Category: Important Questions

Students get through the TN Board 11th Bio Botany Important Questions Chapter 12 Mineral Nutrition which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 12 Mineral Nutrition

Very short answer questions

Question 1.
Differentiate between Micronutrient and Macronutrient.
Answer:

Micronutrient	Macronutrient
Micronutrients are the essential minerals required in less concentrations. Example: Copper	Macronutrients are the essential minerals required in higher concentrations. Example: Nitrogen

Question 2.
State the “Law of minimum” proposed by Liebig.
Answer:
The “law of minimum” states that the productivity of soil depends on the number of essential elements present in minimum quantity.

Question 3.
Mention any four unclassified minerals.
Answer:
Sodium, Cobalt, Silicon and Selenium.

Question 4.
Potassium and Osmotic potential – Comment.
Answer:
Potassium (K) plays a key role in maintaining the osmotic potential of the cell. The absorption of water, movement of stomata and turgidity are due to osmotic potential.

Question 5.
In which form do the following minerals are absorbed by plants.
(a) Nitrogen
(b) Magnesium
(c) Boron
(d) Phosphorous
Answer:

S. No.	Minerals	Absorbable Forms
a.	Nitrogen	Nitrates (NO3)
b.	Magnesium	Mg2+
c.	Boron	Borate (BO3–)
d.	Phosphorous	H2PO4+/HPO4–

Question 6.
Name any two calcium deficiency disease in plants.
Answer:
(a) Blackheart of celery,
(b) Hooked leaf tip in beet, Musa and tomato.

Question 7.
Nitrogen is a Macronutrient – Justify.
Answer:
Nitrogen (N) is required by the plants in the greatest amount. It is an essential component of proteins, nucleic acids, amino acids, vitamins, hormones, alkaloids, chlorophyll and cytochrome.

Question 8.
What are Siderophores?
Answer: Siderophores (iron carriers) are Iron chelating agents produced by bacteria. They are used to chelate ferric iron (Fe³⁺) from the environment and host.

Question 9.
Given below are the plant diseases, name the deficient mineral responsible for the disease.
(a) Khaira disease
(b) Internal cork of apple
(c) Die back of Citrus
(d) Sand drown of tobacco
Answer:

Plant Diseases	Deficient Mineral
Khaira disease of rice	Zinc
Internal cork of apple	Boron
Dieback of Citrus	Copper
Sand drown of tobacco	Magnesium

Question 10.
Define Calmodulin.
Answer:
Calmodulin is a Ca²⁺ modulating protein in eukaryotic cells. It is a heat-stable protein involved in fine metabolic regulations.

Question 11.
When does an essential mineral is considered “toxic”?
Answer:
An increase of mineral nutrients more than the normal concentration causes toxicity. A concentration, at which 10 % of the dry weight of tissue is reduced, is considered toxic.

Question 12.
Give a brief note on Aluminium toxicity.
Answer:
Aluminium toxicity causes precipitation of nucleic acid, inhibition of ATPase, inhibition of cell division and binding of the plasma membrane with Calmodulin.

Question 13.
Define nitrogen fixation. Mention its types.
Answer:
The process of converting atmospheric nitrogen (N₂) into ammonia is termed as nitrogen fixation. Nitrogen fixation can occur by two methods: 1. Biological; 2. Non-Biological.

Question 14.
How does nitrogen fixation occur non-biologically?
Answer:
Non – Biological nitrogen fixation:

• Nitrogen fixation by a chemical process in the industry.
• Natural electrical discharge during lightning fixes atmospheric nitrogen.

Question 15.
Name the types of Autotrophic nutrition.
Answer:
(a) Photosynthetic autotrophs and (b) Chemosynthetic autotrophs.

Question 16.
What are obligate parasites? Give example.
Answer:
Obligate (Total) parasites completely depend on the host for survival. Example: Cuscuta.

Question 17.
What is a Lichen?
Answer:
Lichens is a mutual association of Algae and Fungi. Algae prepare food and fungi absorbs water and provides thallus structure.

Short answer questions

Question 1.
List out the criteria for being essential minerals.
Answer:
Arnon and Stout (1939) gave criteria required for essential minerals:

1. Elements are necessary for growth and development.
2. They should have a direct role in the metabolism of the plant.
3. It cannot be replaced by other elements.
4. Deficiency makes the plants impossible to complete their vegetative and reproductive phase.

Question 2.
Mention the role of Sulphur in plants.
Answer:
Sulphur (S): Essential component of amino acids like cystine, cysteine and methionine, a constituent of coenzyme A, Vitamins like biotin and thiamine, a constituent of proteins and ferredoxin. Plants utilise sulphur as sulphate (SO₄^–) ions.
Deficiency symptoms: Chlorosis, anthocyanin formation, stunted growth, rolling of leaf tip and reduced nodulation in legumes.

Question 3.
Give a short note on EDTA.
Answer:
EDTA – Ethylene Diamine Tetra Acetic acid is a chelating agent.
Plants growing in alkaline soil though supplied with all nutrients including iron show iron deficiency. To rectify this, iron is made into a soluble complex by adding EDTA (Fe-EDTA Complex) which can be easily absorbed by plants.

Question 4.
Write a brief note on Manganese toxicity.
Answer:
Increased Concentration of Manganese will prevent the uptake of Fe and Mg, prevent translocation of Ca to the shoot apex and cause their deficiency. The symptoms of manganese toxicity are the appearance of brown spots surrounded by chlorotic veins.

Question 5.
Write a note on Hydroponics and Aeroponics.
Answer:
Hydroponics or Soilless culture: Von Sachs developed a method of growing plants in the nutrient solution. The commonly used nutrient solutions are Knop solution (1865) and Arnon and Hoagland Solution (1940). Later the term Hydroponics was coined by Goerick (1940) and he also introduced commercial techniques for hydroponics. In hydroponics roots are immersed in the solution containing nutrients and air is supplied with help of a tube.
Aeroponics: This technique was developed by Soifer Hillel and David Durger. It is a system where roots are suspended in air and nutrients are sprayed over the roots by a motor-driven rotor.

Question 6.
Mention the three possible ways by which ammonia is converted into amino acids during nitrogen metabolism.
Answer:
Ammonia is converted into amino acids by the following processes:
(a) Reductive amination, (b) Trans-amination and (c) Catalytic amination.

Question 7.
Write the equation for reductive amination.
Answer:

Question 8.
Transamination – Write a note.
Answer:
Transfer of amino group (NH₃⁺) from glutamic acid glutamate to keto group of a keto acid. Glutamic acid is the main amino acid from which other amino acids are synthesised by transamination. Transamination requires the enzyme transaminase and coenzyme pyridoxal phosphate (derivative of vitamin B6 – pyridoxine).

Question 9.
Give an account on GOGAT – pathway.
Answer:
Glutamate amino acid combines with ammonia to form the amide glutamine.

Glutamine reacts with α ketoglutaric acid to form two molecules of glutamate.

Question 10.
Write a note on Saprophytes.
Answer:
Saprophytes derive nutrients from dead and decaying matter. Bacteria and fungus are the main saprophytic organisms. Some angiosperms also follow the saprophytic mode of nutrition. Example: Neottia. Roots of Neottia (Bird’s Nest Orchid) associate with mycorrhizae and absorb nutrients as a saprophyte. Monotropa (Indian Pipe) grow on humus-rich soil found in thick forests. It absorbs nutrient through mycorrhizal association.

Question 11.
What is Mycorrhizae?
Answer:
Mycorrhizae: Fungi associated with roots of higher plants including Gymnosperms. Example: Pinus.

Long answer questions

Question 1.
Write in detail about functions, mode of absorption and deficiency symptoms of any five macronutrients.
Answer:

Question 2.
Give in detail about the vitality of Boron and Zinc in plant nourishment.
Answer:
Zinc (Zn): Essential for the synthesis of Indole acetic acid (Auxin) activator of carboxylases, alcohol dehydrogenase, lactic dehydrogenase, glutamic acid dehydrogenase, carboxypeptidases and tryptophan synthetase. It is absorbed as Zn²⁺ ions.
Deficiency: Little leaf and mottle leaf due to deficiency of auxin, Inter veinal chlorosis, stunted growth, necrosis and Khaira disease of rice.
Boron (B): Translocation of carbohydrates, uptake and utilisation of Ca⁺⁺, pollen germination, nitrogen metabolism, fat metabolism, cell elongation and differentiation. It is absorbed as borate BO^3- ions.
Deficiency: Death of root and shoot tips, premature fall of flowers and fruits, the brown heart of beet root, the internal cork of apple and fruit cracks.

Question 3.
Explain in detail the symbiotic nitrogen fixation with nodulation.
Answer:
i. Nitrogen fixation with nodulation:
Rhizobium bacterium is found in leguminous plants and fixes atmospheric nitrogen. This kind of symbiotic association is beneficial for both the bacterium and the plant. Root nodules are formed due to bacterial infection. Rhizobium enters into the host cell and proliferates, it remains separated from the host cytoplasm by a membrane.
Stages of Root nodule formation:

1. Legume plants secret phenolics which attracts Rhizobium.
2. Rhizobium reaches the rhizosphere and enters into the root hair, infects the root hair and leads to curling of root hairs.
3. The infection thread grows inwards and separates the infected tissue from normal tissue.
   
4. A membrane bound bacterium is formed inside the nodule and is called a bacteroid.
5. Cytokinin from bacteria and auxin from host plant promotes cell division and leads to nodule formation.

Question 4.
Describe the various stages of the nitrogen cycle.
Answer:
The nitrogen cycle consists of the following stages:
1. Fixation of atmospheric nitrogen: Di-nitrogen molecule from the atmosphere progressively gets reduced by the addition of a pair of hydrogen atoms. A triple bond between two nitrogen atoms (N = N) is cleaved to produce ammonia. The nitrogen fixation process requires a Nitrogenase enzyme complex, Minerals (Mo, Fe and S), anaerobic condition, ATP, electron and glucose 6 phosphates as H⁺ donor. The nitrogenase enzyme is active only in anaerobic condition. To create this anaerobic condition a pigment known as leghaemoglobin is synthesized in the nodules which act as an oxygen scavenger and removes the oxygen. Nitrogen-fixing bacteria in root nodules appear pinkish due to the presence of this leghaemoglobin pigment.

Overall equation:
N₂ + 8e^– + 8H⁺ + 16 ATP → 2NH₃⁺ + H₂ + 16 ADP + 16 Pi
2. Nitrification: Ammonia (NH₃⁺) is converted into Nitrite (NO₂^–) by Nitrosomonas
bacterium. Nitrite is then converted into Nitrate (NO₃^–) by Nitrobacter bacterium. Plants are more adapted to absorb nitrate (NO₃^–) than ammonium ions from the soil.

3. Nitrate Assimilation: The process by which nitrate is reduced to ammonia is called nitrate assimilation and occurs during the nitrogen cycle.

4. Ammonification: The decomposition of organic nitrogen (proteins and amino acids) from dead plants and animals into ammonia is called ammonification. Organism involved in this process are Bacillus ramosus and Bacillus Vulgaris.
5. Denitrification: Nitrates in the soil are converted back into atmospheric nitrogen by a process called denitrification. Bacteria involved in this process are Pseudomonas, Thiobacillus and Bacillus subtilis.

Question 5.
Explain the types of parasitic mode of nutrition in angiosperms.
Answer:
Parasitic mode of nutrition in angiosperms:
Organisms deriving their nutrient from another organism (host) and causing the disease to the host are called parasites.
a. Obligate or Total parasite – Completely depends on the host for their survival and produces haustoria.

1. Total stem parasite: The leafless stem twine around the host and produce haustoria. Example: Cuscuta (Dodder), a rootless plant growing on Zizyphus, Citrus and so on.
2. Total root parasite: They do not have stem axis and grow in the roots of host plants to produce haustoria. Example: Rafflesia, Orobanche and Balanophora.

b. Partial parasite – Plants of this group contain chlorophyll and synthesize carbohydrates. Water and mineral requirements are dependent on a host plant.

1. Partial Stem Parasite: Example: Loranthus and Viscum (Mistletoe) Loranthus grows on fig and mango trees and absorb water and minerals from the xylem.
2. Partial root parasite: Example: Santalum album (Sandalwood tree) in its juvenile stage produces haustoria which grow on roots of many plants.

Higher Order Thinking Skills (HOTs)

Question 1.
In metro cities like Chennai, there is a lack of space for a garden. Suggest a technique by which plants can be grown without the need for soil and land space.
Answer:
The technique which can be used is hydroponics in which nutrient solution is used for growing plants.

Question 2.
Carnivorous plants like Nepenthes and venus fly trap have nutritional adaptation. Which nutrient do they especially obtain and from where?
Answer:
Plants like nepenthes and venus fly trap etc. are found grouping in nitrogen-deficient soil. To compensate, their body parts are modified to trap insects from which they obtain sufficient nitrogen for their survival. Hence, they are carnivorous in nature.

Question 3.
Excess of manganese in the soil leads to deficiency of Mg, Fe and Ca. Justify.
Answer:
Increased concentration of Manganese will prevent the uptake of Fe and Mg and prevent translocation of Ca to the shoot apex and cause their deficiency. This condition is referred to as manganese toxicity.

Question 4.
Name the crucial enzyme found in the root nodules of leguminous plants for nitrogen fixation. Also, name the pigment which is highly essential for the activation of the enzyme.
Answer:
Root nodules of legume plants contained nitrogenase enzyme for nitrogen fixation.
Leghaemoglobin is the pink coloured pigment that creates an anaerobic condition for the activation of the nitrogenase complex.

Question 5.
Plants require nutrients. If we supply these in excess will they be beneficial to plants? If yes, how? If no, why?
Answer:
No, an excess supply of nutrients is not advised for the healthier growth of plants. Due to the fact that increased concentration of nutrients may lead to the toxicity of particular minerals which prevents the uptake of other minerals and also affect various other metabolic activities of the cells, leading to the poor growth of the plant.

Question 6.
X is a primitive, eukaryotic chlorophyllous organism and Y is a eukaryotic, achlorophyllous organism. Both X and Y live mutually together benefitting each other.
(a) Name this mutual association.
(b) What X and Y are?
(c) Explain their relationship?
Answer:
(a) Lichens.
(b) X = Algae, Y = fungi
(c) In lichens, Algae prepare food and fungi absorbs water and provides thallus structure.

Choose the correct answer.

1. Identify the micronutrient.
(a) Potassium
(b) Phosphorous
(c) Manganese
(d) Magnesium
Answer:
(c) Manganese

2. Which mineral is essential for cell wall formation in Equisetuml?
(a) Boron
(b) Silicon
(c) Cellulose
(d) Carbon
Answer:
(b) Silicon

3. ‘Law of Minimum’ was proposed by ……………
(a) Van Helmont
(b) Von Sachs
(c) Wood word
(d) Liebig
Answer:
(d) Liebig

4. Identify the wrong statement(s).
(i) Molybdenum is a micronutrient.
(ii) Carbon, Hydrogen, Nitrogen are skeletal elements.
(iii) Manganese is the activator for RUBP carboxylase.
(iv) Potassium maintains osmotic potential of the cell.
(a) (i) and (iv)
(b) (ii) and (iii)
(c) Only (ii)
(d) Only (iv)
Answer:
(b) (ii) and (iii)

5. Siderophores are ……….. chelators.
(a) B
(b) Fe³⁺
(c) Ca²⁺
(d) Cl^–
Answer:
(b) Fe³⁺

6. Match the following:
(i) Boron                            1. Nitrogen metabolism
(ii) Molybdenum               2. Formation of Porphyrin
(iii) Zinc                             3. Translocation of sugars
(iv) Iron                             4. Biosynthesis of auxin
(a) (i) – 3, (ii) – 1, (iii) – 4, (iv) – 2
(b) (i) – 2, (ii) – 3, (iii) – 4, (iv) – 1
(c) (i) – 3, (ii) – 4, (iii) – 2, (iv) – 1
(d) (i) – 4, (ii) – 1, (iii) – 2, (iv) – 3
Answer:
(a) (i) – 3, (ii) – 1, (iii) – 4, (iv) – 2

7. Exanthema in Citrus is due to ………… deficiency.
(a) Mo
(b) Cu
(c) B
(d) Zn
Answer:
(b) Cu

8. Mottle leaf disease is a ………… deficiency disease.
(a) Gibberellins
(b) Cytokinin
(c) Auxin
(d) Ethylene
Answer:
(c) Auxin

9. Pollen germination requires ………… mineral.
(a) Copper
(b) Molybdenum
(c) Chlorine
(d) Boron
Answer:
(d) Boron

10. ………… toxicity leads to precipitation of nucleic acids.
(a) Manganese
(b) Iron
(c) Aluminium
(d) Boron
Answer:
(c) Aluminium

11. The term soilless culture refers to …………
(a) Aeroponics
(b) Aquaculture
(c) Hydroponics
(d) Drip irrigation
Answer:
(c) Hydroponics

12. The term hydroponics was coined by …………..
(a) Amon
(b) David Durger
(c) Liebig
(d) Goerick
Answer:
(d) Goerick

13. Which of the following is a free-living bacterium?
(a) Rhizobium
(b) Clostridium
(c) Escherichia
(d) Cyanobacteria
Answer:
(b) Clostridium

14. Legume plants secrete………… which attracts Rhizobium.
(a) Toluenes
(b) Phenolics
(c) Octanes
(d) Xylenes
Answer:
(b) Phenolics

15. In N₂, how many bonds are there between two nitrogen atoms?
(a) Two
(b) Four
(c) Three
(d) One
Answer:
(c) Three

16. Leghaemoglobin pigments removes ……….. to activate nitrogenase enzyme.
(a) CO
(b) CO₂
(c) O
(d) N
Answer:
(c) O

17. Which bacterium is NOT involved in Denitrification?
(a) Pseudomonas
(b) Thiobacillus
(c) Bacillus subtilis
(d) Nitrobacter
Answer:
(d) Nitrobacter

18. On reaction with a-ketoglutaric acid, ammonia yields ………….
(a) Glutamate
(b) Malate
(c) Glutamine
(d) Oxoglutarate
Answer:
(a) Glutamate

19. Pyridoxal phosphate is a derivative of vitamin ……………
(a) A
(b) B₁₂
(c) B₆
(d) D
Answer:
(c) B₆

20. Identify the correct pair:
(i) Cuscuta – Stem parasite
(ii) Viscum – Total parasite
(Hi) Dionaea – Saprophyte
(iv) Lichen – Fungi and bacteria
(a) (i) only
(b) (ii) only
(c) Both (i) and (ii)
(d) Both (iii) and (iv)
Answer:
(a) (i) only

21. Assertion: Loranthus is a partial parasite.
Reason: Partial parasite depends on the host for water & minerals only.
(a) A is right, R is wrong
(b) Both A and R are correct
(c) A is right R, explains A
(d) A is the wrong R, explains A
Answer:
(c) A is right R, explain A

22. Pitcher plant is the common name for …………
(a) Drosera
(b) Nepenthes
(c) Utricularia
(d) Dionaea
Answer:
(b) Nepenthes

23. Insectivorous plants are usually seen in ………… deficient soil.
(a) Sulphur
(b) Nitrogen
(c) Carbon
(d) Potassium
Answer:
(b) Nitrogen

24. ………… is a pioneer species in xeric succession.
(a) Lichens
(b) Mycorrhizae
(c) Cyanobacteria
(d) Pteridophyte
Answer:
(a) Lichens

Students get through the TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Very short answer questions

Question 1.
Define transport in plants. Mention the tissues involved in transportation.
Answer:
Transport is the process of moving water, minerals, and food to all parts of the plant body. Conducting tissues such as xylem and phloem play an important role in transport.

Question 2.
What is the need for transport in plants?
Answer:
Water absorbed from roots must travel up to leaves by xylem for food preparation by photosynthesis. Likewise, food prepared from leaves has to travel to all parts of the plant including roots for growth and other processes.

Question 3.
Compare Active transport with Passive transport.
Answer:

Active Transport	Passive Transport
It is an uphill process.	It is a downhill process.
Energy is required.	Energy is not required.
It is a Biological process.	It is a physical process.
(e.g.) Na+ – K+ pump.	(e.g.) Osmosis.

Question 4.
Define the term diffusion.
Answer:
The net movement of molecules from a region of their higher concentration to a region of their lower concentration along a concentration gradient until an equilibrium is attained.

Question 5.
How the diffusing molecules will move?
Answer:
In diffusion, the movement of molecules is continuous and random in order in all directions.

Question 6.
Name the transport proteins of the plasma membrane involved in facilitated diffusion.
Answer:
Channel protein and carrier protein.

Question 7.
Give a brief account of Porin.
Answer:
Porin is a large transporter protein found in the outer membrane of plastids, mitochondria, and bacteria which facilitates smaller molecules to pass through the membrane.

Question 8.
State the role of Channel Protein.
Answer:
Channel protein forms a channel or tunnel in the cell membrane for the easy passage of molecules to enter the cell. The channels are either open or remain closed.

Question 9.
Apart from water, what are the substrates that are transported through aquaporins?
Answer:
Glycerol, Urea, CO₂, NH₃, Metalloids, and Reactive Oxygen Species (ROS).

Question 10.
How does a carrier protein function?
Answer:
Carrier protein acts as a vehicle to carry molecules from outside of the membrane to inside the cell and vice versa. Due to the association with molecules to be transported, the structure of carrier protein gets modified until the dissociation of the molecules.

Question 11.
On which basis, the carrier proteins are classified? Mention its types.
Answer:
There are three types of carrier proteins on the basis of handling of molecules and direction of transport. They are,

1. Uniport,
2. Symport and
3. Antiport.

Question 12.
Mention the drawbacks of diffusion.
Answer:
In diffusion, there is a lack of control over the transport of selective molecules. There is a possibility of harmful substances entering the cell by a concentration gradient.

Question 13.
Co-transport and counter transport differ from each other. Justify.
Answer:
In co-transport, two molecules are transported together in the same direction whereas, in counter¬transport two molecules are transported in opposite directions to each other.

Question 14.
State any two vital roles of water in plants.
Answer:

1. Water maintains the internal temperature of the plant.
2. It maintains the turgidity of the cell.

Question 15.
List out a few imbibing.
Answer:
Gum, starch, proteins, cellulose, agar, and gelatin.

Question 16.
Define Imbibition.
Answer:
The phenomenon in which a substance uptake the water and swell up is called Imbibition. The substance is referred to as Imbibant. Example: Swelling of water-soaked seeds.

Question 17.
List out the substances that are transported by facilitated diffusion.
Answer:
Sugars, amino acids, nucleotides, ions, and cell metabolites.

Question 18.
How imbibition, is important for plants?
Answer:
Significance of imbibition:

1. During the germination of seeds, imbibition increases the volume of the seed enormously and leads to the bursting of the seed coat.
2. It helps in the absorption of water by roots at the initial level.

Question 19.
Define Water potential.
Answer:
Water potential is the potential energy of water in a system compared to pure water when both temperature and pressure are kept the same. Water potential is denoted by the symbol ¥ (psi).

Question 20.
Define Osmotic Pressure.
Answer:
When a solution and its solvent (pure water) are separated by a semipermeable membrane, pressure is developed in the solution, due to the presence of dissolved solutes. This is called osmotic pressure (OP).

Question 21.
Mention the symbolic representation of water potential and osmotic pressure.
Answer:
Water potential – ψ (psi) and Osmotic pressure – π (pi).

Question 22.
Mention the alternate terminologies and symbolic representation of solute potential and Matric potential.
Answer:

S. No.	Potential Energy	Alternate Terminology	Symbolic Representation
(a)	Solute potential	Osmotic potential	ψS
(b)	Matric potential	Imbibition pressure	ψM

Question 23.
Expand and Define TP.
Answer:
TP stands for Turgor pressure. When a plant cell is placed in pure water (hypotonic solution) the diffusion of water into the cell takes place by endosmosis. It creates a positive hydrostatic pressure on the rigid cell wall by the cell membrane. Henceforth the pressure exerted by the cell membrane towards the cell wall is Turgor Pressure (TP).

Question 24.
Mention the pressures that act upon the cell to make it a full turgid.
Answer:
Turgor pressure and wall pressure make the cell fully turgid. TP + WP = Turgid.

Question 25.
Define Diffusion Pressure Deficit (DPD).
Answer:
The difference between the diffusion pressure of the solution and its solvent at a particular temperature and atmospheric pressure is called Diffusion Pressure Deficit (DPD).

Question 26.
Why DPD is also called suction pressure?
Answer:
Increased DPD favors endosmosis or it sucks the water from the hypotonic solution, hence it is called suction pressure.

Question 27.
Define Osmosis.
Answer:
Osmosis represents the movement of water or solvent molecules through a selectively permeable membrane from the place of its higher concentration (high water potential) to the place of its’ lower concentration (low water potential).

Question 28.
State the significance of plasmolysis.
Answer:
Plasmolysis is exhibited only by living cells and so it is used to test whether the cell is living or dead.

Question 29.
What happens if a plant cell is treated with a hypertonic solution?
Answer:
When a plant cell is kept in a hypertonic solution, water leaves the cell due to exosmosis. As a result of water loss, protoplasm shrinks and the cell membrane is pulled away from the cell wall and finally, the cell becomes flaccid. This process is named plasmolysis.

Question 30.
Give an example for plasmolysis and also mention the types of plasmolysis.
Answer:
Wilting of plants noticed under the condition of water scarcity is an indication of plasmolysis. Three types of plasmolysis occur in plants: /) Incipient plasmolysis, ii) Evident plasmolysis, and iii) Final plasmolysis.

Question 31.
Draw a simplified diagram depicting Reverse Osmosis.
Answer:

Question 32.
If a cell in the cortex with DPD of 5 atm is surrounded by hypodermal cells with DPD of 2 atm, what will be the direction of movement of water?
Answer:
Water will move from low DPD to high DPD (hypodermis 2 atm to cortex 5 atm).

Question 33.
Arrange in the correct sequence in concern with the pathway of water, in roots, (cortex, root hair, xylem, epidermal cell, endodermis, and pericycle).
Answer:
Root hairs → epidermal cell → cortex → endodermis → pericycle → xylem.

Question 34.
Mention the possible routes of water movement across root cells.
Answer:
There are three possible routes of water.
They are i) Apoplast, ii) Symplast and iii) Transmembrane route.

Question 35.
What are the objections to the osmotic active absorption theory?
Answer:
Objections to osmotic theory: (i) The cell sap concentration in the xylem is not always high. (ii) Root pressure is not universal in all plants, especially in trees.

Question 36.
Name any two respiratory inhibitors.
Answer:
Potassium cyanide (KCN) and chloroform.

Question 37.
Define Ascent of sap.
Answer:
The water within the xylem along with dissolved minerals from roots is called sap and its upward transport is called ascent of sap.

Question 38.
State Relay Pump theory.
Answer:
Relay pump theory of Godlewski (1884)
Periodic changes in osmotic pressure of living cells of the xylem parenchyma and medullary ray act as a pump for the movement of water.

Question 39.
What do you mean by the term ‘Embolism”?
Answer:
Gas bubbles expanding and displacing water within the xylem element is called cavitation or embolism. .

Question 40.
Which is the most widely accepted theory to prove the Ascent of sap? Who proposed it.
Answer:
Cohesion-tension theory or Transpiration pull theory proposed by Dixon and Jolly (1894- 1924).

Question 41.
Define Transpiration.
Answer:
The loss of excess of water in the form of vapor from various aerial parts of the plant is called transpiration.

Question 42.
State any two theories regarding the mechanism of stomatal movement.
Answer:
Starch-sugar interconversion theory and Active potassium transport ion concept.

Question 43.
Draw and label the structure of stomata.
Answer:

Question 44.
What is an anti transpirant?
Answer:
The term antitranspirant is used to designate any material applied to plants for the purpose of retarding transpiration. An ideal antitranspirant checks the transpiration process without disturbing the process of gaseous exchange.

Question 45.
List out the uses of antitranspirants.
Answer:

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

Question 46.
What are hydathodes?
Answer:
Hydathodes are the stomata-like pores present in plants that grow in moist and shady places.

Question 47.
Define the term “Translocation of organic solutes”.
Answer:
The phenomenon of food transportation from the site of synthesis to the site of utilization is known as the translocation of organic solutes. The term solute denotes food material that moves in a solution.

Question 48.
What do you mean by Phloem loading?
Answer:
The movement of photosynthates (products of photosynthesis) from mesophyll cells to phloem sieve elements of mature leaves is known as phloem loading.

Question 49.
List the merits of the Munch-Mass Flow hypothesis.
Answer:

1. When a woody or herbaceous plant is girdled, the sap contains high sugar-containing exudates from the cut end.
2. A positive concentration gradient disappears when plants are defoliated.

Question 50.
Define – Flux and its types.
Answer:
The movement of ions into and out of cells or tissues is termed transport or flux. Entry of the ion into the cell is called influx and exit is called efflux.

Short answer questions

Question 1.
Draw a Flow Chart illustrating various types of cell-to-cell transport.
Answer:

Question 2.
Enumerate the significance of diffusion.
Answer:
Significance of diffusion in Plants

1. Gaseous exchange of O₂ and CO₂ between the atmosphere and stomata of leaves takes place by the process of diffusion. O₂ is absorbed during respiration and CO₂ is absorbed during photosynthesis.
2. In transpiration, water vapour from intercellular spaces diffuses into the atmosphere through stomata by the process of diffusion.
3. The transport of ions in mineral salts during passive absorption also takes place by this process.

