NCERT MCQ CLASS-9 CHAPTER-8 | ENGLISH NCERT MCQ | | BEEHIVE IMPORTANT MCQ | | REACH FOR THE TOP | EDUGROWN

In This Post we are  providing Chapter-8 Reach For The Top NCERT MCQ for Class 9 English which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON REACH FOR THE TOP

Question 1: Santosh had —- brothers/sisters.
(a) Five brothers
(b) Four sisters
(c) Two Brothers and 1 sister
(d) Four brothers

Answer :  A

Question 2. What happened when Santosh scaled up the mount Everest for the first time?
(a) she won
(b) she hoisted the flag
(c) she became the youngest woman to achieve the feat.correct
(d) None

Answer :  C

Question 3. What does “Santosh” mean?
(a)Bravery
(b)Beautiful
(c)Educated
(d)Contentment

Answer :  D

Question 4. What is considered the roof of the world?
(a) sky
(b) stars
(c) top of Mt Everest
(d) all of the above

Answer :  C

Question 5: At the age of —- Santosh Yadav scaled Mt. Everest for the first time.
(a) Sixteen
(b) Fifteen
(c) Twenty
(d) Thirty

Answer:    C

Question 6. During her training period, as a young tennis player, Maria went to bed at
(a)7pm
(b)9pm
(c)8:30pm
(d)8pm

Answer :  D

Question 7. What is the name of Santosh Yadav’s village?
(a) Joniya
(b) Hariya
(c) Bariya
(d) Jenna

Answer :  A

Question 8: The marital age of that society was
(a) Fifteen
(b) Twelve
(c) Sixteen
(d) Eighteen

Answer:    C

Question 9. When was little Maria packed off to U.S and for what purpose and with whom?
(a) at the age of 9
(b) at the age of 14
(c) at the age of 15
(d) at the age of 10 for training in tennis with her father

Answer:    D

Question 10. Santosh went to the local village school because
(a)That was her desire
(b)To maintain the prevailing family custom
(c)To make her parents happy
(d)None of the above

Answer :  B

Question 11. Why was Santosh under pressure?
(a) she had to study
(b) her parents were asking her to get married
(c) she wanted to work
(d) none

Answer :  B

Question 12: After completing her high school study Santosh went to
(a) MP
(b)Jodhpur
(c) Jaipur
(d) New Delhi

Answer:   C

Question 13 – Why was the holy man who gave blessings to Santosh’s mother surprised?
(a) because grandmother was seeking blessing for a daughter against the common trend of son
(b) because he saw her first time
(c) because he had never seen such a woman
(d) none

Answer:   A

Question 14- When did Santosh leave home for Delhi?
(a) at the age of 10
(b) at the age of 14
(c) at the age of 15
(d) at the age of 16

Answer:   D

Question 15 – What made Santosh’s parents ready to pay her school fees at Delhi?
(a) her plan to work part time to pay her school fee to get education
(b) her shouting
(c) her tears
(d) her obstinacy

Answer:   A

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CLASS 11th CHAPTER -15 Plant Growth and Development |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 15 Plant Growth and Development includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -15 Plant Growth and Development. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -15 Plant Growth and Development | NCERT BIOLOGY SOLUTION |

Exercises

Page No: 253

1. Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.
 
Answer
 • Growth is an irreversible permanent increase in size of an organ or its parts or even of an individual cell.

• Differentiation is the process in which the cells derived from root apical and shoot-apical meristems and cambium differentiate and mature to perform specific functions.
 
• Development is a term that includes all changes that an organism goes through during its life cycle from germination of the seed to senescence.
 
• Dedifferentiation is the process in which permanent plant cells regain the power to divide under certain conditions.
 
• Redifferentiation is the process in which de-differentiated cells become mature again and lose their capacity to divide.
 
• Determinate growth is the type of growth in which growth stops after a certain phase.
 
• Meristem are the specialised regions in the plants where active cell divisions take place.
 
• The increased growth per unit time is termed as growth rate.
 
2. Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant?
 
Answer
Growth, at a cellular level, is principally a consequence of increase in the amount of protoplasm. Measuring the growth of protoplasm involves many parameters such as the weight of the fresh tissue sample, the weight of the dry tissue sample, the differences in length, area, volume, and cell number measured during the growth period. Hence, there cannot be one parameter good enough to demonstrate growth throughout the life of a flowering plant.
 
3. Describe briefly:
(a) Arithmetic growth
(b) Geometric growth
(c) Sigmoid growth curve
(d) Absolute and relative growth rates
 
Answer
(a) In arithmetic growth, following mitotic cell division, only one daughter cell continues to divide while the other differentiates and matures. The elongation of roots at a constant rate is an example of arithmetic growth. On plotting length of the organ against time, a linear curve is obtained. Mathematically, it is expressed as:
L= L0 + rt
Here, Lt is length at time ‘t’, L0 is length at time 0 and r is the rate per unit time.

(b) In most systems, the initial growth is slow and called lag phase, and it increases rapidly thereafter at an exponential rate and called log or exponential phase. Here, both the progeny cells following mitotic cell division retain the ability to divide and continue to do so. However, with limited nutrient supply, the growth slows down leading to a stationary phase. The graph of the geometric growth gives a sigmoid curve.
 
(c) A sigmoid curve is a characteristic of living organism growing in a natural environment. This curve is divided into three phases – lag phase, log phase or exponential phase of rapid growth, and stationary phase.
Sigmoid curve Graph
Exponential growth can be expressed as:
W1= W0ert
W1 = final size (weight, height, number etc.)
W0 = initial size at the beginning of the period
r = growth rate
t = time of growth
e = base of natural logarithms
 
(d) The measurement and the comparison of total growth per unit time is called the absolute growth rate. 
The growth of the given system per unit time expressed on a common basis, e.g., per unit initial parameter is called the relative growth rate.
 
4. List five main groups of natural plant growth regulators. Write a note on discovery, physiological functions and agricultural/horticultural applications of any one of them.
 
Answer
The five main groups of natural plant growth regulators are:
(i) Auxins
(ii) Gibberellic acid
(iii) Cytokinins
(iv) Ethylene
(v) Abscisic acid
A note on discovery, physiological functions and agricultural/horticultural applications of Auxins are:
 
→ Discovery: The first observations regarding the effects of auxins were made by Charles Darwin and Francis Darwin when they observed that the coleoptiles of canary grass responded to unilateral illumination by growing towards the light source (phototropism).
After a series of experiments, it was concluded that the tip of coleoptile was the site of transmittable influence that caused the bending of the entire coleoptile. Auxin was isolated by F.W. Went from tips of coleoptiles of oat seedlings.

→ Physiological Functions:
• They control plant cell-growth.
• They cause the phenomenon of apical dominance.
• They control division in the vascular cambium and xylem differentiation.
• They induce parthenocarpy and prevent abscission of leaves and fruits.

