Liver_ d) Peeled potato_
Answer:
a) Cotton fibre – Cellulose
b) Exoskeleton of cockroach – Chitin
c) Liver – Glycogen
d) Peeled Potato – Starch.

Question 8.
What are primary and secondary metabolites? Give examples.
Answer:
Primary metabolites :
The metabolites have identifiable functions and play known roles in normal physiological processes are called as primary metabolites.
Ex : Hydrogen, Carbon, Oxygen, Nitrogen etc.

Secondary metabolites :
Metabolic products that do not have identifiable functions in the host organism are called as secondary metabolites.
Ex : Alkaloids, Flavonoids, Rubbers, Resins etc.

Short Answer Type Questions

Question 1.
Explain briefly the metabolic basis for ‘living’.
Anwer:
Metabolic pathways can lead to a more complex structure from a simpler structure. (Anabolic pathways). [For example lactic acid becomes cholesterol or sucrose formation from C02 and water in Mesophyll] or lead to a simple structure from a complex structure (catabolic pathways) [glucose becomes lactic acid in our skeletol muscle). Anabolic pathways consume energy whereas catabolic pathways lead to the release of energy.

For example when glucose is degraded to lactic acid in our skeletal muscle, energy is liberated called Glycolysis, Living organisms have learnt to trap this energy and store it in the form of chemical bonds. When ever requires, this energy is utilised for biosynthetic, osmotic and mechanical work that we perform. The most important form of energy currency in living systems is adenorine triphosphate (ATP).

Question 2.
Schematically represent primary, secondary and tertiary structures of a hypothetical polymer using protein as an example.
Answer:
The sequence of amino acids, i.e., the positional information in a protein is called the primary structure of a protein. A protein is imagined as a line, the left end represented by the first amino acid called N- terminal amino acid and the right end represented by the last aminoacid called C-terminal aminoacid In proteins, only, right handed helics are observed. Other regions of the protein thread are folded into other forms in what is called the secondary structure. In addition, the long protein chain is also folded upon itself like a hollow woolen ball giving rise to the tertiary structure.

Question 3.
Nucleic acids exhibits secondary structure. Justify with example.
Answer:
Nucleic acids exhibit a wide variety of seconary structure. For example, one of the secondary structures exhibited by DNA is Watson – Crick model. According to this, DNA exists as a double helix. The two stands of polynucleotides are antiparallel i.e., run in the opposite direction. The back bone is formed by the sugar-phosphate, sugar chain. The nitrogen bases are projected more or less perpendiular to this backbone but face inside. A combines with T by two hydrogen bonds where G combines with C by three hydrogen bonds. One full turn of the helical strand would involve ten base pairs. The length of one coil is 34 A and the distance between base pair is 3.4 A . This type of DNA is called B-DNA.

Question 4.
Comment on the statement living sate is a non-equilibrium steady – state to be able to perform work1.
Answer:
Several chemical compounds are present in a living organism called metabolites or biomolecules are present at concentrations characteristic of each of them. For example the blood concentration of Glucose in a normal healthy individual is 4.5 – 5.0 mp. All living organisms exist in a steady state characterised by concentrations of each of these biomolecules. These are in a metabolic flux. Any chemical or physical process moves spontaneously to equilibrium, The steady state is a non equilibrium state. The systems cannot work at equilibrium. As living organisms work continuously, they cannot afford to reach equilibrium. Hence the living state is a non-equilibrium steady state to be able to perform work.

Question 5.
Is rubber a primary metabolite or a secondary metabolite ? Write four sentences about rubber.
Answer:
Rubber is a secondary metabolite. Rubber is an elastic hydro carbon polymer that was orginally derived from latex, d milky colloid produced-by some plants. The purified form of natural rubber is the chemical polyisoprene. It is used extensively in many products, as a synthetic rubber. It is normally very strechy and flexible and extremely waterproof.

Question 6.
Dynamic state of body constituents is a more realistic concept than the fixed concentrations of body constituents at any point of time. Elaborate.
Answer:
All the living organisms contains biomolecules in certain concentrations, which have a turn over. They are constantly being changed into some other biomolecules and also made from some other biomolecules. This breaking and making occurs through chemical reactions called metabolism. Each of the metabolic reactions results in the transformation of biomolecules.

Example :
Removal of CO₂ from amino acids making into an amine removal of amino group in a nucleotide base. Majority of metabolic reactions do not occur in isolation but are linked to some other reactions. Metabolites are converted into each other by a series of linked reactions called metabolic pathways. Flow of metabolites through metabolic pathways has a definite rate and direction like auto¬mobile traffic. This metabolic flow is called the dynamic state of body constituents.

Long Answer Type Questions

Question 1.
What are secondary metabolites? Enlist them indicating their usefulness to man.
Answer:
Metabolic products that do not have identifiable functions in the host organism are called secondary Metabolites. They are Alkaloids, Flavinoids, Rubber, Essential oils. Antibiotics, Coloured pigments, Scents, Gums and Spices.

Alkaloids :

1. Alkaloids from plant extracts have been used as ingredients in lotions (liquid medicine) and poisons.
2. Ancient people used plant extracts containing alkaloids for treating a large number of ailments including snake bite, fever and insanity.

Flavinoids :
These are a widely distributed group of polyphenolic compounds with health related properties which include anticancer, antiviral, anti inflamatory activities, effects on capillary fregility and can ability to inhibit human platelet aggregation.

Rubber :

1. Uncured rubber is used for cements for adhesive, insulating and friction tapes. The flexibility of rubber is dften used in hose, tires and rollers for a wide variety of devices.
2. Its elasticity makes it suitable for various kinds of shock absorbers.
3. It is impermeable to gases, it is useful in the manufacture of articles such as air hoses, balloons, bulls and cushions.

Essential oils :

1. An essential oil is a concentrated hydrophobic liquid containing volatile aroma compounds from plants.
2. These are also known as volatile oils, etheral oils or aetherolea. These are used in aromatherapy.