Question 3.
How semipermeable and selectively permeable membranes differ from each other?
Answer:

Semi-Permeable	Selectively Permeable
Semi-permeable allow diffusion of solvent molecules but do not allow the passage of solute molecule. Example: Parchment paper.	All biomembranes allow some solutes to pass in addition to the solvent molecules. Example: Plasmalemma, tonoplast, and membranes of cell organelles.

Question 4.
Give an account on Aquaporin.
Answer:
Aquaporin is a water channel protein embedded in the plasma membrane. It regulates the massive amount of water transport across the membrane. Plants contain a variety of aquaporins. Over 30 types of aquaporins are known from maize. Currently, they are also recognized to transport substrates like glycerol, urea, CO₂, NH₃, metalloids, and Reactive Oxygen Species (ROS) in addition to water. They increase the permeability of the membrane to water. They confer drought and salt stress tolerance.

Question 5.
Give an account of Matric’s potential.
Answer:
Matric potential represents the attraction between water and the hydrating colloid or gel-like organic molecules in the cell wall which is collectively termed as matric potential. Matric potential is also known as imbibition pressure. The matric potential is maximum (most negative value) in dry material. Example: The swelling of soaked seeds in water.

Question 6.
How DPD differs in various conditions of a cell?
Answer:

• DPD in normal cell: DPD = OP – TP.
• DPD in the fully turgid cell: Osmotic pressure is always equal to turgor pressure in a fully turgid cell.
• OP = TP or OP-TP =0. Hence DPD of fully turgid cell is zero.
• DPD in the flaccid cell: If the cell is in flaccid condition there is no turgor pressure or TP=0.
  Hence DPD = OP.

Question 7.
Compare Hypertonic, Hypotonic, and Isotonic solutions.
Answer:

Question 8.
Give an account of Endosmosis and Exosmosis.
Answer:

1. Endosmosis: Endosmosis is defined as the osmotic entry of solvent into a cell or a system when it is placed in pure water or hypotonic solution. For example, dry raisins placed in the water, it swells up due to turgidity.
2. Exosmosis: Exosmosis is defined as the osmotic withdrawal of water from a cell or system when it is placed in a hypertonic solution. Exosmosis in a plant cell leads to plasmolysis.

Question 9.
Give an account of Deplasmolysis.
Answer:
The effect of plasmolysis can be reversed, by transferring them back into the water or hypotonic solution. Due to endosmosis, the cell becomes turgid again. It regains its original shape and size. This phenomenon of the revival of the plasmolyzed cell is called deplasmolysis. Example: Immersion of dry raisin in water.

Question 10.
Give an account of Reverse Osmosis and its uses.
Answer:
Reverse Osmosis follows the same principles of osmosis, but in the reverse direction. In this process movement of water is reversed by applying pressure to force the water against a concentration gradient of the solution. In regular osmosis, the water molecules move from the higher concentration (pure water = hypotonic) to the lower concentration (saltwater = hypertonic). But in reverse osmosis, the water molecules move from the lower concentration (saltwater = hypertonic) to higher concentration (pure water = hypotonic) through a selectively permeable membrane.
Uses: Reverse osmosis is used for the purification of drinking water and desalination of seawater.

Question 11.
Differentiate between the types of Plasmolysis.
Answer:
Difference between plasmolysis types.

Incipient Plasmolysis	Evident Plasmolysis	Final Plasmolysis
No morphological symptoms appear in plants.	Wilting of leaves appears.	Severe wiping and drooping of leaves appear.
The plasma membrane separates only at the comer from the cell wall of cells.	The plasma membrane completely detaches from the cell wall.	Plasma membrane completely detaches from cell wall with a maximum shrinkage of volume.
It is reversible.	It is reversible.	It is irreversible.

Question 12.
Write in brief about Non-Osmotic active absorption.
Answer:
Bennet-Clark (1936), Thimann (1951), and Kramer (1959) observed absorption of water even if the concentration of cell sap in the root hair is lower than that of the soil water. Such a movement requires an expenditure of energy released by respiration (ATP). Thus, there is a link between water absorption and respiration. It is evident from the fact that when respiratory inhibitors like KCN, Chloroform are applied there is a decrease in the rate of respiration and also the rate of absorption of water.

Question 13.
Explain the pulsation theory of J.C. Bose.
Answer:
Bose invented an instrument called the Crescograph, which consists of an electric probe connected to a galvanometer. When a probe is inserted into the inner cortex of the stem, the galvanometer showed high electrical activity. Bose believed a rhythmic pulsating movement of the inner cortex like a pump (similar to the beating of the heart) is responsible for the ascent of sap. He concluded that cells associated with xylem exhibit pumping action and pumps the sap laterally into xylem cells.

Question 14.
List out the setbacks of Root Pressure Theory.
Answer:
The following objections have been raised against root pressure theory:

1. Root pressure is totally absent in gymnosperms, which includes some of the tallest plants.
2. There is no relationship between the ascent of sap and root pressure. For example, in summer, the rate of the ascent of sap is more due to transpiration in spite of the fact that root pressure is very low. On the other hand, in winter when the rate of ascent of sap is low, a.high root pressure is found.
3. The ascent of sap continues even in the absence of roots.
4. The magnitude of root pressure is about 2atm, which can raise the water level up to few feet only, whereas the tallest trees are more than 100m high.

Question 15.
Describe the structure of stomata.
Answer:
The epidermis of leaves and green stems possess many small pores called stomata. The length and breadth of stomata are about 10-40 p and 3 – 10 p respectively. Mature leaves contain between 50 and 500 stomata per mm2. Stomata are made up of two guard cells, special semilunar or kidney-shaped living epidermal cells in the epidermis. Guard cells are attached to surrounding epidermal cells known as subsidiary cells or accessory cells. The guard cells are joined together at each end but they are free to separate to form a pore between them. The inner wall of the guard cell is thicker than the outer wall. The stoma opens to the interior into a cavity called the sub-stomatal cavity which remains connected with the intercellular spaces.

Question 16.
Explain the theory of photosynthesis in guard cells and state its demerits.
Answer:
Theory of Photosynthesis in guard cells:
Von Mohl (1856) observed that stomata open in light and close in the night. According to him, chloroplasts present in the guard cells photosynthesize in the presence of light resulting ‘ in the production of carbohydrate (Sugar) which increases osmotic pressure in guard cells.
It leads to the entry of water from other cells and the stomatal aperture opens. The above process vice versa at the night lead to the closure of stomata.

Demerits

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

Question 17.
Draw a diagrammatic representation of the steward scheme of stomatal movement.
Answer:

Question 18.
Why wilting occurs? Explain its types.
Answer:
Excessive loss of water through transpiration leads to wilting. In general, there are three types of wilting as follows,
a. Incipient wilting: The water content of plant cells decreases but the symptoms are not visible.
b. Temporary wilting: On hot summer days, the freshness of herbaceous plants reduces turgor pressure in the daytime and regains it at night.
c. Permanent wilting: The absorption of water virtually ceases because the plant cell does not get water from any source and the plant cell passes into a state of permanent wilting.

Question 19.
Give an account of types of structural modifications of leaves for reducing transpirational – loss.
Answer:
Leaf structure: Some anatomical features of leaves like sunken stomata, the presence of hairs, cuticle, the presence of hydrophilic substances like gum, mucilage help to reduce the rate of transpiration. In xerophytes the structural modifications are remarkable. To avoid transpiration, as in Opuntia the stem is flattened to look like leaves called Phylloclade. Cladode or cladophyll in Asparagus is a modified stem capable of limited growth looking like leaves. In some plants, the petioles are flattened and widened, to become phyllodes example Acacia melanoxylon.

Question 20.
Comment on various chemicals inducing stomatal closure.
Answer:
Carbon-di-oxide induces stomatal closure and acts as a natural antitranspirant. Further, the advantage of using CO₂ as an antitranspirant is its inhibition of photorespiration. Phenyl Mercuric Acetate (PMA), when applied as a foliar spray to plants, induces partial stomatal closure for two weeks or more without any toxic effect. The use of abscisic acid highly induces the closing of stomata. Dodecenyl succinic acid also affects on stomatal closure.

Question 21.
Classify translocation based on direction.
Answer:
Phloem translocates the products of photosynthesis from leaves to the area of growth and storage, in the following directions,
Downward direction: From leaves to stem and roots.
Upward direction: From leaves to developing buds, flowers, fruits for consumption and storage. Germination of seeds is also a good example of upward translocation.
Radial direction: From cells of the pith to cortex and epidermis, the food materials are radially translocated.

Question 22.
What do you understand by the term source and sink in plant physiology?
Answer:
The source is defined as any organ in plants which are capable of exporting food materials to the areas of metabolism or to the areas of storage. Examples: Mature leaves and germinating seeds.

A sink is defined as any organ in plants that receives food from a source. Example: Roots, tubers, developing fruits, and immature leaves.

Question 23.
Define Phloem unloading with its steps.
Answer:
From sieve elements, sucrose is translocated into sink organs such as roots, tubers, flowers, and fruits, and this process is termed phloem unloading. It consists of three steps:

1. sieve element unloading: Sucrose leave from sieve elements.
2. Short distance transport: Movement of sucrose to sink cells.
3. Storage and metabolism: The final step when sugars are stored or metabolized in sink cells.

Question 24.
Enumerate the objections of the Munchi Mass Flow hypothesis.
Answer:

1. Munchi Mass Flow hypothesis explains the unidirectional movement of solute only. However, bidirectional movement of solute is commonly observed in plants.
2. The osmotic pressure of mesophyll cells and that of root hair do not confirm the requirements.
3. This theory gives a passive role to sieve tube and protoplasm, while some workers demonstrated the involvement of ATP.

Question 25.
Donnan equilibrium. Add a note.
Answer:
Within the cell, some of the ions never diffuse out through the membrane. They are trapped within the cell and are called fixed ions. But they must be balanced by the ions of opposite charge. Assuming that a concentration of fixed anions is present inside the membrane, more cations would be absorbed in addition to the normal exchange to maintain the equilibrium. Therefore, the cation concentration would be greater in the internal than in the external solution. This electrical balance or equilibrium controlled by electrical as well as diffusion phenomenon is known as the Donnan equilibrium.

Long answer questions

Question 1.
Write in detail about types of carrier protein.
Answer:
There are three types of carrier proteins classified on the basis of handling of molecules and the direction of transport. They are i. Uniport, ii. Symport, iii. Antiport.

1. Uniport: In this molecule of a single type move across a membrane independent of other molecules in one direction.
2. Symport or co-transport: The term symport is used to denote an integral membrane protein that simultaneously transports two types of molecules across the membrane in the same direction.
3. Antiport or Counter Transport: An antiport is an integral membrane transport protein that simultaneously transports two different molecules, in opposite directions, across the membrane.

Question 2.
Write a note on Solute potential and Pressure Potential.
Answer:
1. Solute Potential (ψ_S): Solute potential, otherwise known as osmotic potential denotes the effect of dissolved solute on water potential. In pure water, the addition of solute reduces its free energy and lowers the water potential value from zero to negative. Thus the value of solute potential is always negative. In a solution at standard atmospheric pressure, water potential is always equal to solute potential (ψ_W = ψ_S).
2. Pressure Potential (ψ_P): Pressure potential is a mechanical force working against the effect of solute potential. Increased pressure potential will increase water potential and water enters cells and cells become turgid. This positive hydrostatic pressure within the cell is called Turgor pressure. Likewise, withdrawal of water from the cell decreases the water potential and the cell becomes flaccid.

Question 3.
With the help of a diagram explain the possible route of water across root cells.
Answer:
There are three possible routes of water. They are i) Apoplast, ii) Symplast and iii) Transmembrane route.

1. Apoplast: The apoplast (Greek: apo = away; plast = cell) consists of everything external to the plasma membrane of the living cell. The apoplast includes cell walls, extracellular spaces, and the interior of dead cells such as vessel elements and tracheids. In the apoplast pathway, water moves exclusively through the cell wall or the non-living part of the plant without crossing any membrane. The apoplast is a continuous system.
   
2. Symplast: The symplast (Greek: sym = within; plast = cell) consists of the entire mass of cytosol of all the living cells in a plant, as well as the plasmodesmata, the cytoplasmic channel that interconnects them. In the symplastic route, water has to cross the plasma membrane to enter the cytoplasm of the outer root cell; then it will move within adjoining cytoplasm through plasmodesmata around the vacuoles without the necessity to cross more membrane, till it reaches xylem.
3. Transmembrane route: In the transmembrane pathway water sequentially enters a cell on one side and exits from the cell on the other side. In this pathway, water crosses at least two membranes for each cell. Transport across the tonoplast is also involved.

Mechanism of Water Absorption Kramer (1949) recognized two distinct mechanisms which independently operate in the absorption of water in plants. They are, i) active absorption ii) passive absorption.

Question 4.
Point out the differences between Active absorption and Passive absorption.
Answer:

Active Absorption	Passive Absorption
Active absorption takes place by the activity of root and root hairs.	The pressure for absorption is not developed in roots and hence roots play a passive role.
Transpiration has no effect on active absorption.	Absorption regulated by transpiration.
The root hairs have high DPD as compared to soil solution and therefore water is taken by tension.	The absorption occurs due to tension created in xylem sap by transpiration pull, thus water is sucked in by the tension.
Respiratory energy needed.	Respiratory energy not required.
It involves the symplastic movement of water.	Both symplast and apoplast movement of water involved.                                          ‘

Question 5.
Explain about Cohesion – Tension Theory.
Answer:
Cohesion-tension theory was originally proposed by Dixon and Jolly (1894) and again put forward by Dixon (1914, 1924). This theory is based on the following features:

1. Strong cohesive force or tensile strength of water: Water molecules have the strong mutual force of attraction called cohesive force due to which they cannot be easily separated from one another. Further, the attraction between a water molecule and the wall of the xylem element is called adhesion. These cohesive and adhesive force works together to form an unbroken continuous water column in the xylem. The magnitude of the cohesive force is much high (350 atm) and is more than enough to ascent sap in the tallest trees.
2. Continuity of the water column in the plant: An important factor that can break the water column is the introduction of air bubbles in the xylem. Gas bubbles expanding and displacing water within the xylem element is called cavitation or embolism. However, the overall continuity of the water column remains undisturbed since water diffuses into the adjacent xylem elements for continuing ascent of sap.
3. Transpiration pull or Tension in the unbroken water column: The unbroken water column from leaf to root is just like a rope. If the rope is pulled from the top, the entire rope will move upward. In plants, such a pull is generated by the process of transpiration which is known as transpiration pull. Water vapour evaporates from mesophyll cells to the intercellular spaces near stomata as a result of active transpiration. The water vapours are then transpired through the stomatal pores. Loss of water from mesophyll cells causes a decrease in water potential. So, water moves as a pull from cell to cell along the water potential gradient. This tension, generated at the top (leaf) of the unbroken water column, is transmitted downwards from the petiole, stem and finally reaches the roots. The cohesion theory is the most accepted among plant physiologists today.

Question 6.
Describe the types of transpiration in plants.
Answer:
Types of Transpiration: Transpiration is of the following three types:

1. Stomatal transpiration: Stomata are microscopic structures present in high numbers on the lower epidermis of leaves. This is the most dominant form of transpiration and being responsible for most of the water loss (90 – 95%) in plants.
2. Lenticular transpiration: In stems of woody plants and trees, the epidermis is replaced by periderm because of secondary growth. In order to provide gaseous exchange between the living cells and the outer atmosphere, some pores which look like lens-shaped raised spots are present on the surface of the stem called Lenticels. The loss of water from lenticels is very insignificant as it amounts to only 0.1 % of the total.
3. Cuticular transpiration: The cuticle is a waxy or resinous layer of cutin, a fatty substance covering the epidermis of leaves and other plant parts. Loss of water through the cuticle is relatively small and it is only about 5 to 10 % of the total transpiration. The thickness of cuticle increases in xerophytes and transpiration is very much reduced or totally absent.

Question 7.
Give an account of Starch – Sugar Interconversion Theory.
Answer:

1. According to Lloyd (1908), the turgidity of guard cells depends on the interconversion, of starch and sugar. It was supported by Loftfield (1921) as he found guard cells containing sugar during the daytime when they are open and starch during the night when they are closed.
2. Sayre (1920) observed that the opening and closing of stomata depend upon the change in pH of guard cells. According to him stomata open at high pH during the daytime and become closed at low pH at night. The utilization of CO₂ by photosynthesis during the light period causes an increase in pH resulting in the conversion of starch to sugar. Sugar increase in cell favors endosmosis and increases the turgor pressure which leads to opening of stomata. Likewise, the accumulation of CO₂ in cells during the night decreases the pH level resulting in the conversion of sugar to starch. Starch decreases the turgor pressure of the guard cell and stomata close.
3. The discovery of enzyme phosphorylase in guard cells by Hanes (1940) greatly supports the starch-sugar interconversion theory. The enzyme phosphorylase hydrolyses starch into sugar and high pH followed by endosmosis and the opening of stomata during light. The vice versa takes place during the night.
   
4. Steward (1964) proposed a slightly modified scheme of starch-sugar interconversion theory. According to him, Glucose-1-phosphate is osmotically inactive. Removal of phosphate from Glucose-1-phosphate converts to Glucose which is osmotically active and increases the concentration of guard cells leading to the opening of stomata.

Objections to Starch-sugar interconversion theory

1. In monocots, the guard cell does not have starch.
2. There is no evidence to show the presence of sugar at a time when starch disappears and stomata open.
3. It fails to explain the drastic change in pH from 5 to 7 by change of CO₂

Question 8.
Describe the K⁺ Transport theory on transpiration.
Answer:
The theory of K⁺ transport theory was proposed by Levit (1974) and elaborated by Raschke (1975). According to this theory, the following steps are involved in the stomatal opening:

In light

1. In the guard cell, starch is converted into organic acid (malic acid).
2. .Malic acid in the guard cell dissociates to malate anion and proton (H⁺).
3. Protons are transported through the membrane into nearby subsidiary cells with the
   exchange of K⁺ (Potassium ions) from subsidiary cells to guard cells. This process involves an electrical gradient and is called ion exchange.
4. This ion exchange is an active process and consumes ATP for energy.
5. Increased K⁺ ions in the guard cell are balanced by CL ions. An increase in solute concentration decreases the water potential in the guard cell.
6. Guard cell becomes hypertonic and favours the entry of water from surrounding cells.
7. Increased turgor pressure due to the entry of water opens the stomatal pore.

In Dark

1. In the dark photosynthesis stops and respiration continues with the accumulation of CO₂ in the sub-stomatal cavity.
2. Accumulation of CO₂ in the cell lowers the pH level.
3. Low pH and a shortage of water in the guard cell activate the stress hormone Abscisic acid (ABA).
4. ABA stops further entry of K⁺ ions and also induces K⁺ ions to leak out to subsidiary cells from guard cells.
5. Loss of water from the guard cell reduces turgor pressure and causes the closure of stomata.

Question 9.
Draw the structure of hydathode and explain Guttation.
Answer:

During high humidity in the atmosphere, the rate of transpiration is much reduced. When plants absorb water in such a condition root pressure is developed due to excess water within the plant. Thus excess water exudates as a liquid from the edges of the leaves and are called guttation. Example: Grasses, tomato, potato, brinjal, and Alocasia. Guttation occurs through stomata-like pores called hydathodes generally present in plants that grow in moist and shady places. Pores are present over a mass of loosely arranged cells with large intercellular spaces called epithem. This mass of tissue lies near vein endings (xylem and Phloem). The liquid coming out of the hydathode is not pure water but a solution containing a number of dissolved substances.

Question 10.
Explain Ganong’s Potometer experiment and its purpose.
Answer:

Ganongs potometer is used to measure the rate of transpiration indirectly. In this, the amount of water absorbed is measured, and assumed that this amount is equal to the amount of water transpired.
The apparatus consists of a horizontal graduated tube that is bent in opposite directions at the ends. One bent end is wide and the other is narrow. A reservoir is fixed to the horizontal tube near the wider end.
The reservoir has a stopcock to regulate water flow. The apparatus is filled with water from the reservoir. A twig or a small plant is fixed to the wider arm through a split cock. The other bent end of the horizontal tube is dipped into a beaker containing colored water. An air bubble is introduced into the graduated tube at the narrow end keep this apparatus in bright sunlight and observe. As transpiration takes place, the air bubble will move towards the twig. The loss is compensated by water absorption through the xylem portion of the twig. Thus, the rate of water absorption is equal to the rate of transpiration.

Question 11.
Describe the ringing experiment with a diagram.
Answer:
The experiment involves the removal of all the tissue outside to vascular cambium (bark, cortex, and phloem) in woody stems except the xylem. Xylem is the only remaining tissue in the girdled area which connects the upper and lower part of the plant. This setup is placed in a beaker of water. After some time, it is observed that swelling on the upper part of the ring appears as a result of the accumulation of food material. If the experiment continues within days, the roots die first. It is because the supply of food material to the root is cut down by the removal of phloem. The roots cannot synthesize their food and so they die first. As the roots gradually die the upper part (stem), which depends on root for the ascent of sap, will ultimately die.
Ringing experiment.

Question 12.
Write in detail about Passive Absorption of minerals salts.
Answer:
1. Ion-Exchange: Ions of external soil solution were exchanged with same charged (anion for anion or cation for cation) ions of the root cells. There are two theories explaining this process of ion exchange namely: i. Contact exchange and ii. Carbonic acid exchange.

1. Contact Exchange Theory: According to this theory, the ions adsorbed on the surface of root cells and clay particles (or clay micelles) are not held tightly but oscillate within a small volume of space called oscillation volume. Due to the small space, both ions overlap each other’s oscillation volume and exchange takes place.
2. Carbonic Acid Exchange Theory: According to this theory, soil solution plays an important role by acting as a medium for ion exchange. The CO₂ released during respiration of root cells combines with water to form carbonic acid (H₂CO₃). Carbonic acid dissociates into H⁺ and HCO₃^– in the soil solution. These H⁺ ions exchange with cations adsorbed on clay particles and the cations from micelles get released into soil solution and gets adsorbed on root cells.
   
   

Question 13.
Describe Lundegardh’s Cytochrome Pump Theory.
Answer:
Lundegardh’s Cytochrome Pump Theory:
Lundegardh and Burstrom (1933) observed a correlation between respiration and anion absorption. When a plant is transferred from water to a salt solution the rate of respiration increases which is called anion respiration or salt respiration. Based on this observation Lundegardh (1950 and 1954) proposed cytochrome pump theory which is based on the following assumptions:

1. The mechanism of anion and cation absorption is different.
2. Anions are absorbed through the cytochrome chain by an active process, cations are absorbed passively.
3. An oxygen gradient is responsible for oxidation at the outer surface of the membrane and reduction at the inner surface.

According to this theory, the enzyme dehydrogenase on the inner surface is responsible for the formation of protons (H⁺) and electrons (e^–). As electrons pass outward through the electron transport chain there is a corresponding inward passage of anions. Anions are picked up by oxidized cytochrome oxidase and are transferred to other members of the chain as they transfer the electron to the next component.
The theory assumes that cations (C⁺) move passively along the electrical gradient created by the accumulation of anions (A^–) at the inner surface of the membrane.
Main defects of the above theory are:

1. Cations also induce respiration.
2. Fails to explain the selective uptake of ions.
3. It explains absorption of anions only.
   

Question 14.
Explain Protein-Lecithin Theory.
Answer:
Bennet-Clark’s Protein-Lecithin Theory:
In 1956, Bennet-Clark proposed that the carrier could be a protein associated with phosphatide called lecithin. The carrier is amphoteric (the ability to act either as an acid or a base) and hence both cations and anions combine with it to form the Lecithinion complex in the membrane. Inside the membrane, the Lecithin-ion complex is broken down into phosphatidic acid and choline along with the liberation of ions. Lecithin again gets regenerated from phosphatidic acid and choline in the presence of the enzyme choline acetylase and choline esterase. ATP is required for the regeneration of lecithin.

Higher Order Thinking Skills (HOTs)

Question 1.
Why during rainy seasons, the wooden doors and windows are difficult to close and open? Give the phenomenon behind this and also define the phenomenon.
Answer:
The swelling of wooden windows, doors due to high humidity during the rainy season is due, to imbibition. Imbibition refers to the uptake of water and swelling of substances that are not water-soluble.

Question 2.
While making dry fishes at home high salt concentration is applied. Why? Name the phenomenon.
Answer:
The high salt concentration is applied to extract the excess water from the fishes thereby inhibiting the microbial growth and increase the shelf life. This phenomenon is called exosmosis.

Question 3.
Observe the histogram and answer the following question.

(a) Which type of transpiration does ‘A’ and ‘C’ represent?
(b) Name ‘B’ and also define the plant part. Which is responsible for ‘B’ type of transpiration?
Answer:
(a) A – Stomatal transpiration, C – Cuticular transpiration and B – Lenticular transpiration. Lenticels are the lend-shaped raised spots present on the surface of the stem.

Question 4.
Plants are highly adaptable to the environment where they survive Opuntia which lives in xeric condition shows phylloclade adaptation. Which part of Opuntia is modified as phylloclade? Why does it modify?
Answer:
In Opuntia, the stem is flattered to look like leaves called phylloclade. It is modified to avoid transpirational loss of water.

Question 5.
Nowadays, water scarcity is becoming a prime problem. To compensate for the need, various strategies are being carried, out by the Governments at national and international levels. One such effective technology is the Desalination of seawater. Which principle is followed in their technology. Define it.
Answer:
Desalination of seawater is being effectively earned out by Reverse Osmosis. In Reverse osmosis, the water molecules are forced by applying pressure from lower concentration to higher concentration through a selectively permeable membrane.

Question 6.
Observe the diagram, it is a plant cell undergoing plasmolysis.
(a) Which stage of plasmolysis does the cell represent?

(b) Whether it is reversible?
(c) What happens if this occurs in the leaf cells?
Answer:
(a) Evident plasmolysis.
(b) Yes, evident plasmolysis is reversible.
(c) Under evident plasmolysis, wilting of leaves appears.

Choose the correct answer.