→ Horticulture Application:
• They help to initiate rooting in stem cuttings, an application widely used for plant propagation.
• 2-4 D is used weedicide to kill broadleaf, dicotyledonous weeds.
• They induce parthenocarpy in tomatoes.
• It promote flowering e.g. in pineapples.

Page No: 254

5. What do you understand by photoperiodism and vernalisation? Describe their significance.
 
Answer
 
The response of plants to periods of day/night is termed as photoperiodism. It is hypothesised that the hormonal substance responsible for flowering is formed in the leaves, subsequently migrating to the shoot apices and modifying them into flowering apices. Photoperiodism helps in studying the response of flowering in various crop plants with respect to the duration of exposure to light.

There are plants for which flowering is either quantitatively or qualitatively dependent on exposure to low temperature. This phenomenon is termed vernalisation. It refers specially to the promotion of flowering by a period of low temperature. It prevents precocious reproductive development late in the growing season, and enables the plant to have sufficient time to reach maturity.

6. Why is Abscisic acid also known as stress hormone?
 
Answer
Abscisic acid stimulates the closure of stomata in the epidermis and increases the tolerance of plants to various kinds of stresses. Therefore, it is also called the stress hormone. It promotes seed dormancy and ensures seed germination during favourable conditions. It helps seeds withstand desiccation. It also helps in inducing dormancy in plants at the end of the growing season and promotes abscission of leaves, fruits, and flowers.

7. ‘Both growth and differentiation in higher plants are open’. Comment.
 
Answer
The higher plants retain the capacity for unlimited growth throughout their life. This ability of the plants is due to the presence of meristems at certain locations in their body. The cells of such meristems have the capacity to divide and self-perpetuate. Therefore, growth in higher plants are open. Also, some of these cells always undergo differentiation after some rounds of cell division. Hence, the differentiation is also open.
 
8. ‘Both a short day plant and a long day plant can flower simultaneously in a given place’. Explain.
 
Answer
Flowering in some plants depends on relative durations of light and dark periods. The short-day plant and long-day plant can flower at the same place, provided they have been given an adequate photoperiod.
 
9. Which one of the plant growth regulators would you use if you are asked to:
(a) Induce rooting in a twig
► Auxins
 
(b) Quickly ripen a fruit
► Ethylene

(c) Delay leaf senescence
► Cytokinins

(d) Induce growth in axillary buds
► Cytokinins

(e) ‘Bolt’ a rosette plant
► Gibberellins

(f) Induce immediate stomatal closure in leaves.
► Abscisic acid
 
10. Would a defoliated plant respond to photoperiodic cycle? Why?

Answer
No, a defoliated plant will not respond to the photoperiodic cycle as leaves are the sites of perception of light/dark duration. Therefore, in the absence of leaves, the plant would not respond to light.
 
11. What would be expected to happen if:
(a) GA3 is applied to rice seedlings
(b) Dividing cells stop differentiating
(c) A rotten fruit gets mixed with unripe fruits
(d) You forget to add cytokinin to the culture medium.
 
Answer
 
(a) If GA3 is applied to rice seedlings, then the rice seedlings will show internode-elongation and increase in height.

(b) If dividing cells stop differentiating, then the plant organs such as leaves and stem will not be formed. The mass of undifferentiated cell is called callus.

(c) If a rotten fruit gets mixed with unripe fruits, then the ethylene produced from the rotten fruits will hasten the ripening of the unripe fruits.

(d) If you forget to add cytokinin to the culture medium, then cell division, growth, and differentiation will be slower.
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CLASS 11th CHAPTER -14 Respiration in Plants |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 14 Respiration in Plants includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -14 Respiration in Plants. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -14 Respiration in Plants | NCERT BIOLOGY SOLUTION |

Page No: 238
 
Exercises
 
1. Differentiate between
(a) Respiration and Combustion
(b) Glycolysis and Krebs’cycle
(c) Aerobic respiration and Fermentation
 
Answer

(a) Respiration and Combustion

Respiration
Combustion
It is a biochemical process.It is a physiochemical process.
It occurs inside the cells.It does not occur inside cells.
It is a biologically-controlled process.It is an uncontrolled process.
No light is producedLight may be produced
ATP is generated.ATP is not generated.
Enzymes are requiredEnzymes are not required

(b) Glycolysis and Krebs’cycle

Glycolysis
Krebs’cycle
The breakdown of glucose to pyruvic acid is called glycolysis.The further processing of pyruvic acid through aerobic route is called Krebs’ cycle.
It is a linear pathway.It is a cyclic pathway.
It occurs in the cell cytoplasm.It occurs in the mitochondrial matrix.
It occurs in all living beings.It occurs in aerobic respiration.
It generates 2 NADH2 and 2 ATP molecules on the breakdown of one glucose molecule.It produces 6 NADH2, 2 FADH2, and 2 ATP molecules on the breakdown of two acetyl-CoA molecules.
No carbon dioxide evolvedCarbon dioxide evolved

(c) Aerobic respiration and Fermentation

Aerobic respiration
Fermentation
The complete oxidation of organic substances in the presence of oxygen.The incomplete oxidation of glucose in the absence of oxygen.
It occurs in the cytoplasm and mitochondriaIt occurs in the cytoplasm
The end products are carbon dioxide and waterThe end products are ethyl alcohol and carbon dioxide
Complete oxidation of the respiratory substrate takes placeIncomplete oxidation of the respiratory substrate takes place
About 36 ATP molecules are producedOnly 2 ATP molecules are produced


2. What are respiratory substrates? Name the most common respiratory substrate.
 
Answer
A compound which is oxidized during respiration is called respiratory substrate. Glucose is the most common respiratory substrate.
 
3. Give the schematic representation of glycolysis?
Answer
 
schematic representation of glycolysis
 
4. What are the main steps in aerobic respiration? Where does it take place?
 
Answer
 
The main steps in aerobic respiration and places of their occurrence are:
→ Glycolytic breakdown of glucose in to pyruvic acid in cytoplasm.
→ Krebs cycle in matrix of mitochondria
→ Electron transport system in inner mitochondrial membrane
→ Oxidative phosphorylation in the inner mitochondrial membrane.
 
5. Give the schematic representation of an overall view of Krebs cycle.
 
Answer
 
Krebs Cycle
 
6. Explain ETS.
 
Answer
The metabolic pathway through which the electron passes from one carrier to another is called the electron transport system (ETS). It is located in the inner mitochondrial membrane. Electrons from NADH produced in the mitochondrial matrix during citric acid cycle are oxidised by an NADH dehydrogenase (complex I), and electrons are then transferred to ubiquinone located within the inner membrane. Ubiquinone also receives reducing equivalents via FADH2 (complex II) that is generated during oxidation of succinate in the citric acid cycle. The reduced ubiquinone (ubiquinol) is then oxidised with the transfer of electrons to cytochrome c via cytochrome bc 1 complex (complex III). The cytochrome c acts as a mobile carrier between complex III and cytochrome c oxidase complex, containing cytochrome a and a3, along with copper centres (complex IV).
During the transfer of electrons from each complex, the process is accompanied by the production of ATP from ADP and inorganic phosphate by the action ATP synthase (complex V). The amount of ATP produced depends on the molecule, which has been oxidized. 2 ATP molecules are produced by the oxidation of one molecule of NADH. One molecule of FADH2, on oxidation, gives 3 ATP molecules.
 
