Category: Class 11

Students get through the MP Board Class 11th Biology Important Questions Chapter 14 Respiration in Plants which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Chapter 14 Respiration in Plants

Respiration in Plants Class 11 Important Questions Very Short Answer Type

Question 1.
What is glycolysis?
Answer:
Glycolysis is the process which occur in the cytoplasm in which one molecule of glucose is converted into two molecules of pyruvic acid.
C₆H₁₂O₆ + 2 ATP + 4ADP + H₃PO₄ + 2NAD → 2CH₃COCOOH + 2ADP + 4ATP + 2NADH₂ + 2H₂O

Question 2.
Where is glycolysis and Krebs cycle completed?
Answer:
Glycolysis: It is completed in the cytoplasm.
Krebs cycle: This process is completed in mitochondria.

Question 3.
How much energy is liberated by oxidation of 1 mole glucose?
Answer:
One mole glucose liberates 673 kcal or 38 ATP by aerobic oxidation and 21 kcal or 2 ATP by anaerobic process.

Question 4.
Why anaerobic respiration releases less energy than aerobic respiration?
Answer:
In anaerobic respiration, due to incomplete oxidation of food material some energy remains within it, whereas in aerobic respiration all energy liberates due to complete oxidation. Due to this, anaerobic respiration liberates less energy than aerobic respiration.

Question 5.
What is the initial and the end product of glycolysis and where does the complete process take place?
Answer:
The initial and the end product of glycolysis is glucose and pyruvic acid and the whole process takes place in the cell plasm.

Question 6.
Where does anaerobic respiration occurs in the cell?
Answer:
Anaerobic respiration occurs in the cytoplasm part of the cell.

Question 7.
What is anaerobic respiration?
Answer:
Type of respiration in which incomplete oxidation of food occurs in absence of oxygen is called as anaerobic respiration. Less energy is released during this respiration.

Question 8.
What is Aerobic respiration?
Answer:
Type of respiration in which complete oxidation of food occurs in the presence of oxygen is called as aerobic respiration. More energy is released during this process.
C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂ + 673 k.cal Energy [38 mol ATP].

Question 9.
Potted plant must not be kept in the room at night while sleeping. Why?
Answer:
During night photosynthesis do not occur thus plants can not use CO₂  produced during respiration. Therefore, concentration of CO₂  increases in the room, which may cause harm to the person sleeping in that room. Thus, potted plants are not kept in the room at night while sleeping.

Plant Growth and Development Class 11 Important Questions Short Answer Type

Question 1.
What is Respiratory Substrate? Name the substrate generally used. (NCERT)
Answer:
Such substances which undergoes decomposition inside the cells for production of energy are called as Respiratory substrate.
e.g. Food materials like carbohydrates, fats, proteins etc.
Substrate which is generally used as respiratory substrate is Glucose.

Question 2.
What is Oxidative phosphorylation? (NCERT)
Answer:
In all living beings ATP is generated during oxidative breakdown of complex food products. This process of synthesis of ATP molecules from ADP and inorganic phosphate by using the energy of food materials is called as oxidative phosphorylation.

Question 3.
What is the importance of energy released during breathing? (NCERT)
Answer:
During oxidation of food inside the cell, total energy produced by the respiratory substrate is not released at a time. It is stored in the form of ATP. Whenever body requires energy it decomposes to give energy. Thus, ATP is also called as energy currency. This energy is used for performing different life activities of the body.

Question 4.
Write differences between following : (NCERT)
(a) Respiration and Combustion.
(b) Glycolysis and Krebs cycle.
(c) Aerobic respiration and fermentation.
Answer:
(a) Differences between Respiration and Combustion used first, they decompose to produce glucose with the help of enzymes. When carbohydrates are not available, fat decomposes into fatty acid and glycerol. When both carbohydrates and fats are not available then protein decomposes to form glucose. Later glucose undergoes oxidation. Respiration is a catabolic process and the respiratory path is catabolic.

Fatty acid first converts into Acetyl Co-A, then enters into Krebs cycle. When fatty acid is to be synthesized in the living organisms Co-A is removed from the respiratory path. In this way respiratory path is used for synthesis (Anabolism) as well as decomposition (Catabolism) of fatty acid. Therefore respiratory pathway can be called as Amphibolic pathway.

(b) differences between Glycolysis and Krebs cycle

Glycolysis	Krebs cycle
1. It is the first step of respiration process.	It is the second step of respiration process.
2. It occurs in the cytoplasm part of the cell.	It occurs in the mitochondria part of the cell.
3. It occurs in aerobic and anaerobic respiration both.	It occurs in the aerobic respiration only.
4. It is linear pathway.	It is cyclic pathway.
5. One molecule of glucose breaks into 2 molecules of pyruvic acid.	One molecule of Acetyl CoA breaks into CO2  and water.
8 molecules of ATP are formed during this process.	30 molecules of ATP are formed during this process

(c) Differences between Aerobic respiration and Fermentation.

Aerobic  respiration	Fermentation
1. Complete oxidation of carbohydrates occurs.	incomplete oxidation of carbohydrates occurs.
2. It occurs in presence of oxygen.	It occurs in absence of oxygen.
3. Quantity of CO2 produced is more.	Quantity of CO2 produced is less.
4. It occurs inside the cell in the presence of specific enzymes.	It occurs outside the cell in the presence of zymase complex enzyme obtained from yeast.
5. It is not a biological process.	it is a biological process.

Question 5.
Respiration pathway is amphibolic. Discuss it. (NCERT)
Answer:
Generally glucose is used as respiratory substrate. If oligo or polysaccharides are used first, they decompose to produce glucose with the help of enzymes. When carbohydrates are not available, fat decomposes into fatty acid and glycerol. When both carbohydrates and fats are not available then protein decomposes to form glucose. Later glucos undergoes oxidation.

Respiration is a catabolic process and the respiratory path is catabolic. Fatty acid first converts into Acetyl Co-A, then enters into Krebs cycle. When fatty acid is to be synthesized in the living organisms Co-A is removed from the respiratory path. in this way respiratory path is used for synthesis (Anabolism) as well as decomposition (Catabolism) of fatty acid. Therefore respiratory pathway can be called as Amphibolic pathway.

Question 6.
Draw glycolysis through ray diagram.
Or
Explain glycolysis with the help of flow chart.
Answer:

Question 7.
What are the main steps of aerobic respiration? Where does it occur?
Answer:
Aerobic respiration is completed in two steps :

1. Glycolysis
2. Krebs cycle.

1. Glycolysis: It occurs in the cytoplasm.

2. Krebs cycle : (TCA OR TRICARBOXYLIC ACID CYCLE)
Pyruvic acid produced by glycolysis undergoes aerobic oxidation in the matrix of mitochondria through the TCA cycle. This cycle serves as a common pathway for carbohydrates, fats and proteins. Before participating in TCA cycle pyruvic acid enters a mitochon¬drion. Here, it is decarboxylated and the remaining 2-carbon fragment is combined with a molecule of coenzyme A to form acetyl CoA. This reaction is an oxidation reduction process and produces H ions and electrons along with carbon dioxide. During the process NAD is reduced to NADH₂.

Acetyl CoA is also formed from the beta-oxidation of fatty acids. Acetyl CoA acts as a connecting link between glycolysis and Krebs cycle. After the conversion of pyruvic acid into Acetyl CoA, acetyl CoA enters into Krebs cycle.

Steps of Krebs cycle are as follow :

1. Condensation: Acetyl CoA (2C compound) combines with oxaloacetic acid (4C compound) in the presence of condensing enzyme citrate synthetase to form citric acid (6C compound). It is the first product of Krebs cycle. CoA is released here.
Citrate synthetase

2. Dehydration: Citric acid in presence of aconitase enzyme is converted into cis- aconitic acid H20 is released here.

3. Hydration: Cis-aconitic acid is converted into isocitric acid with the addition of water in presence of enzyme aconitase.

4. Dehydration: Isocitric acid is oxidized into oxalosuccinic acid by reacting with NAD+ in presence of isocitrate dehydrogenase.

5. Decarboxylation: Oxalosuccinic acid converts into a-ketoglutaric acid by removal of one molecule of CO₂ in presence of carboxylase enzyme.

6. Dehydrogenation and Decarboxylation: a-ketoglutaric acid reacts with NAD⁺ and CoA and get oxidized into succinyl CoA in presence of a-ketoglutarate dehydrogenase enzyme. TPP, Lipoic acid and Mg⁺⁺ are also required for this reaction.

7. Formation of GTP: Succinyl CoA on hydrolysis produces succinic acid and CoA in presence of succinyl thiokinase enzyme. During this process one molecule of GTP is formed from GDP by using energy released during the reaction.

8. Dehydrogenation: Succinic acid is oxidized into fumaric acid by reacting with FAD in presence of succinyl dehydrogenase enzyme. FADH2 is produced in the reaction.
Succinyl dehydrogenase

9. Hydration: Fumaric acid reacts with one molecule of water in presence of fumarase enzyme and forms malic acid.

10. Dehydrogenation: At last Malic acid is oxidized by reacting with NAD in presence of Malate dehydrogenase enzyme to form oxaloacetic acid.

Oxaloacetic acid thus produced combines with Acetyl CoA to form citric acid which again enter into the cycle thus the cycle is repeated.

Question 8.
Explain electron transport chain?
Answer:
Electron transport chain: As you know that, H ions and electrons removed from the respiratory substrate during oxidation do not directly react with oxygen. Instead they reduce acceptor molecules NAD and FAD to NADH₂ and FADH₂ respectively. These molecules then transfer their electron to a system of electron acceptors and transfer molecules. The proteins of the inner mitochondrial membrane act as electron transporting enzymes.

They are arranged in an ordered manner in the membrane and function in a specific sequence. This assembly of electron transport enzymes is known as mitochondrial respiratory chain or the electron transport chain. Specific enzymes of this chain receive electrons (and in a few cases protons also) from reduced prosthetic groups, NADH₂ or FADH₂ produced by glycolysis and the TCA cycle. The electrons are then transported successively from enzyme to enzyme, down a descending ‘stairway ’of energy-yielding reactions.

The electron transport chain includes several cytochromes, such as: Cyt-Q, Cyt-b, Cyt-C_1, Cyt-c, Cyt-a and Cyt-a₃.
At the end of the chain the electrons and the accompanying protons (H+) are combined with O₂ to form water. Oxygen is thus the terminal electron acceptor of the mitochon¬drial respiratory chain.
Three molecules of ATP are formed from ADP by using energy liberated during various steps of electron transport. ATP molecules are produced during given steps :

1. When NADH₂ is oxidized to NAD by reacting with FAD.
2. When electron transfer from cytochrome-b to cytochrome-c₁.
3. When electron transfer from cytochrome-a to cytochrome-a₃.

Now it is clear that oxidation of one molecule of reduced NADH₂ or NADPH₂ results in the formation of 3 molecules of ATP while oxidation of FADH₂ leads to the formation of 2 molecules of ATP.
At each step of electron transport the electron acceptor has a higher electron affinity than the electron donor from which it receives the electron.

The energy from such electron transport is utilized in transporting protons (H⁺) from the matrix across the inner membrane to its outer side. This creates a higher proton concentration outside the inner membrane than in the matrix. The difference in proton concentration across the inner membrane is called proton gradient. Now it is clear that oxidation of one molecule or reduced NADH₂ or NADPH₂ results in the formation of 3 molecule of ATP while Oxidation of FADH₂ leads to the formation of 1 molecules of ATP.

Question 9.
Draw Krebs cycle.
Or,
Explain Krebs cycle with flow chart.
Answer:

Question 10.
What is Respiratory quotient? What is R.Q. value of fat? (NCERT)
Answer:
Respiratory quotient: The ratio of volume of CO₂ released to the volume of O₂ absorbed during respiration is called respiratory ratio or R.Q. The value of R.Q. depends upon the chemical nature of respiratory substrate

\(\text { R.Q. }=\frac{\text { Volume of } \mathrm{CO}_{2} \text { released }}{\text { Volume of } \mathrm{O}_{2} \text { absorbed }}\)
The value of R.Q. varies with substrate. Thus, the measurement of R.Q. gives an idea of the nature of respiratory substrate being respired in a particular tissue. R.Q. is usually measured by Ganong’s respirometer.

Fatty substances: R.Q. of fatty substance is always less than unity (0-7). e.g., Germinating seeds of castor, mustard, linseed, til, etc.
C₁₈H₃O6₂ + 26O₂ → 18CO₂ + 18H₂O
\(\mathrm{R.Q} .=\frac{18 \mathrm{CO}_{2}}{26 \mathrm{O}_{2}}=\frac{18}{26}=0 \cdot 7 \text { (less than unity). }\)

Plant Growth and Development Class 11 Important Questions Long Answer Type

Question 1.
Write differences between following : (NCERT)
(a) Aerobic and Anaerobic respiration.
(b) Glycolysis and Fermentation.
(c) Glycolysis and Citric acid cycle.
Answer:
(a) Differences between aerobic and anaerobic respiration

Aerobic Respiration	Anaerobic Respiration
1. Aerobic respiration requires oxygen and takes place in cytoplasm and mitochondria.	It does not require oxygen and takes place in cytoplasm only.
2. The substrate is completely oxidized.	The substrate is incompletely oxidized.
3. End products are CO2 and H2O.	End products are ethyl alcohol and CO2.
4. 38 ATP (673 kcal) energy is produced.	2 ATP (21 kcal) energy is produced.
5. Harmful substances are not formed.	Harmful substance ethyl alcohol is formed.

(b) Differences between Glycolysis and Fermentation

Glycolysis	Kerbs cycle
1. This process occurs inside the cytoplasm of the cell.	It occurs outside the cell.
2. It is a biochemical process.	2. It is a biochemical process.
3. Enzymes required for this process are found in the cytoplasm of the cell.	Zymase complex enzyme obtained from yeast cells help for this process.
4. Energy is released in the form of ATP.	Energy is released in the form of heat.

(c) Difference between Glycolysis and Citric acid cycle:

(a) Differences between Respiration and Combustion used first, they decompose to produce glucose with the help of enzymes. When carbohydrates are not available, fat decomposes into fatty acid and glycerol. When both carbohydrates and fats are not available then protein decomposes to form glucose. Later glucose undergoes oxidation. Respiration is a catabolic process and the respiratory path is catabolic.

Fatty acid first converts into Acetyl Co-A, then enters into Krebs cycle. When fatty acid is to be synthesized in the living organisms Co-A is removed from the respiratory path. In this way respiratory path is used for synthesis (Anabolism) as well as decomposition (Catabolism) of fatty acid. Therefore respiratory pathway can be called as Amphibolic pathway.

(b) differences between Glycolysis and Krebs cycle

Glycolysis	Krebs cycle
1. It is the first step of respiration process.	It is the second step of respiration process.
2. It occurs in the cytoplasm part of the cell.	It occurs in the mitochondria part of the cell.
3. It occurs in aerobic and anaerobic respiration both.	It occurs in the aerobic respiration only.
4. It is linear pathway.	It is cyclic pathway.
5. One molecule of glucose breaks into 2 molecules of pyruvic acid.	One molecule of Acetyl CoA breaks into CO2  and water.
8 molecules of ATP are formed during this process.	30 molecules of ATP are formed during this process

(c) Differences between Aerobic respiration and Fermentation.

Aerobic  respiration	Fermentation
1. Complete oxidation of carbohydrates occurs.	incomplete oxidation of carbohydrates occurs.
2. It occurs in presence of oxygen.	It occurs in absence of oxygen.
3. Quantity of CO2 produced is more.	Quantity of CO2 produced is less.
4. It occurs inside the cell in the presence of specific enzymes.	It occurs outside the cell in the presence of zymase complex enzyme obtained from yeast.
5. It is not a biological process.	it is a biological process.

Question 2.
What is your imagination while calculating number of ATP molecules pro reduced during oxidation of one molecule of glucose? (NCERT)
Answer:
(A) In glycolysis cycle :

Thus, on complete oxidation of one molecule of glucose, 38 molecules of ATP are produced.

Above calculation is done on the basis of imagination of following points:

• Oxidation of glucose is systematic metabolic path in which from one substrate another intermediate compound is formed starting from glycolysis, Kerb’s cycle then through electron transport system.
• NADH₂ produced during glycolysis enters into mitochondria and then into electron transport system to produce ATP by phosphorylation of ADP.
• Intermediate compounds does not help for formation of any other compound.
• Only glucose is used as substrate for respiration.
  No any other compound can enter into intermediary path unless it converts into glucose.

Plant Growth and Development Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
Glycolysis takes place in :
(a) Cytoplasm
(b) Chloroplast
(c) Ribosomes
(d) Mitochondria.
Answer:
(a) Cytoplasm

Question 2.
How many molecules of ATP are formed in Kreb’s cycle :
(a) 28
(b) 18
(c) 30
(d) 8.
Answer:
(c) 30

Question 3.
How many molecules of ATP are formed from the anaerobic respiration of one molecule of glucose :
(a) 8
(b) 16
(c) 2
(d) 14.
Answer:
(c) 2

Question 4.
What is ATP:
(a) Oxidative enzyme
(b) A hormone
(c) A protein
(d) Molecule of high energy phosphate bonds.
Answer:
(d) Molecule of high energy phosphate bonds.

Question 5.
ATP synthesis in mitochondria require :
(a) o2
(b) NADP
(c) FMN
(d) Pyruvic acid.
Answer:
(a) o2

Question 6.
Alcohol is formed in :
(a) Aerobic respiration
(b) Anaerobic respiration
(c) Photosynthesis
(d) Photorespiration.
Answer:
(b) Anaerobic respiration

Question 7.
Krebs cycle takes place in :
(a) Vesicles of E. R.
(b) Matrix of mitochondria
(c) Dictyosomes
(d) Lysosornes.
Answer:
(b) Matrix of mitochondria

Question 8.
Flowers and fruits maintain their flavour and taste when kept in refrigerators because:
(a) Deficiency of O₂
(b) Respiration is inactive
(c) CO₂ is collected
(d) H₂O is converted into ice.
Answer:
(b) Respiration is inactive

Question 9.
End-product of glycolysis is :
(a) Oxalo acetic acid
(b) Acetic acid
(c) Pyruvic acid
(d) Malic acid.
Answer:
(c) Pyruvic acid

Question 10.
End-product of anaerobic respiration in our muscles is :
(a) Malic acid
(b) Lactic acid
(c) Citric acid
(d) Alcohol.
Answer:
(b) Lactic acid

Question 11.
End product of Krebs cycle is :
(a) CO₂ and H₂O
(b) H₂O and Citric acid
(c) H₂O and OAA
(d) HO₂ and NADPH₂.
Answer:
(a) CO₂ and H₂O

Question 12.
Another name of TCA cycle is :
(a) Krebs cycle
(b) Calvin’s cycle
(c) EMP
(d) Hatch and Slack cycle.
Answer:
(a) Krebs cycle

Question 13.
EMP Energy pack of a living cell is :
(a) Cytochrome
(b) ADP
(c) ATP
(d) Chlorophyll.
Answer:
(c) ATP

Question 14.
Addition of phosphate molecule to ADP to form ATP is called as :
(a) Protein synthesis
(b) Photosynthesis
(c) Phosphorylation
(d) Pinocytosis.
Answer:
(c) Phosphorylation

Question 15.
In which process oxidative phosphorylation occurs :
(a) Protein synthesis
(b) Nitrogen fixation
(c) Respiration
(d) Transpiration.
Answer:
(c) Respiration

Question 16.
Photorespiration is characteristic feature of which type of plant:
(a) C₃
(b) C₄
(c) CAM
(d) None of these.
Answer:
(a) C₃

Question 17.
How much quantity of energy is released during aerobic respiration of one molecule of glucose:
(a) 637 k. cal
(b) 640 k. cal
(c) 673 k. cal
(d) 693 k. cal.
Answer:
(c) 673 k. cal

Question 18.
Link between glycolysis and Krebs cycle is :
(a) Citric acid
(b) Malic acid
(c) Fumaric acid
(d) Acetyl Co-A.
Answer:
(d) Acetyl Co-A.

Question 19.
What is essential for both photosynthesis and respiration :
(a) Chlorophyll
(b) CO₂
(c) Water
(d) Cytochrome.
Answer:
(d) Cytochrome.

Question 20.
How much quantity of energy is released on hydrolysis of ATP into ADP :
(a) 120 cal
(b) 1200 cal
(c) 12000 cal
(d) None of these.
Answer:
(c) 12000 cal

Question 21.
R.Q. of germinating castor seed is :
(a) One
(b) Less than one
(c) More than one
(d) Zero.
Answer:
(b) Less than one

Question 22.
Cytochrome contains:
(a) Mg
(b) Fe
(c) Mn
(d) Cu.
Answer:
(b) Fe

Question 23.
How many molecules of ATP are formed during glycolysis :
(a) 0
(b) 2
(c) 4
(d) 8.
Answer:
(d) 8.

Question 24.
Photorespiration is related to :
(a) Glyoxysome
(b) Lysosome
(c) Mesosome
(d) Ribosome.
Answer:
(a) Glyoxysome

Question 25.
Electron transport system is found in which part of the mitochondria :
(a) Outer membrane
(b) Inner cristae
(c) Inner membrane
(d) Inner membrane sac.
Answer:
(c) Inner membrane

Question 26.
Meaning of photophosphorylation is :
(a) Formation of ATP from ADP
(b) Formation of NADP
(c) Formation of ADP from ATP
(d) Formation of PGA.
Answer:
(a) Formation of ATP from ADP

Question 27.
In glycolysis process which substance converts into end product:
(a) Protein into glucose
(b) Glucose into fructose
(c) Starch into glucose
(d) Glucose into pyruvic acid.
Answer:
(d) Glucose into pyruvic acid.

Question 28.
Power house of the cell is :
(a) Golgi complex
(b) Ribosome
(c) Mitochondria
(d) Lysosome.
Answer:
(c) Mitochondria

Question 29.
Enzymes involved in glycolysis are found in the :
(a) Mitochondria
(b) Cytoplasm
(c) Mitochondria and cytoplasm
(d) Vacuole.
Answer:
(b) Cytoplasm

Question 30.
Which of the following can respire without oxygen :
(a) Root
(b) Seed
(c) Stem
(d) Leaf.
Answer:
(b) Seed

2. Fill in the blanks:

1. C₆H₁₂O₆ + ………….. → 6CO + 6H₂O + ………….. ATP.
Answer:
6O₂, 38,

2. R.Q. of organic acid is always ………….. .
Answer:
More than one,

3. C₆H₁₂O₆ ………….. + 2CO₂.
Answer:
2C, H₅OH,

4. Respiratory quotient is measured by ………….. .
Answer:
Respirometer,

5. Krebs cycle occurs in the ………….. part of the cell.
Answer:
Mitochondria,

6. Energy coin of the cell is ………….. .
Answer:
ATP,

7. In ………….. respiration incomplete oxidation of food occurs.
Answer:
Anaerobic,

8. End product of glycolysis is ………….. .
Answer:
Pyruvic acid,

9. Fermentation is done by ………….. enzyme of yeast cell.
Answer:
Zymase,

10. ………….. is a link between glycolysis and Kerbs cycle.
Answer:
Acetyl CoA,

11. ………….. is oxidised during respiration process.
Answer:
Glucose,

12. The arrangement of electron transport enzymes through which electron transfer from one electron acceptor to other is called as ………….. .
Answer:
Electron transport system.

3. Match the following:

(A)

Column ‘A’	Column ‘B’
1. Glycolysis	(a) Electron transport system
2. Krebs cycle	(b) Cytoplasm
3. Photorespiration	(c) Mitochondria
4. ADP + Pi → ATP	(d) C3 plants
5. Cytochrome	(e) Phosphorylation.

Answer:
1. (b) Cytoplasm
2. (c) Mitochondria
3. (d) C₃ plants
4. (e) Phosphorylation.
5.  (a) Electron transport system.

(B)

Column ‘A’	Column ‘B’
1. Carbohydrates	(a) R.Q, – less than one
2. Protein	(b) R.Q.- Unity (1)
3. Organic acid	(c) R.Q. – 0
4. Fat	(d) R.Q. – 0.7
5. Succulent plants	(e) R.Q. – more than one.

Answer:
1. (b) R.Q.- Unity (1)
2. (a) R.Q, – less than one
3. (e) R.Q. – more than one.
4. (d) R.Q. – 0.7
5. (c) R.Q. – 0

4. Answer in one word:

1. What is net gain of energy from oxidation of one mole of glucose?
Answer:
673 Kcal in aerobic respiration and 21 kcal in anaerobic respiration,

2. Where does anaerobic respiration occur in cell?
Answer:
Cytoplasm,

3. Name the organelle structure where electron transport chain occurs?
Answer:
Inner membrane of mitochondria,

4. Where does glycolysis takes place?
Answer:
Cytoplasm,

5. What is ATP?
Answer:
Energy-rich compound which binds phosphate,

6. Which process is responsible for alcohol formation?
Answer:
Anaerobic respiration,

7. Where does Krebs cycle takes place?
Answer:
In the matrix of mitochondria,

8. How many molecules of ATP are formed during glycolysis?
Answer:
8 ATP,

9. In which form energy is released during respiration?
Answer:
ATP.

Students get through the MP Board Class 11th Biology Important Questions Chapter 8 Cell: The Unit of Life which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Questions Chapter 8 Cell: The Unit of Life

Cell: The Unit of Life Class 11 Important Questions Very Short Answer Type

Question 1.
What is the mesosome of prokaryotic cells? Give its functions.
Answer:
Mesosome is a special fype of membrane produce by the growth of plasma membrane of prokaryotic cell.
Functions of Mesosome:

• Formation of cell wall.
• DNA replication and distribution in the daughter cells.
• Help in cellular secretion and respiration.

Question 2.
Give characteristics of prokaryotic cell.
Answer:
Characteristics of prokaryotic cells :

• Primitive type of cells.
• True nucleus is not found in them.
• Membrane bound cytoplasmic organelles are not found in them.
• 70 S type of ribosomes are found in them.
  e.g. Blue green algae, Bacteria.

Question 3.
What are semiautonomous cell organelles ? Give at least one example.
Answer:
Organelles which can synthesize their protein can reproduce itself and have its own genetic material (DN A and RNA) are called as semiautonomous cell organelles, e.g., Chloroplast, mitochondria, etc.

Question 4.
Why lysosomes are called as the pocket of suicidal bag?
Answer:
In certain pathological conditions, the lysosomes start digesting the various organelles of cell and process is called autolysis. During this process, membrane of lysosome ruptures and enzymes are released into the cell which destroys and decomposes its own contents thus acting as suicidal bag.

Question 5.
Describe the significance of diffusion.
Answer:
Significance of diffusion :

• It involves in the gaseous exchanges during photosynthesis and respiration.
• During transpiration water is evaporated by the process of diffusion.
• Diffusion plays a vital role in the transport of substances in plants.
• It regulates body temperature.

Question 6.
What happens when R.B.Cs. are kept in hypertonic solution?
Answer:
When R.B.Cs. are placed in a hypertonic solution exosmosis of water takes place and they become plasmolysed. In other words, exosmosis causes shrunken appearance or crenation. A solution of NaCl having more than 0.9% concentration is hypertonic because their osmotic pressure is greater than the cytoplasm of the R.B.Cs.

Question 7.
What are nuclear pores? Give its functions.
Answer:
In some parts of nuclear membrane pores or gaps are found called as nucleopores or nucleus pore. It is formed by fusion of outer and inner nuclear membrane.
Function of nucleopores: RNA and proteins from the nucleus reaches to the cytoplasm through these pores.

Question 8.
How does neutral solvent passes through plasma membrane? Does polar solvent passes through plasma membrane in same way? If not then how does it transport it?
Answer:
Neutral solvent passes through plasma membrane according to concentration gradient, i.e. from higher concentration to lower concentration by diffusion process.
No, polar solvent can not pass through plasma membrane in same way. Since polar solvent can not pass through non-polar lipid, It binds with carrier protein to pass through plasma membrane.

Cell: The Unit of Life Class 11 Important Questions Short Answer Type

Question 1.
Define prokaryotic cells and eukaryotic cells with examples.
Answer:
Prokaryotic cells :
These are the primitive and underdeveloped cells. They do not have nucleolus and nuclear membrane. This type of nucleus is called as incipient nucleus. These cells do not possess double membraned cell organelles like mitochondria, chloroplast, golgi body and lysosomes. The genetic material is found only in the form of circular DNA molecule suspended in the cytoplasm itself. Cells bear only ribosomes to synthesize protein in the cytoplasm, e g., Bacteria and blue-green algae.

Eukaryotic cells :
These are the well-organized and well-developed cells. These cells contain nuclear membrane which surrounds the genetic material to form nucleus. Membrane bound cell organelles like mitochondria, golgi body, chloroplast are also present, e.g., cells of all organisms except bacteria and blue-green algae.

Question 2.
Name the scientist who discovered the cell, nucleus, pre-existing cell and organism composed of cell.
Answer:

• Cell was discovered by Robert Hooke in 1665.
• Nucleus was discovered by Robert Brown in 1831.
• Rudolf Virchow (1885) discovered that cell exists from pre-existing cell by division.
• Organism composed of cell was discovered by Theodor Schwann in the year 1839.

Question 3.
Where is energy stored in the cell?
Answer:
Energy of the cell is stored in the high energy bonds of phosphate groups present in ATP (Adenosine triphosphate). After breakage these bonds release energy.

Question 4.
What is totipotency?
Answer:
The capability or potential of every cell of a plant to develop individually into a new plant is known as totipotency and the cells are called as totipotent cells. This idea was first proposed by Haberlandt in the year 1902. Later in 1950 Stecoard developed a new plant from the cells of the root of carrot plant. This technique is nowadays utilized in developing many plants.

Question 5.
How flow of information take place inside the cell? Explain.
Answer:
Flow of information : DNA is the hereditary material of most of the cells. The genetic informations are coded in the DNA. This DNA molecule synthesize RNA by the process of transcription. RNA can synthesize protein through the process of translation.

Question 6.
Draw a well labelled diagram of a prokaryotic cell.
Answer:

Question 7.
What do you understand by cell theory? Explain.
Answer:
In 1838 and 1839, Schleiden and Schwann proposed a theory about cell which is known as cell theory. The main postulates of the cell theory are as follows :

• Each living body is made up of single or many cells.
• New cells arise from pre-existing cells.
• Structural composition of all cells is basically similar and hence they perform similar metabolic activities.
• The activities shown by an organism is the result of total sum of the activities performed by the cell of its body.

Question 8.
Define cell.
Answer:
According to cell theory Schleiden and Schwann, “Cell is the structural and functional unit of all organisms”. In other words, cell is the smallest living unit which is capable of maintaining its independent existence.
In 1963, Loewy and Siekevity defined cell as, “Cell is a biological unit delimited by a semipermeable membrane and capable of self-reproduction in a medium free of other living system.

Question 9.
What is hybridization? What are its advantages?
Answer:
Hybridization is a technique in which the protoplasm of organisms of different species with different genetic constituency are fused to get a new species. In 1977, Y.P.S. Bajaj obtained a new variety of plant by fusing the cells of Petunia and tobacco.
Advantages :

• New improved varieties can be produced in plants.
• Plants of useful breed can be fielded by this technique.

Question 10.
Write down the advantages of Multicellularity.
Answer:
Advantages of multicellularity :

• Multicellular organisms are adapted in better way by division of labour as cells of these organisms organize into kinds of tissues and organs to perform specialized functions such as respiration, digestion, excretion, reproduction, etc.
• In multicellular organisms, in many activities number of cells work in coordination. e.g., heart pumps the blood by well coordinated movement of all its muscles.
• In a multicellular organism, cell has a dual existence as an individual and as a part of community.
• The life span of multicellular organisms is comparatively greater than unicellular organisms due to following reasons :
  • They can work more efficiently due to division of labour.
  • In spite of death of few cells the organism remains alive.
  • They have capacity of replacement of dead cells.
  • They have capacity to adapt themselves according to its environment.
  • They have various developed systems for performing different functions.

Question 11.
Draw a well labelled diagram of PPLO cell.
Answer:

Question 12.
Write down the difference between :
(a) Hydrophilic and Hydrophobic molecule
(b) Passive and Active transport.
(c) Pinocytosis and Phagocytosis.
Answer:
(a) Difference between hydrophilic and hydrophobic molecule :
Molecules which has a very strong liking for water are called as hydrophilic molecules, whereas the molecules which has fear from water molecules are called as hydrophobic molecules. Ex. Bilayered lipid of plasma membrane consists of lipid molecules with two ends hydrophilic and hydrophobics ends. The hydrophilc ends of phospholipid molecules face each other where as hydrophilic ends face the outer protein layer.

(b) Difference between passive and active transport:
Passive transport is kind of physical process and does not involve neither movement of cell nor consumption of energy. e.g., Transport by diffusion and osmosis.
Active transport is the movement of ions or molecules across plasma membrane against concentration gradient. The transport of molecules against concentration gradient involves consumption of energy, e.g., Transport of K⁺ through plasma membrane.

(c) Difference between pinocytosis and phagocytosis :
Pinocytosis is non-specific intake by a cell of a tiny droplets of extracellular fluid which cannot otherwise pass through the cell membrane. It is also called cell drinking.
In this process, a small region of plasma membrane invaginates and a fluid droplet passes into the pocket so formed. The pocket deepens and finally nips off as a fluid-filled vacuole, the pinocytotic vesicle or pinosome.

Phagocytosis :
The process of engulfing large sized particles of solid food by cell through the plasma membrane is known as phagocytosis. This process is easily seen in protozoa and in certain cells of metazoa.

Question 13.
Write down the functions of Active Transport.
Answer:
Functions of Active Transport :

• Active transport helps in maintaining a definite ion concentration and definite osmotic pressure within the living system.
• It helps in maintaining water and ionic balance between cell and extracellular fluids.
• Active transport also helps in maintaining membrane potential by keeping the inner side of the membrane relatively electronegative to its outer side.
• It helps for transport of nutrient materials at higher rate.
• Harmful substances are removed from the cell due to this process.
• As this process is selective, it helps in maintaining the chemical composition of the protoplasm.

Question 14.
Explain the importance of osmosis.
Answer:
Importance of osmosis :

• Water is absorbed by root hairs from soil by the process of osmosis,
• Different organs of plant show growth and turgidity only through osmosis,
• The opening and closing of stomata takes place through endosmosis and exosmosis,
• Water moves in the plant body from cell to cell through osmosis.

Question 15.
Why did shrunken grapes or raisins become swollen when kept in water?
Answer:
When shrunken grapes are placed in water, they absorb water by the process of endosmosis through plasma membrane because of their higher concentration. Here plasma membrane acts as semipermeable membrane.

Question 16.
What is cell-wall, explain it?
Answer:
Cell wall is thick, rigid, non-living envelop which surrounds the plasma membrane. It is composed of network of microfibrils embedded in a gel matrix. The microfibrils are mainly cellulosic in nature, but in bacteria and blue green algae, they are formed of protein and polysaccharides. The gel matrix contains protein, pectin, hemicellulose and lignin. The cell wall is found in plant cells and in bacterial cells.

Question 17.
Write down two chemical components of plasma membrane.
Answer:
Protein and lipid.

