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Students get through the MP Board Class 11th Biology Important Questions Chapter 12 Mineral Nutrition which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Chapter 12 Mineral Nutrition

Mineral Nutrition Class 11 Important Questions Very Short Answer Type

Question 1.
Write the name of symbiotic bacteria present in the roots of leguminous plant.
Answer:
Rhizobium bacteria are present in the roots of leguminous plant which absorb atmospheric nitrogen.

Question 2.
Write the name of two micronutrients.
Answer:
The two micronutrients present in plants are Manganese (Mn) and Molybdenum (Mo).

Question 3.
Write the deficiency symptoms of potassium.
Answer:
Deficiency of potassium inhibits protein synthesis. Scorching of older leaves is the most important symptoms of trees. The leaf tips and margins show general chlorosis. Shortening of internodes also occurs.

Question 4.
Why nitrogen is essential for growth of plant?
Answer:
Protein is essential for growth of plant and for development of protein nitrogen is essential, so for the proper development of plant nitrogen is very much essential.

Question 5.
What are the drawbacks of soilless culture?
Answer:
Drawbacks :

• It will be very expensive.
• Limited production will be obtained.
• Crops will never be grown by this method.

Question 6.
What are micronutrients? Give two examples.
Answer:
The mineral elements which are required in very low concentration for the plant growth are known as micronutrient or microelements.
e.g., Zn, Mo, Cl, Mn, B, Cu, etc.

Question 7.
Describe the deficiency symptoms of nitrogen in plants.
Answer:
Deficiency symptoms of nitrogen:

• Yellowing i.e., Chlorosis of leaves, older leaves become yellow first.
• Acceleration in the rate of formation of purple pigment anthocyanin in the leaves which become purple in colour.
• Flowering is delayed or is completely suppressed.

Question 8.
Which pigment is present in the roots of leguminous plants?
Answer:
Leghaemoglobin.

Question 9.
Write the importance of molybdenum in plants.
Answer:
Molybdenum act as an activator for the enzyme nitrate reductase involved in nitrogen metabolism.

Question 10.
In plants, the organic solutes are transported through which tissue?
Answer:
In plants, the organic solutes are transported through phloem tissue.

Question 11.
What is Mineral nutrition?
Answer:
The utilization of various kinds of absorbed ions by a plant for its growth and development is called as Mineral nutrition.

Question 12.
What are nutrient elements?
Answer:
Elements which are essential for plant growth are called as nutrient elements.

Question 13.
In study of water culture purity of water and mineral nutrient is essential. Why? (NCERT)
Answer:
In study of water culture purity of water and pure mineral nutrients are essential because plants can absorb only pure water and pure minerals and their essentiality for the plants can be determined correctly. Impure mineral nutrient cannot help to determine essentiality of the minerals for plants.

Question 14.
Explain following With examples: Macronutrients, micronutrients, useful elements, essential elements, toxic elements.
Answer:
1. Macronutrients: Essential elements which are required in large quantity for normal growth and development of plants are called as macronutrients.
e.g. C, H, O, N, P, K, Ca, S, Mg, Fe, etc.

2. Micronutrients: The mineral elements which are required in very low concentration for the plant growth are known as micronutrient or microelements.
e.g., Zn, Mo, Cl, Mn, B, Cu, etc.

3. Useful elements: Some essential elements which are essential for higher plants called as useful elements.
e.g., Na, Si, Ca and Se.

4. Essential elements: Out of approximately 30 elements universally present in the plants, 16 elements are essential for plants, i.e., they require them.
e.g. C, H, O, N, P, K, Ca, S, Mg, Fe, Mn, Zn, B, Cu, Cl, Al.

5. Toxic element: Concentration of any mineral ion which decreases dry weight of plant by 10% is called as Toxic element. It is different for different plants.
e.g. As, Cu, Hg etc.

Question 15.
Give four general importance of mineral salt.
Answer:
Importance of Mineral salt:

1. Nutrition of plants: Some minerals like N, P, S, etc. help to produce protein.
2. As catalyst: Some minerals act as catalyst in biochemical reactions. e.g. Mn.
3. Bodybuilding: Some elements help to produce structural part, such as cell wall of the cell.
4. Balancing work: Some mineral elements stop toxic effect of some other elements. e.g., Calcium, Magnesium.

Question 16.
Write difference between Active and Passive absorption.
Answer:
Differences between Active and Passive absorptions

Mineral Nutrition Class 11 Important Questions Short Answer Type 

Question 1.
Mention the name of any five macroelements and their deficiency symptoms. (NCERT)
Answer:
Macronutrients :

1. Nitrogen
2. Phosphorus,
3. Sulphur,
4. Magnesium and
5. Calcium.

1.Deficiency symptoms of nitrogen:

• Stunted growth with yellowish-green leaves in young plants.
• Yellowing and drying of older leaves.
• Flowering reduced.

2. Deficiency symptoms of phosphorus :

• Stunted growth.
• The leaves become purple coloured due to excessive anthocyanin.
• Premature leaf falls and necrotic spots on the fruits.
• Vascular tissues less developed.
• Low rate of protein synthesis.

3. Deficiency symptoms of sulphur :

• Plants stunted in growth and flowering delayed.
• Chloroplast becomes pale green in colour.
• Chlorosis of leaves.
• Hard woody stem.

4. Deficiency symptoms of magnesium :

• Interveinal chlorosis or molted chlorosis with green veins.
• Brittleness of the leaves and necrosis.
• Wilting, drying off and dropping of the tips of plant.
• Necrotic patches on the leaves.

5. Deficiency symptoms of calcium :

• Calcium deficiency causes malformation of the younger leaves.
• Chlorosis also occurs along the margins of younger leaves.

Question 2.
Describe essentiality of potassium in plants and give its deficiency symptoms.
Answer:
Potassium: It is essential for activating enzymes concerned in the synthesis of polypeptides from amino acids and is also essential for the process of photosynthesis, and respiration.

Deficiency symptoms: Deficiency of potassium inhibits protein synthesis. Scorch-ing of older leaves is the most important symptom on trees. The leaf tips and margins show general chlorosis. Shortening of internodes also occurs.

Question 3.
All elements are not essential for plants. Explain it. (NCERT)
Answer:
All elements absorbed by the plants are not essential.
Criteria for Essentiality: According to Arnon and Stout, 1939, the criteria for essentiality of mineral elements are as follows :

• The plant is unable to grow normally and complete its life cycle, in the absence of the elements.
• The element is specific and cannot be replaced by another element, however close it may be in the periodic table.
• The element plays a direct role in the metabolism of the plant.

Question 4.
Give atleast essential elements. Describe them and give their deficiency symptoms. (NCERT)
Answer:
1. Sulphur: It is absorbed in the form of sulphate ions from the soil. It is important constituent of some amino acids. It is also essential for the synthesis of sulphur con¬taining vitamins like biotin, thiamine and coenzyme A. Disulphide bridge (S-S) plays an important role in determining protein structure and sulphydryl groups (SH) are necessary for the activity of many enzymes.

Deficiency symptoms :

• Reduced leaf, stunted growth, a general chlorosis, followed by the production of anthocyanin pigments in some species. The younger leaves are affected first.
• Inward rolling of leaf margins and tips.
• Due to development of sclerenchymatous tissue, stem becomes hard.

2. Phosphorus: It is an important constituent of phospholipids, nucleic acids and various coenzymes like NAD, NADP and ATP. It is absorbed from soil in the form of phosphate. Phospholipids are the constituents of membranes. NAD, NADP and ATP are required for oxidation-reduction reactions of photosynthesis, respiration and fat metabo-lism. ATP is the chief source of energy.

Deficiency symptoms :

• Premature leaf fall.
• Formation of necrotic areas on leaves and fruits.
• Leaves become dark blue in colour.
• The growth of roots and shoots is completely restricted.

3. Calcium: It is a constituent of the middle lamella, in which it is present in the form of calcium pectate. It helps to stabilize the structure of the chromosomes.
Deficiency symptoms: Calcium deficiency causes malformation of the younger leaves. Chlorosis also occurs along the margins of the younger leaves.

4. Magnesium: It is one and only constituent of the chlorophyll molecule. It acts as activator for many enzymes in phosphate transfer reactions particularly in carbohydrate metabolism and nucleic acid synthesis. It is also believed to be an important binding agent in ribosomes, where protein synthesis takes place.
Deficiency symptoms: Magnesium deficiency results in extensive interveinal chlorosis of leaves. The older leaves are affected first. Ultimately, leaves develop necrotic spots.

5. Iron: It is essential for chlorophyll synthesis. It is important constituent of ferredoxin, flavoprotein, iron-porphyrin, catalase, peroxidase and cytochromes. Iron is primarily concerned in the formation of chloroplast protein in the leaves.
Deficiency symptoms: Pronounced interveinal chlorosis occurs due to deficiency of iron.

Question 5.
Which statement is right out of following if it is wrong, give reason : (NCERT)
(a) Deficiency of boron causes formation of sclerenchymatous axis.
(b) AH elements present in the cell are essential.
(c) Nitrogen is not an essential element for plants.
(d) It is very easy to determine essentiality of micronutrients in plants because it is taken in very less quantity.
Answer:
(a) Wrong, because deficiency of Boron effect on activity of membrane, pollen germination, cell elongation, cell differentiation and transport of carbohydrates, (b), (c) and (d) are right statements.

Question 6.
If plant shows deficiency symptoms of two elements how will you determine deficient element practically? (NCERT)
Answer:
Deficiency symptoms of different elements are different. By providing particu¬lar deficient element if deficiency symptoms eradicates then deficient element can be confirmed.

Question 7.
Why deficiency symptoms are observed in newly formed parts of some plants whereas in some other plants it is observed in mature part? (NCERT)
Answer:
Deficiency symptoms in plant parts depends on mobility of elements. Deficiency symptoms are observed in old tissue first as compared to new tissues of young part. For example, symptoms of deficiency of nitrogen, potassium and magnesium are observed in old leaves first of all.

Biomolecules of old leaves decomposes and element move towards the new leaves.
When elements are non-mobile then they cannot reach to young parts thus deficiency symptoms are observed in new leaves.

Question 8.
Describe passive absorption of minerals in plants. (NCERT)
Answer:
Passive Absorption: In this method, the movement of minerals into the cells or tissue takes place without the expenditure of energy on the part of absorbing cells. This is a purely physical phenomenon and takes place in many ways.

1. Diffusion: It can be observed by first placing a plant cell or tissue in a medium of high concentration of salt and then transferring it into lower salt concentration. During this transfer, some of the ions taken up will diffuse into the lower salt concentration from the higher salt concentration, until the concentration of both the solutions get balanced. The whole process is unaffected by temperature or metabolic inhibitors.

2. Mass flow: Plants which have a higher rate of transpiration contain comparatively high concentrations of minerals. It is well-known that increase in transpiration is accompanied by increased absorption of ions from the soil. In this way the ions are absorbed by the plants in a mass flow due to the higher rates of transpiration.

3. Donnan equilibrium: According to Donnan, cells contain some fixed ions which do not exert any diffusion pressure. Therefore, ions similar in charge to that of fixed ions, move inwardly through diffusion. Oppositely charged ions also pass inwardly to balance them. Such a diffusion continues till the multiple of cations and anions inside cells becomes equal to that of ions present in outside medium.

Question 9.
Describe process of formation of Root nodules. (NCERT)
Answer:
Formation of Root nodule: It is a combined activity of roots of nutrient plants (Leguminous plants) and Rhizobium bacterias.

Steps of formation of Root nodules are as follows :
First of all Rhizobium bacteria multiply and are collected surrounding the roots of leguminous plants and get connected to root hair cells.

Root hair turns and bacteria attack them. An infected thread appears which carry Rhizobium bacteria to the cortex, where they begin to produce nodule. Now bacterias are released from the thread and enters into the cells which help for differentiation of nitrogen fixation cells. In this way, root nodules are formed. Leguminous plants provide nutrients to bacteria and Rhizobium bacteria provides nitrogenous compounds to plant by nitrogen fixation.

Mineral Nutrition Class 11 Important Questions Long Answer Type 

Question 1.
What are the materials required for atmospheric nitrogen fixation? Give their role in Nitrogen fixation. (NCERT)
Answer:
Rhizobium bacteria shows symbiotic relationship with the roots of leguminous plants like pea, moong, gram, beans etc. Symbiosis occurs in the root nodules. Rhizobium bacteria obtains food from the leguminous plants and Rhizobium bacteria help for nitrogen fixation and provides nitrogenous compounds to the plants.
Symbiotic Nitrogen fixation: The process of nitrogen fixation is done by nitrogen-fixing bacterias like Rhizobium, Cyanobacteria, nitrogen-fixing fungi etc.
Mechanism of Nitrogen fixation: Following enzymes and materials are required for nitrogen fixation:

• Nitrogenase and Hydrogenase enzymes.
• Leghaemoglobin: It protects the enzyme from oxygen.
• One non-haem iron protein: e.g. Ferredoxin: It is an electron carrier.
• Hydrogen donor substances like: NADPH₂, FMNH₂, Pyruvate, Sucrose, Glucose etc.
• Energy compound ATP.
• Cofactor like TPP (Thiamine pyrophosphate), CoA, inorganic phosphate, Mg++ etc.
• Co and Mo.
• A compound to fix ammonia produced by reduction of nitrogen.
  Following are the steps of natural nitrogen fixation:
  1. Reduction of molecular nitrogen:
  
  

Question 2.
What is Hydroponics? Give its uses.
Answer:
Hydroponics (Soilless culture): Large- scale cultivation of plants in water culture or soilless culture is known as hydroponics. It was discovered by Prof. W. F. Gariak in 1929. The plants are grown in large, shallow pots which are full of nutrient solutions. The tubs or tanks are covered with wire netting to provide support to the seedlings. The solution is aerated at regular intervals by means of an inlet tube.
There are certain advantages and disadvantages of this technique.

Uses:

• Possibility to provide desirable nutrient environment.
• Regulation of pH of the nutrient solution conductive to specific crop.
• It is possible to replace the nutrient, so more healthy plant as compared to soil culture can be obtained.
• On the roof of multistoried buildings of metropolitan cities, vegetables and fruits may be produced in large scale.
• Ripening time of fruit is less in soilless culture.

Question 3.
Give an experiment to demonstrate the essentiality of minerals required by plants.
Answer:
Essentiality of minerals required by plants can be demonstrated by using a normal culture medium such as Sach’s culture solution. Components of this solution are as follows:
Sach’s culture solution :

This experiment is performed to ascertain the role of individual elements in the normal growth and development of plants and the effect produced by their deficiencies. In water culture (solution culture) experimental plants are grown in a medium, the chemical composition of which is known and kept under control.

Method: Take seven wide-mouthed bottles of neutral glass of the same size. Clean and thoroughly sterilize them with nitric acid and distilled water. Fit each bottle with a cork containing two holes; one in the centre for introducing a seedling and the other at the side for a bent glass tube for aerating the culture solution. Make a slit running from the central hole to the rim of the cork for readily taking out the seedling.

Now, prepare a normal culture solution and pour it into the first bottle. Then, prepare six other solutions from which omit by turn Mg, Ca, Fe, K, P and N, and designate them as Mg, Ca, Fe, K, P and N, and fill the remaining seven jars with these solutions. Take seed¬lings of same kind and more or less same size. Introduce a seedling in each bottle through split cork. Wrap the bottles with black paper and expose them to light. Make arrangement

for proper aeration of roots. Renew the culture solution fortnightly. Observe carefully the symptoms of plants appeared due to the deficiency of various elements.