7. Distinguish between the following:
(a) Aerobic respiration and Anaerobic respiration
(b) Glycolysis and Fermentation
(c) Glycolysis and Citric acid Cycle
 
Answer

(a) Aerobic respiration and Anaerobic respiration

Aerobic respiration
Anaerobic respiration
It takes place in the presence of oxygen.It takes place in the absence of oxygen.
It occurs in cytoplasm and mitochondria.It occurs in cytoplasm
The end products of aerobic respiration are carbon dioxide and water.The end products of fermentation are ethyl alcohol and carbon-dioxide.
Complete oxidation of respiratory substrate takes place.Incomplete oxidation of respiratory substrate takes place.
36-38 ATP molecules are produced.Only 2 ATP molecules are produced.

(b) Glycolysis and Fermentation

Glycolysis
Fermentation
Glycolysis occurs during aerobic and anaerobic respiration.Fermentation is a type of anaerobic respiration.
The end products of aerobic respiration is pyruvic acidThe end products of aerobic respiration is ethanol or lactic acid

(c) Glycolysis and Citric acid Cycle

Glycolysis
Citric acid Cycle
It is a linear pathway.It is a cyclic pathway.
It occurs in the cell cytoplasm.It occurs in the mitochondrial matrix.
It occurs in all living beings.It occurs in aerobic respiration.
It generates 2 NADH2 and 2 ATP molecules on the breakdown of one glucose molecule.It generates 6 NADH2, 2 FADH2, and 2 ATP molecules on the breakdown of two acetyl-CoA molecules.

8. What are the assumptions made during the calculation of net gain of ATP?
 
Answer
 
The following assumptions are made during the calculation of net gain of ATP:
• There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
• The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
• None of the intermediates in the pathway are utilised to synthesise any other compound.
• Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.
 
9. Discuss “The respiratory pathway is an amphibolic pathway.”
Answer
The breaking down process occurring within the organism is known as catabolic process and the synthesis process is known as anabolism. The respiratory pathway involves both the catabolic and anabolic process, so it is referred as amphibolic pathway.
 
10. Define RQ. What is its value for fats?
 
Answer
The ratio of the volume of CO2 evolved to the volume of O2 consumed in respiration is called the respiratory quotient (RQ). The RQ for carbohydrates is 1. The RQ for fat and protein is less than 1.
For example: Calculations for a fatty acid, tripalmitin
2(C5H98O6) + 145O2 → 102 CO2 + 98 H2O
RQ = Volume of CO2 evolved/Volume of O2 consumed
      = 102 CO2/145O2 = 0.7

11. What is oxidative phosphorylation?
 
Answer
 
The generation of ATP from ADP during electron transport system by utilizing the energy obtained during oxidative reaction is called oxidative phosphorylation.
 
12. What is the significance of step-wise release of energy in respiration?
 
Answer
 
The significance of step-wise release of energy in respiration are:
→ It facilitates the utilization of the relatively higher proportion of the energy in ATP synthesis.
→ The activities of enzymes for the different steps may be enhanced or inhibited by specific compounds. This provides a mean of controlling the rate of the pathway and the energy output according to the need of the cell.
→ The same pathway may be utilized for forming intermediates used in the synthesis of other biomolecules like amino acids.
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CLASS 11th CHAPTER – 13 Photosynthesis in Higher Plants |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 13 Photosynthesis in Higher Plants includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -13 Photosynthesis in Higher Plants. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -13 Photosynthesis in Higher Plants | NCERT BIOLOGY SOLUTION |

Page No: 224
 
Exercises
 
1. By looking at a plant externally can you tell whether a plant is C3 or C4? Why and how?
 
Answer
 
We can’t tell whether a plant is C3 or C4 by looking at a plant externally. However, plants which which are adapted to dry climates follow the C4 pathway. Unlike C3 plants, the leaves of Cplants have a special anatomy but this difference can only be observed at the cellular level.

2. By looking at which internal structure of a plant can you tell whether a plant is C3 or C4? Explain.
 
Answer
 
As leaves of C4 plants have a special anatomy called Kranz anatomy. This makes them different from C3 plants. Special cells, known as bundle-sheath cells, surround the vascular bundles. These cells have a large number of chloroplasts. They are thick-walled and have no intercellular spaces. Therefore, we can tell whether a plant is C3 or C4 through internal structure.
 
3. Even though a very few cells in a C4 plant carry out the biosynthetic – Calvin pathway, yet they are highly productive. Can you discuss why?
 
Answer

In C4 plants photorespiration does not occur because they have a mechanism that increases the concentration of CO2 at the enzyme site. This takes place when the C4 acid from the mesophyll is broken down in the bundle sheath cells to release CO2 that results in increasing the intracellular concentration of CO2. In turn, this ensures that the RuBisCO functions as a carboxylase minimising the oxygenase activity. Thus, the photosynthesis rate increases and make C4 plants more productive.

Page No: 225

4. RuBisCo is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisCo carries out more carboxylation in C4 plants?
 
Answer
 
The enzyme RuBisCo is absent from the mesophyll cells of C4 plants. It is present in the bundle-sheath cells surrounding the vascular bundles. In Cplants, the Calvin cycle occurs in the bundle-sheath cells. The primary COacceptor in the mesophyll cells is phosphoenol pyruvate -a three-carbon compound. It is converted into the four-carbon compound oxaloacetic acid (OAA). OAA is further converted into malic acid. Malic acid is transported to the bundle-sheath cells, where it undergoes decarboxylation and CO2 fixation occurs by the Calvin cycle. This prevents the enzyme RuBisCo from acting as an oxygenase.
 
5. Suppose there were plants that had a high concentration of Chlorophyll b, but lacked chlorophyll a, would it carry out photosynthesis? Then why do plants have chlorophyll b and other accessory pigments?
 
Answer
 
If there were complete absence of chlorophyll a in a plant, it would not carry out photosynthetic activity at all because chlorophyll a is the chief pigment associated with photosynthesis as it traps light. Other accessory pigments like chlorophyll b, santhophylls and carotenoids are equally essential as they also absorb light and transfer energy to chlorophyll a. They also enable a wider range of wavelength of incoming light to be utilised for photosynthesis and protect chlorophyll a from photo- oxidation.
 
6. Why is the colour of a leaf kept in the dark frequently yellow, or pale green? Which pigment do you think is more stable?
 
Answer
 
Chlorophyll or green pigment is unable to absorb energy in the absence of light therefore loses its stability. Thus, the colour of leaf changes to yellow or pale green. This shows that Carotenoids and Xanthophyll pigments are more stable.
 
7. Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?
 