Antibiotics :

1. Antibiotics are defined as chemicals of natural organic origin that will till certain harmful pathogens.
2. They should not be toxic or have side effects to the host.
3. An antibiotic is a substance that harms or destroys the bacteria that cause infection and disease. We take antibiotics to fight bacterial infections.

Spices :

1. Asafoetida is a good remedy for whooping cough and stomach ache caused due to gas.
2. Cardamon (elachi) helps to control bad breath and digestive disorder.

Question 2.
What are the processes used to analyse elemental composition, organic constituents and inorganic constituents of living tissue? What are the inferences on the most abundant constituents of living tissue? Support the inferences with appropriate data.
Answer:
Take a living tissue (a vegetable or a piece of liver) and grind it in trichloroacetic acid with the help of Mortar and pestle. The thick slurry obtained was strained through a cheese cloth or cotton, two fractions are formed. The first one is filtrate or acid soluble pool which consists of thousands of organic compounds and the second one is retentate or acid insoluble fraction. All the carbon compounds that we get from living tissues can be called biomolecules.

Weigh a small amount of living tissue and dry it. After the evaporation of water the material is burnt to oxidise all the carbon compounds. The remaining ash contains inorganic elements like sodium, potassium, calcium and magnesium and inorganic compounds like sulphate, phosphate etc., therefore chemical analysis gives elemental composition of living tissues in the form of hydrogen, oxygen, chlorine, carbon etc. From a biological point of view the organic constituents are classified into amino acids, nucleotide bases, fatty acids etc.

Question 3.
Nucleic acids exhibit secondary structure. Describe through Watson – Crick Model.
Answer:
Nucleic acids exhibit a wide variety of secondary structure. For example, one of the secondary structures exhibited by DNA is Watson – Crick model. According to this, DNA exists as a double helix. The two stands of polynucleotides are antiparallel i.e., run in the opposite direction. The back bone is formed by the sugar-phosphate, sugar chain. The nitrogen bases are projected more or less perpendiular to this backbone but face inside. A combines with T by two hydrogen bonds where G combines with C by three hydrogen bonds. One full turn of the helical strand would involve ten base pairs. The length of one coil is 34 A° and the distance between base pair is 3.4 A°. This type of DNA is called B-DNA.

Question 4.
What is the difference between a nucleotide and nucleoside? Give two examples of each with their structure.
Answer:

Question 5.
Describe various forms of lipid using a few examples.
Answer:
Lipids are orgnaic compounds which are insoluble in water. They are fats and fatty acids, oils, triglycerides, phospholipids, steroids, waxes etc.,

Fattyacids :
They have a carboxyl group attahed to an R group. The R group could be a methyl or ehtyl or higher number of CH₂ groups. For example pulmitic acid has 16 carbons including arboxyl carbon. Fatty acids could be saturated or unsaturated.

Glycolipids :
They are composed mainly of mono-di and trisubstituted Glycerols, the most well-known being the fatty acid esters of Glycerol called Triglycerides. In these compounds, the three hydroxyl groups of Glycerol are each esterified usually by different fatty acids. They function as a food store.

Phospholipids :
Some lipids have phophorous and phsophorylated organic compound in them called phospholipids. They are found in cell membrane.
Ex : Lecithin.

Intext Questions

Question 1.
What are macromolecules? Give examples.
Answer:
Acid Insoluble, high molecular weight substances of the living tissue are called macromolecules or Biomacromolecules.
Ex : Polysacharides, Polypeptides, Nucleic acid.

Question 2.
Illustrate a glycosidic, peptide and a phospho-diester bond.
Answer:
Glycosidic Bond :
Bond involving carbons of adjacent sugar molecules.

Peptide bond :
Bond between aminoacids in a protein.

Phosphodiester bond :
The bond between the phosphate and hydroxyl group of sugar is called ester bond.

The ester bond an either side of phosphate is called phosphodjester bond.

Question 3.
What is meant by tertiary structure of proteins?
Answer:
3 Dimensional view of a protein essential for many biological activities is called Tertiary structure of proteins.

Question 4.
Find and write down structures of 10 interesting small molecular weight bio-molecules. Find if there is any industry, which manufactures the compounds by isolation. Find out who are the buyers.
Answer:
Aminoacids, Monosacharide and disacharide sugars, fatty acids, Glycerol, Nucleotides, Nucleosides, nitrogen bases are the biomolecules which have low molecular weight.

Question 5.
Proteins have primary sturcture. If you are given a method to known which aminoacid is at either of the two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein?
Answer:
Proteins are the polymers of a-amino acids. The linear sephence of amino acid is called primary structure. Most of proteins starts with an amino acids called Methionine but they are not homologous. Hence, the detection of N termeni (Or) C-termeni amino acid is not give information of homogeneity of a protein.

Question 6.
Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins (eg. Cosmotics etc.,)
Answer:
Erythroproteins, Monoclenal Antibodies. Interferons thrombin and fibrinogen. Insulin, Renin, Proteins are also commonly used in the manufacture of cosmotics, toxins and as biological buffers.

Question 7.
Explain the composition of triglyceride.
Answer:
Triglycerides consist entirely of just three elements, carbon, Hydrogen and oxygen. The molecules of life all have carbon ‘back bones’ meaning the basic shape of the molecule comes from of connected carbon atoms. There are many hydrogen atoms connected to the carbon backbone of a molecule of life. Each triglyceride contains a small amount of O₂.

Triglycerides have four basic parts, one glycerol molecule and three fatty acids. A glycerol is a three carbon molecule the fatty acids are long chains of carbon and hydrogen with two oxygen atoms at one end to produce a Triglyceride.

Question 8.
Can you describe what happens when milk is converted into curd or yoghurt based on your understanding of proteins.
Answer:
Proteins are macromolecules formed by the polymerization of amino acids. Structurally proteins are of 4 types.

1. Primary
2. Secondary
3. Tertiary
4. quaternary.

Structure :
More than are polyreptide chains assemble to form the quaternary structure. When milk is converted to curd, these complex proteins get denatured, thus converting globuler proteins into fibrous proteins. Therefore by the process of denaturation, the secondary and tertiary structures of proteins are destroyed.