1. …………. is a downhill process using physical forces.
(a) Short distance transport
(b) Translocation
(c) Active transport
(d) Passive transport
Answer:
(d) Passive transport

2. The smell of the room spray can be felt everywhere inside a closed room. This is because of ………….
(a) Osmosis
(b) Passive transport
(c) Diffusion
(d) Imbibition
Answer:
(c) Diffusion

3. A mixture of ………… and potassium permanganate is used for fumigation.
(a) Acetaldehyde
(b) Calcium oxide
(c) Formalin
(d) Vinegar
Answer:
(c) Formalin

4. ROS stands for …………
(a) Reduction Oxygen Species
(b) Reactive Oxygen Syndrome
(c) Reducive Oxygen Species
(d) Reactive Oxygen Species
Answer:
(d) Reactive Oxygen Species

5. Peter Agre discovered aquaporin in ………….
(a) RBC
(b) WBC
(c) Platelets
(d) Plasma membrane
Answer:
(a) RBC

6. Protoplasm is made of ……….. of water.
(a) 80-90%
(b) 85-90%
(c) 60-80%
(d) 75-85%
Answer:
(c) 60-80%

7. ………… does not act as an imbibant.
(a) Protein
(b) Starch
(c) Stone
(d) Gum
Answer:
(c) Stone

8. Which physiological process can be observed in a germinating seed?
(a) Diffusion
(b) Osmosis
(c) Imbibition
(d) Facilitated diffusion
Answer:
(c) Imbibition

9. Water potential is measured in ………….
(a) Watt
(b) Joule
(c) Calories
(d) Pascal
Answer:
(d) Pascal

10. Water potential can be determined by solute potential and ………….
(a) Matric potential
(b) Pressure potential
(c) Osmotic potential
(d) Osmotic potential
Answer:
(b) Pressure potential

11. Osmotic pressure is represented by the Greek letter ………….
(a) α
(b) π
(c) ψ
(d) θ
Answer:
(b) π

12. If osmotic pressure has a positive value, the osmotic potential has ………… value.
(a) Positive
(b) Negative
(c) Neutral
(d) Zero
Answer:
(b) Negative

13. Which combination of pressures makes a cell full turgid?
(a) TP + WP
(b) OP -TP
(c) SP + OP
(d) WP + SP
Answer:
(a) TP + WP

14. In a hypertonic solution, which of the following substance concentration will be high?
(a) Solute
(b) Solvent
(c) Both a & b
(d) None
Answer:
(a) Solute

15. Dry raisins kept in the water begin to swell. It is a perfect example for ………….
(a) Osmosis
(b) Plasmolysis
(c) Endosmosis
(d) Final plasmolysis
Answer:
(c) Endosmosis

16. Among the following physiological processes, which one occurs only in the living cells?
(a) Diffusion
(b) Osmosis
(c) Exosmosis
(d) Plasmolysis
Answer:
(d) Plasmolysis

17. To revive a plasmolyzed cell, it should be treated with ………. solution.
(a) Isotonic
(b) Hypertonic
(c) Hypotonic
(d) Neutral
Answer:
(c) Hypotonic

18. Principle behind the desalination of sea water is …………
(a) Endosmosis
(b) Diffusion
(c) Reverse osmosis
(d) Deplasmolysis
Answer:
(c) Reverse osmosis

19. The final destination of water entering the root hair is ………….
(a) Endodermis
(b) Pericycle
(c) Xylem
(d) Cortex
Answer:
(c) Xylem

20. The nature of cell sap is ………….
(a) Hypertonic
(b) Hypotonic
(c) Isotonic
(d) Apoplast
Answer:
(a) Hypertonic

21. Relay pump theory was put forth by ………….
(a) Atkins and Preistley
(b) Strasburger and Overton
(c) Godlewski
(d) J.C. Bose
Answer:
(c) Godlewski

22. Who is called as father of plant physiology?
(a) J.C. Bose
(b) Stephen Hales
(c) Dixon
(d) Unger
Answer:
(b) Stephen Hales

23. An instrument deviced by J.C. Bose for proving the pulsating movement of cortex is ……….
(a) Seismograph
(b) Galvanometer
(c) Crescograph
(d) Radiograph
Answer:
(c) Crescograph

24. In embolism of xylem, water is displaced by …………
(a) Callose
(b) Photosynthates
(c) Gas bubbles
(d) Porin proteins
Answer:
(c) Gas bubbles

25. Transpiration is a kind of ………. occur through plant body.
(a) Condensation
(b) Sublimation
(c) Evaporation
(d) Percipitation
Answer:
(c) Evaporation

26. Rate of water movement through xylem is …………
(a) 65 cm/second
(b) 75 cm/min
(c) 12 cm/min
(d) 82 cm/sec
Answer:
(b) 75 cm/min

27. ………… is a fatty substance covering the epidermis of leaves.
(a) Mucin
(b) Cutin
(c) Porin
(d) Suberin
Answer:
(b) Cutin

28. Approximately, a corn plant transpires ……….. litres of H₂O per day.
(a) 2
(b) 8
(c) 16
(d) 450
Answer:
(a) 2

29. Starch – sugar interconversion theory was proposed by ………..
(a) Loft field
(b) Lloyd
(c) Von Mohl
(d) Hanes
Answer:
(a) Lloyd

30. ABA stands for ……..
(a) Abscisic acid
(b) Ascorbic acid
(c) Acetyl Butyric Acid
(d) Acetic acid
Answer:
(a) Abscisic acid

31. Pick out the natural antitranspirant.
(a) O₂
(b) SO₂
(c) CO₂
(d) CO
Answer:
(c) CO₂

32. Guttation occurs through …………
(a) Stoma
(b) Epithem
(c) Hydathodes
(d) Epidermis
Answer:
(c) Hydathodes

33. CO₂ inhibits …………
(a) Transpiration
(b) Photorespiration
(c) Both a and b
(d) Respiration
Answer:
(c) Both a and b

34. Match the following:

S. No.	Particulars	S. No.	Particulars
1.	Theory of K+transport	(i)	Atkins and Priestley
2.	Osmotic active absorption	(ii)	Unger
3.	Capillary theory	(iii)	Levit
4.	Imbibition theory	(iv)	Boehm

(a) 1 – (iii), 2 – (i), 3 – (iv), 4 – (ii)
(b) 1 – (i), 2 – (ii), 3 – (i), 4 – (iii)
(c) 1 – (ii), 2 – (i), 3 – (iv), 4 – (iii)
(d) 1 – (i), 2 – (iii), 3 – (ii), 4 – (iv)
Answer:
(a) 1 – (iii), 2 – (i), 3 – (iv), 4 – (ii)

35. …………. is used to measure the rate of transpiration.
(a) Ganongs respiroscope
(b) Ganongs potometer
(c) Arc auxanometer
(d) None of these
Answer:
(b) Ganongs potometer

36. Electro-Osmotic theory was propounded by ……….
(a) Mason and Masked
(b) Fenson and Spanner
(c) Curtis
(d) Levit
Answer:
(b) Fenson and Spanner

37. Translocation of photosynthetic products is a ……… movement.
(a) unidirectional
(b) upward
(c) downward
(d) multi-directional
Answer:
(d) multidirectional

Students get through the TN Board 11th Bio Botany Important Questions Chapter 10 Secondary Growth which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 10 Secondary Growth

Very short answer questions

Question 1.
Which meristem is responsible for secondary growth in dicots? Mention its types.
Answer:
Secondary growth in dicots is the responsibility of lateral meristems. They are vascular cambium and cork cambium.

Question 2.
Define vascular cambial ring.
Answer:
The interfascicular cambium joins with the intrafascicular cambium on both sides to form a continuous ring. It is called a vascular cambial ring.

Question 3.
Name the two types of initials developed by vascular cambium.
Answer:
Fusiform initials and ray initials.

Question 4.
What happens to the primary xylem and phloem during secondary growth?
Answer:
Due to the continued formation of secondary xylem and phloem through vascular cambial activity, both the primary xylem and phloem get gradually crushed.

Question 5.
Differentiate between porous and non-porous wood.
Answer:

Porous Wood	Non-porous Wood
Common in angiosperms.	Common in gymnosperms.
Porous because it contains vessels.	Non-porous because it does not contain vessels.
Example: Morus	Example: Pinus

Question 6.
State the types of the lateral meristem, their role in woody plants.
Answer:

Types of Lateral meristem	Function
Vascular cambium	Produces xylem and Phloem
Cork cambium	Produces the bark of the tree

Question 7.
Springwood – comment.
Answer:
In the spring season, cambium is very active and produces a large number of xylary elements having vessels/tracheids with a wide lumen. The wood formed during this season is called springwood or earlywood.

Question 8.
Define the term Dendroclimatoiogy.
Answer:
Dendroclimatology is a branch of dendrochronology concerned with constructing records of past climates and climatic events by analysis of tree growth characteristics, especially growth rings.

Question 9.
Which study does dendrochronology deal with?
Answer:
The determination of the age of a tree by counting the annual rings is called dendrochronology.

Question 10.
What do you mean by the term “Autumn Wood”?
Answer:
In winter, the cambium is less active and forms fewer xylary elements that have narrow vessels/tracheids and this wood is called autumn wood or latewood.

Question 11.
All growth rings are not annual. Why?
Answer:
Sometimes annual rings are called growth rings but all the growth rings are not annual. In some trees, more than one growth ring is formed within a year due to climatic changes.

Question 12.
State the location of tyloses.
Answer:
In many dicot plants, the lumen of the xylem vessels is blocked by many balloon-like ingrowths from the neighboring parenchymatous cells. These balloon-like structures are called tyloses.

Question 13.
List out the substances that are accumulated in tyloses.
Answer:
In fully developed tyloses, starchy crystals, resins, gyms, oils, tannins, or colored substances are found.

Question 14.
What are tylosoids?
Answer:
In angiosperms, the sieve tubes are blocked by tylose-like ingrowths from the neighbouring parenchymatous cells. Example: Bombax. These are called tylosoids.

Question 15.
Why the heartwood is unable to conduct water?
Answer:
The heartwood stops conducting, water. As vessels of the heartwood are blocked by tyloses, water is not conducted through them. Due to the presence of tyloses and their contents, the heartwood becomes colored, dead, and the hardest part of the wood.

Question 16.
Give the proper botanical term for heartwood and sapwood?
Answer:
Heartwood – Duramen and Sapwood – alburnum.

Question 17.
Write a brief note on Haematoxylin.
Answer:
Hematoxylin is a dye, obtained from the heartwood of Haematoxylum campechianum used to stain plant materials for observation under a microscope, especially the nucleus of the cell.

Question 18.
From where does the Canada balsam is obtained? Mention its use.
Answer:
Canada balsam is obtained from the resin ducts of a gymnospermic plant called Abies balsamea. It is used as a mounting medium for microscopic slide preparation.

Question 19.
List out the names of the plants and the types of bast fibers produced from that plant.
Answer:

Plant Name	Bast Fibre
Linum ustitaissimum	Flax
Cannabis sativa	Hemp
Crotalaria juncea	Sun hemp
Corchorus capsularis	Jute

Question 20.
What is periderm? Mention its components.
Answer:
The periderm is a protective tissue of secondary origin that replaces the epidermis and often the primary cortex. The periderm consists of phellem, phellogen, and phelloderm.

Question 21.
What is Phelloderm?
Answer:
Phelloderm (Secondary cortex) is a tissue resembling cortical living parenchyma produced centripetally (inward) from the phellogen as a part of the periderm of stems and roots in seed plants.

Question 22.
What are Lenticels? State its uses.
Answer:
Lenticel is raised opening or pore on the epidermis or bark of stems and roots. It is formed during secondary growth in stems.
Lenticel is helpful in the exchange of gases and transpiration called lenticular transpiration.

Question 23.
Define filling tissue.
Answer:
When phellogen is more active in the region of lenticels, a mass of loosely arranged thin-walled parenchyma cells is formed. It is called complementary tissue or filling tissue.

Question 24.
Enumerate any two commercial barks.
Answer:
(a) Cinchona bark used for antimalarial drug production.
(b) Cinnamon bark used as a spice.

Short answer questions

Question 1.
What do you understand by the term longitudinal growth?
Answer:
The plant organs originating from the apical meristems pass through a period of expansion in length and width. The roots and stems grow in length with the help of apical meristems. This is called primary growth or longitudinal growth.

Question 2.
Compare Intra fascicular & Inter fascicular cambium.
Answer:

Intrafascicular cambium	Interfascicular cambium
Present inside the vascular bundles.	Present in between the vascular bundles.
Originates from the procambium.	Originates from the medullary rays.
Initially, it forms a part of the primary meristem.	From the beginning, it forms a part of the secondary meristem.

Question 3.
Comment on Storied cambium.
Answer:
If the fusiform initials are arranged in horizontal tiers, with the end of the cells of one tier appearing at approximately the same level, as seen in the tangential longitudinal section (TLS), it is called storied (stratified) cambium.

Question 4.
How pseudo-annual rings are formed?
Answer:
Additional growth rings are developed within a year due to adverse natural calamities like drought, frost, defoliation, flood, mechanical injury, and biotic factors during the middle of a growing season, which results in the formation of more than one annual ring. Such rings are called pseudo- or false-annual rings.

Question 5.
List out the differences between springwood and autumn wood.
Answer:

Spring Wood or Early Wood	Autumn Wood or Late Wood
The activity of the cambium is faster.	The activity of the cambium is slower.
Produces a large number of xylem elements.	Produces fewer xylem elements.
Xylem vessels/trachieds have wider lumen.	Xylem vessels/tracheids have narrow lumen.
Wood is lighter in color and has a lower density.	Wood is darker in color and has a higher density.

Question 6.
Classify and explain the types of angiospermic woods based on the diameter of xylem vessels.
Answer:
On the basis of the diameter of xylem vessels, two main types of angiosperm woods are recognized.

1. Diffuse porous woods: Diffuse porous woods are woods in which the vessels or pores are rather uniform in size and distribution throughout an annual ring. Example: Acer.
2. Ring porous woods: The pores of the earlywood are distinctly larger than those of the latewood. Thus rings of wide and narrow vessels occur. Example: Quercus.

Question 7.
Differentiate between diffuse-porous wood and ring-porous wood.
Answer:

Diffuse porous wood	Ring porous wood
This type of wood is formed where the climatic conditions are uniform.	This type of wood is formed where the climatic conditions are not uniform.
The vessels are more or less equal in diameter in any annual ring.	The vessels are wide and narrow within any annual ring.
The vessels are uniformly distributed throughout the wood.	The vessels are not uniformly distributed throughout the wood.

Question 8.
Give an account on Phellogen.
Answer:
Phellogen (Cork Cambium):
It is a secondary lateral meristem. It comprises homogenous meristematic cells, unlike vascular cambium. It arises from the epidermis, cortex, phloem, or pericycle (extrastelar in origin). Its cells divide periclinally and produce radially arranged files of cells. The cells towards the outer side differentiate into phellem (cork) and those towards the inside as phellogen (secondary cortex).

Question 9.
Define the term Rhytidome.
Answer:
Rhytidome is a technical term used for the outer dead bark which consists of periderm and isolated cortical or phloem tissues formed during successive secondary growth. Example: Quercus.

Question 10.
Compare ring barks with scale barks.
Answer:
If the phellogen forms a complete cylinder around the stem, it gives rise to ring barks. Example: Quercus. When the bark is formed in overlapping scale-like layers, it is known as scale bark. Example: Guava.

Question 11.
Enumerate the beneficial aspects of bark.
Answer:
Bark protects the plant from parasitic fungi and insects, prevents water loss by evaporation, and guards against variations of external temperature. It is insect repellent, decay proof, fireproof and is used in obtaining drugs or spices. The phloem cells of the bark are involved in the conduction of food while secondary cortical cells involved in storage.

Question 12.
(i) Draw and label the structure of lenticel.
Answer:

(ii) Draw the C.S. of Wood showing the annual ring.
Answer:

Long answer questions

Question 1.
Differentiate between secondary growth in dicot stem and root.
Answer:

Secondary growth in dicot stem	Secondary growth in dicot root
The cambial ring formed is circular in cross-section from the beginning.	The cambial ring formed is wavy in the beginning and later becomes circular.
The cambial ring is partially primary (fascicular cambium)and partially secondary (Interfascicular cambium) in origin.	The cambial ring is completely secondary in origin.
Generally, periderm originates from the cortical cells (extrastelar in origin).	Generally, periderm originates from the pericyle. (intrastelar in origin).
More amount of cork is produced as the stem is above the ground.	Generally, less amount cork is produced as the root is underground.
Lenticels of periderm are prominent.	Lenticels of periderm are not very prominent.

Question 2.
Depict the tissue lineage during secondary growth in the dicot stem and root.
Answer:

Question 3.
List out the differences between Vascular cambium and cork cambium.
Answer:

Vascular cambium	Cork cambium
Also called cambium.	Also called phellogen.
It arises from procambium and interfascicular parenchyma in stems and from conjunctive parenchyma in roots.	It arises from the epidermis, cortex, phloem, or pericyle in both stems and roots.
It comprises long fusiform and short ray initials.	It comprises homogenous cells.
It produces secondary phloem towards the outer side and secondary xylem towards the inner side.	It produces phellem(cork) towards the outer side and phelloderm (secondary cortex) towards the inner side.

Question 4.
Differentiate between Phellem and Phelloderm.
Answer:

Phellem (Cork)	Phelloderm (Secondary cortex)
It is formed on the outer side of phellogen.	It is formed on the inner side of the phellogen.
Cells are compactly arranged in regular tires and rows without intercellular spaces.	Cells are loosely arranged with intercellular spaces.
Protective in function.	As it contains chloroplast, it synthesizes and stores food.
Consists of non-living cells with suberized walls.	Consists of living cells, parenchyma­tous in nature and does not have suberin.
Lenticels are present.	Lenticels are absent.

Question 5.
Differentiate between Heartwood and Sapwood.
Answer:

Sap Wood (Alburnum)	Heart Wood (Duramen)
Living part of the wood.	The dead part of the wood.
It is situated on the outer side of the wood.	It is situated in the center part of the wood.
It is less colored.	It is dark in colored.
Very soft in nature.	Hard in nature.
Tyloses are absent.	Tyloses are present.
It is not durable and not resistant to microorganisms.	It is more durable and resists microorganisms.

Question 6.
Why the study of Growth rings is significant?
Answer:
Importance of Studying Growth Rings:

• Age of wood can be calculated.
• The quality of timber can be ascertained.
• Radio-Carbon dating can be verified.
• Past climate and archaeological dating can be made.
• Provides evidence in forensic investigation.

Higher Order Thinking Skills (HOTs)

Question 1.
What do heartwood and sap wood stand for?
Answer:
Heartwood refers to the dark central part of the wood.
Sapwood refers to the pale peripheral part of the wood.

Question 2.
Arrange the following in the sequence you would find them in a plant starting from ‘ periphery-Phellem, Phellogen, phelloderm.
Answer:
Phellem (cork) → Phellogen (cork cambium) → Phelloderm (secondary cortex).

Question 3.
A transverse section of the trunk of a tree shows concentric rings known as growth rings.
How these rings are formed? Mention its significance.
Answer:
Growth rings denote the combination of earlywood and latewood formed during the spring season and winter season respectively and the rings are more evident due to the high density of latewood. Growth rings (annual rings) in a cut stem can be used to calculate and determine the age of a tree.

Question 4.
If one debarks a tree, what parts of the plant are being removed?
Answer:
The term bark refers to all the tissues outside the vascular cambium i.e., periderm, cortex, primary phloem, and secondary phloem. If a person debarks a tree all these layers of the plant are removed.

Question 5.
Given down is the list of commercial products obtained from plants. Mention the parts of the plant which are used to obtain them.
(a) Rubber (b) Cinnamon (c) Turpentine
Answer:

Product	Plant Source
Rubber	Latex vessels of the inner bark of Hevea brasiliensis.
Cinnamon	The bark of Cinnamomum zeylanicum.
Turpentine	The resin obtained from the bark of Pinus.

Choose the correct answer.

1. Longitudinal growth refers to ……….. .
(a) Primary growth
(b) Secondary growth
(c) Tertiary growth
(d) Abnormal growth
Answer:
(a) Primary growth

2. Secondary growth is absent in ………… .
(a) Gymnosperms
(b) Angiosperms
(c) Dicots
(d) Monocots
Answer:
(d) Monocots

3. Interfascicular cambium originates from ………… .
(a) Procambium
(b) Vascular cambium
(c) Pith
(d) Medullary rays
Answer:
(d) Medullary rays

4. ……….. is also called as wood.
(a) Primary xylem
(b) Primary Phloem
(c) Secondary xylem
(d) Pith
Answer:
(c) Secondary xylem

5. Soft wood lacks ………… .
(a) Tracheids
(b) Pith
(c) Vessels
(d) Medullary rays
Answer:
(c) Vessels

6. ……….. produces the primary plant body.
(a) Procambium
(b) Apical meristem
(c) Lateral meristem
(d) Vascular cambium
Answer:
(b) Apical meristem

7. Dendrochronology deals with determining the ………… .
(a) Growth of trees
(b) Age of trees
(c) Branching pattern of trees
(d) Growth and climatic influence on trees
Answer:
(b) Age of trees

8. Diffuse porous wood is seen in …………. .
(a) Quercus
(b) Maple
(c) Pinus
(d) Acer
Answer:
(d) Acer

9. The balloon-like swellings blocking the lumen of xylem vessels are called ………….. .
(a) Xyloses
(b) Callose
(c) Tyloses
(d) Tylosoids
Answer:
(c) Tyloses

10. Sapwood can also be termed as ……….. .
(a) Duramen
(b) Porous wood
(c) Phellogen
(d) Alburnum
Answer:
(d) Alburnum

11. Hematoxylin especialls stains ………… .
(a) Mitochondria
(b) Nucleus
(c) Golgi bodies
(d) Ribosomes
Answer:
(b) Nucleus

12. Sun hemp is obtained from ………… .
(a) Crotolaria iuncea
(b) Corchorus capsularis
(c) Linum ustitaissimum
(d) Cannabis sativa
Answer:
(a) Crotolaria iuncea

13. The non-living cork cells are with ………… walls.
(a) Lignified
(b) Suberized
(c) Cutinised
(d) Cellulose rich
Answer:
(b) Suberized

14. ………….. gives rise to the secondary cortex.
(a) Phelloderm
(b) Endoderm
(c) Phellogen
(d) Periderm
Answer:
(a) Phelloderm

15. Scale barks can be observed in …………. .
(a) Guava
(b) Quercus
(c) Coconut
(d) Neem
Answer:
(a) Guava

16. Which of the following is termed as Cork?
(a) Phellem
(b) Periderm
(c) Phellogen
(d) Phelloderm
Answer:
(a) Phellem

17. Polyderm is found in ………..
(a) Stem
(b) Roots
(c) Underground stem
(d) Both (a) and (c)
Answer:
(d) Both (a) and (c)

18. …………… are formed only during secondary growth.
(a) Stomata
(b) Endodermis
(c) Pith
(d) Lenticel
Answer:
(d) Lenticel

19. Turpentine is obtained from ………. species.
(a) Cinchona
(b) Acacia
(c) Pinus
(d) Cinnamomum
Answer:
(c) Pinus

20. Rubber is a ………..
(a) Latex
(b) Resin
(c) Alkaloid
(d) Drug
Answer:
(a) Latex

Students get through the TN Board 11th Bio Botany Important Questions Chapter 9 Tissue and Tissue System which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 9 Tissue and Tissue System

Very short answer questions

Question 1.
How primary meristem differs from secondary meristem?
Answer:

Primary Meristem	Secondary Meristem
Derived during the embryonic stage.	Derived during the later stage of plant development.
Gives rise to primary permanent tissues.	Gives rise to cark cambium & inter fascicular cambium.

Question 2.
What are simple tissues? Mention its types.
Answer:
Simple tissues are composed of one type of cells only. The cells are structurally and functionally similar. It is of three types.

1. Parenchyma,
2. Collenchyma and
3. Sclerenchyma

Question 3.
Define Idioblasts.
Answer:
Parenchyma cells that store resin, tannins, crystals of calcium carbonate, calcium oxalate are called idioblasts.

Question 4.
Mention a few places in plants where collenchyma cells can be observed?
Answer:
Hypodermis of dicot stem, petiole, pedicle.

Question 5.
Classify collenchyma based on cell wall pectinisation.
Answer:
Angular collenchyma, lacunar collenchyma, lamellar collenchyma, Annular collenchyma.

Question 6.
What are Bone cells? Give an example.
Answer:
Rod-shaped sclereids with dilated ends are called bone cells or Osteosclereids.
Example: Seed coat of Pisum.

Question 7.
Fibres are supporting tissues. Justify.
Answer:
Fibres provide mechanical strength and support and protect the plants from strong winds. Hence fibres are called supporting tissues.

Question 8.
Define Wood fibres.
Answer:
Wood Fibres: These fibres are associated with the secondary xylem tissue. They are also called xylary fibres. These fibres are derived from the vascular cambium.

Question 9.
What are Bast fibres?
Answer:
Bast fibres or Extra Xylary Fibres: These fibres are present in the phloem. Natural Bast fibres are strong and cellulosic. Fibres obtaining from the phloem or outer bark of jute, kenaf, flax and hemp plants. The so-called pericyclic fibres are actually phloem fibres.

Question 10.
Write a note on surface fibres.
Answer:
Surface fibres are produced from the surface of the plant organiser Cotton and silk cotton are examples. They occur in the test of seeds.

Question 11.
Define Complex tissue & mention its types.
Answer:
A complex tissue is a tissue with several types of cells but all of them function together as a single unit. It is of two types – xylem and phloem.

Question 12.
What do you understand by Centrarch Xylem?
Answer:
Protoxylem is located in the centre surrounded by the metaxylem is called Centrarch. In this type, only one vascular strand is developed. Example: Selaginella sp.

Question 13.
State the Xylem elements.
Answer:
Xylem consists of Four Types of Cells:

1. Tracheids
2. Vessels or Trachea
3. Xylem Parenchyma and
4. Xylem Fibres.

Question 14.
What is the relation between Haberlandt & complex tissue?
Answer:
Haberlandt in 1914 used the term Leptome for phloem and hadrome for xylem.

Question 15.
When do you call the perforation plate in vessels as simple?
Answer:
In vessels, due to the dissolution of the entire cell wall, a single pore is formed at the perforation plate. Such a perforation is called a simple perforation plate. Example: Mangifera.

Question 16.
Draw the annular and reticulate types of secondary wall thickening in tracheids.
Answer:

Question 17.
Where fibre-tracheids can be seen?
Answer:
Between fibres and normal tracheids, there are many transitional forms that are neither typical fibres nor typical tracheids. The transitional types are designated as fibre-tracheids.

Question 18.
What are axial parenchyma & Ray parenchyma?
Answer:
Parenchyma arranged longitudinally along the long axis is called axial parenchyma. Ray parenchyma is arranged in radial rows.

Question 19.
Mention the components of Phloem.
Answer:
Phloem consists of Four Types of Cells:

1. Sieve elements,
2. Companion cells
3. Phloem parenchyma and
4. Phloem fibres

Question 20.
Companion cells are absent in which groups of plants?
Answer:
Gymnosperms and angiospenns.

Question 21.
Write a brief note on Bast fibres.
Answer:
The fibres of sclerenchyma associated with phloem are called phloem fibres or bast fibres. They are narrow, vertically elongated cells with very thick walls and a small lumen. Among the four phloem elements, phloem fibres are the only dead tissue. These are the strengthening as well as supporting cells.

Question 22.
Define the term Syncyte and give examples.
Answer:
Syncyte: Cell which is formed by fusion of cell is called Syncyte.
Example: Vessels (Dead syncyte), sieve tube (living syncyte).

Question 23.
Mention the cell wall chemicals of parenchyma.
Answer:
Cellulose & Pectin.

Question 24.
Name the three types of tissue system as proponed by Sachs.
Answer:
1. Epidermal tissue system (derived from protoderm), 2. Ground tissue system (derived from ground meristem), 3. Vascular tissue system (derived from procambium).

Question 25.
Mention any two plants having multi seriate epidermis.
Answer:
Ficus & Nerium.

Question 26.
What are passage cells? State its function.
Answer:
The endodermal cells, which are opposite to the protoxylem elements, are thin-walled without Casparian strips. These cells are called passage cells. Their function is to transport water and dissolved salts from the cortex to the protoxylem.

Question 27.
Write a brief note on pericycle.
Answer:
The pericycle is a single or few-layered parenchymatous found inner to the endodermis. It is the outermost layer of the stele. Rarely thick-walled sclerenchymatous. In angiosperms, the pericycle gives rise to lateral roots.

Question 28.
List out the storage products seen in the medulla.
Answer:
Tannins, Phenols, calcium oxalate crystals, fatty substances, starch.

Question 29.
Draw and label the open vascular bundle.
Answer:

Question 30.
The vascular bundle of the monocot stem is said to be closed. Why?
Answer:
The vascular bundle of the monocot stem is said to be closed since there is no cambium present between the xylem & phloem.

Question 31.
Define Stele.
Answer:
Stele: All the tissues present inside the endodermis comprise the stele. It includes the pericycle and vascular system.

Question 32.
Where starch sheath is seen? Why it is called so?
Answer:
In the stem, the innermost layer of the cortex is called the endodermis. Since starch grains are abundant in these cells, it is also called a starch sheath. The starch sheath is homologous to the endodermis of the root.

Question 33.
Define Eustele.
Answer:
In the dicot stem, vascular bundles are arranged in a ring around the pith. This type of stele is called eustele.

Question 34.
What is a Bundle cap?
Answer:
In the stem of the sunflower (Helianthus), a few layers of sclerenchyma cell occur in patches outside the phloem in each vascular bundle. This patch of sclerenchyma cell is called Bundle cap or Hardbast.

Question 35.
Mention the nature of the vascular bundle of the dicot stem.
Answer:
In the dicot stem, the vascular bundle is conjoint, collateral, open & endarch.

Question 36.
What are medullary rays?
Answer:
The pith extends between the vascular bundles. These extensions of the pith between the vascular bundles are called primary pith rays or primary medullary rays.

Question 37.
What are protoxylem lacunae?
Answer:
In the mature vascular bundle of the monocot stem, the lowest protoxylem disintegrates and forms a cavity called a protoxylem lacuna.

Question 38.
Define mesophyll tissue & mention its types.
Answer:
The ground tissue system that lies between the epidermal layers of leaf is known as mesophyll tissue. Often it is differentiated into palisade parenchyma on the adaxial (upper) side and spongy parenchyma on the abaxial (lower) side.

Question 39.
Mention the functions of palisade & spongy parenchyma.
Answer:
Palisade parenchyma performs photosynthesis spongy parenchyma facilitates gaseous exchange.

Question 40.
Where the respiratory cavity is located?
Answer:
The air space that is found next to the stomata is called the respiratory cavity or substomatal cavity.

Question 41.
What is border parenchyma?
Answer:
In the dicot leaf, vascular bundles are surrounded by a compact layer of parenchymatous cells called bundle sheath or border parenchyma.

Question 42.
What are silica cells?
Answer:
Some of the epidermal cells of the grass are filled with silica. They are called silica cells.

Question 43.
Define Kranz Sheath.
Answer:
In C4 grasses, the bundle sheath cells are living and involve in C4 photosynthesis. This sheath is called Kranz sheath.

Question 44.
What do you mean by Hydathode?
Answer:
A hydathode is a type of epidermal pore, commonly found in higher plants. Structurally, hydathodes are modified stomata, usually located at leaf tips or margins, especially at the teeth.

Question 45.
Guttation – Explain.
Answer:
Hydathodes discharge liquid water with various dissolved substances from the interior of the leaf to its surface. This process is called guttation. Example many grasses.

Question 46.
Define Halophiles.
Answer:
Plants that grow in the salty environment are called Halophiles.

Short answer questions

Question 1.
Who proposed the Tunica-Corpus theory? Add a note on it.
Answer:
Tunica-Corpus theory is proposed by A. Schmidt (1924). Two zones of tissues are found in the apical meristem.

1. The tunica: It is the peripheral zone of the shoot apex, that forms the epidermis.
2. The corpus: It is the inner zone of the shoot apex, that forms the cortex and stele of the shoot.

Question 2.
Histogen theory of Stem. Explain.
Answer:
Histogen theory is proposed by Hanstein (1868) and supported by Strassburger. The shoot apex comprises three distinct zones.