Answer
The plants placed in light will have darker leaves as compared to leaves of a plant placed in shade. As leaves in shade get lesser light for photosynthesis so they perform lesser photosynthesis as compared to the leaves or plants kept in sunlight. To increase the rate of photosynthesis, the leaves present in shade have more chlorophyll pigments. This increase in chlorophyll content increases the amount of light absorbed by the leaves, which in turn increases the rate of photosynthesis which makes the leaves or plants in shade greener than the leaves or plants kept in the sun.
 
8. Figure 13.10 shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions:
(a) At which point/s (A, B or C) in the curve is light a limiting factor?

(b) What could be the limiting factor/s in region A?
(c) What do C and D represent on the curve?
 
Answer
 
Figure 13.10
 
(a) At point A
(b) Light is a limiting factor also, water, temperature, and the concentration of carbon dioxide could also be limiting factors in the region A.
(c) C represents the stage beyond which light is not a limiting factor. D represents the stage beyond which intensity of light has no effect on the rate of photosynthesis.
 
9. Give comparison between the following:
(a) C3 and C4 pathways
(b) Cyclic and non-cyclic photophosphorylation
(c) Anatomy of leaf in C3 and C4 plants
 
Answer
 
(a) C3 and C4 pathways

Cpathways
Cpathways
The primary acceptor of CO is RUBP – a five-carbon compound.The primary acceptor of CO2 is phosphoenol pyruvate – a three-carbon compound.
The first stable product is 3 phosphoglycerate.The first stable product is oxaloacetic acid.
It occurs only in the mesophyll cells of the leaves.It occurs in the mesophyll and bundle-sheath cells of the leaves.
It is a slower process of carbon fixation and photo-respiratory losses are high.It is a faster process of carbon fixation and photo-respiratory losses are low.

(b) Cyclic and non-cyclic photophosphorylation

Cyclic photophosphorylation
Non-cyclic photophosphorylation
It occurs only in photosystem I.It occurs both in photosystems I and II.
It involves only the synthesis of ATP.It involves the synthesis of ATP and NADPH2.
In this process, photolysis of water does not occur. Therefore, oxygen is not produced.In this process, photolysis of water takes place and oxygen is liberated.
In this process, electrons move in a closed circle.In this process, electrons do not move in a closed circle.

(c) Anatomy of leaf in C3 and C4 plants

Anatomy of leaf in C3
Anatomy of leaf in C4
Bundle-sheath cells are absentBundle-sheath cells are present
RuBisCo is present in the mesophyll cellsRuBisCo is present in the bundle-sheath cells.
The first stable compound produced is 3-phosphoglycerate – a three-carbon compound.The first stable compound produced is oxaloacetic acid – a four-carbon compound.
Photorespiration occursPhotorespiration does not occur
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CLASS 11th CHAPTER – 12 Mineral Nutrition |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 12 Mineral Nutrition includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -12 Mineral Nutrition. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -12 Mineral Nutrition | NCERT BIOLOGY SOLUTION |

Page No: 205
 
Exercises
 
1. ‘All elements that are present in a plant need not be essential to its survival’. Comment.
 
Answer
Plants tend to absorb different kinds of nutrients from soil but all of them are not essential to its survival because they do not directly involved in the composition of their body. A nutrient is inessential for a plant if it is not directly involved in the plant’s physiology and metabolism. However, if the concentration of micronutrients such as Fe, Mn, Cu, Zn, Cl, etc., rise above their critical values, they appear to be toxic for the plant.

2.Why is purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics?
 
Answer
As in hydroponics, the plants are grown in a nutrient solution in the absence of soil. Since the plant roots are exposed to a limited amount of the solution, there are chances that the concentrations of oxygen and other minerals in the plant roots would reduce. Therefore, in studies involving mineral nutrition using hydroponics, purification of water and nutrient salts is essential so as to maintain an optimum growth of the plants.
 
3. Explain with examples: macronutrients, micronutrients, beneficial nutrients, toxic elements and essential elements.
 
Answer
→ Macronutrients: Elements which are present in large amounts in plant tissues are called macronutrients. They are present in plant tissues in amounts more than 10 mmole kg–1 of dry matter. Carbon, hydrogen, oxygen, nitrogen, phosphorous, sulphur, potassium, calcium and magnesium are examples.

→ Micronutrients: Elements which are present in small amounts i.e. less than 10 mmole per kg of dry matter are called micronutrients. Iron, manganese, copper, molybdenum, zinc, copper, boron, chlorine and nickel are examples.
 
→ Beneficial Elements: The nutrients that may not be essential, but are beneficial to plants are called beneficial elements. Sodium, silicon, cobalt and selenium are examples.
 
→ Toxic elements: Any mineral ion concentration in tissues that reduces the dry weight of tissues by about 10 per cent is considered toxic. For example, when manganese is present in large amounts, it induces deficiencies of iron, magnesium, and calcium by interfering with their metabolism.

→ Essential elements: The elements which are absolutely necessary for plant growth and reproduction are called essential elements. The requirement of these elements is specific and non-replaceable. These elements are further divided into two broad categories based on their quantitative requirements.

4. Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency.
 
Answer
Five different deficiency symptoms in plants are:

→ Chlorosis is the loss of chlorophyll leads to the yellowing of leaves. It is caused by the deficiencies of nitrogen, potassium, magnesium, sulphur, iron, manganese, zinc, and molybdenum.

→ Necrosis is the death of plant tissues as a result of the deficiencies of calcium, magnesium, copper, and potassium.

→ Inhibition of cell division is caused by the deficiencies of nitrogen, potassium, sulphur, and molybdenum.

→ Delayed flowering is caused by the deficiencies of nitrogen, sulphur, and molybdenum.

→ Stunted plant growth is a result of the deficiencies of copper and sulphur.
 
5. If a plant shows a symptom which could develop due to deficiency of more than one nutrient, how would you find out experimentally, the real deficient mineral element?
 
Answer
Every element shows certain characteristic deficiency symptoms in the plants also different plants respond in different ways to the deficiency of the same nutrient. Thus, in order to identify the real deficient mineral element, we need note all the available symptoms in different parts of the plant. Then we have to compare the noted information with the available standard symptom table. 
 
6. Why is that in certain plants deficiency symptoms appear first in younger parts of the plant while in others they do so in mature organs?
 
Answer
Mobility of an element play very important role in determining whether the deficiency symptoms appear first in younger parts or older parts. Elements such as nitrogen, potassium, and magnesium are highly mobile. These elements move from the mature organs to the younger parts of a plant. Therefore, the symptoms for the deficiencies of these elements first appear in the older parts of the plant. Elements such as calcium and sulphur are relatively immobile. These elements are not transported out of the older parts of a plant. Therefore, the symptoms for the deficiencies of these elements first appear in the younger parts of the plant.
 
