NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | BIOLOGY IMPORTANT QUESTIONS | CHAPTER -16 | DIGESTION AND ABSORBTION | EDUGROWN |

In This Post we are  providing Chapter-16 DIGESTION AND ABSORBTION NCERT MOST IMPORTANT QUESTIONS for Class 11 BIOLOGY 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 MOST IMPORTANT QUESTIONS ON DIGESTION AND ABSORBTION

Q1. Correct the statement given below by the right option shown in the bracket against them.
a. Absorption of amino acids and glycerol takes place in the (small intestine/ large intestine).
b. ‘The faeces in the rectum initiate a reflex causing an urge for its removal, (neural /hormonal)
c. Skin and eyes turn yellow in infection, (liver/stomach)
d. Rennin is a proteolytic enzyme found in gastric juice in (infants/adults).
e. Pancreatic juice and bile are released through (intestinepancreatic/ hepato-pancreatic duct).
f. Dipeptides, disaccharides and glycerides are broken down into simple substances in region of small intestine, (jejunum/duodenum)
Ans: a. Absorption of amino acids and glycerol takes place in the small intestine.
b. The faeces in the rectum initiate a neural reflex causing an urge for its removal.
c. Skin and eyes turn yellow in infection of liver. .
d. Rennin is a proteolytic enzyme found in gastric juice in infants.
e. Pancreatic juice and bile are released through hepato-pancreatic duct.
f. Dipeptides, disaccharides and glycerides are broken down into simple substances in duodenum region of small intestine.

Q2. What are three major types of cells found in the gastric glands? Name their secretions.
Ans: The mucosa of stomach has gastric glands. Gastric glands have three major types of cells namely
(i) mucus neck cells which secrete mucus;
(ii) peptic or chief cells which secrete the proenzyme pepsinogen; and
(iii) parietal or oxyntic cells which secrete HC1 and intrinsic factor (factor essential for absorption of vitamin B12).

Q3. How are the activities of gastro-intestinal tract regulated?
Ans: The activities of the gastro-intestinal tract are under neural and hormonal control for proper coordination of different parts. The sight, smell and/or the presence of food in the oral cavity can stimulate the secretion of saliva. Gastric and intestinal secretions are also, similarly, stimulated by neural signals. The muscular activities of different parts of the alimentary canal can also be moderated by neural mechanisms, both local and through CNS. Hormonal control of the secretion of digestive juices is carried out by local hormones produced by the gastric and intestinal mucosa.

Q4. Distinguish between constipation and indigestion. Mention their major causes.
Ans: Constipation: In constipation, the faeces are retained within the rectum as the bowel movements occur irregularly.
Indigestion: In this condition, the food is not properly digested leading to a feeling of fullness. The causes of indigestion are inadequate enzyme secretion, anxiety, food poisoning, over eating, and spicy food.

Q5. Describe the enzymatic action on fats in the duodenum.
Ans: In the duodenum fats are broken down by pancreatic lipases with the help of bile into di- and monoglycerides.
ncert-exemplar-problems-class-11-chapter-16-digestion-absorption-3


Q6. A person had roti and dal for his lunch. Trace the changes in those during its passage through the alimentary canal.
Ans: The process of digestion is accomplished by mechanical and chemical processes. The buccal cavity performs two major functions, mastication of food and facilitation of swallowing. The teeth and the tongue with the help of saliva masticate and mix up the food thoroughly. Mucus in saliva helps in lubricating and adhering the masticated food particles into a bolus. The bolus is then conveyed into the pharynx and then into the oesophagus by swallowing or deglutition. The bolus further passes down through the oesophagus by successive waves of muscular contractions called peristalsis. The gastro-oesophageal sphincter controls the passage of food into the stomach. The saliva secreted into the oral cavity contains electrolytes and enzymes, salivary amylase and lysozyme. The chemical process of digestion is initiated in the oral cavity by the hydrolytic action of the carbohydrate splitting enzyme, the salivary amylase. About 30% of starch is hydrolysed here by this enzyme (optimum pH 6.8) into a disaccharide – maltose.
• The stomach stores the food for 4-5 hours. The food mixes thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall and is called the chyme. The proenzyme pepsinogen, on exposure to hydrochloric acid gets converted into the active enzyme pepsin, the proteolytic enzyme of the stomach. Pepsin converts proteins into proteoses and peptones (peptides).
• The bile, pancreatic juice and the intestinal juice are the secretions
released into the small intestine. Pancreatic juice and bile are released through the hepato-pancreatic duct. The pancreatic juice contains inactive enzymes—trypsinogen, chymotrypsinogen, procarboxypeptidases,
amylases, lipases and nucleases. Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice.
Proteins proteoses and peptones (partially hydrolysed proteins) in the chyme reaching the intestine are acted upon by the proteolytic enzymes of pancretic juice as given below.
ncert-exemplar-problems-class-11-chapter-16-digestion-absorption-4