Question 4.
Name any five micronutrients and give their specific role and deficiency symptoms.
Answer:
Nutrient elements which are required in traces for normal growth and development of the plants are called as Micronutrients.
e.g. Manganese, Boron, Copper, Zinc, Molybdenum and Chlorine. Their specific role and deficiency symptoms are as follows :

Mineral Nutrition Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
In Nitrogen cycle nitrifying bacteria convert:
(a) NH₃ into N₂
(b) Nitrogen fixation
(c) NH₃ into NO₂
(d) Amino acid into NHr
Answer:
(c) NH₃ into NO₂

Question 2.
Function of Zinc is :
(a) Production of Chlorophyll
(b) Production of IAA
(c) Closing of stomata
(d) Oxidation of Carbohydrates.
Answer:
(b) Production of IAA

Question 3.
Symptoms of destruction of Chlorophyll is called as :
(a) Necrosis
(b) Chlorosis
(c) Stunted shoot
(d) Rotting.
Answer:
(c) Stunted shoot

Question 4.
Plants which can prepare their own food are called as :
(a) Heterotrophs
(b) Autotrophs
(c) Mixotrophs
(d) None of these.
Answer:
(b) Autotrophs

Question 5.
Falling of premature leaves, flowers and fruits is called as :
(a) Abscission
(b) Dieback
(c) Stunted
(d) Hypertrophy.
Answer:
(a) Abscission

Question 6.
In which form ‘C’ is available to plants :
(a) ‘C’ element
(b) CO₂
(c) CO₃
(d) Amino acid.
Answer:
(b) CO₂

Question 7.
Which element is required by the plants in large quantity :
(a) N
(b) P
(c) Ca
(d) S.
Answer:
(a) N

Question 8.
Which element out of following is not macroelement:
(a) P
(b) K
(c) Mg
(d) Mo.
Answer:
(d) Mo.

Question 9.
Hydroponics name was given by :
(a) Knop and Sach
(b) Hoagland
(c) Gariak
(d) Sach.
Answer:
(c) Gariak

Question 10.
Soil element available to plant is called as :
(a) Mineral salt
(b) Micro salt
(c) Nutrient substance
(d) None of these.
Answer:
(a) Mineral salt

Question 11.
How much percentage of dry weight of plants is formed by Carbon, Hydrogen and Oxygen
(a) 10-15%
(b) 15-25%
(c) 25-35%
(d) 85-95%.
Answer:
(d) 85-95%.

Question 12.
Element required for synthesis of IAA :
(a) B
(b) Cu
(c) Zn
(d) Mo.
Answer:
(c) Zn

Question 13.
Active absorption of minerals in plants depends on :
(a) Availability of O₂
(b) Light
(c) Temperature
(d) Availability of CO₂.
Answer:
(a) Availability of O₂

Question 14.
Element which is not essential for the plants out of the following is :
(a) Iron
(b) Magnesium
(c) Lead
(d) Phosphorus.
Answer:
(c) Lead

Question 15.
Trace element out of following is :
(a) Ca
(b) Mg
(c) C
(d) Cu.
Answer:
(d) Cu.

2. Fill in the blanks:

1. Monotropa is an example of ………………………………… plant.
Answer:
Saprophyte

2. Orobanche is an example of ………………………………… plant.
Answer:
Total root parasite

3. Zn, Cu, Mn, B, Mo and Cl are ………………………………… elements.
Answer:
Micro

4. Extensive interveinal Chlorosis is caused due to deficiency of ………………………………… .
Answer:
Mg

5. Element act as activator for the enzyme nitrate reductase is ………………………………… .
Answer:
Mo

6. Rhizobium is a ………………………………… nitrogen fixing bacteria.
Answer:
Symbiotic

7.
Answer:
2HNO₂

8. ……………………………….. element participates in various metabolic activities of plant.
Answer:
Essential

9. Tetra-pyrrol porphyrin ring of chlorophyll has ………………………………… element at the centre.
Answer:
Mg

10. Lack of ………………………………… causes the absence of glands in the roots of leguminous plants.
Answer:
Boron

11. Rhizobium is found in the glands of the leguminous plants situated at ………………………………… .
Answer:
Roots

12. The element which are required for the growth of plants are called ………………………………… element.
Answer:
Essential

13.‘Whiptail disease’ in cabbage is due to deficiency of ………………………………… .
Answer:
Molybdenum.

3. Match the following:

(A)

Answer:
1. (e) Insectivorous.
2. (d) Plant ash
3. (a) Minor trace element
4. (b) Hydroponics
5. (c) Critical element

(B)

Answer:
1. (d) Stomatal movement
2.  (e) Chlorophyll.
3.  (a) Nitrogen metabolism
4. (b) Photolysis of water
5.  (c) Amino acid.

(C)

Answer:
1.  (d) Total stem parasite
2. (e) Partial stem parasite.
3. (b) Total root parasite
4. (c) Insectivorous
5. (a) Symbiotic

4. Answer in one word:

1. Write two mineral element essential for production of chloroplast.
Answer:
Mg and Fe

2. Name the pigment found in the root nodule of leguminous plant.
Answer:
Leghaemoglobin

3. Full form of NPK is.
Answer:
Nitrogen, Phosphorus, Potassium

4. Name the enzyme induce Nitrogen fixation.
Answer:
Nitrogenase

5. Name one free-floating aquatic insectivorous plant.
Answer:
Utricularia

6. Name the process in which roots of the plants are kept in nutrient solution for plant growth.
Answer:
Hydroponics

7. Deficiency of which element causes abscission of premature leaves.
Answer:
Phosphorus

8. Name the root of the Cuscuta plant which penetrate into the body of host plant.
Answer:
Haustorial root

9. Name the macroelement which forms structural component of plant body.
Answer:
Carbon

10. From where plants obtain hydrogen.
Answer:
Water

Students get through the MP Board Class 11th Biology Important Questions Chapter 10 Cell Cycle and Cell Division which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Questions Chapter 10 Cell Cycle and Cell Division

Cell Cycle and Cell Division Class 11 Important Questions Very Short Answer Type

Question 1.
What is the average cell cycle duration of cells of mammalia?
Answer:
Average cell cycle duration of cells of mammals is 24-25 hours.

Question 2.
What is the difference between cytokinesis and karyokinesis?
Answer:
Division of the cytoplasm is called as cytokinesis, whereas division of the nucleus is called as karyokinesis.

Question 3.
What is G₀ phase of cell cycle?
Answer:
Cells which do not divide further and reaches to inactive phase, is called as G₀ phase of cell cycle. At this stage cell shows high rate of (active) metabolism but does not divide. They divide depending upon requirement of the organism.

Question 4.
Why Mitosis cell division is called as similar division?
Answer:
At the end of this cell division daughter cells have same number of chromosomes like parent cell thus Mitosis division is called as similar cell division.

Question 5.
Where daughter cells formed by meiosis cell division are same and different?
Answer:
Four daughter cells formed by meiosis division in the testis are the spermatids which are same in size whereas in the ovary of female four daughter cells formed by meiosis are different in size, i.e. one ovum which is larger in size and three polar bodies, which are smaller in size.

Question 6.
Is Mitosis division possible without DNA replication in ‘S’ phase of cell cycle?
Answer:
No, Mitosis division is not possible without DNA replication in ‘S’ phase of cell cycle, because duplication of DNA is essential for Mitosis division.

Question 7.
Name the stage of cell division which involves crossing-over.
Answer:
Crossing-over occurs during pachytene stage of Prophase-I of meiosis cell division.

Question 8.
Name the stain used for staining chromosomes of the cells for study.
Answer:
Acetocarmine stain is used for staining chromosomes of the cell for study.

Question 9.
Name the cells exhibiting mitosis cell division.
Answer:
Mitosis is the characteristic feature of somatic cells.

Question 10.
Name the cells exhibiting meiosis cell division.
Answer:
Meiosis takes place in generative cells to produce gametes.

Question11.
What is chiasmata?
Answer:
The place of chromosomes where they come into contact with each other during crossing-over is called as chiasmata.

Question 12.
List the various phases of the first meiotic division.
Answer:
The following stages are found in the first meiotic division :

1. 1. Prophase-I: It has following substages :
   • Leptotene,
   • Zygotene,
   • Pachytene,
   • Diplotene and
   • Diakinesis.
2. Metaphase-I,
3. Anaphase-I,
4. Telophase-I.

Question 13.
What is synapsis?
Answer:
Pairing of homologous chromosomes in zygotene stage of Prophase-I of meiosis cell division is known as synapsis. The paired structure formed by synapsis is called as bivalent.

Question 14.
What do you mean by Mitotic poison?
Answer:
Chemical substances which inhibit the mitotic division are called as Mitotic poison.
e.g.,

1. Cholchicine : It stops formation of spindles during Metaphase.
2. Ribonuclease : It stops Prophase stage.
3. Mustard gas : It breaks chromosomes.

Question 15.
Name the parts of a flowering plant where meiosis cell division occurs.
Answer:
In the pollen sac of androecium and ovary of gynoecium meiosis division occurs to produce male and female gametes respectively.

Cell Cycle and Cell Division Class 11 Important Questions Short Answer Type

Question 1.
Describe the phenomenon occurs during interphase of cell cycle.
Answer:
Interphase is the period between two successive cell divisions. It has following three stages:

• G_1-phase : RNA and protein synthesis occurs during this stage.
• S-phase : DNA synthesis occurs during this stage.
• G₂-phase : Synthesis of protein and RNA continue in this stage.

Question 2.
What is the significance of Meiosis division?
Answer:
Significance of meiosis : The significance of meiosis are as follows :

• The number of chromosomes is stable generation to generation due to the meiotic division.
• The number of chromosomes is reduced to half in the gametes.
• The characters of parents are transmitted into offspring by this process.
• Crossing-over takes place during this stage which can produce new characters in the organisms.
• Variations are also occurred which help in the study of evolution.
• There are four haploid cells formed from one diploid cell.

Question 3.
Name the stage of cell cycle when following phenomenon occurs :
(i) Movement of chromosomes towards the equator of the cell.
(ii) Breaking of centromere and separation of chromosomal half.
(iii) Pairing of homologous chromosomes.
(iv) Exchange between homologous chromosomes.
Answer:
(i) Metaphase,
(ii) Anaphase,
(iii) Zygotene stage of Prophase-I of Meiosis-I,
(iv) Pachytene stage of Prophase-I of Meiosis-I.

Question 4.
Explain the following :
(i) Synapsis,
(ii) Bivalent,
(iii) Chiasmata.
Answer:
(i) Synapsis :
Pairing of homologous chromosomes in zygotene substage of prophase-I of meiosis cell division is known as synapsis. The paired structure formed as result of synapsis is called as bivalent.

(ii) Bivalent:
The pairs of homologous chromosome formed during zygotene stage of meiosis-I. The chromosomes contract and become short and thick.

(iii) Chiasmata :
In Pachytene stage of Prophase-I of Meiosis-I division, non-sister chromatids of two chromosomes, which come in contact of each other and fixes themselves tightly for crossing-over. These fixed parts are called as Chiasmata.

Question 5.
Write differences between cytokinesis in plant and animal cell.
Answer:
Cytokinesis : Division of cytoplasm after Karyokinesis to produce daughter cells is called as Cytokinesis.

(a) Cytokinesis in plant cell:
It immediately follows karyokinesis. In plant cells, a cell plate of dense cistemae is laid down at the equatorial plate. It grows centrifugally (external laterally) towards plasma membrane until it divides the cell into two parts. The space in between the two membranes of cistemae is filled by the deposition of pectates of calcium and magnesium to become middle lamella.

(b) Cytokinesis in animal cell:
In animal cell, cytokinesis occurs by an invagination of cell membrane almost in the middle of the cell. The furrow gradually depens and ultimately divides the cell into two parts.

Question 6.
Write differences between Anaphase of Mitosis and Meiosis cell division.
Answer:
Differences between Anaphase of Mitosis and Meiosis cell division

Question 7.
Analyse each phase of cell division and explain how does following changes occurs in the cell:
(i) Number of chromosomes (N) in each cell.
(ii) Quantity of DNA in each cell.
Answer:
(i) In all cells during prophase, metaphase and anaphase stage of cell division, number of chromosomes become double and in telophase stage number of chromosomes become half again, i.e., (N).
(ii) During cell division quantity of DNA changes in all stages. During prophase, metaphase and anaphase number of DNA increases, it becomes double, but in telophase stage quantity of DNA becomes half.

Question 8.
Discuss following with your teacher :
(i) When cell division occurs in haploid seed and lower category of plants ?
(ii) Where cell division does not occur in the haploid cells of higher category of plant?
Answer:
(i) Haploid spores of lower category of plant divides by mitosis division to form gametophyte, whereas zygote develops to form sporophyte and meiosis division occurs for formation of haploid spores, e.g., moss, fern etc.
(ii) Synergids and antipodal cells found in the ovules of angiosperms are haploid. They do not show cell division. After completion of life span they dies.

Question 9.
What is crossing-over? Throw light on its significance.
Answer:
Crossing-over : Crossing-over is the exchange of chromosomal segments or genes between homologous chromosomes during diplotene stage of meiosis-I.

Significance of crossing-over :

• New characters are originated by this process,
• Variations occur in organisms,
• It plays an important role in organic evolution,
• It results in the production of adaptation in organisms.

Question 10.
Write the significance of mitosis.
Answer:
Significance of mitosis:

• Growth and development takes place due to mitosis,
• Some micro-organisms use this method for asexual reproduction,
• Dead cells are substituted by new cells by mitotic division,
• General repairing process is result of mitotic division,
• Genetic information are transferred from parents to offspring by this process.

Question 11.
What is the difference between anaphase-1 of meiosis and anaphase of mitosis ? What is its effect on the whole process?
Answer:
In meiotic anaphase-I, the two chromosomes of each bivalent separate from each other and move towards opposite poles of the spindle. But the two chromatids of each chromosome still continue to remain joined with one another with centromere. Thus, only one chromosome of each homologous chromosome pair reaches each pole and the number of chromosomes are reduced to half in daughter nuclei.

In mitotic anaphase-I, centromere of each chromosomes divides and permits the separation of two sister chromatids, each chromatid after separation is called daughter chromosomes. The daughter chromosomes move apart and migrate towards opposite poles. The centromere is pulled towards poles of the spindle and the arms of chromosomes are dragged behind.

Effect of whole process:
Anaphase-I of meiosis resulting in the formation of daughter nuclei having half of the chromosomes to their parents. Thus, the number of chromosomes is constant from generation to generation.

Question 12.
Write a note on necessity of two types of cell divisions in multicellular organisms.
Answer:
Necessity of two types of cell divisions :
There is complexity in structure and physiology of multicellular organisms. Thus, twc types of cell divisions are required in them. The process of meiosis cell division help in the maintenance of chromosome number in the species. Diploid (2x) organism produce haploid (x) gametes, which on fusion form diploid zygote, whereas zygote divides by mitosis division to form diploid organism. Healing of wound in multicellular organisms occurs by mitosis cell division.

Question 13.
Describe the prophase and anaphase of mitosis along with diagram.
Answer:
1. Prophase : (Pro = before, phasis – stage).

• In this stage chromosomes are found in the form of long threads which are called as chromatin net.

• Chromosomes become short and thick and each of them splits lengthwise forming two chromatids.
• At the ending of prophase nuclear membrane and nucleolus start to disappear.

2. Anaphase :

• During this stage the centromere of each chromosome gets divided into two which make the chromatids free from each other.
• Every chromatid with its centromere moves towards its poles.
• Termination of anaphase movement occurs when the chromosomes form a densely packed group at the two poles.

Question 14.
Draw labelled diagram of different stages of Mitosis cell division.
Answer:

Question 15.
Describe the metaphase of mitosis along with diagram.
Answer:

Metaphase : The following events occur during mitotic metaphase: ‘

• Nuclear membrane and nucleolus disappear during this stage.
• Chromosomes move towards equatorial plane of the cell.
• Chromatids of chromosomes become dense and separated from each other except centromere.
• Spindle fibres are formed.
• The centromeres of the chromosomes are arranged on equator and chromatid towards the opposite poles. Fig Metaphase

Question 16.
Write differences between following :
(i) Mitosis and Meiosis cell division,
(ii) Chromatin and Chromatid,
(iii) Centromere and centriole,
(iv) Centromere and Chromomere,
(v) Metaphase-I and Metaphase-II,
(vi) Zygotene and Pachytene,
(vii) Cell furrow and Cell plate.
Answer:
(i) Differences between Mitosis and Meiosis cell division :

(ii) Differences between Chromatin and Chromatid

(iii) Differences between Centromere and Centriole

(iv) Differences between Centromere and Chromomere

(v) Differences between Metaphase-I and Metaphase-II

(vi) Differences between Zygotene and Pachytene

(vii) Differences between Cell furrow and Cell plate

Question 17.
Define the following :
1. Homologous chromosomes,
2. Synapsis,
3. G₂alent,
5. Tetrad.
Answer:
1. Homologous chromosomes :
Homologous chromosomes are paired during zygotene and then form chiasmata at some places where crossing over takes place between them and finally resulting in the production of new characters and possibilities of organic evolution.