Question 9.
Draw the sturcture of the amino acid, alanine.
Answer:

Question 10.
What are gums made of? Is Fevicol different?
Answer:
Gums are hetero polysacharides. They are made from two or more different types of mono sacharides. On the other hand, Fevicol is polyvinyl alcohol (PVA) glue. It is not a polysacharide.

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 9th Lesson Cell: The Unit of Life Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 9th Lesson Cell: The Unit of Life

Very Short Answer Questions

Question 1.
What is the significance of vacuole in a plant cell?
Answer:
In plnats, vacuole contains sap mainly composed of water, metabolic byeproducts, excretions and other waste materials. They also play an important role in osmoregulation.

Question 2.
What does ‘s’ refer to a 70s and 80s ribosome?
Answer:
In both 70 ‘s’ ribosomes, and 80 ‘S’ ribosomes, ‘s’ stands for the sedimentation coefficient, [expressed in Svedberg unit]. It is indirectly a measure of density and size.

Question 3.
Mention a single membrane bound organelle which is rich in hydrolytic enzymes.
Answer:
Lysosome.

Question 4.
What are gas vacuoles 7 State their functions.
Answer:
Gas vacuoles are the aggregates of a number of small hollow cylindrical vesicles present in the cytoplasm of the floating purple and green photosynthetic bacteria. They are peroneable to atmospheric gases and help the bacteria to help floating on the surface of water.

Question 5.
What is the function of a polysome?
Answer:
Several ribosomes may attach to a single m-RNA and form a chain called polyribosomes or polysome. The ribosomes of a polysome translate the m-RNA into proteins.

Question 6.
What is the feature of a metacentric chromosome?
Answer:
The metacentric chromosome has middle centromere forming two equal arms of the chromosome.

Question 7.
What is refered to as satellite chromosome?
Answer:
A few chromosomes have non-staining secondary constrictions at a constant location, which gives the appearence of a small fragment called satellite. The chromosome with satellite is called satellite chromosome.

Question 8.
What are microbodies? What do they contain?
Answer:
Peroxysomes and glyoxisomes are called Microbodies. Peroxysomes are involved in the convertion of fatty acids into carbohydrates and in photorespiration. Glyoxysomes contain the enzymes of glyoxylatic cycle which convert stored lipids to carbohydrates.

Question 9.
What is middle lamella made of? What is its functional significance?
Answer:
Middle lamella is mainly composed of calcium pectate and holds or glues the different . neighbouring cells together.

Question 10.
What is osmosis?
Answer:
Movement of molecules or ions or water from low concentrated place to high concentrated place through semi permeable membrane is called osmosis.

Question 11.
Which part of the Bacterial cell is targeted in gram staining?
Answer:
Chemical composition of the cell envelope.

Question 12.
Which of the following is not correct? a) Robert Brown discovered the Cell. b) Schleden and Schwann formulated the cell theory c) Virchow explained that cells ae formed from pre existing cells d) Aunicellular organism carries out its lofe activities with in a single cell.
Answer:
a) is not correct. The cell was discovered by “Robert Hooke”.
b) is correct.
c) is correct.
d) is correct.

Question 13.
New cells generate from a) Bacterical Fermentation b) regeneration of old cells c) Pre existing cells d) Abiotic materials.
Answer:
Pre existing cells.

Question 14.
Match the following :

Answer:

Question 15.
Which of the following is correct : a) Cells of all living organisms have a nucleus b) Both animals and plant cells have a well defined cell wall, c) In Prokaryotes, there are no membrane bound organcells. d) Cells are formed de novo from abiotic materials.
Answer:
“C” is correct. In Prokaryotes, there are no membrane bound organelles like chloroplast. Mitochondria, ER of Golgi complex.

Short Answer Type Questions

Question 1.
Describe the cell organelle which contains chlorophyll Pitments.
Answer:
The chlorophyll containing cell organelle is chloroplast. They are found in mesophyll cells of the leaves. They are lens shaped, oval, spherical or discoid or ribbon shaped, having 5 – 10 pm length and 2 – 4 pm width. They varies in number from one (chlamydomonus) to 20 – 40 per cell in the masophyll.

Each chloroplast is a double membrane bound structure with periplastidial space in between them. Inner to Inner membrane fluid filled space is present called the stroma. A number of organised flattened membranous sacs called the thylakoids are present in the stroma, which are arranged like a pile of coins called grana. The grana are inter connected by flat membranous tubules called stroma lamellae.

The stroma of the chlorplast contains enzymes required for the synthe is of carbohydrates proteins, small circular double stranded DNA molecules and 70s Ribosomes, photosynthetic pigments are present in thylakoids, involve in high reaction of photosynthesis.

Question 2.
Describe the structure and function of celi organelle that can be considered as power house of the cell.
Answer:
Mitochondria is considered as power house of the cell.

Each mitochondria is sausage shaped or cylindrical having a diameter of 0.2 to 1.0 µm and length of 1.0 to 4.1 µm. Mitochondria is a double membrane bound cell organelle of which outer membrane is smooth and inner membrane forms a number of infoldings towards inside called the cristae.

The space inner to inner membrane is fluid filled called matrix. The cristae contains several stalked particles called oxysomes or elementary particles. Mitochondria are the sites of aerobic respiration. They produce cellular energy in the form of ATP hence they are called power house of the cell. The matrix also contains single circular DNA molecule, 80s RNA molecules and 70s Ribosomes.

Question 3.
Describe the structure of nucleus.
Answer:
The cell organelle that controls all the metabolic activities of the cell is called nucleus. It was first described by Robert Brown in 1831. Each nucleus is a spherical ball like structure, consists of 4 parts. They are A) Nuclear membrane, B) Nucleoplasm, C) Chromatin material and D) Nucleolus.

A) Nuclear membrane :
Nucleus is covered by double layered lipoprotenous membrane with perinuclear space in between them. At certain places, Nuclear pores are present which acts as passages between nucleoplasm and the cyloplasm in both directions.