Q7. What are the various enzymatic types of glandular secretions in our gut helping digestion of food? What is the nature of end products obtained after complete digestion of food?
Ans: Enzymatic types of glandular secretions in our gut:
a. Salivary glands: Saliva is mainly produced by three pairs of salivary glands, the parotids (cheek), the sub-maxillary/sub-mandibular (lower jaw) and the sublinguals (below the tongue). These glands situated just outside the buccal cavity secrete salivary juice into the buccal cavity.
b. Gastric glands: The mucosa of stomach has gastric glands. Gastric glands have three major types of cells namely
(i) mucus neck cells which secrete mucus;
(ii) peptic or chief cells which secrete the proenzyme pepsinogen; and
(iii) parietal or oxyntic cells which secrete HC1 and intrinsic factor (factor essential for absorption of vitamin B12).
c. The bile, pancreatic juice and the intestinal juice are the secretions
released into the small intestine. Pancreatic juice and bile are released through the hepato-pancreatic duct. The pancreatic juice contains inactive enzymes—trypsinogen, chymotrypsinogen, procarboxypeptidases,
amylases, lipases and nucleases. Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice. The bile released into the duodenum contains bile pigments (bilirubin and biliverdin), bile salts, cholesterol and phospholipids but no enzymes.
ncert-exemplar-problems-class-11-chapter-16-digestion-absorption-5

Q8. Discuss mechanisms of absorption.
Ans: Mechanisms of absorption: Absorption is the process by which the end products of digestion pass through the intestinal mucosa into the blood or lymph. It is carried out by passive, active or facilitated transport mechanisms. Small amounts of monosaccharides like glucose, amino acids and some electrolytes like chloride ions are generally absorbed by simple diffusion. The passage of these substances into the blood depends upon the concentration gradients. However, some substances like glucose and amino acids are absorbed with the help of carrier proteins. This mechanism is called the facilitated transport.
Transport of water depends upon the osmotic gradient. Active transport occurs against the concentration gradient and hence requires energy. Various nutrients like amino acids, monosaccharides like glucose, electrolytes like Na+ are absorbed into the blood by this mechanism.

Q9. Discuss the role of hepato-pancrdatic complex in digestion of carbohydrate, protein and fat components of food.
Ans: The bile, pancreatic juice and the intestinal juice are the secretions released into the small intestine. Pancreatic juice and bile are released through the hepato-pancreatic duct. The pancreatic juice contains inactive enzymes— trypsinogen, chymotrypsinogen, procarboxypeptidases, amylases, lipases and nucleases. Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice.
The bile released into the duodenum contains bile pigments (bilirubin and biliverdin), bile salts, cholesterol and phospholipids but no enzymes. Bile helps in emulsification of fats, i.e., breaking down of the fats into very small micelles. Bile also activates lipases.
Proteins proteases and peptones (partially hydrolyzed proteins) in the chime reaching the intestine are acted upon by the proteolytic enzymes of pancreatic juice as given below:
ncert-exemplar-problems-class-11-chapter-16-digestion-absorption-6
Q10. Explain the process of digestion in the buccal cavity with a note on the arrangement of teeth.

Ans: The process of digestion in the buccal cavity: The buccal cavity performs two major functions, mastication of food and facilitation of swallowing. The teeth and the tongue with the help of saliva masticate and mix up the food thoroughly. Mucus in saliva helps in lubricating and adhering the masticated food particles into a bolus. The bolus is then conveyed into the pharynx and then into the oesophagus by swallowing or deglutition. The bolus further passes down through the oesophagus by successive waves of muscular contractions called peristalsis. The gastro-oesophageal sphincter controls the passage of food into the stomach. The saliva secreted into the oral cavity contains electrolytes and enzymes, salivary amylase and lysozyme. The chemical process of digestion is initiated in the oral cavity by the hydrolytic action of the carbohydrate splitting enzyme, the salivary amylase. About% of starch is hydrolysed here by this enzyme (optimum pH 6.8) into a disaccharide—maltose. Lysozyme present in saliva acts as an antibacterial agent that prevents infections.

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NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | BIOLOGY IMPORTANT QUESTIONS | CHAPTER -17 | BREATHING AND EXCHANGE OF GASES | EDUGROWN |

In This Post we are  providing Chapter-17 BREATHING AND EXCHANGE OF GASES NCERT MOST IMPORTANT QUESTIONS for Class 11 BIOLOGY 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 MOST IMPORTANT QUESTIONS ON BREATHING AND EXCHANGE OF GASES

Question 1.
What is chloride shift? Write its significance during respiration.

Answer:
The chloride ions (CI) inside RBC combine with potassium ion (K+) to form potassium chloride (KCL), whereas hydrogen carbonate ions (HCO3) in the plasma combine with Na’ to form sodium hydrogen carbonate (NaHCO3) Nearly 70% of carbon dioxide is transported from tissues to the lungs in this form.