Question 18.
Write down the difference between:
(i) Exosmosis and Endosmosis,
(ii) Exocytosis and Endocytosis,
(iii) Cell wall and Cell membrane,
(iv) Semipermeable and Selective permeable membrane
Answer:
(1) Differences between Exosmosis and Endosmosis:

Exosmosis	Endosmosis
1. Water moves towards outside the solution.	Water moves towards inside the solution.
2. This process is completed in hypertonicsolution.	This process is completed in hypotonie solution.
3 The turgor pressure of the cell is decreased and DPD is increased.	The turgor pressure of the cell is increased and DPD is decreased.
4. Plasmolysis occurs in the cell.	Plasmolysis does not occur in the cell.

(ii) Difference between Exocytosis and Endocytosis : The secretory products moved
outside the cell cytoplasm is called as Exocytosis. This process requires energy from ATP.
e.g., Transpiration, where as in Endocytosis is the process of engulfing large sized particles
of food substances, e.g., movement of cell and cell sap.

(iii) Differences between Cell wall and Cell Membrane:

Cell wall	Cell membrane
1. It is the outermost layer or boundary of cell.	It is the outermost layer or boundary of cytoplasm,
2. It is found only in the plant cells.	It is found in all living cells.
3. It is made up of cellulose and lignin.	It is made up of lipid and proteins.
4. It is hard.	It is thin and elastic.
5. It is permeable.	It is selectively permeable.

(iv) Semipermeable and Selective permeable membrane: Membrane which allow only the solvent to pass through it not the solute are called as semipermeable membrane, where as selective permeable membrane allow only certain selective ions and molecules to pass through it.

Question 19.
Which membrane is known as “Protein iceberg in a sea of lipids” and why?
Answer:
According to Fluid mosaic model, plasma membrane is made up of lipid and protein, in which lipid is present at normal temperature (37°C) in the form of fluid in which solid particles of protein floats. Thus plasma membrane is known as “Protein iceberg in a sea of lipids”.

Question 20.
What are raphides?
Answer:

Raphides are excretory materials of plants which are found in the form of crystals of calcium oxalate.
When calcium oxalate crystals are needle shaped they are called as raphides e.g.,
Pistia. If calcium oxalate crystals are more or less star shaped, they are called as Sphaero raphides e.g., Opuntia.

Question 21.
Write the functions of plasma membrane.
Answer:
Functions of plasma membrane :

• It provides shape of the cell.
• It protects the cell from external forces.
• It helps in the cellular exchange.
• It forms organelles of the cell.
• The inpushing help intake of materials.

Question 22.
Describe plasmolysis.
Answer:
Plasmolysis:
When a living cell is placed in a hypertonic solution then exosmosis of water takes place hence the water of cell comes out into outer solution. Due to exosmosis of water the protoplasm shrinks away from the cell wall and an irregular mass at the centre. This shrinkage of protoplasm is known as plasmolysis.

When plasmolysed cell is placed in water, the water enters into the cell sap, the cell becomes turgid and the protoplasm again assumes its normal sap and position. This phenomenon is called as deplasmolysis.

Question 23.
Write down the difference between Diffusion and Osmosis.
Answer:
Differences between Diffusion and Osmosis :

Diffusion	Osmosis
1. It is the movement of solute particles from higher concentration to lower concentration.	It is the movement of solvent from the region of lower concentration to higher concentration across a semipermeable membrane.
2. It occurs in solids, liquids and gases.	It occurs only in liquids.
3. Semipermeable membrane is not required for this process.	This process requires semipermeable membrane.
4. Diffusion is physical process.	Osmosis is a vital process.

Question 24.
Give exceptions of cell theory.
Answer:
Exception of cell theory :

• Viruses do not have cellular struture.
• In RBCs aerobic respiration do not occur.
• RBCs are incomplete cells, as they do not have nucleus, mitochondria, endoplasmic reticulum etc.
• Well developed neuron can not show cell division.
• Inspite of capacity cells of liver and muscles do not shows cell division but can regenerate lost or damaged part.

Question 25.
Define Bio-membrane.
Answer:
The cytoplasm of every cell is enclosed by a layer of lipo-protein membrane called as plasma membrane. The plasma membrane and the subcellular membranes are collectively known as Biological membranes or Bio-membrane.

Question 26.
Sometime plants wilts after addition of fertilizers to the soil. Why?
Answer:
If excess quantity of fertilizers added to the soil then concentration of soil become more as compared to cell-sap of rootcells, thus water from the plantcells is drawn out by exosmosis process and the plant wilts.

Question 27.
What happens when RBCs are kept in hypotonic solution.
Answer:
When RBCs are kept in hypotonic solution endosomosis occurs, If the concentration of hypotonic solution is less than 0-9%, then water or solvent continously enters into RBCs due to which cells of RBCs may burst.

Question 28.
What are trophoplasm?
Answer:
Part of the cytoplasm which contain metabolically active organelles like mitochondria, golgi complex, plastids, endoplasmic reticulum etc. is called as trophoplasm.

Question 29.
Both Lysosomes and vacuoles are bounded by single unit membrane but functionally both are different. Write a note on it.
Answer:
Lysosomes are formed from gogli complex. It consists of many spherical tiny bags filled with hydrolytic enzymes like hydrolases, lipases, proteases etc. All these enzymes become active in acidic medium and help for digestion of carbohydrates, lipids, proteins and nucleic acids etc.

Where as Vacuoles are also bounded by single membrane called as Tonoplast, which encloses excess water, excretory materials and cellular products. Ions and other substances enters into vacuole through tonoplast against concentration gradient in plant cells. Thus concentration of vacuoles is always greater than the concentration of cytoplasm.

Question 30.
Describe division of labour in multicellular organisms.
Answer:
Multicellular organisms are made up of different types of cells. In multicellular organisms, cells are interdependent on each other for their survival. They interact and cooperate with one another. In a multicellular organism different functions are performed by different tissues. This is known as division of labour whereas in unicellular organism, all the vital functions are performed by a single cell.
Division of labour in multicellular organisms can be explained by following points :

• Some cells produces additional substances which help to connect different organs with each other.
• Some cells help to carry impulses.
• In multicellular organisms different life processes like respiration, excretion and circulation etc. are performed by different organ systems formed by different types of tis¬sues.
• Many cells dies everyday in the body but new cells are continuously formed in the body. For ex. life span of RBCs is 120 days but new RBCs are continuously formed by bone marrow.

Thus worn out and death of cells in multicellular oiganism does not cause any harm to the organism because new cells are regularly formed in them,

Cell: The Unit of Life Class 11 Important Questions Long Answer Type

Question 1.
Give names of two cytoplasmic organelles which are surrounded by double membrane. Give characteristics and functions with labelled diagram.
Answer:
Two cytoplasmic organelles surrounded by double membrane are :
(i) Golgi body,
(ii) Endoplasmic reticulum.
(i) Golgi body :
Golgi bodies are found in all eukaryotic cells except RBCs. Golgi body is made up of double layered unit membrane. Under electron microscope, three components are seen:
1. Cisternae : These are flat sacs having 4 to 7 in numbers. Each sac is 60A thick.
2. Vesicles : These are produced by budding or constriction of sacs and are 60n in diameter.
3. Vacuoles : These develop over the body as rounded structure.

Functions of Golgi body : Some of the most important functions of Golgi body are the following:

• Formation of secretory vesicles.
• Formation of complex molecules of carbohydrate from simple sugars.
• Formation of cell wall and cell plate in plant cell during cell division.
• They form lysosomes which form acrosome after modification.
• They form glycoprotein by aggregation of carbohydrates and proteins.
• Golgi body also helps in the secretion of hormones from endocrine cells.
• It stimulates mitochondria to produce ATP.
• Formation of plasma membrane.

(ii) Endoplasmic reticulum :
Under electron microscope cytoplasmic matrix is seen to be traversed by a complex network of interconnecting membranes of different shapes and sizes. Since this network is concentrated in the endoplasm it is known as endoplasmic reticulum (ER). The endoplasmic reticulum was reported for the first time by Porter and Kallman in 1945.

Endoplasmic reticulum is a complex vacuolar system extending from the nucleus throughout cytoplasm to the cell wall. This vacuolar system is nothing else but spaces enclosed by double membrane.
Types of ER: Depending upon the nature of its membranes, ER is of two main types, i.e., smooth and rough surface.
(i) Smooth surface ER or Agranular ER:
It is formed of smooth membranes. There are no ribosomes attached to its membranes. Smooth surface ER is found in those cells which are almost inactive in protein synthesis. It is generally found in adipose cells, interstitial cells, glycogen storing cells of liver, leucocytes, mature spermatocytes and retinal cells.

(ii) Rough surface ER or Granular ER:
The granular or rough surface endoplasmic reticulum possesses rough walls because the ribosomes are attached with its membranes. Granular ER is found in those cells which are actively involved in protein synthesis, e.g., pancreatic cells, plasma cells, goblet cells and liver cells.

Structure of ER :
Endoplasmic reticulum consists of membrane lined channels or space containing a fluid called endoplasmic matrix. It is composed of three kinds of structures :

Heterochromatin is the dark stained, condensed end region of chromatin fibres whereas ettchromatin is the light stained and diffused regions of the chromatin. Nucleus also contains a large, spherical acidophilic dense granules called as nucleolus.
Functions of nucleus :

• Nucleus regulates all the activities of the cell.
• It synthesizes all types of cellular RNA, which are necessary forprotein synthesis.
• It plays an important role in cell division and controls growth of the cell.
•  It contains genes and chromosomes which carries genetic information from generation to generation.

(ii) Centriole or Centrosome : Centrosomes are found in all the animal cells except mature RBCs but absent in plant cells.
Structure :
Centrioles and basal bodies are cylindrical structures which are 0.15 to 0.25 μ in diameter and usually 0.3-0.7 μ in length. The wall of each centriole and also of the basal body is formed of nine groups of triplet microtubules or fibres. These are equally spaced on the periphery of imaginary cylinder the space between and around the triplet microtubules is filled with the amorphous electron dense material.

All the nine triplet microtubules which form the walls are identical. Each microtubule is formed of three subtubules or subfibrils. The innermost subunit is designated by A, the middle one by B and the outer one by C. Only the A subunit tubule is round, the others are partial and share their wall with the preceding tubule. Often the triplets are thought to run parallel to one another and to the long axis of the cylinder but this is not always true.

The A tube of each triplet is linked with the C-tube of the neighbouring triplet by a dense link. A-C linkers are responsible for the radial tilt of the triplets. At
(a) Cisternae:
These are elongated, flattened sac like unbranched tubules arranged in parallel rows having the diameter of 40 to 50 mμ Cisternae are found in the cells of liver, pancreas, notochord and brain which are actively involved in protein synthesis.
(b) Vesicles :
These are round, ovoid membrane bound vacuolar spaces having the diameter of 25 to 500 mμ. They are also called microsomes. They generally remain isolated in the cytoplasm and occur in most cells but especially abundant in the pancreatic cells.

(c) Tubules: Tubules are small, smooth walled, branched, tubular spaces of 50-100 μ diameter. These are found in the cells which are engaged in the synthesis of steroids like cholesterol, glycerides and hormones.
Functions of Endoplasmic Reticulum:

• It forms ultrastructural framework in the cytoplasm and provides mechanical support to the cell.
• Rough endoplasmic reticulum provides surface for the attachment of ribosome, which are involved in protein synthesis thus indirectly they help for protein synthesis.
• It helps in intracellular transport.
• During cell division, it helps for formation of nuclear membrane.
• It stores protein.
• It helps for exchange of material between inner and outer part of the cell.
• It provides surface for various enzymatic activities.
• Smooth endoplasmic reticulum helps in the lipid and glycogen synthesis with the help of enzymes stored in it.
• In liver cells, it helps in glycogenolysis.

Question 2.
Describe structure of following with labelled diagrams :
(i) Nucleus,
(ii) Centrosome.
Answer:
(i) Nucleus :
All eukaryotic cells normally contain a rounded, circular or oval structure which is called nucleus. It was discovered by Robert Brown in 1831. According to Beller, “Nucleus is a structure surrounded by cytoplasm and which produces chromosomes during cell division”.
Each nucleus is surrounded by a bilayered cell membrane which contains numerous pores.
The outer membrane of nucleus possesses numerous ribosomes. Each nucleus is filled with nucleoplasm, which contains a network of chromatin. During cell division these chromatin fibres condense to form chromosomes. In nucleus two types of chromatin have been identified:

1. Heterochromatin and
2. Euchromatin.

the proximal end of centrioles and basal bodies is found a cartwheel structure composed of central rod or hub in the centre and radiating from it are nine spokes. These spokes end in the dense feet. Cells of plants lack centrioles.
Functions of centriole :

• During the cell division they help in the formation of spindles.
• Centriole serve as foci for the production of new centrioles and basal bodies.
• The microtubules of cilia and flagella originate and bear by basal bodies.

Question 3.
“Cell is the basic unit of life”. Explain the statement.
Answer:
Each cell is an autonomous unit. It independently carries out all fundamental processes like respiration, excretion, energy utilization, etc. It oxidizes food molecules to release energy to perform its functions like synthesis of materials for body movement, secretion and transport. Cell exchange gases like O₂ and CO₂. It synthesizes DNA for its duplication. The cell maintains its internal physico-chemical conditions. Each cell has its own life span.

The above facts will prove cell as self-contained autonomous unit.
In multicellular organisms cells are interdependent on each other for their survival. They interact and co-ordinate with one another, but they shows the capacity of independent existance and multiplication. This can be demonstrated by isolating cells of multicellular organism and growing them in a culture medium under controlled conditions, then like unicellular orgnanisms these cells can perform all life activities independently control and coordination is done by the materials present within the cell.

Each cell is found surrounded by plasma membrane, which encloses protoplasm, thus separate protoplasm from the environment and acts as an indepdent unit. Thus cell is called as “basic unit of life”.

Question 4.
Write differences between Prokaryotic and Eukaryotic cells.
Or,
Write five differences between Prokaryotic and Eukaryotic cells on the basis of complexity.
Answer:

Question 5.
Distinguish between Plant and Animal cell.
Or,
Write five differences between Plant cell and Animal cell.
Answer:
Differences between Plant and Animal cells

Question 6.
What is centromere? How classification of chromosomes is done on the basis of location of centromere? Describe with diagrams.
Answer:
Centromere :
The chromatids of chromosomes are held together at a point is called Centromere. It divides chromosome into two equal or unequal halves.
There are four types of chromosomes on the basis of position of centromere.

1. Acrocentric : These are rod shaped chromosomes which has terminal centromere, so they possess only one arm.
2. Telocentric : These are rod shaped chromosomes having subterminal centromere. One arm of the chromosome is very long and other is very short.
3. Sub-metacentric : These are ‘L’ shaped chromosomes with centromere slightly away from the mid point so that two arms are unequal.
4. Metacentric : These are ‘V’ shaped chromosomes in which centromere is located at the middle of the chromosome so that the two arms are almost equal.

Cell: The Unit of Life Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
Cell is discovered by :
(a) Robert Hooke
(b) Robert Brown
(c) Schleiden
(d) Schwann.
Answer:
(a) Robert Hooke

Question 2.
Nucleus is discovered by :
(a) Schleiden
(b) Schwann
(c) Robert Brown
(d) Mendel.
Answer:
(c) Robert Brown

Question 3.
Cell theory was proposed by :
(a) Schleiden and Schwann
(b) Lamarck and Traviranus
(c) Muir and associates
(d) Maheshwari and Guha.
Answer:
(a) Schleiden and Schwann

Question 4.
Scientist associated with totipotency is :
(a) Lamarck
(b) Haberlandt
(c) Schleiden
(d) Schwann.
Answer:
(b) Haberlandt

Question 5.
Formation of a well developed organism from a vegetative cell is called :
(a) Somatic hybridization
(b) Somatic reproduction
(c) Totipotency
(d) Plasmolysis.
Answer:
(c) Totipotency

Question 6.
Cell is an autonomous unit because :
(a) It possesses the capacity of locomotion
(b) It can reproduce
(c) It contains nucleus
(d) It possesses the capacity of regulation and direction of all vital processes.
Answer:
(d) It possesses the capacity of regulation and direction of all vital processes.

Question 7.
Prokaryotic cell does not contain :
(a) Cell wall
(b) Nuclear membrane
(c) Plasma membrane
(d) Vacuole.
Answer:
(b) Nuclear membrane

Question 8.
Which prokaryotic cell having very minute cell structure :
(a) Rhizobium
(b) Mycoplasma
(c) Nostoc
(d) Bacillus.
Answer:
(b) Mycoplasma

Question 9.
Which statement is not correct:
(a) Cell was discovered by Robert Brown.
(b) Cell theory was proposed by Schleiclen and Schwann.
(c) According to Virchow new cells develop from the division of pre-existing cells, [omnis cellula e cellula]
(d) Unicellular organisms perform all the activites in single cell.
Answer:
(a) Cell was discovered by Robert Brown.

Question 10.
New cells are formed :
(a) By bacterial fermentation
(b) By reproduction of old cells
(c) From pre-existing cells
(d) From non-living substances.
Answer:
(c) From pre-existing cells

Question 11.
Which statement is correct:
(a) All genes are found in the nucleus
(b) Plasma membrane is not found in both plant and animal cell
(c) Membrane bound organelles are found, in the cell membrane of prokaryotic cell.
(d) Production of cell occurs from non-living materials.
Answer:
(a) All genes are found in the nucleus

Question 12.
Ingestion of solid particles by plasma membrane is called :
(a) Pinocytosis
(b) Exocytosis
(c) Endocytosis
(d) Phagocytosis.
Answer:
(d) Phagocytosis.

Question 13.
What is cell omitting :
(a) Active transport
(b) Diffusion
(c) Osmosis
(d) Inactive transport.
Answer:
(a) Active transport

Question 14.
Ingestion of liquids by cell membrane is called :
(a) Endocytosis
(b) Pinocytosis
(c) Osmosis
(d) Diffusion.
Answer:
(b) Pinocytosis

Question 15.
Amount of proteins in plasma membrane is about:
(a) 30-40%
(b) 10-20%
(c) 5%
(d) 60-80%.
Answer:
(d) 60-80%.

Question 16.
Percentage of lipid in plasma membrane is about:
(a) 20-40%
(b) 60-80%
(c) 5%
(d) 10-20%.
Answer:
(a) 20-40%

Question 17.
Outer and inner layer of plasma membrane is made up of:
(a) Lipid
(b) Protein
(c) Carbohydrate
(d) None of these.
Answer:
(b) Protein

Question 18.
Desmosomes are:
(a) Pores present in plasma membrane
(b) Special area of plasma membrane where other plasma membrane attached
(c) Substance take part in the constitution of plasma membrane
(d) Chemical which j oin two cells.
Answer:
(b) Special area of plasma membrane where other plasma membrane attached

Question 19.
Lamellar model of plasma membrane was proposed by :
(a) Danielli and Davson
(b) Robertson
(c) Robert Brown
(d) Singer and Nicholson.
Answer:
(a) Danielli and Davson

Question 20.
The cell membrane is composed of:
(a) Phospholipid
(b) Nucleoprotein
(c) Polysaccharides
(d) Lipoprotein.
Answer:
(d) Lipoprotein.

Question 21.
All are membrane bounded cell organelles :
(a) Mitochondria
(b) Lysosomes
(c) Spherosomes
(d) Ribosomes.
Answer:
(d) Ribosomes.

Question 22.
Robertson’s model of cell membrane is similar to that of Danielli and Davson’s
in:
(a) Types of protein
(b) Lamellar structure
(c) Permeases
(d) Carrier particles.
Answer:
(b) Lamellar structure

Question 23.
‘Power house of the cell’ is:
(a) Nucleus
(b) Mitochondria
(c) Golgi body
(d) Chioroplast.
Answer:
(b) Mitochondria

Question 24.
Which of the following is called as ‘Suicidal bag of the cell’:
(a) Lysosome
(b) Mitochondria
(c) Peroxisomes
(d) Golgi body.
Answer:
(a) Lysosome

Question 25.
The word chromosome was coined by :
(a) Balbiani
(b) Waldeyer
(c) Sutton
(d) Purkinje.
Answer:
(b) Waldeyer

2. Fill in the blanks:

1. Cell theory was proposed for the first time ………………..
Answer:
Schleiden and Schwann

2. Nucleus was discovered by …………………..
Answer:
Robert Brown

3. ……………….. is present in animal cell not in plant cell.
Answer:
Centrosome

4. …………………. is the main constituent of cell wall.
Answer:
Cellulose

5. ……………….. are called suicidal bags of cell.
Answer:
Lysosomes

6. Polymorphism phenomenon occurs in a cell organelle called as ……………………
Answer:
Lysosome

7. In plant cells, golgi complex is formed of unconnected units called ……………………
Answer:
Dictyosome

8. …………………. is engaged in the synthesis of glycogen, lipids and steroids.
Answer:
Photorespiration

9. The tubular arrangement in cilia and flagella is …………………………..
Answer:
9 + 2

10. …………………. form spindle fibres at the time of cell division.
Answer:
Centrosome

11. The thickness of cell membrane is about ……………….
Answer:
75 Å

12. …………….. is called a protein factory of the cell.
Answer:
Ribosome

3. Match the following:
(A)

Column ‘A’	Column ‘B’
1. Mitochondria	(a) Cellulose
2. Nucleus	(b) Lysosoine
3. Cell wall	(c) R.N.A. and Protein
4. Engine of cell	(d) Power house of cell
5. Nucleolus	(e) Director of the cell

Answer:
1. (d) Power house of cell
2. (e) Director of the cell
3. (a) Cellulose
4. (b) Lysosoine
5. (c) R.N.A. and Protein

(B)

Column ‘A’	Column ‘B’
1. Cistemae	(a) Green Algae
2. Stroma	(b) Dictyosome
3. Animal cell	(c) Endoplasmic Reticulum
4. Centriole	(d) R.N.A. + Protein
5. Ribosome	(e) Sperm.

Answer:
1. (b) Dictyosome
2. (a) Green Algae
3. (d) R.N.A. + Protein
4. (e) Sperm.
5. (c) Endoplasmic Reticulum

(C)

Column ‘A’	Column ‘B’
1. Amyloplast	(a) Polyribosomes
2. Golgi complex	(b) Passive transport
3. Known as Bioblast	(c) Starch grains
4. Cluster of Ribosomes	(d) Secretion
5. Saves energy	(e) Mitochondria.

Answer:
1. (c) Starch grains
2. (d) Secretion
3. (e) Mitochondria.
4. (a) Polyribosomes
5. (b) Passive transport

(D)

Column ‘A’	Column ‘B’
1. Semiautonomous organelle	(a) Ribosome
2. Suicidal bag	(b) Chioroplast and Mitochondria
3. Physical basis of life	(e) Mitochondria
4. Protein factory	(d) Lysosome
5. Power house of the cell	(e) Protoplasm
6. Cell theory	(f) Schleiden and Schwann.

Answer:
1. (b) Chioroplast and Mitochondria
2. (d) Lysosome
3. (e) Protoplasm
4. (a) Ribosome
5. (e) Protoplasm
6. (f) Schleiden and Schwann
(E)

Column ‘A’	Column ‘B’
I. Cristae	(a) Disc like flat membranous component in grana
2. Vacuole	(b) Finger like projection in the rnìtochondria
3. Thylakoid	(c) Large specious rounded sac like structure of Golgi complex.

Answer:
1. (b) Finger like projection in the rnìtochondria
2. (c) Large specious rounded sac like structure of Golgi complex.
3. (a) Disc like flat membranous component in grana

4. Write true or false:

1. Small bodies have more surface unit volume.
Answer:
True

2. Golgi body help in protein synthesis.
Answer:
False

3. Osmoregulatory organelle is vacuole.
Answer:
True

4. Skeleton of cell are endoplasmic reticulum.
Answer:
True

5. Ribosome was discovered by Palade.
Answer:
True

6. The outermost layer of cellwall is called middle lamella.
Answer:
True

7. The chief role of nucleolus in a nucleus concerns with DNA replication.
Answer:
False

8. Cilia are formed by lysosome.
Answer:
False

9. Mitochondria are called the Power house of the cell.
Answer:
True

10. Centrosome forms asters during cell division.
Answer:
True

5. Answer in one word:

1. What will you call a cell not having E.R., golgi body, mitochondria, nuclear membrane etc.?
Answer:
Prokaryotic cell

2. In which cell, plastids and cell wall are present?
Answer:
Plant cell

3. What will you call the fusion product of protoplast of two somatic cells?
Answer:
Heterokaryon

4. Which of the following is dead-plasma membrane or cell wall?
Answer:
Cell wall

5. Which mature body cell are/is capable of reproduction?
Answer:
Nerve cells

6. RER occurs more abundantly in cells which are synthesizing .
Answer:
Protein

7. Inner membrane of cristae which bears enzymes necessary for electron transport chain.
Answer:
Oxisomes

8. The protein present in microtubules are ………………
Answer:
Tubulin

9. Proteins associated with chromosomes are ………………….
Answer:
Histone and Non-histone

10. Recent name of bioblasts is ………………..
Answer:
Mitochondria

Students get through the MP Board Class 11th Biology Important Questions Chapter 15 Plant Growth and Development which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Chapter 15 Plant Growth and Development

Plant Growth and Development Class 11 Important Questions Very Short Answer Type

Question 1.
What is known as growth? Describe in brief different phases of growth.
Answer:
According to Miller, ‘Growth is the permanent change in the size, weight and volume in the cell and organs of the body’.
Phases of growth :
1. Ceil division phase: In this phase one cell divides to form many cells.
2. Cell enlargement phase: The newly formed cell increases in size and attains maturity. Large vacuoles develop within the cell.
3. Cell differentiation and morphogenesis: Here cells differentiate into tissue and formation of different organs begin.

Question 2.
What are plant hormones?
Answer:
Plant hormones are the organic chemical substances which regulate growth and other physiological functions in plants at a site away from its place of production and active in very minute quantity e.g., Auxin, Gibberellin.

Question 3.
What is Florigen?
Answer:
Florigen is a flowering hormone which is produced in presence of proper photoperiod by the phytochromes present in the plasma membrane of leaf cells. The hormone is then transported to the floral bud region where it induces flowering.

Question 4.
Write the name of hormone which accelerates the cell division.
Answer:
Cytokinin.

Question 5.
What is 2,4-D? Also elaborate it.
Answer:
It is an artificial growth hormone. 2,4-D stands for 2,4-dichlorophenoxyacetic acid.

Question 6.
Write the name of apparatus used for measuring plant growth.
Answer:
Auxanometer.

Question 7.
What are growth inhibitor hormones? Write two functions.
Answer:
Abscisic acid (ABA) and Ethylene are top two plant growth inhibitor hormones. For their function

Functions of abscisic acid :
1. It induces dormancy in buds and seeds as opposed to gibberellic acid which breaks dormancy.
2. It promotes senescence of leaves, appearance of abscission layer, leaf-fail and ageing which can be effectively reverted by application of cytokinins.
3. It inhibits lettuce seed germination but this can also be reversed by kinetin.
4. It inhibited gibberellin induced growth in various tests and is believed to be a powerful gibberellin antagonist.

Ethylene hormone was discovered by scientist Burg (1962) which is a gaseous hormone.
Main functions of ethylene :

1. Triple response: Ethylene gas inhibits stem growth, helps in swelling of stem and destroys geotropism.
2. Flowering: It decreases flowering but in pineapple it induces flowering.
3. Sex modification: It increases number of female flower and decreases number of male flowers in plants.
4. Ripening of fruits: It is used in ripening of fruits, therefore nowadays ethephone (schloroethyl phosphoric acid which produces ethylene gas) is used for ripening fruits at industrial level.

Question 8.
Elaborate I.A.A. and describe at least one function of it.
Answer:
I. A. A. is Indole Acetic Acid. It controls the growth of plants.

Question 9.
Write the name of hormones which initiate flowering.
Answer:

• Florigen,
• Vemalin.

Question 10.
Name the hormone that exists in gaseous state.
Answer:
Ethylene.

Question 11.
Why it is beneficial in some plants to cut the tip of newly formed branches?
Answer:
Cutting of branches, initiates growth of lateral buds which results into formation of more new branches and the plant becomes more dense.

Question 12.
Name two artificial plant hormones.
Answer:

1. Naphthalene acetic acid (N. A. A.).
2. 2,4, 5-Trichlorophenoxy acetic acid (2, 4, 5-T).

Plant Growth and Development Class 11 Important Questions Short Answer Type

Question 1.
What do you mean by growth-regulating substance? Name any three growth regulator substances.
Answer:
The chemical substances which regulates growth and development in organisms are called as growth regulating substances. Actually, these are the organic substances which are naturally produced in the plants to control the growth and other physiological functions at a site away from its place of production and are active in extremely minute quantities and called as hormones. Three plants hormones are :

1. Auxin,
2. Gibberellin,
3. Cytokinin.

Question 2.
Give four functions of cytokinin hormone.
Answer:
Functions of cytokinin :

1. Activates cell division.
2. Helps in the expansion of green leaves.
3. Accelerates protein synthesis.
4. Inhibition of the breakdown of leaf pigments and nucleic acid. Chemically they are derivatives of adenine with furfuryl group at 6th position.

Question 3.
Is any leafless plant can react with photoperiodism? Yes or No? Why? (NCERT)
Answer:
No, because flowering in some plants not only depend on light or dark period but it depends on duration of light. At the apical bud, floral but develops but it can not experience photoperiodism. Leaves can experience light or dark period. A hormone florigen is responsible for flowering. It travels from leaf to floral bud only when leaf is exposed to required period of light, Gibberellin and Anthesin hormones together induce flowering.

Question 4.
Give four functions of auxin.
Answer:
Functions of auxin :

1. Helps to increase in height of plants.
2. It induces root development.
3. It induces production of parthenocarpic fruits.
4. It prevents defoliation of leaves and induces flowering.
5. Weed eradication.

Question 5.
Give importance of abscisic acid.
Answer:
Functions of abscisic acid :
1. It induces dormancy in buds and seeds as opposed to gibberellic acid which breaks dormancy.
2. It promotes senescence of leaves, appearance of abscission layer, leaf-fail and ageing which can be effectively reverted by application of cytokinins.
3. It inhibits lettuce seed germination but this can also be reversed by kinetin.
4. It inhibited gibberellin induced growth in various tests and is believed to be a powerful gibberellin antagonist.

Question 6.
Give four functions of gibberellin hormone.
Answer:
Functions of gibberellin :

1. Parthenocarpic fruits may be produced by their application in tomato, pear, apple etc.
2. Induces stem elongation by promoting growth of intemodes.
3. Flower development in lettuce, barley etc.
4. Seed germination by breaking their dormancy.
5. Buds development.

Question 7.
Give four main functions of ethylene hormone.
Answer:
Ethylene hormone was discovered by scientist Burg (1962) which is a gaseous hormone.
Main functions of ethylene :

1. Triple response: Ethylene gas inhibits stem growth, helps in swelling of stem and destroys geotropism.
2. Flowering: It decreases flowering but in pineapple it induces flowering.
3. Sex modification: It increases number of female flower and decreases number of male flowers in plants.
4. Ripening of fruits: It is used in ripening of fruits, therefore nowadays ethephone (schloroethyl phosphoric acid which produces ethylene gas) is used for ripening fruits at industrial level.

Question 8.
Write short note on apical dominance?
Answer:
Apical dominance: In many plants where apical bud grows, axillary bud do not grow i.e., apical bud dominates growth of axillary bud. Actually apical bud produces a hormone which is transported to different parts of phloem and it inhibits growth of axillary bud. The hormone is auxin. On the other hand cytokinin induces growth of axillary bud.

Question 9.
What is phytochrome? Give its importance in plants.
Answer:
Phytochrome: It is well established that short day plant does not produce flower if the dark period is interrupted by a brief flash of light. It has been observed that the wavelength of 660 in the orange red colour is the most effective wavelength for inhibiting the process of flowering. Far red light on the contrary, does not break up a long night into two short nights. Besides, far red radiation of a wavelength of 740 has been found to reverse the effect of red light by Borthwick et al (1952) and Downs (1956) and is termed as red-far red reversible photoreaction.

If a brief flash of red light in the mid-night is followed by a brief flash of far red radiation, its inhibitory effect is counteracted and flowering takes place. If far red radiation is followed further in sequence by red light, flowering will again be inhibited, i.e., the radiation is last used in the sequence, will determine the response of the plant. This discovery ultimately resulted in the discovery of the pigment is called as phytochrome by Butter et al (1959), since light energy cannot be effected unless it is absorbed by a pigment.
The main characteristics of the pigment are :

1. Proteinaceous in nature.
2. Located in plasma membrane.
3. Found in all green plants.
4. Exists in two different forms :(a) Red light-absorbing forms designated as P_R and
   (b) Far-red light-absorbing forms designated as P_FR.
5. Both of these forms are photochemically interconvertible.
6. On absorbing red light (660-665 nm) P_R form is converted into P_FR form.
7. On absorbing far-red light (730-735 nm) P_FR form is converted into P_R form.
8. The P_FR form of the pigment gradually changes into P_R form in dark.

Spectrophotometric examination of the pigment by Briggs suggests that most of the phytochrome is found in inactive form. According to Hartman the biologically active form of phytochrome is some unknown derivative of P_FR. He has suggested a different scheme of phytochrome action.

Question 10.
Why Abscisic acid is called as tension hormone? (NCERT)
Answer:
Abscisic acid is called as tension hormone due to following reasons :

• It prevents growth in plants.
• It induces leaf fall making the leaves weak.
• It prevents effect of gibberellin hormone.
• It prevents seed germination.

Question 11.
Short day plant and long-day plants show flowering together in one place. Explain it. (NCERT)
Answer:
1.  Short-day plants: Need a short daylight period ranging between 8-10 hrs and a continuous dark period exceeding 12hours(14to \6his.),i.e., Xanthium, Soyabeans, Tobacco, Gossipium, Coffee etc. The salient features of such plants are given below :

• These plants require a relatively long period of darkness for flowering. It is generally longer than a certain critical length. If dark period is less, no flowering would result. If this period is interrupted even with a small exposure, the plants will not flower.
• No flowering occurs if a weak intensity of light is given to those plants.
• No flowering under alternating cycles of short dark and light period.
• They are also termed as long night plants because length and continuity of night determine the flowering.

2. Long day plants: These plants require a longer daylight (14 -16hrs) for flowering, e.g., Supuracea etc. They are characterised by the following :

• They require a photoperiod of more than a critical length. They require either a small period of darkness or no darkness for flowering.
• Flowering is full in continuous light.
• Darkness has inhibitory impact on these plants.
• They can be induced to flowering by short photoperiods but accompanied by shorter dark period.

Question 12.
Growth in flowering plants can not be described under one parameter. Why? (NCERT)
Answer:
Growth at cellular level in flowering plants is the result of growth in the proto¬plasm, which can not be measured. Thus growth in plants is measured by some other ways.
Some way of measurements are: Increase in fresh weight, dry weight, lengthwise area, volume and number of cells. At the root tip meristems produces more than 17,500 new cells by cell division per hour, whereas in watermelon at cell division stage growth is slow because cells increases in number without increasing in volume by multiplication, later growth become 3,50,000 times than previous time due to cell elongation. Increase in length of pollen tube per unit time can be measured easily. In dorsiventral leaves, growth in surface area of leaves is measured.
Thus, from above examples, it is clear that growth in plants can not be described under one parameter.