1. Dermatogen: It is a outermost layer. It gives rise to the epidermis.
2. Periblem: It is a middle layer. It gives rise to the cortex.
3. Plerome: It is the innermost layer. It gives rise to stele.

Question 3.
Explain briefly Korper Kappe theory.
Answer:
Korper Kappe theory is proposed by Schuepp. There are two zones in the root apex:

1. The Korper zone forms the body.
2. Kappe zone forms the cap. This theory is equivalent to the tunica corpus theory of shoot apex. The two divisions are distinguished by the type of T (also called Y divisions). Korper is characterised by inverted T divisions and kappe by straight T divisions.

Question 4.
Write a note on the Quiescent centre concept.
Answer:
The quiescent centre concept was proposed by Clowes (1961) to explain root apical meristem activity. This centre is located between the root cap and differentiating cells of the roots. The apparently inactive region of cells in the root promeristem is called the quiescent centre. It is the site of hormone synthesis and also the ultimate source of all meristematic cells of the meristem.

Question 5.
Enumerate the functions of parenchyma.
Answer:
Parenchyma may store various types of materials like water, air, ergastic substances. The turgid parenchyma cells help in giving rigidity to the plant body. Partial conduction of water is also maintained through parenchymatous cells.

Question 6.
How buoyancy is maintained in aquatic plants?
Answer:
In aquatic plants like Nymphae, the parenchyma cells contain air in their intercellular spaces which provides aeration and buoyancy.

Question 7.
Compare angular collenchyma with lacunar collenchyma.
Answer:
Types of Collenchyma

1. Angular collenchyma: It is the most common type of collenchyma with the irregular arrangement and thickening at the angles where cells meet. Example: Hypodermis of Datura and Nicotiana.
2. Lacunar collenchyma: The collenchyma cells are irregularly arranged. The cell wall is thickening on the walls bordering intercellular spaces. Example: Hypodermis of Ipomoea.

Question 8.
Draw and label (a) Branchysclereids, (b) Osteosclereids.
Answer:

Question 9.
Differentiate between Exarch and Endarch condition.
Answer:

Exarch Xylem	Endarch Xylem
In exarch condition, the protoxylem lies towards the periphery & the metaxylem lies towards the centre.	In endarch condition, the protoxy­lem lies towards the centre & the meta­xylem lie towards the periphery.
It is seen in roots.	It is seen in stems.

Question 10.
Compare primary phloem with secondary phloem.
Answer:

Primary Phloem	Secondary Phloem
Primary phloem is derived from procambium.	Secondary phloem is derived from vascular phloem.

Question 11.
Point out the angiospermic families that do not possess xylem vessels.
Answer:
Winteraceae, Tetracentraceae and Trochodendraceae are the vesselless angiospermic families.

Question 12.
Write a short note on Phloem parenchyma.
Answer:
The parenchyma cells associated with the phloem are called phloem parenchyma. These are living cells. They store starch and fats. They also contain resins and tannins in some plants. The primary phloem consists of axial parenchyma and the secondary phloem consists of both axial and ray parenchyma. They are present in Pteridophytes, Gymnosperms and Dicots.

Question 13.
Classify meristem based on function.
Answer:
Based on function, meristems are classified into 3 types.

1. Peridenn giving rise to epidermis.
2. Procambium giving rise to primary vascular tissues.
3. Ground meristem giving rise to cortex a pitch.

Question 14.
Give an account of the piliferous layer.
Answer:
The outer layer of the root is known as the piliferous layer or epiblema. It is made up of a single layer of parenchyma cells which are arranged compactly without intercellular spaces. It is devoid of epidermal pores and cuticle. Root hair is always single-celled, it absorbs water and mineral salts from the soil. Another important function of the piliferous layer is protection.

Question 15.
What are bulliform cells? How it helps the plants?
Answer:
In leaves, some cells of the upper epidermis (Example: Grasses) are larger and thin-walled. They are called bulliform cells or motor cells. These cells are helpful for the rolling and unrolling of the leaf according to the weather change.

Question 16.
Draw the stoma with a dumb-bell shaped guard cell.
Answer:

Question 17.
Define Trichoblasts.
Answer:
The Piliferous layer of the root has two types of epidermal cells, long cells and short cells.’ The short cells are called trichoblasts. Trichoblasts are elongate into root hairs.

Question 18.
What are Prickles? Mention its benefit to plants.
Answer:

(a) Prickles: Prickles, are one type of epidermal emergences with no vascular supply. They are stiff and sharp in appearance. (Example: Rose), (b) Prickles also provide protection against animals and they also check excessive transpiration.

Question 19.
Distinguish between Extrasteiar & Intrastelar ground tissue.
Answer:

Extrastellar Ground Tissue	Intrastelar Ground Tissue
Ground tissues present outside to the stele is called Extrasteiar ground tissue. Example: Cortex	Ground tissue present inside the stele is called Intrastelar ground tissue. Example: Pericycle, Medullary rays. Pith

Question 20.
Define Pith.
Answer:
The central part of the ground tissue is known as the pith or medulla. Generally, this is made up of thin-walled parenchyma cells with intercellular spaces. The cells in the pith generally store starch, fatty substances, tannins, phenols, calcium oxalate crystals, etc.

Question 21.
Draw & label bicollateral vascular bundle.
Answer:

Question 22.
What are Casparian strips?
Answer:
The radial and the inner tangential walls of endodermal cells are thickened with suberin and lignin. This thickening was first noted by Robert Caspary in 1965. So these thickenings are called Casparian strips.

Question 23.
Differentiate between Protoxylem & Metaxylem.
Answer:

Proto xylem	Meta xylem
First formed primary xylem.	Later formed primary xylem.
Found in developing organs.	Found in developed primary organs.
Elements relatively smaller in size.	Elements relatively larger in size.

Question 24.
Draw & label the ground plan of T.S. of Dicot root.
Answer:

Question 25.
What is the role of Casparian strips in roots?
Answer:
The main function of Casparian strips in the endodermal cells is to prevent the re-entry of water into the cortex once water entered the xylem tissue.

Question 26.
State the importance of cambium.
Answer:
Cambium consists of brick-shaped and thin-walled meristematic cells. It is one to four layers in thickness. These cells are capable of forming new cells during secondary growth.

Question 27.
What are dorsiventral leaves? Give an example.
Answer:
In dorsiventral leaves, the mesophyll is differentiated into palisade and spongy parenchyma, the former occurring on the upper side and the latter on the lower side. Example: Sunflower.

Question 28.
What are isobilateral leaves? Give an example.
Answer:
In the isobilateral leaf, palisade is present on both sides of the leaf and in between them, spongy parenchyma is present. Example: Nerium.

Question 29.
Compare the characters of Palisade parenchyma & spongy parenchyma.
Answer:

Palisade Parenchyma	Spongy Parenchyma
Located beneath the upper epidermis.	Located beneath palisade parenchyma.
Made up of vertically elongated cylindrical cells.	Made up of irregularly shaped cells.
Performs photosynthesis.	Performs gaseous exchange.

Long answer questions

Question 1.
Enumerate the characters of meristematic tissue.
Answer:
The characters of meristematic tissues:

• The meristematic cells are isodiametric and may be, oval, spherical or polygonal in shape.
• They have generally dense cytoplasm with a prominent nucleus.
• Generally, the vacuoles in them are either small or absent.
• Their cell wall is thin, elastic and essentially made up of cellulose.
• These are the most actively dividing cells.
• Meristematic cells are self-perpetuating.

Question 2.
Classify meristem based on position with a simplified diagram.
Answer:

1. Apical meristem: Present in apices of root and shoot. It is responsible for the increase in the length of the plant, it is called primary growth.
2. Intercalary meristem: Occurs between the mature tissues. It is responsible for the elongation of internodes.
3. Lateral meristem: Occurs along the longitudinal axis of stem and root. It is responsible for secondary tissues and the thickening of the stem and root. Example: vascular cambium and cork cambium.
   Different types of meristems

Question 3.
Elaborate on the different types of economically important fibres.
Answer:
Economically fibres may be grouped as follows:

1. Textile Fibres: Fibres utilized for the manufacture of fabrics, netting and cordage etc.
   a. Surface Fibres: Example: Cotton.
   b. Soft Fibres: Example: Flax, Jute and Ramie
   c. Hard fibres: Example: Sisal, Coconut, Pineapple, Abaca etc.
2. Brush fibre: Fibres utilized for the manufacture of brushes and brooms.
3. Rough weaving fibres: Fibres utilized in making baskets, chairs, mats etc.
4. Papermaking fibres: Wood fibres utilized for paper making.
5. Filling fibres: Fibres used for stuffing cushions, mattresses, pillows, furniture etc. Example: Bombax and Silk cotton.

Question 4.
List out the differences between Meristematic tissue & permanent tissue.
Answer:

Meristematic tissue	Permanent tissue
Cells divide repeatedly	Do not divide
Cells are undifferentiated	Cells are fully differentiated
Cells are small and Isodiametric	Cells are variable in shape and size
Intercellular spaces are absent	Intercellular spaces are present
Vacuoles are absent	Vacuoles are present
Cell walls are thin	Cell walls maybe thick or thin
Inorganic inclusions are absent	Inorganic inclusions are present

Question 5.
Sieve cells & Sieve tubes both are phloem elements. Yet they differ. How?
Answer:

Sieve cells	Sieve tubes
Have no companion cells	Have companion cells
The sieve areas do not form sieve plates	The sieve areas are confined to sieve plates
The sieve areas are not well differentiated	The sieve areas are well differentiated
They are elongated cells and are quite long with tapering end walls	They consist of vertical cells placed one above the other forming long tubes connected at the walls by sieve pores
The sieve are smaller and numerous	The sieve pores are longer and fewer
Found in Pteridophytes and Gymnosperms	Found in Angiosperms

Question 6.
How tracheids differ from fibres?
Answer:

Tracheids	Fibres
Not much elongated	Very long cells
Possess oblique end walls	Possess tapering end walls
Cell walls are not as thick as Fibres	Cell wall are thick and lignified
Possess various types of thickenings	Possess only pitted thickenings
Responsible for the conduction and also mechanical support	Provide only mechanical support

Question 7.
Point out the functions of the Epidermal tissue system.
Answer:
Functions of Epidermal Tissue System:

1. This system in the shoot checks excessive loss of water due to the presence of a cuticle.
2. The epidermis protects the underlying tissues.
3. Stomata is involved in transpiration and gaseous exchange.
4. Trichomes are also helpful in the dispersal of seeds and fruits and provide protection against animals.
5. Prickles also provide protection against animals and they also check excessive transpiration.
6. In some rose plants, they also help in climbing.
7. Glandular hairs repel herbivorous animals.

Question 8.
Draw and label the transverse section of the monocot stem.
Answer:

Question 9.
Draw a labelled diagram of T.S. of Dicot leaf.
Answer:

Question 10.
Differentiate between stoma & hydathode.
Answer:

Stomata	Hydathodes
Occur in the epidermis of leaves, young stems.	Occur at the tip or margin of leaves that are grown in a moist shady place.
The stomatal aperture is guarded by two guard cells.	The aperture of hydathodes is surrounded by a ring of cuticularized cells.
The two guard cells are generally surrounded by subsidiary cell.	Subsidiary cells are absent.
The opening and closing of the stomatal aperture are regulated by guard cells.	Hydathode pores remain always open.
These are involved in the transpiration and the exchange of gases.	These are involved in guttation.

Higher Order Thinking Skills (HOTs)

Question 1.
Protoxylem is the first formed Xylem. If the protoxylem is surrounded by phloem what kind of arrangement of xylem would you call it? Give an example.
Answer:
If the protoxylem is surrounded by phloem the vascular bundle is said to concentric and amphicribral. Example: Ferns.

Question 2.
Observe the diagram given below and answer the questions.
(a) Mention the parts A, B, C
(b) Name the part of the plant where we can see this structure more in number.

Answer:
(a) A – Stoma, B – Guard cell and C – Subsidiary cell.
(b) Stomata are seen more in number on the lower epidermis of leaves.

Question 3.
In a Biology practical class, your subject teacher has placed a glass slide showing the transverse section of the dicot stem. State any two possible reasons to call the slide is a T.S of dicot stem.
Answer:
In the dicot stem, the stele is Eustele (he. vascular bundles are arranged in the form of a ring). Presence of endarch xylem and central pith.

Question 4.
While eating pear fruit it is usually seen that some stone-like structures get entangled in the teeth, what are these stone-like structures called?
Answer:
The stone-like structures that we feel while eating pear fruit are the branchy sclereids or stone cells which are isodiametric sclereids with the hand cell wall.

Question 5.
A certain tissue in a green plant somehow get blocked and the leaves wilted. What was the tissue that got blocked?
Answer:
The tissue that got blocked may be the xylem. It is through the xylem the water and other minerals are transported from root to leaves and other parts. So if the xylem is blocked, the water and mineral supply is stopped, leading to wilting of leaves that eventually may lead to the death of the plant.

Question 6.
The cross-section of a plant material shown the following features on viewing under the microscope.
(a) Radically arranged vascular bundles.
(b) Four xylem arms with protoxylem facing outer side. To which organ should it be
arranged.
Answer:
Dicot roots show radically arranged vascular hurdles with tetrarch and exarch xylem.

Question 7.
Write the precise function of:
(a) Aerenchyma (b) Collenchyma (c) Sieve tube
Answer:
(a) Aerenchyma – Provides buoyancy in aquatic plants.
(b) Collenchyma – Mechanical strength & support.
(c) Sieve tube – Food conduction

Choose the correct answer.

1. The term meristem is coined by
(a) Schmidt
(b) Clowes
(c) Nageli
(d) Schleiden
Answer:
(c) Nageli

2. The study of tissues is called as
(a) Anatomy
(b) Cytology
(c) Histology
(d) Embryology
Answer:
(c) Histology

3. Primary growth is the responsibility of meristem.
(a) Apical
(b) Primary
(c) Intercalary
(d) Lateral
Answer:
(a) Apical

4. Identify the mismatch pair:
(i) Nageli – Xylem
(ii) Leptome – Phloem
(iii) Tunica Corpus theory – Hanstein
(iv) Tracheids – Sanio
(a) (i) – only
(b) (iii) – only
(c) Both (i) and (iii)
(d) None
Answer:
(b) (iii) – only

5. Who proposed Korper kappe theory?
(a) Schmidt
(b) Hanstein
(c) Nageli
(d) Schuepp
Answer:
(d) Schuepp

6. Parenchyma storing calcium carbonate crystals are called …..
(a) Leucoplasts
(b) Elaioplasts
(c) Idioblasts
(d) Chromoplasts
Answer:
(c) Idioblasts

7. ………. is seen in all organs of plant.
(a) Chlorenchyma
(b) Sclerenchyma
(c) Parenchyma
(d) Collenchyma
Answer:
(c) Parenchyma

8. ………… is absent in roots.
(a) Chlorenchyma
(b) Sclerenchyma
(c) Parenchyma
(d) Collenchyma
Answer:
(d) Collenchyma

9. ………….. sclereids are seen in the seed coat of pisum.
(a) Macro
(b) Osteo
(c) Tricho
(d) Branchy
Answer:
(a) Macro

10. Fibres forming mesocarp of coconut is called as
(a) Surface fibres
(b) Soft fibres
(c) Mesocarp fibres
(d) Septate fibres
Answer: (c) Mesocarp fibres

11. ………… is called as Leptome.
(a) Parenchyma
(b) Phloem
(c) Xylem
(d) Fibres
Answer:
(6) Phloem

12. Multiple perforation plate is seen in …………. .
(a) Lepidodendron
(b) Limnophyton
(c) Mangifera
(d) Liriodendron
Answer: (d) Liriodendron

13. ………… species of gymnosperms have vessels.
(a) Cycas
(b) Selaginella
(c) Gnetum
(d) Conifer
Answer:
(c) Gnetum

14. ………. Only living cells among Xylem elements are
(a) Vessels
(b) Tracheids
(c) Xylem Parenchyma
(d) Xylem fibres
Answer: (c) Xylem Parenchyma

15. In sieve elements, mature sieve plates are blocked by ………… .
(a) Suberin
(b) Lignin
(c) Callose
(d) Pectin
Answer: (c) Callose

16. Campanion cells are present only in …………. .
(a) Pteridophytes
(b) Angiosperms
(c) Gymnosperms
(d) Dicots
Answer:
(b) Angiosperms

17. Rolling & unrolling of leaves due to whether change are controlled by ………… .
(a) Sensory cells
(b) Motor cells
(c) Subsidory cells
(d) Trichomes
Answer:
(b) Motor cells

18. Sunken stoma is seen in ………. .
(a) Cycas
(b) Nerium
(c) Both (a) and (b)
(d) None
Answer:
(c) Both (a) and (b)

19. Root hairs originated from ………… .
(a) Trichomes
(b) Epidermis
(c) Pericycle
(d) Trichoblasts
Answer:
(d) Trichoblasts

20. Which is NOT a part of intrastelar ground tissue?
(a) Pericycle
(b) Medullary ray
(c) Pith
(d) Cortex
Answer:
(d) Cortex

21. Usually ………… form the hypodermis.
(a) Chlorenchyma
(b) Sclerenchyma
(c) Collenchyma
(d) Parenchyma
Answer:
(c) Collenchyma

22. In angiosperms, ……….. gives rise to lateral roots.
(a) Pith
(b) Endodermis
(c) Pericycle
(d) Trichoblasts
Answer:
(c) Pericycle

23. In cucurbitaceae, the vascular bundles are
(a) Bicollateral
(b) Collateral closed
(c) Concentric
(d) Radial
Answer:
(a) Bicollateral

24. Major function of epiblema is
(a) Transport
(b) Support
(c) Protection
(d) Conduction of food
Answer:
(c) Protection

25. Innermost layer of cortex is
(a) Epiblema
(b) Endodermis
(c) Pericycle
(d) Medullary rays
Answer:
(b) Endodermis

26. In T.S. of bean root, the xylem is
(a) Polyarch
(b) Hexarch
(c) Tetrarch
(d) Wedge shaped
Answer:
(c) Tetrarch

27. Phellogen arises from ……….., in dicot roots.
(a) Endodermis
(b) Piliferous layer
(c) Pericycle
(d) Periderm
Answer:
(c) Pericycle

28. Eustele is seen in …………. stem.
(a) Maize
(b) Grass
(c) Paddy
(d) Sunflower
Answer:
(d) Sunflower

29. In monocot stem, the xylem vessels are in the form of ………… .
(a) X
(b) Y
(c) W
(d) +
Answer:
(b) Y

30. Hardbast is composed of …………. .
(a) Parenchyma
(b) Collenchyma
(c) Sclerenchyma
(d) Prosenchyma
Answer:
(c) Sclerenchyma

31. Silica cells seen in epidermis contain
(a) Magnesium
(b) Calcium
(c) Silica
(d) Sand
Answer:
(c) Silica

32. Guttation occurs through
(a) Stomata
(b) Lenticles
(c) Cuticle
(d) Hydathodes
Answer:
(d) Hydathodes

33. Kranz sheath in granes perform photosynthesis.
(a) C₃
(b) C₄
(c) C₂
(d) C₁₂
Answer:
(b) C₁₂

34. Halophiles survive in environment.
(a) Dry
(b) Aquatic
(c) Saline
(d) Cold
Answer:
(c) Saline

35. Who coined the term Hadrome?
(a) Hanstein
(b) Nageli
(c) Hofmeister
(d) Haberlandt
Answer:
(d) Haberlandt

36. Match the following:

(a)	Dermatogen	(0	Cortex
Cb)	Periblem	(ii)	Root cap
(c)	Plerome	(iii)	Stele
id)	Calyptrogen	(iv)	Epidermis

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

Students get through the TN Board 11th Bio Botany Important Questions Chapter 8 Biomolecules which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 8 Biomolecules

Answer the following short answers.

Question 1.
What is meant by the cellular pool?
Answer:
The cell components are made of a collection of molecules called a cellular pool, which consists ’ of both inorganic and organic compounds.

Question 2.
What are Micronutrients? Give any two examples.
Answer:
Micronutrients, which are required in trace amounts, eg. Cobalt, zinc, boron, copper, molybdenum and manganese) and are essential for enzyme action.

Question 3.
Mention any four properties of water.
Answer:

1. Adhesion and cohesion property.
2. High latent heat of vaporisation.
3. High melting and boiling point.
4. Universal solvent.

Question 4.
Define primary metabolites.
Answer:
Primary metabolites are those that are required for the basic metabolic processes like photosynthesis, respiration, protein and lipid metabolism of living organisms.

Question 5.
Match the following:

(i) Enzymes	(a) Abrin
(ii) Amino acid	(b) Morphine
(iii) Alkaloids	(c) Peroxidase
(iv) Toxins	(d) Leucine

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

Question 6.
Define monosaccharide.
Answer:
Monosaccharides are relatively small molecules constituting a single sugar unit. Glucose has a chemical formula of CgH^Og. It is a six-carbon molecule and hence is called hexose.

Question 7.
Define glycosidic bond.
Answer:
The bond formed between the glucose and fructose molecule by the removal of water is called a glycosidic bond. This is another example of a strong covalent bond.

Question 8.
What is chitin made up of?
Answer:
Chitin is a homo polysaccharide with amino acids added to form mucopolysaccharide. The ’ basic unit is a nitrogen-containing glucose derivative known as N-acetyl glucosamine. It forms the exoskeleton of insects and other arthropods. It is also present in the cell walls of fungi.

Question 9.
How does Herbivores digest cellulose?
Answer:
Herbivores can digest them With the help of bacteria present in the gut which produces en2yme cellulase. This is an example of mutualism.

Question 10.
What are steroids? Explain with an example.
Answer:
These are complex compounds commonly found in the cell membrane and animal hormones, eg. Cholesterol which reinforces the structure of the cell membrane in animal cells and in an unusual group of cell wall deficient bacteria —Mycoplasma.

Question 11.
Define the term amphoteric.
Answer:
(NH₂), an acidic carboxylic group (COOH) and a hydrogen atom (H) and side-chain or variable R group* The amino acid is both an acid and a base is called amphoteric.

Question 12.
What do you know about the primary structure of a protein?
Answer:
The primary structure is the linear arrangement of amino acids in a polypeptide chain.

Question 13.
What is meant by an ionic bond?
Answer:
It is formed between any charged groups that are not joined together by a peptide bond. It is stronger than a hydrogen bond and can be broken by changes in pH and temperature.

Question 14.
Explain briefly anabolic reactions.
Answer:
Anabolic (building up of organic molecules). Synthesis of proteins from amino acids and synthesis of polysaccharides from simple sugars are examples of anabolic reactions.

Question 15.
What are the lock and key mechanism?
Answer:
The substrate binds to the specially formed pocket in the enzyme – the active site, this is called the lock and key mechanism of enzyme action. As the enzyme and substrate form an ES complex, the substrate is raised in energy to a transition state and then breaks down into products plus an unchanged enzyme.

Question 16.
What are competitive inhibitors? Give an example.
Answer:
Molecules that resemble the shape of the substrate and may compete to occupy the active site of the enzyme are known as competitive inhibitors, eg. the enzyme that catalyses the reaction between carbon dioxide and the CO₂ acceptor molecule in photosynthesis, known as ribulose biphosphate carboxylase oxygenase (RUBISCO) is competitively inhibited by oxygen/carbon-di-oxide in the chloroplast.

Question 17.
Define the terms nucleotide and nucleoside?
Answer:
DNA and RNA are polymers of monomers called nucleotides, each of Which is composed of a nitrogen base, a pentose sugar and a phosphate. A purine or a pyrimidine and ribose or deoxyribose sugar are called nucleoside.

Question 18.
Mention any two sulphur-containing amino acids.
Answer:
Methionine and cysteine.

Question 19.
What is meant by Plectonemic coiling of DNA?
Answer:
Plectonemic coiling – the two strands of the DNA are wrapped around each other in a helix, making it impossible to simply move them apart without breaking the entire structure.

Question 20.
Give a short note on RNA.
Answer:
tRNA (transfer RNA): Translates the code from mRNA and transfers amino acids to the ribosome to build proteins. It is highly folded into an elaborate 3D structure and comprises about 15% of total RNA. It is also called soluble RNA.

Answer In brief.

Question 1.
Explain Polysaccharide with an example.
Answer:

1. These are made of hundreds of monosaccharide units. Polysaccharides also called “Glycans”.
2. Long-chain of branched or unbranched monosaccharides is held together by glycosidic bonds.
3. Polysaccharide is an example of a giant molecule, a macromolecule and consists of only one type of monomer.
4. Polysaccharides are insoluble in water and are sweet. Cellulose is an example built from repeated units of glucose monomer.
5. Depending on the function, polysaccharides are of two types: Storage Polysaccharide and structural Polysaccharide.

Question 2.
Describe the test for reducing sugars.
Answer:
Aldoses and ketoses are reducing sugars. This means that, when heated with an alkaline solution of copper (II) sulphate (a blue solution called benedict’s solution), the aldehyde or ketone group reduces Cu²⁺ ions to Cu⁺ ions forming a brick-red precipitate of copper (I) oxide. In the process, the aldehyde or ketone group is oxidised to a carboxyl group (- COOH). This reaction is used as a test for reducing sugar and is known as Benedict’s test. The results of benedict’s test depend on the concentration of the sugar. If there is no reducing sugar it remains blue.

1. Sucrose is not a reducing sugar.
2. The greater the concentration of reducing sugar, the more is the precipitate formed and greater is the colour change.

Question 3.
Define triglycerides. Explain with examples.
Answer:
Triglycerides are composed of a single molecule of glycerol bound to 3 fatty acids. These include fats and oils. Fatty acids are long-chain hydrocarbons with a carboxyl group at one end which binds to one of the hydroxyl groups of glycerol, thus forming an ester bond. Fatty acids are structural unit of lipids and are carboxylic acid of long-chain hydrocarbons. The hydrocarbon can vary in length from 4-24 carbons and the fat may be saturated or unsaturated. In saturated fatty acids, the hydrocarbon chain is single-bonded.
eg. Palmitic acid, Stearic acid and unsaturated fatty acids.
eg. Oleic acid, linoleie acid) the hydrocarbon chain is double bonded (one/two/three). In general solid fats are saturated and oils are unsaturated, in which most are globules.

Question 4.
Enumerate the properties of enzymes.
Answer:
Properties of Enzyme:

1. All are globular proteins.
2. They act as catalysts and effective even in small quantity.
3. They remain unchanged at the end of the reaction.
4. They are highly specific.
5. They have an active site where the reaction takes place.
6. Enzymes lower the activation energy of the reaction they catalyse.

Question 5.
What are Allosteric enzymes? Explain with a suitable example.
Answer:
They modify enzyme activity by causing a reversible change w the structure of the enzyme active site. This in turn affects the ability of the substrate to bind to the enzyme. Such compounds are called Allosteric inhibitors, eg. The enzyme hexokinase which catalysis glucose to glucose-6 phosphate in glycolysis is inhibited by glucose 6 phosphate. This is an example of a feedback allosteric inhibitor.

Question 6.
Distinguish between nucleoside and nucleotide with two examples.
Answer:

Nucleoside	Nucleotide
It is a combination of base and sugar.	It is a combination of nucleoside and phosphoric acid.
eg. Adenosine = Adenine + Ribose Guanosine = Guanine + Ribose	eg. Adenylic acid = Adenosine + phosphoric acid Guanylic acid = Guanosine + phosphoric acid

Question 7.
Explain ribosomal RNA. Add a note on its function.
Answer:
RNA (ribosomal RNA): Single-stranded, metabolically stable, make up the two subunits of ribosomes. It constitutes 80% of the total RNA. It is a polymer with varied length from 120-3000 nucleotides and gives ribosomes their shape. Genes for rRNA are highly conserved and employed for phylogenetic studies.

Question 8.
Explain the process of negative feedback inhibition with a schematic diagram.
Answer:
Negative Feedback Inhibition: When the end product of a metabolic pathway begins to accumulate, it may act as an allosteric inhibitor of the enzyme controlling the first step of the pathway. Thus the product starts to switch off its own production as it builds up. The process is- self – regulatory. As the product is used up, its production is switched on once again. This is called end-product inhibition.

Question 9.
Explain the Michaelis-Menton Constant (km) with graphical representation.
Answer:
The rate of reaction is directly proportional to the enzyme concentration.

When the initial rate of reaction of an enzyme is measured over a range of substrate concentrations (with a fixed amount of enzyme) and the results plotted on a graph. With increasing substrate concentration, the velocity increases – rapidly at lower substrate concentration.

Question 10.
What is the three types of chemical bond in protein structure? Explain them with an example.
Answer:
A hydrogen bond is formed between some hydrogen atoms of oxygen and nitrogen in a polypeptide chain. The hydrogen atoms have a small positive charge and oxygen and nitrogen have a small negative charge. Opposite charges attract to form hydrogen bonds. Though these bonds are weak, a large number of them maintains the molecule in 3D shape.
Ionic Bond: It is formed between any charged groups that are not joined together by a peptide bond. It is stronger than a hydrogen bond and can be broken by changes in pH and temperature.
Disulfide Bond; Some amino acids like cysteine and methionine have sulphur. These form a disulphide bridge between sulphur atoms and amino acids.
Hydrophobic Bond: This bond helps some protein to maintain structure. When globular proteins are in solution, their hydrophobic groups point inwards away from water.