2. Synapsis:
Pairing of homologous chromosomes in zygotene substage of prophase-I of meiosis cell division is known as synapsis. The paired structure formed as result of synapsis is called as bivalent.

3. G₂-Phase :
In this phase, RNA and protein continues to be synthesised. The cell is now ready to enter the mitotic phase. The cells of this stage contain double amount of DNA.

4. Bivalent:
The pairs of homologous chromosome formed during meiosis-I. The chromosomes contract and become short and thick.

5. Tetrad:
In Pachytene stage of Prophase-I of Meiosis-I, Chromosomes split lengthwise but still remain attached at the point of centromere to form four chromatids called as Tetrad.

Question 18.
Which of the following statement is associated with :
(a) Prophase,
(b) Metaphase,
(c) Anaphase,
(d) Telophase,
(e) Interphase of mito-sis.
1. The nuclear membrane reappears.
2. Chromosomes are thickest and shortest.
3. Chromosomes begin to coil.
4. Centromere divides into two.
5. Nucleus is active, but chromosomes are not distinct.
6. Followed by cytokinesis.
7. Each chromosome consists of two chromatids.
Answer:
1. (d) Telophase
2. (b) Metaphase
3. (c) Anaphase
4. (c) Anaphase
5. (e) Interphase of mito-sis
6. (d) Telophase
7. (c) Anaphase

Cell Cycle and Cell Division Class 11 Important Questions Long Answer Type

Question 1.
Write important differences between Mitosis and Meiosis cell division.
Answer:
Differences between Mitosis and Meiosis

Question 2.
Describe different stages of Meiosis and its significance also.
Answer:
Meiosis-I division :
1. Prophase-I (Gk., pro = before, phasis = stage) : The prophase of first meiotic division is a complex phase of longer duration which is further divided into following five substages:
(i) Leptotene or Leptonema (Gk., leptos = thin, nema = thread) :
At this stage condensation of chromatin starts and the fine chromatin fibres appear with granule like chromomeres on them. The chromomeres are the regions where chromatin fibres are highly coiled.

(ii) Zygotene or Zygonema (Gk., zygo = paired, tiema = thread):
At zygotene stage two homologous chromosomes (one paternal and other maternal) lie side by side which is known as pairing of homologous chromosomes or synapsis. Each pair is called as a bivalent, they are similar in length and in position of their centromere.

The process of synapsis can take place in many ways. It may start from the ends of homologous chromosomes and proceed towards the centromere or it may start from centromere and proceed towards the ends. It may also begin at one point or many points simultaneously. It is believed that synapsis is initiated by some sort of attraction force.

(iii) Pachytene or Pachynema (Gk., pachus = yolked or thick, nema = thread): At pachytene chromosomes contract longitudinally, resulting in shorter and thicker threads. Each unit is a bivalent or tetrad which is composed of two homologous chromosomes in close longitudinal union and which contains four chromatids. Each chromatid has its own centromere. The chromatids of each homologue is called sister chromatids.

During pachytene two of the homologous chromosome exchange their segments. This process is called as crossing-over. During this process, two non-sister chromatids, one from each bivalent partially coil around each other. Then both chromatids break at contact place and reunite in this way that exchange of chromosomal segments take place. This event is called crossing-over. The points of crossing-over are called chiasmata.

(iv) Diplotene or Diplonema (Gk., diplos = double, nema = thread):
In diplotene, the two chromosomes of each bivalent move away from each other and the chiasmata finally disappear. This event is terminalization of chiasmata. The two homologous chromosomes, thus separate from each other, however not completely because both remain united at the point of interchange or chiasmata.

(v) Diakinesis (Gk., dia = through, kinesis = division or movement):
During this stage the bivalents condense further, i.e., continue shortening and also become more thicker. Homologous chromosomes move apart. The nucleolus and nuclear membrane begin to dis¬appear. In animal cell both the centrioles travel to opposite poles. Astral rays emerge out from the centriole and the structure appear like aster or star and spindle formation starts.

2. Metaphase-I (Gk., meta = between) :
This stage begins with disappearance of nuclear membrane and nucleolus. The bivalents arrange themselves at the equatorial plane in such a way that one chromosome faces one pole of the spindle apparatus and the other one another pole. The spindle fibres get attached to bivalents at the centromere.

3. Anaphase-I (Gk,, ana = up,phasis = stage):
Due to contraction of spindles the two chromosomes of each bivalent separate from each other and move towards opposite poles of the spindle. But the two chromatids of each chromosome still continue to remain jointed with one another with centromere. Thus, only one chromosome of each homologous pair reaches at each pole and the number of chromosomes are reduced to half in daughter nuclei.

4. Telophase-I (Gk, telo = end,phasis = stage):
As soon as the chromosomes reach the poles decondensation starts and they lose their identity. Nuclear envelope is formed around the daughter nuclei and nucleolus also reappears.

Cytokinesis :
Telophase-I is generally followed by cytokinesis (similar to mitosis) to form two haploid (n) daughter cells.

Interphase:
A short interphase called interkinesis may intervene between meiosis-I and meiosis-II. However, replication of DNA does not take place in this phase. Sometimes the daughter nuclei are not fully constituted in telophase-I. Instead the chromosomes reachng the opposite poles may progress directly to prophase-II.

Meiosis-II division:
It is similar to mitosis and is also known as equational division. It consists of following phases:
1. Prophase-II:
Sister chromatids of each chromosome begin to condense and chromosomes reappear. Nuclear envelope and nucleolus disappear. Spindle fibres begin to disappear.

2. Metaphase-II:
The haploid number of chromosomes become arranged at the equatorial plane of spindle apparatus, each chromosome gets attached to spindle fibre by its centromere.

3. Anaphase-II:
The centromere divides and sister chromatids by the shortening of spindle fibres are pulled apart to their respective poles. Each half chromosome (chromatid) move towards the opposite pole.

4. Telophase-II:
The chromosomes on the respective poles uncoil and form chromatin network again. Nuclear membrane and nucleolus reappear. Finally, at the end of meio- sis-II four haploid nuclei are produced.

Cytokinesis : It is similar to mitosis, in animal cell by furrow formation and in plant cell by cell plate formation.
This way each daughter cell of meiosis-I produces two haploid daughter cells at the end of meiosis-II. In other words we can say that four haploid cells are formed from a diploid cell by the process of meiosis.

Significance of meiosis :

• It plays an important role in sexual reproduction of both the lower as well as higher organisms. Gametes are produced by this process.
• The process of meiosis helps in the maintenance of chromosome number in the species. Diploid (2n) organisms produce haploid (n) gametes which on fusion form diploid organism again.

Question 3.
Give necessity and significance of cell cycle and describe the phenomenon occurs in its various stages.
Answer:
Cell cycle :
Complete life cycle of any cell is called as cell cycle. Cell cycle includes various phases of the growth, development and reproduction of the cell. It is a continuous process.

Significance of cell cycle :
The cell, after its formation from pre-existing cell takes sometime to undergo further division. In this ‘resting period’ the cell grows in size, its nuclear materials increases and thus cell get ready for the next division. Hence, cell cycle is a cyclic process from the existence of a new cell to the division of cell. The total duration of a cell cycle constitutes one generation time.

Duration of cell cycle :
The duration of cell cycle varies from 25-30 hours in different cells. In case of E. coli the cell cycle is completed in 20 minutes, in rat 22-25 hours, in root tip of onion 20 hours and in human it is completed in 26 hours.

Stages of cell cycle :
Howard and Pele in 1953 discovered various stages of cell cycle. According to them cell cycle is completed in two phases :

1. Interpbase: Interphase is the interval period between the two successive divisions, when cell does not show any division. But it prepares itself for it by synthesizing new proteins and nucleic acids. In interphase chromosomes are not distinguished, they are in the fonn of chromatin network. This phase is also called as preparatory phase.

Interphase is divided into following three subphases, i.e., G₁, S and G₂phase :
(i) Presynthetic or G₁– phase : This phase is also known as the period of initial growth as the young daughter cell grow in size during this phase. Synthesis of new proteins and RNA needed for various activities of growing cell also occurs in this phase. G₁ -phase in most of the cells lasts for about 10-12 hours. Non-dividing cells remain permanently in this stage. There is no change in the DNA contents of the cell at this stage.
Mitochondria, chloroplast, lysosome, Golgi body, vacuoles, ribosomes, etc. are formed in this phase.

(ii) Synthetic phase or S-phase :
This phase is characterized by the replication of DNA molecules. Thus, each chromosome now carries a duplicate set of genes. Each chro¬mosome is now composed of two sister chromatids held together by a common centromere. The cell thus retains the original diploid (2n) number of chromosomes but it has duplicate set of genes. It lasts for about 6-8 hours.

(iii) Post-synthetic or G₂-phase :
In this phase nucleus increases in its size. The synthesis of protein and RNA continues in this phase, r RNA and m RNA required for cell division are synthesized in this phase and the cell prepares itself for the cell division. Mito¬chondria and chloroplasts are duplicated in this phase. Centriole begins to form proteins required for the formation of spindle fibres.

2. Phase of cell division or M-Phase :
In this phase, a cell tends to divide and generating two daughter cells. It is completed in following two stages :

• (i) Karyokinesis : It is the stage of division of nucleus, which is completed in four stages :
  • Prophase,
  • Metaphase,
  • Anaphase and
  • Telophase.
• (ii) Cytokinesis : It is the stage of division of cytoplasm to form daughter cells.

Question 4.
Write short note on Karyotype and Idiogram.
Answer:
Karyotype:
We have well studied earlier that each and every eukaryotic organism including plants and animals is characterised by a set of chromosomes which have certain
constant features. The chromosomes of a particular organism or species are identified on the basis of following important features :

• Number of chromosomes,
• Shape and size of chromosomes,
• Position of centromere,
• Length and ratio of two arms of a chromosome, (v) Presence of secondary constriction,
• Size of satellite bodies on chromosomes.
  Example : Karyotype of Ginkgo biloba.

The term karyotype has been proposed for “the group of characteristics that identifies a chromosome set (haploid) of a particular species.” This karyotype is represented by a diagram called idiogram.

Idiogram :
The idiogram of an organism is prepared by arranging the chromosomes of somatic cells in a descending order of size keeping their centromeres in a straight line (Fig.).

Thus, the longest chromosome is placed on the extreme left and the smallest one on the extreme right. In organisms having sex chromosomes, the sex chromosomes are sometimes placed at the extreme right but more commonly they are placed at their appropriate position according to their size and marked by X and Y. Each chromosome in a karyotype is designated by a serial number according to its position. (44 + XY) (44 + XX).

Cell Cycle and Cell Division Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
Chiasmata formation takes place during:
(a) Diplotene
(b) Leptotene
(c) Pachytene
(d) Diakinesis.
Answer:
(c) Pachytene

Question 2.
Duplication of chromosomes takes place during mitosis in:
(a) Early prophase
(b) Late prophase
(c) Interphase
(d) Late telophase.
Answer:
(b) Late prophase

Question 3.
Centromere in the first metaphase of meiosis:
(a) Divide
(b) Do not divide
(c) Divide but do not separated
(d) Are not similar.
Answer:
(c) Divide but do not separated

Question 4.
Spindle fibres are attached with chromosomes at:
(a) Telomere
(b) Chromomere
(c) Centromere
(d) Kinetochore.
Answer:
(c) Centromere

Question 5.
Part of the chromosome where crossing-over takes place is:
(a) Chromomeres
(b) Bivalent
(c) Chiasmata
(d) Centromere.
Answer:
(c) Chiasmata

Question 6.
Number of chromosomes becomes half during:
(a) Prophase-1
(b) Anaphase-I
(c) Metaphase-1
(d) Metaphase-lI.
Answer:
(b) Anaphase-I

Question 7.
Reduction division takes place ¡n:
(a) Pollen grains
(b) Pollen tube
(c) Pollen mother cell
(d) Generative cell.
Answer:
(c) Pollen mother cell

Question 8.
Spindle fibres are formed from :
(a) Protein
(b) Lipid
(c) Cellulose
(d) Pectin.
Answer:
(a) Protein

Question 9.
Synapsis takes place between :
(a) Similar chromosomes
(b) Two homologous chromosomes
(c) Non-homologous chromosomes
(d) None of these.
Answer:
(b) Two homologous chromosomes

Question 10.
In animals, spindle fibres are formed from :
(a) Centriole
(b) Centromere
(c) Nucleus
(d) Mitochondria.
Answer:
(a) Centriole

Question 11.
Formation of synaptical complex takes place during :
(a) Leptotene
(b) Zygotene
(c) Pachytene
(d) Diplotene.
Answer:
(b) Zygotene

Question 12.
Diploid number of chromosomes in an organism is 8. Then find out the number of chromatids in each daughter cell will be :
(a) 2
(b) 4
(c) 8
(d) 16.
Answer:
(c) 8

Question 13.
Stage of meiosis in which synapsis takes place is :
(a) Leptotene
(b) Pachytene
(c) Zygotene
(d) Metaphase-I.
Answer:
(c) Zygotene

Question 14.
Stage of meiosis in which centromere divide :
(a) Diplotene
(b) Metaphase-I
(c) Pachytene
(d) Anaphase-II
Answer:
(d) Anaphase-II

Question 15.
During interphase RNA and protein are synthesized :
(a) In S-phase
(b) In G₁phase
(c) In G₂-phase
(d) Both (b) and (c).
Answer:
(d) Both (b) and (c).

Question 16.
In which stage of meiosis centromere divides :
(a) Early prophase
(b) Early metaphase
(c) Early anaphase
(d) Post anaphase.
Answer:
(d) Post anaphase.

Question 17.
In which stage chromonemata begins to form paired chromosome :
(a) Zygotene
(b) Leptotene
(c) Pachytene
(d) Diplotene.
Answer:
(a) Zygotene

Question 18.
In which type of cell division number of chromosomes reduces :
(a) Meiosis
(b) Mitosis
(c) Amitosis
(d) Cleavage.
Answer:
(a) Meiosis

Question 19.
Which organ of the cell disappear during mitosis cell division :
(a) Plastid
(b) Plasma membrane
(c) Nucleus
(d) None.
Answer:
(c) Nucleus

Question 20.
How many chromosomes are present in a cell of human being after meiosis : (a) 46
(b) 23
(c) 20
(d) 48.
Answer:
(b) 23

Question 21.
Life of multicellular organisms begin from :
(a) Sperm
(b) Egg
(c) Zygote
(d) None of these.
Answer:
(c) Zygote

Question 22.
Diploid chromosome number in human is :
(a) 23
(b) 46
(c) 92
(D) 48
Answer:
(b) 46

Question 23.
Synapsis between chromosomes takes place during :
(a) Leptotene
(b) Pachytene
(c) Zygotene
(d) Diakinesis.
Answer:
(c) Zygotene

Question 24.
Stage of mitosis in which centromere divides :
(a) Anaphase
(b) Prophase
(c) Metaphase
(d) Telophase.
Answer:
(a) Anaphase

Question 25.
Which of the following divide during karyokinesis :
(a) Pollen grain
(b) Nucleus
(c) Cell wall
(d) Cytoplasm.
Answer:
(b) Nucleus

2. Fill in the blanks:

1. Somatic cells multiply by ……………….
Answer:
Mitosis

2. Mitosis results in the formation of nuclei having ………………. number of chromosomes.
Answer:
Same (identical)

3. The second division of meiosis can be described as ………………. division.
Answer:
Mitotic

4. The region of attachment of chromosomes to spindle fibres is called ………………….
Answer:
Centromere

5. The haploid condition is reached by ……………. stage.
Answer:
Anaphase

6. In parent and daughter cell, number of chromosomes are same, thus this type of cell division is called as ………………….
Answer:
Mitosis division

7. Chromosomes arrange themselves during metaphase at the ………………… of the cell.
Answer:
Equator

3. Match the following:

1. (d) Crossing-over
2. (e) Homologous pairing,
3. (a) Mitosis,
4. (b) Metaphase,
5. (c) S-phase.