Plant Growth and Development Class 11 Important Questions Long Answer Type

Question 1.
What is growth? Describe various stages of cell growth.
Answer:
Growth: Growth is a vital process in which organism increase in its size, weight, volume and structure.
Stages of Growth :
1. Cell division: During this stage the cells divide and redivide by mitotic division for a definite period depending upon the organ of the plant in which growth is taking place.

2. Cell elongation phase: During this phase the newly formed cell increase in size due to their internal metabolic activities. In this phase cell wall materials and water increases 5 to 10 times of their original value

3. Cell differentiation phase: During its structural, qualitative and quantitative changes takes place and cells attain final definite shape, structure, function and properties.

Question 2.
What are photoperiodism and vernalization? Give their economic importance. (NCERT)
Answer:
1. Photoperiodism: The growth and development of large number of plants is dependent upon the duration of light availability called Photoperiodism. The reproductive growth and flowering is mainly controlled by light period and temperature. The flowering requires a certain day length, i.e., the relative length of day and night which is called as photoperiod.

Economic importance of photoperiodism :

• The knowledge regarding photoperiodism is important for hybridization.
• Characterization and structure of florigen is useful for industrial purposes.
• It is an excellent example of physiological preconditioning of plants.
• Induction to flowering may be used in horticulture.
• By this process plants which produces fruits once in a year, can produce fruits twice in a year.

2. Vernalization: It has been observed that if seeds of winter seasonal plants are kept at 0-5° C temperature for some days and then sown during spring season, flowering occurs in them like other spring seasonal plants. This phenomenon is called as vernalization.

If vernalized seeds are kept at high temperature for some time then vernalization effect is lost. This process is called as devernalization. It is believed that stimuli is received by apical meristem and vernalin hormone which is a gibberellin type hormone is secreted, which is transported to growth region.
In Siberia where soil remain cover by ice for 10 months, wheat is produced by this process in two months.
Economic importance of vernalization :

1. By this process winter plants can be converted into spring plants.
2. Crops can be protected from natural harm effects.
3. Flowering can be done in plants in short period of time.
4. Crops can be produced in short period of time by this process.

Question 3.
In higher plants growth and differentiation is open. Explain it. (NCERT)
Answer:
In higher plants growth can be represented by a sigmoid curve (S-shaped curve).
There are three phases of this curve :

1. Lag phase: When the rate of growth is very slow.
2. Log phase: It is the phase of rapid growth.
3. Steady phase: Growth slows down at this stage or reaches to an equilibrium as cell division stops.
   When cell losses its division capacity, they proceed towards cell differentiation to produce different tissues to perform different functions. The growth of a cell, tissue, organ, organ system or an organism follow same pattern of growth, thus it is open, whereas development is flexible. Actually, development is the sum total of growth and differentiation.

Question 4.
Write short note on following :
(a) Arithmetic growth (NCERT)
(b) Geometrical growth
(c) Sigmoid growth rate
(d) Absolute and relative growth rate.
Answer:
(a) Arithmetic growth: After mitosis division one cell undergoes arithmetic growth by regular cell division, whereas the second cell differentiate and mature. It occurs in a constant rate.

(b) Geometrical growth: In geometrical growth both the cells produced by mitosis undergoes cell division. But due limited nutrient supply gradually growth rate become slow, at last it reaches to steady state.

(c) Sigmoid growth rate: If a graph drawn between increase in size of the organism versus time then ‘S’ shaped curve is obtained called as sigmoid curve. It has three phases :

1. Lag phase: When the rate of growth is slow in initial stage.
2. Log phase: When the rate of growth is fast or rapid.
3. Steady phase: When rate of growth reaches to an equilibrium because rate of production of cells become equal to cell death.

(d) Absolute and relative growth rate: Measurement and comparison of complete development in unit time is called as Absolute growth rate, whereas measurement of growth per unit time by any other given way is called as relative growth rate.

Question 5.
Define Growth, Differentiation, Development, De-differentiation, Re-differentiation, Meristem and growth rate. (NCERT)
Answer:

1. Growth: It is the irreversible or progressive increase in size, shape, volume and weight of an organism.
2. Cell differentiation: After cell division and cell elongation a cell undergoes differentiation. The cell differentiates, change in shape, position and get modified, matured as permanent cells to perform specific function. This phenomenon is called as differentiation.
3. Development: Development is the gradual growth of the living body in its life cycle.
4. De-differentiation: Differentiated cells regain their cell division capacity in some specific condition. This phenomenon is called as De-differentiation.
5. Re-differentiation: New cells formed by de-differentiated cells losses their cell division capacity again and become permanent tissue by differentiation to perform specific function. This is called as re differentiation.
6. Meristems: Group of similar, immature plant cells, which can show regular cell division to form new cells are called as meristems.
7. Growth rate: Increase in size, dry weight, volume or number of cells of the living organism per unit time is called as growth rate.

Question 6.
If you are asked to do following works which plant hormone can be used?
(a) To produce root in a twig of plant. (NCERT)
(b) To ripe fruit fast.
(c) To prevent falling of premature leaves.
(d) For growth of axillary bud.
(e) To promote flowering in rose.
(f) To close stomata soon.
Answer:
(a) Auxin,
(b) Ethylene,
(c) Gibberellin,
(d) Cytokinin,
(e) Gibberellin,
(f) Abscisic acid.

Question 7.
What will happen if: (NCERT)
(a) GA₃ is provided to seedlings of rice.
(b) Dividing cells stops cell division.
(c) Keeping a decay fruit with unripen fruits.
(d) If you forget to add cytokinin in the culture medium of plant.
Answer:
(a) Rice plant will increase in height.
(b) Cells when loss capacity to divide, differentiate to form different tissues to perform specific function.
(c) If a decay fruit is kept with unripen fruits then all fruits will decay.
(d) If cytokinin is not added to culture medium of plant then it may affect formation of chlorophyll in the young leaves, growth of lateral branches, growth of adventitious branches. In absence of cytokinin abscission of leaves may occur.

Question 8.
What are synthetic growth hormone? What are their importance in agriculture?
Answer:
Synthetic growth hormone or growth regulator :
As hormone regulates growth in organisms therefore they are called as growth hormone. Following are few examples of growth regulators, their importance in agriculture are as follows:
1. Morpactins: It is a synthetic hormone and derivative of fluorine-carboxylic acids. It induces growth of axillary bud by inhibiting growth of stem, leaf lamina etc. It increases production of oranges.

2. Malic Hydrazide (MH): It is a synthetic hormone which inhibit growth of grass, shrubs and trees. It inhibits germination of potato and onion thus they can be kept for long period of time.

3.  Cycocel (CCC): Chemically it is chloroethyl-trimethyl ammonium chloride. It is used to kill weeds.
4. Synthetic Auxin (IBA and N.A.A.): It prevents defoliation of fruits and leaves.
5. Alpha naphthalene acetic acid: It is a synthetic hormone used for inhibiting growth of buds in godown of potato. Therefore potatoes can be kept for long time.
6. 2, 4 Dichlorodiphenoxy acetic acid: It is used to kill weeds.
7. Ethaphone: Chemically it is 2-chloroethyl phosphoric acid. It is used for ripening fruits like banana, grapes, mango at industrial level.

Question 9.
What is flowering hormone? Describe various types of flowering hormones in plants.
Answer:
Flowering hormones: Flowering hormones are the hormones which induces flowering by the effect of temperature and light.
There are two types of flowering hormones :

1. Vemalin,
2. Florigen.

1.Vernalin: It induces flowering by regulating vernalization process. As a result of vernalization, i.e. when apical bud receives winter stimuli produces vernalin hormone which acts like gibberellin and induces flowering.

2. Florigen: Phytochrome pigments found in the green leaves after absorbing light rays produces florigen hormone which is transported to growing region, where it induces flowering process.

Question 10.
Describe various factors which affect growth.
Answer:
Factors affecting growth :
1. Food supply: It affects the rate of growth firstly because it provides growth material to the growing region and secondly because it provides potential energy to the growing region.

2. Water supply: It has a direct relationship with the rate of growth because it is necessary for all the metabolic activities of protoplasm and for increasing the turgidity of the cell for cell enlargement.

3. Oxygen supply: Oxygen increases growth as it helps in respiration to convert potential energy into kinetic energy required for the vital activities of protoplasm.

4. Temperature: It affects growth in a way that growth occur between 4-45°C, optimum activity takes place at 28-33°C.

5. Light: All the three aspects of light intensity, quality and periodicity affects growth. High intensity of light induce dwarfing of the plant and increases the loss of water. Weak light intensity reduces the rate of overall growth and also photosynthesis. Different colour (wavelength) also affects the growth of plants. In blue-violet colour light, intermodal growth is pronounced while green colour light reduces the expansion of leaves. The red colour favours elongation.

Infrared and ultraviolet are detrimental to growth. There is a remarkable effect of the duration of light on the growth of vegetative as well as reproductive structure.
6. Growth hormones: Now it is well established that the growth of plant is controlled by certain organic compounds present in very minute quantities. These compounds are called hormones, phytohormones or growth-promoting substances.

Plant Growth and Development Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
Gibberellins was first extracted from :
(a) Gibberella fujikuroi
(b) Gelidium
(c) Gracelaria
(d) Aspergillus.
Answer:
(a) Gibberella fujikuroi

Question 2.
Storage sprouting of potato can be prevented by :
(a) IAA
(b) Malic hydrazide
(c) Cytokinins
(d) Gibberellins.
Answer:
(b) Malic hydrazide

Question 3.
The following is a naturally occurring growth inhibitor :
(a) IAA
(b) ABA
(c)NAA
(d) GA₃.
Answer:
(b) ABA

Question 4.
The following hormone is concerned chiefly with cell division in plants :
(a) IAA
(b) Kinin (zeatin)
(c) GA₃
(d) 2,4-D.
Answer:
(b) Kinin (zeatin)

Question 5.
Gibberellic acid has been successfully employed to induce flowering :
(a) In short-day plants under long-day conditions
(b) In long-day plant under short-day conditions
(c) For some plants
(d) None of these.
Answer:
(b) In long-day plant under short-day conditions

Question 6.
The leaves of Mimosa pudica drop down on touch because :
(a) The plant has nervous system
(b) The leaves are very tender
(c) The leaf tissues are injured
(d) The turgor of the leaf changes.
Answer:
(d) The turgor of the leaf changes.

Question 7.
Vernalization is :
(a) Growth curve in response to light
(b) Recurrence of day and night
(c) Effect of day length on plant growth
(d) Acceleration of the ability of flower by low-temperature treatment.
Answer:
(d) Acceleration of the ability of flower by low-temperature treatment.

Question 8.
Effect of length of the day on flowering is called :
(a) Phototropism
(b) Photoperiodism
(c) Photorespiration
(d) Photo-oxidation.
Answer:
(b) Photoperiodism

Question 9.
In plants the hormone associated with cell division is :
(a) GA
(b) 2,4-D
(c) IAA
(d) Kinin.
Answer:
(d) Kinin.

Question 10.
Cytokinin:
(a) A hormone that stimulate cell division
(b) A process of cell division
(c) A form of cell movement ‘
(d) A substance that produces dormancy.
Answer:
(a) A hormone that stimulate cell division

Question 11.
Three main growth inducing hormones in plants are :
(a) Auxin, Gibberellin and Ethylene
(b) Gibberellin, Cytokinin and Abscisic acid
(c) Ethylene, Abscisic acid and cytokinin
(d) Auxin, Gibberellin and cytokinin.
Answer:
(d) Auxin, Gibberellin and cytokinin.

Question 12.
Cytokinesis induces:
(a) Cell division
(b) Cell elongation
(c) Stem elongation
(d) Parthenocarpy.
Answer:
(d) Parthenocarpy.

2. Fill in the blanks:

1. First auxin isolated from human urine is ……………………….. .
Answer:
IAA

2. ……………………….. is used in the ripening of fruits.
Answer:
Ethylene

3. ……………………….. hormone is reponsible for flowering.
Answer:
Florigen

4. ……………………….. hormone is used to increase length of genetically dwarf plants.
Answer:
Gibberellin

5. ……………………….. hormone increases abscissing and senescence.
Answer:
Abscisic acid

6. ……………………….. hormone is responsible for cell division.
Answer:
Cytokinin

7. Germination of seeds over parent plant is called ……………………….. .
Answer:
Vivipary

8. Bakani disease of rice is caused by a fungus known as ……………………….. .
Answer:
Gibberella fuji kuroi

9. Growth occurs per unit time period in the living organisms is called as ……………………….. .
Answer:
Growth rate

10. In ……………………….. growth after mitosis division only one daughter cell shows regular cell division.
Answer:
Arithmetic.

3. Match the following:

(A)

Column ‘A’	Column ‘B’
1. Florigen	(a) IAA
2. Abscission	(b) Protein
3. Delay in senescence	(c) Cytokinin
4. Aleuron layer	(d) ABA
5. Auxin	(e) Flowering.

Answer:
1. (e) Flowering.
2. (d) ABA
3. (c) Cytokinin
4. (b) Protein
5. (a) IAA.

(B)

Column ‘A’	Column ‘B’
1. Dormin	(a) Auxin
2. GA3	(b) Abscisic acid
3. Zeatin	(c) Gibberellin
4. 2,4-D	(d) Cytokinin
5. Termination of seed dormancy	(e) Gibberellin.

Answer:
1. (b) Abscisic acid
2. (c) Gibberellin,
3. (d) Cytokinin
4. (a) Auxin
5. (e) Gibberellin.

4. Answer in one word:

1. Name the tissues responsible for growth in plants.
Answer:
Meristematic tissue

2. Write the name of various phases of growth.
Answer:
Cell division stage, Cell elongation stage, Cell maturation stage

3. Elaborate the term 2,4-D. Also write one function of it.
Answer:
2, 4 di phenoxy acetic acid, it is a growth hormone and used as weedicide

4. Elaborate the term IAA and write one function of it.
Answer:
Indole acetic acid, used in development of seedless fruits

5. Name the hormone which is found in gaseous state.
Answer:
Ethylene

6. Name the apparatus used for measurement of plant growth.
Answer:
Auxanometer

7. Name the growth regulator which help to stop germination of potato and onion.
Answer:
Malic hydrazide

8. Name any two synthetic Auxins.
Answer:
(a) I. B. A. (Indole buteric acid),
(b) N. A.A. (Naphthol acetic acid).

Students get through the MP Board Class 11th Biology Important Questions Chapter 7 Structural Organisation in Animals which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Questions Chapter 7 Structural Organisation in Animals

Structural Organisation in Animals Class 11 Important Questions Very Short Answer Type

Question 1.
What is polycythemia?
Answer:
Polycythemia is an abnormality in human beings in which the number of R.B.Cs. of blood is abnormally increased.

Question 2.
What are ligaments? Write their functions.
Answer:
Ligaments are the tufts of elastin filaments that connect bones with bones.

Question 3.
Define sarcomere.
Answer:
Distance between two adjacent ‘Z’ lines of a muscle fibre of striped muscle is called as sarcomere. The sarcomeres are the functional units of striped muscle fibres.

Question 4.
What is haemopoiesis?
Answer:
Formation of blood in body is called haemopoiesis.

Question 5.
Name the glands found in the skin of human.
Answer:
Three types of glands are found in the skin of human :

1. Sweat gland,
2. Sebaceous gland and
3. Mammary gland.

Question 6.
What is known as haversian system?
Answer:
In mammalian bone, the calcified matrix is deposited in the form of lamellae which arranged around numerous canals. These canals are called haversian canals and lamella are called haversian lamella. Each haversian canal and the lamellae forms
haversian system.

Question 7.
Write any two similarities between blood and lymph.
Answer:

1. Blood and lymph both contain W.B.Cs.
2. Both contain sugar, urea, amino acids and both having the capacity of coagulation.

Question 8.
Name the tissue which joined muscle and bone.
Answer:
Tendon joined muscles with bone.

Question 9.
Where are mast cells situated? Name the substances secreted by them.
Answer:
Mast cells are found in the interstitial cells around the blood vessels. These cells secrete histamine, heparin, serotonin, etc. which are found in blood plasma.

Question 10.
Write the name of two specific proteins that form contractile muscle fibres.
Answer:
Actin and myosin.

Question 11.
Draw the diagram of adipose tissue.
Answer:

Question 12.
Draw a labelled diagram of unstriped muscles.
Answer:

Question 13.
Write the names of the proteins found in bone and cartilage.
Answer:
Ossein protein in bone and chondrin in cartilage.

Question 14.
Name the tissue found in the inner lining of blood vessels.
Answer:
Simple epithelium and pseudostratified epithelium tissues are found in the inner lining of blood vessels.

Question 15.
Name the tissue found in the inner lining of fallopian tube and bronchioles.
Answer:
Ciliated epithelium.

Question 16.
Name the part of human body which can be repaired easily if damaged.
Answer:
Skin.

Question 17.
Name the corpuscles found in the blood.
Answer:
Corpuscles found in the blood are ;

• Red blood corpuscles,
• White blood corpuscles.

Question 18.
What make the top layers of cells of the stratified epithelium water proof?
Answer:
The top layer of the cells of the stratified epithelium replace their cytoplasm with a water proof protein keratin and become dead and water proof.

Question 19.
What is the life span of erythrocytes in human ? Where are R.B.Cs. pro-duced during embryonic stage and in adult stage?
Answer:
In human life span of erythrocytes is 120 days.
In embryonic stage erythrocytes are mainly formed in the liver and spleen. But from birth onwards, these are formed in the red bone marrow.

Question 20.
Define tissue.
Answer:
Group of cells which are similar in their origin, structure and function are called as tissue.

Question 21.
Define organ and organ system.
Answer:
A collection of various tissues to form a structure, which perform specific function is called as organ.
e.g. Stomach, intestine, liver etc.
Various organs held together to perform a specific function is galled as organ system.
e.g. Digestive system, respiratory system etc.

Question 22.
Why fatty people feel less cold as compared to thin people?
Answer:
In fatty people more quantity of adipose tissue is found, which acts as a heat insulator under the skin, thus fatty people feel less cold as compared to thin people.

Question 23.
Give functions of Histamin, Heparin and Seratonin.
Answer:
Mast cells of Areolar tissue secretes Histamin, Heparin and Seratonin. Their functions are:

• Histamin : Dilates the blood vessels and controls the blood flow.
• Heparin : It prevents coagulation of blood inside the blood vessels, thus acts as anticoagulant.
• Seratonin : It constricts blood vessels and increases blood pressure.

Question 24.
Name a plasma protein which helps for coagulation of blood in vertebrates.
Answer:
Fibrinogen.

Question 25.
Why coagulation of blood does not occur inside the blood vessels?
Answer:
Coagulation of blood does not occur inside the blood vessels due to presence of an anticoagulant substance heparin.

Question 26.
What is node of Ranvier?
Answer:
Part of the myelinated nerve fibre where myelin sheath is not found is called as
Node of Ranvier.

Question 27.
In the tissue of which animals haversian system is found?
Or,
In which tissues of mammalia haversian canals are found?
Answer:
Haversian system is found in the bones of mammals.

Structural Organisation in Animals Class 11 Important Questions Short Answer Type

Question 1.
Give types and functions of plasma protein.
Answer:
Plasma protein can be divided into three types :
(1) Serum albumin,

• Serum globulin,
• Fibrinogen.

Functions of plasma protein :

1. Body resistant: Globulin protein functions as antibodies.
2. Prevent loss of blood : Fibrinogen which is produced in the liver helps in clotting of blood.
3. Keeps blood in fluid form : Albumin and globulin has capacity to hold water in themselves, thus helps to maintain liquid form of blood.
4. It helps in circulation or transportation.
5. It maintains pH value of the blood.
6. It maintains equal temperature in the whole body.
7. It conducts heat.

Question 2.
What are following and where they are found :

• Osteocyte,
• Nissil’s granules,
• Haemoglobin,
• Haversian system,
• Canaliculi,
• Ciliated epithelium,
• Lacunae.

Answer:

• Osteocyte : The cells found in the matrix of bones are called as Osteocyte.
• Nissil’s granules : These are found in the cytoplasm of the nerve cells which are concerned with impulse conduction.
• Haemoglobin : A complex protein found in red blood corpuscles and helps to carry oxygen during respiration.
• Haversian system : These are group of tubules found in the bones of mammals.
• Canaliculi: Lacunae are found united with each other by means of Canaliculi in the bones of mammals.
• Ciliated epithelium : When at free end of columnar or cuboidal cells there is fine thread like cilia found, they are called as ciliated epithelium. They are found in the inner lining of trachea, oviduct, ureter.
• Lacunae : Small bag like structures found in the matrix of bones and cartilage called as lacunae. Bone or cartilage cells are found inside them.

Question 3.
Describe any four differences between Tendon and Ligament.
Answer:
Differences between Tendon and Ligament

Tendon	Ligament
1. They are the tufts of collagen fibres that connect muscles with bones.	They are the tufts of elastin filaments that connect bones with bones.
2. The filaments of tendon are found in the form of thick bundles.	The filaments of ligament are thin and separated with each other.
3. Yellow fibres are not found in the tendons.	Yellow fibres are present.
4. Tendon cells are arranged in a sequence.	Ligament cells are irregularly arranged.

Question 4.
What is lymph? Describe any two functions of lymph.
Answer:
The blood secreted in the tissues from blood corpuscles is called as lymph. It contains white blood corpuscles and R.B.Cs. are absent in it. Thus it is a colourless fluid found in the intercellular spaces and is made up of blood plasma, W.B.Cs., O₂, nutritious and excretory substances.
Functions :

• W.B.Cs. present in lymph destroy pathogens by the process of phagocytosis.
• It plays an important role in circulation of fluids.
• It functions as a medium between blood vessels and cells.

Question 5.
Write down the name of muscles participating an important role in the following processes :

• Movement of legs,
• Movement of food in oesophagus,
• Contraction in blood capillaries,
• Closing of one eye.

Answer:

Muscular movement – Types of muscles

• Movement of legs – Striated muscles
• Movement of food in oesophagus – Unstriated muscles
• Contraction in blood capillaries – Unstriated muscles
• Closing of one eye – Circular unstriated muscles.

Question 6.
What are voluntary and involuntary muscles?
Answer:
(i) Voluntary muscles:
The muscles which are innervated by motor nerves from brain and spinal cord and are under the control of will of the animals are known as voluntary muscles. Example : Striped or skeletal muscles like biceps, triceps are voluntary muscles.

(ii) Involuntary muscles:
The muscles which are innervated by autonomous nervous system and whose contraction is not controlled by the will of animal are known as involuntary muscles. Example : Smooth, visceral or unstriated muscles are involuntary muscles.

Question 7.
Describe the differences between Blood and Lymph. Or, Write the four differences between Blood and Lymph.
Answer:
Differences between Blood and Lymph

Blood	Lymph
1. Blood contains R.B.Cs.	R.B.Cs. absent.
2. W.B.Cs. are fewer in number and neutrophils are more in number.	W.B.Cs. are more in number and lymphocytes are also more.
3. Amount of soluble protein is higher than the amount of insoluble proteins.	Amount of soluble protein is lesser than the amount of insoluble proteins.
4. The amount of O2 and nutrients is more in blood.	Amount of O2 and nutrients is lesser in lymph.
5. The amount of CO2 and excretory substances is lesser.	The amount of CO2 and excretory substance is more.
6. It is red coloured fluid.	It is a colourless fluid.

Question 8.
Write differences between following :
(a) A-band and I-band,
(b) Plasma and Serum.
Answer:
(a) Differences between A-band and I-band

A-band	I-band
1. Dark coloured band found in striped muscle is called as A-band.	Light coloured band found in striped muscle is called as I-band.
2. A light line present at the middle of A- band is called as H-line or Hensen’s line.	A dark line present at the middle of I-band is called as Z-line or membrane of Kranse. (Distance between two adjacent Z-lines is called as sarcomere.)
3. A-band consists of myosin protein.	I-band consists of actin protein.

(b) Differences between Plasma and Serum :

Plasma	Serum
1. A granulated liquid part of the blood is called as plasma.	The liquid remain after clotting of blood is called as serum.
2. Plasma word is also used for body fluid, e.g. Lymph.	Serum word used sometimes for such fluid which is antibody against any specific disease.
3. Plasma is the blood plasma which contains fibrinogen protein.	Serum means that blood plasma which do not contain fibrinogen protein.

Question 9.
Find out odd from the following :
(a) Areolar tissue, blood, nerve cell (Neuron), tendon.
(b) Red blood corpuscles, white blood corpuscles, platelets, cartilage.
(c) Exocrine, endocrine, salivary gland, ligament.
(d) Maxilla, mandible, labrum, antenna
(e) Prothorax, mesothorax, metathorax, coxa.
Answer:
(a) Neuron,
(b) Cartilage,
(c) Ligament,
(d) Antenna,
(e) Coxa.

Question 10.
Draw labelled diagram of Reproductive system of Earthworm.
Answer:

Question 11.
Draw labelled diagram of Alimentary canal of Cockroach.
Answer:

Question 12.
Write differences between Bone and Cartilage.
Answer:
Differences between Bone and Cartilage

Bone	Cartilage
1. It is rigid, non-flexible, strong structure.	Soft and flexible structure.
2. Matrix is composed of protein called ossein.	Matrix is made up of protein called chondrin.
3. Matrix occurs in concentric lamellae.	Matrix occurs in homogenous mass.
4. Osteocytes (bone cells) lie singly in lacunae.	Chondrocytes (cartilage cells) lie in lacunae singly or in groups of two or four.
5. Bone cells (osteocytes) are irregular and give off branching processes.	Cartilage cells (chondrocytes) are oval and without processes.
6. Matrix of bone cells have the deposition of inorganic salts.	There is no deposition of inorganic salts.
7. Bones are richly supplied with blood.	Soft and flexible structure.
8. Bones of mammals show haversian system.	Matrix is made up of protein called chondrin.

Question 13.
Describe body segmentation in earthworm.
Answer:
Body segmentation in earthworm :
The entire elongated body is made up of a series of distinct segments. These are called metameres which are separated from each other by intersegmental grooves. The number of segments varies from 100-120. The metameres can be recognised externally in the form of circular annuli. These are internally divided by septa to divide the body cavity into compartments. This type of segmentation in earthworms which is both external and internal is known as metamerism or metameric segmentation.

However, in P. posthuma internal septa are absent. The first segment at the anterior end is called peristomium or buccal segment. It is characterised by the presence of a small projection over the crescent shaped mouth. It is known as prostomium. Anus is found at the posterior end of the last segment.

A prominent circular band of glandular tissue surrounds segments from 14-16. These segments are fused and collectively known as clitellum and are not distinct externally. The clitellum secretes mucus and albumen. It divides the whole body into

• preclitellar,
• clitellar and
• postclitellar regions.

The secretion of clitellum helps in the formation of ootheca which is the site of fertilization of eggs.
On the mid dorsal surface of earthworm a blue line is observed called as dorsal blood vessel.

Question 14.
Answer the following questions :
(i) Give function of Nephridia.
(ii) How many types of nephridias are found in earthworm on the basis of their location?
Answer:
(i) Function of Nephridia :
Nephridia are the excretory organs of earthworm, which consists of three parts : (a) Ciliated funnel like structure, (b) Body and (c) Nephri- diopore.
Ciliated funnel like structure collect excess fluid from the coelomic cavity and through its body tube transfer the fluid to the alimentary canal.

(ii) Types of Nephridia on the basis of their locations :
1. Integumentary Nephridia : These are found in all the body segments except the first two on the inner surface of body wall. About 200-250 these nephridia are found in each segment. These lack nephrostome and open to Mouth outside through independent nephridiopores. Due to this reason, integumentary nephridia are also known as Exonephric nephridia.

2. Pharyngeal Nephridia:
These nephridia are found on both the sides of pharynx and oesophagus in 4^th, 5^th and 6^th body segment in the form of paired tufts. Each tuft has many nephridia with branched tubule without mouth. The common duct formed by the union of terminal ducts of each ne- phridium opens in the alimentary canal. There¬fore, these are also known as enteronephric nephridia.

3. Septal Nephridia :
These are found on both the surfaces of each intersegmental septum behind 15th segment. These have typical nephridial structure and open into the intestine and thus pour their excreta into intestine itself. Septal nephridia are also known as Enteronephric nephridia.

The main nitrogenous product excreted by earthworm is urea. Therefore, it is a ureotelic animal but in addition to urea, a little amount of ammonia, amino acids and other nitrogenous compounds are also contained in the excretory fluid of earthworm.

Question 15.
Draw labelled diagram of alimentary canal of earthworm.
Answer:

Question 16.
Differentiate the following :
(a) Prostomium and Peristomium,
(b) Septal nephridia and pharyngeal nephridia.
Answer:
(a) Prostomium and Peristomium :
Prostomium is a lobe like structure which cover the mouth. It is a sensory organ which help to open crack of the soil and help to crawl inside it, whereas peristomium is the first segment of the body of earthworm. Mouth is found in the peristomium.

(b) Septal nephridia and Pharyngeal nephridia:
Septal nephridia are the excretory organs found on both the surfaces of intersegmental septum behind 15^th segment. They have typical nephridia structure open into intestine and thus pour their excreta into intestine itself, whereas pharyngeal nephridia are found on both the sides of pharynx and oesophagus in 4^th 5^thand 6^th body segments in the form of paired tufts. Each tuft has many nephridia with branched tubule without mouth. Each nephridia opens into alimentary canal.

Question 17.
What are following and give their location in the body :
(a) Chondrocyte,
(b) Nerve axon,
(c) Ciliated epithelium.
Answer:
(a) Chondrocyte :
Chondrocytes are the cells found in the cartilages. Matrix of chondrin is semisolid made up of chondromucoprotein called as chondrin. Externally the cartilage is bound by a layer of densely arranged connective tissue called as perichondrium. Cells of cartilage are called as chondrocytes. Chondrocytes are found in the inter-cellular space called as lacunae. Each lacunae contains 2 to 4 chondrocytes. Chondrocytes secretes chondrin.

(b) Nerve axon :
A single nerve cell is called as neuron. Cell body of neuron is called as cyton. Many processes arises from cyton are called as dendrons. One process become long and thick called as axon. At the end, it forms branches called as nerve endings, which are connected to dendrites of next neuron through synapsis. Axon may give side branches called as collateral fibre.

Axon is found surrounded by neurilemma membrane. Many axon contains additional sheath called as Myelin sheath. Part of the axon where myelin sheath is not found is called as Node of Ranvier. Axon helps to transmit message to next neuron, muscles and glands.

(c) Ciliated epithelium :
It consists of columnar or cuboidal cells bearing cilia at their free ends. They are found at the roof of the buccal cavity of frog, surface of the gills of molluscs and respiratory passages of vertebrates.
Its main function is to maintain the flow of mucus in one direction, i.e., it helps to push mucus towards the pharynx, it also helps to maintain circulation of cerebrospinal fluid.

Question 18.
Draw labelled diagram of Digestive system of frog.
Answer:

Question 19.
Answer in one word or one line :

• Give common name of Periplaneta americana.
• How many spermathecae are found in earthworm?
• Give location of ovary of cockroach.
• How many segments are found in the abdomen part of cockroach?
• Where Malpighian tubules are found?

Answer:

• Cockroach.
• 4 pairs of spermathecae lies in segments 6 to 9.
• In female cockroach, a pair of ovaries are found, which extend between 2nd to 6th segments.
• Abdomen is the largest part of the body of cockroach which is dorsoventrally flattened. It consists of 11 segments in embryos while in adults it has 10 segments.
• Malpighian tubules are the excretory organs of cockroach, which are found at the juncture of mid gut and hind gut of the alimentary canal.

Structural Organisation in Animals Class 11 Important Questions Long Answer Type

Question 1.
Describe various types of epithelium tissue with labelled diagrams.
Answer:
Epithelium tissue :
The tissue that forms lining of various internal organs and also cover the body surface of an organism are called as epithelium tissue. There are fol-lowing types of epithelium tissue :

(A) Simple epithelium tissue :
It consists of a single layer of cells over a basement membrane. It is of following types :

• Simple squamous epithelium: It consists of thin, broad, flat cells with prominent nucleus. e.g. Outer lining of skin, blood vessels etc.
• Simple cuboidal epithelium: Cuboid shaped cells with granular protoplasm. e.g. Thyroid gland and liver.
• Simple columnar epithelium: Cells are much longer looks like column, nucleus is usually located at the base of the cells. e.g. Lining of alimentary canal, uterus, fallopian tube etc.
• Simple ciliated epithelium : It consists of columnar or cuboidal cells bearing cilia at their free ends. e.g. Roof of the buccal cavity of frog gills of molluscs and respiratory passage of vertebrates etc.
• Simple pseudostratifled epithelium : All the cells of this type of tissue lies over the basement membrane, only tall cells reach the free surface thus it gives an appearance of two layered cellular structure thus called as pseudostratifled. e.g. Trachea, large bronchi, male urethra etc.

(B) Compound epithelium :
It is formed of several layer of epithelial cells over the basement membrane. It may be of two types :
(i) Transitional epithelium : In this outermost layer of cells are highly elastic.
e.g. Inner lining of the urinary bladder and ureter.
(ii) Stratified epithelium : Cells of stratified epithelium are arranged in many layers one above the other. ITie nature of this layer depends upon the shape of cells of the top layer which may be keratinized (due to deposition of water proof keratin) or non-keratinised (due to absence of keratin).
e.g. Epidermis of skin, pharynx, buccal cavity, oesophagus, vagina etc.

Question 2.
Write differences between following :
(a) Simple epithelium and Compound epithelium,
(b) Heart muscle and Striated muscle,
(c) Dense regular and Dense irregular connective tissue,
(d) Adipose tissue and Blood tissue,
(e) Simple gland and Compound gland,
(f) White collagen fibre and Yellow elastin fibre.
Answer:
(a) Differences between Simple and Compound epithelium

Simple Epithelium	Compound Epithelium
1. A single layer of cells are found over basement membrane.	More than one layer of cells are found over basement membrane.
2. All cells are similar.	Similar, dissimilar both the types of cells are found.
3. All cells are in contact of basement membrane.	Only lower layer of cells are in contact of basement membrane.
4. Its main function is protection.	It provides mechanical support to lower tissues.

(b) Differences between Heart muscle and Striated muscle

Heart Muscle	Striated Muscle
1. They forms wall of the heart.	They are found attached to the bones and arranged in bundles.
2. Muscle fibres are united by bridges to form network or branched structure.	Muscle fibres are long cylindrical but non-tapering cells.
3. Fibres are multinucleated.	Fibres are generally uninucleated.
4. Nucleus lies at the centre of the cell.	Nucleus lies at the periphery.
5. Capable of rhythmic and autonomic contraction.	Capable of quick and sustained contraction.
6. They are involuntary in function.	They are voluntary in function.
7. They are immune to fatigue.	Fibres are subjected to fatigue when overworked

(c) Differences between Dense Regular and Dense Irregular Connective Tissue

Dense Regular Connective Tissue	Dense Irregular Connective Tissue
1. Collagen fibres are found in between other parallel fibres and are regularly distributed and fibroblast lies in continuous rows. e.g. Tendon (Bundle of collagen fibres gives strength to the tendon and makes it tough.)	Dense irregular connective tissue contains fibroblast and many other fibres, which are irregularly distributed (scattered) in the connective tissue. e.g. Cartilage, bone, blood etc.