Answer In detail.

Question 1.
Describe the structure of a protein with a neat diagram.
Answer:

1. Protein is synthesised on the ribosome as a linear sequence of amino acids which are held together by peptide bonds. After synthesis, the protein attains conformational change into a specific 3D form for proper functioning. According to the mode of folding, four levels of protein’ organisation have been recognised namely primary, secondary, tertiary and quaternary.
2. The primary structure is the linear arrangement of amino acids in a polypeptide chain.
3. Secondary structure arises when various functional groups are exposed on the outer surface of the molecular interaction by forming hydrogen bonds. This causes the amino acid chain to twist into a coiled configuration called a-helix or to fold into flat β-pleated sheets.
4. Tertiary protein structure arises when the secondary level proteins fold into a globular structure called domains.
5. Quaternary protein structure may be assumed by some complex proteins in which more than one polypeptide forms a large multi-unit protein. The individual polypeptide chains of the protein are called subunits and the active protein itself is called a multimer.
6. eg. Enzymes serve as a catalyst for chemical reactions in the cell and are non-specific. Antibodies are complex glycoproteins with specific regions of attachment for various organisms.

Question 2.
What are the types of cofactors? Explain each of them.
Answer:

1. Many enzymes require non-protein components called cofactors for their efficient activity. Cofactors may vary from simple inorganic ions to complex organic molecules. They are of three types: inorganic ions, prosthetic groups and coenzymes.
2. Holoenzyme: Active enzyme with its non-protein component.
3. Apoenzyme: The inactive enzyme without its non-protein component.
4. Inorganic ions help to increase the rate of the reaction catalysed by enzymes. Example: Salivary amylase activity is increased in the presence of chloride ions.
5. Prosthetic groups are organic molecules that assist in the catalytic function of an enzyme. Flavin adenine dinucleotide (FAD) contains riboflavin (Vit B2), the function of which is to accept hydrogen. ‘Haem’ is an iron-containing prosthetic group with an iron atom at its centre.
6. Coenzymes are organic compounds that act as cofactors but do not remain attached to the enzyme. The essential chemical components of many coenzymes are vitamins, eg. NAD, NADP, Coenzyme A, ATP.
   

Question 3.
Explain the structure of DNA with Watson and Crick modal.
Answer:

1. Watson and Crick shared the Nobel Prize in 1962 for their discovery, along with Maurice Wilkins, who had produced the crystallographic data supporting the model, Rosalind Franklin (1920-1958) had earlier produced the first clear crystallographic evidence of a helical structure. James Watson and Francis Crick of Cavendish laboratory in Cambridge built a scale model of the double-helical structure of DNA which is the most prevalent form of DNA, the B-DNA. This is the secondary structure of DNA.
2. As proposed by James Watson and Francis Crick, DNA consists of a right-handed double helix with 2 helical polynucleotide chains that are coiled around a common axis to form a right-handed B form of DNA. The coils are held together by hydrogen bonds which occur between complementary pairs of nitrogenous bases. The sugar is called 2-deoxyribose because there is no hydroxyl at position 2′. Adenine and thiamine base pairs have two hydrogen bonds while guanine and cytosine base pairs have three hydrogen bonds.
   Chargaff’s Rule:
   • A = T; G = C
   • A+G = T + C
   • A : T = G : C =1
3. As published by Erwin Chargaff in 1949, a purine pairs with pyrimidine and vice versa. Adenine (A) always pairs with Thymine (T) by a double bond and Guanine (G) always pairs with Cytosine (C) by a triple bond.

Question 4.
Explain any two factors affecting the rate of enzyme reaction, with the help of graphical representation.
Answer:
Enzyme Reactions: Enzymes are sensitive to environmental condition. It could be affected by temperature, pH, substrate concentration and enzyme concentration. The rate of enzyme reaction is measured by the amount of substrate changed or amount of product formed, during a period of time.
Temperature: Heating increases molecular motion. Thus the molecules of the substrate and enzyme move more quickly resulting in a greater probability of occurrence of the reaction. The temperature that promotes maximum activity is referred to as optimum temperature.

pH: The optimum pH is that at which the maximum rate of reaction occurs. Thus the pH change leads to an alteration of enzyme shape, including the active site. If extremes of pH are encountered by an enzyme, then it will be denatured.

Substrate Concentration: For a given enzyme concentration, the rate of an enzyme reaction increases with increasing substrate concentration.
Enzyme Concentration: The rate of reaction is directly proportional to the enzyme concentration:

Question 5.
Describe the structure and functions of various other polysaccharides.
Answer:

Other Polysaccharides	Structure	Functions
Inulin	The polymer of fructose.	It is not metabolised in the human body and is readily filtered through the kidney.
Hyaluronic acid	Heteropolymer of d glucuronic acid and D-N acetyl glucosamine.	It accounts for the toughness and flexibility of cartilage and tendon.
Agar	Mucopolysaccharide from red algae.	Used as a solidifying agent in culture medium in the laboratory.
Heparin	Glucosamine glycan contains variably sulphated disaccharide unit present in the liver.	Used as an anticoagulant.
Chondroitin sulphate	Sulphated glycosaminoglycan composed of altering sugars (N-acetylglucosamine and glucuronic acid).	Dietary supplement for treatment of osteoarthritis.
Keratan sulphate	Sulphated glycosaminoglycan and is a structural carbohydrate	Acts as a cushion to absorb mechanical shock.

Choose the correct answer.

1. Which is the most abundant component in living organisms.
(a) Minerals
(b) Macromolecules
(c) Water
(d) Protein
Answer:
(c) Water

2. In a water molecule, the hydrogen and oxygen atom stick together by:
(a) Monovalent bond
(b) Covalent bond
(c) Hydrogen bond
(d) None of the above
Answer:
(b) Covalent bond

3. Morphine is the first alkaloid to be found from a plant called:
(a) Vinca rosea
(b) Sweet pea
(c) Delonix regia
(d) Papaver somniferum
Answer:
(d) Papaver somniferum

4. Indicate a macromolecule:
(a) Amino acid
(b) Protein
(c) Nucleotide
(d) Glucose
Answer:
(b) Protein

5. The number of sugar units present in oligosaccharides:
(a) 14 to 15
(b) 6 to 8
(c) 2 to 10
(d) 11 to 12
Answer:
(c) 2 to 10

6. Sucrose is a
(a) Polysaccharide
(b) Disaccharide
(c) Monosaccharide
(d) Triglyceride
Answer:
(b) Disaccharide

7. A test for the presence of starch by adding a solution of iodine gives:
(a) Greenish blue colour
(b) Reddish green colour
(c) blue-black colour
(d) Violet-pink colour
Answer:
(c) blue-black colour

8. Glycogen is not seen in the organs of the human body:
(a) Muscle fibre
(b) Liver
(c) Brain
(d) Kidney
Answer:
(c) Brain

9. Chitin is composed of
(a) Mucopolysaccharides
(b) Oligopolysaceharides
(c) Glycoprotein
(d) Dipolysaccharides
Answer:
(a) Mucopolysaccharides

10. Match the following.

(i) Inulin	(a) heteropolymer of D glucose
(ii) Hyaluronic acid	(b) Mucopolysaccharides
(Hi) Heparin	(c) Polymer of fructose
(iv) Agar	(d) Glycosamine glycon

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

11. Lipids do not include:
(a) Steroid
(b) Waxes
(c) Enzymes
(d) phospholipids
Answer:
(c) Enzymes

12. A molecule of glycerol bound to have:
(a) 5 fatty acids
(b) 6 fatty acids
(c) 4 fatty acids
(d) 3 fatty acids
Answer:
(d) 3 fatty acids

13. Indicate saturated fatty acids:
(a) Palmitic acid
(b) Oleic acid
(c) Linoleic acid
(d) None of the above
Answer:
(a) Palmitic acid

14. Phospholipids serve as a major structural component of:
(a) Feathers
(b) Cell membrane
(c) Leaves
(d) Skin
Answer:
(b) Cell membrane

15. Cholesterol is an example of:
(a) Membrane lipids
(b) Triglycerides
(c) Steroids
(d) Adipose tissue
Answer:
(c) Steroids

16. The term ‘protein’ was coined by:
(a) Watson
(b) Gerardus Johannes Mulder
(c) Erwin Chargaff
(d) Maurice Wilkins
Answer:
(b) Gerardus Johannes Mulder

17. First protein insulin was sequenced by;
(a) Fred Sanger
(b) Robert Brown
(c) Robert Hooke
(d) Christian Anfinsen
Answer:
(a) Fred Sanger

18. Protein is synthesized in:
(a) Mitochondria
(b) Golgi body
(c) Lysosome
(d) Ribosome
Answer:
(d) Ribosome

19. A linear arrangement of amino acids is a polypeptide chain is seen in:
(a) Secondary structure of the protein
(b) Primary structure of the protein
(c) Tertiary structure of the protein
(d) Quaternary protein structure
Answer:
(b) Primary structure of the protein

20. Protein denaturation is due to:
(a) Exposure to pressure
(b) Exposure to light
(c) Exposure to heat
(d) None of the above
Answer:
(c) Exposure to heat

21. In a polypeptide chain hydrogen bone is formed between some hydrogen atoms of:
(a) Oxygen and methane
(b) Ethylene and nitrogen
(c) Nitrogen and methane
(d) Oxygen and nitrogen
Answer:
(d) Oxygen and nitrogen

22. Disulfide bond is seen between some amino acids like:
(a) Glycine and alanine
(b) Serine and proline
(c) Cysteine and methionine
(d) Aspartate and glutamate
Answer:
(c) Cysteine and methionine

23. Synthesis of polysaccharides from simple sugars is termed as:
(a) Catabolic reaction
(b) Anabolic reaction
(c) Hydrolytic reaction
(d) Oxidative reaction
Answer:
(b) Anabolic reaction

24. Indicate the correct statement:
(a) The rate of reaction is indirectly proportional to the enzyme concentration.
(b) The rate of reaction is directly proportional to the enzyme concentration.
(c) The rate of reaction is indirectly proportional to an increase in temperature
(d) None of the above.
Answer:
(c) The rate of reaction is indirectly proportional to an increase in temperature

25. The increased concentration of malonate inhibits the reaction of the enzyme, succinic dehydrogenase. This type of inhibitors is termed as:
(a) Competitive inhibitors
(b) Non-competitive inhibitors
(c) Irreversible inhibitors
(d) None of the above
Answer:
(a) Competitive inhibitors

26. NADP serves as:
(a) Apoenzyme
(b) Holoenzyme
(c) Coenzyme
(d) None of the above
Answer:
(c) Coenzyme

27. Formation of new chemical bonds using ATP as a source of energy is the mode of action of the enzymes.
(a) Hydrolase
(b) Isomerase
(c) Lyase
(d) Ligase
Answer:
(d) Ligase

28. DNA and RNA are polymers of monomers called:
(a) Nucleoside
(b) Nucleotide
(c) Pyrimidine
(d) Dinucleotide
Answer:
(b) Nucleotide

29. Which of the RNA constitutes 80% of the total RNA:
(a) mRNA
(b) tRNA
(c) rRNA
(d) None of the above
Answer:
(c) rRNA

30. Who got a noble prize for the finding of the helical structure of DNA?
(a) Rosalind Franklin and Erwin Chargaff
(b) Maurice Wilkins and Rosalind Franklin
(c) James Watson and Francis Crick
(d) Robert Hooke and Robert Brown
Answer:
(c) James Watson and Francis Crick

Students get through the TN Board 11th Bio Botany Important Questions Chapter 7 Cell Cycle which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 7 Cell Cycle

Answer the following short answers.

Question 1.
Who coined the word “Cell” and “protoplasm”?
Answer:
Robert Hooke, Jan Evangelista purkyne (J.E.Purkinje)

Question 2.
Mentions any two roles of the nucleus.
Answer:

1. Control activities of the cell.
2. Genetic information copied from cell to cell while the cell divides.

Question 3.
Define the term haploid.
Answer:
In meiosis, the daughter cells contain half the number of chromosomes of the parent cell and are known as a haploid state (n).

Question 4.
Define the term Cytokinesis.
Answer:
Whichever division takes place, it is normally followed by division of the cytoplasm to form separate cells, called cytokinesis.

Question 5.
What is C – Value?
Answer:
C-Value is the amount in picograms of DNA contained within a haploid nucleus.

Question 6.
What do you know about clone dolly?
Answer:
Since the DNA of cells in G₀, does not replicate. The researcher is able to fuse the donor cells from a sheep’s mammary glands into G₀ state by culturing in the nutrient-free state. The G₀ donor nucleus synchronized with the cytoplasm of the recipient egg, which developed into the clone, Dolly.

Question 7.
Define maturation promoting factor (MPF).
Answer:
One of the proteins synthesized only in the G₂ period is known as Maturation Promoting’Factor (MPF). It brings about the condensation of interphase chromosomes into the mitotic form.

Question 8. Define Amitosis.
Answer:
Amitosis is also called direct or incipient cell division. Here there is no spindle formation and chromatin material does not condense.

Question 9.
Mention two drawbacks of amitosis.
Answer:

1. Causes unequal distribution of chromosomes.
2. Can lead to abnormalities in metabolism and reproduction.

Question 10.
Explained closed mitosis.
Answer:
In closed mitosis, the nuclear envelope remains intact and chromosomes migrate to opposite poles of a spindle within the nucleus.

Question 11.
How does an aster form?
Answer:
In an animal cell, the centrioles extend a radial array of microtubules towards the plasma membrane when they reach the poles of the cell. This arrangement of microtubules is called an aster. Plant cells do not form asters.

Question 12.
What do you mean by anaphase-promoting complex cyclosome (APC/C)
Answer:
A ubiquitine ligase is activated called the anaphase-promoting complex cyclosome (APC/C) leads to degradation of the key regulatory proteins at the transition of metaphase to anaphase. APC is a cluster of proteins that induces the breaking down of cohesion proteins which leads to the separation of chromatids during mitosis.

Question 13.
List out various stages of prophase I of Meiotic cell division
Answer:
Leptotene, zygotene, pachytene, diplotene and diakinesis.

Question 14.
Define Chiasmata.
Answer:
The homologous chromosomes remain attached at one or more points where crossing over has taken place. These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called Chiasmata.

Question 15.
What is meant by metaphase plate?
Answer:
Bivalent (pairs of homologous chromosomes) aligned at the equator of the cell known as the metaphase plate.

Question 16.
Define interkinesis.
Answer:
The stage between the two meiotic divisions is called interkinesis which is short-lived.

Question 17.
Mention any two of the significance of meiosis.
Answer:
This maintains a definite constant number of chromosomes in organisms. Crossing over takes place and the exchange of genetic material leads to variations among species. These variations are the raw materials for evolution. Meiosis leads to genetic variability by partitioning different combinations of genes into gametes through independent assortment.

Question 18.
Explain the term ‘Mitogen’.
Answer:
The factors which promote cell cycle proliferation are called mitogen. Plant mitogens include gibberellin, ethylene, Indole acetic acid, kinetin. These increase the mitotic rate.

Question 19.
Define Anastral condition in cell division.
Answer:
This is present only in plant cells! No asters or centrioles are formed only spindle fibers are formed during cell division.

Question 20.
Explain Amphiastral condition.
Answer:
Aster and centrioles are formed at each pole of the spindle during cell division. This is found in animal cells.

Answer In brief.

Question 1.
Enumerate any three important features of the chromosome.
Answer:

1. The shape of the chromosome is specific: The long, thin, lengthy structured chromosome contains a short, constricted region called a centromere. A centromere may occur anywhere along the chromosome, but it is always in the same position on any given chromosome.
2. The number of chromosomes per species is fixed: for example, the mouse has 40 chromosomes, the onion has 16 and humans have 46.
3. Chromosomes occur in pairs: The chromosomes of a cell occur in impairs, called homologous pairs. One of each pair come originally from each parent, eg. human has 46 chromosomes,
4. coming originally from each parent in the process of sexual reproduction.

Question 2.
Tabulate the time duration in hours of the different phases of the cell cycle.
Answer:

Phase	Time duration (in hrs)
G1	11
S	8
G2	4
M	1

Question 3.
Explain G₁ phase of cell cycle.
Answer:
The first gap phase-2 C amount of DNA in cells of G₁. The cells become metabolically active and grow by producing proteins, lipids, carbohydrates, and cell organelles including mitochondria and endoplasmic reticulum. Many checkpoints control the cell cycle. The checkpoint called the restriction point at the end of G₁ determines a cell’s fate whether it will continue in the cell cycle and divide or enter a stage called G₀ as a quiescent stage and probably as specified cell or die. Cells are arrested in G₁ due to:

1. Nutrient deprivation.
2. Lack of growth factors or density-dependent inhibition.
3. Undergo metabolic changes and enter into G₀ state.

Biochemicals inside cells activate cell division. The proteins called kinases and cyclins activate, genes and their proteins to perform cell division. Cyclins act as major checkpoint which operates in G₁ to determine whether or not a cell divides.

Question 4.
Distinguish between Karyokinesis and cytokinesis.
Answer:

Karyokinesis	Cytokinesis
Involves division of the nucleus.	Involves division of cytoplasm.
The nucleus develops a constriction at the center and becomes dumbell shaped.	Plasma membrane develops a constriction along with nuclear constriction.
Constriction deepens and divides the nucleus into two.	It deepens centripetally and finally divides the cell into two cells.

Question 5.
Describe the prophase of Mitotic cell division.
Answer:

1. Prophase is the longest phase in mitosis. Chromosomes become visible as long thin thread-like structure, condenses to form compact mitotic chromosomes. In-plant cells initiation of spindle fibers takes place, the nucleolus disappears. The nuclear envelope breaks down. Golgi apparatus and endoplasmic reticulum are not seen.
2. In the animal cell, the centrioles extend a radial array of microtubules towards the plasma membrane when they reach the poles of the cell. This arrangement of microtubules is called an aster. Plant cells do not form asters.

Question 6.
How Do you calculate the length of the S period?
Answer:
A Culture of animal cells in which the cell cycles were asynchronous was incubated with 3H-Thymidine for 10 minutes. Autoradiography showed that 50% of the cells were labeled. If the cell cycle time (generation time) was 16 hrs how long was the S period?
Length of the S period Fraction of cells in DNA replication x generation time
Length of the S period == 0.5 x 16 hours = 8 hours

Question 7.
Explain the process of cytokinesis in plant cells.
Answer:
Division of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from center towards lateral walls – centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments, and vesicles from the Golgi apparatus and ER. The Golgi Vesicles contain carbohydrates such as pectin, hemicellulose which move along the microtubule of the phragmoplast to the equator fuse, forming a new plasma membrane arid the materials which are placed their becomes new cell wall. The first stage of cell wall construction is a line dividing the newly forming cells called a cell plate. The Cell plate eventually stretches right across the cell forming the middle lamella. Cellulose builds up on each side of the middle lamella to form the cell walls of two new plant cells.

Question 8.
Explain the sequences of Anaphase I Telophase I in Meiotic cell division.
Answer:
Anaphase I: Homologous chromosomes are separated from each other. Shortening of spindle fibers takes place. Each homologous chromosome with its two chromatids and undivided centromere move towards the opposite poles of the cells. The actual reduction in the number of chromosomes takes place at this stage. Homologous chromosomes which move to the opposite poles are either 1 paternal or maternal in origin. Sister chromatids remain attached with their centromeres.
Telophase I: Haploid set of chromosomes are present at each pole. The formation of two daughter cells, each with the haploid number of chromosomes. Nuclei are reassembled. Nuclear envelope forms around the chromosome and the chromosomes become uncoiled. Nucleolus reappears.

Question 9.
Give the differences between mitosis in plant and animal cells.
Answer:

Plants	Animals
Centrioles are absent.	Centrioles are present.
Asters are not formed.	Asters are formed.
Cell division involves the formation of a cell plate.	Cell division involves furrowing and cleavage of the cytoplasm.
Occurs mainly at meristem.	Occurs in tissues throughout the body.

Question 10.
What is Endomitosis? Explain with example.
Answer:
The replication of chromosomes in the absence of nuclear division and cytoplasmic division resulting in numerous copies within each cell is called endomitosis. Chromonema does not separate to form chromosomes but remains closely associated with each other. The nuclear membrane does not rupture. So no spindle formation. It occurs notably in the salivary glands of Drosophila and other flies. Cells in these tissues contain giant chromosomes (polyteny), each consisting of thousands of intimately associated, or synapsed, chromatids, eg. Polytene chromosome.

Answer In detail.

Question 1.
Draw and label the events of the cell cycle.
Answer:

Question 2.
Draw the schematic diagram of anaphase-promoting complex cyclosome and explain briefly.
Answer:
A ubiquitine ligase is activated called the anaphase-promoting complex cyclosome (APC/C) leads to degradation of the key regulatory proteins at the transition of metaphase to anaphase. APC is a cluster of proteins that induces the breaking down of cohesion proteins which leads to the separation of chromatids during mitosis.

Question 3.
Explain the events that take place at metaphase and anaphase stages of somatic cell division – with diagram.
Answer:
Metaphase: Chromosomes (two sister chromatids) are attached to the spindle fibers by kinetochore of the centromere. The spindle fibers is made up of tubulin. The alignment of the chromosome into the compact group at the equator of the cell is known as the metaphase plate. This is the stage where chromosome morphology can be easily studied.
The kinetochore is a DNA-Protein complex present in the centromere DNA where the microtubules are attached. It is a trilaminar disc-like plate.
The spindle assembly checkpoint decides the cell to enter anaphase.
For diagram refer to Figure 7.2 (Metaphase only).
Anaphase: Each chromosome split simultaneously and two daughter chromatids begin to migrate towards two opposite poles of a cell.
Each centromere splits longitudinally into two, freeing the two sister chromatids from each other. Shortening of spindle fiber and longitudinal splitting of centromere creates a pull that divides the chromosome into two halves.
Each half receives two chromatids (that is sister chromatids are separated). When the sister chromatids separate the actual partitioning of the replicated genome is complete.
For diagram refer to Figure 7.2 (Anaphase only).

Question 4.
Explain the different stages of prophase I of meiotic cell division.
Answer:

1. Prophase I: Prophase I is of longer duration and it is divided into 5 substages – Leptotene, Zygotene, Pachytene, Diplotene, and Diakinesis.
2. Leptotene: Chromosomes are visible under a light microscope. Condensation of chromosomes takes place. Paired sister chromatids begin to condense.
3. Zygotene: Pairing of homologous chromosomes takes place and it is known as synapsis. Chromosome synapsis is made by the formation of the synaptonemal complex. The complex formed by the homologous chromosomes is called bivalent (tetrads).
4. Pachytene: At this stage, bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consist of 4 chromatids and 2 centromeres. Synapsis is completed and, recombination nodules appear at a site where crossing over takes place between non-sister chromatids of the homologous chromosome. Recombination of homologous chromosomes is completed by the end of the stage but the chromosomes are linked at the sites of crossing over. This is mediated by the enzyme recombinase.
5. Diplotene: Synaptonemal complex disassembled and dissolves, the homologous chromosomes remain attached at one or more points where crossing over has taken place.
   These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called Chiasmata.
6. Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialized chromosomal structures that hold the homologous chromosomes together.
   Sister chromatids remain closely associated whereas the homologous chromosomes tend to separate from each other but are held together by chiasmata. This substage may last for days or years depending on the sex and organism.
7. The chromosomes are very actively transcribed in females as the egg stores up materials for use during embryonic development. In animals, the chromosomes have prominent loops called the lampbrush chromosome.
8. Diakinesis: Terminalisation of chiasmata. Spindle fibers assemble. The nuclear envelope breaks down. Homologous chromosomes become short and condensed. Nucleolus disappears.

Question 5.
Enumerate the differences between mitosis and meiosis.
Answer:

Mitosis	Meiosis
One division	Two divisions
A number of chromosomes remain the same.	The number of chromosomes is halved.
Homologous chromosomes line up separately on the metaphase plate.	Homologous chromosomes line up in pairs at the metaphase plate.
Homologous chromosome do not pair up	Homologous chromosomes pair up to form bivalent.
Chiasmata do not form and crossing over never occurs.	Chiasmata form and crossingover occurs.
Daughter cells are genetically identical.	Daughter cells are genetically different from the parent cells.
Two daughter cells are formed.	Four daughter cells are formed.

Question 6.
Draw and label the various stages of mitosis cell division.
Answer:

Choose the correct answer.

1. The word cell coined by:
(a) Robert brown
(b) Leeuwenhoek
(c) Anton
(d) Robert Hooke
Answer:
(d) Robert Hooke

2. Which of the following is incorrect about the role of the nucleus?
(a) Control activities of the cell
(b) Genetic information copied from cell to cell
(c) Gametic cells fused together in sexual reproduction
(d) Characters passed on to new individuals
Answer:
(c) Gametic cells fused together in sexual reproduction

3. Match the following with phase and time duration:

Phase	Time duration (in hrs)
G1	11
S	8
G2	4
M	1

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

4. Name the types of nuclear divisions?
(a) Mitosis
(b) Mitosis and Meiosis
(c) A Mitosis
(d) Meiosis
Answer:
(b) Mitosis and Meiosis

5. The checkpoint is otherwise called:
(a) Important point
(b) collecting point
(c) restriction point
(d) Controlling point
Answer:
(c) restriction point

6. In S phase DNA count increases from 2C to:
(a) 3C
(b) 4C
(c) 5C
(d) 6C
Answer:
(b) 4C

7. Which of the following, Microtubules are formed?
(a) G₀
(b) G₁
(c) S
(d) G₂
Answer:
(d) G₂

8. Which animal regenerates the parts of its body?
(a) starfish
(b) amoeba
(c) shark
(d) jellyfish
Answer:
(a) starfish

9. Pick the incorrect statement of the significance of meiosis:
(a) maintain a definite constant number of chromosome
(b) Adaption of an organism to various stress
(c) crossing over takes place
(d) The chromosome number increases in the Organism.
Answer:
(d) The chromosome number increases in the Organism.

10. Which one of the following is not a mitotic poison?
(a) Cyanide
(b) A zide
(c) 2, 4, dinitrophenol
(d) Polyamines
Answer:
(d) Polyamines

11. Insulin and steroid hormones are examples of:
(a) Inhibiting factors
(b) Growth factors
(c) Limiting factors
(d) Synthetic factors
Answer:
(b) Growth factors

12. Identify the correct statement for equational division:
(a) The number of chromosomes in the parent and daughter cells remains the same
(b) Chromosome in the parent and daughter are different
(c) Double the number of chromosomes
(d) It depends upon the division
Answer:
(a) The number of chromosomes in the parent and daughter cells remains the same

13. The number of chromosomes in humans are:
(a) 36
(b) 40
(c) 46
(d) 16
Answer:
(c) 46

14. The chromosomes of a cell occur in pairs called:
(a) haploid pair
(b) homologous pair
(c) diploid pair
(d) Tetraploid
Answer:
(b) homologous pair

15. The amount in picograms of DNA contained within a haploid nucleus is called:
(a) B-value
(b) X – value
(c) Y-value
(d) C – value
Answer:
(d) C – value

16. In meiosis, the daughter cells contain half the number of chromosomes of the parent cell and is known as:
(a) Triploid
(b) Haploid
(c) Tetraploid
(d) Diploid
Answer:
(b) Haploid

17. The major checkpoint which operates in G1 to determine whether or not a cell divides is:
(a) Kinase
(b) protease
(c) cyclins
(d) pepsin
Answer:
(c) cyclins

18. A cell division in which no spindle formation and non condensation of chromatin material is called:
(a) Mitosis
(b) Amitosis
(c) Meiosis
(d) None of the above
Answer:
(b) Amitosis

19. Karyokinesis involves in:
(a) division of cytoplasm
(b) division of cells
(c) division of nucleus
(d) none of the above
Answer:
(c) division of nucleus

20. The sequence of stages of mitotic cell division is as follows:
(a) Metaphase, Anaphase, Prophase and Telophase
(b) Telophase, Anaphase, Prophase and Metaphase
(c) Prophase, Anaphase, Telophase and metaphase
(d) Prophase, Metaphase, Anaphase and Telophase
Answer:
(d) Prophase, Metaphase, Anaphase and Telophase

21. The longest phase in mitosis is:
(a) Metaphase
(b) Telophase
(c) Prophase
(d) Anaphase
Answer:
(c) Prophase

22. The cells, which do not form asters during cell division are:
(a) Animal cells
(b) Plant cells
(c) Virus
(d) None of the above
Answer:
(b) Plant cells

23. Mitosis cell division occurs during:
(a) Cogenesis
(b) Gametogenesis
(c) Somatic growth
(d) Spermatogenesis
Answer:
(c) Somatic growth

24. The Golgi vesicles contain carbohydrates such as:
(a) Pectin
(b) Actin
(c) Glucose
(d) Fructose
Answer:
(a) Pectin

25. A sexual reproduction is prominent is:
(a) Starfish
(b) Dolphin
(c) Yeast
(d) Delonix regia
Answer:
(c) Yeast

26. Synapsis takes place in meiosis during:
(a) Pachytene
(b) Metaphase
(c) Diplotene
(d) Zygotene
Answer:
(d) Zygotene

27. Chiasmata is the point where:
(a) Grossing over takes place
(b) Nuclear division takes place
(c) Cytokinesis occur
(d) Nucleolus disappears
Answer:
(a) Grossing over takes place

28. The random distribution of homologous chromosomes in a cell in metaphase I of meiotic cell division is called:
(a) Segregation
(b) Independent assortment
(c) Linkage
(d) None of the above
Answer:
(b) Independent assortment

29. The stage between two meiotic divisions is called:
(a) Karyokinesis
(b) cytokinesis
(c) Interphase
(d) Interkinesis
Answer:
(d) Interkinesis

30. The factors which promote cell cycle proliferation is called:
(a) Mitotic poison
(b) action
(c) mitogen
(d) recombinase
Answer:
(c) mitogen

Students get through the TN Board 11th Bio Botany Important Questions Chapter 6 Cell: The Unit of Life which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 6 Cell: The Unit of Life

Answer the following short answers.