4. Write true or false:

1. Interphase is the stage in between Prophase-I and Metaphase-I.
Answer:
False

2. Mitosis is also called as Reductional Division.
Answer:
False

3. Meiosis results in the formation of two haploid cells.
Answer:
False

4. Meiosis takes place in reproductive cells.
Answer:
True

5. During Prophase-I of Meiosis-I, the nuclear membrane and nucleolus disappear in Diakinesis stage.
Answer:
True

5. Answer in one word:

1. Short Interphase between Meiosis-I and Meiosis-II.
Answer:
Interkinesis

2. Division which takes place in somatic cells.
Answer:
Mitosis

3. The process of interchange of genetic material between non-sister chromatid.
Answer:
Crossing-over

4. The term given to each bivalent containing four chromatids at pachytene stage.
Answer:
Tetrad

5. The name of cell division taking place in reproductive cells.
Answer:
Meiosis

Students get through the MP Board Class 11th Biology Important Questions Chapter 13 Photosynthesis in Higher Plants which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Chapter 13 Photosynthesis in Higher Plants

Photosynthesis in Higher Plants  Class 11 Important Questions Very Short Answer Type

Question 1.
What is Photosynthesis? Explain it with chemical reaction.
Answer:
Photosynthesis: Photosynthesis is a chemical process by which green plants synthesize carbohydrate like simple food materials from carbon dioxide and water in the presence of sunlight. Water molecules are oxidized and CO₂ molecules are reduced by this process. Photosynthesis is expressed with the help of following chemical reaction :

Question 2.
Name the products of light reaction of photosynthesis process.
Answer:
The products of light reaction of photosynthesis process are ATP and NADPH₂.

Question 3.
What are Hill reagents or Hill’s oxidizing agents?
Answer:
Hill observed that evolution of O₂ was possible when chloroplast of Stellaria media suspended in water in absence of CO₂ were exposed to light and supplied with salts such as ferricyanides, benzoquinones, dichlorophenol, which served as hydrogen acceptors. These substances are called as Hill reagents.

Question 4.
Name the photosynthetic pigments found in bacteria.
Answer:
Photosynthetic pigments found in the bacteria are :

• Bacteriochlorophyll,
• Bacterioviridin
• Carotenoids etc.

Question 5.
Name any two insectivorous plants.
Answer:

1. Nepenthes,
2. Drosera.

Question 6.
Write any two differences between CAM and C₄ plants.
Answer:
Differences between CAM and C₄ plants

Photosynthesis in Higher Plants  Class 11 Important Questions Short Answer Type 

Question 1.
Give one example to prove that oxygen released during photosynthesis pro-cess is oxygen of water, not of CO₂.
Answer:
The oxygen liberated during photosynthesis is a part of water molecule. This is proved by Rubem and Camanis experiment. For this process such type of water is taken whose oxygen contains O¹⁸ isotopes and after equation the oxygen liberated is also O¹⁸.

By this experiments it is proved that after photosynthesis the liberated O₂ is the part of water molecule.

Question 2.
What is compensation point of Light intensity?
Answer:
Afternoon when intensity of light is high, rate of photosynthesis is high so plants releases O₂ whereas at night when there is no sunlight, photosynthesis process does not occur but during morning and evening when there is less intensity of light rate of photosynthesis become very low, equal to rate of photosynthesis.

Therefore, during this time O₂ released during photosynthesis process is totally used for respiration process and CO₂ produced during respiration is toally used up for photosynthesis process by the plant, thus there is no net release of any gas during this time. This condition is called as compensation point.

Question 3.
Describe Hill Reaction in 50 words and give required chemical reactions.
Answer:
Hill reaction : It is the first step of photosynthesis process, which occurs in the grana part of the chloroplast.
It occurs in the presence of light and it was discovered by scientist Hill. It is completed in following steps:
1. Photophosphorylation :

2. Photolysis of water :
12H₂O → 12H⁺ +12OH^–
12OH^– -12e^– →12OH
12OH → 6H₂O + 3CO₂

3. Hydrogenation :

Question 4.
Give significance of Photosynthesis process.
Answer:
Significance of Photosynthesis process :

• By this process green plants not only produces food for themselves but also for all other organisms. Thus, green plants are also called as producers.
• Plants uses harmful CO₂ from the atmosphere to prepare organic food and releases 02 into the atmosphere, thus help to maintain O₂-CO₂ balance in the nature.
• It regulates temperature by absorbing sunlight.
• By absorbing CO₂ from the atmosphere it helps to reduce global warming.

Question 5.
Give affect of Light and CO₂ on photosynthesis process.
Answer:

• Affect of Light: Rate of photosynthesis increases with increase in the intensity of light and decreases in less intensity of light. Colour of light also affect rate of photosynthesis. Rate of photosynthesis is highest in red light, in blue light rate of photosynthesis is second highest. Greenlight is reflected back by chlorophyll molecule, thus rate of photosynthesis is nil in green light.
• Affect of CO₂: Rate of photosynthesis increases with increase in the quantity of CO₂ but up to a certain limit (according to law of limiting factor) beyond which there is no effect.

Question 6.
Explain any two internal factors which affect photosynthesis process.
Answer:
Internal factors affecting photosynthesis process :

1. Chlorophyll: The amount of chlorophyll present in the plant part has a direct relationship with the rate of photosynthesis because it is the pigment which is photoreceptive and is directly involved in trapping the light energy.
2. Accumulation of end products: The rate of translocation of food manufactured in the leaves decreases in the afternoon and therefore it starts accumulating in the mesophyll cells. The accumulation of the end product decreases the rate of photosynthesis in the afternoon.

Question 7.
If a leaf is kept in a dark room then its colour gradually become yellow and greenish-yellow. Which pigment is more stable according to your view ? (NCERT)
Answer:
Chlorophyll-b is more stable which is yellow-green coloured pigment and Xanthophyll (Yellow pigment) in the leaves are more stable.

Question 8.
Two sides of dorsiventral leaves are different in colour, Le., upper part is dark as compared to lower part or compare leaves of the plants kept in the shade with plants kept in light. Which one is dark coloured and why? (NCERT)
Answer:
In dorsiventral leaves upper surface of the leaves contain more chlorophyll-a. Place where chlorophyll-a is found increases rate of absorption of light as compared to lower part. Part of the leaf where more chlorophyll-a is found appears bright green coloured whereas part of the leaves which remain in the shade appears less green due to less quantity of chlorophyll-a.

Question 9.
What is Emmerson effect?
Answer:
Emmerson effect: During working on photosynthesis process Robert Emmerson determined quantum production (number of O₂ molecules produced by absorption of one quantum light) of light of various wavelengths and found that quantum production is maximum in red light of680 nm. wavelength but when wavelength of this red light is increased more, then quantum production falls all of a sudden. This is called as Red drop.
Emmerson also observed that when less wavelength light is provided with red light of 680 nm. wavelength then quantum production increases again. This is called as Emmerson enhancement effect.

Question 10.
Can you identify observing external features whether the plant is C_{3 Or} C₄ plant?
Answer:
Generally C₄ plants grow at high intensity of light at day temperature 30-35°C, i.e, in dry atmosphere whereas C₃ plants grows in moderate temperature. C₄ plants has special anatomy called as Kranz anatomy. Thus, unless we observe internal anatomy it is not possible to identify a plant whether it is C₃  or C₄ plant.

Question 11.
Can you identify observing internal structure whether the plant is C₃ or C₄ plant?
Answer:
Anatomy of C₄ shows that it has Kranz anatomy, i.e., (a) Presence of a double spiral bundle sheath cells tightly packed around the vascular bundle.
(b) Bundle sheath cells are connected with mesophyll cells by plasmodesmata.
(c) Leaves of C4 plants contain two types of chloroplasts :

1. Mesophyll chloroplast: It is smaller, grana is present there and starch grains are absent.
2. Bundle sheath chloroplast: It is larger in size, lacking grana and possessing starch grains.
   In C₄ plants, CO₂ fixation occurs by C₃ and C₄ cycle. C₃ cycle occurs in the bundle sheath whereas C₄ cycle occurs in the mesophyll cells.

Question 12.
RuBisCo is an enzyme which acts as carboxylase and oxygenase enzyme. Why you assume that RuBisCo is found in C4 plants does carboxylation in more quantity?
Answer:
RuBisCo enzyme shows more affinity to CO₂ as compared to O₂. Both CO₂ and O₂ may bind to the active site of this enzyme. This binding is competitive type. Binding of O₂ or CO₂ to the enzyme depends on concentration of these substances.
C₄ plants has such a system that increases CO₂ concentration in them. Thus, it acts as RuBisCo carboxylase in them and it cannot act like oxygenase.

Question 13.
Assume that here plants having higher concentration of chlorophyll-b were found, whereas chlorophyll-a deficient plants were less in number, were capable to do photosynthesis? Then why chlorophyll-b is found in the plants and what are the role of other minor pigments ? (NCERT)
Answer:
Chromatography of chlorophyll shows that green colour of leaves is due to presence of four pigments:

1. Chlorophyll-a,
2. Chlorophyll-b,
3. Xanthophyll and
4. Carotene.

Following diagram proof that maximum photosynthesis occurs in the area of blue and red light of the spectrum. Some photosynthesis occurs in other waves of the spectrum. Though chlorophyll-a is the main pigment which absorbs light, then also other pigments found in the thylakoid, such as Chlorophyll-b, Xanthophyll and carotene absorb light and transfer it to the chlorophyll-a. Thus, they are called as accessory pigments.
These pigments increases wave area which also prevent chlorophyll-a from photooxidation.

Question 14.
Rate of photosynthesis is affected by light.

Answer following questions on the basis of graph:
(a) At which points (A, B, C or D) of the graph light is the limiting factor.
(b) Which limiting factor lies at point A.
(c) What is represented by C and D in the graph.
Answer:
(a) At point B and C light is the limiting factor.
(b) At point A of the graph sunlight, temperature, CO₂ concentration, water etc. fac-tors of photosynthesis are present and they altogether affecting photosynthesis at a time. Out of them anyone become main factor affecting photosynthesis process.

(c) C and D of the graph represent that there is linear relationship between low intensity of light and rate of CO₂ assimilation.
At high intensity of light there is no increase in this rate but other factors become limiting.
After a limit incident, ray causes decomposition of chlorophyll molecule, due to which rate of photosynthesis decreases.

Question 15.
Write differences between following : (NCERT)
(a) C₃  and C₄ Pathways.
(b) Cyclic and Non-Cyclic Photophosphorylation.
(c) Morphology of leaves of C₃  and C₄ plants.
Answer:
(a) Differences between C₃  and C₄ Pathways

(b) Differences between Cyclic and Non-Cyclic Photophosphorylation

(c) Differences between Morphology of leaves of C₃ and C4 plants

Question 16.
Describe Photorespiration in brief.
Answer:
Photorespiration or C2 Cycle: It was normally believed that rate of respiration is equal in day and night. Recently it has been observed that light affects respiration and the rate of respiration in light maybe three to five times higher than the respiration in darkness. Such type of respiration is called photorespiration and is marked as one of the new discoveries of plants physiology. In photorespiration, temperature plays a very vital role, its rate being very high in between 25-35°C. It also depends upon the concentration of oxygen and increases with increasing oxygen concentration even up to 100%. However, the normal respiration is independent of oxygen concentration. In normal respiration the respiratory substrate is glucose while in photorespiration glycolic acid (2-carbon compound) serves as a substrate.

The diagnostic features of photorespiration are as follows :

• The respiratory substrate is glycolate a 2-carbon compound.
• The substrate is always recently formed.
• The entire process of photorespiration occurs in between chloroplast, cytoplasm, peroxisome and mitochondria.
• It shows positive correlationship with O₂ concentration. The ribulose diphosphate reacts with O₂ to produce one molecule of 3-carbon, phosphoglyceric acid (PGA) and one molecule of 2-carbon phosphoglycolic acid.

RuDP + O₂ → phosphoglyceric acid + phosphoglycolic acid

Phosphoglycolic acid is immediately dephosphorylated into glycolate. The latter en-ters the peroxisome, where glycolate is catalyzed to produce glyoxylate and H₂O₂.
The glyoxylate is transmitted to glycine which is also obtained from serine.
The glycine molecule enters the mitochondria where they lose CO₂ to form serine. Serine diffuses into the peroxisome where it is converted into glycerate which then enters the chloroplast and the Calvin cycle to produce sugars.

• In this process ATP molecules are not formed.
• The rate of photorespiration is highly accelerated in between 25° to 35°C.
• This is found only in presence of light and only in green cells.

Question 17.
What do you mean by Total parasitic? Explain with example.
Answer:
Total Parasitic Nutrition: It is a type of nutrition in which parasitic plant totally depend on the host plant for food as well as shelter. Such plants are non-green e.g., Cuscuta orobanche. The total parasites develop haustoria or sucking roots which go into the vascular bundles of the host plant and absorb prepared food and water from there.

Question 18.
“CO₂ is very essential for the production of carbohydrates by the plants.” Prove this statement by describing the experiment with labelled diagram.
Or,
Give one experiment to prove that CO₂ is necessary for photosynthesis pro¬cess.
Answer:
Experiment to demonstrate that CO₂ is necessary for photosynthesis process: It can be demonstrated by Moll’s experiment:

Method: Take a potted plant having long leaves. Keep it in darkness for 24 to 48 hours to make it starch free. In the morning before sunrise, insert the apex of a leaf (through a split cork) in a wide mouthed bottle which has been laid on its side on the table, and in which caustic potash has been introduced.

Now expose the apparatus to sunlight for some-time. Then pluck the leaf and decolourize it with the help of hot alcohol and test with weak iodine solution. It is found that the apical portion of the leaf lying inside the bottle does not give the starch reaction, whereas the portion outside the bottle turns blue. This proves that starch is formed in those parts only which get CO_2.

Question 19.
Write short note on Facultative or Semi-parasitic nutrition.
Answer:
Facultative or Semi-parasitic nutrition: The type of nutrition in which plant may become parasite under certain conditions is called as facultative nutrition. These parasitic plants have chlorophyll and therefore, synthesize their organic food but they depend on the host plants for their water and mineral requirements.

Question 20.
Write differences between Light reaction and Dark reaction.
Answer:
Differences between Light Reaction and Dark Reaction

Question 21.
What are CAM plants? Give characteristic features of CAM plants and explain mechanism of CAM cycle.
Answer:
Crassulacean Acid Metabolism or CAM Cycle: Crassulacean acid metabolism occurs in certain succulent plants of the family Crassulaceae. As they exhibit a special type of CO₂ assimilation, they are called as CAM plants.

1. Dicotyledonous family:

• Crassulaceae (e.g.,Sedum, Opuntia),
• Cactaceae,
• Chenopodiaceae,
• Compositae,
• Convolvulaceae,
• Euphorbiaceae,
• Caryo- phyllaceae.

2. Monocotyledonous family :

• Liliaceae,
• Orchidaceae.

3. Pteridophytic family: Polypodiaceae.
Characteristic features of CAM plants: CAM plants exhibit the following characteristic features:

• Normally stomata are open during night (dark) and closed during day (light)
• CO₂ fixation takes place during day time and malic acid is formed during night.
• Malic acid is stored in the large vacuoles, which is the characteristic of the cells of CAM plants.
• In day light, decarboxylation of malic acid takes place and CO₂ is formed which is used in C₃ cycle to produce storage glycans.

Malic acid + NADP⁺ → Pyruvic acid + CO₂ + NADPH₂

• In next dark period storage glycans are catabolized through glycolysis and produces PEP as CO₂ acceptor molecule.
  
• Stomata remain closed during day and open at night.
• CO₂ is fixed via. PEP-carboxylase initially into OAA and then to other 4 carbon acids i.e., malic acid. This is called as dark
• CO₂ fixation and is more efficient at 10-15°C.
• The larger vacuoles are used for storing malic and other acids in large amount.

Photosynthesis in Higher Plants  Class 11 Important Questions Long Answer Type 

Question 1.
Describe light reaction of photosynthesis process.
Answer:
Light Reaction: In this reaction photolysis of water and production of the reducing power take place.
In this step, which takes place in the presence of sunlight and chlorophyll there is a break down of the water molecules. Oxygen escapes to the air in the molecular form while hydrogen is accepted by Nicotinamide adenine dinucleotide phosphate (NADP). The whole process can be represented by following summary equation.

This part of the photosynthetic mechanism which takes place in light is known as the Light Reaction or Hill Reaction, after its discoverer. Not only is the reducing agent produced but some solar energy get stored up as the chemical energy of NADPH₂ (about 18 molecules of ATP).