(d) Differences between Adipose Tissue and Blood Tissue

Adipose Tissue	Blood Tissue
1. It is a semisolid connective tissue.	It is a liquid connective tissue.
2. It is located under the skin.	It is located inside the blood vascular system.
3. It stores fat and prevents heat loss.	It helps to transport various materials in the body.

(e) Differences between Simple and Mixed (Compound) gland

Simple gland	Mixed (Compound) gland
1. Duct of glands are unbranched.	Duct of glands are branched.
2. These glands are formed by tubular cells which open at the body surface.	These glands are compound structure have a compound shape of tubular and alveolar gland. Main tube opens at the surface of the body.

(f) Differences between White collagen fibres and Yellow elastin fibres

White collagen fibre	Yellow elastin fibre
1. They are thick, ribbon like and arranged in bundles.	They are less in number, branched and scattered.
2. They are unstretchable.	They are stretchable.
3. They are formed of a substance called as collagen.	They are formed of a substance called as elastin.
4. They are hard and strong.	They are elastic in nature.

Question 3.
Describe circulatory system of earthworm.
Answer:
Circulatory system of earthworm :
Closed type of circulatory system is found in earthworm, i.e. blood is restricted to blood vessels only.
The blood consists of plasma and colourless nucleated cells. Haemoglobin is dis-solved in plasma giving red colour to the blood, which help to transport O₂and CO₂ Circu-latory system consists of following parts :
1. Hearts: In earthworm lateral hearts (4 pairs) are found in the 7^th, 9^th 12^th and 13,^thsegment. These are in the form of large, thick muscular contractile vessels.

2. Main blood vessels :
(i) Dorsal vessels : It runs on mid dorsal region of alimentary canal run from anterior to posterior end. It helps to distribute blood except 13^th segment.
(ii) Ventral vessels: It is found in the mid ventral region of alimentary canal extended from anterior to posterior end. The vessel function as distributing vessel.
(iii) Lateral oesophageal vessels : A pair of these vessels are found on the 13^th seg-ment which help for collection.
(iv) Sub-neural vessels : It is found beneath the nerve cord and extends from 14^thsegment to the last segment of the body. It acts as collecting vessels.

Question 4.
Give functions of the following :
(a) Ureters of frog.
(b) Malpighian tubule.
(c) Body wall of earthworm.
Answer:
(a) Function of ureter of frog :
A pair of ureter are found in frog which arises from kidneys. Ureter opens into cloaca. In frog nitrogenous waste is urea, which reaches to kidney through blood, after filtration it forms urine, which passes through ureter to the cloaca. In male frog ureter acts as urinogenital duct, which carries sperms as well as urine. Cloaca helps to remove sperms, urine as well as faecal matter out of the body.

(b) Function of Malpighian tubule :
Malpighian tubules are the excretory organs of insects like cockroach. It helps to excrete metabolic waste out of the body from haemolymph. They are 80-90 in number and found in 6-8 groups at the juncture of mid gut and hind gut of alimentary canal. One end of these tubules opens in haemocoel and absorbs the excretory substances like K⁺, Na⁺, sodium urate, excess water and CO²
(c) Function of body wall of earthworm :
The body of earthworm is thin and slimy. Its body wall consists of following layers :

(i) Cuticle:
It is secreted by epidermal cells. It is non-cellular, thin, flexible and made up of collagen fibres.

(ii) Epidermis :
It is situated beneath the cuticle, single layered and made up of columnar cells. It has many types of cells like supporting cells, basal cells, globlefor mucous cells and albumen cells. Setal sac are fonned due to invagination of epidermis.The basal cell of setal sac functions as formative cell of seta.

(iii) Musculature:
Beneath epidermis, there is a continuous layer of circular muscles. It is followed by a thick layer of longitudinal muscles which extend all along the length of body. Muscles are unstriped or smooth. Longitudinal muscles are arranged in the form of bundles which are separated from each other by connective tissues.

(iv) Coelomic epithelium:
It is thin layer of parietal peritoneum situated below musculature which is membranous and mesodermal in origin.

Question 5.
Write differences between Male and Female cockroach.
Answer:
Differences between Male and Female cockroach

Male cockroach	Female cockroach
1. The abdomen is long and broad.	The abdomen is short and narrow.
2. Anal styles are present.	Anal styles are absent.
3. Brood pouch is absent. – Brood pouch is present.	Brood pouch is present.
4. 8th teigum is covered by 7thtergum. – Both 8thand 9th terga are covered with 7thtergum.	Both 8thand 9thterga are covered with 7th tergum.
5. Males have longer antennae.	Antennae of females are short.
6. All 9 sterna are visible.	Only 7 sterna are visible.
7. Gonapophysis are 3 in numbers.	Gonapophysis are 3 pairs in number.

Question 6.
Explain the mouth parts of cockroach along with diagram.
Or,
Write the name and function of mouth parts present in cockroach.
Answer:
Mouth parts:
At the anterior end of the head there is a mouth which is provided with many appendages collectively called as mouth parts. The components of mouth parts are as follows:
(i) Labrum:
It forms the upper lip of the mouth. It is broad and chitinous. A thin plate called epipharynx is fused to the inner surface of the labrum. It holds the food particles during feeding.

(ii) Mandibles :
A pair of stout and chitinous mandibles are found on the sides of the head capsule. It lies below the genae. Teeth are present on its inner margin while on its outer margin a soft lobe prostheca is present. It helps in cutting the food material into pieces.

(iii) Maxillae:
Two maxillae lie beneath mandibles. Each maxilla consists of following parts:
(a) Protopodite : It is the basal portion made up of cardo and stipes.
(b) Endopodite : Arises from stipes and consists of galea (hood like) and lacinia (plate like). The latter also bears numerous strong sensory bristles at its inner surface.
(c) Exopodite : It consists of a small basal sclerite, the palpifer and a five jointed maxillary palp with sensory bristles. It holds the food and gives to the mandibles for masti-cation. It is also used in cleaning the antennae and legs.

(iv) Labium :
It represents the fused second pair of maxillae lying behind the mouth. It consists of following parts :
(a) Protopodite : It again consists of sub-mentum (proximal large part), mentum (middle smaller part) and a pre-mentum. The sub-mentum and mentum together are called post-mentum. It represents the fused cardons while pre-mentum is the fused portion of two stipes.
(b) Endopodite : It is represented by ligula, glossa and paraglassa.
(c) Exopodite : It consists of two parts
(i) palpiger and
(ii) labial palp. The palpiger arise from pre-mentum while labial palp is three jointed structure bearing sensory bristles.
The labium does not take part in feeding. It is sensory in nature.

(v) Hypopharynx :
It is a hanging structure between two maxillae in front of labium. It is small, conical in shape. An efferent salivary duct carrying saliva from salivary glands opens at the base of hypopharynx. It directs salivary juice towards the food. It functions like tongue.

Structural Organisation in Animals Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
With which of the following animals protandry is associated:
(a) Rat
(b) Frog
(c) Cockroach
(d) Earthworm.
Answer:
(d) Earthworm.

Question 2.
Tympanum is found in :
(a) Earthworm
(b) Cockroach
(c) Frog
(d) Rat.
Answer:
(c) Frog

Question 3.
In which of the following animal, the body is divided into head, thorax and
abdomen:
(a) Cockroach
(b) Earthworm
(c) Rat
(d) Frog.
Answer:
(a) Cockroach

Question 4.
Frog’s cloacal aperture is the outlet for :
(a) Urine
(b) Faecal material
(c) Gametes
(d) All of these.
Answer:
(d) All of these.

Question 5.
Number of spiracles in cockroach is :
(a) 10 pairs
(b) 12 pairs
(c) 20 pairs
(d) 8 pairs.
Answer:
(a) 10 pairs

Question 6.
In which of the following animals, the gizzard is associated with alimentary
canal:
(a) Rat
(b) Frog
(c) Cockroach
(d) Snakes.
Answer:
(c) Cockroach

Question 7.
Hibernation and aestivation can be observed in :
(a) Earthworm
(b) Cockroach
(c) Frog
(d) Lizard.
Answer:
(c) Frog

Question 8.
Nephrons are associated with :
(a) Kidney
(b) Liver
(c) Brain
(d) Testes.
Answer:
(a) Kidney

Question 9.
A radial symmetry is found in :
(a) Starfish
(b) Sponges
(c) Annelida
(d) Echinoderms.
Answer:
(a) Starfish

Question 10.
In frog special connecting veins found between liver and alimentary canal
and between kidney and lower parts of the body are called as:
(a) Hepatic portal system
(b) Hepatic portal and Renal portal system
(c) Renal portal system
(d) None of these.
Answer:
(b) Hepatic portal and Renal portal system

(B) Choose the correct answers :

Question 1.
Which of the following is responsible for the coagulation of blood :
(a) Prothrombin
(b) Ca
(c) Fibrinogen
(d) All of these.
Answer:
(d) All of these.

Question 2.
R.B.Cs. are nucleated in :
(a) Human
(b) Rabbit
(c) Lion
(d) Frog.
Answer:
(d) Frog.

Question 3.
Ciliated cells are found in :
(a) Bronchus
(b) Pancreas
(c) Liver
(d) Uterus.
Answer:
(a) Bronchus

Question 4.
Ligaments connect the:
(a) Bone from muscles
(b) Muscles with muscles
(c) Bone with bone
(d) Muscles with cartilage.
Answer:
(c) Bone with bone

Question 5.
Mast cells are found in :
(a) Muscles
(b) Nerve fibres
(c) Connective tissue
(d) Ligaments.
Answer:
(c) Connective tissue

Question 6.
The lifespan of R.B.Cs. mammals is :
(a) 120-127 days
(b) 60 days
(c) Few hours
(d) 80 days.
Answer:
(a) 120-127 days

Question 7.
Cells present in cartilage are called :
(a) Histiocytes
(b) Chondrocytes
(e) Lymphocytes
(d) Osteocytes.
Answer:
(b) Chondrocytes

Question 8.
Croup of cells having similar structure and functions are called as :
(a) Organ
(b) System
(c) Tissue
(d) Organ system.
Answer:
(c) Tissue

Question 9.
The hump of camel is made up of:
(a) Skeleton tissue
(b) Muscular tissue
(c) Cartilagenous tissue
(d) Adipose tissue.
Answer:
(d) Adipose tissue.

Question 10.
Functional hormone present in nervous system during impulse generation is: (a) Acetylcholine
(b) Parathormone
(c) Cortisteron
(d) Corticosterone.
Answer:
(a) Acetylcholine

Question 11.
In vertebrate animals, striated muscles are found in :
(a) Lungs
(b) Blood vessels
(c) Gall bladder
(d) Hind limbs.
Answer:
(d) Hind limbs.

Question 12.
Voluntary muscles are found ¡n:
(a) Lungs
(b) Liver
(c) Arms
(d) Heart.
Answer:
(c) Arms

Question 13.
Ciliated epithelium is found in:
(a) Taste buds
(b) Inner lining of stomach
(c) Trachea
(d)None of these.
Answer:
(c) Trachea

2. Fill in the blanks:

(A)
1. Cells found in the cartilage are ………………..
Answer:
Chondrioblast

2. Nervous tissue help for …………….. and ……………..
Answer:
Control, Co-ordination

3. Presence of ……………. is specific characteristic of Phylum Chordata.
Answer:
Notochord

4. ……………… and …………………. help for locomotion in earthworm.
Answer:
Setae, muscles

5. Frog is …………….. oiganism.
Answer:
Amphibian

(B)
1. ………………. is an anticoagulant, found in the blood.
Answer:
Heparin

2. The red blood corpuscles of ……………….. contain nucleus.
Answer:
Camel

3. Cholesterol is synthesised by the ………………
Answer:
Liver

4. The abnormal rise in the total blood R. B. Cs. count is known as …………………..
Answer:
Polycythemia

5. ………………. tissue works as a packing tissue.
Answer:
Areolar.

3. Match the following:
(A)

Column ‘A’	Column ‘B’
1. Haversian system	(a) Adipose tissue
2. Adipocytes	(b) Areolar tissue
3. Mast cells	(c) Mammalian bone
4. Platelets	(d) Nerve cell
5. Axon	(e) Blood.

Answer:
1. (c) Mammalian bone
2. (a) Adipose tissue
3. (b) Areolar tissue
4. (e) Blood.
5. (d) Nerve cell

(B)

Column ‘A’	Column ‘B’
1. Stratified epithelium	(a) Erythrocytes
2. Polycythemia	(b) Myelinated nerve fibres
3. Blood coagulation	(c) Trachea
4. Dendrite	(d) Collagen fibres
5. White fibrous tissue	(e) Prothrombin.

Answer:
1. (d) Collagen fibres
2. (a) Erythrocytes
3. (e) Prothrombin
4. (b) Myelinated nerve fibres
5. (c) Trachea

(C)

Column ‘A’	Column ‘B’
1. Compound epithelium	(a) Alimentary canal
2. Compound eye	(b) Cockroach
3. Septal nephridia	(c) Skin
4. Open circulatory system	(d) Mosaic vision
5. Intestinal caecum	(e) Earthworm
6. Osteocyte	(f) Penis
7. Copulatory organ	(g) Bone.

Answer:
1. (c) Skin
2. (d) Mosaic vision
3. (e) Earthworm
4. (b) Cockroach
5. (a) Alimentary canal
6. (g) Bone
7. (f) Penis

(D)

Column ‘A’	Column ‘B’
1. Phagocytosis of harmful germs	(a) Bone
2. Connection at the joint of bones	(b) Bone marrow
3. Haversian canals in mammals	(c) Sarcomere
4. Structural and functional unit of muscle	(d) Lymphocyte
5. Production of red blood corpuscles	(e) Ligament.

Answer:
1. (d) Lymphocyte
2. (e) Ligament.
3. (a) Bone
4. (c) Sarcomere
5. (b) Bone marrow

(E)

Column ‘A’	Column ‘B’
1. Less protein containing matrix	(a) Serum
2. Liquid squeezes out of the clot	(b) Leukemia
3. Increase in number of W.B.Cs.	(c) Haemophilia
4. No blood clotting occurs	(d) Anticoagulent
5. Sodium oxalate	(e) Lymph.

Answer:
1. (e) Lymph
2. (a) Serum
3. (b) Leukemia
4. (c) Haemophilia
5. (d) Anticoagulent

4. Write true or false:

1. In the foetus, R.B.Cs. are mainly formed in liver and spleen.
Answer:
True

2. The hump of Camel is made up of skeletal tissue.
Answer:
False.

3. Protein present in the matrix of cartilage is known as chondrin.
Answer:
True,

4. Cytoplasm of muscle cell is called neuroplasm.
Answer:
False.

5. The fibrous tissue which connects the two bones is Ligament.
Answer:
True,

6. Body of earthworm is metamerically segmented.
Answer:
True

7. Head of cockroach consists of five segments.
Answer:
False.

8. Jointed leg is the characteristic of the class Insecta.
Answer:
True

9. Male and female cockroach may be identified on the basis of anal style.
Answer:
True

10. Main function of clitellum is copulation.
Answer:
False.

5. Answer in one word:

1. Supporting cells which hold together the nerve cells.
Answer:
Neuroglia cells,

2. Life span of R.B.Cs.
Answer:
120 days

3. R,B.Cs. are found in.
Answer:
Bone marrow and Spleen

4. Thick filaments made up of protein.
Answer:
Myosin

5. Tells present in cartilage are called.
Answer:
Chondrocytes

6. Which type of corpuscles is responsible for coagulation of blood.
Answer:
Prothrombin, Ca, Fibrinogen and trypnase enzyme or thrombin

7. Write the pH value of blood.
Answer:
7.4

Students get through the MP Board Class 11th Biology Important Questions Chapter 16 Digestion and Absorption which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Questions Chapter 16 Digestion and Absorption

Digestion and Absorption Class 11 Important Questions Very Short Answer Type

Question 1.
What is dental formula ? Give dental formula of human. (NCERT)
Answer:
Dental Formula: Formula representing number of different types of teeth found in the upper and lower half of the jaws is called as Dental formula.
Dental Formula = \(\frac{\text { Number of teeth in one half of upper jaw }}{\text { Number of teeth in one half of lower jaw }} \times 2\)
= \(\frac{\mathrm{I}, \mathrm{C}, \mathrm{P}, \mathrm{M}}{\mathrm{I}, \mathrm{C}, \mathrm{P}, \mathrm{M}}\)
Whereas,
I = Incisor,
C = Canine,
P = Premolar and M = Molar teeth
Dental formula of human = \(\frac{2,1,2,3}{2,1,2,3} \times 2=\frac{16}{16}=32\)

Question 2.
Name different types of teeth and give number of them in an adult person. (NCERT)
Answer:
In human four types of teeth are found :

1. Incisor (Chisel shaped): helps to cut the food.
2. Canine (Dagger shaped): helps for tearing of food.
3. Premolar and Molar: helps for grinding food.
4. Dental formula of permanent teeth of adult human :
   \(\frac{I_{2}, C_{1}, P_{2}, M_{3}}{I_{2}, C_{1}, P_{2}, M_{3}} \times 2\)
   or
   \(=\frac{2,1,2,3}{2,1,2,3} \times 2=\frac{8}{8} \times 2=\frac{16}{16}=32 \text { teeth }\)

Question 3.
Name bile pigments.
Answer:
Bile contains two pigments :

1. Bilirubin: Yellow pigment.
2. Biliverdin : Geen pigment.

Question 4.
Write the name of disease caused by deficiency of protein and iodine.
Answer:

• Disease caused due to deficiency of protein: Kwashiorkor and Marasmus.
• Disease caused due to deficiency of iodine: Goitre.

Question 5.
Write down the difference between chyme and chyle of man’s digestive system.
Answer:

• Chyme is the food paste formed due to peristaltic movement of the walls of stomach, whereas chyle is the food paste formed due to peristaltic movement of duodenum.
• Chyme is acidic in nature whereas chyle is alkaline in nature.

Question 6.
Name two functions of small intestine.
Answer:

1. Absorption of fatty substances.
2. Walls of intestine produce mucous which makes food sticky.

Question 7.
What is anaemic stage? How it is cured?
Answer:
Anaemia is a stage in which the amount of haemoglobin in blood is decreased. The chief cause of this condition is the deficiency of iron. It is also caused due to deficiency of vitamin B₆ and B₁₂. It can be cured by taking iron elements, vitamin B₆ and vitamin B₁₂.

Question 8.
Is digestion of protein possible when trypsinogen is transferred into stomach?
Answer:
The digestion of protein is not possible when enzyme trypsinogen is transferred into stomach because stomach contains HCl and thus, solution of stomach becomes acidic while trypsinogen enzyme can function only in alkaline medium.

Question 9.
Write the name of any four enzymes present in pancreatic juice.
Answer:

1. Amylase,
2. Lipase,
3. Trypsin and
4. Chymotrypsin.

Question 10.
If carbohydrate digesting enzymes are added in stomach then digestion of carbohydrate is possible. Why?
Answer:
Digestion of carbohydrate increases because amount of carbohydrate digesting enzyme increases.

Question 11.
Microvilli are found in intestine but are absent in stomach, why?
Answer:
Microvilli present in intestine perform the absorption of food materials. These villi increase the absorbing surface of intestine. Absorption is not accompanied in stomach hence villi are not found within it.

Question 12.
How vitamin A affects the vision? Explain it.
Answer:
Rods of retina are covered by a photosensitive pigment. These pigments decompose in sunlight but they are also synthesized in presence of vitamin A. Rhodopsin is not synthesized in the absence of vitamin A hence the vision of the person is affected and resulting night blindness.

Question 13.
What is lacteales? Where it is found?
Answer:
Lacteales is a short lymph duct found in villi of intestine. It absorbs the fatty acid and glycerol in small intestine. Because of fat particle the colour of lymph is white.

Question 14.
How ingested fat is absorbed in the body?
Answer:
First of all bile juice form the emulsion of fat then lipase enzyme hydrolyses emulsified fat into fatty acids and glycerol. Then it is absorbed by the microvillies of small intestine and passes into lacteal tubules and then through lymphatic system distributed to all parts of the body.

Question 15.
What is rickets? Name the vitamin, the deficiency of which is responsible for it.
Answer:
Rickets disease occurs in the children. Bones become weak and bent. This disease is caused due to the deficiency of vitamin D.

Question 16.
How milk is digested in the stomach?
Answer:
The gastric gland of the stomach secretes gastric juice which contains dil HCl and two inactive enzymes pepsinogen and prorenin. Inactive prorenin converts into active renin in the presence of dilute, hydrochloric acid. Renin hydrolyse milk protein (casein) into calcium paracaseinate.

Digestion and Absorption Class 11 Important Questions Short Answer Type

Question 1.
Define diphyodont, thecodont and heterodont.
Answer:
1. Diphyodont: Two sets of teeth are produced :

1. Temporary or milk teeth,
2. Permanent teeth.
3. Temporary teeth are replaced by permanent teeth during the age period of 6-12 years.

2. Thecodont: Teeth are embedded into the socket of jawbones.
3. Heterodont: Teeth are of various sizes, shapes and structures. In man these are of 4 types: Incisors, canines, premolars and molars.

Question 2.
How does digestion of polysaccharide and disaccharide occurs in the alimentary canal?
Answer:
(A) Digestion of Polysaccharide: Digestion of Polysaccharide like starch is completed in following three steps:
1. In the Buccal cavity: Salivary gland of the buccal cavity secretes saliva, which contains an enzyme ptyalin. Ptyalin hydrolyses starch into maltose.

2. In the Duodenum: Pancreas secretes pancreatic juice, which reaches to the duodenum, through pancreatic duct. Pancreatic juice contains Amylase enzyme, which hydrolyses starch into maltose.

3. In the Intestine: Intestinal gland of the intestine secretes intestinal juice (succus entericus) which contains maltase enzyme. Maltase hydrolyses maltose into two molecules of glucose.

(B) Digestion of Disaccharides: Disaccharides like Maltose, sucrose and lactose are digested in the intestine. Intestinal gland of intestine secretes Intestinal juice (succus entericus), which contains 3 enzymes for digesting disaccharides. They are Maltase, Sucrose and Lactose. They hydrolyse disaccharides in following ways :

Question 3.
Describe digestion of protein in the alimentary canal. (NCERT)
Answer:
Digestion is physicochemical process which decompose complex chemical sub¬stances of food into simple substances so that they are absorbed by the wall of alimentary canal. Digestion of protein occurs by the help of gastric juice, pancreatic juice and intestinal juice in the different parts of the alimentary canal in the following way :

1. Buccal cavity: Digestion of protein does not occur in this part.
2. Stomach: In the stomach, gastric juice is found which contains HCl, pepsinogen and prorenin enzymes. These enzymes convert into their active form by the help of HCl.

Pepsin converts protein into proteases and peptone.

Rennin converts milk protein into calcium paracaseinate.

3. Duodenum: In the duodenum, pancreatic juice helps for the digestion of protein. Trypsin and chymotrypsin enzymes present in the pancreatic juice convert peptones into polypeptides.

4. Intestine: In the intestine, intestinal juice is found which contains erepsin enzyme which converts di, tri and polypeptides into amino acids.

Question 4.
Bile juice does not contain any digestive enzyme even then it is important for digestion, why? (NCERT)
Answer:
Bile is a secretion of liver which emulsifies fat and activate some enzymes. Bile juice does not contain any digestive enzymes but still, it plays an important part like :

• It changes acidic food into alkaline one.
• It emulsifies fat so that lipase may act upon it.
• It converts inactive trypsin into active one.
• It instigates walls of intestine to start peristaltic movement.
• It’kills harmful bacteria of food.
• It neutralizes acidic nature of the food which comes from the stomach.

Digestion and Absorption Class 11 Important Questions Long Answer Type

Question 1.
How does digestion of butter occurs in your body? How its absorption occurs? Describe in details.
Answer:
(A) Digestion of fat in our body: Digestion of fat in our body occurs in following steps:
1. Digestion in the stomach: Gastric gland of the stomach secretes gastric juice, which contains enzyme gastric lipase. It hydrolyses emulsified fat into fatty acid and glycerol. Gastric lipase is very slow in function.

2. Digestion in the Duodenum: Duodenum receivers bile juice from the liver and pancreatic juice from the pancreas.
Bile juice helps for emulsification of fat

Pancreatic juice contains lipase enzyme, which hydrolyses emulsified fat into fatty acid and glycerol.

3. Digestion in the Ileum : Ileum part of the intestine contains intestinal gland, which secretes intestinal juice (succus entericus). It hydrolyses emulsified fat into fatty acid and glycerol.

(B) Absorption of fat: After completion of digestion process in the intestine, still some fats remain undigested and remain in the form of mono and diglycerides. These undi¬gested fats combine with bile salt to form small globules called as micelles. These are absorbed by the cells of microvilli by diffusion, where they forms large globules by forming phospholipids.

Now, these large globules are called as chylomicron. Chylomicron reaches to the lymph capillaries present in the microvilli become milky in colour due to presence of large globules. Milky lymph is called as chyle and now these fine lymph capillaries are called as Lacteals, which opens into lymph vessels. These lymph vessels carries digested fat to the thoracic lymph duct from where it reaches to the hepatic portal system.

Question 2.
Give role of pancreatic juice in digestion of protein. (NCERT)
Answer:
Pancreatic juice is secreted by the pancreas, which reaches to the duodenum through pancreatic duct. Pancreatic juice contains 6 enzymes but out to them three enzymes help for digestion of protein.

1. Inactive enzyme trypsinogen is converted into active trypsin by the intestinal enzyme enterokinase.

2. Inactive chymotrypsinogen is converted into active chymotrypsin by the activity of Trypsin.

3. Trypsin and Chymotrypsin both help to hydrolyse peptone into peptides.

4. Carboxypeptidase: It hydrolyses peptides into amino acids.

Question 3.
How does digestion of protein occurs in the stomach ?
Or
Explain the process of digestion in the stomach. (NCERT)
Answer:
Digestion in stomach: Stomach is the widest part of the alimentary canal where food is kept for 1- 5 hours. Stomach contains a number of microscopic gastric glands. Secretion of gastric gland is called gastric juice. It contains proenzyme pepsinogen, HCl, mucus, a weak enzyme gastric lipase and intrinsic factor. In calf a weak enzyme rennin is also present.

Mucus protects the stomach wall against HCl and proteolytic enzyme pepsin. It also lubricates the food. HCl converts proenzyme pepsinogen into proteolytic enzyme pepsin. Pepsin is also autocatalytic. Other functions of HCl includes providing acidic pH for working of pepsin, stoppage of ptyalin action and prevention of bacterial growth. The enzymes of gastric juice helps in the digestion as follows :
1. Digestion by pepsin: It is secreted in the form of inactive pepsinogen and is activated by HCl to digest protein molecules.

2. Digestion by mucin: It is forming a protective sheath on mucous membrane after decreasing the acidic effect of gastric juice. Water and mucin of saliva moisten the food and make the mucous membrane slippery. Thus it helps in the process of digestion.

3. Digestion by rennin : It is secreted in the form of prorennin and is activated by HCl. It converts milk into curd and takes part in the process of digestion as follows :

4. Digestion by gastric lipase: It is responsible for hydrolysis of fats into fatty acids and glycerols.
At the end of its stay in stomach, the food is converted into a pulpy mass called chyme. The pyloric valve opens at intervals and chyme passes into duodenum part of small intestine.

Question 4.
Describe about the gland which contains the islets of Langerhans.
Answer:
The islets of Langerhans are found in pancreas. It is a pink coloured small gland situated in the ‘U’ shaped portion of duodenum. It is covered by a membrane which is filled with a connective tissue. It contains a number of the lobules held together by connective tissue.

A lobule has several alveoli or acini. An alveolus possess a narrow cavity and a layer of glandular cells. The latter secrete pancreatic juice. The narrow cavity forms a ductule. The ductules form ducts, which join to produce a single pancreatic duct. The latter opens into duodenum either independently or along with common bile duct.

Question 5.
Give role of chymotrypsin. Name the gland from which it is secreted and also name two more enzymes belongs to same category by this gland. (NCERT)
Answer:
Chymotrypsin is a proteolytic enzyme secreted as inactive form in the pancreatic juice called as chymotrypsinogen.
Chymotrypsinogen is activated by trypsin into active chymotrypsin.

Chymotrypsin is produced in inactive form as chymotrypsin in the pancreatic juice which is secreted by Pancreas. It reaches to the duodenum through pancreatic duct.
Two other enzyme belongs to some category, i.e. proteolytic enzymes secreted by pancreas are :

1. Trypsinogen and
2. Carboxypeptidase.

Question 6.
What will happen if hydrochloric acid is not secreted in the stomach? (NCERT)
Answer:
Functions of HCl in the stomach are as follow :

1. Stops activity of ptyalin as due to HCl, medium of food become acidic.
2. Enzymes of gastric juice become active in acidic medium.
3. It destroys harmful germs present in the food.
4. It activates inactive enzymes pepsinogen and prorenin into active pepsin and renin respectively.
5. Helps to digest bones.
6. Regulates opening and closing of pyloric aperture.
   If hydrochloric acid is not secreted in the stomach then all above functions will be effected.

Question 7.
Answer in brief: (NCERT)
(a) Villi are found in the intestine but not in the stomach, why?
(b) How does pepsinogen converts into active form?
(c) What are the basic layers of wall of the alimentary canal?
(d) How does bile help to digest fat?
Answer:
(a)Microvilli present in intestine perform the absorption of food materials. These villi increase the absorbing surface of intestine. Absorption is not accompanied in stomach hence villi are not found within it.

(b) Gastric juice secreted by gastric gland of the stomach contains dil HCl, Which converts inactive pepsinogen into active pepsin.

(c) Basic layers of the wall of the alimentary canal are – Serosa, Muscular layer, Submucosa and Mucosa.
(d) Bile juice secreted by the liver reaches to the duodenum through bile duct. It do not contain any enzyme but help for digestion in following ways :

• It helps for emulsification of fat.
• It helps to neutrilize acidic nature of the food which comes from the stomach, because bile juice itself is alkaline in nature.

Question 8.
Give functions of Liver. (NCERT)
Answer:
Functions of Liver: Liver is the most important organ of the body, it performs the following function:

1. Excess quantity of glucose in blood in liver cells is converted into glycogen.
2. Excess quantity of amino acids in hepatic cells are converted into urea.
3. Synthesis of albumin from amino acids.
4. Production of blood proteins like prothrombin and fibrinogen which are essential for blood coagulation.
5. Production of Heparin, an enzyme which prevents clotting of blood in the blood vessels.
6. Production of R.B.C. during foetal life.
7. Storage of Vitamin A, B₁₂ and D.
8. Storage of inorganic substances like copper and iron.
9. It charges acidic food into alkaline one.
10. Excess quantity of glucose present in the blood reaches to liver and converted into fat by lipogenesis process and stored in the adipose tissue.
11. When there is lack of glucose in the blood Glycogen of the liver decomposes by glycogenolysis process to proudce glucose and supplied to the blood.
12. Kupffer cells found in the cells of liver kills harmful bacterias by phagocytic action.
13. It helps for digestion of fat by emulsification by its bile juice. It also help to neutralize acidic nature of the food as bile juice is alkaline.
14. It helps to store copper and iron.

Digestion and Absorption Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
The function of bile is:
(a) Emulsification of fat
(b) Excretion of waste materials
(c) Digestion of fat by enzymes
(d) None of these.
Answer:
(a) Emulsification of fat

Question 2.
Amylase enzyme act on:
(a) Carbohydrate
(b) Protein
(c) Fat
(d) Sugar.
Answer:
(a) Carbohydrate

Question 3.
Lactase is found in:
(a) Saliva
(b) Bile
(c) Pancreatic juice
(d) Intestinal juice.
Answer:
(d) Intestinal juice.

Question 4.
Goitre affects:
(a) Metabolism
(b) Vision
(c) Excretion
(d) Speech.
Answer:
(a) Metabolism

Question 5.
Which of the following pair is correctly matched:
(a) Renin-protein
(b) Amylase-lactose
(c) Trypsin-starch
(d) Invertase-sucrose.
Answer:
(a) Renin-protein

Question 6.
Enzyme arginase is found in:
(a) Buccal cavity
(b) Stomach
(c) Intestine
(d) Liver.
Answer:
(d) Liver.

Question 7.
Bile salts are secreted into alimentary canal, where they involve in the absorption of:
(a) Na and Ca ions
(b) Fat-soluble vitamins
(c) Amino-acids and monosaccharides
(d) All the nutrients present in chyme.
Answer:
(b) Fat-soluble vitamins

Question 8.
Found in the gastric juice: (NCERT)
(a) Pepsin, Lipase, Renin
(b) Trypsin, Lipase, Renin
(c) Trypsin, Pepsin, Lipase
(d) Trypsin, Pepsin, Renin.
Answer:
(a) Pepsin, Lipase, Renin

Question 9.
Succus entericus name is given for: (NCERT)
(a) Joining part of the Ileurn and Large intestine
(b) Intestinal juice
(c) Swelling in the alimentary canal
(d) Appendix.
Answer:
(b) Intestinal juice

Question 10.
Rickets can be prevented by taking:
(a) Oranges
(b) Carrots
(c) Green vegetables
(d) Calciferol.
Answer:
(d) Calciferol.

Question 11.
Amylolytic enzymes are formed in:
(a) Salivary glands and liver
(b) Stomach and liver
(c) Stomach and pancreas
(d) Salivary glands and pancreas.
Answer:
(d) Salivary glands and pancreas.

Question 12.
islets of Langerhans produce:
(a) Insulin
(b) Rennin
(c) Ptyalin
(d) HCl.
Answer:
(a) Insulin

Question 13.
The hardest substance of vertebrátes body is:
(a) Keratin
(b) Enamel
(c) Dentine
(d) Chondrin.
Answer:
(a) Keratin

Question 14.
Pepsin is secreted in:
(a) Intestine
(b) Liver
(c) Gonads
(d) Stomach.
Answer:
(d) Stomach.

Question 15.
Peyer’s patches contain:
(a) Mucus
(b) Sebum
(c) Lymphocytes
(d) Red blood cells.
Answer:
(d) Red blood cells.

Question 16.
The pH of succus entericus is:
(a) 78
(b) 66
(c) 56
(d) 20.
Answer:
(c) 56

Question 17.
Digestion of protein takes place in :
(a) Duodenum and stomach
(b) Stomach and oesophagus
(c) Small and large intestine
(d) Intestine and rectum.
Answer:
(a) Duodenum and stomach

Question 18.
The food present in stomach consisting of acidic medium is called :
(a) Chegle
(b) Chyme
(c) Bolus
(d) All of these.
Answer:
(b) Chyme

Question 19.
Goblet cells secretes:
(a) Enzymes
(b) Mucus
(c) Hormones
(d) HCl.
Answer:
(b) Mucus

Question 20.
Glycogen is stored in the:
(a) Blood
(b) Liver
(c) Lungs
(d) Kidney.
Answer:
(b) Liver

Question 21.
Herbivorous organisms can digest cellulose because :
(a) In the gastric juice enzymes for this are found
(b) Bacterias present in the caecum help for this
(c) Alimentary canal is long
(d) Molar and premolar teeth help for chewing.
Answer:
(b) Bacterias present in the caecum help for this

Question 22.
Pancreatic juice helps for digestion of:
(a) Protein
(b) Protein and fat
(c) Carbohydrates and Protein
(d) Protein, fat, carbohydrates.
Answer:
(d) Protein, fat, carbohydrates.