In response to chloride ions (CI) diffuse from plasma into erythrocytes to maintain the ionic balance. This is called the chloride shift.

Significance: It maintains electrochemical neutrality during respiration.

Question 2.
What is the role of the carbonic anhydrase enzyme in the transport of gases during respiration?

Answer:
Carbon dioxide produced by the tissues diffuses passively into the bloodstream and passes into the red blood corpuscles where it reacts with water to form carbonic acid (H2CO3). This reaction is catalysed by the enzyme, carbonic anhydrase found in the erythrocytes and takes less than one second to complete the process. Immediately after its formation, carbonic acid dissociates into hydrogen (H+) and bicarbonate (HCO3) ions. The majority of bicarbonate ions (HCO3 ) formed within the erythrocytes diffuse out into the plasma along a concentration gradient. These combine with haemoglobin to form the haemoglobin acid (H.Hb).
Breathing and Exchange of Gases Class 11 Important Extra Questions Biology 1

Question 3.
What is partial pressure? How does it help in gaseous exchange during respiration?

Answer:
During inspiration and expiration, gases move freely by the process of diffusion. Diffusion of any molecule takes place from high to low concentration. The process of diffusion is directly proportional to the pressure caused by the gas alone. The pressure exerted by an individual gas is called partial pressure. It is represented as PO2, PCO2, and PN2, for oxygen, carbon dioxide and nitrogen respectively.

The inspired air ultimately reaches the alveoli of the lung, which in turn receives the blood supply of the pulmonary circulation. At this stage the oxygen of the inspired air is taken in by the blood and carbon dioxide is released into the alveoli for expiration.

In this way, the gases exchange takes place due to partial pressure.

Question 4.
How does hemoglobin help in the transport of oxygen from the lung to tissues?

Answer:
Blood is the medium for the transport of oxygen from the respiratory organ to the different tissues and carbon dioxide from tissues to the respiratory organs. 97% of the oxygen is transported from the lungs to the tissues in combination with hemoglobin (Hb + O2 — HbO2), oxyhemoglobin and 3% is transported in dissolved condition by the plasma.

Under high partial pressure oxygen easily binds with hemoglobin in the pulmonary capillaries. When this oxygenated blood reaches the different tissues, the partial pressure of oxygen declines and the bonds holding oxygen to hemoglobin become unstable. As a result, oxygen is released from the capillaries.

Question 5.
Explain breathing disorders in brief.

Answer:

  1. Asthma is caused by an allergic reaction. There is difficulty in breathing.
  2. Pneumonia is caused by bacterial infection. There are fever, pain and severe cough.
  3. Tuberculosis is an infectious bacterial disease of the lungs and in serious cases, blood may come out while coughing.

Question 6.
In what form O2 is carried in blood? What happens to it when blood reaches the tissue?

Answer:
O2 is carried in combination with the hemoglobin of RBCs and forms oxyhemoglobin.

In tissues, there is the dissociation of oxyhaemoglobin and release of Or It diffuses into the tissue cells where it is used in oxidation.

Question 7.
Explain gas transport in the blood.

Answer:
It may be explained in two steps.
(a) Transport of O2 from lungs to tissues.
(b) Transport of CO2 from tissues to lungs.

A. Oxygen Transport

  1. O2 is transported in the blood via haemoglobin.
  2. O2 diffuses into RBC and combines with haemoglobin to form oxyhaemoglobin.
  3. Oxyhaemoglobin breaks into haemoglobin and oxygen at the tissues, where there are high PCO2 and PO2.
  4. In the lungs, oxyhaemoglobin is formed due to high PO2 and low PCO2.

B. CO2 Transport: CO2 is transported in 3 ways with blood.

  1. 70% of CO2 in RBC reacts with H2O to form H2CO3
    Breathing and Exchange of Gases Class 11 Important Extra Questions Biology 2
  2. The rest 30% CO2 combining with Hb to form carbon haemoglobin. (HCO3 carried by RBC and plasma)
  3. Some CO2 dissolves in plasma on reaching the lungs.
    HCO3 + H + H2CO3
    H2CO3 CO2 + 2H2O
    And this CO2 is expelled out through the lungs.

Question 8.
Name and explain the respiratory organs of the following,

(i) Insect

Answer:
Insect: The integument of insects is thick and highly impermeable to minimise the loss of water through the environment. The exchange of gases cannot take place through the skin covering of these insects. These insects have a highly developed complex system called the tracheal. This mode of respiration is called tracheal respiration.

(ii) Neries
Answer:
Neries: Parapodia is the respiratory oxygen in neries. In this organism respiratory occurs through the skin covering the parapodia (Locomotory organs), which is again very thin, moist, permeable and highly vascular.

(iii) Prawn
Answer:
Prawn: Gills, in the animals like prawns, certain molluscs, fishes, tadpoles, the process of gaseous exchange occur by special respiratory organs called gills. These are richly supplied with blood and readily absorb oxygen found dissolved in water and release CO2 back into the water.