B) Nucleoplasm :
Inner to nuclear membrane, nucleoplasm in present. It consists of chromatin material, nucleolus and RNA molecules nucleoplasm is also called karyoplasm.

C) Chromation material:
Inter phase nucleus has a loose and indistinct network of nucleoprotein fibres called chromatin. It contains DNA and some basic proteins called histones some on histone proteins and also RNA. The chromation is furthur classified into Heterochromatin and Euchromatin of these, Euchromatin is delicate, less condensed and active.

D) Nucleolus :
It is a dense, spherical shaped structure present inside the nucleus. It plays an indirect role in protein synthesis by producing rib somes.

Functions :

1. It controls the heriditary characterestics of an organism.
2. It is responsible for the protein synthesis, cell division, growth and differentiation.
3. It controls all the metabolic activities of the cell.

Question 4.
Comment on the cartwheel Structure of centriole.
Answer:
Centrosome is an organelle usually containing two cylindrical structure called centrioles. They are surrounded by amorphous pericentriolar materials. Both the centrioles in a centrosome lie perpendicular to each other in which each has an organisation like “cartwheel”. They are made up of nine evenly spaced peripheral fibrils of tubulin. Each of the peripheral fibril is a triplet. The adjacent triplets are also linked.

The central part of the centriole is also proteinaceous and called the ‘Hub’, which is connected with tubules of the peripheral triplets by radial spokes made of protein. The centrioles form the basal body of cilia and flagella and spindle fifres that give rise to spindle apparatus during cell division in animal cells.

Question 5.
Briefly describe the cell theory.
Answer:
Schwann proposed the hypothesis that the bodies of animals and plants are composed of cells and products of cells. Schleiden, examined a large number of plants and observed that all plants are composed of different kinds of cells which form the tissues of the plant. Schleiden and Schwann together formulated cell theory. But this theory did not explain as to how new cells were formed. Rudolf virchow first explained that cells divided and new cells are formed from pre-existing cells. He modified the hypothesis of Schleiden and Schwann and give a definite shape to cell theofy- Which states that,

1. all living organisms are composed of cells and products of cells
2. all cells arise from the pre-existing cells.

Question 6.
Differentiate between Rough Endoplasmic Reticulum (RER) and Smooth Endoplasmic Reticulum (SER).
Answer:

Question 7.
Give the biochemical composition of plasma membrane. How are lipid molecules arranged in the membrane?
Answer:
Plasma membrane is made up of lipids that are arranged in fila. Later biochemical investigations clearly revealed that the cell membrane also posses protein and Carbohydrate. The lipids are arranged with in the membrane, with the polar (hydrophilic) head towards the outer sides and the tail (hydrophobic)towards the inner part. With in the lipids, proteins are classified into integral or peripheral proteins.

An improved model Nicholson (1972) widely accepted as “Fluid Mosaic Model”, According to this, the quasi fluid nature of lipid enables lateral movement of proteins within the overall bilayer.

Question 8.
What are plasmids? Describe their role in Bacteria.
Answer:
Plasmids are small circular DNA molecules outside the genomic DNA. The plasmid DNA conferms certain unique phenotypic characters to Bacteria i.e„ resistance to antibiotics, the plasmid DNA is used to monitor bacterial transformation with foreign DNA.

Question 9.
What are histones? What are their functions?
Answer:
Histones are the proteins closely associated with DNA molecules. They are responsible for structure of chromatin and play an important role in the regulation of gene expression. Five types of histone have been identified H1, H2A, H2B, H3and H4. The other 4 types of histones associate with DNA to form Nucleosomes. They plan an instrumental role in the regulation of many important biological process involving DNA such as transcription, DNA repair & cell cycle.

Question 10.
What is cytoskeleton? What functions is it involved in?
Answer:
An elaborate network of filamentous proteinaceous structures present in the cytoplasm is called cytoskeleton. Eukaryotic cells contain three major components of cytoskeleton, microfilaments, intermediate filaments and microtubules. Cytoskeleton is involved in mechanical support, maintanance of cell shape, cell motility, intracellular transport, signaling across the cell and Karyokinesis.

Question 11.
What is endomembrane system? What cell organelles are not included in it? Why?
Answer:
A group of cell organelles with coordinate functions is called endomembrane system. Mitochondria, chloroplast and peroxisomes are not associated with endomembrane system. Because their functions are not coordinated with the ER, Golgicomplx, lysosomes and vacuoles.

Question 12.
Distinguish between Active transport and Passive transport?
Answer:

Question 13.
What are mesosomes? What do they help in?
Answer:
Extensions of the plasma membrane into the cell are called mesosomes. These extensions are in the form of vesicles, tubules and cisternae. They help in cell wall formation, DNA replication and its distribution to daughter cells, help in respiration, secretion processes to increase the surface area of the plasma membrane to absorb nutrients and enzymetic content.

Question 14.
What are nucleosomes? What are they made of?
Answer:
Under electron Microscope, chromation appears as “beads on string”. These beads are known as Nucleosomes. Atypical nucleosome contains 20,0 bp of DNA double helix wrapped around a core of histone octamer having two copies of each of four types of histone proteins viz H2A, H2B, H3 and H4. HI Histone lies outside the nucleosome core and seals the two turns of DNA by binding at the point where DNA enters and leaves the core. The DNA continues between two nucleosomes is called linker DNA.

Question 15.
How do neutral solutes move across the Plasma membrane? Can the polar molecules also move a cross it in the same way? If not then how are these transported across the membrane.
Answer:
Neutral solutes move a cross the plasma membrane by the process of simple diffusion along the concentration gradient i.e., from higher concentration to the lower concentration. Polar molecules can not pass through the non polar lipid bilayer. Beacuse they require a carriers protein of the membrane to facilitate their transport across the membrane.