Question 1.
What is primary magnification?
Answer:
The first magnification of the microscope is done by the objective lens which is called primary magnification.

Question 2.
Name the two kinds of an electron microscope.
Answer:
There are two kinds of electron microscopes namely:

1. Transmission Electron Microscope (TEM).
2. Scanning Electron Microscope (SEM).

Question 3.
What is the main use of TEM?
Answer:
This is the most commonly used electron microscope that provides two-dimensional images. The components of the microscope are as follows: (i) Electron Generating System, (ii) Electron Condenser, (Hi) Specimen Objective, (iv) Tube Lens, (v) Projector. It is used for studying the detailed structure of viruses, mycoplasma, cellular organelles, etc.

Question 4.
List out the name of microscopes in which the light of illumination is visible light.
Answer:
Light Microscope, Darkfield Microscopy, Phase contrast Microscope.

Question 5.
What are the constituents of protoplasm?
Answer:
It is primarily made of water contents and various other solutes of biological importance such as glucose,fatty acids, amino acids, minerals,vitamins, hormones and enzyme.

Question 6.
Define cohesiveness of protoplasm.
Answer:
Particles or molecules of protoplasm are adhered with each other by forces, such as Vander Waal’s bonds, that hold long chains of molecules together.

Question 7.
Nome the three types of cells.
Answer:
The three types of cells are Prokaryotes, Mesokaryotes, and Eukaryotes.

Question 8.
Define mesokoryotes.
Answer:
These organisms share some of the characters of both prokaryotes and eukaryotes. In other words, these are organisms intermediate between pro and eukaryotes.

Question 9.
Explain primary cell wall.
Answer:
It is the first layer of inner to middle lamellae, primarily consisting of a loose network of cellulose microfibrils in a gel matrix. It is thin, elastic, and extensible.

Question 10.
How does a ‘glycocalyx’ form?
Answer:
The Carbohydrate molecules of the cell membrane are short-chain polysaccharides. These are either bound with ‘glycoproteins’ or ‘glycolipids’ and form a glycocalyx.

Question 11.
Define Endocytosis and exocytosis.
Answer:
Cell surface membranes are able to transport individual molecules and ions. There are processes in which a cell can transport a large number of solids and liquids into the cell (endocytosis) or out of the cell (exocytosis).

Question 12.
What do you understand by the term phagocytosis?
Answer:
The particle is engulfed by a membrane, which folds around it and forms a vesicle. The enzymes digest the material and products are absorbed by cytoplasm.

Question 13.
Define cristae of mitochondria.
Answer:
The inner membrane is convoluted (infoldings), called the crista (plural: cristae). Cristae contain most of the enzymes for the electron transport system.

Question 14.
Name any three odors plastids.
Answer:
The three odors plastids are Chloroplast, Phaeoplast, Rhodoplast.

Question 15.
What is the main function of polysome?
Answer:
The function of polysomes in the formation of several copies of a particular polypeptide during protein synthesis.

Question 16.
What are Glyoxysomes?
Answer:
Glyoxysome is a single membrane-bound organelle. It is a sub-cellular organelle and contains enzymes of glyoxylate pathway, fi-oxidation of fatty acid occurs in glyoxysomes of germinating seeds, eg. Castor seeds.

Question 17.
What is the main function of .plant vacuoles?
Answer:
The major function of plant vacuole is to maintain water pressure known as turgor pressure, which maintains the plant structure. Vacuoles organize themselves into a storage/sequestration compartment.

Question 18.
What are the reserve materials present in prokaryotes?
Answer:
In prokaryotes, reserve materials such as phosphate granules, cyanophycean granules, glycogen granules, poly-hydroxy butyrate granules, sulfur granules, carboxysomes, and gas vacuoles are present.

Question 19.
Explain briefly the Holocentric chromosomes.
Answer:
Holocentric chromosomes have centromere activity distributed along the whole surface of the chromosome during mitosis. The holocentric condition can be seen in Caenorhabditis elegans (transparent nematode) Mid many insects.

Question 20.
What do you know about microphotographs?
Answer:
Images of structures observed through microscopes can be further magnified, projected, and saved by attaching a camera to the microscope by a microscope coupler or; eyepiece adaptor. Picture taken using an inbuilt camera in a microscope is called microphotography or microphotograph.

Answer In brief.

Question 1.
Describe briefly the Darkfield microscope.
Answer:

1. The darkfield microscope was discovered by Z. Sigmondy (1905). Here the field will be dark but the object will be glistening so the appearance will be bright.
2. A special effect in an ordinary microscope is brought about by means of a special component called Patch Stop Carrier’.
3. It is fixed in the metal ring of the condenser component
4. Patch top is a small glass device that has a dark patch at the center of the disc leaving a small area along the margin through which the light passes.
5. The light passing through the margin will travel oblique like a hollow cone and strikes the object in the periphery, therefore the specimen appears glistening in a dark background.

Question 2.
Explain the principle and function of a scanning electron microscope.
Answer:

1. This is used to obtain the three-dimensional image and has a lower resolving power than TEM.
   In this, electrons are focused by means of lenses into a very fine point.
2. The interaction of electrons with the specimen results in the release of different forms of radiation (such as Auger electrons, secondary electrons, backscattered electrons) from the surface of the specimen.
3. These radiations are then captured by an appropriate detector, amplified, and then imaged on a fluorescent screen.
4. The magnification is. 2,00,000 times and resolution is 5-20 nm.

Question 3.
Enumerate the functions of the cell wall.
Answer:

1. Offers definite shape and rigidity to the cell.
2. Serves as a barrier for several molecules to enter the cells.
3. Provides protection to the internal protoplasm against mechanical injury.
4. Prevents the bursting of cells by maintaining the osmotic pressure.
5. Plays a major role by acting as a mechanism of defense for the cells.

Question 4.
What do you know about signal transduction in cells? Explain briefly.
Answer:

1. The process by which the cell receives information from outside and responds is called signal transduction.
2. Plants, fungi, and animal cells use nitric oxide as one of the many signaling molecules.
3. The cell membrane is the site of chemical interactions of signal transduction.
4. Receptors receive the information from the first messenger and transmit the message through series of membrane proteins.
5. It activates a second messenger which stimulates the cell to carry out specific functions.

Question 5.
List out the functions of Golgi bodies.
Answer:

1. Glycoproteins and glycolipids are produced.
2. Transporting and storing lipids.
3. Formation of lysosomes.
4. Production of digestive enzymes.
5. Cell plate and cell wall formation
6. Secretion of Carbohydrates for the formation of plant cell walls and insect cuticles.

Question 6.
How do the grana in chloroplast form? Mention their structure and function.
Answer:

1. Grana (singular: Granum) are formed when many of these thylakoids are stacked together
   like a pile of coins.
2. Light is absorbed and converted into chemical energy in the granum, which is used in the stroma to prepare carbohydrates. Thylakoids contain chlorophyll pigments.
3. The chloroplast contains osmophilic granules, 70s ribosomes, DNA (circular and non-histone), and RNA.
4. This chloroplast genome encodes approximately 30 proteins involved in photosynthesis including the components of photosystem I & II, cytochrome bf complex, and ATP synthase.
5. One of the subunits of Rubisco is encoded by chloroplast DNA.
6. It is the major protein component of chloroplast stroma, the single most abundant protein on earth.
7. The thylakoid contains small, rounded photosynthetic units called quantosomes.
8. It is a semi-autonomous organelle and divides by fission.

Question 7.
Explain the types of chromosomes based on the position of the centromere.
Answer:

1. Based on the position of the centromere, chromosomes are called telocentric (terminal centromere), Acrocentric (terminal centromere capped by telomere), Sub metacentric
2. The eukaryotic chromosomes may be tod-shaped (telocentric and acrocentric), L-shaped (sub-metacentric), and V-shaped (metacentric).
   

Question 8.
Describe the structure of flagellum in bacteria.
Answer:
The gram-positive bacteria contain only two basal rings. S-ring is attached to the inside of peptidoglycan and M-ring is attached to the cell membrane. In Gram-negative bacteria, two pairs of rings proximal and distal ring are connected by a central rod. They are L- Lipopolysacchride ring P- Peptidoglycan ring, S-Super membrane ring, and M-membrane ring. The outer pair of L and P rings are attached to the cell wall and the inner pair of S and M rings attached to the cell membrane.

Question 9.
Explain the structure and function of Cilia.
Answer:

Cilia (plural) are short cellular, numerous microtubule bound projections of the plasma membrane.
Cilium (singular) is a membrane-bound structure made up of basal body, rootlets, basal plate, and shaft. The shaft or axoneme consists of nine pairs of microtubule doublets, arranged in a circle along the die periphery with two central tubules, (9 + 2) arrangement of microtubules is present. Microtubules are made up of tubulin. The motor protein dynein connects the outer microtubule pair and links them to the central pair. Nexin links the peripheral doublets of microtubules.

Answer In detail.

Question 1.
Enumerate the physical properties of protoplasm.
Answer:
Physical Properties of Protoplasm: The protoplasm exists either in a semisolid (jelly-like) state called ‘gel’ due to suspended particles and various chemical bonds or maybe a liquid state called ” ‘sol’.
The colloidal protoplasm which is in gel form can change into sol form by solation and the sol can change into a gel by gelation. These gel-sol conditions of the colloidal system are the prime basis for the mechanical behavior of cytoplasm.

1. Protoplasm is translucent, odorless, and polyphasic fluid.
2. It is a crystal colloid solution which is a mixture of chemical substances forming crystalloid
   i.e. true solution (sugars, salts, acids, bases) and others forming colloidal solution (Proteins and lipids).
3. It is the most important property of the protoplasm by which it exhibits three main phenomena namely Brownian movement, amoeboid movement, and cytoplasmic streaming or cyclosis.
   The viscosity of protoplasm is 2-20 centipoises. The Refractive index of the protoplasm is 1.4.
4. The pH of the protoplasmic around 6.8, contains 90% water (10% in dormant seeds)
5. Approximately 34 elements are present in protoplasm but only 13 elements are main or universal elements i.e. C, H, O, N, Cl, Ca, P, Na, K, S, Mg, I, and Fe. Carbon, Hydrogen, Oxygen, and Nitrogen form 96% of protoplasm.
6. Protoplasm is neither a good nor a bad conductor of electricity. It forms a delimiting the membrane on coming in contact with water and solidifies when heated.

Cohesiveness: Particles or molecules of protoplasm are adhered with each other by forces, such as Vander Waal’s bonds, that hold long chains of molecules together. This property varies with the strength of these forces.
Contractility: The contractility of protoplasm is important for the absorption and removal of water especially stomatal operations.
Surface tension: The proteins and lipids of the protoplasm have less surface tension, hence they are found at the surface forming the membrane. On the other hand, the chemical substances (NaCl) have high surface tension, so they occur in deeper parts of the cell protoplasm.

Question 2.
Describe the structure and function of mitochondria.
Answer:

1. It was first observed by A. Kolliker (1880). Altmann (1894) named it as Bioplasts. Later Benda (1897, 1898), named as mitochondria. They are ovoid, rounded, rod shape and pleomorphic structures.
2. Mitochondrion consists of a double membrane, the outer and inner membrane.
3. The outer membrane is smooth, highly permeable to small molecules and it contains proteins called Porins, which form channels that allow free diffusion of molecules smaller than about 1000 daltons and the inner membrane divides the mitochondrion into two compartments, the outer chamber between two membranes and the inner chamber filled with matrix.
4. The inner membrane is convoluted (infoldings), called the crista (plural: cristae). Cristae contain most of the enzymes for the electron transport system.
5. The inner chamber of the mitochondrion is filled with proteinaceous material called mitochondrial matrix.
6. The inner membrane consists of stalked particles called elementary particles or Fernandez Moran particles, F1 particles, or Oxysomes.
7. Each particle consists of a base, stem, and round head. In the head, ATP synthase is present for oxidative phosphorylation.
8. The inner membrane is impermeable to most ions, small molecules and maintains the proton gradient that drives oxidative phosphorylation.
   
9. Mitochondria contain 73% of proteins, 25-30% of lipids, 5-7 % of RNA, DNA (in traces), and enzymes (about 60 types).
10. Mitochondria are called the Powerhouse of a cell, as they produce energy-rich ATP.
11. All the enzymes of Kreb’s cycle are found in the matrix except succinate dehydrogenase. Mitochondria consist of circular DNA and 70S ribosome.
12. They multiply by fission and replicates by strand displacement model. Because of the presence of DNA, it is a semi-autonomous organelle.
13. The unique characteristic of mitochondria is that they are inherited from female parents only. Mitochondrial DNA comparisons are used to trace human origins.
14. Mitochondrial DNA is used to track and date recent evolutionary times because it mutates 5 to 10 times faster than DNA in the nucleus.

Question 3.
Explain the structure of Ribosomes. Mention the types of ribosomes.
Answer:
Ribosomes were first observed by George Palade (1953) as dense particles or granules in the electron microscope. Electron microscopic observation reveals that ribosomes are composed of two rounded subunits, united together to form a complete unit. Mg²⁺ is required for the structural cohesion of ribosomes. Biogenesis of ribosomes is de nova formation, auto replication, and nucleolar origin. Each ribosome is made up of one small and one large subunit. Ribosomes are the sites of protein synthesis in the cell. The ribosome is not a membrane-bound organelle.

Types of Ribosomes
70S Ribosomes (subunit 30S and 50S)	80S Ribosomes (subunits 40S and 60S)
3 RNA molecule.	4 RNA molecule.
(i) 16SrRNA in 30S subunit.	(i) 18SrRNA in 40S small subunit.
(ii) 23S and 5Sin SOS large subunit. Prokaryotic cells of Blue-green algae Bacteria, Mitochondria, and Chloroplast of many Algae and higher plants.	(ii) 28S, 5.8S, and 5S in larger 60S subunit. Eukaryotic cells of Plants and animals

Question 4.
Give details of the special types of chromosomes in plant and animal cells.
Answer:

1. Special types of chromosomes are found only in certain special tissues. These chromosomes are larger in size and are called giant Chromosomes in certain plants and they are found in the suspensors of the embryo.
2. The polytene chromosome and lampbrush chromosome occur in animals and are also called giant chromosomes.
3. Polytene chromosomes observed in the salivary glands of Drosophila (fruit fly) by C.G. Balbiani in 1881. In larvae of many flies, midges (Diptera), and some insects the interphase chromosomes duplicate and reduplicate without nuclear division.
4. A single chromosome that is present in multiple copies forms a structure called a polytene chromosome which can be seen in the light microscope.
5. They are genetically active. There are distinct alternating dark bands and light inter-bands. About 95%of DNA are present in bands and 5% in inter-bands. The polytene chromosome has extremely large puff called Balbiani rings which is seen in chironortious larvae. It is also known as chromosomal puff. Puffing of bands are the sites of intense RNA synthesis,
6. As this chromosome occurs in the salivary gland it is known as salivary gland chromosomes, Polyteny is achieved by repeated replication of chromosomal DNA several times without nuclear division and the daughter chromatids aligned side by side and do not separate (called endomitosis).
7. Gene expression, transcription of genes, and RNA synthesis occur in the bands along the polytene chromosomes. Maternal and paternal homologs remain associated side by side is called somatic pairing.
8. Lampbrush chromosomes occur at the diplotene stage of the first meiotic prophase in oocytes of an animal Salamandar and in the giant nucleus of the unicellular alga Acetabularia.
9. It was first observed by Flemming in 1882. The highly condensed chromosome forms the chromosomal axis, from which lateral loops of DNA extend as a result of intense RNA synthesis.

Question 5.
Describe the structure of eukaryotic flagella and explain the movement of the flagellum.
Answer:

1. Eukaryotic Flagella are enclosed by a unit membrane and it arises from a basal body Flagella is composed of outer nine pairs of microtubules with two microtubules in its center (9+2 arrangement).
2. Flagella are microtubule projections of the plasma membrane. The flagellum is longer than cilium (as long as 200pm). The structure of the flagellum has an axoneme made up of microtubules and protein tubulin.
   
3. Movement: Outer microtubule doublet is associated with axonemal dynein which generates force for movement. The movement is ATP-driven. The interaction between tubulin and dynein is the mechanism for the contraction of cilia and flagella. Dynein molecules use energy from ATP to shift the adjacent microtubules. This movement bends the cilium or flagellum.

Question 6.
Describe the structure and functions of Lysosomes.
Answer:
Lysosomes were discovered by Christian de Duve (1953), these are known as suicidal bags. They are spherical bodies enclosed by a single unit membrane. They are found in eukaryotic cells. Lysosomes are small vacuoles formed when small pieces of the Golgi body are pinched off from its tubules.
They contain a variety of hydrolytic enzymes, that can digest material within the cell. The membrane around the lysosome prevents these enzymes from digesting the cell itself.

Functions:

Intracellular digestion: They digest carbohydrates, proteins, and lipids present in the cytoplasm.
Autophagy: During the adverse condition, they digest their own cell organelles like mitochondria and endoplasmic reticulum
Autolysis: Lysosome causes self-destruction of cells on the insight of disease they destroy the cells.
Ageing Lysosomes have autolytic enzymes that disrupt intracellular molecules.
Phagocytosis: Large cells or contents are engulfed and digested by macrophages, thus forming a phagosome in the cytoplasm. These phago some fuse with lysosome for further digestion.
Exocytosis: Lysosomes release their enzymes outside the cell to digest other cells.

Choose the correct answer.

1. The word cell was first used by:
(a) Robert brown
(b) Robert Hooke
(c) Zemike
(d) Robert schwann
Answer:
(b) Robert Hooke

2. Idea of cell theory was first proposed by:
(a) Matthias Schleiden
(b) Theodor schwann
(c) H.J. Dutrochet
(d) Rudolf Virchow
Answer:
(c) H.J. Dutrochet

3. Phase the contrast microscope was invented by:
(a) Zemike
(b) Robert brown
(c) Sigmondy
(d) Robert Hooke
Answer:
(a) Zemike

4. Source of illumination for image formation in dark field microscope is:
(a) Electron
(b) ultraviolet light
(c) X- rays
(d) Visible light
Answer:
(d) Visible light

5. Who coined the term “protoplasm”?
(a) Corti
(b) Van mohl
(c) Purkinje
(d) Fisher
Answer:
(c) Purkinje

6. Indicate the wrong statement:
(a) All organisms are made up of cells
(b) All metabolic reactions take place inside the cell
(c) The structure and function of the cell is controlled by DNA
(d) Group of cells with different structures are called tissue
Answer:
(d) Group of cells with different structures are called tissue

7. Approximately the number of elements present in protoplasm is:
(a) 28
(b) 34
(c) 38
(d) 24
Answer:
(b) 34

8. These organisms with the primitive nucleus are called:
(a) Mesokaryotes
(b) prokaryotes
(c) Eukaryotes
(d) None of the above
Answer:
(b) prokaryotes

9. Cell wall is not present in:
(a) Bacteria
(b) Fungi
(c) Animal cell
(d) plant cell
Answer:
(c) Animal cell

10. The carbohydrate molecules of the cell membrane are:
(a) Long-chain polysaccharides
(b) Short-chain polysaccharides
(c) Long-chain glycoproteins
(d) Short-chain glycolipids.
Answer:
(b) Short-chain polysaccharides

11. One of the many signaling molecules used by plants, fungi, and animal cell is:
(a) sodium chloride
(b) cupric oxide
(c) Nitric oxide
(d) None of the above
Answer:
(c) Nitric oxide

12. Cytoplasm helps the movement of cellular materials around die cell through a process called:
(a) Cytoplasmic streaming
(b) Brownian movement
(c) Active movement
(d) none of the above
Answer:
(a) Cytoplasmic streaming

13. The name endoplasmic reticulum was given by:
(a) Camillo
(b) K.R.Porter
(c) Nickolson
(d) S.B.Roberts
Answer:
(b) K.R.Porter

14. The functions of the Golgi body includes:
(a) helping cell division
(b) initialing protein synthesis
(c) Transporting and storing lipids
(d) None of the above
Answer:
(c) Transporting and storing lipids

15. Mitochondria are called the powerhouse of the cell as they:
(a) Synthesis lipid
(b) Involve in protein synthesis
(c) initiate oxidation metabolism
(d) Produce energy-rich ATP
Answer:
(d) Produce energy-rich ATP

16. Mitochondrial DNA mutates …………. times faster than DNA in the nucleus.
(a) 15 to 20
(b) 30 to 40
(c) 5 to 10
(d) 10 to 15
Answer:
(c) 5 to 10

17. Plastids were classified into various types according to their structure pigments and function:
(a) Robert Hooke
(b) A.F.tJ. Schimper
(c) A. Kolliker
(d) Altmann
Answer:
(b) A.F.tJ. Schimper

18. B – oxidation of fully acids occurs in glyoxysomes of germinating seeds of:
(a) Paddy
(b) Brinjal
(c) Ladies finger
(d) Caster
Answer:
(d) Caster

19. Centriole consists of nine triplet peripheral fibrils made up of
(a) tubulin
(b) fibrils
(c) annuli
(d) none of the above.
Answer:
(a) tubulin

20. Match the following:

Column-I	Column-II
(i) Thylakoids	(a) Disc-shaped sacs in Golgi apparatus.
(ii) Cristae	(b) Condensed structure of DNA.
(iii) Cistemae	(c) Flat membranous sacs in stroma.
(iv) Chromatin	(d) Infoldings in mitochondria.

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

21. During cell division, chromation is condensed into an organised form called:
(a) nucleolus
(b) Euchromation
(c) Nuclear pores
(d) Chromosomes
Answer:
(d) Chromosomes

22. Chromatin is made up of
(a) DNA, Polysaccharides, and RNA
(b) DNA, lipids, and RNA
(c) DNA, protein, and RNA
(d) RNA, Glucose, and lipid
Answer:
(c) DNA, protein, and RNA

23. Chromosomes having terminal centromere are called:
(a) Acrocentric
(b) Telocentric
(c) Metacentric
(d) Submetacentric
Answer:
(b) Telocentric

24. Polytene chromosome in the salivary glands of drosophila was first observed by:
(a) Flemming
(b) C.G. Balbiani
(c) Harry Beevers
(d) A. Kolliker
Answer:
(b) C.G. Balbiani

25. Lampbrush chromosome occur at the diplotene stage of first meiotic prophase in oocytes of:
(a) Frog
(b) Moth
(c) Sherk
(d) Salamandra
Answer:
(d) Salamandra

26. The main function of Bacterial flagellum is:
(a) Locomotion
(b) Protection
(c) Feeding
(d) none of the above
Answer:
(a) Locomotion

27. The molecules involved in the mechanism for the contraction of cilia and flagella are:
(a) Aetin and peptin
(b) Flagellin
(c) Tubulin and dynein
(d) none of the above
Answer:
(c) Tubulin and dynein

28. The technique of staining the cells and tissue is called:
(a) microphotography
(b) histochemistry
(c) anatomy
(d) geochemistry
Answer:
(b) histochemistry

29. The stain used for staining mitochondria of the cell is:
(a) Sudan black
(b) Eosin
(c) cotton blue
(d) Janus green
Answer:
(d) Janus green

30. Cilia are short cellular, numerous microtubules bound projections of:
(a) mitochondrial membrane
(b) Cell wall
(c) Plasma membrane
(d) Nuclear membrane
Answer:
(c) Plasma membrane

Students get through the TN Board 11th Bio Botany Important Questions Chapter 5 Taxonomy and Systematic Botany which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 5 Taxonomy and Systematic Botany

Answer the following short answers.

Question 1.
Define taxonomy.
Answer:
Davis and Heywood (1963) defined taxonomy as “the science dealing with the study of classification including the bases, principles, rules, and procedures”.

Question 2.
Describe a genus with an example.
Answer:
Genus consists of multiple species which have similar characters but differ from toe species of another genus, eg. Helimthmt Tridax.

Question 3.
What are the types of species?
Answer:
There are different types of species and they are as follows:

1. Process of evolution – Biological Species
2. Product of evolution – Morphological Species and Phylogenetic Species

Question 4.
Differentiate between Anamorph and Telomorph.
Answer:

Anamorph	Asexual reproductive stage of fungus.
Telomorph	Sexual reproductive stage of fungus.

Question 5.
Write Common name and Scientific name of any two plants.
Answer:

Common name	Scientific name
Paddy	Oryza Sativa
Groundnut	Arachis hypogea

Question 6.
What is holotype nomenclature?
Answer:
A specimen or illustration originally cited by the author in the protologue. It is a definitive reference source for identity. Citation of holotype and submission of it is one of the criteria for valid publication of a botanical name.

Question 7.
Define Flora.
Answer:
Flora is the document of all plant species in a given geographic area. Flora consists of a total number of plant species in an area and gives information about the flowering season, fruiting season, and distribution for the given geographic area.

Question 8.
Mention any two international botanical gardens.
Answer:

1. New York Botanic garden, USA.
2. Royal Botanic Garden, Kew – England.

Question 9.
Define Chemotaxonomy.
Answer:
Chemotaxonomy is the scientific approach to the classification of plants on the basis of their biochemical constituents.

Question 10.
What are the aims, of chemotaxonomy?
Answer:

1. To develop taxonomic characters which may improve the existing system of plant classification.
2. To improve present-day knowledge of phytogeny of plants.

Question 11.
Define Serotaxonomy.
Answer:
The classification of very similar plants by means of differences in the proteins they contain, to solve taxonomic problems is called serotaxonomy.

Question 12.
What is meant by DNA barcodes?
Answer:
The genetic sequence used to identify a plant is known as “DNA tags” or “DNA barcodes”.

Question 13.
What do you know about cladistics?
Answer:
The method of classifying organisms into monophyletic group of a common ancestor based on shared apomorphic characters is called cladistics (from Greek, klados- branch).

Question 14.
Explain monadelphous stamens.
Answer:
In Aeschynomene Aspera, the stamens are fused to form two bundles each containing five stamens (5)+5. Stamens are monadelphous.

Question 15.
Mention any two oil plants with their scientific name.
Answer:

1. Arachis hypogea (Groundnut).
2. Pongamia pinnata (Pungam).

Question 16.
Mention any two medicinal plants with their useful parts of the plant.
Answer:

Medicinal plant	Useful part
Atropa belladonna (deadly nightshade)	Roots
Datura stramonium (Jimsonweed)	Leaves and roots

Question 17.
What is meant by scapigerous Inflorescence?
Answer:
The Inflorescence axis (peduncle) arising from the ground bearing a cluster of flowers at its apex. Pedicels are of equal length, arising from the apex of the peduncle which brings all flowers at the same level.

Question 18.
List any two economic uses of plants under the family Liliaceae.
Answer:

Question 19.
List any two ornamental plants under the family Fabaceae.
Answer:
Butea frondosa (Flame of the forest), Clitoria ternatea, Lathyrus odoratus (Sweet pea), and Lupinus hirsutus (Lupin).

Question 20.
What is meant by root module?
Answer:
Taproot system, roots are nodulated, have tubercles containing nitrogen-fixing bacteria (Rhizobium leguminosarum)

Answer In brief.

Question 1.
Explain the differences between taxonomy and systematics.
Answer:

Taxonomy	Systematics
The discipline of classifying organisms into taxa.	The broad field of biology that studies the diversification of species.
Governs the practices of naming, describing, identifying, and specimen preservation.	Governs the evolutionary history and phylogenetic relationship in addition to taxonomy.
Classification + Nomenclature = Taxonomy	Taxonomy + Phylogeny = Systematics

Question 2.
What is Botanical nomenclature? Explain the international code of botanical nomenclature.
Answer:
Assigning a name for a plant is known as Nomenclature. This is based on the rules and recommendations of the International Code of Botanical Nomenclature. ICB deals with the names of existing (living)and extinct (fossil) organisms. The elementary rule of the naming of the plant was first proposed by Linnaeus in 1737and 1751 in his Philosophia Botanica. In 1813 a detailed set of rules regarding plant nomenclature was given by A.P. de Candolle in his famous ‘ work “Theorie elementaire de la botanique”. Then the present ICBN was evolved by following the same rules of Linnaeus, A.P. de Candolle, and his son Alphonse de Candolle.

Question 3.
Mention any three roles of Botanical gardens.
Answer:

1. Gardens with an aesthetic value attract a large number of visitors. For example, the Great Banyan Tree (Ficus benghalensis) in the Indian Botanical Garden at Kolkata.
2. Gardens have a wide range of species and supply taxonomic material for botanical research.
3. It can integrate information from diverse fields like Anatomy, Embryology, Phytochemistry, Cytology, Physiology, and Ecology.