(iv) Birds
Answer:
Birds: (lungs). In birds and mammals, the skin is impermeable. These have a high metabolic rate and their oxygen requirement is very high. Birds have spongy lungs to have a more extensive respiratory surface. These lungs always remain in the body to keep the respiratory surface moist, which is necessary for the exchange of respiratory gases.

(v) Fishes
Answer:
Oxygen and carbon dioxide dissolves in water, and most fishes exchange dissolved oxygen and carbon dioxide in water by means of the gills.

(vi) Earthworm.
Answer:
Earthworms do not have lungs. They breathe through their skin. Oxygen and carbon dioxide pass through the earthworm’s skin by diffusion

Question 9.
Define the following terms:

(a) Anaerobic respiration,

Answer:
Anaerobic respiration: It is a process that does not involve the use of molecular oxygen. Food is not completely oxidised to CO2 and water. Less energy is present in anaerobic respiration.

(b) Breathing,
Answer:
Breathing: It is a physical process, which brings in fresh air to the respiratory surface and removes foul impure airs from the outside. It occurs outside the cells and is thus an extracellular process.

(c) Vital capacity,
Answer:
Vital capacity: It is defined as an important measure of pulmonary capacity. It is the maximum amount of air a person can expel from the lungs after first filling the lungs to their maximum extent.

Vital capacity is the sum total of inspiration reserve volume, tidal volume and expiratory reserve volume.
(1 + 1 + VC = IRV = TV/ERV)

(d) Tidal volume,
Answer:
Tidal volume: It is defined as the volume of air normally inspired or expired in one breath without doing any effort. It is about 500 ml in an adult person. It represents the volume of air, which is renewed in the respiratory system during every breathing.

(e) Respiratory centre.
Answer:
Respiratory centre: A number of groups of neurons located bilaterally in the medulla oblongata control the respiratory. These are called respiratory centres. These centres are named the dorsal respiratory group. Ventral respiratory group and pneumatic centre.

Question 10.
Write the role of the diaphragm and its Costals muscles in the breathing process.

Answer:
During breathing, when the lungs contract their volumes decrease resulting in the increase of air pressure in the lungs. Hence, the air is exhaled from the lungs. These two processes are called inspiration and expiration. During normal breathing, the downward and upward movement of the diaphragm takes place. When the diaphragm, contracts, the lower surface of the lung is pulled downward consequently the volume of the lungs increases.

This causes the inhalation of air or inspiration. When the diaphragm relaxes, lungs are compressed and air exhaled, expiration takes place. The demand for extra oxygen is fulfilled by the expansion of the rib cage, during exercise when the rate of breathing increases.

During expiration, high pressure is generated in the lungs and air moves out. The upward movement of the rib cage is caused mainly by the external intercostals muscles present between the ribs along with the assistance of few other adjacent muscles.

Similarly, the downward movement of the rib cage is facilitated by the internal intercostals, external oblique and internal oblique muscles, position of the diaphragm, ribs and sternum during breathing as shown in the diagram
Breathing and Exchange of Gases Class 11 Important Extra Questions Biology 3
Position of diaphragm, ribs and sternum during breathing

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NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | BIOLOGY IMPORTANT QUESTIONS | CHAPTER -15 | PLANT GROWTH AND DEVELOPMENT | EDUGROWN |

In This Post we are  providing Chapter-15 PLANT GROWTH AND DEVELOPMENT NCERT MOST IMPORTANT QUESTIONS for Class 11 BIOLOGY 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 MOST IMPORTANT QUESTIONS ON PLANT GROWTH AND DEVELOPMENT

Question 1:

Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.

ANSWER:

(a) Growth

It is an irreversible and permanent process, accomplished by an increase in the size of an organ or organ parts or even of an individual cell.

(b) Differentiation

It is a process in which the cells derived from the apical meristem (root and shoot apex) and the cambium undergo structural changes in the cell wall and the protoplasm, becoming mature to perform specific functions.

(c) Development

It refers to the various changes occurring in an organism during its life cycle – from the germination of seeds to senescence.

(d) De-differentiation

It is the process in which permanent plant cells regain the power to divide under certain conditions.

(e) Re-differentiation

It is the process in which de-differentiated cells become mature again and lose their capacity to divide.

(f) Determinate growth

It refers to limited growth. For example, animals and plant leaves stop growing after having reached maturity.

(g) Meristem

In plants, growth is restricted to specialised regions where active cell divisions take place. Such a region is called meristem. There are three types of meristems – apical meristem, lateral meristem, and intercalary meristem.

(h) Growth rate

It can be defined as the increased growth in plants per unit time.

Question 2:

Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant?

ANSWER:

In plants, growth is said to have taken place when the amount of protoplasm increases. 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. Measuring the growth of plants using only one parameter does not provide enough information and hence, is insufficient for demonstrating growth.