Question 16.
Name two cell organelles that are double membrane bound. What are the .characteristics of these two organelles. State their functions and draw labelled diagrams of both.
Answer:
Mitochondria and chloroplasts are double membrane found cell organelles. Mitochondria is a rod shaped cell organelle or sausage shaped or cylindrical having a diameter of 0.2 to 1.0 mµ and length of 1.0 to 4 mµ. They are the sites of aerobic respiration and produce cellular energy in the form of ATP hence they are called “Power houses of the cell”.

Chloroplasts are lens shaped, oval, spherical or ribbon like bodies having 5-10 mp length and 2.4 mµ width. They contain enzymes required for the synthesis of carbohydrates and proteins.

Question 17.
What are the characteristics of a prokaryotic cell?
Answer:

1. Nuclear envelope is absent. Cell shows a single circular, supercoiled naked DNA called nucleoid.
2. Nucleous is absent.
3. Endomembrane system is absent.
4. Mitochondria, plastids, lysosomes, peroxisomes and cytoskeleton are absent.
5. Respiratory enzymes are located in cell membrane.
6. 70 s type of ribosomes are present.
7. Cell is surrounded by a cell wall, made up of polysaccharides, lipids and proteins.
8. Cells divide amitotically.

Question 18.
Cell is the basic unit of life. Discuss in brief.
Answer:
Schleiden explained a large number of plants and observed that all plants are composed Of different kinds of cells which form the tissues of the plant. All living organisms are composed of cells and products of cells. All cells arise from pre-existing cells. A cell is the basic unit of life. It is the smallest unit (or building block) of a living organism that is capable of life. Within cells, there are structures that work togehter to allow the cell to live. Some structures transport materials throughout the cell other structures make food and others release energy for the cell to use.

When you look at a green plant, there is a green chemical called chlorophyll. This chemical enables’ plants to use the sun’s energy to make food for themselves. This is found in the cell part called the chloroplast. The nucleus of plant cells is the control centre of the cells. It directs everything a cell’does. The cytoplasm’ is a fluid inside the cell and the cell parts float in it.

Question 19.
Both lysosomes and vacuoles are endomembrane structures, yet they differ in terms of their functions. Give a critical comment.
Answer:
Lysosomes are the membrane found vesicular structures, filled with hydrolytic enzymes capable of digesting proteins, lipids and nucleic acids. These enzymes are optimally active at the acidic pH. Under starvation, lysosomes digest cellular contents by releasing hydrylyzing enzymes and cause death of cells called acetolysis.

Vacuole is the membrane found space present in the cytoplasm, contains mainly water, metabolic bye products, excretions and other waste materials. In some plant cells, vacuolar sap contains some pigments like anthocyanin which impart colour to the plant part; The vacuole is covered by single membrane called Tonoplast. The tonoplast facilitates the transport of number of ions and other materials against concentration gradients, into the vacuole. Vacuoles play an important role in osmoregulation.

Question 20.
Briefly give the contributions of the following scientists in formulating the cell theory, a) Rudolf Virchow b) Schleiden and Schwann.
Answer:
a) Rudolf Virchow :
He first explained that cells divided and new cells are formed from pre existing cells (Orhnis cellula – e cellula). He modified the hypothesis of Schleiden and Schwann to give the cell theory a final shape cell theory states that all living organisms are composed of cells and porducts of cells. All cells arise from pre-existing cells.

b) Schleiden and Schwann :
In 1838, Mathias Schleiden, examined a large number of plants and observed that cell plants are composed of different kinds of cells which form the tissues of the plant. Theodoe Schwann (1839), a British Zoologist, studied different types of animal cells and reported that cells had a thin outer layer which is called plasma membrane. He also concluded, based on his studies ‘ on plant tissues, the presence of cell wall is a unique character of the plant cells. Based on this, Schwann proposed the hypothesis that the bodies of animals and plants are composed of cells and products of cells.

Question 21.
Is extra genomic DNA present in prokaryotes and eukaryotes? If yes, indicate their location in both the types of organisms.
Answer:
Yes, Extra genome DNA is present in prokaryotes and Eukaryotes. In Prokaryotes, extra genome DNA (plasmid) is present along with Nucleoid and floats freely with in the cytoplasm. In Eukaryotes well defined nucleus is present with specific number of chromosomes. But a extra genome DNA is present in cell organelles like Mitochondria and chloroplast. These two organelles are believed to have originally been independent prokaryotes.

Question 22.
Structure and function are correlatabie in living organisms. Can you justify this by taking plasma membrane as an example?
Answer:
Plasma membrane is made up of bilayered lipids. Later biochemical investigations clearly revealed that the plasma membrane also possess protein and carbohydrate. The lipid component of the membrane is phosphoglycerides. The proteins present in the membrane are integral or peripheral type. Integral proteins are burried in the membrane. Peripheral proteins are on the surface of the membrane. Plasma membrane play an important role in transport of the molecules into and out of cells. This membrane is selectively permeable to some molecules present on either side of it. Many molecules can move freely across the membrane without utilising energy is called passive transport of ions.

As the polar molecules cannot pass through non-polar lipid bilayer, they require carrier protein of the membrane to facilitate their transport across the membrane against concentration gradient i.e, from lower to higher concentration. Such a transport by utilising energy is called active transport of ions.
Ex : Na⁺/K⁺ Pump.

Question 23.
Discuss briefly the role of nucleolus in the cells actively involved in protein synthesis.
Answer:
The nucleoplasm contains nucleous and chromatin material. The nucledi are spherical structures, involved in the synthesis of ribosomal RNA. Longer and more numerous nucledi are present in cells actively carrying out protein synthesis. The nucleous develops from the secondary constriction of a specialized chromosome known as the “nucleolar organiser”. The nucleolus consists of RNA, protein and small amount of DNA. It disappears during the end of prophase and reappears at the end of Telophase. It is also called Plasmosome or ‘Ribosomal factory1. –

Question 24.
Explain the association of Carbohydrate to the plasma membrane and its significance.
Answer:
Plasma membrane is made up of lipids and proteins. Later, biochemicals investigations clearly revealed that the cell membrance also possess protein and carbohydrate. They are present s short, unfranched or branched chains of sugars (oligosaccharides) attached either to exterior ectoproteins or the polar ends of phospholipids at the external surface of the plasma membrane. All types of oligosaccharides of the plasma membrane are formed by various combinations of six principle sugars like D-galactose, D-mannose, L-fructose, N-acetyl neuramic acid (sialic acid), N-acetyl D-glucosamine and N-acetyl – D – gaiactosamine.