Question 23.
Disease caused due to deficiency of vitamin B₁₂ :
(a) Beri-beri
(b) Pellagra
(c) Colossi
(d) Scurvy.
Answer:
(a) Beri-beri

Question 24.
What is called as thiamine :
(a) Vitamin-B
(b) Vitamin-A
(c) Vitamin-B,
(d) Vitamin B complex.
Answer:
(c) Vitamin-B,

Question 25.
Enterokinase induces:
(a) Pepsinogen
(b) Trypsin
(c) Pepsin
(d) Trypsinogen.
Answer:
(d) Trypsinogen.

Question 26.
All digestive enzymes are:
(a) Lipases
(b) Hydrolase
(c) Transferase
(d) Oxidases.
Answer:
(b) Hydrolase

Question 27.
In the caecum of rabbit digestion of which substance occurs
(a) Fat
(b) Starch
(c) Cellulose
(d) Protein.
Answer:
(c) Cellulose

Question 28.
Disease caused due to deficiency of vitamin-D is:
(a) Rickets
(b) Scurvy
(c) Night blindness
(d) Pellagra.
Answer:
(a) Rickets

Question 29.
Pellagra is caused due to deficiency of
(a) Vitamin-B5
(b) Vitamin-C
(c) Vitamin-D
(d) Vitamin-E.
Answer:
(a) Vitamin-B5

Question 30.
Ptyalin hydrolyses:
(a) Fat
(b) Protein
(c) Lipid
(d) Starch.
Answer:
(d) Starch.

Question 31.
islet of Langerhans produces:
(a) Insulin
(b) Trypsin
(c) Lipase
(d) Amylase.
Answer:
(a) Insulin

Question 32.
Pepsin is produced in the:
(a) Intestine
(b) Liver
(c) Gonads
(d) Stomach.
Answer:
(d) Stomach.

2. Fill in the blanks:
1. The secretion of liver is …………………………. .
Answer:
Bile,

2. Marasmus disease is caused due to deficiency of …………………………. in food.
Answer:
Protein,

3. Villi are found in …………………………. and …………………………. .
Answer:
Jejunum, ileum,

4. The secretion of HCl and gastric juice is controlled by …………………………. hormone.
Answer:
Enterogestron,

5. Herbivorous animal can digest cellulose because their …………………………. is well-developed.
Answer:
Caecum,

6. Rickets caused due to deficiency of …………………………. .
Answer:
Vitamin D.

7. Pepsin formation takes place in …………………………. .
Answer:
Stomach,

8. In digestion emulsification of fat takes place by …………………………. .
Answer:
Bile juice,

9. Food reaches to stomach from oesophagus by …………………………. movement.
Answer:
Peristalsis,

10. Digestive system of human consists of …………………………. and …………………………. .
Answer:
Alimentary canal, Digestive glands,

11. Saliva is produced by …………………………. pairs of salivary glands.
Answer:
Three,

12. Wall of the alimentary canal from oesophagus to rectum consists of …………………………. layers.
Answer:
Three or four.

3. Match the following:

(A)

Column ‘A’	Column ‘B’
1. Enterokinase	(a) Casein
2. Ruminant	(b) Vitamin C
3. Rennin	(c) HCl
4. Collagen	(d) Reticulum
5. Pepsin	(e) Trypsin.

Answer:
1. (e) Trypsin.
2. (d) Reticulum
3. (a) Casein
4. (b) Vitamin C
5. (c) HCl.

(B)

Column ‘A’	Column ‘B’
1. Night blindness	(a) Vitamin C
2. Pelagra	(b) Vitamin B
3. Beriberi	(c) Vitamin B12
4. Scurvy	(d) Vitamin A
5. Pemitious anaemia	(e) Nicotinamide.

Answer:
1. (d) Vitamin A
2. (e) Nicotinamide.
3. (b) Vitamin B
4. (a) Vitamin C
5. (c) Vitamin B₁₂.

(C)

Column ‘A’	Column ‘B’
1. Bilirubin and Biliverdin	(a) Parotid
2. Hydrolysis of starch	(b) Bile juice
3. Digestion of fat	(c) Lipase
4. Salivary gland	(d) Amylase.

Answer:
1. (b) Bile juice
2. (d) Amylase.
3. (c) Lipase
4. (a) Parotid.

4. Answer in one word:

1. Write the name of bile pigments.
Answer:
Bilirubin and Biliverdin,

2. Name the vitamin known as anti sterility vitamin.
Answer:
Vitamin E,

3. Write the name of disease caused due to deficiency of vitamin B.
Answer:
Beriberi,

4. Write the name of any four water-soluble vitamins.
Answer:
Vitamin B₁, B₂, B₆ and Vitamin C,

5. Write the name of any four enzymes found in pancreatic juice.
Answer:
Trypsin, Amylase, Lipase, Nuclease,

6. Write the name of any two diseases of children caused due to the deficiency of protein.
Answer:
Kwashiorkor, Marasmus,

7. Write the name of organ functioning as endocrine as well as exocrine gland.
Answer:
Pancreas,

8. Write the name of any two protein-digesting enzymes.
Answer:
Trypsin, Chymotrypsin,

9. Name the vitamin responsible for night blindness.
Answer:
Vitamin A,

10.Which provides more energy, carbohydrate or fat?
Answer:
Fat,

11. Name the disease caused due to deficiency of Vitamin-C.
Answer:
Scurvy,

12.Name the disease caused due to deficiency of Vitamin-D.
Answer:
Rickets,

13.Name the disease caused due to deficiency of iodine.
Answer:
Goitre,

14.Name the hormone induce secretion of gastric juice.
Answer:
Gastrin,

15. Which component of food is digested by bile juice?
Answer:
Lipid and fat,

16. Which enzyme is secreted by salivary gland?
Answer:
Ptyalin,

17. Where gastric juice is secreted?
Answer:
By gastric gland of stomach,

18. Which type of stomach is found in ruminate animal?
Answer:
Ruminant stomach,

19. Pepsin is secreted in which medium?
Answer:
Acid medium,

20. Insulin is secreted by which part?
Answer:
islets of Langerhans of pancre.

Students get through the MP Board Class 11th Biology Important Questions Chapter 17 Breathing and Exchange of Gases which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Questions Chapter 17 Breathing and Exchange of Gases

Breathing and Exchange of Gases  Class 11 Important Questions Very Short Answer Type

Question 1.
How smoking affects the process of respiration ?
Answer:
Regular smoking resulting in the deposition of tar of smoke on the walls of trachea. Due to which their cells are always irritated and resulting cough. The cells of respiratory organs are irritated due to substances of smoke which produces cancer in neck, mouth and lungs. This smoke is also collected in the air chambers of lungs which reduces their volume and resulting congestion.

Question 2.
What is Vital capacity ? Give its importance. (NCERT)
Answer:
Volume of air which can be maximally inspired and also maximally expired by the lungs is termed as Vital capacity.
In males, it is about 4500 ml while in females, it is about 3000 ml. (The vital capacity of the lungs in mountain dwellers and athletes is always high in comparison to others.)
Importance: On the basis of vital capacity of the lungs of a patient defect or disease of the lungs can be diagnosed by the doctors.

Question 3.
Name the condition of the body occurs due to deficiency of oxygen.
Answer:
Condition arises in the body due to deficiency of oxygen is called as Hypoxia. This condition arises due to height or deficiency of air pressure or due to cyanide poison or carbon monoxide.

Question 4.
What is Tidal volume? Determine Tidal value of a healthy adult person in one hour. (NCERT)
Answer:
Volume of the air inspired and expired by the lungs during effortless normal breathing in a healthy adult person is called as Tidal volume. It is about 500 ml in adult normal person. It increases during excitement and activity.
Tidal value of lungs in one breathing is 500 ml.
In one minute normal person respire 16 times and in one hour 16 x 60 times. Therefore, Tidal value of a healthy person in one hour is 500 mix 16 x 60 = 4,80,000 ml..

Question 5.
How does concentration of CO₂ and dissociation of oxyhaemoglobin are related with each other ?
Answer:
Concentration of CO₂ regulate the process of dissociation of oxyhaemoglobin. Both the processes are proportional to each other. Blood become acidic when CO₂ concentration increases in the blood, as CO₂ reacts with water to form carbonic acid (H₂CO₃).
CO₂+H₂O → H₂CO₃ Carbonic acid
Acidic medium of the blood stimulate dissociation of oxyhaemoglobin into oxygen
and haemoglobin.

Question 6.
Write differences between Respiration and Breathing.
Answer:
Differences between Respiration and Breathing

Respiration	Breathing
1. Respiration is a biochemical process in which food materials are oxidized by oxygen.	It is a physical process in which O2 is taken during inspiration and CO2 is released during expiration.
2. ATP (energy) is released in this process.	ATP is not released in this process.
3. This process is always intracellular.	It is an extracellular process.

Question 7.
It is better to respire through nose than mouth. Why ?
Answer:
Respiration through nose is better because of the following reasons :

• It is a natural process.
• Air of the atmosphere contain dust particles, bacterias and other harmful substances, which are filtered by the hairs present in the nose wall.
• Turbinal bone of the nose heat the air, thus it acts as air conditioner.

Question 8.
Respiratory organs are not essential for protozoans ?
Answer:
Protozoans are unicellular animals like Amoeba, Paramoecium, Euglena. Gaseous exchange occurs through plasma membrane by diffusion process. Thus, respiratory organs are not essential for protozoans.

Question 9.
Write names of respiratory organs of leech and insect.
Answer:
Animals Respiratory organs
Leech – Skin
Insect – Trachea.

Question 10.
How much volume of the air is left in the lungs at the end of forcible expiration ? (NCERT)
Answer:
At the end of forcible expiration volume of the air left in the lungs is 1200 ml. It is called as residual volume.

Question 11.
Name the respiratory organs of the following ;
(1) Fish,
(2) Amphibians,
(3) Birds,
(4) Earthworm.
Answer:

Organisms	Respiratory organs
(1) Fish	Gills
(2) Amphibians	Skin, lungs, buccal cavity
(3) Birds	Lungs and air sacs
(4) Earthworm	Skin.

Breathing and Exchange of Gases Class 11 Important Questions Short Answer Type

Question 1.
What is the significance of intercostal muscles in expiration?
Answer:
Intercostal muscles on contraction reduce the volume of the thorax. Decrease in the volume of thorax will also cause the decrease in the volume of lungs. Thus air pressure is increased within lungs and air is passed out from the lungs. Thus intercostal muscles helps in the process of expiration.

Question 2.
What is Respiration? It is of how many types? Explain in brief.
Answer:
Respiration: Respiration is an oxidation reaction inside the cell in the presence of specific enzymes in which complex organic substances are oxidized to produce C02, H20, other organic substances and energy.
C₆H_12O6 +6O₂ → 6H₂O+ 6CO₂ + 673 kcal energy or38ATP
It is of two types :
1. Anaerobic respiration: In this process, food materials are incompletely oxidized in the absence of 02 to produce C02 and simple organic substances like ethyl alcohol, citric acid, lactic acid, malic acid, butyric acid and oxalic acid. It is found in micro-organisms, bacteria, yeasts and some animal tissues.

C₆H_12O6 → 2C₂H₅OH +2C0₂ +21 kcal or 2ATP

2. Aerobic respiration: It is a respiration in which food materials are completely oxidized in the presence of 02 to produce C02, H20 and energy. It is found in plants and animals.
C₆H_12O6 +6O₂ → 6CO₂ + 6H₂O+ 673kcal or38ATP

Question 3.
Describe the chemical nature and properties of haemoglobin.
Answer:
Red blood corpuscles contain a red pigment known as haemoglobin. Haemoglo¬bin possessing a specific protein known as globulin, in which iron is associated with homegroup. Haemoglobin is a micromolecule with imperial formula: C₅₈₇ H₁₂₁₃ N₁₉₅ O₂₁₄ S₃ Fe. Haemoglobin combine with oxygen and form oxyhaemoglobin. It is dissociated in tissues into O₂ and haemoglobin and thus it plays an important role in the conduction of O₂ in the cells.

Question 4.
Differentiate between external and internal respiration.
Answer:
Differences between external and internal respiration

External Respiration	Internal Respiration
1. It is a physical process.	It is both a physical as well as physiological process.
2. It involves lungs, skin and other respiratory surfaces.	It involves every cell.
3. 02 is absorbed and CO2 is released in this process.	Food is oxidized and CO2 is released in this process.
4. It results in exchange of gases between inspired air and blood.	It results in the exchange of gases between living cells and the blood across the tissue fluid.

Question 5.
What do you understand by chloride shift?
Answer:
Chloride shift: During respiration food materials are oxidized and releases CO₂ and energy (ATP). Released CO₂ is then transported into cells through tissue fluid. In tissues, this CO₂ combines with water and forms carbonates in R.B.C. which increases their pH. To maintain electrical concentration balance bicarbonate ions (HCO^–₃) are diffused into
plasma and equal amount of Cl^– ions are diffused into erythrocytes (R.B.C.). Thus, sodium bicarbonate is collected in the plasma and potassium chloride is collected in R.B.C. This whole process is termed as chloride shift.

Question 6.
Write differences between following :
(a) Tracheoles and Bronchioles,
(b) Carbamino haemoglobin and Oxyhaemoglobin.
Answer:
(a) Differences between Tracheoles and Bronchioles

Tracheoles	Bronchioles
1. In insects each trachea branches extensively to form tracheoles which ends intracellularly in the tissue cells.	In mammals each bronchi divides into bronchioles.
2. Through tracheoles exchange of gases occurs between air and cell fluid.	Exchange of gases do not occur through bronchioles but air passes through it.

(b) Differences between Carbamino haemoglobin and Oxyhaemoglobin

1. It is formed by combination of CO2 with amino-haemglobin.	It is formed by combination of oxygen with haemoglobin.
2. CO2 is transported in this form through blood.	O2 is transported in this form through blood
3. Carbamino haemoglobin is carried through blood towards the lungs.	Oxyhaemoglobin is carried through blood to all parts of the body.
4. Blood containing carbamino haemoglobin is called as impure blood.	Blood containing oxyhaemoglobin is called as pure blood.

Question 7.
What will be the salient features for any respiratory surface for gaseous exchange?
Answer:
The following are the characteristics for any respiratory surface :

• The surface must be thin and permeable,
• It must be moist so that gases may dissolve in it.
• The surface area must be vast so that more gaseous exchange may take place,
• It must be well-supplied with capillary network of blood,
• It must be always in contact with oxygen.

Question 8.
How the concentration of CO₂ and dissociation of oxyhaemoglobin are related with each other?
Answer:
The concentration of CO₂regulate the process of dissociation of oxyhaemoglobin. Both processes are proportional with each other. The blood becomes acidic when CO₂ concentration is increased in blood because CO₂ forms carbonic acid (CO₂+H₂O → H₂O^–₃)when it combines with the water of blood plasma. Acidic medium of blood stimulates the dissociation of oxyhaemoglobin into oxygen and haemoglobin. Thus, it is dissociated into O₂and haemoglobin.

Question 9.
Mitochondria is regarded as powerhouse of cells, why?
Answer:
Mitochondria is a cell organelle in which food materials are oxidized to form ATP (energy) which stores higher amount of energy. When energy is required then the dissociation of ATP takes place and a large amount of energy liberated, hence mitochondria is called as powerhouse of the cell.

Question 10.
Differentiate Aerobic and Anaerobic respiration.
Answer:
Differences between Aerobic and Anaerobic respiration

Aerobic respiration	Anaerobic respiration
1. Aerobic respiration requires oxygen takes place in cytoplasm and mitochondria.	It does not require oxygen and takes place in cytoplasm only.
2. The substrate is completely oxidized.	The substrate is incompletely oxidized.
3. End-products are CO2 and H2	End-products are ethyl alcohol and CO2.
4. 38 ATP (673 kcal) energy is produced.	2 ATP (21 kcal) energy is produced.
5. Harmful substances are not formed.	Harmful substance ethyl alcohol is forme.

Question 11.
Write difference between Respiration and Combustion.
Answer:
Differences between Respiration and Combustion

Respiration	Combustion
1. It is a vital process	It is a physical process.
2. Rate of oxidation is slow.	Rate of oxidation is faster.
3. It is controlled process.	It is uncontrolled process.
4. Energy is released in the form of ATP.	Energy is released in the form of light and heat.
5. It is controlled process.	It is an uncontrolled process.

Question 12.
What is oxygen debt?
Answer:
Oxygen debt: During strenuous work, the muscle does not get sufficient amount of oxygen to meet its energy needs immediately. Hence, it contracts anaerobically and accumulates lactic acid produced by anaerobic glycolysis. During recovery stage, the accumulated lactic acid is oxidized by utilizing oxygen in addition to the normal regular required manner. The extra oxygen consumed during recovery stage is called oxygen debt of the muscles. It is used in oxidizing lactic acid and in restoring the depleted creatine phosphate and ATP in the muscle fibre.

Question 13.
Why does muscles get tired while doing more work?
Answer:
While doing active or fast work, muscles need fast supply of energy, for which anaerobic oxidation of food occurs in the muscles to supply energy at fast rate. Accumulation of lactic acid occurs in the muscle due to this muscles become tired and stop functioning. This is called as Muscle fatigue. After taking rest for sometime lactic acid converts into pyruvic acid again and muscles become normal.

Question 14.
Define oxygen dissociation curve. (NCERT)
Answer:
Oxygen dissociation curve: It is a sigmoid or ‘S’ shaped curve which represents percentage of oxygen saturation in the haemoglobin due to _pO2 pressure (partial). Sigmoid curve is formed due to saturation of haemoglobin.

Breathing and Exchange of Gases Class 11 Important Questions Long Answer Type

Question 1.
Explain the mechanism of C_O2 transport through blood. (NCERT)
Answer:
Mechanism of transport of C_O2 through blood: Respiraion is defined as “A biochemical catabolic process which involves stepwise degradation or oxidation of glucose in mitochondria, with the release of energy and formation of ATP, inside the living cell at body temperature”.

In our body the circulation of CO₂ with blood takes place by the following methods :
1. Dissolved in plasma: The solubility of CO₂ in blood is twenty times more than the solubility of _O2 in blood.
But only 7% of the total C0₂ is transported through plasma. When CO₂ comes in contact with plasma then it forms carbonic acid by the use of water of plasma. This carbonic acid is transported up to lungs and dissociated to release water and CO₂ is transported into alveoli of the lungs.

CO₂ +H₂O ⇌  H₂C0₃(Carbonic acid)

2. As carbamino compounds: About 30% of CO₂ passing into erythrocytes, combine with haemoglobin to form a stable compound known as carbamino-haemoglobin. After reaching the alveolar surfaces these compounds release CO₂ on it which is then diffused into alveoli.

HbNH₂ +CO₂ →HbNHCOOH (in tissue) (Carbamino compound)

3. As bicarbonates: Remaining 70% of CO₂ is transported in the form of bicarbonates. Bicarbonates are formed by the reaction of C02 with the carbonates of sodium and potassium. These bicarbonates are transported by plasma or erythrocytes. After reaching alveoli these compounds release C0₂ on it which is then diffused into alveoli. Carbonic anhydrase enzyme catalyses this reaction.
In this way CO₂ and is released from bicarbonate. At the same time oxygenation of haemoglobin simultaneously releases CO₂. In this way CO₂ is released in the lungs. This CO₂ thus released, is forced out of the body through expiration.
H₂O+ CO₂ + Na₂CO₃ → 2NaHC0₃ (Sodium bicarbonate)

Question 2.
Describe any five diseases related to respiration along with two symptoms.
Or,
Describe some disorders of respiratory system.
Answer:
Some common disorders of respiratory system of human beings :
1.Asthma: It is a type of allergic reaction in which mucous of respiratory canal become dense and thus expiration is affected. This disease is generally caused due to pollen grains of flowers and some food substances.

2. Common cold: It is a viral disease in which swelling occurs in mucous membrane and secretion of mucous is also increased. Nose becomes closed and thus intake of oxygen is difficult.

3. Emphysema: It is the condition where the breaking up of the walls separating the alveoli occurs reducing the gas exchange area of the lungs.

4. Lung cancer: The disease is generally found in smokers. The lung tissues growing abnormally and ultimately resulting in the death of the suffering person.

5.Tuberculosis: It is a bacterial disease of lungs and is caused by Mycobacterium tuberculosis. These bacteria releasing tuberculin a toxin which causes fever, loss of weight and general weakness in the patient.

6. Pneumonia: It is also a bacterial disease, caused by Diplococcus pneumonia. The symptoms are accumulation of lymph and mucous in the alveoli and bronchioles. This seriously effects the breathing.

7. Pleurisy: The lungs are enveloped by a thin sheet of smooth epithelium called pleura. Sometimes this pleura gets inflamed because of certain reasons. The disease thus results is called pleurisy.

Question 3.
Explain the process of inspiration in normal condition. (NCERT)
Answer:
Breathing: Breathing or ventilation is the movement that brings in fresh air into the lungs for gaseous exchange and takes out the foul air from them. There are two types of muscles which play an important role in breathing movement.

1. External intercostal muscles: A pair of these muscles arising from upper part of the ribs and is attached with the lower part of the backward rib.
2. Internal intercostal muscles: A pair of these muscles arising from lower part of each rib which goes up to the upper part of the backward rib and joined with it.

Mechanism of breathing: The process of breathing involves in the following two processes:
1. Inspiration or Inhalation: Inspiration is a process in which air taken into the lungs. During inspiration the external
intercostal muscle contract, the front ends of the ribs move upward and outward. At the same time the radial muscles of diaphragm contracts making it somewhat straight. This whole process increases the volume of the airtight chest cavity. This lowers the partial pressure inside the lungs. To make up the pressure air enters from outside through trachea and its branches to the air sacs and alveoli of the lungs.

2. Expiration or Exhalation:
Expulsion of the air to the outside from the body is termed as expiration. In this process internal intercostal muscle contracts due to which ribs return to their normal position.
The diaphragm relax, becomes convex and the abdominal organs place in its original position. Consequently, the capacity of the chest cavity decreases. The lungs become compressed, the pressure in them rises with the result the air from the lungs rushes out through air passage. Both of these processes are repeated alternatively. During resting period a healthy person breaths 16-20 times per minute.

Question 4.
Write differences between following: (NCERT)
(a) IRV and ERV
(b) Inspiratory capacity (IC) and Expiratory capacity (EC)
(c) Vital capacity and Total lung capacity.
Answer:
(a) Differences between IRV and ERV

Inspiratory reserve volume (IRV)	Expiratory reserve volume (ERV)
1. It is the amount of air that can be inhaled forcefully by the lungs after normal inspiration.	It is the amount of air that can be exhaled out forcefully by the lungs after normal expiration.
2. It is about 2500 ml in adult normal man and 1900 ml in normal woman.	It is about 1000 ml in adult normal man and 700 ml in normal woman.

(b) Differences between Inspiratory capacity (IC) and Expiratory capacity (EC)

Inspiratory capacity (IC)	Expiratory capacity (EC)
1. It is the volume of air which can be maximally inspired.	It is the volume of air which can be maximally expired.
2. Its value is equal to sum of Tidal volume and Inspiratory reserve volume (TV + IRV).	Its value is equal to sum of Tidal and Expiratory reserve volume (TV + ERV).
3. In normal male its value is (500 + 3000) ml i.e. 3500ml whereas in normal female its value is (500 + 1900) ml i.e. 2400ml.	In normal male its value is (500 +1000) ml i.e. 1500ml whereas in normal female its value is (500 +700) ml i.e. 1200 ml.

(c) Differences between Vital capacity and Total lung capacity

Vital capacity	Total lung capacity
1. Volume of air which can be maximally inspired and also maximally expired by the lungs in a normal healthy person is called as vital capacity.	It is the sum of Vital capacity and residual volume of the air in the lungs of a normal healthy person.
2. In normal man, it is about 4500 (3000 + 500 + 1000) ml whereas in normal woman it is about 3100 (1900 + 500 + 700) ml.	In normal man, it is about 5700 (4500 +1200) ml whereas in normal woman it is about 4200 (3100+ 1100) ml.

Question 5.
How is respiration process regulated? (NCERT)
Answer:
Regulation of Respiration: Normally, the rate of respiration is determined by both, nervous and chemical control.
1. Nervous control: Diaphragm and intercostal muscles are innervated by nerves from the central nervous system. Nerves are influenced by the respiratory centre located in medulla oblongata. Medulla oblongata consists of two parts inspiratory centre and expiratory centre, both the centres do not work together, i.e., when inspiratory centre is active, expiratory centre is inactive and when expiratory centre is active, inspiratory centre is inactive. This makes respiration rhythmic process, which is essential for effective gaseous exchange. In the pons, there is another centre called pneumotaxic centre which acts on the inspiratory centre and produces periodic inhibition. This inspiratory centre is said to be in a constant state of excitability and it sends impulses to inspiratory muscles resulting inspiration.

Pneumotaxic centre receives impulses from the inspiratory centre during inspiration which in turn transmits impulses to the expiratory centre. The expiratory centre is stimulated by the pneumotaxic centre.
Expiratory centre is connected to the vagus nerves, which innervates the lungs. During inspiration when lungs expand, stretch receptors present in the lung are stimulated and send impulses to the expiratory centre through vagus. Expiratory centre then sends inhibitory impulses to the inspiratory centre to stop inspiration. It is called Herring-Breuer reflex.

2. Chemical control: The main object of respiration is to maintain appropriate oxygen and carbon dioxide concentration and also hydrogen ion concentration in the blood. Pulmonary ventilation is highly sensitive to changes in oxygen, carbon dioxide and hydrogen ions. None of the three areas expiratory, inspiratory and pneumatic, are affected directly by changes in ion concentration of CO₂ and hydrogen in blood. One more centre chemosensitive area is believed to be located bilaterally and ventrally in the medulla.

It lies in the area where glossopharyngeal and vagus nerves enter the medulla. This area is very sensitive to blood, C02 and hydrogen ion concentration and in turn, stimulates other parts of the respiratory system. The sensory neurons of the chemosensitive area are especially excited by hydrogen ions which provide direct stimulus for these neurons. However, hydrogen ions can not easily cross the blood-brain barrier or blood-cerebrospinal fluid barrier.

So, changes in H⁺ ions concentration have little effect in exciting the neurons of the chemosensitive area than do the changes in carbon dioxide. CO₂ can cross blood-brain barrier easily. If there is increase in CO₂ concentration in the blood, there also occurs increase in _pCO₂ both in the interstitial fluid of the medulla and the cerebrospinal fluid. In these fluids, CO₂ reacts with water to form carbonic acid which dissociates to form hydrogen ions and bicarbonate ions. Thus, more hydrogen ions are there in the chemosensitive area of the respiratory centre when the blood CO₂ increases. This stimulates the chemosensitive area thereby causing increased strength of both inspiratory and expiratory signals to the respiratory muscles.

Oxygen concentration in blood does not have a significant direct effect on the respiratory centre of the brain, which controls respiration. Special chemoreceptors located outside the central nervous system are responsive to changes in blood O₂, CO₂ and hydrogen ion concentration. These chemical receptors are located in the carotid and aortic bodies.

Question 6.
What is the affect of _pCO₂ on transport of oxygen? (NCERI)
Answer:
The gaseous exchange in the lungs depends on the difference between the oxygen and carbon dioxide pressures in the pulmonary alveoli and the venous blood flowing to the lungs. The partial pressure of oxygen (_pO₂) in the air sacs of lungs is higher than that of the lung capillaries, while the partial pressure of carbon dioxide (_pCO₂) in the lung capillaries is higher than that in air sacs. In this way oxygen passes from air into the blood and C02 from blood into air and exhaled.
In the inspired air, the partial pressure of oxygen (_pO₂) is normally 158 mm Hg, while that of carbon dioxide (_pCO₂) is 0.3 mmHg.

The lungs and alveoli already contain some air even after expiration and so this air contains \more carbon dioxide and less oxygen in comparison to inspired air. Thus, when this air mixes Vith inspired air, the partial pressure of oxygen comes down to 100 mm Hg and that of CO₂ increases to 40 mm Hg.

The pulmonary artery contains deoxygenated blood in which partial pressure of oxygen,(_pO₂) is very low (about 40 mm Hg) than that in the alveoli. So because of this difference in PP, alveolar oxygen diffuses in the blood capillaries around it. It is called oxygenation. This oxygenated blood is collected from alveolar capillaries into the pulmonary veins. _pO₂, in it is about 95 mm Hg, at this pressure the oxygenated blood contains 19 8% oxygen. Similarly, pCO₂ in the deoxygenated blood of alveolar capillaries is 46 mm Hg which is higher than the _pO₂ of the alveolar air i.e., 40 mm Hg. So the difference in the _pCO₂ causes CO₂ to diffuse from alveolar capillaries to alveoli till blood carbon dioxide partial pressure comes down to 40 mm Hg.

At this partial pressure, the contents of carbon dioxide in the arterial blood comes down to 49% from 52-7% in the venous blood. In this way alveolar air gives away oxygen (received from inspired air) to the blood in pulmonary vein and gets CO₂ from the blood of pulmonary artery.

Question 7.
Describe gaseous exchange in insects or arthropoda.
Answer:
The insects have acquired thick, impermeable integument to minimize the loss of water from the body surface by evaporation and adaptation to terrestrial life. They have evolved a complex system of intercommunicating air tubes called tracheae. The tracheae are formed by the invagination of integument.


Fig. (A) Insect tracheal system (Dorsal tracheae white, ventral trachea black); (B) Tracheole at resting stage; (C) Tracheole at active stage These tracheal tubes opens out by paired lateral apertures called spiracles or stigmata in the body wall. The stigmata lead into wider atrium. Each trachea branches extensively to form tracheoles. These fine tracheoles end blindly intracellularly in the tissue cells.

The tissue fluid can diffuses into the tracheoles and the oxygen present in the air in tracheoles gets dissolved in it directly and hence into the cells. The blood of cockroach is called haemolymph. It is colourless and does not contain any respiratory pigment, thus O₂ comes directly into tissue cells through trachea.

When the insects is at rest, it has low oxygen requirement. This is met without much effort from the oxygen dissolved in tissue fluid which has diffused into tracheoles. When the insect is moving or flying, it has high oxygen requirement. To meet this, respiratory movements take place. They involve alternate relaxation and contraction. The former draws fresh air into the tracheal system through the spiracles (inspiration) while the later expels air (expiration). In such situation, fluid is withdrawn from the tracheoles into the cells and the air goes deep into the tracheoles and reaches the cells directly.

Question 8.
Explain the following respiratory organs with suitable diagrams:
(1) Larynx,
(2) Trachea,
(3) Bronchi,
(4) Lungs.
Answer:
The respiratory system of mammals, chiefly made up of larynx, trachea, bronchi and lungs.
1. Larynx: It is situated in neck on the level of fourth to sixth cervical vertebrae. The larynx is formed with the help of various cartilages thyroid, cricoid, paired arytenoid and epiglottis out of these thyroid cartilage is the largest. Epiglottis is situated behind the tongue. Its main function is to cover the entrance of the larynx so that food may not enter it during swallowing. The larynx is lined with numerous membrane covered with ciliated epithelium. When air is passed through it sound is produced.

2. Trachea: The larynx at the level of the seventh cervical vertebra joins the trachea which is a tube about 12 cm long. It is supported by half rings of cartilages which are articulated with the help of ligaments. The trachea is internally lined with mucous membrane. Externally it is covered by connecting tissue membrane. Trachea enters the thoracic cavity. At the level of fourth or fifth vertebra it bifurcates into bronchi.

3. Bronchi: The two bronchi, right and left, are termed as primary bronchi. These enter the lung from their own side. Each after entering the lung divides into smaller branches, which ultimately profusely branch in the surface of lung. The structure of these smaller branches called bronchiole is the same as that of trachea.

4. Lungs: The lungs are two in number, situated in the thoracic cavity. They are dark red coloured, spongy bag like structures. Each lung is distinguished into three surfaces i.e., coastal, spongy, bag-like structures. Each lung is mediastinal surface, a groove is present through which bronchi, pulmonary nerves, pulmonary artery, veins and lymph vessels enter. All these structures are bound together by a connective tissue, into a bundle called root of the lung.

Breathing and Exchange of Gases Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
In cellular respiration breakdown of glucose into C02 and HzO takes place in:
(a) Nucleus
(b) Cytoplasm
(c) Mitochondria
(d) Lysosome.
Answer:
(c) Mitochondria

Question 2.
Rate of respiration will be lower during :
(a) Playing
(b) Sleeping
(c) Eating
(d) Speaking.
Answer:
(b) Sleeping

Question 3.
Bicarbonate ions can be generated in :
(a) Lymphocytes
(b) Neutrophil
(c) Basophil
(d) R.B.Cs.
Answer:
(d) R.B.Cs.

Question 4.
The blood of animal is light blue in colour due to the presence of:
(a) Cytochrome
(b) Haemoglobin
(c) Haemocyanin
(d) Cobalt.
Answer:
(c) Haemocyanin

Question 5.
The conductor of O₂ or respiratory pigment of vertebrates is :
(a) Haemocyanin
(b) Chloroquine
(c) Haemoglobin
(d) Haematin.
Answer:
(c) Haemoglobin

Question 6.
Which of the following is capable of carrying O₂ :
(a) Lymph
(b) Serum
(c) Blood
(d) Plasma.
Answer:
(c) Blood

Question 7.
The causes of sudden and deep expiration is :
(a) Increase of CO₂ concentration in blood
(b) Increase in the concentration ofO₂
(c) Decrease in the concentration of CO₂
(d) Increase in the concentration of O₂ and decrease in the concentration of CO₂ in blood.
Answer:
(a) Increase of CO₂ concentration in blood

Question 8.
After inspiration diaphragm becomes :
(a) Relaxed
(b) Constricted
(c) Expanded
(d) Remain intact.
Answer:
(b) Constricted

Question 9.
The process of respiration is regulated by :
(a) Central nervous system
(b) Lungs
(c) Trachea
(d) Peripheral nervous system.
Answer:
(a) Central nervous system

Question 10.
Invertebrates O₂ is conducted :
(a) After absorption on the surface of R.B.Cs.
(b) Through the formation of a complex with Hb
(c) Through plasma
(d) All of these.
Answer:
(d) All of these.

Question 11.
ATP is:
(a) Enzyme
(b) Protein
(c) Hormone
(d) High energy molecule.
Answer:
(d) High energy molecule.