Question 3:

Describe briefly:

(a) Arithmetic growth

(b) Geometric growth

(c) Sigmoid growth curve

(d) Absolute and relative growth rates

ANSWER:

(a) Arithmetic growth

In arithmetic growth, one of the daughter cells continues to divide, while the other differentiates into maturity. The elongation of roots at a constant rate is an example of arithmetic growth.

(b) Geometric growth

Geometric growth is characterised by a slow growth in the initial stages and a rapid growth during the later stages. The daughter cells derived from mitosis retain the ability to divide, but slow down because of a limited nutrient supply.

(c) Sigmoid growth curve

The growth of living organisms in their natural environment is characterised by an S-shaped curve called sigmoid growth curve. This curve is divided into three phases – lag phase, log phase or exponential phase of rapid growth, and stationary phase.

Exponential growth can be expressed as:

Where,

W1 = Final size

W0 = Initial size

r = Growth rate

t= Time of growth

e = Base of natural logarithms

(d) Absolute and relative growth rates

Absolute growth rate refers to the measurement and comparison of total growth per unit time.

Relative growth rate refers to the growth of a particular system per unit time, expressed on a common basis.

Question 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:

Plant growth regulators are the chemical molecules secreted by plants affecting the physiological attributes of a plant. There are five main plant growth regulators. These are:

(i) Auxins

(ii) Gibberellic acid

(iii) Cytokinins

(iv) Ethylene

(v) Abscisic acid

(i) Auxins

Discovery:

The first observations regarding the effects of auxins were made by Charles Darwin and Francis Darwin wherein they saw the coleoptiles of canary gross bending toward a unilateral source of light.

It was concluded after a series of experiments that some substance produced at the tip of coleoptiles was responsible for the bending. Finally, this substance was extracted as auxins from the tips of coleoptiles in oat seedlings.

Physiological functions:

1. They control plant cell-growth.

2. They cause the phenomenon of apical dominance.

3. They control division in the vascular cambium and xylem differentiation.

4. They induce parthenocarpy and prevent abscission of leaves and fruits.

Horticultural applications:

1. They are used as the rooting hormones in stem cuttings.

2. 2-4 D is used weedicide to kill broadleaf, dicotyledonous weeds.

3. They induce parthenocarpy in tomatoes.

4. They promote flowering in pineapple and litchi.

(ii) Gibberellic acid

Discovery:

Bakane or the “foolish rice seedling” disease was first observed by Japanese farmers. In this disease, rice seedlings appear to grow taller than natural plants, and become slender and pale green. Later, after several experiments, it was found that this condition was caused by the infection from a certain fungus Gibberella fujikuroi. The active substance was isolated and identified as gibberellic acid.

Physiological functions:

1. It causes elongation of internodes.

2. It promotes bolting in rosette plants.

3. It helps in inducing seed germination by breaking seed dormancy and initiating the synthesis of hydrolases enzymes for digesting reserve food.

Horticultural applications:

1. It helps in increasing the sugar content in sugarcane by increasing the length of the internodes.

2. It increases the length of grape stalks.

3. It improves the shape of apple.

4. It delays senescence.

5. It hastens maturity and induces seed-production in juvenile conifers.

(iii) Cytokinins

Discovery:

Through their experimental observations, F. Skoog and his co-workers found that the tobacco callus differentiated when extracts of vascular tissues, yeast extract, coconut milk, or DNA were added to the culture medium. This led to the discovery of cytokinins.

Physiological functions:

1. They promote the growth of lateral branches by inhibiting apical dominance.

2. They help in the production of new leaves, chloroplasts, and adventitious shoots.

3. They help in delaying senescence by promoting nutrient mobilisation.

Horticultural applications:

1. They are used for preventing apical dominance.

2. They are used for delaying senescence in leaves.

(iv) Ethylene

Discovery:

It was observed that unripe bananas ripened faster when stored with ripe bananas. Later, the substance promoting the ripening was found to be ethylene.

Physiological functions:

1. It helps in breaking seed and bud dormancy.

2. It promotes rapid internode-elongation in deep-water rice plants.

3. It promotes root-growth and formation of root hairs.

4. It promotes senescence and abscission of leaves and flowers.

5. It hastens the respiration rate in fruits and enhances fruit ripening.

Horticultural applications:

1. It is used to initiate flowering and synchronising the fruit set in pineapples.

2. It induces flowering in mango.

3. Ethephon is used to ripen the fruits in tomatoes and apples, and accelerate the abscission of flowers and leaves in cotton, cherry, and walnut.

4. It promotes the number of female flowers in cucumbers.

(v) Abscisic acid

Discovery:

During the mid 1960s, inhibitor-B, abscission II, and dormin were discovered by three independent researchers. These were later on found to be chemically similar and were thereafter called ABA (Abscisic acid).