Significance :

1. Glycophorins are found to contain certain antigenic determinants for the ABO Blood groups and MN flood groups.
2. Sialic acid contents a high negative change to the cell surface of erythrocyte.

Question 25.
Multicellular organisms have division of labour. Explain.
Answer:
Multicellular organisms are made up of millions and trillions of cells. All these cells perform specific functions. All the cells specialised for performing similar functions are grouped together as tissues in the body. Hence a particular function is carried out by a group of cells at a definite plae in the body. Similarly different functions are carried out by different groups of cells in an organism. By dividing labour, multicelled organisms are able to complete more complex tasks.

Question 26.
What are nuclear pores? State their function.
Answer:
The nuclear membrane is interupted by a minute pores. Which are formed by the fusion of its two membranes. These pores are called Nuclear pores. These regulate the transportation of molecular, between the nucleus and cytoplasm. In eukaryotic cells, the nucleus is separated from the cytoplasm by nuclear envelope. The envelope safeguards the DNA contained in the nucleus. Inspite of this barrier, there is still communication between the nucleus and cytoplasm by nuclear pores.

Each nuclear pore is a large complex of proteins that allows small molecules and ions to freely pass of diffuse into or out of the nucleus. Nuclear pores also allow necessary proteins to enter the nucleus from the cytoplasm ; if the proteins have special sequences that indicate, they belong to the nucleus. These sequences tags are known as nuclear localization signals. Similarly RNA transcribed in the nucleus and proteins are destined to enter the cytoplasm have nuclear export sequences that tag them for release through the nuclear pores.

Long Answer Type Questions

Question 1.
What structural and functional attributes must a cell have to be called a living cell?
Answer:
All living organisms are made up of cells and product of cells. Cell is the basic structural and functional unit of living organism. Each cell have different organelles and perform different functions.

1. Cells obey laws of energetics i.e., they transform energy.
2. Cells highly structured with emergent properties.
3. Cells have an evolutionary origin.
4. Cells metabolize means possess metabolic pathways, process nutrients, self adjust to environment.
5. Cells self replicate – nucleic acids, ribosomes.
6. Cells osmoregulate – vacuoles, vesicles.
7. Cells communicate – Glycoproteins.
8. Cells shows animation Cyclosis.
9. Cell grow, divide and differentiate.
10. Cells die.

Question 2.
Eukaryotic cells have organelles which may
a) Not be bound by a membrane
b) Bound by a single membrane
c) Bound by a double membrane.
Give the various sub-cellular organelles into these three categories.
Answer:
a) Not be bound by a membrane : Nucleolus
b) Bound by a single membrane : Lysosomes, vacuoles.
c) Bound by a double membrane : Mitochondria, chloroplast, nucleus..

Question 3.
The genomic content of the nucleus is constant for a given species where as the extra chromosomal DNA is found to be variable among the members of a population. Explain.
Answer:
In Prokaryotes (Bacteria) in addition to the genomic DNA, small circular DNA molecules are present in the cytoplasm. These small DNA molecules are called plasmids. They confers unique phenotypic character to such bacteria (i.e.,) resistance to antibiotics. It is also used to monitor bacterial transformation with foreign DNA.

In Eukaryotes, extra DNA molecules are present both in chloroplast (stroma) and mitochondria (matrix). Because of the presence of this DNA molecules, they are treated as self – autonomous cell organelles.

Question 4.
Justify the statement. “Mitochondria are power houses of the cell.
Answer:
Mitochondria is sausage – shaped or cylindrical structure having a diameter of 0.2 to 1.0 µm and length 1.0 to 4.1 µm. Each mitochondrion is a double membrane bound structure with outer membrane and inner membrane dividing its lumen distinctly into two aqueous compartments. The inner compartment is called the matrix. The outer membrane forms the continuous limiting boundary of the organelle. The inner membrane forms a number of infoldings called the cristae towards the matrix.

The cristae increase the surface area. The mitochondria are the sites of aerobic respiration. They produce cellular energy in the form of ATP, hence they are called “power houses” of the cell. The matrix also possesses a single circular DNA molecule, a few RNA molecules, ribosomes (70s) and the compartments required for the synthesis of proteins.

Question 5.
Is there a species specific or region specific type of plastids? How does on distinguish one from the other?
Answer:
Plastids are species specific and are found in all plant cells and in euglenoides. They bear some specific pigments thus imparting specific colours to the part of the plant which posseses them. Based on the type of pigments, plastids are classified into three types. They are Leucoplasts, Chromoplasts, Chloroplasts.

Leucoplasts :
They are the colourless plastids which store food materials. Based on the storage product, they are of 3 types namely Amyloplasts (store starch), elaioplasts (store oils) and aleuroplasts (store proteins).

Chromoplasts :
They are coloured pigments which were in yellow, orange or red in colour. In these plastids, fat soluble carotenoids like carotene and xanthophylls are present which imparts orange, red or yellow colour.

Chlorplasts :
These are green coloured plastids which help in synthesis of food materials by photosynthesis. They contain chlorophyll and carotenoid pigments which trap light energy. Each chloroplast is a oval or spherical, double membrane bound cell organelle. The space present inner to inner membrane is called stroma. A number of organised flattened membranas sacs called thyloukoids are present in the stroma.

Thylakoids are arranged in stacks like the piles of coins called grana. The thylakoids of the different grana are connected by membranous tubules called the stroma lamellae. The stroma of the chloroplast contains enzymes required for the syntheis of carbohydrates and proteins.