According to Robert Emmersion absorption of water in light reaction occurs through two pigment systems :

(a) Pigment System I: This pigment system absorbs only the light of 680 nm wavelength of sunlight. P700 on absorbing photon of light expels two-electron, P700 becomes oxidized. The expelled electron is picked up by ferredoxin. They transfer the electron to cytochrome b6 which supply it to cytochrome f. They transfer the electron to plastocyanin. From here it is handed over back to P700.

The synthesis of ATP in this cyclic electron transport scheme is possible at two locations. Synthesis of ATP may occur between ferredoxin and cytochromes bg and the other when the electron travels from cytochromes fg to plastocyanin.
In this case the electron expelled out by an excited chlorophyll trap is returned to the oxidized chlorophyll after passing through a series of electrons carriers, therefore this process is also called as Cyclic photophosphorylation.

(b) Pigment System II: This pigment system absorbs only the light of wavelength less than 680 nm of sunlight.
In this process pigment chlorophyll-a673 become excited by absorbing 2 photons of light and emit 2 electrons. It then moves through a series of electron carriers, some of which are plastoquinone, cytochrome bg, cytochrome f and plastocyanin. Most probably energy for ATP synthesis is given out between cytochrome bg and cytochrome f. Plastocyanin hand over the electron to P700.

When pigment system II absorbs light then 24 molecules of water dissociates into H⁺ and OH^– ions.
24H₂O → 24OH^– + 24H⁺

OH^– ion from this goes to pigment system II and oxidized to reduce to produce water and oxygen.

24OH^– – 24e^– → 240H
24OH → 12H₂O + 6O₂

24 H⁺ produced by photolysis of water obtain electrons from ferredoxin and reacts with NADP to form NADPH₂.

24H⁺ + 24e^– +12NADP →12NADPH₂
In this electron transfer, one ATP is formed when it transfer from Cyt. bg to Cyt.f. It is the normal process of photophosphorylation in higher plants. The electron expelled by the excited chlorophyll trap is never returned to the same chlorophyll. Therefore this process also called as Acyclic phosphorylation.

In this way in six complete cycle of cyclic phosphorylation 12ATP are formed and in six complete acyclic phosphorylation 6ATP and 12NADPH₂ are formed. Whereas in one acyclic phosphorylation 2NADPH₂ are formed and 24H₂O dissociates. Therefore both the systems of light reaction together produces 18ATPand 12NADPH₂ molecules.
Both the reactions occurs in light reaction simultaneously. To complete photosynthesis process both the pigment systems are essential.

Question 2.
Explain dark reaction of photosynthesis process in brief.
Or,
What is Calvin cycle? Describe it in brief.
Answer:
Dark reaction or Blackman’s reaction or Biosynthesis phase or Calvin’s cycle : The non-photochemical reaction or dark reaction of photosynthesis was first of all established by Blackman (1905). The reactions of dark phase of photosynthesis occur in the stroma of the chloroplast. In this phase the carbon dioxide is converted to carbohydrate through a series of enzyme catalyzed reactions.

The path of carbon in photosynthesis is popularly called Calvin cycle or C₃ cycle. For this great piece of work Calvin was awarded Noble prize in 1961. This cycle is also called as Bassham and Calvin cycle, Blackman reaction, Carbon assimilation, Path of carbon in photosynthesis etc.

The process can be represented by following reaction :

6CO₂ + 12NADPH₂ +18 ATP→ C₆H₁₂O₆ + 12NADP + 18ADP + 6H₂O + 18Pi

Most of the intermediate compounds of Calvin cycle are 3 carbon compounds. Therefore this cycle is also called as C₃ cycle.
Calvin cycle may be explained by following two steps :

1. Synthesis of carbohydrates
2. Regeneration of ribulose diphosphate.

1. Synthesis of carbohydrates: 6 molecules of CO₂ are first accepted by 6 molecules of Ribulose-1, 5-diphosphate and forms an unstable compound which soon breaks into 12 molecules of 3 carbon compound called as Phosphoglyceric acid. The phosphoglyceric acid is then reduced to 12 molecules of phosphoglyceraldehyde by 12 molecules of NADPH₂. 5 molecules from 12 molecules of phosphoglyceraldehyde converts into Dihydroxyacetone phosphate.

Now dihydroxyacetone phosphate reacts with 3 molecules of phosphoglyceraldehyde to form Fructose-1, 6-diphosphate. One molecule from it converts into Fructose-6-phosphate which converts into glucose or starch.

2. Regeneration of ribulose diphosphate: The regeneration of ribulose-1, 5- diphosphate is essential to carry on the process of photosynthesis. Fructose 6 phosphate and phosphoglyceraldehyde combine and break into 4-carbon compound erythrose-4-phosphate and 5-carbon compound xylulose-5-phosphate.

The former combines with a molecule of dihydroxyacetone phosphate to form sedoheptulose-1,7-diphosphate from which one phosphate is removed to form sedoheptulose-7- phosphate. The latter and phospho¬glyceraldehyde combine to produce one molecule each of xylulose-5-phosphate and ribulose-5-phosphate. Both of these compounds get converted into ribulose-5-phosphate which ultimately forms ribulose-1,5-diphosphate using ATP which is obtained from the photophosphorylation process.

Question 3.
Write a short essay on Bacterial photosynthesis.
Answer:
Bacterial photosynthesis: The anaerobic photosynthetic bacteria may be rods, cocci, or spirilla depending on their colouration. They are known as green or purple bacteria. They use sunlight as source of energy for photosynthesis but like other eukaryotic cells they do not ‘split water’ to transfer the energy or to obtain reducing power. Thus no oxygen is evolved by them. This process is therefore called anoxygenic (without producing oxygen) photosynthesis.

In place of water these bacteria obtain reducing power from hydrogen sulphide, thiosulphate, hydrogen or even some organic compound. They possess a pigment called bacterio-chlorophyll which is different from the chlorophyll pigment found in higher plants. This can be summarized as below :

There are three types of photosynthetic bacteria :

1. Green sulphur bacteria: They contain bacterial chlorophyll. These bacterias are found in H2S medium. In presence of light they reduces CO₂ Reaction is exergonic.
   
   Example: Chlorobacteria, chlorobium.
2. Purple sulphur bacteria : Bacteriochlorophyll is found. They can survive as heterotrophs in organic compounds in absence of inorganic sulphur compound.
   
   Example: Chromacium.
3. Purple non sulphur bacteria : These bacteria uses simple organic compounds such as alcohol, organic acid etc.
   
   As these bacterias uses visible light therefore they are also called as photo-autotrophic.

Question 4.
Write differences between photorespiration, true respiration and photosynthesis.
Answer:
Differences between Photorespiration, True Respiration and Photosynthesis

Question 5.
Explain Blackman’s principle of limiting factor.
Answer:
According to Blackman’s principle of limiting factor: When a physiological process is conditioned by number of factors then the rate of physiological process is limited by the factor found in least quantity.
For example, photosynthesis process is conditioned by many factors such as CO₂, water, light etc. If the concentration of CO₂ is least, then the rate of photosynthesis increases with increase in the quantity of CO₂. As the CO₂ supply increases gradually some other factor become limiting factor i.e., in such condition increased quantity of CO₂ does not help to increase rate of photosynthesis.

The principle was explained by Blackman as follows :

Suppose a leaf is exposed to such a light intensity which can allow the leaf to utilize 5mg of CO₂ per hour in photosynthesis. If 1 mg of CO₂ enters the leaf in one hour, the rate of photosynthesis is limited due to CO₂ factor. If the concentration of the CO₂ is increased from 1 to 5mg per hour, the rate of photosynthesis also increases along the line AC. Thus, the increase in photosynthetic rate will be proportionate with the increase in CO₂ concentration up to 5mg. Any further increase in the CO₂ concentration will have no effect on the rate of photosynthesis, which has become constant along the line CD. It is because the light factor (low intensity) has now become the limiting factor.

Now the rate of photosynthesis will increase further along the line CE only if the light intensity is also increased from low to a medium. At point E, the medium light intensity again becomes limiting factor and the rate of photosynthesis will again be- Fig Graphical representation of Blackman„’S come constant along the line EF. In the same way, law of limiting factor, when light intensity is increased from medium to higher, an increase in the rate of photosynthesis takes place along the EG by adding CO₂. When the rate attains maximum at G, further increase in CO₂ will not increase the rate of photosynthesis which become constant along the line GH. Here, higher light intensity again becomes a limiting factor.

Thus, it is quite evident from the above illustration that the rate of photosynthesis cannot be increased by increasing only one factor. The other factor should also be increased in proper proportion for favourable effect. Besides CO₂ and light, other factors such as temperature, water, etc. may also become limiting under certain conditions.

Photosynthesis in Higher Plants Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
Which method was used by Calvin to explain path of photosynthesis :
(a) Chromatography
(b) Electrophoresis
(c) Spectrophotometry
(d) Histochemistry.
Answer:
(a) Chromatography

Question 2.
Highest effective wavelength of light for photosynthesis is of which colour :
(a) Green
(b) Yellow
(c) Red
(d) Violet.
Answer:
(c) Red

Question 3.
Light reaction occurs in :
(a) Stroma
(b) Grana
(c) Membrane of Chloroplast
(d) Endoplasmic reticulum.
Answer:
(b) Grana

Question 4.
After absorption of radiation by pigment system-I electrons are emitted by:
(a) Chlorophyll-683
(b) Chlorophyll-673
(c) Chlorophyll-695
(d) P-700.
Answer:
(d) P-700.

Question 5.
First step of photosynthesis is :
(a) ATP synthesis
(b) Excitement of chlorophyll and emission of electrons
(c) Photolysis of water
(d) Release of oxygen.
Answer:
(b) Excitement of chlorophyll and emission of electrons

Question 6.
Photophosphorylation is the method in which :
(a) Aspartic acid is formed
(b) CO₂ and H₂O combine
(c) Light energy converts into chemical energy
(d) PGA is formed.
Answer:
(c) Light energy converts into chemical energy

Question 7.
Rate of photosynthesis depends on :
(a) Photoperiod
(b) Intensity of light
(c) Quality of light
(d) Temperature.
Answer:
(a) Photoperiod

Question 8.
CO₂ acceptor of C₃ cycle is :
(a) Phosphoglyceraldehyde
(b) Ribulose diphosphate
(c) Phosphoglyceric acid
(d) Pyruvic acid.
Answer:
(c) Phosphoglyceric acid

Question 9.
C₄ cycle of photosynthesis is absent in :
(a) Zea mays
(b) Triticum vulgare
(c) Oryza sativa
(d) Euphorbia.
Answer:
(d) Euphorbia.

Question 10.
During photosynthesis :
(a) Reduction of water and oxidation of CO₂
(b) Reduction of CO₂ and oxidation of water
(c) Oxidation of both CO₂ and H₂O
(d) Reduction of both CO₂ and H₂O.
Answer:
(b) Reduction of CO₂ and oxidation of water

Question 11.
Electron acceptor required for photophosphorylation and oxidative phosphorylation is:
(a) O₂
(b) CO₂
(c) Cytochrome
(d) Water.
Answer:
(c) Cytochrome

Question 12.
First stable compound of C₃ plant during photosynthesis is :
(a) PGA
(b) Starch
(c) Pyruvic acid
(d) Ribulose diphosphate.
Answer:
(a) PGA

Question 13.
Product of photophosphorylation is :
(a) NADP
(b) ADP from ATP
(c) ATP from ADP
(d) PGA.
Answer:
(c) ATP from ADP

Question 14.
COz acceptor of C₃ cycle is :
(a) RuDP
(b) PGA
(c) OAA
(d) PEPA.
Answer:
(a) RuDP

Question 15.
Kranze anatomy is found in the leaves of:
(a) C₂ plants
(b) C₃ plants
(c) C₄ plants
(d) Fleshy plants.
Answer:
(c) C₄ plants

Question 16.
CO2 acceptor of C₄ cycle is :
(a) PEP
(b) RuDP
(c) PGA
(d) OAA.
Answer:
(a) PEP

Question 17.
Solar energy converts into chemical energy in which process :
(a) Digestion
(b) Respiration
(c) Transpiration
(d) Photosynthesis.
Answer:
(d) Photosynthesis.

Question 18.
NADP is reduced to NADPH₂ during :
(a) Cyclic photophosphorylation
(b) Acyclic photophosphorÿlation
(c) Calvin’s cycle
(d) PS – I.
Answer:
(b) Acyclic photophosphorÿlation

Question 19.
In pigment system-il:
(a) Fixation of CO₂ occurs
(b) Reduction of CO₂ occurs
(c) Decomposition of water occurs
(d) All of these.
Answer:
(c) Decomposition of water occurs

Question 20.
Every green plant which do photosynthesis have:
(a) Chlorophyll-a
(b) Chlorophyl1b
(c) Chlorophyll-c
(d) Chlorophyll-d.
Answer:
(a) Chlorophyll-a

Question 21.
Photorespiration is the special quality of these plants:
(a) C₃ plants
(b) C₄ plants
(c) CAM plants
(d) None of these.
Answer:
(a) C₃ plants

Question 22.
in photorespiration the first process is:
(a) Carboxylation
(b) Decarboxylation
(c) Oxygenation
(d) Phosphorylation.
Answer:
(c) Oxygenation

Question23.
In photorespiration glycolate is changed into CO₂ and serin amino acid in:
(a) Chioroplast
(b) Peroxisorne
(c) both (a) and (b)
(d) Mitochondria.
Answer:
(d) Mitochondria.

Question 24.
The end product of the photosynthesis is:
(a) CO₂ and °2
(b) Carbohydrates
(c) CO₂ and carbohydrates
(d) CO₂ and starch.
Answer:
(b) Carbohydrates

Question 25.
The principle of limiting factor was proposed by:
(a) Blackman
(b) Arnon
(c) Living
(d) Hatch and Slack.
Answer:
(a) Blackman

2. Fill in the blanks:

1. In photorespiration the formation of glycine from glycolic acid is done in ……………. .
Answer:
Peroxisome,

2. Photorespiration is also called ……………. metabolism.
Answer:
Glycolate

3. The process of intake or synthesis of food material is called ……………. .
Answer:
Nutrition

4. Ribose diphosphate works in high concentration of ……………. as oxygenate.
Answer:
Oxygenase

5. Monatropa is a main ……………. plant.
Answer:
Saprophytic

6. In pigment system I ……………. photophosphorylation occurs.
Answer:
Cyclic

7. Products of light reaction are ……………. and ……………. .
Answer:
ATP, NADPH2

8. Chemical formula of chlorophyll-a is ……………. .
Answer:
C₅₅H₇₂O₃₅N₄Mg

9. Main pigment required for photosynthesis is ……………. .
Answer:
Chlorophyll

10. ……………. is reduced during photosynthesis.
Answer:
CO₂

3. Match the following:

(A)

Answer:
1. (c) Photorespiration
2. (d) Plants which need light
3. (a) Green chlorophyll
4. (b) C₄
5. (f) Autotroph.
6. (e) Symbiont

(B)

Answer:
1. (c) C₄ cycle
2. (d) Photophosphorylation
3. (e) Principle of limiting factor.
4. (a) C₃ cycle
5. (b) Red drop

4. Answer in one word:

1. Where does light reaction occurs in the chloroplast?
Answer:
Grana

2. What is the main source of oxygen evolved during photosynthesis?
Answer:
Water

3. Write the names of end-products of light reaction.
Answer:
ATP and NADPH₂

4. Give chemical formula of chlorophyll.
Answer:
C₅₅H₇₂O₃₅N₄Mg

5. Name the element found in the middle of chlorophyll molecule.
Answer:
Magnesium

6. Name the raw materials of photosynthesis process.
Answer:
CO₂, water, chlorophyll, light

7. Which colour of light shows highest rate of photosynthesis?
Answer:
Red light

8. Where does photosynthesis process occur.
Answer:
Chloroplast

9. Name first product of the C₃ cycle.
Answer:
Phosphoglyceric acid (PGA)

10. Name photosynthetic pigments found in the bacteria.
Answer:
Bacteriochlorophyll, Bacterioviridin, carotenoids

11. Name one enzyme found in the cells of C₄ plants.
Answer:
Phosphoenol-pyruvate carboxylase

12. Write full name of CAM.
Answer:
Crassulacean acid metabolism.

Students get through the MP Board Class 11th Biology Important Questions Chapter 9 Biomolecules which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Questions Chapter 9 Biomolecules

Biomolecules Class 11 Important Questions Very Short Answer Type

Question 1.
What is keratin?
Answer:
Keratin is a simple albuminoid protein and is the main constituent of hair, skin, nails, horns, feathers and wool. It is also known as scleroprotein. It is insoluble in neutral solvents but soluble in strong acids and bases.