Physiological functions:

1. It acts as an inhibitor to plant metabolism.

2. It stimulates stomatal closure during water stress.

3. It induces seed dormancy.

4. It induces abscission of leaves, fruits, and flowers.

Horticultural application:

It induces seed dormancy in stored seeds.

Question 5:

What do you understand by photoperiodism and vernalisation? Describe their significance.

ANSWER:

Photoperiodism refers to the response of plants with respect to the duration of light (i.e., period of day and night). On the basis of its response to the duration of light, a plant is classified as a short-day plant, a long-day plant, or a day-neutral plant. Short-day plants flower when they are exposed to light for a period less than the critical day-length (for example: Chrysanthemum). Long-day plants flower when they are exposed to light for a period more than the critical day-length (for example: radish). When no marked correlation is observed between the duration of exposure to light and the flowering response, plants are termed as day-neutral plants (for example: tomato).

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.

Vernalisation is the cold-induced flowering in plants. In some plants (such as the winter varieties of wheat and rye and biennials such as carrot and cabbage), exposure to low temperature is necessary for flowering to be induced. The winter varieties of rye and wheat are planted in autumn. They remain in the seedling stage during winters and flower during summers. However, when these varieties are sown in spring, they fail to flower. Similar response is seen in cabbage and radish.

Question 6:

Why is Abscisic acid also known as stress hormone?

ANSWER:

Abscisic acid is called stress hormones as it induces various responses in plants against stress conditions.

It increases the tolerance of plants toward various stresses. It induces the closure of the stomata during water stress. 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.

Question 7:

Both growth and differentiation in higher plants are open’. Comment.

ANSWER:

Growth and development in higher plants is referred to as being open. This is because various meristems, having the capacity for continuously dividing and producing new cells, are present at different locations in these plant bodies.

Page No 254:

Both a short day plant and a long day plant can flower simultaneously in a given place’. Explain.

ANSWER:

The flowering response in short-day plants and long-day plants is dependent on the durations for which these plants are exposed to light. The short-day plant and long-day plant can flower at the same place, provided they have been given an adequate photoperiod.

Question 9:

Which one of the plant growth regulators would you use if you are asked to:

(a) Induce rooting in a twig

(b) Quickly ripen a fruit

(c) Delay leaf senescence

(d) Induce growth in axillary buds

(e) ‘Bolt’ a rosette plant

(f) Induce immediate stomatal closure in leaves.

ANSWER:

(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 – Gibberellic acid

(f) Induce immediate stomatal closure in leaves – Abscisic acid

Question 10:

Would a defoliated plant respond to photoperiodic cycle? Why?

ANSWER:

A defoliated plant will not respond to the photoperiodic cycle.

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. Therefore, in the absence of leaves, light perception would not occur, i.e., the plant would not respond to light.

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NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | BIOLOGY IMPORTANT QUESTIONS | CHAPTER -14 | RESPIRATION IN PLANTS | EDUGROWN |

In This Post we are  providing Chapter-14 RESPIRATION IN PLANTS NCERT MOST IMPORTANT QUESTIONS for Class 11 BIOLOGY 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 MOST IMPORTANT QUESTIONS ON RESPIRATION IN PLANTS

Question 1.
Why does anaerobic respiration produce less energy than aerobic respiration?

Answer:
Because it does not involve the use of molecular oxygen. Food is not completely oxidized to CO2 and waste. So a lot of energy is still retained with the food. Less energy is produced on anaerobic respiration. It yields only about 5% of the energy available in glucose. Then it is a wasteful process or respiration. In anaerobic respiration, yeast metabolizes glucose to ethanol and CO2 without any use of molecular oxygen.

Question 2.
Explain the effects of temperature on the rate of respiration.

Answer:
Respiration is reduced at very high (above 50°c) and very low (near freezing temp.) temperatures. This is because enzymes can work best between 30°c – 40°c and get inactivated at very high and very’ low tempera¬tures.

Question 3.
How does the exchange of respiratory gases occur in plants?

Answer:
The gaseous exchange takes place through

  1. General body surface
  2. Lenticels.
  3. Stomata present in leaves and young stems. Oxygen becomes transported from cell to cell by diffusion.

Question 17.
Explain RQ significance.
Answer:
RQ value for carbohydrates is 1. It is less than one if proteins are being burnt and more than one if fats are being burnt. So RQ values are important in identifying the kind of substrate used in respiration.

Question 4.
Describe in detail the aerobic oxidation of pyruvic acid.

Answer:
Pyruvic acid generated in the crystal is transported to mitochondria and initiates the second phase of respiration. Before pyruvic acid enters the Kreb’s cycle, one of its three carbon atoms is oxidized to carbon dioxide in the reaction called oxidative decarboxylation.

Pyruvate is first decarboxylated and then oxidized by the enzyme pyruvate dehydrogenase. The combination of the remaining two-carbon acetate unit is readily accepted by a sulfur-containing compound coenzyme A to form acetyl COA. During the process, NAD is reduced to NADH. This process is represented as.
Respiration in Plants Class 11 Important Extra Questions Biology 3
During this process, two molecules of NADH are produced, and thus, it results in a net gain of 6ATP molecules.