Question 6.
Write the functions of the following.
a) Centromere b) Cell wall c) Smooth ER d) Golgi complex e) Centrioles
Answer:
a) Centromere :
It is required for proper chromosome seggregation. The centromere keeps the two sister chromatids together. It is also where the chromosomes attaches to the spindle apparatus during Mitosis and Meiosis.

b) Cell wall:
It gives a definite shape to the cell and protects the cell from mechanical damage and infection. It also helps in cell-to-cell interaction and acts as a barrier to undesirable macromolecules.

c) Smooth ER :
It helps in synthesis of lipids, metabolism of carbohydrates and calcium concentration, drug detoxification and attachment of receptors on cell membrane proteins. The smooth ER also contains the enzyme Glucose – 6 – Phosphatase which converts Glucose – 6 – Phosphate to Glucose.

d) Golgi complex :
It is the important site for the formation of glycoproteins and glycolipids. It is also involved in the synthesis of cell wall materials and also plays a main role in the formation of cell plate during cell division.

e) Centrioles :
They form the basal body of cilia and flagella and spindle fibres that give rise to spindle apparatus during cell division in animal cells.

Question 7.
Are the different types of plastids interchangeable? If yes, give examples where they are getting converted from one type to another.
Answer:
Yes. Plastids are inter changeable in form. Generally three types of plastids are present in plant cells namely, leucoplasts (storage), chromoplasts (coloured, attraction) and chloroplasts (synthesis of food). Chromoplasts are coloured plastids (orange, yellow or red), occurs in the cells of petals, fruits etc.

They contain less chlorophylls.and more carotenes (orange) or (red) and xanthophylls (yellow). The red colour of tomato is due to the presence of lycopene in the chromoplasts. The chromoplasts of red algae contain phycocyanin and phycoerythrin. The chromoplasts of Brown algae contain fucoxanthin.

Depending upon circumstances, one type of plastid may be converted into another type.
For Ex :

1. The leucoplasts in stem tubers of potato, on exposure to sunlight transofrm into chloroplasts.
2. In capsicum, the cells of ovary consists of leucoplasts. When ovary changes into fruit, leucoplasts are transformed into chloroplasts. When the fruit ripens chloroplasts are changed into chromoplasts.

Question 8.
Describe the structure of the following with the help of labelled diagrams.
i) Nucleus ii) Centrosome.
Answer:
i) Nucleus :
It was first discovered by Robert Brown. Later the material of the nucleus stained by the basic dyes was given the name chromatin by Flemming. The interphase nucleus has highly extended and elaborate nucleoprotein fibres called chromatin, nuclear matrix and one or two nucleoli. Electron micrsocopy has revealed that the nucleus is covered by double layered nuclear envelope, with a space between called perinuclear space, forms a barrier between materials present inside the nucleus and the cytoplasm.

The outer nuclear envelope remains continuous with the endoplasmic reticulum and also bears ribosomes. At a number of places, the nuclear envelope is interrupted by nuclear pores, which allows the passage of RNA and protein molecules in both directions between the nucleus and cytoplasm. The nucleoplasm contains nucleolus and chromatin. The nucleoli and spherical structures, involved in active ribosomal RNA synthesis. The chromatin, in different stages of cell division, become chromosomes. They contain DNA and some basic proteins called histones and also RNA. Nucleus plays an important role in biogensis of ribosomes. It plays a significant role in mitosis.

ii) Centrosome :
Centrosome is an organelle usually containing two cylindrical structures called centrioles. They are surrounded by amorphous pericentriolar materials. Both the centrioles in a centrosome lie perpendicular to each other in which each has an organisation like cartwheel. They are made up of nine evenly spaced peripheral fibrils of tubulin. Each of the peripheral fibril is a triplet. The adjacent triplets are also linked. The central part of the centriol is also proteinaceous and called the ‘hub’, which is connected with tubules of the peripheral triplets by radial spokes made of protein. The centrioles form the basal body of cilia and flagella and spindle fibres that give rise to spindle apparatus during cell divirsion in animal cells.

Question 9.
What is a centromere? How does the position of centromere form the basis of classification of chromosomes. Support your answer with a diagram showing the position of centromere on different types of chromosomes.
Answer:
Centromere is the region of chromosome that becomes attached to spindle fibres. Special proteins surrounded the centromere. They form a disc shaped structures called kinetochores.

Each chromosome shows centromere at a specific position. Basing on the position of centromere, four types of chromosomes are recognised. They are :
1) Metacentric :
“If the centromere is situated at mid point of a choromosome”. It is “V” shaped and consists of two equal arms.

2) Sub-metacentric :
“If the centromere is situated slightly away from mid point of a chromosome”. It is ‘L’ shaped. It consists of two unequal arms.

C) Acrocentric :
If the centromere is situated at the sub terminal position of a chromosome. It is rod shaped or “J” shaped. It consists of very long arm and a very small arm.

D) Telocentric :
“If the centromere is situated at the terminal position of a chromosome”. It is “i” shaped and has only one arm.

Intext Questions

Question 1.
What is a mesosome in a prokaryotic cell? Mention the functions that it performs.
Answer:
Extensions of plasma membrane into the cell in the form of vesicles, tubules and lamellae are called mesosomes. They help in cell wall formation, DNA Replication and its distribution to daughter cells, in respiration, secretion and to increase the surface area of plasma membrane (absorption of nutrients) and enzymatic content.

Question 2.
How do neutral solutes move across the plasma membrane? Can the polar molecules also move across it in the same way? If not, then how are these transported across the membrane?
Answer:
Neutral solutes move across the plasma membrane by the process of simple diffusion along the concentration gradient. Polar molecules cannot pass through the non polar lipid bilayer, they require a carrier protein of the membrane to facilitate their transport across the membrane.

Question 3.
Name the two cell-organelles that are double membrane bound. What are the characteristics of these two organelles? State their functions and draw labelled diagrams of both.
Answer:
Chloroplast and Mitochondria.

1. Chlorplast is a lens shaped, oval or spherical or even ribbon like organelle, involves in photosynthesis.
2. Mitochondria is a sausage shaped or cylindrical cell organelle, involved in Respiration.