7. How are the minerals absorbed by the plants?
 
Answer
 
The absorption of minerals by plants takes by the roots of plants in two main phases: apoplast and symplast.
In the first phase, there is a rapid uptake of nutrients from the soil into the free spaces of plant cells. This process is passive and it usually occurs through trans-membrane proteins and ion-channels.
In the second phase, the ions are taken slowly into the inner spaces of the cells. This pathway generally involves the expenditure of energy in the form of ATP.
 
8. What are the conditions necessary for fixation of atmospheric nitrogen by Rhizobium. What is their role in N2 -fixation?
 
Answer
 
Rhizobium bacteria need symbiotic association with legume plants to carry out nitrogen fixation. Root nodules contain the necessary enzymes for nitrogen fixation and thus enable rhizobium to fix nitrogen. The enzyme nitrogenase facilitates the conversion of nitrogen into ammonia which is the first stable product of nitrogen fixation. Ammonia is then converted into glutamic acid. Glutamic acid is then utilised by plants to make amino acids which are ultimately utilised to make protein.
 
9. What are the steps involved in formation of a root nodule?
 
Answer
Nodule formation involves a sequence of multiple interactions between Rhizobium and roots of the host plant. These steps are:
→ Rhizobium bacteria contact a susceptible root hair and divides near it.
→ After successful infection, the root hairs get curled and are invaded by the bacteria.
→ This invasion is followed by the formation of an infection thread that carries the bacteria into the cortex of the root.
→ The bacteria get modified into rod-shaped bacteroides and cause inner cortical and pericycle cells to divide. Division and growth of cortical and pericycle cells lead to nodule formation.
→ The nodules finally get connected with the vascular tissues of the roots for nutrient exchange.
 
10. Which of the following statements are true? If false, correct them:
(a) Boron deficiency leads to stout axis. 
► True

(b) Every mineral element that is present in a cell is needed by the cell. 
► Every mineral element that is present in a cell is not needed by the cell.

(c) Nitrogen as a nutrient element, is highly immobile in the plants.
► Nitrogen as a nutrient element,is highly mobile in plants.

(d) It is very easy to establish the essentiality of micronutrients because they are required only in trace quantities.
► True
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CLASS 11th CHAPTER – 11 Transport in Plants |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 11 Transport in Plants includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -11 Transport in Plants . NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -11 Transport in Plants | NCERT BIOLOGY SOLUTION |

Exercises

Page No: 193

1. What are the factors affecting the rate of diffusion?
 
Answer
Factors affecting the rate of diffusion are:
→ Gradient of concentration
→ Permeability of the membrane separating them
→ Temperature and pressure
 
2. What are porins? What role do they play in diffusion?
 
Answer
Porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria allowing molecules up to the size of small proteins to pass through. They help in facilitating the passive transport of small-sized protein molecules.
 
3. Describe the role played by protein pumps during active transport in plants.
 
Answer
Proteins pumps are made up of specific proteins called trans-membrane proteins which use energy to carry substances across the cell membrane. These pumps can transport substances from a low concentration to a high concentration. The rate of transport reaches the maximum when all the protein transporters are being used or are saturated.
 
4. Explain why pure water has the maximum water potential.
 
Answer
Water molecules possess kinetic energy. In liquid and gaseous form, they are in random motion that is both rapid and constant. The greater the concentration of water in a system, the greater is its kinetic energy or water potential. Pure water has the highest concentration of water molecules. Therefore, it has the highest water potential. When some solute is dissolved in water, the water potential of pure water decreases.
 
5. Differentiate between the following:
(a) Diffusion and Osmosis
(b) Transpiration and Evaporation
(c) Osmotic Pressure and Osmotic Potential
(d) Imbibition and Diffusion
(e) Apoplast and Symplast pathways of movement of water in plants.
(f) Guttation and Transpiration.
Answer
 
(a) Diffusion and Osmosis
Diffusion
Osmosis
It is a movement of molecules from high concentration to low concentration.It is a movement of molecules from high concentration to low concentration through semi permeable membrane.
It can occur in solids, liquids, and gases.It occurs in the liquid medium.
It does not require any driving Force.It occurs in response to a driving force.

(b) Transpiration and Evaporation

Transpiration
Evaporation
It occurs in plants.It occurs from any free surface and involves living and non-living surfaces.
It occurs mainly through the stomatal pores on plant leaves.It occurs through any free surface.
It is controlled by environmental factors as well as physiological factors of plantsIt is entirely driven by environmental factors.

(c) Osmotic Pressure and Osmotic Potential

Osmotic Pressure
Osmotic Potential
It is expressed in bars with a positive sign.It is expressed in bars with a negative sign.
It is a positive pressure.It is a negative pressure.
Its value increases with an increase in the concentration of solute particles.Its value decreases with an increase in the concentration of solute particles.

(d) Imbibition and Diffusion

Imbibition
Diffusion
Imbibition is a special type of diffusion when water is absorbed by solids and colloids causing them to enormously increase in volume.Diffusion is the passive movement of particles, ions, and molecules along the concentration gradient.
It usually involves water.It involves solids, liquids, and gases.

(e) Apoplast and Symplast pathways of movement of water in plants.

Apoplast pathways
Symplast pathways
It is the system of adjacent cell walls that is continuous throughout the plant, except at the casparian strips of the endodermis in the rootsIt is the system of interconnected protoplasts.
Water moves through the intercellular spaces and the walls of cellsWater travels through the cytoplasm of cells and intercellular movement is through plasmodesmata.
It is a faster process of water movement and water moves through mass flow.It is a slower process of water movement.

(f) Guttation and Transpiration.

Guttation
Transpiration
It occurs through the vein endings of leaves.It occurs through the stomata.
Water is lost from the leaves in the form of liquid droplets.Water is lost from the leaves in the form of water vapour.
It occurs usually at night.It occurs usually during the day.
It is regulated by humidity, temperature and presence of water in soil.It is regulated by a number of external and internal factors such as relative humidity, temperature opening and closing of stomata, etc.

6. Briefly describe water potential. What are the factors affecting it?
 
Answer
 
Water potential is the potential energy of water to move from one part to the other during various cellular processes such as diffusion, osmosis, etc. It is denoted by the Greek letter Psi or Ψ and is expressed in Pascals (Pa). A system with higher concentration of water has a higher kinetic energy or water potential. Pure water has the highest water potential while Solutions have lower water potential than it.
Solute potential and pressure potential are the two factors which affect water potential.
 
7. What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer
If a pressure greater than atmospheric pressure is applied to pure water or a solution than its ater potential increases because of pressure potential which develops due to increased pressure. It is equivalent to pumping water from one place to another.

8. (a) With the help of well-labelled diagrams, describe the process of plasmolysis in plants, giving appropriate examples.

(b) Explain what will happen to a plant cell if it is kept in a solution having higher water potential.
 