Question 2.
What is chitin?
Answer:
Chitin is an important polysaccharide of invertebrates. It is found in the exoskeleton of insects and crustaceans and in the cell wall of fungi.

Question 3.
What do you understand by denaturation of proteins?
Answer:
The bonds of proteins are broken at higher temperature and thus structure and specific characters of proteins are changed. This phenomenon is called denaturation of protein.

Question 4.
What is nucleic acid?
Answer:
Nucleic acids are the polymer of nucleotides made up of C, H, O, N and P which controls the basic functions of the cell.

Question 5.
What are catalysts?
Answer:
Substance which can accelerate the rate of a chemical reaction are called as
catalysts.

Question 6.
What is enzyme? Explain.
Answer:
Enzymes are proteinic substances also called as biocatalysts, synthesized in living cells and initiate or accelerate metabolic reactions.

Question 7.
Define isoenzymes.
Answer:
Enzymes having different molecular configuration but having similar functions are known as isoenzymes, e.g., Lactic acid dehydrogenase (L.D.H.) which may occur in 5 possible forms in the blood sera and tissues of most vertebrates.

Question 8.
What is activation energy?
Answer:
The amount of energy required for the activation of a substance is called as activation energy. Enzymes lowering the energy of activation of the substrate so that biochemical reaction can take place at normal body temperature (at 37°C).

Question 9.
What is enzyme inhibition?
Answer:
The substance which reduces or inhibits the activity of enzyme are called enzyme inhibitor and the whole phenomenon is called as enzyme inhibition.

Question 10.
Give any two biological significance of enzymes.
Answer:

1. Enzymes play an active role in all rnetabolic and catabolic reactions and it is required for all the types of activities of the body.
2. Enzymes play an important role in biosynthesis, growth and repair of the body.

Question 11.
What are cofactors?
Answer:
A non-protein moiety attached to the protein part of enzyme is called as cofactor. If the cofactor is of inorganic nature like potassium, calcium, magnesium, manganese it is called prosthetic group. If cofactor attached to an enzyme protein is organic moiety like NAD, NADP or FAD it is known as coenzyme.

Question 12.
Define prosthetic group.
Answer:
The non-protein part of enzyme is called as cofactor. When the cofactor is of inorganic nature like K, Ca, Mg, Mn it is known as prosthetic group. Prosthetic group is generally tightly bounded to the protein part of enzyme.

Question 13.
Write differences between Apoenzyme and Hoioenzyme.
Answer:
All enzymes are proteins. However, there are some enzymes conjugated with a non-protein moiety (cofactor). Protein part of these conjugated enzymes are called as apoenzymes.
Enzymes with prosthetic group and apoenzyme are called as holoenzymes.

Question 14.
Write the names of any two polysaccharides and give their functions.
Answer:

1. Cellulose : It forms cell wall of plant cell. ,
2. Chitin : It forms exoskeleton of insects.

Question 15.
Write one difference in the following pair of words :

• Purineland pyrimidine,
• Starch and glycogen,
• Nucleoside and nucleotide,
• Fibrous and globular proteins.

Answer:

• Difference between purine and pyrimidine : Purine is two ringed heterocyclic nitrogenous base whereas pyrimidine is single ringed nitrogenous base.
• Difference between starch and glycogen : Starch is the storage food product of plants and glycogen of animals.
• Difference between nucleoside and nucleotide : Nucleotides are made up of nitrogenous base, pentose sugar and phosphate group whereas nucleosides are made up of only nitrogenous base and sugar.
• Difference between fibrous and globular proteins : Fibrous proteins are elongated molecules that may be constituted of several coiled peptide chains which are lightly linked whereas globular proteins are spherical or ovoid in shape due to folded peptide chains.

Question 16.
What do you understand by derived proteins?
Answer:
Derived proteins : Proteins which are derived from simple and conjugated proteins are called derived protein. They are not found in nature as such. They are obtained by the partial hydrolysis of natural proteins. These are formed by either change in the structure or breaking of their bonds, e.g., Peptones, protease, dipeptides, tri and tetrapeptides etc.

On complete hydrolysis of any protein the intermediate products are derived proteins whereas end-products are amino acids.

Question 17.
What is Dextrin? Give its importance.
Answer:
Dextrin is obtained by partial hydrolysis of starch, i.e., it is an intermediary substance of starch synthesis. It is used as sticking material in the cell.

Question 18.
What is agar-agar? Explain.
Answer:
Agar-agar is a polysaccharide, obtained from algae Gelidium, Gracilaria, which is mainly used in production of medium for culturing bacteria. It is neutral in nature. Iris- agar obtained from algae Chondrus crispes is used in preparation of chocolate, paint, toothpaste, ice-cream etc.

Question 19.
Write names of macromolecules of cell.
Answer:
Macromolecules of cell are:

• Polysaccharide,
• Protein,
• Nucleic acid etc.

Question 20.
Why are enzymes called biocatalysts?
Answer:
Enzymes are called as biocatalysts because they are the proteinous chemical substances which catalyse or accelerate rate of biochemical reactions occurs in the living body.

Question 21.
Give two important special characteristics of enzymes.
Answer:

1. Catalytic properties : Enzymes are biological catalysts. The small quantity of enzyme may catalyse metabolism of larger quantity of substances.
2. Specificity of enzymes : Enzymes are highly specific in nature, i.e., a particular enzyme can catalyse a particular reaction, e.g., Enzyme sucrase can catalyse only hydrolysis of sucrose.

Question 22.
What is Active site? Explain it.
Answer:
Part of the enzyme to which substrate molecule fit for biochemical reaction in the body of a living organism is called as active site. It is a three-dimensional structure to which substance fit. If geometry of active site changes then substrate fails to fit into the active site of enzyme. Actually active site is the binding site of the enzyme.

Biomolecules Class 11 Important Questions Short Answer Type

Question 1.
Explain any four factors affecting enzyme activity.
Answer:
The following four factors affect the activity of enzyme:
1. Temperature :
Metabolic reaction catalysed by enzymes increases the rate of activity at higher temperature because as temperature increases the kinetic energy of molecules also increases. But as the kinetic energy of molecules increases weak hydrogen bonds are more rapidly ruptured. The loss of hydrogen bonds modify the tertiary structure of enzymes and its catalytic activity, such alteration is called denaturation of enzyme.

2. Enzyme concentration:
As concentration of enzyme increases, the rate of enzyme activity also increases.

3. Substrate concentration:
In general, an increase in substrate concentration increases the rate of reaction but high concentration of the substrate, inhibits the activity of enzyme.

4. Hydrogen ion concentration (p^H) :
An enzyme have an optimum pH, but the changes in hydrogen ion concentration causes change in enzyme activity. Every enzyme requires a specific pH for the maximum activity, e.g., Trypsin in duodenum acts in alkaline medium (p^H = 8.5).

Question 2.
Explain exoenzymes and endoenzymes with suitable examples.
Answer:
Exoenzymes :
Enzymes which act outside the cells of their origin are called as exoenzymes. Enzymes secreted by the alimentary canal and used for digestion such as pepsin, trypsin, etc. are exoenzymes.

Endoenzymes :
Those enzymes which are produced inside the cell and participate in various reactions inside the cell are called intercellular or endoenzymes. They are active only inside the cells. Enzymes form mitochondria which takes part in cellular respiration.

Question 3.
What is the importance of enzymes?
Answer:
Importance of enzymes :

• Enzymes play an active role in all metabolic and catabolic reactions and it is required for every type of activities of the body,
• Enzymes catalyse and regulate the process of digestion, assimilation and absorption like chemical reactions,
• Enzymes play an important role in biosynthesis, growth and repair of the body,
• Enzymes regulate the energy release in respiration which is used in respiration, muscular, physical and mental activities.

Question 4.
Write down the similarities between enzymes and inorganic catalysts.
Answer:
Similarities between enzymes and inorganic catalysts :

• Enzymes and inorganic catalysts both of them remain unchanged at the end of the reaction chemically and quantitatively and can be used again to catalyse another reaction.
• Both are required in small quantity.
• They do not alter the equilibrium of the reversible reaction.
• Both of them increase the rate of reaction by lowering the activation energy but they do not initiate the reaction.
• The complexes formed with the reactants by both of them are short lived.

Question 5.
What is peptide bond?
Answer:
The bond formed between the carboxylic group (- COOH) of one amino acid and amino group (- NH₂) of another amino acid is called as peptide bond. A molecule of water is released during the formation of peptide bond.

Each peptide chain of considerable length may possess 50 to millions of amino acid units. A peptide may be dipeptide (with 2 amino acid units), a tripeptide (with 3 units) and so on. Beyond 10 amino acid units, a peptide is called polypeptide.

Question 6.
What is deoxyribonucleoside? How it is converted into nucleotides?
Answer:
The substances which are formed from deoxyribose sugar and nitrogenous bases are called deoxyribonucleosides on the basis of nitrogenous bases they are of four types.
Nucleosides can be converted into nucleotides after their reaction with phosphoric acid. After condensation they form nucleic acids.

Question 7.
Name the constituents of the nucleotides found in DNA.
Answer:
The following four types of nucleotides are found in DNA :

Question 8.
Write down the function of proteins.
Answer:
Function of proteins :

• Proteins form various structures of our body such as biomembranes, organelles (ribosome), hair and nails.
  Skin, horns, wool (all contain keratin), cartilage (contains collagen) and bones (contain ossein protein).
• Proteins function as biocatalyst (enzyme) in our body, e.g., Amylase and pepsin.
• The important cellular structures like cell membranes, connective tissue have proteins in their structure in which proteins functioning as a carrier molecule of different elements and compounds.
• Contractile proteins (e.g., Actin and myosin) participate in cellular movements and locomotion.

Question 9.
Write any four differences between DNA and RNA.
Or,
Write five differences between DNA and RNA on the basis of molecular composition and function.
Answer:
Differences between DNA and RNA

Question 10.
Define proteins and write their uses.
Answer:
Proteins :
Proteins are the polymer of amino acids. It is complex nitrogenous compound of the protoplasm. About 20 amino acids are joined together by means of peptide bonds form a protein molecule. About 20% part, the body of organisms is made up of proteins.
Uses of proteins :
Tertiary structure: Linear arrangement of amino acids through peptide linkage to form a polypeptide chain is called as Primary structure of protein. Folding of a linear polypeptide chain into specific coiled structure is referred as Secondary structure of protein.
1. Insulin : It is used for treatment of Diabetes mellitus.

2. Immunoglobulin : Immunoglobulin of the blood plasma in mammals and other animals act as antibodies which neutralize the harmful effect of foreign agents like viruses, bacterias etc.

3. Albumin : Albumin protein of egg (egg – white) is used for hair care.

4. Structural components of cells : Some proteins are essential structural compo-nents of cell membranes, organelles, cytoplasm, extracellular material and fibres. Keratin is the main constituent of hair, skin, nails, horns, feathers and wool. Cartilage is made of collagen. In plants protein is found in the walls of pollen grains.

5. Enzymes (Biocatalysts): The most important feature of proteins is their ability to function within the living cell as reaction catalysing enzymes. Enzymes usually increase the reaction rate by manifold. Enzymes play a key role in the metabolism.

6. Hormones : Proteins serve as hormones also. Hormones from pancreatic islets of Langerhans, pituitary, parathyroid and gastrointestinal mucosa are of peptide nature. Hormones of thyroid and adrenal medulla are derivatives of amino acid tyrosine.

7. As carriers : Haemoglobin, the respiratory pigment of many animals is a conju-gated protein composed of colourless basic proteins, the globin and haem. It has unique ability to bind oxygen in a loose and easily reversible combination. Myoglobin transports oxygen in muscles. In plants p-protein is involved in the transport of organic compounds through phloem.

8. Growth and repair : Proteins play a key role in general body growth and in the repair of wear and tear of the cells and body as a whole.

9. Formation of rhodopsin : The visual purple rhodopsin is made up of retinene an aldehyde derivative of vitamin-A and a protein opsin.

10. Melanin synthesis : Melanin, the skin pigment is derived from the tyrosine.

11. Formation of urea : In ureotelic animals, ornithine, citrulline and arginine (all amino acids) take part in the formation of urea by a cyclic process.

12. In some organisms protein is stored food, such as Albumin of egg, yolk of egg, in wheat as Glutein etc.

13. Fibroin protein secreted by silkworm produces silk on drying.

14. It may release energy by decomposing when required.

Question 11.
Draw a well labelled diagram of Watson and Crick model of DNA.
Answer:

Question 12.
Write short note on structural polysaccharides.
Answer:
Structural Polysaccharides : These polysaccharides take part in forming the structural framework of the cell wall in plants and skeleton of animals. They are of two main types:
1. Cellulose :
Cellulose is an important constituent of cell walls of plant, which provides mechanical support to plant cell. It is polymer of glucose and made up of unbranched chain of approximately 6,000 β-D-glucose units linked by 1,4-glycosidic linkage. It consti¬tutes 20-40% of cell wall.

2. Chitin :
Chitin is an important heteropolysaccharide of invertebrates. It is found in the hard exoskeleton of insects and crustaceans and in the cell wall of fungi. It is polyglycol consisting of N-acetyl-D-glucosamine units connected throughβ-1, 4-glycosidic linkage. Although chitin is soft and leathery it becomes hard when calcium carbonate or certain proteins are deposited in it. The insolubility of structural polysaccharides in water helps to retain the form and to strengthen the structure of the organism. In fungal cell wall chitin is often called as fungus cellulose.

Question 13.
Explain DNA replication.
Or,
Explain DNA duplication in short.
Answer:
Watson and Crick after giving the double helix model of DNA, also postulated the mechanism of DNA duplication, also known as replication. According to them, during duplication, the weak hydrogen bonds between the nitrogenous base of the nucleotides get separated, so that two polynucleotide chains of DNA also separate and uncoil. The chains thus, separated are complementary to one another.

These strands act as template and because of the specificity of base pairing each nucleotide of separated chain attracts its complementary nucleotide from the cell cytoplasm. Once the nucleotides are attached by their hydrogen bonds their sugar radicals write through their phosphate components completing the formation of a new polynucleotide chain. This results in the formation of two double helixes of DNA where in each molecule has one old strand contributed by parent DNA and one synthesized new. This method of DNA duplication is known as semi-conservative method.

Question 14.
Describe the functions of nucleic acids.
Or,
Explain the utility of nucleic acids.
Answer:
Utility of nucleic acids:

• Nucleic acids are the hereditary materials of organisms which involve in the transfer of hereditary characters from one generation to the next.
• DNA controls the synthesis of enzymes which control the various activities of the body.
• Nucleic acids also control protein synthesis,
• Nucleic acids form maximum portion of chromatin network,
• It causes mutation in living beings,
• They form enzymes.

Question 15.
Elaborate the term RNA. Also describe the types and functions of RNA.
Or,
Write location and kinds of RNA in the cell.
Answer:
RNA : Ribonucleic acid (RNA) located in the nucleus, cytoplasm, ribosomes and in some other cell organelles.
Types of RNA and their functions : RNA are of three types :

1. Messenger RNA (mRNA):
It makes a small fraction 5-10%. This RNA directs the sequence of amino acids in protein synthesis after joining with ribosomes. It carries the genetic information contained in DNA. It is short lived and has rapid turn over. It is formed of 700-1500 nucleotides and has a molecular weight from 5,00,000 to 20,00,000. The sequence of three nitrogenous bases of tnRNA forms a codon which is responsible for coding of one amino acid.

2. Ribosomal RNA (rRNA) :
It makes 80% of total cell RNA. It is the most stable type of RNA and is associated with ribosomes.

3. Transfer RNA (tRNA) :
It makes a small fraction (10-15%) of RNA. These are smallest molecules formed of 73-93 nucleotides with molecular weight ranging between 25,000 to 30,000. tRNA works as adaptor molecules for carrying amino acids to the TMRNA template during protein synthesis.