Question 4.
List out the uses of the herbarium.
Answer:

1. Herbarium provides resource material for systematic research and studies.
2. It is a place for the orderly arrangement of voucher specimens.
3. Voucher specimen serves as a reference for comparing doubtful newly collected fresh specimens.
4. Voucher specimens play a role in studies like floristic diversity, environmental assessment, ecological mechanisms, and survey of unexplored areas.
5. Herbarium provides an opportunity for documenting biodiversity and Studies related to the field Of ecology and conservation biology.

Question 5.
Explain Bentham and the hooker system of Glassification.
Answer:

1. A widely followed natural system of classification Considered the best was proposed by two English botanists George Bentham (1800 – 1884) and Joseph Dalton Hooker (1817-1911).
2. The classification was published in a three-volume work as “Genera Plantation”
   (18624 883) describing 202 families and 7569 genera and 97,205 Species.
3. In this system, the seeded plants were classified into 3 major classes such as Dicotyledonae, Gymnospermae, and Monocotyledonae.

Question 6.
What do you know about karyotaxonomy?
Answer:

1. Chromosomes are the carriers of genetic information. Increased knowledge about the chromosomes has been used for extensive biosystematic studies and resolving many taxonomic problems.
2. Utilization of the characters and phenomena of cytology for the explanation of the taxonomic problems is known as cytotaxonomy or karyotaxonomy.
3. The characters of chromosomes such as number, size, morphology, and behavior during meiosis have proved to be of taxonomic value.

Question 7.
What is RFLP? Explain briefly.
Answer:
RFLPs are a molecular method of genetic analysis that allows the identification of taxa based on unique patterns of restriction sites in specific regions of DNA. It refers to differences between taxa in restriction sites and therefore the lengths of fragments of DNA following cleavage with restriction enzymes.

Question 8.
What is the significance of molecular taxonomy?
Answer:

1. It helps to identify a very large number of species of plants and animals by the use of conserved molecular sequences.
2. Using DNA data evolutionary patterns of biodiversity are now investigated.
3. DNA taxonomy plays a vital role in phytogeography, which ultimately helps in genome mapping and biodiversity conservation.
4. DNA- based molecular markers used for designing DNA-based molecular probes, have also been developed under the branch of molecular systematics.

Question 9.
Write briefly about the diagnostic features of Solanaceae family.
Answer:

1. Leaves alternate, exstipulate
2. Flowers actinomorphic, pentamerous
3. Calyx often persistence / accrescent
4. Stamens 5, epipetalous, poricidal in dehiscence
5. Carpels 2, ovary superior, 2 chambered, obliquely placed, falsely four-chambered placenta swollen, ovule numerous,
6. Fruits berry or capsule, vascular bundles with both outer and inner phloem (Bicollateral vascular bundle)

Question 10.
Write down the floral formula of the following plants.
Answer:

1. Datura metal,
2. Solanum nigrum,
3. Allium cepa.
   

Answer In detail.

Question 1.
What is the concept of Species? Explain briefly the various concepts.
Answer:
Species are the fundamental unit of taxonomic classification. The Greek philosopher Plato proposed the concept of “eidos” or species and believed that all objects are shadows of the “eidos”. According to Stebbins (1977) species is the basic unit of the evolutionary process. A species is a group of individual organisms which have the following characters.

1. A population of organisms that closely resemble each other more than the other population.
2. They descend from a common ancestor.
3. In sexually reproducing organisms, they interbreed freely in nature, producing fertile offspring.
4. In asexually reproducing organisms, they are identified by their morphological resemblance.
5. In the case of fossil organisms, they are identified by morphological and anatomical resemblance. Species concepts can be classified into two general groups. The concept emphasizing the process of evolution that maintains the species as a unit and that can result in evolutionary divergence and speciation. Another concept emphasizes the product of evolution in defining a species.

Question 2.
Describe the method of preparation of herbarium specimen.
Answer:

1. Plant Collection: Plant specimen with flower or fruit is collected.
2. Documentation of field site data: Certain data are to be recorded at the time of plant collection. It includes date, time, country, state, city, specific locality information, latitude, longitude, elevation, and landmark information. These data will be typed onto a herbarium label.
3. Preparation of plant specimen: Plant specimen collected from the field is pressed immediately with the help of a portable field plant press. Plant specimen is transferred to a standard plant press (12” x 18”) which between two outer 12” * 18” frames and secured by two straps.
4. Mounting herbarium specimen: The standard size of herbarium sheet is used for mounting the specimen (29cm x 41cm). specimens are affixed to herbarium sheet with standard white glue or solution of Methylcellulose.
5. Herbarium label: Herbarium label size is generally 4-5” wide and 2-3” tall. A typical label contains all information like habit, habitat, vegetation type, landmark information, latitude, longitude, image document, collection number, date of collection, and name of the collector.
6. Protection of herbarium sheets against mold and insects: Application of 2% Mercuric chloride, Naphthalene, DDT, carbon disulfide. Fumigation using formaldehyde. Presently deep freezing(-20°C) method is followed throughout the world.

Question 3.
Explain the basis of molecular taxonomy with its uses.
Answer:

1. Molecular Taxonomy is the branch of phytogeny that analyses hereditary molecular differences, mainly in DNA sequences, to gain information and to establish genetic relationships between the members of different taxonomic categories.
2. The advent of DNA cloning and sequencing methods has contributed immensely to the development of molecular taxonomy and population genetics over the years.
3. These modern methods have revolutionized the field of molecular taxonomy and population genetics with improved analytical power and precision.
4. The results of a molecular phylogenetic analysis are expressed in the form of a tree called a phylogenetic tree.
5. Different molecular markers like allozymes, mitochondrial DNA, microsatellites, RFLP (Restriction Fragment Length Polymorphism), RAPD (Random amplified polymorphic DNA), AFLPs (Amplified Fragment Length Polymorphism), single nucleotide polymorphism-SNP, microchips, or arrays are used in the analysis.

Uses-of molecular taxonomy:

1. Molecular taxonomy helps in establishing the relationship of different plant groups at the DNA level.
2. It unlocks the treasure chest of information on the evolutionary history of organisms.

Question 4.
Write briefly about Cladistic analysis.
Answer:

1. Cladistics is one of the primary methods of constructing phylogenies or evolutionary histories. Cladistics uses shared, derived characters to group organisms into clades.
2. These clades have at least one shared derived character found in their most recent common ancestor that is not found in other groups hence they are considered more closely related to each other.
3. These shared characters can be morphological such as, leaf, flower, Suit, seed, and so on; behavioral, like the opening of flowers nocturnal/diurnal; molecular like DNA or protein sequence, and more.
4. Cladistic accepts only monophyletic groups. Paraphyletic and polyphyletic taxa are occasionally considered when such taxa conveniently treated as one group for practical purposes.
   eg. dicots, Sterculiaceae. Polyphyletic groups are rejected by cladistic.

(a) Monophyletic group: Taxa comprising all the descendants of a common ancestor.

(b) Paraphyietic group: Taxon that includes an ancestor but not all of the descendants of that ancestor.

(c) Polyphyletic group: Taxa that includes members from two different lineages.

Need for Cladistics:

1. Cladistics is now the most commonly used and accepted method for creating a phylogenetic system of classifications.
2. Cladistics produces a hypothesis about the relationship of organisms to predict the morphological characteristics of organisms.
3. Cladistics helps to elucidate the mechanism of evolution.

Question 5.
Explain the general characters of the family Solanaceae.
Answer:

1. Distribution: Family Solanaceae includes about 88 genera and about 2650 species, of these Solanum, is the largest genus of the family with about 1500 species. Plants are worldwide in distribution but more abundant in South America.
2. Habit: Mostly annual herbs, shrubs, small trees (Solanum violaceum) lianas with prickles (Solanum trilobatum) present in some taxa, many with stellate trichomes; rarely vines (Lycium, Sinensis)
3. Root: Branched tap root system.
4. Stem: Herbaceous or woody; erect or twining, or creeping; sometimes modified into tubers (Solanum tuberosum) often with collateral vascular bundles.
5. Leaves: Alternate, simple, rarely pinnately compound (Solanum tuberosum and (Lycopersicon esculentum) exstipulate, opposite or sub-opposite in the upper part, unicostate reticulate venation.
6. Inflorescence; Generally axillary or terminal cymose (Solanum) or solitary flowers (Datura stramonium). Extra axillary scorpioid cyme called rhiphidium (Solanum nigrum) solitary and axillary (Datura and Nicotiana) umbellate cyme (Withania somnifera).
7. Flowers: Bracteate (Petunia), or ebracteate (Withania) pedicellate, bisexual, heterochlamydeous, actinomorphic, or weakly zygomorphic due to oblique position of ovary pentamerous, hypogynous. ,
8. Calyx: Sepals 5, synsepalous, valvate, persistent rarely the sepals are 4 or 6. Often enlarging to envelop the fruit (Physalis, Withania).
9. Corolla: Petals 5, sympetalous, rotate, tubular (Solanum) or bell-shaped (Atropa) or infundibuliform YPemwi’aj usually alternate with sepals; rarely bilipped and zygomorphic (Schizanthus) usually valvate, sometimes convolute (Datum).
10. Androecium: Stamens 5, epipetalous, filaments usually Unequal in length, stamens only 2 in Schizanthus, 4 and didynamous in (Salpiglossis) Anthers dithecous, dehisce longitudinally or poricidal. …
11. Gynoecium: Bicarpeliary, syncarpous obliquely placed, ovary superior, bilocular but looks tetralocular due to the formation of false septa, numerous ovules in each locule on axile placentation.
12. Fruit: A capsule or berry. In Lycopersicon esculentum, Capsicum, the fruit is a berry and in species of Datura and Petunia, the fruit is a capsule.
13. Seed: Endospermous.

Question 6.
Give the floral characters of Pisum sativum.
Answer:

Activity

Text Book Page No. 128

Write common names and scientific names of 10 different plants around your home.
Answer:

Common name	Scientific name
Neem tree	Azadirachta indica
Tamarind	Tamarindus indica
Coconut	Cocos nucifera
Palmyra	Borassusflabellifer
Flame of the forest	Delonix regia
Mango	Mangifera indica
Fishtail palm	Caryota urens
Royal palm	Bresdoxa regia
Gliricidia	Erythrina Indica
Agni maram	Sesbania grandiflora

Text Book Page No,136

Prepare a herbarium of 5 common weed plants found inside your school campus /nearby garden/wasteland.
Answer:
Common weed plants: i) Tridax procumbence, (ii) Vinca rosea, (Hi) Lamium amplexicaula, . (iv) Poa annua, (v) Stellaria media.

Text Book Page No.160

Can you identify this?

(a) Name the family.
Answer:
Asparagales

(b) Write the binomial.
Answer:
Aloe vera.

(c) List the economic uses.
Answer:
(i) Cosmetic and medicine.
(ii) Soothing and moisturing agent.
(iii) Protection for human from sunburn.
(iv) Treatment of wounds and bums.
(v) Aloe vera gel is used commercially as an ingredient in yogurts.
(vi) As dietary supplement.
(vii) Skin treatment in Ayurvedic medicine.

Choose the correct answer.

1.18th International Botanical congress was held in 2011 at:
(a) London, U.K
(b) Melbourne, Australia
(c) Newyork, U.S.A
(d) Sydney, Australia
Answer:
(b) Melbourne, Australia

2. The lowest of classification is:
(a) Genus
(b) Kingdom
(c) species
(d) Family
Answer:
(c) species

3. Flora is the term used for:
(a) the document of all plant species.
(b) the document of single species in a given geographic area.
(c) the document of only endemic species of plants in a given area.
(d) none of the above.
Answer:
(a) the document of all plant species.

4. The first botanical garden was established by Theophrastus at:
(a) London
(b) Sydney
(c) Athens
(d) Singapore
Answer:
(c) Athens

5. Which is the largest Botanical garden in the world.
(a) Indian Botanical Garden, Kolkata, India
(b) Botanical Garden at Athens.
(c) National Botanical Garden, Lucknow, India
(d) Royal Botanical Garden Kew, England
Answer:
(d) Royal Botanical Garden Kew, England

6. Who was called as Father of Taxonomy?
(a) E.K. Janaki Ammal
(b) Carolus Linnaeus
(d) Theophrastus
(c) Heywood
Answer:
(b) Carolus Linnaeus

7. Plants with one stamen are grouped under:
(a) Tetrandria
(b) Diandria
(c) Monandria
(d) Pentandria
Answer:
(c) Monandria

8. The family coniferae is included under the class:
(a) Dicotyledonae
(b) Monocotyledonae
(c) Gymnospermae
(d) None of the above
Answer:
(c) Gymnospermae

9. Chemotaxomony is mainly based on:
(a) The chemical characters of the plant
(b) the morphological characters of the plant
(c) The phylogenetic characters of the plant
(b) none of the above
Answer:
(a) The chemical characters of the plant

10. The classification based on the characters of chromosome, such as number, size, morphology and behaviour during meiosis is known as:
(a) serotaxonomy
(b) Chemotaxonomy
(c) karyotaxonomy
(d) none of the above
Answer:
(c) karyotaxonomy

11. RAPD( Random amplified polymorphic DNA) is a method:
(a) to identify the morphomatic character of a plant
(b) to identify specific regions of DNA
(c) to identify genetic sequence of a plant
(d) to identify genetic markers using a randomly synthesised primer.
Answer:
(d) to identify genetic markers using a randomly synthesised primer.

12. Genetic sequence used to identify a plant is known as:
(a) DNA tags
(b) Polymorphic DNA
(c) Fragment of DNA
(d) Genome
Answer:
(a) DNA tags

13. Metformin used for treatment of diabetes is exacted from:
(a) Hibiscus rosasinensis
(b) Galega officinalis
(c) Arachis hypogea
(d) none of the above
Answer:
(b) Galega officinalis

14. Earlier classification emphasized on:
(a) Reproductive characters
(b) Vegetative characters
(c) both
(d) Anatomical characters
Answer:
(c) both

15. Naming a plant is called:
(a) systematic botany
(b) Taxonomy
(c) nomenclature
(d) Cytology
Answer:
(c) nomenclature

16. Bentham and Hooker’s classification is:
(a) Phylogenetic system of classification
(b) artificial system of classification
(c) natural system of classification
(d) sexual system of classification
Answer:
(c) natural system of classification

17. Carolus Linnaeus proposed classification based on:
(a) Artificial system
(b) natural system
(c) phylogenetic system
(d) modem system
Answer:
(a) Artificial system

18. The largest group of plant kingdom is
(a) Cryptogams
(b) Angiosperms
(c) Gymnosperms
(d) phanerogams
Answer:
(b) Angiosperms

19. Number of volumes in genera plantorum
(a) two
(b) three
(c) four
(d) five
Answer:
(b) three

20. The family gnetaceae is included under
(a) monochlamydae
(b) monotyledons
(c) dicotyledons
(d) Gymnosperms
Answer:
(d) Gymnosperms

21. Bentoam and Hooker are associated with
(a) Indian botanical garden
(b) Royal botanical garden
(c) American botanical garden
(d) French botanical garden
Answer:
(b) Royal botanical garden

22. The division of angiosperme is achieved by:
(a) floral characters
(b) anatomy
(c) physiology
(d) ecology
Answer:
(a) floral characters

23. Classification of plants into different groups is known as:
(a) Morphology
(b) Physiology
(c) Pathology
(d) Systematic botany
Answer:
(d) Systematic botany

24. Syngenesious anthers and epipetalous stamens are found in:
(a) Liliaceae
(b) malvaceae
(c) Solanaceae
(d) cruciferae.
Answer:
(d) cruciferae.

25. Synandrous condition is common in the family:
(a) Lumbelliferae
(b) Rosaceae
(c) malvaceae
(d) Cucurbitaceae
Answer:
(d) Cucurbitaceae

26. Polyadelphous condition is found in:
(a) Leguninaceae
(b) rutaceae
(c) compositae
(d) Liliaceae
Answer:
(b) rutaceae

27. Indefinite stamens are characteristic of the family:
(a) malvaceae
(b) graniceae
(c) labiatac
(d) cruciferae
Answer:
(a) malvaceae

28. Classical taxonomy is also termed as:
(a) B taxonomy
(b) systematics
(c) description and taxonomy
(d) experimental taxonomy
Answer:
(c) description and taxonomy

29. Match the following:

(i) Pulses	(a) Cluster bean
(ii) Medicinal plants	(b) Arachis hypogea
(Hi) Food and plants	(c) Cowpea
(iv) Oil plants	(d) Mucunapruriens

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

30. FAO declared the year for pulses as:
(a) 2017
(b) 2015
(c) 2018
(d) 2016
Answer:
(d) 2016

Students get through the TN Board 11th Bio Botany Important Questions Chapter 4 Reproductive Morphology which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 4 Reproductive Morphology

Answer the following short answers.

Question 1.
Define floriculture.
Answer:
Floriculture is a branch of Horticulture. It deals with the cultivation of flowers and ornamental crops.

Question 2.
What is axillary inflorescences?
Answer:
Inflorescence present in the axile of the nearest vegetative leaf.

Question 3.
What is meant by spadix inflorescences?
Answer:
An inflorescence with a fleshy or thickened central axis that possesses many unisexual sessile flowers in acropetal succession.

Question 4.
Define Capitulum.
Answer:
The capitulum is a determinate or indeterminate, group of sessile or subsessile flowers arising on a receptacle, often subten&d by an involucre.

Question 5.
What do you know about Helicoid?
Answer:
Axis develops on only one side and forms a coil structure at least at the earlier development stage. Example: Hamelia, potato.

Question 6.
Define Cymule.
Answer:
A small, simple dichasium is called a cymule

Question 7.
Explain briefly about coenanthium.
Answer:
Circular disc-like fleshy open receptacle that bears pistillate flowers at the center and staminate flowers at the periphery. Example: Dorstenia

Question 8.
Define bisexual flower.
Answer:
When a flower contains both androecium and gynoecium is called a perfect flower.

Question 9.
What is meant by polygamous plants?
Answer:
The condition in which bisexual and unisexual (staminate/pistillate) flowers occur in the same plaint is called polygamous.

Question 10.
Explain Actinomorphic flower.
Answer:
The flower shows two mirror images when cut in any plane or radius through the center.
Normally there are more than two planes of symmetry.

Question 11.
What do you know about merosity?
Answer:
A number of floral parts per whorl are called merosity. Perianth merosity is the number of perianth parts per whorl.

Question 12.
Describe briefly deciduous calyx.
Answer:
Calyx that falls after the opening of flower (anthesis) eg. Nelumbo.

Question 13.
What type of calyx is present in Ocimum?
Answer:
Two lipped calyces is present in Ocimum.

Question 14.
Define infundibuliform corolla.
Answer:
Petals fused to form funnel-shaped corolla. Tube gradually widens into limbs, eg. Datura, Ipomoea.

Question 15.
Mention three parts of the stamen.
Answer:

1. Filament,
2. Anther,
3. Connective.

Question 16.
Define the term “Connation”.
Answer:
Connation refers to the fusion of stamens among themselves.

Question 17.
Explain briefly about the Apocarpous ovary.
Answer:
A pistil contains two or more distinct carpels, eg. Annona

Question 18.
Describe Gynobasic style with an example.
Answer:
Arising from the base of the ovary, eg. Lamiaceae (Ocimum), characteristic of Boraginaceae.

Question 19.
What is meant by pomology?
Answer:
The branch of horticulture that deals with the study of fruits and their cultivation are called pomology.

Question 20.
Write down two important functions of seed.
Answer:

1. The seed encloses and protects the embryo for the next generation.
2. It contains food for the development of the embryo.

Answer In brief.

Question 1.
Distinguish between racemose and cymose inflorescence.
Answer:

Racemose	Cymose
The main axis of unlimited growth.	The main axis of limited growth.
Flowers arranged in acropetal succession.	Flowers arranged in a basipetal succession.
The opening of flowers is centripetal.	The opening of flowers is centrifugal.
Usually the oldest flower at the base of the inflorescence axis.	Usually the oldest flower at the top of the inflorescence axis.

Question 2.
Explain the term “Hypanthodium” with a suitable example.
Answer:
A receptacle is a hollow, globose structure consisting of unisexual flowers present on the inner wall of the receptacle. The receptacle is closed except for a small opening called the ostiole which is covered by a series of bracts. Male flowers are present nearer to the ostiole, female and neutral flowers are found in a mixed manner from the middle below, eg. Ficus sp. (Banyan and Pipal).

Question 3.
Describe briefly the three types of sympetalous zygomorphic corolla.
Answer:

1. Bilabiate: Corolla with two lips. eg. Ocimum, Leucas, Adhatoda. Tubular corolla with a single strap-shaped limb. eg. Ray floret of Helianthus.
2. Personate: Corolla made up of two lips with the upper arched and the lower protruding into the corolla throat, eg. Antirrhinum, Linaria.
3. Ligulate: Tubular corolla with a single strap-shaped limb, eg. Ray floret of Helianthus.

Question 4.
What is meant by anther dehiscence? Explain the different kinds of anther dehiscence.
Answer:
It refers to the opening of anther to disperse pollen grains.

1. Longitudinal: Anther dehisces along, a suture parallel to the long axis of each anther lobe, eg. Datura, china rose, cotton.
2. Transverse: Anther dehisces at right angles to the long axis of anther lobe. eg. Malvaceae.
3. Poricidal: Anther dehisces through pores at one end of the thecae, eg. Ericaceae, Solanum, potato, brinjal, Cassia.
4. Valvular: Anther dehisces through a pore covered by a flap of tissue, eg. Lauraceae, Cinnamomum.

Question 5.
Explain the different types of stigma with suitable examples.
Answer:
A stigma is a structure present at the tip of a pistil which receives the pollen grains,

1. Discoid: A disk-shaped stigma is called discoid.
2. Capitate: Stigma appearing like ahead, eg. Alchemilla.
3. Globose: Stigma is spherical in shape is called globose.
4. Plumose stigma: Stigma feathery which is unbranched or branched as in Asteraceae, Poaceae.

Question 6.
Describe the salient features of the floral diagram.
Answer:
A floral formula is a simple way to explain the salient features of a flower. The floral diagram is a representation of the cross-section of the flower, It represents floral whorls arranged as viewed from above. The floral diagram shows the number and arrangement of bract, bracteoles, and floral parts, fusion, overlapping, and placentation.

Question 7.
Distinguish between true fruit and false fruit.
Answer:

True Fruit	False Fruit
The ovary develops into the fruit without any non-carpellary part. eg. Tomato, Mango.	In addition to the ovary the non- carpellary (floral) parts like thalamus (Apple), perianth (jack fruit) and involucre, and perianth (English walnut) develop into a fruit.

Question 8.
Mention any three functions of fruit.
Answer:

1. The edible part of the fruit is a source of food, energy for animals.
2. They are a source of many chemicals like sugar, pectin, organic acids, vitamins, and minerals.
3. The fruit protects the seeds from unfavorable climatic conditions and animals.

Question 9.
Write briefly about the types of multiple fruits for example.
Answer:
A Multiple or composite fruit develops from the whole inflorescence along with its peduncle on which they are borne.
Sorosis: A fleshy multiple fruits that develop from a spike or spadix. The flowers fused together by their succulent perianth and at the same time, the axis bearing them become fleshy or juicy and the whole inflorescence forms a compact mass. eg. Pineapple, Jack fruit, Mulberry.
Syconus: A multiple fruits that develops from hypanthodium inflorescence. The receptacle develops further and converts into fleshy fruit which encloses a number of true fruit or achenes which develop from female flower of hypanthodium inflorescence, eg. Ficus.

Question 10.
What are the types of seed, based on the presence and absence of the endosperm? Explain with a suitable example.
Answer:
Based on the presence or absence of the endosperm the seed is of two types.

1. Albuminous or Endospermous seed: The cotyledons are thin, membranous, and mature seeds have endosperm persistent and nourish the seedling during its early development, eg. Castor, sunflower, maize.
2. Ex-albuminous or non-endospermous seed: Food is utilized by the developing embryo and so the mature seeds are without endosperm; In such seeds, cotyledons store food and become thick and fleshy, eg. Pea, Groundnut.

Answer In detail.

Question 1.
Write an essay on cymose inflorescence.
Answer:
The central axis Stops growing and ends in a flower, further growth is by means of axillary buds. Old flowers present at the apex and young flowers at the base.

1. Simple cyme (solitary): Determinate inflorescence consists of a single flower. It may be terminal or axillary, eg. terminal in Trillium grandiflorum and axillary in Hibiscus.
2. Monochasial Cyme (uniparous): The main axis ends with a flower. From two lateral bracts, only one branch grows further. It may be helicoid (bostryx) or Scorpioid (cincinnus).
   (a) Helicoid: Axis develops on only one side and forms a coil structure at least at the earlier development stage, eg. Hamelia, potato.
   (b) Scorpioid: Axis develops on alternate sides and often becomes a coil structure, eg. Heliotropium.
3. Simple dichasium (Biparous): A central axis ends in a terminal flower; further growth is produced by two lateral buds. Each cymose unit consists of three flowers of which the central one is an old one. This is true cyme. eg. Jasminum.
4. Compound dichasium It has many flowers. A terminal old flower develops lateral simple dichasial cymes on both sides. Each compound dichasium consists of seven flowers, eg. Clerodendron. A small, simple dichasium is called a cymule.
5. Polychasial Cyme (multiparous): The central axis ends with a flower. The lateral axes branches repeatedly, eg. Nerium.
6. Sympodial Cyme: In monochasial cyme, successive axes at first develop in a zigzag manner, and later it develops into a straight pseudo axis. eg. Solanum americanum.

Question 2.
Explain the parts of a flower with a diagram.
Answer:
In a plant, which part would you like the most? Of course, it is a flower, because of its colour and fragrance. The flower is a significant diagnostic feature of angiosperms. It is a modified condensed reproductive shoot. The growth of the flower shoot is determined. There are two whorls, accessory and essential. Accessory whorl consists of calyx and corolla and the essential whorl comprises androecium and gynoecium. The flower is said to be complete when it contains all four whorls. An Incomplete flower is devoid of one or more whorls.

1. Pistil The female reproductive organ of a flower is Gynoecium or pistil. Each member is carpel.
2. Petal Innermost accessory whorl of the flower is corolla. Each member is called a petal.
3. Sepal: Outermost whorl of the flower is calyx. Each member is called sepal.
4. Perianth (perigonium) Undifferentiated calyx and corolla. Individual members are called tepal.
5. Bract Subtending leaf or leaf-like the structure of any flower is called Bract.
6. Stamen Male organ of a flower is the androecium. Each member is the stamen.
7. Thalamus (torus or receptacle) The part of the flower on which other floral parts are attached.
8. Bracteole: A smaller bract present on the side of the pedicel is called bracteole or bractlet. A whorl of bracteoles at the base of the calyx is called epicalyx.
9. Pedicel Stalk of the flower. The flower is pedicellate or sessile depending upon presence or absence. The flowers with a short, rudimentary pedicel are called subsessile flowers.

Question 3.
What is Aestivation? Explain different types of aestivation.
Answer:

1. Valvate Margins of sepals or petals do not overlap but just touch each other, eg. Calyx in members of Malvaceae, Calotropis, Annona.
2. Twisted or convolute or contorted One margin of each petal or sepal overlapping on the other petal, eg. Petals of china rose.
3. Imbricate: Sepals and petals irregularly overlap on each other; one member of the whorl is exterior, one interior, and the rest of the three having one margin exterior and the other interior, eg. Cassia, Delonix. There are three types: (i) Ascendingly imbricate, (ii) Quincuncial, (iii) Vexillary.
4. Quincuncial It is a type of imbricate aestivation in which two petals are external and two internal and one petal with one margin internal and the other margin external, eg. Guava, calyx of Ipomoea, Catharanthus.
5. Vexillary Large posterior petals both margins overlap lateral petals. Lateral petals other margin overlaps anterior petals, eg. Pea, bean.
   

Question 4.
Describe the difference between anthers in plants.
Answer:
Anther types:

1. Monothecal: One lobe with two microsporangia. They are kidney-shaped in a cross-section, eg. Malvaceae.
   Some other types: Hfcplostemonous stamens are uniseriate and equal in number to the petals and opposite the sepals (antisepalous).
   Obhaplostemonous: Stamens are uniseriate, a number equal to petals and opposite the petals (antipetalous)
   Diplostetnonous: Stamens are biseriate, outer antisepalous, inner antipetalous. eg. Murraya.
   Obdiplostemonous Stamens are biseriate, outer antipetalous, inner antisepalous. eg. Caryophyllaceae.
   Poiystemonou Numerous stamens are normally many more than the number of petals.
2. Dithecal: It is a typical type, having two lobes with four microsporangia. They are butterfly-shaped in cross-section, eg. Solanaceae.

Anther attachment:

1. Basifixed (Innate) Base of the anther is attached to the tip of filament, eg. Brassica, Datura.
2. Dorsifixed Apex of filament is attached to the dorsal side of the anther, eg. Citrus, Hibiscus.
3. Versatile: Filament is attached to the anther at the midpoint, eg. Grasses.
4. Adnate: Filament is continued from the base to the apex of anther, eg. Verbena, Ranunculus, Nelumbo.

Anther dehiscence: It refers to the opening of anther to disperse pollen grains.