Question 4.
What are the characteristics of prokaryotic cells.
Answer:
i) Nuclear envelope is absent. Cell shows a single circular, supercoiled naked DNA called nucleoid.
ii) Nucleous is absent.
iii) Endomembrane system is absent.
iv) Mitochondria, plastids, lysosomes, peroxisomes and cytoskeleton are absent.
v) Respiratory enzymes are located in cell membrane.
vi) 70 s type of ribosomes are present.
vii) Cell is surrounded by a cell wall, made up of polysaccharides, lipids and proteins.
viii) Cells divide amitotically.

Question 5.
Multicellular organisms have division of labour. Explain.
Answer:
Multicellular organisms are made up of millions and trillions of cells. All these cells perform specific functions. All the cells specialised for performing similar functions are grouped together as tissues in the body. Hence a particular function is carried out by a group of cells at a definite plae in the body. Similarly different functions are carried out by different groups of cells in an organism. By dividing labour, multicelled organisms are able to complete more complex tasks.

Question 6.
Cell is the basic unit of life. Discuss in brief.
Answer:
Schleiden explained a large number of plants and observed that all plants are composed Of different kinds of cells which form the tissues of the plant. All living organisms are composed of cells and products of cells. All cells arise from pre-existing cells. A cell is the basic unit of life. It is the smallest unit (or building block) of a living organism that is capable of life. Within cells, there are structures that work togehter to allow the cell to live. Some structures transport materials throughout the cell other structures make food and others release energy for the cell to use.

Question 7.
What are nuclear pores ? State their function.
Answer:
The nuclear membrane is interupted by a minute pores. Which are formed by the fusion of its two membranes. These pores are called Nuclear pores. These regulate the transportation of molecular, between the nucleus and cytoplasm. In eukaryotic cells, the nucleus is separated from the cytoplasm by nuclear envelope. The envelope safeguards the DNA contained in the nucleus. Inspite of this barrier, there is still communication between the nucleus and cytoplasm by nuclear pores.

Each nuclear pore is a large complex of proteins that allows small molecules and ions to freely pass of diffuse into or out of the nucleus. Nuclear pores also allow necessary proteins to enter the nucleus from the cytoplasm ; if the proteins have special sequences that indicate, they belong to the nucleus. These sequences tags are known as nuclear localization signals. Similarly RNA transcribed in the nucleus and proteins are destined to enter the cytoplasm have nuclear export sequences that tag them for release through the nuclear pores.

Question 8.
Both lysosomes and vacuoles are endomembrane structures, yet they differ in terms of their functions. Comment.
Answer:
Lysosomes are the membrane found vesicular structures, filled with hydrolytic enzymes capable of digesting proteins, lipids and nucleic acids. These enzymes are optimally active at the acidic pH. Under starvation, lysosomes digest cellular contents by releasing hydrylyzing enzymes and cause death of cells called acetolysis.

Vacuole is the membrane found space present in the cytoplasm, contains mainly water, metabolic bye products, excretions and other waste materials. In some plant cells, vacuolar sap contains some pigments like anthocyanin which impart colour to the plant part; The vacuole is covered by single membrane called Tonoplast. The tonoplast facilitates the transport of number of ions and other materials against concentration gradients, into the vacuole. Vacuoles play an important role in osmoregulation.

Question 9.
Describe the structure of the following with the help of labelled diagrams.
(i) Nucleus (ii) Centrosome
Answer:
i) Nucleus :
It was first discovered by Robert Brown. Later the material of the nucleus stained by the basic dyes was given the name chromatin by Flemming. The interphase nucleus has highly extended and elaborate nucleoprotein fibres called chromatin, nuclear matrix and one or two nucleoli. Electron micrsocopy has revealed that the nucleus is covered by double layered nuclear envelope, with a space between called perinuclear space, forms a barrier between materials present inside the nucleus and the cytoplasm.

The outer nuclear envelope remains continuous with the endoplasmic reticulum and also bears ribosomes. At a number of places, the nuclear envelope is interrupted by nuclear pores, which allows the passage of RNA and protein molecules in both directions between the nucleus and cytoplasm. The nucleoplasm contains nucleolus and chromatin. The nucleoli and spherical structures, involved in active ribosomal RNA synthesis. The chromatin, in different stages of cell division, become chromosomes. They contain DNA and some basic proteins called histones and also RNA. Nucleus plays an important role in biogensis of ribosomes. It plays a significant role in mitosis.

ii) Centrosome :
Centrosome is an organelle usually containing two cylindrical structures called centrioles. They are surrounded by amorphous pericentriolar materials. Both the centrioles in a centrosome lie perpendicular to each other in which each has an organisation like cartwheel. They are made up of nine evenly spaced peripheral fibrils of tubulin. Each of the peripheral fibril is a triplet. The adjacent triplets are also linked. The central part of the centriol is also proteinaceous and called the ‘hub’, which is connected with tubules of the peripheral triplets by radial spokes made of protein. The centrioles form the basal body of cilia and flagella and spindle fibres that give rise to spindle apparatus during cell divirsion in animal cells.

Question 10.
What is a centromere? How does the position of centromere form the basis of classification of chromosomes. Support your answer with a diagram showing the position of centromere on different types of chromosomes.
Answer:
Centromere is the region of chromosome that becomes attached to spindle fibres. Special proteins surrounded the centromere. They form a disc shaped structures called kinetochores.

Each chromosome shows centromere at a specific position. Basing on the position of centromere, four types of chromosomes are recognised. They are :
1) Metacentric :
“If the centromere is situated at mid point of a choromosome”. It is “V” shaped and consists of two equal arms.

2) Sub-metacentric :
“If the centromere is situated slightly away from mid point of a chromosome”. It is ‘L’ shaped. It consists of two unequal arms.

C) Acrocentric :
If the centromere is situated at the sub terminal position of a chromosome. It is rod shaped or “J” shaped. It consists of very long arm and a very small arm.

D) Telocentric :
“If the centromere is situated at the terminal position of a chromosome”. It is “i” shaped and has only one arm.