Answer
 
(a) Plasmolysis occurs when water moves out of the cell and the cell membrane of a plant cell shrinks away from its cell wall. This happens when the plant cell is placed in a hypertonic solution (a solution having more solute concentration than the cell cytoplasm). This causes the water to move out of the cell and toward the solution. The cytoplasm of the cell shrinks and the cell is said to be plasmolysed. This process can be observed in an onion peel kept in a highly concentrated salt solution.
Plasamolysis
 
(b) A hypotonic solution has higher water potential. When a plant cell is placed in a hypotonic solution or a solution having higher water potential, water diffuses into the cell causing the cytoplasm to build up a pressure against the wall, called turgor pressure. This leads to the cell getting swollen. But the rigidity of the cell wall, prevents the cell from bursting.
 
9. How is the mycorrhizal association helpful in absorption of water and minerals in plants?
 
Answer
Mycorrhiza is a symbiotic association of a fungus with a root system. This fungal hyphae either form a dense network around the young roots or they penetrate the cells of the roots. The large surface area of the fungal hyphae is helpful in increasing the absorption of water and minerals from the soil. In return, they get sugar and nitrogenous compounds from the host plants.
 
10. What role does root pressure play in water movement in plants?
 
Answer
Root pressure is the positive pressure which can be responsible for pushing up water to small heights in the stem by the active absorption of nutrients from the soil. However, root pressure does not play a major role in water movement in tall plants. Root pressure contributes towards reestablishment of continuous chains of water molecules in the xylem which often break under enormous tensions created by transpiration pull.
 
11. Describe transpiration pull model of water transport in plants. What are the factors influencing transpiration? How is it useful to plants?
 
Answer
Transpiration creates a suction force inside the xylem. This suction force is called transpiration pull. This is powerful enough to pull the water column from beneath. Adhesion, cohesion and surface tension are the important physical properties of water which further help in the upward movement of water through xylem.
Factors that influence transpiration are temperature, light, humidity, wind speed, number and distribution of stomata, number of stomatal aperture with guard cells open, water status of the plant, canopy structure, etc.
It is useful to plants in following ways:
• creates transpiration pull for absorption and transport of plants
• supplies water for photosynthesis
• transports minerals from the soil to all parts of the plant
• cools leaf surfaces, sometimes 10 to 15 degrees, by evaporative cooling
 

• maintains the shape and structure of the plants by keeping cells turgid.

 
12. Discuss the factors responsible for ascent of xylem sap in plants.
 
Answer
 
Transpirational pull is responsible for the ascent of water in the xylem. This ascent of water is dependent on the following physical factors:
→ Cohesion – Mutual attraction between water molecules
→ Surface tension – Responsible for the greater attraction between water molecules in liquid phase than in gaseous phase
→ Adhesion – Attraction of water molecules to polar surfaces
→ Capillarity -Ability of water to rise in thin tubes
These physical properties give water high tensile strength, i.e., an ability to resist a pulling force and high capillarity, i.e., the ability to rise in thin tubes. The thin tubes of xylem work like capillary tubes.
 
13. What essential role does the root endodermis play during mineral absorption in plants?
 
Answer
In plants, nutrients are absorbed through the active and passive transports. The endodermis of roots have many transport proteins embedded in their plasma membrane. Transport proteins in endodermis cells enable plant cells to adjust the quantity and types of solutes to be absorbed from the soil. It regulates the quantity and type of minerals and ions that reach the xylem tissue of plants.
 
14. Explain why xylem transport is unidirectional and phloem transport bi-directional.
 
Answer
 
Water transported through xylem is utilised in photosynthesis and most of the water is lost through transpiration. Renewed demand for water is once again supplied through the same channel. Hence, transport through xylem is unidirectional.
 
In phloem, food is transported from source to sink. Leaves are the usual source and storage organs are the usual sink. But the storage organs become source when new buds emerge during early spring. Thus, a reverse flow of food is required. Hence, movement through phloem is bi-directional.
 
15. Explain pressure flow hypothesis of translocation of sugars in plants.
 
Answer
 
According to the pressure flow hypothesis, food is prepared in the plant leaves in the form of glucose. Before moving into the source cells present in the phloem, the prepared food is converted into sucrose. Water moves from the xylem vessels into the adjacent phloem, thereby increasing the hydrostatic pressure in the phloem. Consequently, the sucrose moves through the sieve cells of the phloem. The sucrose already present in the sink region is converted into starch or cellulose, thereby reducing the hydrostatic pressure in the sink cells. Hence, the pressure difference created between the source and the sink cells allows sugars to be translocated from the former to the latter. This starch or cellulose is finally removed from the sink cells through active transport.
 
16. What causes the opening and closing of guard cells of stomata during transpiration?
 
Answer

The immediate cause of the opening or closing of the Stomata is a change in the turgidity of the guard cells. The inner walls of the guard cells are thick and elastic, while the outer walls are thin. When turgidity increases within the two guard cells flanking each stomatal aperture or pore, the thin outer walls bulge out and force the inner walls into a crescent shape. The stomatal opening is facilitated by the radial arrangement of the microfibrils. At the time of the closing of the stomata, the guard cells lose their turgidity, the outer and inner walls retain their original shapes, and the microfibrils get arranged longitudinally.
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NCERT MCQ CLASS-9 CHAPTER-7 | ENGLISH NCERT MCQ | | BEEHIVE IMPORTANT MCQ | | PACKING | EDUGROWN

In This Post we are  providing Chapter-7 Packing NCERT MCQ for Class 9 English which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON PACKING

Question 1:

Packing is taken from the extract

(a) Three men in a Boat

(b) Three men in a train

(c) Two women in a Boat

(d) Three men in a car

Answer:

(a) Three men in a Boat.

Question 2:

The writer was proud of himself for his skill of

(a) Packing

(b) Reading

(c) Driving

(d) Telling

Answer:

(a) Packing.

Question 3:

The writer had —– friends with him.

(a) Two

(b) Four

(c) Three

(d) One

Answer:

(a) Two.

Question 4:

—- had spread himself over the easy chair.

(a) Jerome k Jerome

(b) George

(c) Harris

(d) Tom

Answer:

(b) George.

Question 5:

He felt irritated to see others when he was doing all the works

(a) Singing

(b) Working

(c) Sitting

(d) Dancing

Answer:

(c) Sitting.

Question 6:

After finishing his packing the writer sat on the

(a) Bag

(b) Bed

(c) Sofa

(d) Stool

Answer:

(a) Bag.

Question 7:

Who had reminded the writer about the boots, which he had forgotten to pack?

(a) Jerome

(b) Harris

(c) George

(d) Mark

Answer:

(b) Harris.

Question 8:

—– used to haunt the writer while travelling.

(a) Boots

(b) Butter

(c) Toothbrush

(d) T-shirt

Answer:

(c) Toothbrush.

Question 9:

The writer used to carry his toothbrush to the railways station wrapped it in a

(a) Paper

(b) Plastic bag

(c) Cotton

(d) Handkerchief

Answer:

(d) Handkerchief.

Question 10:

Finally the writer found his tooth brush inside the

(a) Bag

(b) Book

(c) Boot

(d) Pocket

Answer:

(c) Boot.