Question 12.
The lobes formed in right and left lungs of man :
(a) 3 and 2
(b) 2 and 3
(c) 3 and 3
(d) 2 and 2.
Answer:
(a) 3 and 2

Question 13.
Decomposition of oxyhaemoglobin into oxygen and deoxyhaemoglobin is due to :
(a) The reduction of oxygen pressure in tissue
(b) Increase in the pressure of oxygen
(c) Same pressure inside and outside the cell
(d) None.
Answer:
(a) The reduction of oxygen pressure in tissue

Question 14.
The exchange of gases in alveoli due to :
(a) Osmosis
(b) Simple diffusion
(c) Transport
(d) Active transport.
Answer:
(b) Simple diffusion

Question 15.
In the red blood cells of Mammals there is no gaseous exchange due to the absence of:
(a) Haemoglobin
(b) Nucleus
(c) Mitochondria
(d) None.
Answer:
(c) Mitochondria

Question 16.
The poisonous effect of carbon monoxide is due to its bondage with the haemo-globin. How much more its bondage with CO₂ as compared to oxygen :
(a) Two times
(b) Twenty times
(c) 200 times
(d) 1000 times.
Answer:
(c) 200 times

Question 17.
The capacity of air for lungs of human is:
(a) 3000 ml
(b) 1500 ml
(c) 1000 ml
(d) 500 ml.
Answer:
(a) 3000 ml

Question 18.
Respiratory centre is situated in:
(a) Medulla oblongata
(b) Central part
(c) Front brain
(d) None.
Answer:
(a) Medulla oblongata

Question 19.
Respiration process occurs in the:
(a) Alveoli of the lungs
(b) Cells of the alimentary canal
(c) Cells of the brain
(d) In all cells of the body.
Answer:
(d) In all cells of the body.

Question 20.
Outer membrane covering of the lungs. is:
(a) Pleural membrane
(b) Pericardium
(c) Pentonium membrane
(d) Mucous membrane.
Answer:
(a) Pleural membrane

2. Fill in the blanks:

1. Lungs are situated in …………………….. .
Answer:
Thoracic cavity,

2. In insects respiration takes place by …………………….. whereas in prawn respiration takes place by ……………………..
Answer:
Spiracles, gills,

3. …………………….. pigment formed in RBCs.
Answer:
Haemoglobin,

4. …………………….. disease is caused due to smoking.
Answer:
Ephysema,

5. The respiratory organs of fishes are …………………….. .
Answer:
Gills,

6. In blood corpuscles, CO₂ transport takes place in the form of …………………….. .
Answer:
Bicarbonate,

7. Krebs cycle takes place in …………………….. of cell.
Answer:
Mitochondria,

8. Yeast fermentation results in the formation of …………………….. and …………………….. .
Answer:
C₂H₅OH and CO₂

9. Lungs are situated in …………………….. cavity.
Answer:
Pulmonary cavity,

10. The partial oxidation of glucose takes place in …………………….. respiration.
Answer:
Anaerobic.

3. Match the following:

(A)

Column ‘A’	Column ‘B’
1. Glycolysis	(a) PPP
2. Krebs cycle	(b) Mitochondria
3. Cristae	(c) Krebs cycle
4. 30 ATP	(d) CO2, ATP
5. 36ATP	(e) Pyruvic acid.

Answer:
1. (e) Pyruvic acid.
2. (d) CO₂, ATP
3.  (b) Mitochondria
4. (c) Krebs cycle
5. (a) PPP.

(B)

Column ‘A’	Column ‘B’
1. Cockroach	(a) Cutaneous respiration
2. Starfish	(b) Trachea
3. Neris	(c) Parapodia
4. Earthworm	(d) Water vascular system
5. Insect	(e) Tracheole.

Answer:
1. (b) Trachea
2. (d) Water vascular system
3. (c) Parapodia
4. (a) Cutaneous respiration
5. (e) Tracheole.

4. Answer in one word:

1. What is the breathing rate of man?
Answer:
16-20 per minute,

2. What is the number and area of the alveoli in human lungs?
Answer:
7,50,000,000 alveoli and 100 sq. meter,

3. Name the respiratory organ of fish.
Answer:
Gills,

4. Name the respiratory organs of leech and insects.
Answer:
Skin, tracheal system,

5. Which part of the body of Hydra takes part in gaseous exchange?
Answer:
Ectoderm,

6. What term is used for deficiency of O₂ in body?
Answer:
Hypoxia,

7. Pigment help for transport of oxygen.
Answer:
Haemoglobin,

8. Which part of the brain regulate respiration process?
Answer:
Medulla oblongata,

9. Name the main respiratory organ of human where gaseous exchange occurs.
Answer:
Lungs,

10. Volume of the gas normally inhaled and exhaled by the lungs in human.
Answer:
Tidal volume.

Students get through the MP Board Class 11th Biology Important Questions Chapter 6 Anatomy of Flowering Plants which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Questions Chapter 6 Anatomy of Flowering Plants

Anatomy of Flowering Plants Class 11 Important Questions Very Short Answer Type

Question 1.
What is tissue?
Answer:
A group of cells which are similar in origin and performing a particular function is collectively called as tissue.

Question 2.
Write main elements present in phloem.
Answer:
The main elements present in phloem are :
Sieve tubes, companion cells, phloem fibres and phloem parenchyma.

Question 3.
What is meristematic tissue?
Answer:
It is the simple type of tissue, composed of similar and immature cells, which continuously divide and form new cells. This type of cells mostly present in the apical portion and help in the growth of plant body.

Question 4.
Vascular bundles are formed by which tissue?
Answer:
Vascular bundles are formed by xylem and phloem.

Question 5.
Differentiate exarch and endarch vascular bundles.
Or,
What do you mean by endarch and exarch vascular bundles?
Answer:

• Endarch V.B.: Vascular bundles, the protoxylem of which is arranged towards the periphery and metaxylem in the central region are called endarch V.B. e.g., V.B. of roots.
• Exarch V.B.: Vascular bundles, the protoxylem of which is aiTanged in the centre and metaxylem towards the peripheral region are called as exarch V.B. e.g., V.B. of stems.

Question 6.
Write the main functions of xylem.
Answer:
The main functions of xylem is to conduct the absorbed water and minerals by root from the soil and to provide mechanical support to the plant.

Question 7.
Where is xylem vessel found ? Write their functions.
Answer:
Xylem vessels are found in the xylem tissues of the vascular bundles. Xylem vessels are responsible for the conduction of absorbed substances.

Question 8.
Write the functions of pericycle.
Answer:

• It forms lateral roots.
• When composed of parenchymatous cells, it may act as storage region.
• In dicot roots, their cells become meristematic and forms part of the cambium.
• Thick walled pericycle gives rise to lateral and adventitious roots.

Question 9.
What is stomata?
Answer:
These are small openings on the surface of the leaf through which exchange of vapour and gases takes place. Stomata performs transpiration.

Question 10.
Which type of vascular bundles are found in the stem of sunflower plant?
Answer:
Conjoint, collateral, open type of vascular bundles are found in the stem of sunflower plant.

Question 11.
What is cambium? Give its function.
Answer:
Cambium is a meristematic tissue. During secondary growth, cells of cambium are divided to form secondary xylem towards inner side and secondary phloem towards outer side of it, thus plant part increases in thickness.

Question 12.
Where is cambium found?
Answer:
Cambium is found in the vascular bundle of dicot stem in between xylem and phloem as meristematic tissue.

Question 13.
Where guard cells are found? Give its function.
Answer:
Stomata (pores) found in the epidermis of leaves are surrounded by two kidney shaped guard cells. Guard cells regulate opening and closing of stomata.

Question 14.
What are Bast fibres?
Answer:
Sclerenchyma cells (fibres) found associated with phloem are called as Bast fibres or phloem fibres.

Question 15.
Write main elements present in xylem.
Answer:
The main elements present in xylem are :

• Xylem tracheids,
• Xylem tracheae (vessels),
• Xylem fibres (wood),
• Xylem parenchyma.

Question 16.
What is tissue system?
Answer:
In flowering plants group of tissues derived from meristematic tissue perform one or many functions together. This system of tissues is called as tissue system.

Question 17.
When transverse section of a part of a plant is observed under compound microscope following structures are observed :
(a) Conjoint vascular bundles, scattered and surrounded by sclerenchymatous sheath.
(b) Phloem parenchymatous.
Identify the plant part.
Answer:
This is transverse section of monocot stem.

Anatomy of Flowering Plants Class 11 Important Questions Short Answer Type

Question 1.
How is study of Anatomy useful to us?
Answer:
Study of external morphology helps us to find out similarity and differences between shape, structure and colour of the organism. Many organisms may look similar externally but may be different in structure internally.

Study of internal structure of plant or animal is called as Anatomy. It helps us to understand about different types of cells and tissues found in them and their arrangement in the plant part. It also helps us to understand about their functions. Such as epidermis helps for protection. Xylem and Phloem are located at the central part which help for conduction of water, minerals and food respectively.

Question 2.
How does structure and location of epidermal cells help the plants to per-form specific function?
Answer:
Epidermis is the outermost protective layer of plant organs. It is usually single layered. In stems and leaves this layer is protected by cuticle (made up of cutin a fatty substance).
Epidermal cells perform several functions, e.g. protection of stems and leaves, absorption (in roots), excretion, secretion, gaseous exchange and regulation of transpiration.
In some monocot leaves (such as grasses), epidermal cells become larger, thin walled and have vacuoles. Such cells are called as bulliform cells or motor cells. These cells brings about rolling of leaves during dry season and this reduces transpiration as in Ammophila, Zea mays etc.

Question 3.
What are the various types of meristems based on their locations? Give their functions.
Answer:
Following are the various types of meristems based on their location :
(i) Apical meristem :
It lies at the apex of the stem and the root of vascular plants, they may also be present at the apices of leaves. Due to its activity the organs increase in length. The growth begins by one or more cells found at the tip of the organ, these cells are called apical cells or apical initials, as they initiate the growth.

(ii) Intercalary meristem :
These are the portions of apical meristems that separate from the apex during development by layers of more mature or permanent tissues and left behind as the apical meristem moves on in growth. The intercalary meristems are intemodal in position. They are found lying in between masses of permanent tissues either at the leaf base or at the base of intemode. This type of meristem is found in stems of monocots. based on their position.

(iii) Lateral meristem :
It is composed of such initials which divide mainly in one plane (periclinal) and increase the diameter of an organ. They may add to the existing tissue or give rise to new tissues. The cambium and cork cambium are the examples of this type.

Question 4.
Write differences between following based on their anatomy :
(a) Dicot and monocot root
(b) Dicot and monocot stem.
Answer:
(a) Differences between Anatomy of Dicot and Monocot Roots

(b) Differences between Diocot and Monocot Stems

Dicot Stem	Monocot Stem
1. Epidermis contains multicellular hairs.	It does not contain multicellular hairs.
2. Hypodermis is collenchymatous.	Hypodermis is sclerenchymatous.
3. Endodermis and pericycle are present.	Endodermis and pericycle are absent.
4. Vascular bundles are few and arranged in a ring.	Vascular bundles are many and scattered.
5. Vascular bundles are open.	Vascular bundles are closed.
6. Bundle sheath is absent.	Vascular bundles are covered by a prominent bundle sheath.
7. Medullary rays and pith are present.	Medullary rays and pith are absent.
8. Secondary growth takes place.	Secondary growth does not take place.
9. Xylem vessels are arranged in radial rows.	Xylem vessels are arranged in the form of‘Y’.
10. Phloem is made up of sieve tubes, companion cells and phloem parenchyma.	Phloem is made up of sieve tubes and companion cells only.
11. Lysigenous cavity is absent.	Lysigenous cavity is present.

Question 5.
Give four characteristics of meristematic tissue.
Answer:
A group of cells which is capable of showing division and redivisions is called as meristematic tissue. Following are the characteristics of meristematic tissue :

1. Cells contain homogeneous thin wall.
2. They contain large nuclei associated with abundant cytoplasm.
3. Intercellular spaces are not present in between them.
4. Cells contain dense protoplasm and do not contain vacuoles. These cells are found in active metabolic state.

Question 6.
Write differences between Vascular Bundles of Root and Stem.
Or,
Write differences between Radial and Conjoint Vascular Bundles.
Answer:
Differences between Radial and Conjoint Vascular Bundles

Radial Vascular Bundle	Conjoint Vascular Bundle
1. Xylem and phloem lie radially, alternately.	Xylem and phloem are present on same radius in one bundle.
2. They lie on separate radii.	They lie on same radius.
3. This type of vascular bundles are found in roots.	This type of vascular bundles are found in stems.

Question 7.
What is Periderm? How is periderm formed in dicot stem?
Answer:
In many woody plants further increase in girth takes place by the formation of new tissues in extrastelar region (cortical region). These new tissues are called periderm or cork which were formed from some meristematic tissues called cork cambium or phellogen.

Periderm is made up of following three types of tissues :
1. Phellogen (cork cambium):
It is secondary lateral meristem arising from hypo- dermis or outer cortex. It consists of a single layer of narrow, thin walled rectangular mer¬istematic cells. The cork cambium cuts-off cells to its both the sides.

2. Phellem (cork):
The cells formed as a result of tangential and periclinal divisions of phellogen cells towards outer side, are called phellem or cork. The cork cells are
suberized and impervious to water and air. Cork cells also possess thermal insulating properties.

3. Phelloderm (secondary cortex):
The cells formed on inner side of the phellogen are called phelloderm or secondary cortex. It consists of parenchymatous cells which often contain chloroplast. Due to rapid growth of secondary cortex sometimes the general cortex is crushed.

Question 8.
Write short note on lenticel.
Answer:
Lenticels:
These are aerating pores formed in the bark, through which exchange of gases takes place. Externally they appear as scars on the surface of the stem. In the region of a lenticel, the cells cut off by the cork cambium towards the outside become rounded to form a loose mass with plenty of intercellular spaces. The tissue is known as complementary tissue. Through the lenticel, tissues inside the cork can communicate with the outside atmosphere.

Question 9.
Write difference between anatomy of Dicot and Monocot leaves.
Answer:
Differences between Anatomy of Dicot and Monocot leaves

Dicot Leaf	Monocot Leaf
1. It is dorsiventral type and the number of stomata differ on both the surface of the leaf.	It is isobilateral type and number of stomata are more or less uniform on both the surface of the leaf.
2. Mesophyll is differentiated into palisade parenchyma and spongy parenchyma.	Not differentiated.
3. Bulliform cells are absent.	Bulliform cells are present in its epidermis and help in folding of leaves.
4. On both the sides of larger vascular bundles parenchyma cells are found.	On both the sides of larger vascular bundles sclerenchyma cells are found.
5. Bundle of sheath is generally single layered, cells are colourless	Bundle of sheath may be single or double. layered, cells generally possesses chloroplast.
6. The stomata have kidney shaped guard cells.	The stomata have dumb-bell shaped guard cells.

Question 10.
Why Xylem and Phloem are called as complex tissues?
Answer:
Complex permanent tissues:
Vascular tissues of plant are complex tissue, which help for conduction of various substances from one part of the plant to other part. Therefore, these are also called as conductive tissues.
Complex permanent tissue is a group of many types of the tissues, which functions as a unit together. Xylem and phloem are main examples of complex tissues.
Tissues which help in the conduction of water and minerals against gravitational force towards upsides are called xylem or wood and tissues which helps to translocate prepared food to different parts of plant are called as phloem or bast. As these tissues consist of more than one type of tissues, therefore they are called as complex tissues.

Anatomy of Flowering Plants Class 11 Important Questions Long Answer Type

Question 1.
Describe anatomy of dicot leaf (Dorsiventral) with labelled diagram.
Answer:
Anatomy of dicot (Dorsiventral) leaf: This type of leaves are commonly horizontal in orientation which contain distinct upper and lower surface. The upper surface which faces the sun is darker than the lower surface. The different parts are :
1. Upper epidermis :
It is made up of tightly packed rectangular barrel shaped, transparent parenchymatous cells in which chloroplast is absent. In this layer stomata are generally absent.

2. Mesophyll :
It is the ground tissue which is situated between upper and lower epidermis. It can be differentiated into two distinct layers.
(i) Palisade parenchyma :
This portion lies just next to the upper epidermis and are roughly rectangular in shape. They are remarkably placed at right angles to the surface of the leaf in order to absorb maximum sunlight for the process of photosynthesis. They are arranged in many parallel rows like palisade anji therefore also called as palisade mesophyll. They contain abundant chloroplast and help in photosynthesis.

(ii) Spongy parenchyma :
The cells of this region are irregular in shape and are not very compact in arrangement. They enclose large air spaces between them which give them spongy characteristic. These cells are simple parenchyma and hence this tissue is known as spongy parenchyma. The intercellular spaces of this tissue communicate with the external atmosphere through the stomata. These cells may develop chloroplast also.

3. Vascular bundles:
They are meant for distributing water and minerals to the lamina of the leaf. These vascular bundles are conjoint, collateral and endarch. The phloem is situated towards the abaxial side and protoxylem points towards the adaxial side. Each vascular bundle has its own vascular sheath around it.

4. Lower epidermis :
This layer is made of a single layer of compactly arranged rectangular, transparent parenchymatous cells. Lower epidermis contains large number of pores called stomata.

Question 2.
What is stomatal system? Describe structure of stomata with labelled diagram.
Answer:
Stomata are minute pores found on the epidermal layer of leaves and other green aerial parts of the plant body. Each stoma or aperture is bounded by two kidney shaped cells, called Guard cells. Guard cells are living and contain chloroplast. The outer wall of guard

cell adjacent to epidermal cell wall is thin whereas the inner wall lying towards the stomata is thicker. In certain plants, some specialized epidermal cells are present. These cells are called Subsidiary or Accessory cells. Each stomata open below in a large cavity, called Sub-stomatal cavity.

In xerophytes, stomata are found sunken in grooves. This greatly reduces rate of transpiration. Function of stomata is exchange of gases and control of transpiration.
Stomatal pore, guard cells and subsidiary or accessory cells together form stomatal system.

Question 3.
Describe the process of secondary growth in the stem of woody angiosperms with labelled diagrams. Give its importance.
Answer:
Secondary growth :
Increase in thickness of the root and stem of dicot plants due to addition of secondary tissues by the activity of cambium and cork cambium is called as secondary growth.
Dicot plants convert into trees due to secondary growth and monocot plants cannot develop to form a tree due to absence of secondary growth. Except a few plants monocot do not shows secondary growth whereas dicot stem as well as root both shows secondary growth.

Secondary growth in dicot stem:
In a typical dicot stem secondary growth occurs in following ways:
1. Activity of cambium:
The vascular bundles in the dicot stem are open type, that is to say that bundles contain intrafascicular cambium in between xylem and phloem which is a primary meristem. The parenchymatous cells found in between the intrafascicular cambium of the adjacent bundles become meristematic to form strip of interfascicular cambium, which laterally join to the intrafascicular cambium to complete a circular ring of cambium. This is known as cambium ring.
Cambium ring cuts secondary phloem towards outside and secondary xylem towards inside.
Generally secondary xylem is produced in more quantity as compared to secondary phloem therefore cambium ring moves towards the periphery.

The primary phloem becomes crushed. Not only this, but there is a continuous collapsing and disintegration of the secondary phloem in its outer region and continuous addition in the inner region just outside the cambium ring. However, the amount of secondary xylem goes on increasing continuously.

2. Formation of annual rings :
In annual herbs, there is little secondary growth. However, in the perennial plants, especially the arborescent ones, secondary growth is more or less continuous process. The process usually starts in the spring, the amount of wood which is formed during the spring and summer is much more than that formed during the winter. Further, the xylem elements which are formed during the spring and subsequent summer are larger in diameter than those formed during the winter.

The spring (or summer) wood is, therefore, lighter in colour and lesser in density in comparison to the winter wood which is comparatively darker in colour and greater in density. The sequence of spring and winter wood on a cut log of a wood is seen in the form of dark and light coloured concentric rings. One light and a dark coloured zone represents one years growth and is known as a growth ring or annual ring.
Thus, by counting the number of growth rings, one can estimate the age of the wood with a fair degree of correctness.

3. Cork cambium or Phellogen and its activity :
With the progressive increase in diameter of the secondary wood, the primary tissues outside the secondary phloem, including the cortex and the epidermis becomes greatly stretched and may even radially crack open. However before this eventuality arises, another cambium arises, in the cortex. This is the
cork cambium.

The cork cambium is completely secondary in nature.
The cells of the cork cambium are brick-shaped and tangentially elongated.
Derivatives are cut off both towards the outside, as well as towards the inside. The outer derivatives become suberized and lead to form the cork or phellem. The inner derivatives may slightly round up but remain arranged in radial rows. They develop plenty of chloroplasts and constitute the secondary cortex or Phelloderm. The cells of the secondary cortex can be distinguished from those of primary cortex by their arrangement in radial rows and presence of chloroplasts.

With the formation of cork which is impervious, both to water and air, all primary tissues of the stem outside the cork cambium are cut off from water supply, dry up and are ultimately shed.
The cork, cork cambium and secondary cortex are collectively known as periderm. Phellem tissue is impervious to water and provides protection to the underlying secondary tissues. In this cork cells certain areas of the cork remain thin walled and loosely arranged. They help in gaseous exchange and are called as lenticels. They are generally produced at the site of the stomata which were present in the young stem.

Question 4.
Differentiate the following :
(a) Vessel and TVacheids,
(b) Parenchyma and Collenchyma,
(c) Heart wood and Sap wood,
(d) Open and Closed Vascular bundle.
Answer:
(a) Differences between Vessels and Tracheids

Vessels	Tracheids
1. These are elongated tube like cells which are placed end to end with each other to form a tube.	These are single elongated cells.
2. End of cells are round.	These are cells with tapering ends.
3. Septa are found between two cells.	Septa between adjacent cells dissolve to form hollow pipes.
4. Vessels are 1 to 6 mm in length.	These are only 1 mm in length.
5. Its lumen is broad. Its lumen is narrow.	Its lumen is narrow.

(b) Differences between Parenchyma and Collenchyma

Parenchyma	Collenchyma
1. Cells are spherical, oval or polygonal in shape.	Cells are polygonal, elongated or oval.
2. Cells are always living.	Living.
3. Cell wall is thin and made up of cellulose.	Cell wall contains high deposition of pectic materials.
4. Cell wall thickening and intercellular space are absent.	Angular thickening on cell wall is present but intercellular space is absent.
5. It is present in any tissue system of the plant body.	It is generally found in the cortex or hypodermis of dicot stem and also in leaves.
6. Parenchymatous cells are involved in photosynthesis, storage and also in mechanical support in the form of ground tissue.materials.	It provides tensile strength to plant axis and may also be involved in photosynthesis and storage of food materials.
7. Cells have the capacity of division.	Cells cannot divide.

(c) Differences between Heartwood and Sapwood

Heartwood	Sapwood
1. It is found in the central part of the stem.	It is found towards the periphery surrounding heart wood.
2. It is dark coloured.	It is light coloured.
3. Cells are dead.	Cells are living.
4. It provides mechanical strength to the plant. It helps in life activities.	It helps in life activities.
5. It is very hard.	It is soft.
6. It contains gum, oil, tannins and resins etc. in more quantity.- These substances are found in less quantity.	These substances are found in less quantity.
1. It is found in the central part of the stem.	It is found towards the periphery surrounding heart wood.

(d) Differences between Open and Closed Vascular Bundle

Open Vascular bundle	Closed Vascular bundle
1. When in a collateral vascular bundle cambium is present in between xylem and phloem, the bundle is called as Open Vascular Bundle.	When in a collateral vascular bundle cambium is absent, it is called as Closed Vascular Bundle.
2. This type of vascular bundles are found in dicot stem.	This type of vascular bundles are found in monocot stem.

Question 5.
You observe transverse section of young stem of a plant under compound microscope. How will you identify that it is dicot stem or monocot stem? Give reason for it.
Answer:
Dicot stem and Monocot stem can be identified by following characteristics :
The T.S. of a dicot stem shows following structures :
(1) Epidermis :
It is single layered, which produces multicellular hairs and are covered by cuticle.

(2) Cortex:
It is below epidermis. The outer cortex is known as hypodermis and are generally made up of collenchyma. The inner cortex is made up of parenchymatous cells with intercellular space. Cortex is followed by single layered endodermis.

(3) Pericycle:
Below endodermis, parenchymatous pericycle is present.

(4) Vascular bundle:
Vascular bundles are made up of xylem, phloem and cambium arranged in the same radius. V.B. are conjoint, collateral, endarch and open. Xylem is made of xylem vessels, tracheids, wood fibres and xylem parenchyma. Phloem is made up of sieve tubes, companion cells and phloem paren¬chyma. A cambium ring is present between xylem and phloem.

(5) Pith :
It is very elaborated and occupies the major central portion of the stem and is composed of rounded or polygonal thin walled cells with intercellular spaces.
The anatomy or T.S. of monocot stem (maize stem) shows the following structures :
(i) Epidermis: It is an outermost layer of the stem which is single layered and without hairs. A layer of cuticle is present outside the epidermis.
(ii) Hypodermis : It is situated just below the epidermis and may be made up of 2 to 4 layers of sclerenchymatous cells.
(iii) Ground tissue: It is made up of parenchymatous cells with plenty of intercellular spaces. These cells are situated below the hypodermis and scattered up to the centre of stem.

(iv) Vascular bundle :
In monocot stem, the vascular bundles are found to be scattered in the ground tissues. They are conjoint, collateral and closed. The xylem is endarch and the vessels are arranged in the form of the letter ‘Y\ the phloem being situated in between the two limbs of the ‘Y’ and consists of sieve tubes and companion cells only. Lysigenous cavity is found below the protoxylem in each vascular bundle.
(v) Pith : Pith is absent.

Question 6.
Describe various types of vascular bundles.
Answer:
Types of vascular bundles: According to the arrangement of xylem and phloem, the vascular bundles are of the following types :
1. Radial vascular bundle :
When xylem and phloem form separate bundles and lie on different radii alternating with each other, is said to be radial vascular bundles as in roots.

2. Conjoint vascular bundle :
When xylem and phloem combine into one bundle. There are different types of conjoint bundles :
(i) Collateral: When xylem and phloem lie together on the same radius, xylem being internal and phloem external. When in a collateral bundle the cambium is present, as in dicotyledonous stems, the bundle is said to be open and when the cambium is absent it is said to be closed, as in monocotyledonous stems.

(ii) Bicollateral vascular bundle :
When in a collateral bundle both phloem and cambium occur twice, once on the outer side of xylem and then again on its inner side. The sequence is outer phloem, outer cambium, xylem, inner cambium and inner phloem. Bicollateral bundle is characteristic of the ground family. It is always open.

3. Concentric vascular bundles:
These are those vascular bundles where one type of the tissue encircles or envelops the other tissue. They are again of two types :

1. Amphivasal vascular bundles: When xylem surrounds the phloem as in Dracena and Yucca.
2. Amphicribral vascular bundles : When phloem encircles the xylem as in Ferns.

Anatomy of Flowering Plants Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
Radial vascular bundles are found in :
(a) Dicot stem
(b) Monocot stem
(c) Dicot root
(d) Monocot root.
Answer:
(d) Monocot root.

Question 2.
Bicollateral vascular bundles are found in :
(a) Liliaceae
(b) Cucurbitaceae
(c) Compositae
(d) Malvaceae.
Answer:
(a) Liliaceae

Question 3.
In amphivasal vascular bundles :
(a) Phloem is covered by xylem
(b) Xylem is covered by phloem
(c) Both are radially arranged
(d) None of these.
Answer:
(d) None of these.

Question 4.
In dicot vascular bundles are :
(a) Open, collateral and endarch
(b) Closed, collateral and endarch
(c) Open, collateral and exarch
(d) Closed, collateral and exarch.
Answer:
(a) Open, collateral and endarch

Question 5.
The chief function of velamen tissue is :
(a) Absorption of water from host plant
(b) Absorption of water from atmosphere
(c) Absorption of mineral salts
(d) None of these.
Answer:
(c) Absorption of mineral salts

Question 6.
The lateral meristem is responsible for :
(a) Increasing height
(b) Increasing thickness
(c) Increasing permanent tissue
(d) None of these.
Answer:
(b) Increasing thickness

Question 7.
Companion cells are associated with :
(a) Sieve tubes
(b) Xylem
(c) Collenchyma
(d) Cambium.
Answer:
(a) Sieve tubes

Question 8.
Tissue responsible for secondary growth is :
(a) Cambium
(b) Cortex
(c) Pericycle
(d) Endodermis.
Answer:
(a) Cambium

Question 9.
Secondary growth is found in :
(a) Monocot plants
(b) Dicot plants
(c) Both (a) and (b)
(d) Bryophytes.
Answer:
(b) Dicot plants

Question 10.
Phellogen forms:
(a) Phloem
(b) Xylem
(c) Cork
(d) Cork and secondary cortex.
Answer:
(d) Cork and secondary cortex.

Question 11.
Lenticel found in bark is :
(a) Air pore
(b) Vessel
(c) Tissue
(d) Stele.
Answer:
(a) Air pore

Question 12.
Annual ring is made up of:
(a) Spring wood
(b) Autumn wood
(c) Both (a) and (b)
(d) None of these.
Answer:
(c) Both (a) and (b)

Question 13.
Brick-shaped suberized cells found outside the cork cambium is :
(a) Secondary cortex
(b) Phellem
(c) Epidermis
(d) Endodermis.
Answer:
(b) Phellem

Question 14.
Balloon shaped structures present inside the vessels are called :
(a) Histogen
(b) Tyloses
(c) Phellogen
(d) Tunica.
Answer:
(b) Tyloses

Question 15.
Periderm includes :
(a) Cork cambium (phellogen), cork (phellem) and secondary cortex (phelloderm)
(b) Cork cambium and cork
(c) Cork
(d) Cork and secondary phloem.
Answer:
(a) Cork cambium (phellogen), cork (phellem) and secondary cortex (phelloderm)

Question 16.
The best method to determine the age of tree is to :
(a) Measure its diameter
(b) Count the number of leaves
(c) Count the number of annual rings at the base of main stem
(d) Find out the number of branches.
Answer:
(c) Count the number of annual rings at the base of main stem

Question 17.
Wood is common name of:
(a) Cambium
(b) Vascular bundles
(c) Phloem
(d) Secondary xylem.
Answer:
(d) Secondary xylem.

Question 18.
Where do casparian strips occur :
(a) Epidermis
(b) Endodermis
(c) Pericycle
(d) Phloem.
Answer:
(b) Endodermis

Question 19.
Dendrochronology is the study of:
(a) Height of tree
(b) Diameter of a tree
(c) Age of the tree by counting the number of annual rings in the main stem
(d) None of these.
Answer:
(b) Diameter of a tree

Question 20.
In roots, lateral branches grow from :
(a) Epiblema
(b) Pericycle
(c) Cortex
(d) Endodermis.
Answer:
(b) Pericycle

Question 21.
Sunken stomata occur in :
(a) Mesophytes
(b) Xerophytes
(c) Hygrophytes
(d) Hydrophytes.
Answer:
(b) Xerophytes

Question 22.
Polyarch and exarch condition is found in:
(a) Monocot stem
(b) Monocot root
(c) Dicot stem
(d) Dicot root.
Answer:
(b) Monocot root

Question 23.
Meristem present in a vascular bundle is:
(a) Fascicular/intrafascicular cambium
(b) Interfascicular cambium
(c) Phellogen
(d) Procambium.
Answer:
(b) Interfascicular cambium

Question 24.
Cork cambium is also called :
(a) Phelloderm
(b) Phellem
(c) Periderm
(d) Phellogen.
Answer:
(d) Phellogen.

Question 25.
Periderm is produced by:
(a) Vascular cambium
(c) Phèllogen
(b) Fascicular cambium
(d) Intrafascicular cambium.
Answer:
(c) Phellogen

Question 26.
Endodermis of dicot stem is also called :
(a) Bundle sheath
(b) Starch sheath
(c) Mesophyll
(d) Water channel.
Answer:
(b) Starch sheath

Question 27.
Tyloses are thickenings seen in :
(a) Ray parenchyma
(b) Collenchyma
(c) Phloem cells
(d) Ray parenchyma and xylem cells.
Answer:
(d) Ray parenchyma and xylem cells.

Question 28.
Casparian strips contain :
(a) Cutin
(b) Pectin
(c) Suberin
(d) Wax.
Answer:
(c) Suberin

Question 29.
A component of xylem is:
(a) Seive tube
(b) Medullary ray
(c) Sclereid
(d) Tracheid.
Answer:
(d) Tracheid.

Question 30.
Vascular bundles occur in a ring in :
(a) Monocot stem
(b) leaf
(c) Root
(d) Dicot stem.
Answer:
(d) Dicot stem.

Question 31.
Secondary xyLem is:
(a) Bast
(b) Bark
(c) Cork
(d) Wood.
Answer:
(d) Wood.

Question 32.
Gymnosperm wood is soft as it:
(a) Lacks cambium
(b) Lacks vessels
(e) Does not yield timber
(d) None of the above.
Answer:
(b) Lacks vessels

Question 33.
Branch of Botany dealing with internal organization of plants is :
(a) Physiology
(b) Anatomy
(c) Ecology
(d) Cytology.
Answer:
(b) Anatomy

Question 34.
Phellogen is also knwon as :
(a) Vascular cambium
(b) Periderm
(c) Cork cambium
(d) Apical meristem.
Answer:
(c) Cork cambium

Question 35.
In an annual ring, the light coloured part in :
(a) Heartwood
(b) Sap wood
(c) Early wood
(d) Late wood.
Answer:
(b) Sap wood

Question 36.
Which is not a part of periderm :
(a) Phellogen
(b) Cork
(c) Secondary cortex
(d) Wood.
Answer:
(d) Wood.

Question 37.
Palisade parenchyma is absent in leaves of: (CBSE 2009)
(a) Gram
(b) Soyabean
(c) Sorghum
(d) Mustard.
Answer:
(c) Sorghum

2. Fill in the blanks:
(A)
1. Phellogen form the ………………….
Answer:
Cork and secondary cortex

2. Arenchyma is a kind of ………………….
Answer:
Parenchyma

3. Bicollateral vascular bundles are found in …………………..
Answer:
Liliaceae

4. Radial vascular bundles are found in …………………..
Answer:
Monocot root

5. Living member of the phloem tissue is ……………………
Answer:
Companion cells

(B) Fill in the blanks :
1. Dermatogen, Periblem and Plerome are found in …………………
Answer:
Root apex and stem apex

2. Bladder like ingrowths in the xylem vessels are called as ………………..
Answer:
Tyloses

3. Latex vessels are found in ………………….
Answer:
Cortex

4. Increase in ……………… of stem and root of dicot plants is called as secondary growth.
Answer:
Thickness

5. Secondary growth occurs due to activity of ………………..
Answer:
Cambium and cork cambium

6. Cortex consists of ………………. tissue.
Answer:
Parenchyma

7. Annual ring in plants is formed due to ………………. and ……………..
Answer:
Spring wood, autumn wood

8. ……………….. and …………… are vascular tissues.
Answer:
Xylem, phloem

9. Companion cells are found in …………………
Answer:
Phloem

10. Radial vascular bundles are found in ……………….
Answer:
Roots

3. Match the following:
(A)

Column ‘A’	Column ‘B’
1.Heartwood	(a) Phellogen
2. Sieve tube	(b) Bark
3. Periderm	(c) Functional wood
4. Lenticels	(d) Non-functional wood
5. Sapwood	(e) Phloem.

Answer:
1. (d) Non-functional wood
2. (e) Phloem
3. (a) Phellogen
4. (b) Bark
5. (c) Functional wood

(B)

Column ‘A’	Column ‘B’
1. Dicot stem	(a) Conjoint, collateral, closed vascular bundle
2. Dicot root	(b) Bulliform cells
3. Monocot stem	(c) Conjoint, collateral, open vascular bundle
4. Monocot root	(d) Radial, less than 6 vascular bundles
5. Monocot leaf	(e) Radial, more than 6 vascular bundles.