(2 NADH + 3 = 6 ATP). 2 molecules of pyruvic acid produced during glycolysis,

Question 5.
Describe the net gain of ATP during respiration.

Answer:
There is a gain of 36 ATP molecules during aerobic respiration of one molecule of glucose. The detail is given in the table.

In most eukaryotic cells, 2 molecules of ATP are required for transporting the NADH produced in glycolysis into mitochondrial for further oxidation. Hence, the net gain of ATP is 36 molecules.

Table ATP molecules produced during respiration.
Respiration in Plants Class 11 Important Extra Questions Biology 4

Question 6.
Define the following:

(a) Respiration
Answer:
Respiration: It is defined as the phenomenon of the release of energy by oxidation of various organic molecules, for cellular use is known as respiration.

(b) Respiratory substrate
Answer:
Respiratory substrate: The compounds that are oxidized during the process of respiration are called respiratory substrates.

(c) Respiratory quotient
Answer:
Respiratory quotient: During respiration oxygen is used and CO2 is released. The ratio of the volume of CO2 evolved to the volume of O2 consumed in respiration is called respiratory quotient (RQ).

(d) Anaerobic respiration
Answer:
Anaerobic respiration: The type of respiration, in which the carbohydrate is incompletely oxidized into some carbonic compounds in absence of oxygen, is called Anaerobic respiration.

(e) Aerobic respiration
Answer:
Aerobic respiration: It is that process of respiration which leads to complete oxidation of organic compound in the presence of oxygen.

This type of respiration is common in higher organisms.

(f) Fermentation.
Answer:
Fermentation: In anaerobic respiration yeasts metabolize glucose to ethanol and CO2 without any use of molecular oxygen. This process is called fermentation in yeasts.

Question 7.
Describe the pentose phosphate pathway.

Answer:
Sometimes oxidation of glucose takes place by another pathway, which is called the pentose phosphate pathway (PPP). In the pentose pathway, glucose-6 phosphate (6C) produced during the early stages of glycolysis or the photosynthates produced during photosynthesis are oxidized to give rise to 6-phosphogluconate. This reaction takes place in the enzyme glucose 6phosphale dehydrogenase and generates NAD PH.

The 6-phosphoglucose molecules are further oxidized by the enzyme 6-phosphogluconate dehydrogenase. As a result of this, one molecule of each ribose-5 phosphate, carbon dioxide, and NADPH is produced, which in turn undergoes many changes to produce glycolytic intermediate. These reactions take place in the cytoplasm.
(From glycolysis)
Respiration in Plants Class 11 Important Extra Questions Biology 5

Question 8.
Calculate the efficiency of respiration in the living system.

Answer:
During aerobic respiration, O2 is consumed and CO2 is released. The ratio of the volume of CO2 evolved to the volume of O2 consumed in respiration is called respiratory quotient (RQ) or respiratory ratio.
Respiration in Plants Class 11 Important Extra Questions Biology 1
The RQ (respiratory quotient) depends upon the respiratory substrate. When a carbohydrate is used as a substrate.

C6H12O6 – 2C2H5OH + 2CO + Energy (247 K.J) on complete oxidation. RQ will be
Respiration in Plants Class 11 Important Extra Questions Biology 6
The total energy yield from 38 ATP molecules comes to 1298 kJ. The energy released by one molecule of glucose on complete oxidation is 2870 kJ. Thus, the efficiency is 45%. Much of the energy generated during respiration is released in the form of heat.

Question 9.
Illustrate the mechanism of the electron transport system.

Answer:
The glucose molecule is completely oxidized by the end of the citric acid cycle. But the energy is not released unless NADH and FADH are oxidized through the electron transport system. Here oxidation means the removal of electrons from it.

The metabolic pathway through which the electron passes from one carrier to another is called the electron transport system (ETS) and it is operative in the inner mitochondrial membrane. Electrons from NADH produced in the mitochondrial matrix are oxidized by an NADH dehydrogenase (Complex I) and electrons are then transferred to ubiquinone.

The ubiquinone located within the inner membrane also receives reducing equivalents via FADH, which is generated during the oxidation of succinate, through the activity of the enzyme, succinate dehydrogenase (complex II). The reduced ubiquinone is then oxidized with the transfer of electrons to the cytochrome complex (Complex III).

Cytochrome is a small protein attached to the outer surface of the inner membrane and acts as a mobile carrier for the transfer of electrons between complex III and complex IV.

(Complex IV) is cytochrome.

When the electrons pass from one carrier to another via complex 1 to IV in the electron transfer chain, they are coupled to ATP synthase (Complex V) for the production of ATP from ADP and inorganic phosphate. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP, while that of one molecule of FADH, produces 2 molecules of ATP.