Question 11:

Who gave the reminder whether the soap was packed or not?

(a) George

(b) Jerome K Jerome

(c) Harris

(d) Tom

Answer:

(a) George.

Question 12:

Finally, the writer finished his packing at

(a) 10:05 AM

(b) 11:00 PM

(c) 10:05 PM

(d) 9:05 PM

Answer:

(c) 10:05 PM.

Question 13:

—— was the worst packer in this world.

(a) Jerome K Jerome

(b) Harris

(b) George

(d) Marh

Answer:

(b) Harris.

Question 14:

They started packing by breaking the

(a) Plate

(b) Jar

(c) Cup

(d) Bottles

Answer:

(c) Cup.

Question 15:

Harris packed the — on the top of a tomato and squashed it.

(a) Bottle

(b) Cup

(c) Strawberry jam

(d) Plate

Answer:

(c) Strawberry jam.

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NCERT MCQ CLASS-9 CHAPTER-6 | ENGLISH NCERT MCQ | | BEEHIVE IMPORTANT MCQ | | MY CHILDHOOD | EDUGROWN

In This Post we are  providing Chapter-6 My Childhood NCERT MCQ for Class 9 English which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON MY CHILDHOOD

Question 1. Who was Samsuddin?
(a) Kalam’s brother
(b) Kalam’s friend
(c) Kalam’s neighbor
(d) Kalam’s cousin

Answer :  D

Question 2. The name of Abdul Kalam’s father was
(a) Jainulabdeen
(b) Samsuddin
(c) Rahimddin
(d) None of the above

Answer :  A

Question 3. Indians will build their own India”- Who said this?
(a) Sivasubramania Iyer
(b) APJ Abdul Kalam
(c) Netaji Subhas Chandra Bose
(d) Mahatma Gandhi

Answer :  D

Question 4. Abdul’s friend Sivapraasan became
(a) A teaching assistant
(b) A newspaper distributor
(c) A scientist
(d) Catering contractor of Southern Railways

Answer :  D

Question 5: The family of Abdul Kalam used to particular in –
(a) Rama Kalyanam Ceremony
(b) Laxmi Puja
(c) Dasseruh
(d) Ponam

Answer:    A

Question 6. How was the wife of Abdul’s science teacher initially?
(a) Poor
(b) Rich
(c) Conservative correct
(d) Beautiful

Answer:    C

Question 7. What bedtime stories did Abdul’s mother and grandmother tell the children of the family?
(a)Ramayana
(b)Mahabharata
(c)Life of the Prophet
(d)Both (a) and (c)

Answer :  D

Question 8. Abdul’s father _______ Abdul for higher studies
(a) Encouraged
(b) Restricted
(c) Gave money
(d) Didn’t support

Answer :  A

Question 9: A.P.J Abdul Kalam was born in the year 
(a) 1938
(b) 1931
(c) 1939
(d) 1930

Answer:   B

Question 10. What bedtime stories did Abdul’s mother and grandmother tell the children of the family?
(a)Ramayana
(b)Mahabharata
(c)Life of the Prophet
(d)Both (a) and (c)

Answer :  D

Question 11. Why did the teacher separate Kalam from his friend?
(a) they were talking in the class
(b) they were not doing their work
(c) they were disturbing the class
(d) because of his own notion of social ranking, religious discrimination

Answer :  D 

Question 12: Kalam inherited faith in goodness and deep kindness from his 
(a) Mother
(b) Brother-in-law
(c) Father
(d) Grandfather

Answer:   A

Question 13. What was the name of Abdul’s science teacher?
(a) Ramanujan Iyer
(b) Madhavan Iyer
(c) Sivasubramania Iyer
(d) Mukhesh Iyer

Answer :  C

Question 14. Why did Sivasubramania serve himself and sat with him to eat?
(a) to avoid any disturbance because of biased behavior of his conservative wife and
(b) to show equal respect for everybody
(c) because he wanted to prove himself great
(c) to demoralize his wife
(d) none

Answer :  A

Question 15. What was Kalam’s age during the second world war?
(a) 9 years
(b) 10 years
(c) 11 years
(d) 8 years

Answer :  D

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NCERT MCQ CLASS-9 CHAPTER-5 | ENGLISH NCERT MCQ | | BEEHIVE IMPORTANT MCQ | | THE SNAKE AND THE MIRROR | EDUGROWN

In This Post we are  providing Chapter-5 The Snake and The Mirror NCERT MCQ for Class 9 English which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON THE SNAKE AND THE MIRROR

Question 1.
While looking at the mirror, the doctor smiled, and he found his smile

(a) Unattractive
(b) Attractive
(c) Wide
(d) Ugly

Answer: (b) Attractive

Question 2.
The doctor described the snake as the one

(a) Very poisonous
(b) Taken with its own beauty
(c) Thick and ugly
(d) That had bitten the thief

Answer: (b) Taken with its own beauty

Question 3.
According to the doctor, the thief had a sense of

(a) Good clothes
(b) Cleanliness
(c) Logic
(d) None of the above

Answer: (b) Cleanliness

Question 4.
Why did the doctor have very little things to carry?

(a) Because the storm destroyed most of the things
(b) He was too poor to have many things
(c) The room was too small to keep a lot of things
(d) Some thief had stolen most of the things

Answer: (d) Some thief had stolen most of the things

Question 5.
What did the doctor do immediately after reaching his friend’s house?

(a) Smeared oil all over his body and took a bath
(b) Drank a glass of water
(c) Took medicines
(d) Narrated the incident of the snake

Answer: (a) Smeared oil all over his body and took a bath

Question 6.
How did the doctor’s wife actually look like?

(a) Thin
(b) Fat
(c) Short
(d) None of the above

Answer: (a) Thin

Question 7.
After the snake moved towards the mirror, the doctor was

(a) A stone in a flesh
(b) A man of flesh and blood
(c) A crying man
(d) A paralyzed man

Answer: (b) A man of flesh and blood

Question 8.
From the doctor’s arm, the snake slithered into his

(a) Leg
(b) Lap
(c) Head
(d) Back

Answer: (b) Lap

Question 9.
The snake looked at the mirror. Seeing that, the doctor wandered

(a) If the snake was admiring its own beauty
(b) If it’s going to bite him
(c) If it’s going to make a noise
(d) If it’s going to smile

Answer: (a) If the snake was admiring its own beauty

Question 10.
While sitting wrapped around by a snake, the doctor felt

(a) The presence of God
(b) The presence of the ghost
(c) The presence of rats
(d) The presence of devils

Answer: (a) The presence of God

Question 11.
The doctor turned to

(a) Stone
(b) Ghost
(c) Snake
(d) Devil

Answer: (a) Stone

Question 12.
What did the snake to after landing on the doctor’s shoulder?