Answer:
1. (c) Conjoint, collateral, open vascular bundle
2. (d) Radial, less than 6 vascular bundles
3. (a) Conjoint, collateral, closed vascular bundle
4. (e) Radial, more than 6 vascular bundles.
5. (b) Bulliform cells

(C)

Column ‘A’	Column ‘B’
1. Isobilateral leaf	(a) Dicot plant
2. Tyloses	(b) Broad and large vessels
3. Dorsiventral leaf	(c) Narrow and small vessels
4. Autumn wood	(d) Conjoint, collateral, closed vascular bundle
5. Spring wood	(e) Ingrowth in the xylem.

Answer:
1. (d) Conjoint, collateral, closed vascular bundle
2. (e) Ingrowth in the xylem.
3. (a) Dicot plant
4. (c) Narrow and small vessels
5. (b) Broad and large vessels

(D)

Column ‘A’	Column ‘B’
1. Cambium	(a) Periderm
2. Cork cambium	(b) Secondary vascular bundle
3. Xylem vessels	(c) Companion cells
4. Sieve tubes	(d) Tyloses.

Answer:
1. (b) Secondary vascular bundle
2. (a) Periderm
3. (d) Tyloses.
4. (c) Companion cells

(E)

Column ‘A’	Column ‘B’
1. Apical meristem	(a) Cork cambium
2. Intercalary meristem	(b) Sunflower stem
3. Secondary meristem	(c) Maize root
4. Collateral and Open	(d) Root apex
5. Radial	(e) Internode.

Answer:
1. (d) Root apex
2. (e) Internode.
3. (a) Cork cambium
4. (b) Sunflower stem
5. (c) Maize root

4. Write true or false:

1. Intercalary meristem occurs in Mint and Grasses.
Answer:
True

2. The xylem having protoxylem towards the centre is called exarch.
Answer:
False

3. Casparian strip is found in passage cells.
Answer:
True

4. The complex tissue includes scierenchyma.
Answer:
False

5. In dicot vascular bundles are open, collateral and endarch.
Answer:
True

6. Secondary growth is not found in monocot plants.
Answer:
True

7. Cork cambium is formed in vascular bundles.
Answer:
False

8. Cork cambium is also known as phellogen.
Answer:
True

9. Heartwood is commercially more important.
Answer:
True

10. Autumn ring is more prominent than spring ring.
Answer:
False

11. Radial vascular bundles are found in dicotyledonous stem.
Answer:
False

12. Lenticels are the pores on the surface of bark of stem.
Answer:
True

13. Isobilateral leaves are found in dicot plants.
Answer:
False

5. Answer in one word:

1. The cell wall of xylem is rich in …………………..
Answer:
Lignin

2. Tissue responsible for secondary growth is ………………..
Answer:
Cambium

3. Tyloses thickening are seen in …………………
Answer:
Ray Parenchyma and Xylem cells

4. Which tissues have only living cells?
Answer:
Parenchyma

5. Brick shaped suberized cells found outside the cork cambium is …………………
Answer:
Cork

6. Abnormal secondary growth is found in ………………..
Answer:
Dracena

7. The tissue responsible for translocation of food material is ……………….
Answer:
Sieve tubes

8. In trees, the growth rings represent………………..
Answer:
Age

9. Vascular cambium and cork cambium are…………………
Answer:
Meristemâtic tissues

10. Maximum activity of cambium is during………………….
Answer:
Summer

11. Write the name of vascular tissue of plants which is non-living.
Answer:
Xylem

Students get through the MP Board Class 11th Biology Important Questions Chapter 5 Morphology of Flowering Plants which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Questions Chapter 5 Morphology of Flowering Plants

Morphology of Flowering Plants Class 11 Important Questions Very Short Answer Type

Question 1.
What are respiratory roots?
Answer:
There is lack of air in saline swampy areas. Roots of the plants grows in such areas grows vertically upward and comes out of the ground, which bears lenticels (some pores) called as pneumatophores through which gaseous exchange occurs during respiration e.g., Rhizophora.

Question 2.
From which part of seed, root develops?
Answer:
Root develops from radicle.

Question 3.
What is mycorrhizal root?
Answer:
These roots are commonly found in gymnosperms. In this root the portion of tap Toot and its branches becomes closely invested by the hyphae of a fungus. These fungal hyphae develop a close mat around the external surface of the root, and root hairs are absent e.g., Pinus.

Question 4.
What is taproot?
Answer:
The root which develops from the radicle and form main root. The lateral branches arising from the tap root or main root are termed as secondary roots. This root goes deep into the soil. It is characteristic of most of dicots, e.g., Mustard.

Question 5.
What is adventitious root?
Answer:
A root arising from any part of the plant other than radicle such as from stem or nodes is called as adventitious root. These roots are present just below the soil and they are present in group.

Question 6.
Write the name of two rootless plants.
Answer:
Wolffia and Utricularia are the two rootless plants.

Question 7.
Cuscuta plant do not bear leaves, how does they obtain their food?
Answer:
Cuscuta (Amarbel) is a parasitic plant. The stem and branches of this plant possessing specialized parasitic roots called as sucking or haustorial roots, through which they absorb food from host plant.

Question 8.
Write the two main characters of root.
Answer:

1. Roots are usually positively geotropic, hydrotropic and negatively phototropic.
2. The tip of a root usually covered by a small conical root cap, which protects root tip.

Question 9.
How do roots transfer water and minerals to various parts of the plant?
Answer:
Root hair zone of the root helps in transfer of water and minerals to various parts of the plant.

Question 10.
What is root cap? Write its importance.
Answer:
Root cap :
It is cap like parenchymatous multicellular structure which covers the root meristem. The cells of the root cap secrete mucilage. The latter lubricates the passage of root through the soil. Cells of root cap also possess starch grains. Meristematic tissue known as calyptrogen, help in the formation of root cap.

Question 11.
Write the name of root which helps in photosynthesis.
Answer:
Assimilatory root, e.g., Tenosporea, Trapa.

Question 12.
What are lateral roots?
Answer:
Lateral roots are endogenous in origin. Lateral roots arise from pericycle.

Question 13.
What are buttress roots?
Answer:
Buttress roots are the roots found around the base of stem and are formed by the branches of main root in which primary growth is unsymmetrical. These look like extra walls built to support any fort’s wall, e.g., Ficus.

Question 14.
Give examples of Napiform root and Conical root.
Answer:

• Napiform root: e.g., Turnip, beet root.
• Conical root: e.g., Carrot.

Question 15.
Stem develops from which part of the plant?
Answer:
Stem develops from plumule of seed of the plant.

Question 16.
Write the names of subaerial modification of stem.
Answer:

• Runner,
• Stolon,
• Offset,
• Sucker.

Question 17.
Distinguish between stolon and runner.
Answer:

• Stolon is a slender lateral branch that arises from the base of the main axis which afterwards grow above soil whereas runner is growing horizontally above ground.
• Stolon is relatively thick whereas runner is thin.

Question 18.
Write the difference between vascular bundle of root and stem.
Answer:
The vascular bundle of root is radial and exarch and in stem vascular bundles are conjoint, collateral and endarch.

Question 19.
What is cladode?
Answer:
These are green stems branches of limited growth which have taken over the function of photosynthesis from the leaves, e.g., Asparagus, Ruscus etc.

Question 20.
Bulb of Onion is stem. Why?
Answer:
Bulb of Onion is stem, because :

• They have nodes and reduced intemodes.
• Adventitious roots arise from its base.
• Scale leaves are present.

Question 21.
Write the edible part of Potato and Onion.
Answer:
The edible part of Potato is stem and that of Onion is fleshy scale leaves.

Question 22.
Write the names of aerial modification of stem.
Answer:

• Stem tendril,
• Stem thorn,
• Phylloclade,
• Cladode,
• Bulbils.

Question 23.
Write differences between Rhizome, Corm and Bulbs.
Answer:
Differences between Rhizome, Corm and Bulbs

Rhizome	Corm	Bulbs
1. They are thick and flat tened.	Corm is thickened, stout and fleshy.	Bulb is reduced and disc shaped stem.
2. Grow horizontally.	Grow in vertical direction.	Grow vertically.
3. Nodes, intemodes, scale leaves and buds are present.	Nodes are shorter. It bears only one intemode.	Nodes and intemodes are absent.

Question 24.
Write the three main difference between root and stem.
Answer:

1. Nodes and intemodes are absent in root and nodes and intemodes are present in stem.
2. Roots contain root cap and root hairs but they are absent in stem.
3. Apical bud is present in stem but it is absent in root.

Question 25.
Write important functions of the stem.
Answer:
Functions of stem :

• Stem plays an important role in the transportation of water, mineral salts and food materials.
• It helps in photosynthesis in young stage.
• Stem forms aerial system of the plant body.
• Stem provides mechanical support to the plant.
• Some stems help in vegetative propagation.

Question 26.
Write differences between Runner, Sucker and Offset.
Answer:
Differences between Runner, Sucker and Offset

Question 27.
Write differences between Stolon and Runner.
Answer:
Differences between Stolon and Runner

Stolon	Runner
1. It is a slender lateral branch that arises from the main axis.	There are narrow green, prostate branches develop at the base of erect shoot.
2. Size of internodes are longer.	Size of internodes are shorter.
3. It grows under the ground upto certain distance, then come up of the soil. e.g., Colocasia.	It grows above the ground. e.g., Doob grass.

Question 28.
What is pseudo stem?
Answer:
In some monocot stem leaf base becomes broad and flat and covers a part of the node of stem and sheathing of many leaves jointly make a stem like structure known as false stem or pseudo stem e.g., Banana.

Question 29.
Write name of two monocot plants whose leaves contain reticulate venation.
Answer:
Smilax and Dioscorea.

Question 30.
Write short note on phyllotaxy.
Answer:
Phyllotaxy is the arrangement of leaves on the stem and branches. The leaves of each plant are arranged in a definite pattern on the stem or branches. Main purpose of phyllotaxy is to provide sufficient light to the leaves. When a single leaf arises at each node and two leaves at adjacent nodes are held opposite to each other the leaves are called alter¬nate. The phyllotaxy is opposite when two leaves arise from each node in opposite directions but when more than two leaves arise from each node and form a whorl around it, they are called whorled.

Question 31.
What is Isobilateral leaf?
Answer:
In Isobilateral leaf both the surfaces are equally illuminate as both the surfaces can face the sun. So, the two surfaces can face the sun. e.g., wheat, rice.

Question 32.
Why insectivorous plants catches insect as a food?
Answer:
Insectivorous plants are present in swampy areas where deficiency of nitrogen occur. For fulfil this deficiency the plant catches insects.
e.g., Pitcher plant, Utricularia.

Question 33.
What are stipules?
Answer:
Leaves of some plants have lateral outgrowths on each side of leaf base, which are called to be as stipules. The leaves without stipules are known as exstipulate. Main function of stipules is to protect the leaf in the bud. When green, these synthesize food also.

Question 34.
What are dorsiventral leaves?
Answer:
A leaf is called as dorsiventral when it grows in such a way that the ventral surface gets more sunlight and is bright green in colour, whereas the dorsal surface gets less light and dull-green in colour.

Question 35.
Write differences between Parallel and Reticulate venation.
Answer:
Differences between Parallel and Reticulate Venation

Parallel Venation	Reticulate Venation
1. Veins in the lamina run almost parallel to one another.	Midrib is centrally placed and veinlets irregularly distributed forming a network.
2. This type of venations are found in the leaves of Monocot plants. e.g., wheat, rice etc.-	This type of venations are found in the leaves of Dicot plants, such as rose, mango etc.

Question 36.
What is a flower?
Answer:
Flower is a modified shoot, the leaves of which are modified into specialized structures called as floral leaves. A typical flower consists of two accessory whorls Calyx and Corolla and two essential whorls Androecium and Gynoecium.

Question 37.
Write floral formula of a flower, which is actinomorphic, hermaphrodite, hypogynous, 5 sepals, gamosepalous, 5 petals, polypetalous, 5 stamens, polyandrous, bicarpellary syncarpous with superior ovary.
Answer:
Floral formula:

Question 38.
What is floral diagram?
Answer:
The diagrammatic representation of no. of individual members of different floral whorls and their arrangement is called floral diagram.

Question 39.
Explain floral formula along with example.
Answer:
The symbolic representation of the number of individuals present in different floral whorls, their arrangement and general characteristics of the flower is called floral formula, e.g., The structure of the flowers of mustard plant can be expressed by the following floral formula:

Question 40.
Write the floral formula of Cruciferae family.
Answer:

Question 41.
What is androphore and gynophore?
Answer:
In flowers, in between corolla and androecium, structure is present called androphore and in between androecium and gynoecium a floral structure is called gynophore. Both are collectively called gynandrophore.

Question 42.
Write the floral formula of ray and disc florets present in sunflower.
Answer:
Ray florets :
Disc florets :

Question 43.
What do you mean by hypogynous flower?
Answer:
When the ovary occupies the highest position and the stamen, petals and sepals are developed below the ovary, the condition is known as hypogyny and the flowers are called hypogynous. In a hypogynous flower, therefore the ovary is superior, e.g., Mustard, Brinjal, China rose, etc.

Question 44.
Write floral formula of Chinarose flower.
Answer:

Question 45.
What is actinomorphic flower?
Answer:
The flower which can be divided into two equal half portions by any plane are called actinomorphic flower.

Question 46.
Write differences between Superior and Inferior ovary.
Answer:
Differences between Superior and Inferior Ovary

Superior Ovary	Inferior Ovary
1. Thalamus is conical or ball shaped.	Thalamus is flask shaped.
2. Position of ovary is superior.	Position of ovary is inferior.
3. Flower is hypogynous. e.g., China rose.	Flower is epigynous. e.g., Sunflower.

Morphology of Flowering Plants Class 11 Important Questions Short Answer Type

Question 1.
Give general characteristics of stem.
Answer:
General characteristics of stem :

• It is negatively geotropic and positively phototropic.
• It develops from the plumule portion of the embryo during seed germination.
• It is differentiated into nodes and intemodes. Leaves are borne on the nodes.
• Axillary buds are found at the axil of leaves.
• Apical bud is found at the terminal end of the stem, which helps to increase size of the stem.
• The branches of the stem are exogenous in origin.
• Multicellular stem hairs are found on the stem.

Question 2.
Write any four differences between Root and Underground stem.
Answer:
Differences between Root and Underground Stem

Root	Underground Stem
1. Nodes and intemodes are absent.	Nodes and intemodes are present.
2. Scale leaves are absent.	Scale leaves are present.
3. Axillary buds and terminal buds are absent.	Axillary buds and terminal buds are present.
4. Adventitious roots do not arise from it.	Adventitious roots grows from its nodes.
5. Root hairs and root caps are found.	Root hairs and root caps are not found.

Question 3.
Give functions of the stem.
Answer:
Functions of the stem :

• The stem acts as the main support for the branches and leaves.
• It prepares food by photosynthesis process in young stage.
• It helps in conduction of water and minerals absorbed by the roots.
• It helps in translocation of food materials synthesized by the leaves to different parts of the plant.
• In desert plants, it helps to prepare and store food.
• In some plants, it helps for vegetative propagation. Example : Potato, Ginger.
• In some plants, stem is modified into thorn which is mainly defensive in function.

Question 4.
Give general functions of leaf.
Answer:
General functions of leaf:

• It manufactures food material by photosynthesis process.
• It helps for exchange of gases through stomata.
• It helps for transpiration through stomata.

Question 5.
Give special functions of leaf.
Answer:
Special functions of leaf:

• Some leaves store food materials in them.
• Some leaves are modified for defence, climbing, reproduction etc.
• Some leaves changes their colour and help for pollination by attracting insects and animals.
• Some leaves help for vegetative propagation.
• Some leaves are modified into roots and help for absorption of water and minerals.

Question 6.
Distinguish between Phylioclade and Phyllode.
Answer:
Differences between Phylioclade and Phyllode

Phylioclade	Phyllode
1. Flattened green aerial stem performs photosynthesis.	Flattened green petiole performs photosynthesis.
2. Nodes and intemodes are well developed.	Nodes and intemodes are totally absent.
3. They take part in vegetative reproduction.	They never take part in vegetative
4. It may bear flowers, scale leaves, bristles or spines.	Phyllode never bear such structures.
5. It originates from axil of leaf.	It originates from axil of bud.
6. It may store food material, mucilage and large quantity of water for future use. e.g., Opuntia.	It generally never stores such substances. e.g., Australian Acacia.

Question 7.
Give functions of calyx.
Answer:
Functions of calyx :

• Protection of floral organs in bud condition.
• Provide support to the base of mature flower.
• Attraction of insects and help in pollination when they are coloured.
• Dispersal of fruits and seeds.
• Help in photosynthesis because they contain chlorophyll.

Question 8.
What do you mean by adhesion of stamens? Describe its various types.
Answer:
Adhesion : Fusion among members belonging to different whorls of flower is called as Adhesion.
Following are the different types of adhesion of stamen to other floral whorls :

1. Epipetalous : When the filaments are fused to the petals to a greater or lesser extent.
   Example: Datura.
2. Epiphyllous : When the filaments are fused to the petals of perianth.
   Example: Onion.
3. Gynandrous : When the stamens adhere to the gynoecium either loosely or very intimately.
   Example: Calotropis.
4. Gynostegium : A protective covering of gynoecium formed by the androecium and not involving any fusion between the androecium and the gynoecium.
   Example: Orchidaceae.

Question 9.
What do you mean by insertion of floral leaves?
Or,
What do you mean by hypogynous, perigynous and epigynous ovary?
Answer:
Insertion of floral leaves : Position of the gynoecium and other floral whorls like calyx, corolla and androecium on the thalamus is called as insertion of floral leaves. It is of following three types :
1. Hypogynous :
When thalamus swells to form a ball like structure then the ovary of the gynoecium occupies the highest position on the thalamus and the stamen petals and sepals develops below the ovary on the thalamus. This type of insertion is called as Hypogynous. The ovary in this case is superior, e.g., Mustard.

2. Perigynous :
In this type the periphery of the thalamus expand to varying degrees making the thalamus disc shaped, cup shaped or flask shaped. The apex of the thalamus lies at the centre to which ovary of the gynoecium is found attached. The remaining parts are borne on the periphery.
Example : Pisum sativum. The ovary in this case is half inferior.

3. Epigynous :
The thalamus is hollowed out in the form of a cup or flask but its internal walls are fused with the wall of the ovary. Only the style and stigma of the pistil are observable in this condition. The other parts of the floral leaves are borne at the top of the ovary.
The ovary is termed as inferior and the thalamus is called as Epigynous.

Question 10.
What do you understand by modification of roots ? Which type of modifications are found in following :
(a) Banyan,
(b) Turnip,
(c) Mangrove?
Answer:
Modification in roots:
Besides the common functions like absorption and fixation the roots are further modified to perform specialized functions such as storage of food, support, respiration etc. This is known as modification.
(a) Banyan :
In banyan trees adventitious roots arise from horizontal branches, grow vertically downwards and penetrate into the soil. These branches increase in thickness by secondary growth and give mechanical support to the horizontal branches. This type of modification of root in banyan tree is called as Prop root.

(b) Turnip :
In turnip tap root is swollen to form spherical shaped structure, which is tapering at the lower part due to storage of food. This type of modification of root in turnip is called as Napiform root.

(c) Mangrove :
Mangrove plants grow in saline swamps or marshy areas, which develop special kind of roots called as respiratory roots or pneumatophores. Roots of these plants do not have proper gaseous exchange due to saline and marshy areas, thus some roots (pneumatophores) grow vertically up and become negatively geotropic, which are provided with air pores called lenticels. Lenticels help for gaseous exchange during respiration.

Question 11.
On the basis of external features give reason for following statements :
(a) All underground parts of plants are not roots.
(b) Flower is a modified shoot.
Answer:
(a) All underground parts of plants are not roots :
It is true that root always
grows down into the earth but some stems are exceptions. Such as ginger, onion and potato. These are examples of underground modification of stems.
Tuber of potato is an underground modification of stem. When a tip of a positively geotropic lower branch of potato stem swells-up due to the accumulation of reserve foods then a potato tuber is formed.
We can differentiate tuber of potato from root by following points :

• Nodes and intemodes are found in it.
• Scale leaves are found in it.
• Buds are found in the form of eyes.
• It is a modified structure for storage of food.

(b) Flower is a modified shoot:
Following points prove that flower is a modified shoot:

• Thalamus of the flower looks like a stem in which nodes and intemodes are condensed.
• The phyllotaxy of floral leaves is similar to those of vegetative leaves. Actually they are modified leaves to perform special functions.
• Sepals of some plants look like leaves which proves that sepals are modified leaves, e.g., Water lily.
• In some plants thalamus elongates and exhibits clear nodes and intemodes. It con¬firms the view that thalamus is a modified branch.
• Like stems flower also develops from a bud.

Question 12.
How Pinnate compound leaf differs from Palmate compound leaf?
Answer:
There are two types of compound leaves :
1. Pinnate compound leaf and
2. Palmate compound leaf, which differs from each other in following ways :

1. Pinnate compound leaf:
The leaf blade is divisible into several leaflets, arranged laterally on the midrib portion. It is of four types :
(a) Unipinnate : When the rachis directly bears the leaflets. In this, rachis acts as midvein and the leaflets are found on its either side.
e.g., Rose, Cassia.
When, there are even numbers of leaflets on the midrib, it is known as paripinnate, e.g., tamarind, and when there are odd number of the leaflets on the midrib it is called as imparipinnate. e.g., Rose, Neem.
(b) Bipinnate : In this type rachis branches only once and bears the leaflets.
Example: Gulmohar.
(c) Tripinnate: When the secondary branches divides once again bearing the leaflets.
Example: Moringa.
(d) Decompound : When there are more than three incisions of the lamina.
Example: Coriander.

2. Palmate compound leaf:
In this type margin of leaf is incised to divide the base of the blade or the petiole in such a manner so that the leaf blade is divided into a number of segments or leaflets. The arrangement of the leaflets look like the fingers of human palm and hence this type of the compound leaf is known as Palmate compound leaf. Depending upon the number of leaflets they are divided into the following types :
(a) Unifoliate : A unifoliate compound leaf is the result of reduction of all leaflets of a multifoliate compound leaf, except the terminal leaflets as in Citrus sp.
(b) Bifoliate : When there are only two leaflets as in Hardwickia.
(c) Trifoliate : When there are three leaflets as in Aegel marmelos.
(d) Quadrifoliate : When there are four leaflets as in Oxalis and Marsileas.
(e) Multifoliate or Digitate: When more than four leaflets are present as in Salmalia sp.

Question 13.
Describe various kinds of phyllotaxy with example.
Answer:
Phyllotaxy is defined as the arrangement pattern of leaves on the stem and branches in order to provide sufficient light and air. It is interesting to note that leaves are never arranged haphazard but with mathematical precision at definite and regular intervals.

Type of phyllotaxy:
This is usually of three types:
1. Alternate : From each node, single leaf arises, two leaves arise from the adjacent nodes and are held opposite to each other, e.g., Sunflower, Mango. These may be of the following types:

• Distichous,
• Tristichous,
• Pentastichous.

2. Opposite :
Two leaves are present on a single node but are opposite to each other that is at 180° to each other. It may be of two types :

• Opposite superposed : When the successive pairs of leaves are placed one above another forming two vertical rows when seen from the top. e.g., Guava.
• Opposite decussate : When are pair of leaf is placed at right angle to the next e.g.
  Tulsi.

3. Whorled or Verticillate :
When there are more than two leaves at each node,arranged in a circle or whorled.

Morphology of Flowering Plants Class 11 Important Questions Long Answer 

Question 1.
Define following terms :
(a) Aestivation,
(b) Placentation,
(c) Actinomorphic,
(d) Zygomorphic,
(e) Superior,
(f) Perigynous,
(g) Epipetalous.
Answer:
(a) Aestivation : Aestivation is the arrangement of floral leaves (Calyx, corolla or perianth) in the bud condition. The various types of the aestivation are :

1. Valvate,
2. Twisted,
3. Imbricate,
4. Descending imbricate,
5. Quincuncial.

(b) Placentation :
The arrangement of the ovules on the placenta of ovary is called as Placentation. The various types of placentations are :

1. Marginal,
2. Parietal,
3. Basal,
4. Axile and
5. Free central type.

(c) Actinomorphic :
When the flower can be divided into two equal halves along more than one vertical plane passing through the middle of the flower then the flower is called as Actinomorphic. e.g., China rose.

(d) Zygomorphic:
When flower can be divided into two equal halves along only one plane then flower is called as Zygomorphic. e.g., Pea.

(e) Superior :
When thalamus of the flower is ball or conical shaped then position of ovary is superior to all other parts of the flower. Such ovary is called as Superior ovary. e.g., China rose, Mustard, Brinjal etc.

(f) Perigynous :
When thalamus of the flower is cup shaped and position of the ovary as compared to other floral parts is half inferior then the flower is called as Perigynous. e.g., Pea, rose, bean etc.

(g) Epipetalous :
When the filaments of stamens are fused to the petals to a greater or lesser extent then stamens are called as Epipetalous. e.g., Dhatura.

Question 2.
Write differences between following :
(a) Racemose and Cymose inflorescence
(b) Tap root and Adventitious root.
(c) Apocarpous and Syncarpous.
Answer:
(a) Differences between Racemose and Cymose Inflorescence

Racemose Inflorescence	Cymose Inflorescence
1. The main axis shows indefinite growth.	The main axis shows definite growth.
2. Main axis gives off flowers laterally.	The main axis and its branches terminate in flower.
3. Flowers grow in acropetaf succession.	Flowers grow in basipetal succession.
4. The order of opening of flowers is centripetal.	The order of opening of flowers is centrifugal.

(b) Differences between Tap and Adventitious Root Systems

Tap Root System	Adventitious Root System
1. Tap roots arise from the radicle of embryo.	These roots are arising either from stem or from the leaves.
2. The primary root formed from the elongation of radicle, is long-lived.	It is short-lived.
3. Tap roots are deep feeder.	It may usually be surface feeder.
4. There is a single main root.	There are a number of roots arising in clusters.
5. Main root is very thicker than other roots.	All roots are fibrous.
6. This root system is usually found in dicotplants.	This root system is usually found in monocot plants.
7. It is always underground and meant for anchorage.	Adventitious roots may be underground for anchorage or aerial for performing specialized functions like assimilation, support, etc.

(c) Differences between Apocarpous and Syncarpous

Apocarpous	Syncarpous
1. The gynoecium with many free carpels is described as Apocarpous. e.g., Family Rannunculus.	The gynoecium with two or more carpels which are fused is described as Syncarpous. e.g., China rose, Dhatura.

Question 3.
What is flower? Describe various parts of a typical angiosperm flower.
Answer:
Flower is a modified shoot which performs the function of sexual reproduction and arises either at the tip of a branch or at the axil of a leaf. It leads to formation of fruits and seeds.

A typical flower has following four parts :

1. Calyx,
2. Corolla,
3. Androecium,
4. Gynoecium.

(A) Accessory parts of flower :
(1) Calyx :
Outermost whorl of the flower is called as Calyx. Unit structure of calyx is called as Sepal. It protects the flower in bud condition. Generally it is green in colour due to presence of chlorophyll, thus can prepare food by photosynthesis process.

(2) Corolla :
It is the second whorl of the flower, which consists of unit structures called as Petals. Generally it is large and colourful to attract insects for pollination.

(B) Essential parts of the flower :
(3) Androecium :
It is the male reproductive part of the flower which forms third whorl of the flower. Unit structure of androecium is called as Stamen. Each stamen consists of three parts :

1. Filament,
2. Anther and
3. Connective. Pollen grains are formed inside the anther. Male gametes are formed in the pollen grains.

(4) Gynoecium :
It is the innermost and fourth whorl of the flower. It is the female reproductive part of the flower. Unit structure of gynoecium is called as Carpel or Pistil. Each pistil consists of three parts:

1. Stigma,
2. Style and
3.  Ovary. Egg is formed in the ovules present in the ovary.

Question 4.
Describe various types of placentations found in flowering plants.
Answer:
Placentation : The arrangement of the ovules on the placenta of the ovary is called as placentation. It is of the following types :

(i) Marginal:
It is found in monocarpellary pistils which bears a longitudinal placenta on the ventral surface. The placenta bears one or two rows of ovules.
Example: Pea.

(ii) Parietal:
The parietal placenta shows two or more than two placentae attached to the wall of ovary. It is found in syncarpous pistils with unilocular ovary. The ovules are borne near the periphery of the ovary and the placenta are situated on the fused margins of the adjacent carpels.
Example: Poppy, Mustard.

(iii) Basal:
The pistil is usually polycarpellary and syncarpous. The ovary is unilocular and possesses one placenta which bears a single ovule towards the base of the ovary.
Example: Helianthus (Sunflower).

(iv) Superficial:
Here the ovules are not restricted to the margins of the carpels but are found on the whole inner surface of the ovary.
Example: Nymphaea.

(v) Axile :
It is found in polycarpellary syncarpous pistils. The ovary is partitioned into two or more chambers or loculi depending upon the number of the fusing carpels. The placentas are found in the centre of the ovary at the place of union of the septa so that a central or axile column is formed which bears the ovules.
Example: China rose.

(vi) Free central:
In this case the pistil is polycarpellary and syncarpous but ovary is unilocular. The ovules are found all around a central column which runs throughout the middle of the ovary.
Example: Dianthus.

Question 5.
How modifications of leaves are helpful to plants?
Answer:
Main function of leaves are photosynthesis and transpiration. In order to perform certain peculiar function the leaves change their shape accordingly. However the main types of modifications are as follows :

1. Leaf spines :
When leaf is modified into sharp pointed slender structures called as spines. They provide protection to the leaf and also checks excessive transpiration.
Example : Asparagus, Ulex etc.

2. Leaf tendrils :
Tendrils are slender wiry often closely coiled structures which helps in climbing of the plant on its support, Whenever they come in contact with a neighbouring living or non-living object they coil around it and help the plant to rise up. For this purpose either full leaf or some leaflets are changed into tendrils.
Example : Pisum sativum, Begonia etc.

3. Leaf hooks :
Here leaf is modified into thick sharp, stiff and curved structures called as hooks. These hooks cling to the bark of a tree and support the climbing plant.
Example: Begonia.

4. Scale leaves :
When leaf is reduced in the form of thin, dry membranous structure to protect the axillary buds in their axil. They are frequently observed in case of underground modification of stem.
Example : Potato, Ginger, Onion etc.

5. Storage leaves :
Leaves of some plants become fleshy due to storage of water and food material, such leaves are called as storage leaves.
Example : Agave, Aloe, Bryophyllum etc.

6. Leafy roots :
When the leaf modified into a root like structure.
Example: Salvinia.

7. Leaf pitcher:
When the entire or a part of the leaf is changed into a dilated pitcher like structure to catch insects to fulfil their nitrogen requirement.
Example : Nepenthes, Dischidia rafflesiana.

8. Leaf bladders:
When some leaves are modified into small oval or circular bladder into a trap-door entrance. This modification is most conspicuous in insectivorous but not in hydrophytic plant species like Utricularia.
Example: Utricularia.

Question 6.
Define inflorescence. Describe on the basis of types of inflorescence found in flowering plants.
Answer:
Inflorescence:
Arrangement of flowers on the floral axis of a plant is known as inflorescence. The central axis of an elongated inflorescence is called as Peduncle.
In some plants flowers arise singly at the apex and are called as Solitary terminal flowers as in Poppy. In some cases like Oxalis, guava etc., single flowers arise in the axil of the leaf and are called as Solitary axillary flowers. The inflorescence may be of following types:
(A) Simple inflorescence,
(B) Compound inflorescence,
(C) Special inflorescence and
(D) Mixed inflorescence.

(A) Simple inflorescence : When flowers are directly bom on the main axis. It is of two types :
(1) Racemose or indefinite : When the growth of the inflorescence continues indefinitely and produces flowers laterally in acropetal succession. It may be of following types :
(a) Typical raceme : e.g.,Mustard
(b) Spike : e.g., Amaranthus
(c) Spikelet: e.g., Wheat
(d) Spadix : e.g., Banana
(e) Catkin : e.g., Mulberry
(f) Umbel: e.g., Coriander
(g) Corymb : e.g.Jberis amara
(h) Capitulum or Head: e.g., Sunflower.

(2) Cymose or definite :
When the growth of the main axis is limited as the terminal bud is modified into a flower. The subsequent flowers are produced in basipetal succession. It may be of following types:
(a) Uniparous or Monochasial cyme : It is of two types :
(i) Helicoid cyme : e.g., Begonia
(ii) Scorpioid cyme : e.g., Heliotropium.

(b) Biparous or Dichashial cyme : e.g., Ixora
(c) Multiparous or Multichashial cyme : e.g., Calotropis.

(B) Compound inflorescence : When the main axis of the inflorescence is branched and the branches bear flowers of the inflorescence, it is called as compound inflorescence. It is of following types :
(a) Compound raceme : e.g.,Goldmohar,
(b) Compound spike : e.g., Spinach,
(c) Compound spadix : e.g., Coconut,
(d) Compound corymb : e.g., Cauliflower,
(e) Compound umbel: e.g., Coriander.

(C) Special inflorescence :
There are certain inflorescences which do not agree to the characteristics already described above. So these may be grouped under the special types and are of the following types :
(a) Cyathium : e.g., Euphorbia
(b) Verticillaster : e.g., Oscimum (Tulsi)
(c) Hypanthodium : e.g., Peepal.

(D) Mixed inflorescence: When the inflorescence is neither true racemose nor cymose but a mixture of two, it is said to be mixed inflorescence. It is of following types :
(a) Spadix of cyme : e.g., Musa sp.
(b) Cyme of umbels : e.g., Lantana
(c) Cyme of corymb : e.g., Oldenlandia, corymbosa.
(d) Cyme of capitula : e.g., Vernonia.