Question 16.
Write names of the nucleosides which help in energy transfer.
Answer:

1. ATP: Adenosine triphosphate,
2. ADP: Adenosine diphosphate,
3. GTP : Guanosine triphosphate,
4. GDP : Guanosine diphosphate,

Question 17.
Describe structure of RNA.
Answer:
Ribonucleic acid (RNA) :
These are second type of nucleic acids which are found in the nucleus, cytoplasm, ribosomes and in some other cell organelles.
RNA occurs normally as long unbranched polymeric molecule in the form of a single chain. Generally non-genetic RNA of eukaryotic or prokaryotic cells have single strand except fRNA which has double strand but without helical structure. But the wound tumour virus (a plant virus) and Rheo virus (animal virus) have double strand in their RNA molecule.

Chemical composition of RNA :
RNA molecules are polymers of ribonucleotides. Each ribonucleotide is made up of three different molecules namely a ribose pentose sugar instead of deoxyribose sugar as in DNA, a phosphoric acid and a nitrogenous base.
Two types of nitrogenous bases are found in RNA :

1. Purines : Adenine (A) and guanine (G).
2. Pyrimidines : Uracil (U) and cytosine (C). Thus, uracil substitutes thymine of DNA. In RNA above four nitrogenous bases join with the ribose sugar to form ribonucleoside.

Nucleosides of RNA : There are four types of ribonucleosides are present in RNA :

1. A + Ribose sugar = Riboadenosine
2. G + Ribose sugar = Riboguanosine
3. C + Ribose sugar = Ribocytidine ‘
4. U + Ribose sugar = Ribouridine.

Nucleotides of RNA : Phosphoric acid joins with each ribonucleoside to form ribonucleotide.

• A + Ribose sugar + Phosphoric acid = Riboadenylic acid
• G + Ribose sugar + Phosphoric acid = Riboguanylic acid
• C + Ribose sugar + Phosphoric acid = Ribocytidylic acid
• U + Ribose sugar + Phosphoric acid = Ribouridylic acid.
  These four ribonucleotides are linked together to form a long strand of ribopolynucleotide.

Question 18.
What is enzyme? Explain the characteristic features of enzymes.
Answer:
Enzymes are proteinic substances also called as biocatalysts synthesized in living cells and initiate or accelerate metabolic reactions by lowering activation energy.
Characteristic features of enzymes :
1. Physical properties :
Physically enzymes behave as colloids or as substance of high molecular mass. They are destroyed or inactivated at temperature below the boiling point of water. At 50°C most enzymes in a liquid medium are inactivated. Dried enzyme extract can endure temperature of 100°C to 120°C or even higher. Thus, enzymes are thermolabile.

2. Chemical properties:
(i) Catalytic properties :
Enzymes are biological catalyst. Catalysts are substances which accelerate the speed of chemical reaction without undergoing any permanent changes. The small quantity of enzyme catalyses the larger quantities of substrates.

(ii) Specificity of enzymes :
Enzymes are highly specific in nature, i.e., a particular enzyme can catalyse only a particular reaction at a particular temperature, e.g., Enzyme sucrase, can catalyse only hydrolysis of sucrose.

3. General properties:

• They initiate and accelerate the rate of biochemical reaction.
• The activity of enzymes depends upon the acidity of the medium (p^H specific). Each enzyme is most active at a particular acidity (p^H).
• Enzyme can accelerate the reaction in either directions :

• All enzymes possess active sites which participate in the biochemical reactions.
• They are very unstable compounds mostly soluble in water, dilute glycerol, NaCl and dilute alcohol.
• They act actively at optimum temperature.
• All enzymes are proteins but all proteins are not enzymes.

Question 19.
Give the composition of Triglyceride.
Answer:
Triglycerides are a group of stored lipid which include oil and fats. Chemically, lipids are the esters or glycerides of fatty acid and glycerol. It is made up of three molecules of fatty acid and one molecule of glycerol. Their general formula is CH₃(CH₂)_n„COOH. The carboxylic group (- COOH) of each fatty acid react with alcoholic group (-OH) of glycerol and form ester linkage.

Question 20.
Write short note on Globular protein.
Answer:
Globular protein:
These are spherical or oval in shape because of the presence of peptide folded or coiled chain which forms compact structure, it is called as globular protein. It forms a three-dimensional structure, which is formed by relatively weak non-covalent bond. It is a secondary structure of protein, soluble in water. These include enzymes, protein, hormones like ACTH, oxytocin, glucagon and insulin etc.

Question 21.
Of which substance gum is made up? Is fevicol different than this?
Answer:
Gum is a carbohydrate, such as D-Galactose, D-Galactonic acid, which are natural substances, whereas fevicol is an artificial product which is used as gum.

Question 22.
Can you explain process of conversion of milk into curd or yogurt on the basis of concept of protein?
Answer:
Protein present in the milk is casein, which is a white coloured substance and has nutritional value. It consists of essential amino-acids required for the growth of human body. Conversion of milk into curd is a chemical process. Lactobacillus bacteria ferment lactose sugar of milk into lactic acid and help to settle casein. Settled casein is called as curd or yogurt.

Question 23.
Can you prepare biomolecule model using commercial model of atom (Ball and stick model)?
Answer:
Yes, we can prepare biomolecule model using commercial model of atom.

Question 24.
Which functional groups are released on titration of amino acid with weak base?
Answer:
On titration of amino acid with weak base two functional groups are released :

• – COOH group (Carboxylic group)
• – NH₂ group (Amino group)

Question 25.
Give structural formula of amino acid Alanine.
Answer:

Question 26.
Write short note on conjugated protein and give two examples of it.
Answer:
Conjugated Proteins : These are complex protein molecules, which is formed by combination of simple protein molecule with characteristics non-protein substance called as prosthetic group, i. e., on hydrolysis conjugated protein yield a protein and a non-protein substance.
Protein + Prosthetic group → Conjugated protein
Conjugated proteins are of following types :

1. Nucleoprotein,
2. Mucoprotein,
3. Glycoprotein,
4. Chromoprotein,
5. Lipoprotein,
6. Metalloprotein and,
7. Phosphoprotein.

e.g., Haemoglobin, Lipoprotein
Globin + Haem → Haemoglobin
Protein + lipid → Lipoprotein.

Biomolecules Class 11 Important Questions Long Answer Type

Question 1.
If a given protein has primary structure, in which amino acids are present at both the ends, then can you connect this information with purity or homogeneity of protein?
Answer:
The primary structure of protein refers to the linear sequence of number of amino acids belonging to the polypeptide chain. The amino acids link together by peptide bond only. It is formed by binding of amino group (-NH₂) of one amino acid with carboxylic group (-COOH) of another amino acid. Therefore, a single polypeptide chain has first amino acid at one end and the last amino acid at the other end. Amino group of first amino acid is called as ‘N’ terminal and carboxylic group of last amino acid is called as ‘C’ terminal.
By above information purity or homogeneity of a protein cannot be determined because many other amino acids (50 to millions) are found in between them.

Question 2.
Find out the proteins used as therapeutic agents and make a list. Give other uses of protein (as cosmetics) etc.
Answer:
1. Insulin : It is used for treatment of Diabetes mellitus.

2. Immunoglobulin : Immunoglobulin of the blood plasma in mammals and other animals act as antibodies which neutralize the harmful effect of foreign agents like viruses, bacterias etc.

3. Albumin : Albumin protein of egg (egg – white) is used for hair care.

4. Structural components of cells : Some proteins are essential structural compo-nents of cell membranes, organelles, cytoplasm, extracellular material and fibres. Keratin is the main constituent of hair, skin, nails, horns, feathers and wool. Cartilage is made of collagen. In plants protein is found in the walls of pollen grains.

5. Enzymes (Biocatalysts): The most important feature of proteins is their ability to function within the living cell as reaction catalysing enzymes. Enzymes usually increase the reaction rate by manifold. Enzymes play a key role in the metabolism.

6. Hormones : Proteins serve as hormones also. Hormones from pancreatic islets of Langerhans, pituitary, parathyroid and gastrointestinal mucosa are of peptide nature. Hormones of thyroid and adrenal medulla are derivatives of amino acid tyrosine.

7. As carriers : Haemoglobin, the respiratory pigment of many animals is a conju-gated protein composed of colourless basic proteins, the globin and haem. It has unique ability to bind oxygen in a loose and easily reversible combination. Myoglobin transports oxygen in muscles. In plants p-protein is involved in the transport of organic compounds through phloem.

8. Growth and repair : Proteins play a key role in general body growth and in the repair of wear and tear of the cells and body as a whole.

9. Formation of rhodopsin : The visual purple rhodopsin is made up of retinene an aldehyde derivative of vitamin-A and a protein opsin.

10. Melanin synthesis : Melanin, the skin pigment is derived from the tyrosine.

11. Formation of urea : In ureotelic animals, ornithine, citrulline and arginine (all amino acids) take part in the formation of urea by a cyclic process.

12. In some organisms protein is stored food, such as Albumin of egg, yolk of egg, in wheat as Glutein etc.

13. Fibroin protein secreted by silkworm produces silk on drying.

14. It may release energy by decomposing when required.

Question 3.
Describe Glycosidic, Peptide and Phosphodiester bonds.
Answer:
1. Glycosidic bond :
During the union of monosaccharide units water molecule is eliminated and the units are linked through an oxygen bridge known as glycoside linkage. Depending upon the steric configuration at carbon-1 of monosaccharide unit the bond is called α and β-bond.

By this bond monosaccharides are converted into oligo and polysaccharides. On hydrolysis they yield monosaccharides.

2. Peptide bond :
The bond formed between the carboxylic group (- COOH) of one amino acid and amino group (- NH₂) of another amino acid is called as peptide bond. A molecule of water is released during the formation of peptide bond.

Each peptide chain of considerable length may possess 50 to millions of amino acid units. A peptide may be dipeptide (with 2 amino acid units), a tripeptide (with 3 units) and so on. Beyond 10 amino acid units, a peptide is called polypeptide.

3. Phosphodiester bond :
Nucleic acid, like DNA molecules are polymers of deoxyribonucleotides. Each deoxyribonucleotides is made up of three different molecules :
(i) a decxyribose sugar (Pentose sugar),
(ii) a phosphoric acid and
(iii) a nitrogenous base. Nitrogenous base may be of two types :
1. Purines : Adenine (A) and Guanine (G).
2. Pyrimidines : Cytosine (C) and Thymine (T).
3′ carbon of one nucleotide combine with 5′ carbon of next (adjacent) nucleotide by phosphate group. Ester bond is formed between phosphate of sugar and hydroxyl group. Ester bond is formed on both the sides of phosphate group, thus it is called as Phosphodiester bond.

Nucleotides unite with each other by phosphodiester bond to form long chain of polynucleotide of DNA or RNA, but in RNA instead of thymine, uracil is found.

Question 4.
What do you mean by Tertiary structure of protein?
Answer:

Tertiary structure: Linear arrangement of amino acids through peptide linkage to form a polypeptide chain is called as Primary structure of protein. Folding of a linear polypeptide chain into specific coiled structure is referred as Secondary structure of protein.
The arrangement and interrelationship of twisted chains of proteins into specific loops and bends are called the tertiary structure of protein.
Tertiary structure of protein is maintained by four kinds of bonds :

• Hydrogen bonds,
• Hydrophobic bonds,
• Ionic or electrostatic bonds and
• Disulphide bonds.

Question 5.
Write structural formula of 10 micro biomolecules of low molecular weight. Find out the industries involved in their production. Who are the customer of them? Find out.
Answer:
Structural formula of low molecular weight micro-biomolecules are as follows: (1) Monosaccharides :

(2) Amino acids:

(3) Oil and fat:

(4) Nitrogenous bases:

(5) Nucleoside and Nucleotide

Question 6.
Give analytical test of protein, fat, oil and sugar and test presence of them
in any fruit juice, saliva, sweat and urine sample. (NCERT)
Answer:
(1) Test for Protein in sample of urine

(2) Test for presence of Oil and Fat in food sample

(3) Test for presence of Starch in food sample

(4) Presence of Sugar in sample of fruit juice .

Question 7.
What do you mean by cellular macromolecules? Write the name of different macromolecules and describe any one in detail.
Answer:
Cellular macromolecules : The molecules formed by the process of polymerization are called as macromolecules. Macromolecules of the cell possess the following characteristic features:

• Chemically they are larger in size,
• Their molecular weight is very high,
• Their solubility is very low.
• Their molecular structure may be simple branched or unbranched and coiled,
• They are the polymers of micromolecules, e.g., (I) Polysaccharides, (II) Proteins and (III) Nucleic acids.

Polysaccharides :
They are polymers of monosaccharides and is made up of 10 to several thousand units of monosaccharides joined by glycosidic linkage. After hydrolysis they produce monosaccharides. Their empirical formula is (C₆H₁₀O₅)_n where n ranges from 10 to 10,000. Their molecular weight is very high.
Polysaccharides are of following two types :
1. Homopolysaccharides : These polysaccharides are made up of one kind of monosaccharides as in starch, cellulose, glycogen and insulin.
2. Heteropolysaccharides : These polysaccharides are made up of two or more kinds of monosaccharides as in agar, pectin and chitin.
On the basis of their functions polysaccharides are classified into following two groups:
(a) Storage polysaccharides and
(b) Structural polysaccharides.

(a) Storage polysaccharides : Polysaccharides which are synthesized by the process of photosynthesis and are stored in the form of food materials in plants and animals are called as storage polysaccharides, e.g., Starch, glycogen, gum, insulin, dextrin.
(i) Starch :
It is the storage carbohydrate of plants. It is formed during the process of photosynthesis and serves as an energy storing material. Starch is a polymer of glucose unit linked together in an α-1,4-glycosidic linkage. It yields only glucose on hydrolysis. In fact it is formed by condensation of amylase and amylopectin. Starch grains may be oval, spherical, lens shaped or irregular.

(ii) Glycogen :
Glycogen is the storage polysaccharide of animals and is called as animal starch. It is a branched polymer of glucose. It is stored mostly in muscles and liver of animals. The glycogen of liver supplies glucose to all tissue through blood. It is also found as storage product in blue-green algae, slime, moulds, fungi and bacteria. Their general formula is (C₆H₁₀O₅)_n)„. It consists of α-D glucose units mostly linked by 1, 4 but highly branched via 1,6-linkage each of which starts other α-1,4 chain of some 8-T 2 glucose units.

(b) Structural polysaccharides :
Polysaccharides which take part in the structural organization and protection mechanism of the body are called structural polysaccharides, e.g., Cellulose, chitin, etc.

Utility of polysaccharides :

• Polysaccharides serves as an important structural component in some animals and in all plants where they constitute cellulose framework.
• They are stored as storage food material and whenever required these can be hydrolyzed to yield energy.
• Cellulose forms protective covering as cell wall in plant cells. In some animals chitin forms protective exoskeleton.
• Mucilage acts as binding substance of cell walls and connective tissue.
  Polysaccharides are also used in textile, furniture, printing, varnish, alcohol and film
  industries.

Question 8.
How many kinds of nucleic acids are found in the cell? Give a detailed account of DNA.
Or,
Explain the structure of DNA.
Answer:
Every type of plant and animal cell (except virus) generally possess following two kinds of nucleic acids :
(i) DNA (Deoxyribonucleic acid),
(ii) RNA (Ribonucleic acid).
Where viruses possess only one type of nucleic acid from the above.
Structure of DNA : Structurally DNA is the chain of polydeoxyribonucleotides. It means that DNA is formed by the polymerization of deoxyribonucleotides. Every DNA molecule possesses the following four kinds of nucleotides :

• Deoxyadenylic acid,
• Deoxyguanylic acid,
• Deoxycytidylic acid,
• Deoxythymidyhc acid.

Watson and Crick (1953) suggested a ‘double helix’ model of DNA. According to them, the DNA molecule consists of two long polynucleotide helically twisted strands which are complementary running in opposite direction (3′ – 5′, 5′ – 3 ‘) and connected together by steps.

The vertical bars are formed by alternate phosphate and sugar groups joined by phosphodiester bonds. These strands are formed together by hydrogen bonds which are established between specific nitrogen base in the sequence A = T, T = A, C ≡ G and G ≡ C. The length of one complete turn is 34Å. There are 10 nitrogenous base pairs in each turn. The distance between two nucleotides is 34 Å. The width of DNA is 20Å. Each successive nucleotide turns 36 degree in the horizontal plane. The twisting of strands results in the formation of deep and shallow spiral groove.

Significance of DNA:

• DNA is the genetic material which transmits parental characters from generation to generation,
• DNA together with protein form chromosomes,
• It is the constituent of ribosomes, mitochondria and plastids.
• It is responsible for mutation and organic evolution,
• It synthesizes enzymes required for various biological activities.