(a) Coiled around his left arm above the elbow
(b) Coiled around his right arm above the elbow
(c) Coiled around his neck
(d) Coiled around his head

Answer: (a) Coiled around his left arm above the elbow

Question 13.
Suddenly, the doctor heard a dull sound that resembled

(a) Sound of frogs
(b) Rubber tube falling on the ground
(c) An opening of a box
(d) Hails falling on the roof

Answer: (b) Rubber tube falling on the ground

Question 14.
Whom did the doctor want to marry?

(a) A scientist
(b) A teacher
(c) A woman doctor
(d) A singer

Answer: (c) A woman doctor

Question 15.
The doctor believed in making himself

(a) Thinner
(b) Honest
(c) Hardworking
(d) Handsome

Answer: (d) Handsome


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CLASS 11th CHAPTER -10 Cell Cycle and Cell Division |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 10 Cell and Cell Division includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -10 Cell and Cell Division . NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -10 Cell and Cell Division| NCERT BIOLOGY SOLUTION |

Page No: 171

Exercises

1. What is the average cell cycle span for a mammalian cell?

Answer
The average cell cycle span for a mammalian cell is approximately 24 hours.
 
2. Distinguish cytokinesis from karyokinesis.
 
Answer
Cytokinesis
Karyokinesis
Cytokinesis is the biological process involving the division of a cell’s cytoplasm during mitosis or meiosis.Karyokinesis is the biological process involving the division of a cell’s nucleus during mitosis or meiosis.
Stages such as prophase, metaphase, anaphase, and telophase are not present in cytokinesis.It is divided into four stages: prophase, metaphase, anaphase, and telophase

3. Describe the events taking place during interphase.

Answer
The interphase, though called the resting phase, is the time during which the cell is preparing for division by undergoing cell growth and DNA replication. The interphase is-divided into three further phases:

→ G1 phase – It is the stage during which the cell grows and prepares its DNA for replication. In this phase, the cell is metabolically active.

→ S phase – It is the stage during which DNA synthesis occurs. In this phase, the amount of DNA (per cell) doubles, but the chromosome number remains the same.

→ G2 phase – In this phase, the cell continues to grow and prepares itself for division. The proteins and RNA required for mitosis are synthesised during this stage.

4. What is G0(quiescent phase) of cell cycle?

Answer
G0 or quiescent phase is the stage in which cells are metabolically active but no longer proliferate unless called on to do so depending on the requirement of the organism.

Page No: 172

5. Why is mitosis called equational division?

Answer
During mitosis, the genetic constitution of the daughter cells is identical to that of parent cells. A diploid cell produces four diploid cells. Since, the chromosome number of the daughter cells remains identical to that of the parent cell, it is called as Equational division.

6. Name the stage of cell cycle at which one of the following events occur:
(i) Chromosomes are moved to spindle equator
(ii) Centromere splits and chromatids separate
(iii) Pairing between homologous chromosomes takes place
(iv) Crossing over between homologous chromosomes takes place

Answer
(i) Metaphase
(ii) Anaphase
(iii) Zygotene of meiosis I
(iv) Pachytene of meiosis I

7. Describe the following:
(a) synapsis (b) bivalent (c) chiasmata
Draw a diagram to illustrate your answer.
Answer

(i) The pairing of homologous chromosomes is called synapsis. This occurs during the second stage of prophase I or zygotene.
Synapsis

(ii) Bivalent or tetrad is a pair of synapsed homologous chromosomes. They are formed during the zygotene stage of prophase I of meiosis.
 
Bivalent

(iii) Chiasmata is the site where two non-sister chromatids of homologous chromosomes have crossed over. It represents the site of cross-over. It is formed during the diplotene stage of prophase I of meiosis.
Chiasmata

 
8. How does cytokinesis in plant cells differ from that in animal cells?

Answer
Cytokinesis
Animal Cells
The division of the cytoplasm takes place by cell plate formation.The division of the cytoplasm takes place by cleavage.
Cell plate formation starts at the centre of the cell and grows outward, toward the lateral walls.Cleavage starts at the periphery and then moves inward, dividing the cell into two parts.

9. Find examples where the four daughter cells from meiosis are equal in size and where they are found unequal in size.

Answer
Examples are:

→ During microsporogenesis, four daughter cells formed in flowering plants are equal in size while during megasporogenesis, four daughter cells formed in flowering plants are unequal in size.

→ During Spermatogenesis or the formation of sperms in human beings occurs by the process of meiosis, four daughter cells formed are equal in size while during Oogenesis or the formation of ovum in human beings occurs by the process of meiosis, four daughter cells formed are unequal in size.

10. Distinguish anaphase of mitosis from anaphase I of meiosis.

Answer
Anaphase of mitosis
Anaphase I of meiosis
Anaphase is the stage during which the centromere splits and the chromatids separate. The Chromatids move to opposite poles.The homologous chromosomes separate, while sisterchromatids remain associated at their centromeres

11. List the main differences between mitosis and meiosis.

Answer
Mitosis
Meiosis
This type of division takes place in somatic cells This type of division takes place in gametic cells
Two daughter cells are formedFour daughter cells are formed
Number of chromosomes remains diploid in daughter cells Number of chromosomes becomes haploid in daughter cells are formed
Mitosis is necessary for growth and repairMeiosis is necessary for sexual reproduction
Crossing over does not take placeCrossing over takes place 

12. What is the significance of meiosis?

Answer
Significance of meiosis are:
→ Conservation of specific chromosome number of each species is achieved across successive generations in sexually reproducing organisms through meiosis.
→ It also increases the genetic variability in the population of organisms from one generation to the next.

13. Discuss with your teacher about
(i) haploid insects and lower plants where cell-division occurs, and
(ii) some haploid cells in higher plants where cell-division does not occur.

Answer
(i) Male bees, wasps and ants are haploid as they are produced from unfertilized eggs.

(ii) Cell division does not happen in synergids and antipodal cells in ovule.

14. Can there be mitosis without DNA replication in S phase?

Answer
No, there cannot be mitosis without DNA replication in S phase because additional DNAs are required for the formation of new cells. DNA duplication is important as it maintains the chromosome number in the daughter cells and hence Mitosis is an equational division. Therefore, the duplication of DNA is an essential step and without it, no mitosis can take place.

15. Can there be DNA replication without cell division?

Answer

DNA replication can take place without cells division as in case of formation of new mitochondria and chloroplasts. During cell division, the parent cell gets divided into two daughter cells. However, if there is a repeated replication of DNA without any cell division, then this DNA will keep accumulating inside the cell. This would increase the volume of the cell nucleus, thereby causing cell expansion.

16. Analyse the events during every stage of cell cycle and notice how the following two parameters change
(i) Number of chromosomes (N) per cell
(ii) Amount of DNA content (C) per cell

Answer
(i) The number of chromosomes (N) remains constant throughout the different stages in mitosis. In meiosis the number of chromosome is reduced to half that is from diploid to haploid. During the anaphase I stage this half number is carried on till the end of meiosis.

(ii) The amount of DNA (C) becomes double during the interphase stage which is followed by mitosis. This doubling occurs due to the replication of the DNA.
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