1. Longitudinal: Anther dehisces along a suture parallel to long axis of each anther lobe, eg. Datura, china rose, cotton.
2. Transverse: Anther dehisces at right angles to the long axis of anther lobe. eg. Malvaceae.
3. Poricidal Anther dehisces through pores at one end of the thecae, eg. Ericaceae, Solarium, potato, brinjal, Cassia.
4. Valvular: Anther dehisces through a pore covered by a flap of tissue, eg. Lauraceae, Cinnamomum.

Anther dehiscing direction: It shows the position of anther opening relative to the anther of the flower.

1. Introrse: Anther dehisces towards the center of the flower, eg. Dianthus.
2. Extrorse: Anther dehisces towards the periphery of the flower, eg. Argemone.

Question 5.
Explain the floral diagram and floral formula of Hibiscus Rosa Sinensis.
Answer:

Question 6.
Write an essay on fleshy fruits and their kind.
Answer:
The fruits are derived from single pistils where the pericarp is fleshy, succulent, and differentiated
into epicarp, mesocarp, and endocarp. It is subdivided into the following.

1. Ben Fruit develops from bi carpellary or multi carpellary, syncarpous ovary. Here the epicarp is thin, the mesocarp and endocarp remain undifferentiated. They form a pulp in which the seeds are embedded, eg. Tomato, Date Palm, Grapes, Brinjal.
2. Drupe: Fruit develops from the monocarpellary, superior ovary. It is usually one-seeded. The pericarp is differentiated into outer skinny epicarp, fleshy and pulpy mesocarp, and hard and stony endocarp around the seed. eg. Mango, Coconut.
3. Pepo: Fruit develops from tri carpellary, inferior ovary. Pericarp terns leathery or woody which encloses, fleshy mesocarp and smooth endocarp. eg. Cucumber, Watermelon, Bottle gourd, Pumpkin.
4. Hesperidiur Fruit develops from the multi carpellary, multilocular, syncarpous, superior ovary. The fruit wall is differentiated into a leathery epicarp with oil glands, a middle fibrous mesocarp. The endocarp forms distinct chambers, containing juicy hairs, eg. Orange, Lemon. ,
5. Pome: it develops from multi carpellary, syncarpous, inferior ovary. The receptacle also develops along with the die ovary and becomes fleshy, enclosing the true fruit. In pome the epicarp is thin skin-like and the endocarp is cartilaginous. eg. Apple, Pear.
6. Balausta: A fleshy indehiscent fruit developing from multi carpellary, multilocular inferior ovary whose pericarp is tough and leathery. Seeds are attached irregularly with testa being the edible portion, eg. Pomegranate.

Activity

Text Book Page No. 121

Prepare a diet chart to provide a balanced diet to an adolescent (a school going child) which includes food items (fruits, vegetable, and seeds) which are non – expensive and are commonly available.
Answer:
Diet chart for school going child

Carbohydrate	Cereals and grains etc – 33%
Vitamin and minerals	Various fruits and vegetables – 33%
Meat protein	Fish, meat, and egg -12%
Milk protein	Diary products -15%
Fat and sugar	Fatty food and sweets etc – 7%

Choose the correct answer.

1. Axillary inflorescence is present in:
(a) Nerium oleander
(b) Theobroma cocoa
(c) Hibiscus rosa sinensis
(d) Couropita guianensis
Answer:
(c) Hibiscus rosa sinensis

2. An unbranched indeterminate inflorescence with sessile flowers is categorized as:
(a) spikelet
(b) spike
(c) simple raceme
(d) none of the above
Answer:
(b) spike

3. Umbeltypeofinflorescenceisseenin:
(a) Allium cepa
(b) Caesal pinia
(c) Cauliflower
(d) Opuntia
Answer:
(a) Allium cepa

4. Cyathium inflorescence consists of:
(a) Small bisexual flowers
(b) small unisexual flower
(c) small anthers
(d) none of the above
Answer:
(b) small unisexual flower

5. Which of the following is a monoecious plant?
(a) Musa
(b) Coconut
(c) Magnifera
(d) Papaya
Answer:
(b) Coconut

6. A piant with both male flowers and bisexual flower is termed as:
(a) androdioecious
(b) gynodioecious
(c) andromonoecious
(d) trioecious
Answer:
(c) andromonoecious

7. Flowers that lack any plane of symmetry and cannot be divided into equal halves in any plane are seen in:
(a) bean
(b) Datura
(c) Cassia
(d) Canna indica
Answer:
(d) Canna indica

8. A flower which is composed of distinct outer calyx and the inner corolla is termed as:
(a) Homochlamydous
(b) dichlamydeous
(c) achlamydeous
(d) none of the above
Answer:
(b) dichlamydeous

9. Match the following:

(i) Anisomerous	(a) Allium
(ii) Bimerous	(b) Annona
(iii) Trimerous	(c) Hibiscus
(iv) Pentamerous	(d) Ixora

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

10. Synsepalous condition is present in the flowers of:
(a) Annona
(b) Papaya
(c) Datura
(d) palmyra
Answer:
(c) Datura

11. Crciform type of corolla is present in:
(a) radish
(b) dianthus
(c) tea
(d) rose
Answer:
(a) radish

12. Flowerpetals fused to form a bell-shaped corolla are termed as:
(a) tubular
(b) Rotate
(c) Infundibuliform
(d) campanulate
Answer:
(d) campanulate

13. Corolla with two lips is present in:
(a) adhatoda
(b) linaria
(c) helianthus
(d) allium
Answer:
(a) adhatoda

14. Aestivation is the term used for:
(a) arrangement of flowers in the inflorescence
(b) arrangement of leaves in the stem.
(c) arrangement of sepals and petals in the flower bud
(d) none of the above
Answer:
(c) arrangement of sepals and petals in the flower bud

15. Match the following:

(i) Monadelphous	(a) Asteraceae
(ii) Diadelphous	(b) Malvaceae
(Hi) Polyadelphous	(c) Fabaceae
(iv) Syngenesious	(d) Bombax

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

16. Didynamous condition in which two stamens with long filaments and two with short filaments and two with short filaments, is seen in:
(a) Ipomoea
(b) Ocimum
(c) Mimosa
(d) Datura
Answer:
(a) Ipomoea

17. Innate condition of anther attachment is named for the condition:
(a) the base of anther is attached to the tip of the filament
(b) the apex of the filament is attached to the dorsal side of the anther.
(c) the filament is continued from the base to the apex of anther
(d) none of the above
Answer:
(a) the base of anther is attached to the tip of the filament

18. Poricidal type of anther dehiscence is present in:
(a) Cinnamomum
(b) Nelumbo
(c) Brinjal
(d) Datura
Answer:
(c) Brinjal

19. A condition in which the gynoecium has four carpels is termed as:
(a) bicarpellary
(b) multicarpellary
(c) unicarpellary
(d) tetracarpellary
Answer:
(d) tetracarpellary

20. Match the following.

(i) Anthophore	(a) Gynandropsis
(ii) Androphore	(b) Capparis
(iii) Gynandrophore	(c) Grewia
(iv) Gynophore	(d) Silene canoidea

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

21. In Malvaceae family the sepals, petals, and stamens are attached at the base of a superior ovary. This condition is known as:
(a) Perigynous
(b) Epigynous
(c) Epiperigynous
(d) Hypogynous
Answer:
(d) Hypogynous

22. Axile placentation is present in:
(a) Hibiscus
(b) mustard
(c) marigold
(d) cucumber
Answer:
(a) Hibiscus

23. Drupe fruit develops from:
(a) tricatpellary inferior ovary
(b) monocarpellary, superior ovary
(c) Bicarpellery ovary
(d) bicarpellary, syncarpous ovary
Answer:
(b) monocarpellary, superior ovary

24. Multiple or composite fruit develops from:
(a) the whole flower
(b) Single flower
(c) the whole inflorescence
(d) none of the above
Answer:
(c) the whole inflorescence

25. Match the following.

(i) Berry	(a) Mango
(ii) Drupe	(b) Tomato
(Hi) Pome	(c) Cucumber
(iv) Pepo	(d) Apple

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

26. In an inflorescence where flowers are borne laterally in an accropeted succession, the position of the youngest floral bud shall be:
(a) proximal
(b) Distal
(c) Intercalary
(d) Anywhere
Answer:
(a) proximal

27. Edible part of jack fruit is:
(a) whole fruit
(b) mesocarp
(c) perianth and seeds
(d) perianth and rachis
Answer:
(c) perianth and seeds

28. Non-endospermous seeds are present in:
(a) Groundnut
(b) Maize
(c) sunflower
(d) castor
Answer:
(a) Groundnut

29. Indicate the correct statement.
(a) Edible part of the fruit is poisonous to animals
(b) Edible part of the fruit is a source of food energy for animals.
(c) Edible part of the fruit is the exclusive source of medicine.
(d) Edible part of the fruit is the only source of fodder.
Answer:
(b) Edible part of the fruit is a source of food energy for animals.

30. In angiosperms embryo seed represents:
(a) female gametophyte
(b) male gametophyte
(c) sporophyte
(d) none of the above
Answer:
(a) female gametophyte

Students get through the TN Board 11th Bio Botany Important Questions Chapter 3 Vegetative Morphology which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 3 Vegetative Morphology

Answer the following short answers.

Question 1.
Define reproductive morphology.
Answer:
It includes Flower/inflorescence, Fruit, and Seed.

Question 2.
Write down any two primary functions of the stem.
Answer:

1. Provides support and bears leaves, flowers, and fruits.
2. It transports water and mineral nutrients to the other parts from the root.

Question 3.
Mention three important components of vegetative morphology.
Answer:

1. Habit,
2. Habitat and
3. Life span.

Question 4.
What are herbs? Give two examples.
Answer:
Herbs are soft-stemmed plants with less wood or no wood. Phyllanthus amarus, Cleome viscosa.

Question 5.
What is the two plant habitat?
Answer:
Depending upon where plants grow habitats may be classified into major categories:

1. Terrestrial and
2. Aquatic.

Question 6.
Define perennial plants give two examples.
Answer:
A plant that grows for many years that flowers and set fruits for several seasons during the life span.

Question 7.
What is meant by Angiosperms?
Answer:
Flowering plants are called “Angiosperms” or Magnoliophyta.

Question 8.
Give two primary functions of the root system.
Answer:

1. Absorb water and minerals from the soil.
2. Help to anchor the plant firmly in the soil.

Question 9.
Explain fusiform root with one example.
Answer:
These roots are swollen in the middle and tapering towards both ends. eg. Raphanus sativus.

Question 10
What is meant by thorn climbers? Give one example.
Answer:
Climbing or reclining on the support with the help of thorns as in Bougainvillea and Carissa.

Question 11.
Explain Phylloclade.
Answer:
Phylloclade is a characteristic adaptation of xerophytes where the leaves often fall off early and modified into spines or scales to reduce transpiration.

Question 12.
Define Bulb.
Answer:
It is a condensed conical or convex stem surrounded by fleshy scale leaves.

Question 13.
What is meant by Cladode?
Answer:
Cladode is a flattened or cylindrical stem similar to Phylloclade but with one or two internodes only.

Question 14.
Define Venation.
Answer:
The arrangement of veins and veinlets on the leaf blade or lamina is called venation. Internally, the vein contains vascular tissues.

Question 15.
What is meant by Phyllotaxy?
Answer:
The mode of arrangement of leaves on the stem is known as phyllotaxy (Gk. Phyllon = leaf; taxis = arrangement).

Question 16.
Give two examples for parallel venation.
Answer:

1. Canna,
2. Bamboo,

Question 17.
Explain opposite phyllotaxy.
Answer:
In this type, each node possesses two leaves opposite to each other.

Question 18.
What is meant by phyllode?
Answer:
Phyllodes are flat, green-colored leaf-like modifications of petioles or rachis. The leaflets or lamina of the leaf are highly reduced or caducous.

Question 19.
Define Heterophylly.
Answer:
The occurrence of two different kinds of leaves in the same plant is called heterophylly.

Question 20.
Give two examples of evergreen plants.
Answer:
Mimiisops, Calophyllum.

Answer In brief.

Question 1.
What is plant morphology? Explain the types of morphology.
Answer:
The study of various external features of the organism is known as morphology. Plant morphology is also known as external morphology deals with the study of the shape, size, and structure of plants and their parts (roots, stems, leaves, flowers, fruits, and seeds). The study of morphology is important in taxonomy. Morphological features are important in determining the productivity of crops. Morphological characters indicate the specific habitats of living as well as the fossil plants and help to correlate the distribution in space and time of fossil plants. Morphological features are also significant for phylogeny.
Plant Morphology can be studied under two broad categories:

1. Vegetative morphology – It includes the shoot system and root system.
2. Reproductive morphology – It includes Flower/inflorescence, Fruit, and Seed.

Question 2.
Explain the adventitious root system with an example.
Answer:

1. Root developing from any part of the plant other than the radicle is called adventitious root. It may develop from the base of the stem or nodes or internodes. eg. Monstera deliciosa, Ficus benghalensis, Piper nigrum.
2. In most of monocots, the primary root of the seedling is short-lived and lateral roots arise from various regions of the plant body. These are a bunch of thread-like roots equal in size which are collectively called fibrous root systems generally found in grasses, eg. Oryza sativa, Eleusine coracana, Pennisetum americanum.

Question 3.
What are the storage roots? Explain each type with a suitable example.
Answer:
Tuberous root: These roots are swollen without any definite shape. Tuberous roots are produced singly and not in clusters, eg. Ipomoea batatas.
Fasciculated root: These roots are in the cluster from the base of the stem. eg. Dahlia, Asparagus, Ru&ltia.
Nodulose root: In this type of root swelling occurs only near the tips. eg. Mtmnfct (arrow root) Curcuma amada (mango ginger), Curcuma loHga (turmeric).
Monilifonn or Beaded root: These roots swell at frequent intervals giving them a beaded appearance, eg. Vitis, Portulaca, Momordica Indian spinach.
Annulatedroot: These roots have a series of ring-like swelling on their surface at regular intervals. eg. Ipecac (Psychotria)

Question 4.
List out the characteristic features of this stem.
Answer:

1. The stem is usually the aerial portion of the plant
2. It is positively phototropic and negatively geotropic.
3. It has nodes and internodes.
4. The stem bears vegetative bud for vegetative growth of the plant, and floral buds for reproduction, and ends in a terminal bud.
5. The young stem is green and thus carries out photosynthesis.
6. During reproductive growth, the stem bears flowers and fruits.
7. Branches arise exogenously.
8. Some stems bear multicellular hairs of different kinds.

Question 5.
What is the secondary function of the stem?
Answer:

1. Food storage – eg. Solanum tuberosum, Colocasia, and Zingiber officinale.
2. Perennation / reproduction – eg. Zingiber officinale, Curcuma longa.
3. Water storage – eg. Opuntia.
4. Buoyancy – eg. Neptunia.
5. Photosynthesis – eg. Opuntia, Ruscus, Casuarina, Euphorbia, Caralluma.
6. Protection – eg. Citrus, Duranta, Bougainvillea, Acacia, Fluggea, Carissa.
7. Support – eg. Passiflora,Bougainvillea, Vitis, Cissus quadrangularis.

Question 6.
What are Bulbils? Explain different types with suitable examples.
Answer:

1. Bulbils are modified and enlarged buds, meant for propagation.
2. When bulbils detach from the parent plant and fall on the ground, they germinate into new plants and serve as a means of vegetative propagation.
3. In Agave and Allium, proliferum floral buds get modified into bulbils.
4. In Lilium bulbiferum and Dioscorea bulbifera, the bulbils develop in the axil of leaves.
5. In Oxalis, they develop just above the swollen root.

Question 7.
What is meant by Rhizome? Give at least three examples.
Answer:
This is an underground stem that grows horizontally with several lateral growing tips. Rhizome possess conspicuous nodes and internodes covered by scale leaves, eg. Zingiber officinale, Canna, Curcuma longa, Maranta arundinacea, Nymphaea, Nelumbo.

Question 8.
List out the primary function of the leaf.
Answer:
Primary functions:

1. Photosynthesis,
2. Transpiration,
3. Gaseous exchange,
4. Protection of buds,
5. Conduction of water and dissolved solutes.

Question 9.
Describe Palmately reticulate venation. Mention its types for example.
Answer:
Palmately reticulate venation (multicostate): In this type of venation there are two or more principal veins arising from a single point and they proceed outward or upwards. The two types of palmate reticulate venation are

1. Divergent type: When all principal veins originate from the base and diverge from one another towards the margin of the leaf as in Cucurbita, Luffa, Carica papaya, etc.
2. Convergent: When the veins converge to the apex of the leaf, as in Indian plum (Zizyphus),
   bay leaf (Cinnamomum)
   
   (a) Pinnately reticulate, (b) Palmately reticulate (Divergent), (c) Palmately reticulate (Convergent)

Question 10.
Give an account of storage leaves with suitable examples.
Answer:
Some plants of saline and xerophytic habitats and members of the family Crassulaceae commonly have fleshy or swollen leaves. These succulent leaves store water, mucilage or food material. Such storage leaves resist desiccation, eg. Aloe, Agave, Bryophyllum, Kalanchoe, Sedum, Sueada, Brassica oleracea (cabbage-variety capitata).

Answer In detail.

Question 1.
Describe the regions of the root with a suitable diagram.
Answer:

The root tip is covered by dome-shaped parenchymatous cells called root caps. It protects the meristematic cells in the apex. In Pandanus multiple root cap is present. In Pistia instead of a root cap, a root pocket is present. A few millimeters above the root cap the following three distinct zones have been classified based on their meristematic activity.

1. Meristematic Zone,
2. Zone of Elongation,
3. Zone of Maturation.

Question 2.
What are the vital functions of root? Explain each function with a suitable example.
Answer:
Epiphytic or velamen root: Some epiphytic orchids develop a special kind of aerial roots which hang freely in the air. These roots develop a spongy tissue called velamen which helps in the absorption of moisture from the surrounding air. eg. Vanda, Dendrobiutn, Aerides.
Foliar root: Roots are produced from the veins or lamina of the leaf for the formation of the new plant, eg. Bryophyllum, Begonia, Zamioculcas.
Sucking or Haustorial roots: These roots are found in parasitic plants. Parasites develop adventitious roots from the stem which penetrate into the tissue of the host plant and suck nutrients, eg. Ciiscuta (dodder), Cassytha, Orobanche (broomrape), Viscum (mistletoe), Dendrophthoe.
Photosynthetic or assimilatory roots: Roots of some climbing or epiphytic plants develop chlorophyll and turn green which help in photosynthesis, eg. Tinospora, Trapa natans (water chestnut), Taeniophyllum.

Question 3.
Explain different types of stem.
Answer:
The majority of angiosperm possess upright, vertically growing erect stem. They are (i) Excurrent, (ii) Decurrent, (iii) Caudex, (iv) Culm

1. Excurrent: The main axis shows continuous growth and the lateral branches gradually becoming shorter towards the apex which gives a conical appearance to the trees, eg. Polyalthia longifolia, Casuarina.
2. Decurrent: The growth of the lateral branch is more vigorous than that of the main axis. The tree has a rounded or spreading appearance, eg. Mangifera indica, Azadirachta indica, Tamarindus indicus, Aegle marmelos.
3. Caudex: It’s an unbranched, stout, cylindrical stem, marked with scars of fallen leaves, eg. Cocus Nucifera, Borassusflabellifermis, Areca catechu.
4. Culm: Erect stems with distinct nodes and usually hollow internodes clasped by leaf sheaths.
   eg. Majority of grasses including Bamboo.

Question 4.
Describe the various sub-aerial stem modifications with suitable examples.
Answer:
Subaerial stem found in plants with the weak stem in which branches lie horizontally on the ground. These are meant for vegetative propagation. They may be subaerial or partially subterranean.
Runner: This is a slender, prostrate branch creeping on die ground and rooting at the nodes, eg. Centella (Indian pennywort), Oxalis (wood sorrel), lawn grass (Cynodon dactylon).
Stolon: This is also a slender, lateral branch originating from the base of the stem. But it first grows obliquely above the ground, produces a loop and bends down towards the ground. When touches the ground it produces roots and becomes an independent plantlet. eg. Mentha piperita (peppermint), Fragaria indica (wild strawberry).
Sucker: Sucker develops from an underground stem and grows obliquely upwards and gives rise to a separate plantlet or new plant, eg. Chrysanthemum, Musa, Bambusa.
Offset: Offset is similar to runner but found in aquatic plants especially in rosette leaved forms.
A short thick lateral branch arises from the lower axil and grows horizontally leafless for a short distance, then it produces a bunch of rosette leaves and roots at nodes, eg. Eiehhornia (water hyacinth), Pistia (water lettuce).

Question 5.
Explain the parts of the leaf with a suitable diagram.
Answer:
Three main parts of a typical leaf are:
(i) Leaf base (Hypopodium), (ii) Petiole (Mesopodium), (iii) Lamina (Epipodium).

1. The part of the leaf attached to the node of the stem is called the leaf base. Usually, it protects growing buds at its axil. In legumes, the leaf base becomes broad, thick, and swollen which is known as a pulvinus. eg. Clitoria, Lablab, Cassia, Erythrina, Butea, Peltophorum.
   In many monocot families such as Arecaceae, Musaceae, Zingiberaceae, and Poaceae the leaf base extends, into a sheath and clasps part or whole of the internode. Such leaf base also leaves permanent scars on the stem when they fall. eg. Arecaceae.
   
2. Petiole (stipe or mesopodium): Tt is the bridge between leaf and stem. Petiole or leaf stalk is a cylindrical or subcylindrical or flattened structure of a leaf that joins the lamina with the stem. A leaf with petiole is said to be petiolate. eg. Ficus, Hibiscus, Mangifera, Psidium. Leaves that do not possess petiole is said to be sessile, eg. Calotropis, Gloriosa.
3. The expanded flat green portion of the leaf is the blade or lamina. It is the seat of photosynthesis, gaseous exchange, transpiration, and most of the metabolic reactions of the plant. The lamina is traversed by the midrib from which arise numerous lateral veins and thin veinlets. The lamina shows great variations in its shape, margin, surface, texture, color, venation, and incision.
4. In most of the dicotyledonous plants, the leaf base bears one or two lateral appendages called the stipules. Leaves with stipules are called stipulate. The leaves without stipules are called exstipulate or estipulate. The stipules are commonly found in dicotyledons. In some grasses (Monocots) an additional outgrowth is present between leaf base and lamina. It is called Ligule. Sometimes, small stipule-like outgrowths are found at the base of leaflets of a compound leaf. They are called stipels. The main function of the stipule is to protect the leaf in the bud condition.

Question 6.
What are the types of pinnately compound leaves? Explain each type with, suitable example.
Answer:
A pinnately compound leaf is defined as one in which the rachis, bears laterally a number of leaflets, arranged alternately or in an opposite manner, as in tamarind, Cassia.

1. Unipinnate: The rachis is simple and unbranched which bears leaflets directly on its sides in alternate or opposite manner, eg. Rose, Neem. Unipinnate leaves are of two types.
   When the leaflets are even in number, the leaf is said to be paripinnate. eg. Tamarind, Abrus, Sesbania, Saraca, Cassia.
   When the leaflets are odd in number, 1he leaf is said to be imparipinnate. eg. Rose, Neem \ (Azadirachta), Chinese box (Murraya).
2. Bipinnate: The primary rachis produces secondary rachis which bears the leaflets. The secondary rachii are known as pinnae. Number of pinnae varies depending on the species, eg. Delonix, Mimosa, Acacia nilotica, Caesalpinia.
3. Tripinnate: When the rachis branches thrice the leaf is called tripinnate. (i.e) the secondary rachii produce the tertiary rachii which bear the leaflets, eg. Moringa, Oroxylum.
4. Decompound: When the rachis of the leaf is branched several times it is called decompound, eg. Daucus carota, Coriandrum sativum, Foeniculum vulgare.

Choose the correct answer.

1. Vegetative morphology of plant includes:
(a) Shoot system, root system, and inflorescence
(b) the root system, flower, and seed
(c) shoot system and root system
(d) flower, fruit, and seed.
Answer:
(c) shoot system and root system

2. Phyllanthus amarus belongs to the group:
(a) Slmibs
(b) Herbs
(c) Climbers
(d) Trees
Answer:
(b) Herbs

3. The root system of the plant is generally:
(a) positively geotropic and negatively phototrophic in nature
(b) negatively geotropic and positively phototrophic in nature
(c) Positively geotropic and negatively phototrophic in nature
(d) negatively geotropic and negatively phototrophic in nature
Answer:
(c) Positively geotropic and negatively phototrophic in nature

4. Otym Sativa has:
(a) Tab toot system
(b) fibrous root system
(c) Adventitious and tap root system
(d) taproot with the secondary root system
Answer:
(b) fibrous root system

5. Match the following:

(i) Conical root	(a) Mechanical support
(ii) Tuberous root	(b) Orchids
(in) Climbing root	(c) Food storage
(iv) Epiphytic root	.(d) Daucus carota

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

6. Foliar root is present in:
(a) randa
(b) Bryophyllum
(c) Delonix regia
(d) piper betel
Answer:
(b) Bryophyllum

7. Match the following:

(i) Pothos	(a) Stem climber
(ii) Ipomoea	(b) Thom climber
(iii) Bignonia	(c) Root climber
(iv) Carissa	(d) Hook climber

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

8. cladode is present in:
(a) Bambusa
(b) Musa
(c) Asparagus
(d) Citrus
Answer:
(c) Asparagus

9. Rhizome is the modification of:
(a) Stem
(b) Root
(c) Undergroimd stem
(d) Undergroimd bulb
Answer:
(c) Undergroimd stem

10. Petiole is present in:
(a) Calotropis
(b) Hibiscus
(c) Gloriosa
(d) None of the above
Answer:
(b) Hibiscus

11. Ficus religiosa has:
(a) Pinnately parallel venation
(b) Palmately reticulate venation
(c) Multicostate venation
(d) Pinnately reticulate venation
Answer:
(d) Pinnately reticulate venation

12. Pinnately compound leaf is present in:
(a) Cassia
(b) Cucurbita
(c) begonia
(d) acalypha
Answer:
(a) Cassia

13. The part of the root which is most active in water absorption is called:
(a) root cap
(b) maturation zone
(c) meristematic zone
(d) zone of elongation
Answer:
(b) maturation zone

14. Venation is a term used to describe the pattern of arrangement of:
(a) floral organs
(b) veins and veinlets in a lamina
(c) flower in inflorescence
(d) all of them
Answer:
(b) veins and veinlets in a lamina

15. Epiphytic roots are found in;
(a) Indian rubber
(b) Orchid
(c) Tinospora
(d) Cuscuta
Answer:
(b) Orchid

16. Potatoes are borne on:
(a) Primary roots
(b) lateral roots
(cl Adventitious roots
(d) axil of scaly leaves
Answer:
(d) axil of scaly leaves

17. Winged petiole is found in:
(a) citrus
(b) radish
(c) acacia
(d) peepal
Answer:
(a) citrus

18. Fibrous root in ficus benghalensis develop from:
(a) stem
(b) node
(c) intemode
(d) none of the above
Answer:
(a) stem

19. Foliar roots are present in:
(a) vanda
(b) Bombax
(c) Bryophyllum
(d) Ficus pumila
Answer:
(c) Bryophyllum

20. Which one of the following is not a characteristic of the root.
(a) presence of root cap
(b) presence of chlorophyll
(c) absence of buds
(d) presence of unicellular hair
Answer:
(b) presence of chlorophyll

21. ………… are the vegetative organs of the flowering plant.
(a) Leaves, stem, fruits
(b) Roots, stem, flowers
(d) Roots, stem, leaves.
(c) Roots, leaves, flowers
Answer:
(d) Roots, stem, leaves.

22. Which is not a stem modification?
(a) Rhizome of ginger
(b) Corn of colocasia
(c) Pitcher of nepenthes
(d) Tuber of potato
Answer:
(c) Pitcher of nepenthes

23. In One of the following stem performs the function of storage and propagation,
(a) Wheat
(b) Ginger
(c) Radish
(d) Paddy
Answer:
(b) Ginger

24. An underground specialised sheet with a reduced disc-like stem covered by fleshy leaves is:
(a) Rhizome
(b) Rhizosphere
(c) Bulb
(d) Bulbil
Answer:
(c) Bulb

25. A phyllode i

s modified:
(a) leaf
(b) Stem
(c) Root
(d) Branch
Answer:
(a) leaf

26. A fibrous root system is better adapted than a tap root system for:
(a) Storage food
(b) Anchorage plant to soil
(c) absorption of water and organic food
(d) Transport of water and organic food
Answer:
(b) Anchorage plant to soil

27. Arrangement of leaves on a stem is called:
(a) Venation
(b) Vernation
(c) Phyllotaxy
(d) Axis
Answer:
(c) Phyllotaxy

28. The pitcher in nepenthes is a modification of:
(a) Stem
(b) root
(c) branch
(d) leaf
Answer:
(d) leaf

29. Leaf spines are present in:
(a) bombax
(b) asparagus
(c) mango
(d) citrus
Answer:
(b) asparagus

30. Heterophylly is found in:
(a) Limnophila heterophylla
(b) Calophyllum
(c) Erythrina
(d) Cabbage
Answer:
(a) Limnophila heterophylla