Question 9.
Describe important properties of enzyme.
Answer:
Important properties of enzymes are as follows :
(i) Catalytic properties :
Enzymes are biological catalyst. Catalysts are substances which accelerate the speed of chemical reaction without undergoing any permanent changes. The small quantity of enzyme catalyses the larger quantities of substrates,

(ii) Specificity of enzymes :
Enzymes are highly specific in nature, i.e., a particular enzyme can catalyse only a particular reaction at a particular temperature, e.g., Enzyme sucrase, can catalyse only hydrolysis of sucrose.

(iii) Colloidal state :
Physically enzymes behave as colloids or as substances of high molecular mass. They are hydrophilic in nature. They are amphoteric in nature. Thus, behave as alkali with acid and as acid with alkali.

(iv) Reversible in action : Enzymes can accelerate the reaction in either directions.
e.g., Sucrase enzyme hydrolyse sucrose into glucose and fructose. It also catalyse backward reaction.

(v) Sensitivity :
Each enzyme is catalytically active within a limited range of pH. Most enzymes operate effectively in a range of about 5 to 9 pH. For example, the optimal range for the activity of malt amylase is 5-2 pH, for salivary amylase between 6-7 to 6-8 pH, for trypsin 8 to 11 pH. In all these cases the activity gradually declines on either side of the optimum pH. Most intracellular enzymes function best at neutral pH 7.

(vi) High molecular weight:
Enzymes are the proteins of high molecular weight, e.g., molecular weight of feredoxin of bacteria is 6000 and pyruvate dehydrogenase is 46,00,000.

(vii) High reactivity :
Small quantity of enzyme may catalyse reaction at fast rate. Total number of substrate molecule which may convert into product in one minute by one enzyme is called as Turnover number. Turnover number of different enzymes are as follows:

• Carbonic anhydrase : 36 million,
• Catalase : 5 million,
• Sucrase or Invertase : 10,000,
• Flavo protein : 90.

Question 10.
Explain biological significance of enzymes.
Answer:
Following are the biological importance of enzymes :
1. Healing of wound :
A protein digesting enzyme obtained from pancreas of the pig is very useful for skin diseases and healing of wound. These enzymes destroys protein digesting enzymes of human body thus help for healing of wound.

2. Dehairing of hides :
Enzymes obtained from pancreas destroys hair follicles of hair, thus help for removal of hair from hide.

3. Dissolving the blood clot:
An enzyme called as urokinase obtained from urea helps to dissolve blood clot which are formed in the brain or arteries, thus saves our life.

4. Changing blood group :
Prof. Ken Furunkawa in 1981 explained the blood group is determined by the presence of sugar found in R.B.Cs. for which specific enzymes are found. If these enzymes are destroyed, then blood group ‘A’ and ‘B’ can change into blood group‘O’.

5. Analysis of biochemicals :
Some enzymes are used for quantitative analysis of some specific substances found in blood. For example, uricase, urease and sucrase enzymes are used to determine quantity of uric acid, urea and sucrose present in the blood respectively.

6. Cheese making :
Animal renin (milk protein digesting enzyme) is used in cheese making.
7. Manufacture of fruit juice : Some enzymes are used for processing of some fruit juices. Pectic enzymes are added to these fruit juices to make it clean and clear.

Biomolecules Class 11 Important Questions Objective Type

1. Choose the correct answers:
Question 1.
Lipids contain carbohydrates and N₂ are called :
(a) Phospholipids
(b) Chromolipids
(c) Aminolipids
(d) Glycolipids.
Answer:
(d) Glycolipids.

Question 2.
Lecithin is :
(a) Phospholipids
(b) Chromolipids
(c) Aminolipids
(d) Glycolipids.
Answer:
(a) Phospholipids

Question 3.
Fats and oils are decomposed by bases and produce :
(a) Soap
(b) Vegetable ghee
(c) Saturated fat
(d) Unsaturated fat.
Answer:
(a) Soap

Question 4.
Compounds with free amino and carboxylic group are called :
(a) Glucose
(b) Nucleotide
(c) Amino acid
(d) None of these.
Answer:
(c) Amino acid

Question 5.
Nucleotide responsible for energy transfer is :
(a) NAD
(b) FAD
(c) FMN
(d) ATP.
Answer:
(d) ATP.

Question 6.
Unit of protein is :
(a) Fatty acids
(b) Monosaccharides
(c) Amino acids
(d) Glycerol.
Answer:
(c) Amino acids

Question 7.
Nucleic acids are the polymers of:
(a) Amino acids
(b) Nucleosides
(c) Nucleotides
(d) Globulin.
Answer:
(c) Nucleotides

Question 8.
Peptide bonds are found in:
(a) Proteins
(b) Fats
(c) Nucleic acids
(d) Carbohydrates.
Answer:
(a) Proteins

Question 9.
Glycosidic bonds are found in:
(a) Nucleic acids
(b) Proteins
(c) Polysaccharides
(d) Monosaccharides.
Answer:
(c) Polysaccharides

Question 10.
A nucleoside differs from nucleotide ¡s not having:
(a) Sugar
(b) Nitrogen base
(c) Phosphate
(d) Both (a) and (e).
Answer:
(c) Phosphate

Question 11.
A unit composed of sugar and nitrogen base linked by glycosidic bond is:
(a) Purine
(b) Glycoside
(c) Nucleoside
(d) Nucleotide.
Answer:
(c) Nucleoside

Question 12.
Building block of nucleic acid is:
(a) Amino acid
(b) Nucleoprotein
(c) Nucleotide
(d) Nucleoside.
Answer:
(c) Nucleotide

Question 13.
Lactose molecule ¡s composed of:
(a) Glucose + Fructose
(b) Glucose + Glucose
(c) Glucose + Galactose
(d) Fructose + Fructose.
Answer:
(c) Glucose + Galactose

Question 14.
Heredity is controlled by:
(a) DNA
(b) RNA
(c) Generally DNA and sometimes RNA in few organisms
(d) None of these.
Answer:
(c) Generally DNA and sometimes RNA in few organisms

Question 15.
DNA is made up of:
(a) Deoxyribonucleotide
(b) Ribonucleotide
(c) Both (a) and (b)
(d) Polysaccharides.
Answer:
(a) Deoxyribonucleotide

Question 16.
Which base is absent in RNA:
(a) Adenine
(b) Guanine
(c) Thymine
(d) Uracil.
Answer:
(c) Thymine

Question 17.
Protein controlling the chemical reactions of the body is:
(a) Enzyme
(b) Vitamin
(e) Hormone
(d) Globulin.
Answer:
(a) Enzyme

Question 18.
Snake poison is a type of protein:
(a) Fibrous protein
(b) Globular protein
(c) Lipoprotein
(d) None of these.
Answer:
(b) Globular protein

Question 19.
Nucleic acids were discovered by:
(a) Watson
(b) Crick
(c) Miescher
(d) Beadle.
Answer:
(c) Miescher

Question 20.
Wood and cotton having higher amount of:
(a) Cellulose
(b) Cutin
(c) Chitin
(d) All of these.
Answer:
(a) Cellulose

Question 21.
Insulin is a type of:
(a) Pigment
(b) Food
(c) Protein
(d) Gas.
Answer:
(c) Protein

Question 22.
Unit of polysaccharide is:
(a) Monosaccharides
(b) Oligosaccharides
(o) Peptides
(d) Fatty acids.
Answer:
(a) Monosaccharides

Question 23.
Doctor generally prescribing to take oil in place of fats because:
(a) Oil contains unsaturated fatty acids
(b) It reduces the amount of cholesterol in blood
(c) They are quickly digested
(d) All of these.
Answer:
(d) All of these.

Question 24.
Double stranded model of DNA was proposed by:
(a) Watson and Crick
(b) Miescher
(c) Muller
(d) None of these.
Answer:
(a) Watson and Crick

Question 25.
How RNA is synthesized:
(a) By DNA
(b) By RNA
(c) By ribosomes
(d) By lysosomes.
Answer:
(a) By DNA

Question 26.
Energy providing substance in our body is:
(a) Carbohydrate
(b) Protein
(c) Fat
(d) Vitamin.
Answer:
(a) Carbohydrate

Question 27.
Heparin of the blood is a:
(a) Glycoprotein
(b) Nucleoprotein
(c) Chromoprotein
(d) Glycolipid.
Answer:
(a) Glycoprotein

Question 28.
Term protein was coined by:
(a) Camilo Golgi
(b) Bergellius
(c) Bloor
(d) Funk.
Answer:
(b) Bergellius

Question 29
Mucin of saliva is:
(a) Nucleoprotein
(b) Glycoprotein
(c) Lipoprotein
(d) Chromoprotein.
Answer:
(b) Glycoprotein

Question 30.
The unit formed by β-glycosidic linkage between sugar and base is called as:
(a) Nucleotide
(b) Nucleoside
(c) Glycoside
(d) Punne.
Answer:
(b) Nucleoside

Question 31.
mRNAispolymerof:
(a) Deoxyribonucleoside
(b) Ribonucleoside
(c) Deoxyribonuc1eotk
(d) Ribonucleotide.
Answer:
(d) Ribonucleotide.

Question 32.
Purine of RNA are:
(a) Guanine and adenine
(b) Uracil and thymine
(c) Adenine and cytosine
(d) Uracil and guanine.
Answer:
(a) Guanine and adenine

Question 33.
Enzyme which convert the starch into sugar is:
(a) Hydrolases
(b) Lipases
(c) Amylases
(d) Nucleases.
Answer:
(c) Amylases

Question 34.
Protein part 01 enzyme is called:
(a) Apoenzyme
(b) Coenzyme
(c) Holoenzyme
(d) Cofactor.
Answer:
(a) Apoenzyme

Question 35.
Peptide bonds are formed between:
(a) Amino acids
(b) Glucose molecule
(c) Sucrose molecule
(d) Glycerol molecule.
Answer:
(a) Amino acids

Question 36.
How enzymes differ from organic catalysts :
(a) Their higher rate of diffusion
(b) Active under high temperature
(c) Active under low temperature
(d) Protein nature.
Answer:
(d) Protein nature.

Question 37.
Which ¡s not a character of enzyme:
(a) They are protein
(b) They increase the rate of biochemical processes
(c) They are specific for a reaction
(d) They are used and consumed in reaction
Answer:
(d) They are used and consumed in reaction

Question 38.
Protein synthesis occurs in:
(a) Ribosomes
(b) Mitochondria
(c) Chromosomes
(d) Centrosomes.
Answer:
(a) Ribosomes

Question 39.
Digestive enzymes in cells are found in:
(a) Ribosomes
(b) Lysosomes
(c) Golgi body
(d) Cell wall.
Answer:
(b) Lysosomes

Question 40.
Who said that enzymes are protein:
(a) Pasteur
(b) Leeuwenhoek
(c) Miller
(d) Sumner.
Answer:
(d) Sumner.

Question 41.
Non-protein part of enzyme ¡s called:
(a) Apoenzyme
(b) Holoenzyme
(c) Prosthetic group
(d) All of these.
Answer:
(c) Prosthetic group

Question 42.
Inorganic prosthetic group of enzyme is called:
(a) Coenzyme
(b) Activator
(c) Hormone
(d) All of these.
Answer:
(b) Activator

Question 43.
Substrate of the amylase enzyme is:
(a) Fat
(b) Protein
(c) Starch
(d) Sucrose.
Answer:
(c) Starch

Question 44.
Diastase enzyme digest:
(a) Starch
(b) Protein
(c) Fat
(d) Amino acid.
Answer:
(a) Starch

Question 45.
The percentage of enzyme in mitochondria is:
(a) 19%
(b) 70%
(c) 14%
(d) 10%.
Answer:
(b) 70%

Question 46.
Esterase enzyme belong to group known as:
(a) Oxidation-reduction
(b) Carhoxylating
(c) Hydrolyzing
(d) Transferase.
Answer:
(c) Hydrolyzing

Question 47.
NADPisa:
(a) Enzyme
(b) Part of sRNA
(c) Coenzyme
(d) Part of tRNA.
Answer:
(c) Coenzyme

Question 48.
Coenzyme is often a :
(a) Carbohydrate
(b) Protein
(c) Vitamin
(d) Fatty acid.
Answer:
(c) Vitamin

Question 49.
Temperature is increased from 3°C to 40°C. The rate of enzyme controlled
biochemical reaction will:
(a) Not change
(b) Increase
(c) Increase initially and then decrease
(d) Decrease.
Answer:
(c) Increase initially and then decrease

Question 50.
Enzymes, vitamins and hormones are common in :
(a) Being proteinaceous
(b) Being synthesized in the body of organisms
(c) Enhancing oxidative metabolism
(d) Regulating metabolism.
Answer:
(d) Regulating metabolism.

2. Fill in the blanks:

1. The molecule on which an enzyme acts is known as ……………….
Answer:
Substrate

2. The inactive form of an enzyme is known as ……………….
Answer:
Proenzyme

3. ……………… is the respiratory pigment of higher animals.
Answer:
Haemoglobin

4. Uracil nitrogenous bases is only found in ……………….
Answer:
RNA

5. A.T.P. is a ……………… nucleotide.
Answer:
Higher

6. Proteins are molecules of C, H, O and ………………….
Answer:
N(Nitrogen)

7. A vitamin is often associated as a ……………….. with an enzyme.
Answer:
Coenzyme

8. Biochemical reactions are regulated by catalysts called ……………….
Answer:
Enzymes

9. Enzymes which breakdown compounds without the involvement of water are
called ………………..
Answer:
Lyases

10. A compound with almost similar structure to the substrate can act as a ………………
Answer:
Competitive inhibitor

3. Match the following:

(A)

Answer:
1. (c) Genes
2. (d) Roughage,
3. (a) Mucin
4. (e) Milk.
5. (b) Lipid

(B)

Answer:
1. (b) Nucleus,
2. (d) Retinol
3. (e) Algae
4. (a) Protein synthesis
5. (c) Schleiden-Schwann

(C)

Answer:
1. (d) NADP
2. (e) Starch
3. (a) Cofactor
4. (b) Carbonic anhydrase
5. (c) Non-competitive inhibitor

4. Write true or false:

1. The wood contains sufficient amount of cellulose.
Answer:
True

2. An element, not present in a nitrogen base is carbon.
Answer:
False

3. Glucose is stored as glycogen in liver.
Answer:
True

4. The steroid, which has antifertility properties is protein.
Answer:
False

5. Most water present in mature plant cell is found in nucleus.
Answer:
False

6. The catalytic activity of inorganic catalyst cannot be changed by any regulating molecule.
Answer:
True
7. Salivary amylase act best at 6.6 p^H.
Answer:
False

8. Phosphohexose isomerase enzyme is classified under transferases.
Answer:
False

9. The enzyme pyruvic decarboxylase catalyses, the removal of C0² from the substrate pyruvic acid.
Answer:
True

10. Enzymes retain their activity even when extracted from cells.
Answer:
True

5. Answer in one word:

1. Process by which edible oils are converted into hard fats is ……………..
Answer:
Hydrogenation

2. Glycogen is stored in animal body in ………………
Answer:
Liver

3. Molecule in the cell, which undergo self replication is ………………..
Answer:
DNA

4. Glycosidic bond is found in ……………….
Answer:
Disaccharide

5. The genetic material in Tobacco Mosaic Virus is …………….
Answer:
RNA

6. Protein part of enzymes is ……………………….
Answer:
Apoenzyme

7. Chemically enzymes are ………………..
Answer:
Proteins

8. Inorganic prosthetic group of enzyme is ………………..
Answer:
Activator

9. Protein synthesis occurs in ………………
Answer:
Ribosome

10. Substrate of amylase enzyme is ………………..
Answer:
Starch

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

(c) Differences between Aerobic respiration and Fermentation.

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

(b) Differences between Glycolysis and Fermentation

(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

(c) Differences between Aerobic respiration and Fermentation.

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)

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

(B)

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:

(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:

(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 :

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)

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)

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

(C)

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

(D)

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

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)

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

(B)

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

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

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

(b) Differences between Plasma and Serum :

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

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

(b) Differences between Heart muscle and Striated muscle

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

(d) Differences between Adipose Tissue and Blood Tissue

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

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

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

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)

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

(B)

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

(C)

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)

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

(E)

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)

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

(B)

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

(C)

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.