Question 7.
Give one-one example of Fabaceae and Solanaceae families and describe them in biotechnical language and draw their floral diagrams.
Answer:

(A) Family Fabaceae : Family Fabaceae was earlier called as Family Papilio- naceae. Pea plant represent Family Fabaceae or Papilionaceae.
Description of Pea flower (Pisum sativum):

1. Inflorescence: Generally racemose. In Crotolaria, it is terminal raceme, in Melilotus it is corymbose raceme whereas in Cicer arietinum it is solitary axillary.
2. Flowers : Pedicellate, bracteate, bisexual, zygomorphic, hypogynous or perigy- nous, complete and pentamerous.
3. Calyx : 5 sepals, gamosepalous, campanulate or tubular, imbricate aestivation.
4. Corolla: 5 petals, polypetalous, papilionaceous, posterior petal is largest and known as standard. Two lat¬eral petals are known as wings. Two anterior petals fuse to form boat shaped keel. The aestivation is vexillary.
5. Androecium: Generally, 10 stamens are present in two groups 9 + 1. It is known as diadelphous condition. The 9 stamens are fused through filament. In Crotolaria and Pongamia, it is monoadelphous. Anthers are dithecous, basifixed and introse.
6. Gynoecium : Monocarpellary, unilocular, su¬perior or half inferior. Marginal placentation. Style simple, stigma capitate.
7. Fruits : Legumes or Pods.
8. Seeds: Non-endospermic.
9. Floral formula:
   Pisum sativum : im

(B) Family Solanaceae :
Dhatura flower belongs to Family Solanaceae.
Description of Datura alba (Dhatura) flower :

1. Inflorescence : Inflorescence seen in the plants is of cymose type. Datura shows dichasial cyme, Solarium shows helicoid cyme, Atropa belladonna shows scorpioid cyme and Nicandra shows solitary axillary type of inflorescence.
2. Flower : Usually the flowers are pedicellate, bisexual, complete, pentamerous, actinomorphic, hypogynous and bracteate or ebracteate. Flower is zygomorphic in Hyosymus niger and Salpiglossis. The flowers of Salpiglossis are always cleistogamous.
3. Calyx : 5 sepals, gamosepalous, hairy, persistent with twisted aestivation.
4. Corolla : 5 petals, gam’opetalous, usually white in colour with valvate aestivation.
5. Androecium : 5 stamens, polyandrous, epipetalous, filament hairy with long anthers.
6. Gynoecium : Bicarpellary, syncarpous, bilocular, oblique ovary, superiorly placed. Placentation is axile with numerous swollen ovules found on the placenta. Style is simple and stigma is bilobed.
7. Fruits : Fruits are either of berry type e.g., Tomato, Brinjal etc. or capsule type e.g., Durantum.
8. Seeds : Seeds are endospermic.
9. Floral Formula :

Question 8.
Explain modification of adventitious roots for storage of food.
Answer:
Modification of adventitious roots for storage of food : Following modifications are found in adventitious roots for storage of food :
1. Fasciculated roots : These are like the tuberous roots but occur in a fascicle or cluster at the base of the stem.
Example : Dahlia, Asparagus etc.

2. Tuberous roots: The root is swollen and fleshy and may assume any shape, rounded, fusiform or any irregular shape, e.g. Sweet potato.
3. Nodulated roots : In this type, the fibrous adventitious roots abruptly become swollen at the tip.
Example : Curcuma amada (Haldi).
4. Moniliform roots : In this type, the root shows alternately swollen and thin portions.
Example: Pasella, Vitis.
5. Annulated roots : In this type, the root shows a series of ring like swellings separated by ring like grooves.
Example : Cephaelis, Ipecac uanha.

Question 9.
Explain modification of adventitious roots for providing mechanical support to the plant.
Answer:
Modification of adventitious roots for providing mechanical support: Following types of modifications are found in adventitious roots for providing mechanical strength:

1. Prop root:
These roots give support to the branches of a tree as the props in a tent. In the Banyan tree (Ficus benghalensis) the prop roots arise from the horizontally growing branches. The prop roots hang down vertically for quite a number of years and then enter the soil. The branches rest on them. The Great Indian Banyan Tree of the Indian Botanical Garden, Sibpur, Calcutta has produced more than 800 such prop roots. It is 200 years old and has occupied 300 sq.m. area.

2. Stilt root:
Stem of monocot plants are very weak and can not stand straight due to their weight. Some roots arise from the basal nodes of such stem and grow obliquely downward, ultimately strike the ground and thus support the plant. These roots are called as stilt root.

3. Buttress root:
The buttress roots are plate like structure formed by the bases of the main roots of tropical trees. It provides support to the plant.
Example : Prombab malabaricum.
Fig. Buttress root.

4. Climbing root:
These roots arise from the nodes of a climbing stem and cling very tenaciously to the substratum which may be a wall or the trunk of tree and thus help the plant in climbing.
Example : Money plant, Black pepper, Pan.

5. Clinging root:
These roots are found in epiphytes. These roots secrete a sticky substance which quickly dries up and aids in the sticking of the root to substratum.
Example: Vanda.

Question 10.
Write down the modification of tap root.
Or,
Explain modification of tap root for food storage with diagram.
Answer:
Modification of tap root:
(I) For food storage:
The tap root becomes swollen and fleshy with the stored food. The secondary roots remain thin. Hypocotyl (embryonic region between cotyledons and radicle) may also join the tap root in storing food. Stem is reduced and discoid in the beginning and bears radicular leaves. Depending upon the shape, the fleshy tap roots are of the following types:
1. Fusiform roots :
When the root is swollen in the middle and gradually tapering towards the apex and base it is called a fusiform root. It is more or less spindle shaped, e.g., Radish (Raphanus sativus).

2. Conical roots:
When the root is broad at the base and gradually tapers towards the apex like a cone, it is said to be conical root. It bears many secondary roots throughout the length, e.g., Carrot (Daucus carota).

3. Napiform roots :
When the root is considerably swollen at the upper part and becoming almost spherical and shaiply tapering at the lower part, it is said to be napiform. e.g., Turnip or Saljam (Brassica rapa), Beet root (Beta vulgaris). Base of root shows radicular leaves. In turnip most of the swollen part is formed hypocotyl. The lower narrow, pointed apical region bears threadlike secondary roots. In beet both the tap root and hypocotyl are swollen.

4. Tuberous roots :
These roots are thick and fleshy and do not maintain any definite shape, e.g.. Four o’clock (Mirabilis jalapa), Parwal (Trichosanthes), etc.

(II) For respiration (Pneumatophore or Respiratory roots):
Many plants growing in estuaries, saline swamps, marshes and salt lakes develop special kinds of roots called respiratory roots or pneumatophores. The plants which develop pneumatophores are called halophytes. These plants do not have proper gaseous exchange due to saline conditions, as a result their growth becomes hindered.

Pneumatophores are helpful in respiration. From the underground roots of the plant, pneumatophores grow vertically upwards and come out of the water in the form of conical spikes. They occur in large numbers around the tree trunks and are provided with air pores called lenticels. Air pores help in proper respiration. e.g., Rhizophora, Avicenna, Sonneratia, Heritiera (vem. Sundri), Kutch vegetation, Mangroves, etc.

Question 11.
Describe various underground modifications of the stem.
Answer:
Underground modifications of the stem : Such stems which grows under the ground are called as underground modifications of stems. It helps plants to survive in unfavourable condition. Stem become modified in various ways for certain special functions, i.e., storage of food material and vegetative propagation.
Following types of modifications are found in underground stem :
1. Rhizome,
2. Stem tuber,
3. Bulb,
4. Corm.

1. Rhizome:
It is a perennial underground stem which shows unlimited growth and develops in favourable condition to form upright shoot and leaves. It is a prostrate, thickened stem, creeping horizontally under the surface of the soil. It is provided with distinct nodes and long or short intemodes. It possesses a bud in the axil of scaly leaf and it ends in a terminal bud. From underside adventitious roots grows.
Example: Ginger, Fern.

2. Stem tuber:
The tubers are solid, thick¬ened stems or branches serving for storage of food and also vegetative propagation. When it develops from root it is called root tuber but when it develops from stem it is called as stem tuber. In Potato plant, tip of a positively geotropic, fleshy stem growing in a vertical direction swells due to the accumulation of reserve foods. These tubers bear temporary scale leaves with buds in their axils called ‘eyes’. Even a single detached bud may grow into a new plant. In this way it helps for vegetative propagation.
Example : Potato.

3. Bulb :
The bulb is a specialized underground reduced stem. In this the stem remains comparatively small and the food material is stored in the large, fleshy scales which invest and overlap the stem. However the stem is no more than a thick disc or very flat cone, but it has an apical bud on the upper side, and adventitious roots are formed in an annual plant from the marginal portion of the underside.

There are two types of bulbs:
(a) Tunicated bulb : In this type of bulb, one scale leaf totally overlap another. Towards the outside cover of dry scale leaves are found.
Example: Onion.

(b) Scaly bulb : In scaly bulb, the fleshy scale composing the main bulk overlap at their margin. No tunic is found to enclose the whole buds.
Example: Garlic.

4. Corm:
Corm is st, fleshy and thickened underground em growing in
vertical directions. There are many buds present in the axil of scale leaves which develop in daughter corms. At the bases or even from sides of stem the adventitious roots develop.Corms are also called condensed rhizomes, help in vegetative propagation and storage.
e.g., Arbi or Ghuiyan (Colocasia), Zimikand (Amorphophallus), Crocus.

Question 12.
Explain the aerial modifications of stem.
Answer:
Aerial Modifications or Metamorphosed Stems :
Sometimes vegetative and floral buds instead of growing into branches and flowers undergo modifications to form specialized structures like tendrils, thorns, phylloclades or bulbils to perform specific functions. These stems are called as metamorphosed stems.

1. Stem tendril:
It is a modification of stem in which axillary bud of leaves modifies, to form a filamentous structure called tendril. They help the plant to attach itself to the support and climb it. They are found in weak stemmed plants. The tendrils are leafless, coiled structures with sensitive tactile pits and adhesive glands to promote fixation of weak stem.
Stem tendrils are of four types :
(a) Axillary : e.g., Passiflora. »
(b) Extra axillary: e.g., Cucurbita, Lagenaria, Luffa.
(c) Leaf opposed : e.g., Grapevine.
(d) Floral bud or Inflorescence tendril: e.g., Antigonon,
Cardiospermum (Balloonvine).

2. Stem thorn :
In this modification, axillary bud is modified into thorn (Duranta) known as stem thorn. Leaves, branches and flowers are developed on thorns. It clarifies that thorn is a modified stem, e.g., Citrus, Bougainvillaea, Pome¬granate, Alhagi, etc.

3. Phylloclade:
It is a modified stem in which stem gets modified into leaflike structures and leaves get modified into spines. The phylloclade is green and flattened (e.g., Opuntia or Nagphani), cylindrical (e.g., Casuarina) or ribbon like (e.g., Zygocactus or Muhlenbeckia) and has distinct nodes and intemodes.

They are green in colour and capable for the synthesis of food by photosynthesis. Formation of phylloclade reduces loss of water through transpiration and therefore conserve water and food. In Opuntia, the phylloclades are fleshy and segmented. Phylloclade develops from branches of unlimited growth. Internally, phylloclades contain many mucilage glands to promote retention of water.

4. Cladode:
These are green stems branches of limited growth (usually one intemode long) which have taken over the function of photosynthesis from the leaves. The true leaves are reduced to scales or spines, e.g., Asparagus, Ruscus, etc. In Ruscus aculeatus, the cladodes are leaflike in appearance with spiny tip, ovate outline and roughly parallel veins. The cladodes are borne in the axils of scale leaves. The stem nature of the cladode of Ruscus is clear from :

• Origin in the axil of scale leaf,
• Bearing a scale leaf in the middle,
• Formation of a bud in the axil of the scale leaf bore in the middle,
• Development of flowers from the axillary bud produced in the middle of cladode.

5. Bulbils :
When axillary bud becomes fleshy and round due to storage of food (as
carbohydrate) then bulbils are formed. These fall down and form new plants, e.g., Aloe, Agave, Onion, Garlic.

Question 13.
What is Venation? Describe its various types.
Answer:
Venation:
Arrangement of veins and veinlets in the lamina (leaf blade) is known as venation. Veins are the conducting tissues consist of xylem and phloem and are found in continuation with the conducting tissue of petiole.
They are concerned with the conduction of water, mineral, salts and food and they form structural framework of the lamina. There are mainly two types of venations :
(A) Reticulate venation, (B) Parallel venation.

(A) Reticulate venation :
Venation in which veins are arranged in such way that they form a network of vein, is called reticulate venation. It is the characteristic feature of dicot leaves. It is of two types :
1. Unicostate: Venation in which leaflets contain a strong midrib or costa, from which lateral veins are arising and these veins form a network is called unicostate venation, e.g., Mango, Guava, Peepal.

2. Multicostate :
Venation in which many strong veins arising from terminal portion of the stalk and then form a network is called multicostate venation. It is also of two types :
(a) Divergent: Venation in which veins arise from the centre and runs towards margin apart from each other is called divergent venation, e.g., Castor, Cotton.
(b) Convergent: Venation in which many veins arising from the centre of stalk, moving towards the margin of the leaf and finally fused is called convergent venation, e.g., Zizyphus (Ber), Smilax.

(B) Parallel venation :
In this type of venation veins in lamina run almost parallel to one another, e.g., Maize, Grass, etc. It is of the following two types :
1. Unicostate : In this type only one midvein is found from which veinlets arise parallel to each other, e.g., Banana, Canna.
2. Multicostate : In this type more than one mid veins are present. It is of two types :

• Convergent: In this type, all midveins run parallel to each other from the base of lamina and unite at the apex, e.g., Grass, Rice, Bamboo, etc.
• Divergent : In this type, all the midveins, instead of going towards the apex, spread out towards the periphery away from each other, e.g., Borassus (Taad).

Question 14.
What do you mean by aestivation of flower? Describe various types of aestivations found in flower.
Answer:
Aestivation :
Aestivation is the arrangement of floral leaves (calyx, corolla or perianth) in the bud condition. The various types of the aestivations are:

• Valvate: Margins of the adjacent petals touch each other but without overlapping.
  Example: Datura.
• Twisted: In this type one margin of the petal overlaps regularly the margins of the adjacent petal, the other margin being overlapped by the margin of another adjacent petal.
  Example: China rose.
• Imbricate : In this type irregular overlapping of petals by one another. It has three subtypes beside the imbricate proper.
  Example: Gulmohar.
• Quincuncial: In this type two petals are external, two internal and one with one margin external while its other margin is internal.
  Example: Pumpkin.
• Descending imbricate or Vexillary: In this type the posterior vexillum or standard petal overlapping the two lateral wings or alae, the latter being external to the two anterior petals forming keel or carina.
  Example: Pea family.

Question 15.
Explain the types of germination.
Answer:
Types of Seed Germination : There are two types of seed germination takes place in higher plants :
1. Hypogeal germination :
At the time of seed germination when the epicotyl elongates and becomes curved, it brings the plumule above the soil. Cotyledons remain underground. Such types of seed germination is called as hypogeal.
Example: Gram, Pea, Maize, Mango, Jackffuit, Paddy, Wheat, Coconut, Date palm etc.

2. Epigeai germination :
In bean, hypocotyl grows actively and becomes curved. It brings the seed above the soil. After coming above the surface of the soil the hypocotyl straightens. The loosened seed coats fall down and cotyledons turn green. Such a type of seed germination is called epigeai. Now epicotyl grows and the plumule gives rise to green leaves. The cotyledons fall down ultimately.

Question 16.
Describe structure of gram seed with labelled diagram.
Answer:
Structure of gram seed :
It is dicotyledonous and non-endospermic seed and is formed in green pod or legume fruit produced on the plant.

External structure :
It is light or dark brown coloured seed. The seed is broad at one end and pointed at the other end. A furrow is found on one side of the seed. A narrow ridge called raphe runs in the furrow. At the middle of raphe an oval shaped chalaza is found. A small oval shaped scar lies on the same side called as hilum. It is the part of the seed by which it was attached to the funicle. In between the hilum and pointed end of seed a small pore is found called as micropyle.

Internal structure :
The gram seed is covered over on all sides by two seed coats. The outer seed coat is brown and hard structure and is called as testa while the inner seed coat is in the form of a white membranous structure called as tegmen. The seed coat encloses embryo containing a short embryo axis. Embryo consists of following parts :
(i) Cotyledons :
Two large and broad structure attached to the embryo axis are called as cotyledons. Food materials are stored in the cotyledons.

(ii) Embryo axis :
The tip of embryo axis towards the micropyle end is called radicle(the future root). The outer end of the embryo axis found deeper in between the cotyledons is the plumule (the future shoot). The part of the embryo axis between the point of attach¬ment of the cotyledons and the plumule is known as epicotyl and that between the point of attachment of cotyledons and radicle as hypocotyl.

Question 17.
Draw labelled diagram of Longitudinal section (L.S.) of following :
(i) Gram seed,
(ii) Maize seed.
Answer:
(i) Gram seed :

(ii) Maize seed:

Morphology of Flowering Plants Class 11 Important Questions Objective Type

1. Choose the correct answers:

(A)
Question 1.
Respiratory roots are found in :
(a) Betel
(b) Chestnut
(c) Jussiaea
(d) Maize.
Answer:
(c) Jussiaea

Question 2.
Roots that absorb oxygen from environment are called :
(a) Hygroscopic
(b) Suckers
(c) Digestive roots
(d) Pneumatophores.
Answer:
(d) Pneumatophores.

Question 3.
Velamen cells are found in :
(a) Respiratory roots
(b) Stilt roots
(c) Fleshy roots
(d) Aerial roots of Orchids
Answer:
(d) Aerial roots of Orchids

Question 4.
Which one of the following is not a Rhizome :
(a) Ginger
(b) Colocasia
(c) Lotus
(d) Turmeric.
Answer:
(c) Lotus

Question 5.
Pneumatophores are found in:
(a) Xerophytes
(b) Epiphytes
(c) Mangrove plants
(d) Hydrophytes.
Answer:
(c) Mangrove plants

Question 6.
Respiratory roots help in:
(a) Transpiration
(b) Carbohydrate metabolism
(e) Protein synthesis
(d) Respiration.
Answer:
(d) Respiration.

Question 7.
Which of the following is partial root parasite :
(a) Cuscuta
(b) Orobanche and Balanophora
(c) Striga and Santalum
(d) Loranthus.
Answer:
(c) Striga and Santalum

Question 8.
Sweet potato is a modification of:
(a) Leaf
(b) Adventitious root
(c) Tap root
(d) Stem.
Answer:
(b) Adventitious root

Question 9.
Type of roots observed in Pandanus is :
(a) Prop roots
(b) Stilt roots
(c) Tuberous roots
(d) Climbing roots.
Answer:
(b) Stilt roots

Question 10.
Adventitious roots:
(a) Develop from radicle
(b) Develop from any part of plant body except the radicle
(c) Develop from flowers
(d) Develop from embryo.
Answer:
(b) Develop from any part of plant body except the radicle

Question 11.
The thick roots that hang down from the Banyan tree are called :
(a) Buttress roots
(b) Pneumatophores
(c) Prop roots
(d) Stilt roots.
Answer:
(c) Prop roots

(B)

Question 1.
Largest (tallest) plants of the world are associated with :
(a) Dicots
(b) Monocots
(c) Angiosperms
(d) Pteridophytes.
Answer:
(c) Angiosperms

Question 2.
Coms in:
(a) A horizontal underground stem
(b) An underground erect stem
(c) An underground root
(d) An underground stem.
Answer:
(b) An underground erect stem

Question 3.
The main function of the stem is:
(a) Storage
(b) To produce branches and leaves
(c) Conduction
(d) All of these.
Answer:
(d) All of these.

Question 4.
Tuber of potato is a modified:
(a) Root
(b) Stem
(c) Creeper
(d) Bulb.
Answer:
(b) Stem

Question 5.
The underground swollen edible part of onion is:
(a) Root
(b) Bulb
(c) Underground stem
(d) Corm.
Answer:
(b) Bulb

Question 6.
In aerial stems food is stored in :
(a) Large roots
(b) Inflorescence
(c) Swollen stems
(d) Swollen leaf bases.
Answer:
(c) Swollen stems

Question 7.
Stem is enlarged in :
(a) Stem tuber
(b) Corm
(c) Bulbs
(d) Rhizome.
Answer:
(b) Corm

Question 8.
The leafy stem of onion which is growing to bear flower is called:
(a) Peduncle
(b) Rachis
(c) Scape
(d) Floral axis.
Answer:
(b) Rachis

Question 9.
The leaves of Bryophyllum produce small plantlets on its margin. After
separation these plantlets can grow into a new plant. It is a form of:
(a) Sexuaireproduction
(b) Fission
(c) Vegetative propagation
(d) Bisexuality.
Answer:
(c) Vegetative propagation

Question 10.
Adnate stipules are found in:
(a) China rose
(b) Cotton
(c) Rose
(d) Polygonum.
Answer:
(c) Rose

Question 11.
Tendrillar stipules are found in:
(a) Smilax
(b) Bean
(c) Acacia
(d) Magnolia.
Answer:
(a) Smilax

Question 12.
When petiole is modified into a leafy structure, ¡t is caLled: (CPMT 1987)
(a) Phylloclade
(b) Leafy petiole
(c) Cladode
(d) True leaf.
Answer:
(b) Leafy petiole

Question 13.
Accessory buds are developed on:
(a) Apex of branch
(b) Stem apex
(c) The margins of axillary buds
(d) The axis of leaves.
Answer:
(c) The margins of axillary buds

Question 14.
Leaves are modified into hooks in:
(a) Bignonia
(b) Duranta
(c) Antigonon
(d) Ruscus.
Answer:
(a) Bignonia

Question 15.
Leaf without petiole is called:
(a) Petiolate
(b) Sessile
(c) Pedicellate
(d) Asparagus.
Answer:
(c) Pedicellate

Question 16.
Stipules get modified into tendrils in:
(a) Smilax
(b) Ascodilus
(c) Gloriosa
(d) Asparagus.
Answer:
(a) Smilax

Question 17.
Phyllode represents:
(a) Green stem
(b) Leaf base
(c) Green root
(d) Petiole.
Answer:
(d) Petiole.

Question 18.
Leaflet tendrils are found in :
(a) Clematis
(b) Pea
(c) Gloriosa
(d) Smilax.
Answer:
(b) Pea

Question 19.
Leaf bladders are found in :
(a) Nepenthes
(b) Utricularia
(c) Drosera
(d) Casuarina.
Answer:
(b) Utricularia

Question 20.
Parallel venation is found in the leaves of:
(a) Castor
(b) Grasses
(c) Moss
(d) Ferns.
Answer:
(b) Grasses

Question 21.
Leaf apex is modified in tendril in :
(a) Pisum
(b) Smilax
(c) Bryophyllum
(d) Rose
Answer:
(d) Rose

Question 22.
The pinnate leaf of Tamarind is :
(a) Tripinnate
(b) Bipinnate
(c) Imparipinnate
(d) Paripinnate.
Answer:
(d) Paripinnate.

Question 23.
In Utricularia, leaf segments are modified into :
(a) Spines
(b) Tendrils
(c) Thorns
(d) bladders.
Answer:
(d) bladders.

Question 24.
Petioles are modified into tendrils in :
(a) Antigonon
(b) Passiflora
(c) Gloriosa
(d) Clematis.
Answer:
(d) Clematis.

Question 25.
A leaf is modified into phyllode in :
(a) Potato
(b) Australian acacia
(c) Citrus
(d) Dalbergia.
Answer:
(b) Australian acacia

(C)

Question 1.
Tissue which connects ovules with ovary is called :
(a) Placenta
(b) Chalaza
(c) Funicle
(c) Hilum.
Answer:
(a) Placenta

Question 2.
Inflorescence which bears sessile and unisexual flower is called :
(a) Spadix
(b) Catkin
(c) Achene
(d) Panicle.
Answer:
(b) Catkin

Question 3.
Inflorescence of Onion is :
(a) Corymb
(b) Cymose umbel
(c) Racemose umbel
(d) Catkin.
Answer:
(b) Cymose umbel

Question 4.
The function of tapetum, innermost layer of the anther is :
(a) Dehiscence
(b) Mechanical
(c) Protection
(d) Nutritional.
Answer:
(c) Protection

Question 5.
Epicalyx is :
(a) Bract whorl
(b) An additional cycle of corolla
(c) A whorl of bracteoles
(d) A whorl of bracts.
Answer:
(c) A whorl of bracteoles

Question 6.
Tetradynamous stamens having :
(a) 6 stamens-2 short and 4 long
(b) 6 stamens-2 small stamens on outer whorl and 4 larger stamens on inner whorl
(c) 6 stamens-2 long and 4 short
(d) 4 stamens-2 long and 2 short.
Answer:
(b) 6 stamens-2 small stamens on outer whorl and 4 larger stamens on inner whorl

Question 7.
Unilocular ovary, in which placenta develops directly on thalamus and bears
a single ovule at the base of ovary is called:
(a) Marginal
(b) Free central
(c) Basal
(d) Parietal.
Answer:
(c) Basal

Question 8.
Tissue which connects ovules with ovary is called:
(a) Placenta
(b) Chalaza
(e) Funicle
(d) Hilum.
Answer:
(a) Placenta

Question 9.
Ovule is jointed with placenta by means of:
(a) Hilum
(b) Funiculus
(c) Petiole
(d) Pedreel
Answer:
(b) Funiculus

Question 10.
In angiosperms an ovule shows:
(a) Megasporophyll
(b) Megasporangium
(c) Megaspore
(d) Megaspore mother cell.
Answer:
(b) Megasporangium

Question 11.
In Monoadeiphous stamen:
(a) Anther and filament are fused
(b) Anthers are fused and filaments are free
(c) Filaments of all stamens are fused and anthers are free
(d) None of these.
(c) Filaments of all stamens are fused and anthers are free

Question 12.
A floral formula will show:
(a) The functions of flower
(b) Symmetry of flower
(c) Graphical representation of the characteristics of flowers
(d) Position of flower.
Answer:
(c) Graphical representation of the characteristics of flowers

Question 13.
Largest flower of the world is:
(a) Cuscuta
(b) Rafflesia
(c) Lora nihus
(d) Drosera.
Answer:
(b) Rafflesia

Question 14.
The national flower of India is:
(a)Rafflesia
(b) Lotus
(c) Rose
(d) Wolfia.
Answer:
(b) Lotus

Question 15.
Presence of unilocutar, monocarpellary, syncarpous ovary having marginal
placentation is the characteristic feature of:
(a) Solanaceae
(b) Cruciferae
(c) Leguminosae
(d) Compositae.
Answer:
(c) Leguminosae

Question 16.
Morphologically the edible part of the apple is :
(a) Cotyledon
(b) Thalamus
(c) Endocarp
(d) Mesocarp.
Answer:
(b) Thalamus

Question 17.
Pollen receiving part of the gynoecium is :
(a) Style
(b) Stigma
(c) Ovule
(d) Ovary.
Answer:
(b) Stigma

Question 18.
The aestivation of corolla of pea is called :
(a) Contorted
(b) Valvate
(c) Vexillary
(d) imbricate.
Answer:
(c) Vexillary

Question 19.
Basal placentation is present in :
(a) Compositae
(b) Solanaceae
(c) Malvaceae
(d) Mimosidae.
Answer:
(a) Compositae

Question 20.
The edible part of cauliflower is :
(a) Fruit
(b) Bud
(c) Inflorescence
(d) Flower
Answer:
(c) Inflorescence

Question 21.
The condition in which stamens are attached with corolla is called :
(a) Episepalous
(b) Gynandrous
(c) Epipetalous
(d) All of these.
Answer:
(c) Epipetalous

Question 22.
Tetradynamous condition is associated with :
(a) Androecium
(b) Gynoecium
(c) Inflorescence
(d) Epicalyx.
Answer:
(a) Androecium

Question 23.
Flower is a shoot with :
(a) Elongated nodes
(b) Condensed shoot
(c) Elongated intemodes
(d) Condensed intemodes.
Answer:
(d) Condensed intemodes.

Question 24.
Smallest flower is found in :
(a) Colocasia antiquorum
(b) Wolffia microscopia
(c) Rosa indica
(d) Ranunculus scleratus.
Answer:
(b) Wolffia microscopia

(D)

Question 1.

is the floral formula of:
(a) Brassica sp.
(b) Pisum sativum
(c) Candytuft
(d) Datura alba.
Answer:
(a) Brassica sp.

Question 2.
In Cruciferae family, inflorescence is generally of the following type :
(a) Solitary axillary
(b) Dichasial cyme
(c) Spike of spikelet
(d) Racemose.
Answer:
(d) Racemose.

Question 3.
Ray florets and disc florets are parts of this type of inflorescence :
(a) Spike of spikelet
(b) Capitulum inflorescence
(c) Corymb inflorescence
(d) Dichasial cyme.
Answer:
(b) Capitulum inflorescence

Question 4.
Oblique ovary is found in this family :
(a) Graminae
(b) Solanaceae
(c) Cruciferae
(d) Liliaceae.
Answer:
(b) Solanaceae

Question 5.
Type of fruits usually found in Cruciferae family :
(a) Siliqua and siliqula
(b) Legume or pod
(c) Caryopsis
(d) Achenal or cypsela.
Answer:
(a) Siliqua and siliqula

2. Fill in the blanks:
(I) Root, Stem And Leaves

1. The type of roots normally found in monocotyledonous plants are ………………….
Answer:
Fibrous

2. The roots in Banyan tree are named ………………..
Answer:
Prop roots

3. Tuber is a modified …………………. and the example is ………………
Answer:
Stem, Potato

4. ………………… in the arrangement of leaves on the nodes of a stem and its branches.
Answer:
Phyllotaxy

5. The plant is which haustoria are found is ………………..
Answer:
Cuscuta

6. The plants which complete their life cycle in many seasons are called ……………………
Answer:
Perennials

7. …………………… are the leafy outgrowths on either side of the leaf base.
Answer:
Stipules

8. The structure that develops from radicle of germinating seed is ……………….
Answer:
Root

9. ……………….. are small pore present in the bark of stem, where exchange of air and water takes place.
Answer:
Lenticel

10. The tip of root usually remains covered by a small conical cap like structure, called ………………..
Answer:
Root cap

(II) FLOWER

1. If a bisexual flower does not open in its life cycle is called ………………………..
Answer:
Cleistogamous

2. Tissue which connects ovules with ovary is ………………..
Answer:
Placenta

3. Ovule is jointed with placenta by means of …………………
Answer:
Funiculus

4. Catkin or verticillaster is a type of …………………….
Answer:
Racemose

5. The inflorescence of Onion is ………………….
Answer:
Cymose umbel

6. Cyathium inflorence is found in ……………………..
Answer:
Euphorbia

7. Embryo is developed from …………………
Answer:
Ovule

8. The central stalk present in each spikelet is called ………………..
Answer:
Rachis

9. Tetradynamous condition is associated with ………………….
Answer:
Androecium

10. The placentation of Cruciferae is ……………………
Answer:
Parietal

11. ………………… pollination occurs in Pinus.
Answer:
Anemophilous type.

(III) PLANT FAMILIES

1. In Solanaceae family, ovary is …………………
Answer:
Oblique

2. Example of Cruciferae family is ………………….
Answer:
Mustard

3. Rice, Wheat, Maize etc. belong to family …………………
Answer:
Poaceae

4. Ray florets and disc florets are parts of ……………….. type of inflorescence.
Answer:
Capitulum

5. Androecium are of ………………… type in family Compositae.
Answer:
Syngenesious

3. Match the following:
(I) Root, Stem And Leaves

(A)

Column ‘A’	Column ‘B’
1. Ginger	(a) Phloem
2. Phylloclade	(b) Root
3. Companion cell	(c) Conical
4. Exarch	(d) Rhizome
5. Carrot	(e) Opuntia.

Answer:
1. (d) Rhizome
2. (e) Opuntia.
3. (a) Phloem
4. (b) Root
5. (c) Conical

(B)

Column ‘A’	Column ‘B’
1. Hydathodes	(a) Lenticel
2. Mint	(b) Stilt root
3. Dahlia	(c) Guttation
4. Bark	(d) Sucker
5. Maize, Sugarcane	(e) Fasiculated

Answer:
1. (c) Guttation
2. (d) Sucker
3. (e) Fasiculated
4. (a) Lenticel
5. (b) Stilt root

(II) FLOWER
(A)

Column ‘A’	Column ‘B’
1. Solanaceae	(a) Oily endosperm
2. Hypanthodium	(b) Cymose umbel
3. Coconut	(c) Mesocarp
4. Inflorescence of Onion	(d) Swollen placenta
5. Mango	(e) Inflorescence.

Answer:
1. (d) Swollen placenta
2. (e) Inflorescence.
3. (a) Oily endosperm
4. (b) Cymose umbel
5. (c) Mesocarp
(B)

Column ‘A’	Column ‘B’
1. Perianth	(a) Thalamus
2. Family	(b) Superior
3. Apple	(c) Cruciform
4. Ovary	(d) Lodicules
5. Corolla	(e) Capitulum.

Answer:
1. (d) Lodicules
2. (e) Capitulum.
3. (a) Thalamus
4. (b) Superior
5. (c) Cruciform

(III) PLANT FAMILIES

Answer:
1. (d) Onion
2. (e) Mustard
3. (a) Datura
4. (b) Pea
5. (c) Composite

4. Write true or false:
(I) Root, Stem And Leaves

1. Radish is an example of phylloclade.
Answer:
False

2. Nepenthes (Pitcher plant) is an example of carnivorous plant.
Answer:
True

3. Casparian strips are found in epidermis.
Answer:
False

4. Secondary growth results in the growth of the girth of the plant and not in the length only.
Answer:
True

5. Scleroids and fibres are the components ofXylem tissue.
Answer:
False

(II) FLOWER

1. An inflorescence which contains gall flower is hypanthodium.
Answer:
True.

2. Water is not necessary for fertilization in Vallisnaria plant.
Answer:
False

3. The protein storing layer of endosperm is called aril.
Answer:
True.

4. The inflorescence of compositae is spike.
Answer:
False

5. Caryopsis fruits are found in family Graminae.
Answer:
True.

(III) PLANT FAMILIES

1. Perianth means calyx and corolla are present.
Answer:
False

2. Botanical name of Cabbage is Brassica rapa.
Answer:
False

3. Replum is characteristic feature of family Cruciferae.
Answer:
True

4. Family Graminae and Cruciferae belong to monocotyledons.
Answer:
False

5. Caryopsis fruits are found in family Cruciferae.
Answer:
False

5. Answer in one word:
(I) Root, Stem And Leaves

1. Tuber of potato is modified.
Answer:
Stem

2. Stem which is green and leaf like is called.
Answer:
Phylloclade

3. In Opuntia, spines are modified.
Answer:
Leaves

4. The function of prop roots of Banyan is.
Answer:
To provide support

5. In aerial stems food is stored in.
Answer:
Swollen roots

6. Name one negatively phototropic organ of the plant.
Answer:
Root

7. The structure that develops from plumule of the germinating seed.
Answer:
Stem

8. Phylloclade of single intemode is.
Answer:
Cladode

9. Name of one nitrogen fixing bacteria.
Answer:
Rhizobium

10. Give the example of assimilatory root.
Answer:
Trapa

11. Write the example of respiratory root.
Answer:
Jussiaea

12. Which plants have reticulate venation ?
Answer:
Dicot leaves.

(II) FLOWER

1. Largest flower of the world is.
Answer:
Rafflesia

2. Spadix inflorescence is found in.
Answer:
Banana

3. The endosperm in angiospermic plant is.
Answer:
Triploid

4. Monosporic and nucleate embryo sac is referred as.
Answer:
Polygonum type

5. The edible part of cauliflower is.
Answer:
Inflorescence

6. Which type of inflorescence found in Tulsi?
Answer:
Verticillaster inflorescence

7. Which type of corolla is found in Mustard?
Answer:
Crossed or Cruciform

(III) PLANT FAMILIES

1. Corolla arranged in the form of a cross in family.
Answer:
Cruciferae

2. Rice, Wheat, Maize etc. belong to family.
Answer:
Poaceae

3. Oblique ovary is found in family.
Answer:
Solanaceae

4. Spike of spikelet type of inflorescence is found in family.
Answer:
Poaceae

5. Type of fruits usually found in Cruciferae family.
Answer:
Siliqua

6. Write the botanical name of Tomato.
Answer:
Lycopersicum esculentum