The electrons are earned by the cytochromes and recombine with their protons before the final stage when the hydrogen atom is accepted by oxygen to form water. Oxygen acts as the final hydrogen acceptor. The whole process by which oxygen allows the production of ATP by phosphorylation of ADP is called oxidative phosphorylation.

Note: There are two routes by which hydrogen from the substrate molecule passes. In route 1.3 ATP molecules are formed for every pair of hydrogen atoms. In route 2, only 2ATP molecules are formed from one pair of hydrogen atoms.

Oxygen acts as the final hydrogen acceptor and forms water.
NAD = nicotinamide adenine dinucleotide.
MN = flavin mononucleotide,
FAD = flavin adenine dinucleotide.

ETC produces 32 ATP molecules per glucose molecule and is the major source of cell energy.
Respiration in Plants Class 11 Important Extra Questions Biology 7
Electron Transport Chain.

Question 10.
Describe the process and role of the citric acid cycle in living organisms.

Answer:
In the process of respiration, the carbohydrates are converted into pyruvic acid through a series of enzymatic reactions. These reactions are known as glycolysis and take place in the cytosol. The pyruvic acid thus formed enters in mitochondria where O2 and necessary enzymes are available; the pyruvic acid is finally converted into CO2 and H2O. This reaction series is known as Krebs Cycle or Citric acid cycle or Tricarboxylic acid (TCA) cycle.

During this cycle, 3 molecules of NAD and one molecule of FAD (Flavin Adenine Dinucleotide) are reduced to produce NADH and FADH respectively. NADH and FADH, so produced during the citric acid cycle are linked with the electron transport system and produce ATP by oxidative phosphorylation, The summary equation for this phase of respiration may therefore be written as follows:
Respiration in Plants Class 11 Important Extra Questions Biology 8
Respiration in Plants Class 11 Important Extra Questions Biology 9
Kreb’s cycle. It follows glycolytic reactions shown in and pyruvate oxidation.

It involves two processes

  1. removal of hydrogen and
  2. the breaking off of carbon dioxide units one by one.
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NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | BIOLOGY IMPORTANT QUESTIONS | CHAPTER -13 | PHOTOSYNTHESIS IN HIGHER PLANTS | EDUGROWN |

In This Post we are  providing Chapter-13 PHOTOSYNTHESIS IN HIGHER PLANTS NCERT MOST IMPORTANT QUESTIONS for Class 11 BIOLOGY 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 MOST IMPORTANT QUESTIONS ON PHOTOSYNTHESIS IN HIGHER PLANTS

Question 1.
Give four important differences between photosynthesis and respiration.

Answer:

PhotosynthesisRespiration
1. Occurs only in green cells.1. Occurs in all Living cells.
2. Occurs in chloroplasts in the cell.2. Occurs in mitochondria.
3. Needs light to occur.3. Does not need light.
4. Uses CO2 and water.4. Uses glucose and oxygen releases CO2 and water.
5. It ïs a constructive process.5. It is a destructive process.

Question 2.
Explain the principle of limiting factor with a suitable graph.

Answer:
Law of limiting factor: When a chemical reaction is conditioned as to its rapidity by a number of separate factors then the rate of reaction is as rapid as the slowest factor permits.

As is clear that as the light intensity is increased the rate of photosynthesis increases proportionately until some other factor like CO, becomes limiting. Ultimately the plant becomes light-saturated indicating that light is no more the limiting factor. If now the cone, of CO, is increased the rate of photosynthesis increases until the light becomes a limiting factor.

Question 3.
(i) What does chlorophyll do to the light falling on it?

Answer:
Chlorophyll molecule becomes excited as soon as light falls on it. It was given out an electron acceptor which returns to release gradually energy in the form of ATP.

(ii) Which pigment system absorbs the red wavelength of light?

Answer:
Photosystem I.

Question 4.
Name the two main sets of reactions in photosynthesis in which light energy is required write down the reaction.

Answer:
(i) Photolysis for breaking down of water molecule.
energy
Photosynthesis in Higher Plants Class 11 Important Extra Questions Biology 4
(ii) Photophosphorylation for the conversion of light energy into chemical energy.
ADP + iP → ATP

Question 5.
What is photorespiration? Describe the process in detail and link it with the Calvin cycle.

Answer:
Enzyme Rubisco catalyzes the carboxylation reaction where CO2 combines with RuBP. This enzyme catalyzes the combination of O2 with RuBP called oxygenation. Respiration that is initiated in chloroplasts and occurs in light only is called photorespiration.

The oxygenation of RuBP in presence of O2 is the first of photorespiration, which leads to the formation of one molecule of phosphoglycolate, a two-carbon compound, and one molecule of PGA. While PGA is used up in the Calvin cycle, the phosphoglycolate is dephosphorylated to form glycolate in the chloroplast and in turn diffused to peroxide, where it is oxidized to glyoxylate.

In the peroxide, the glyoxylate is used to form amino acid and glycine-calycine enters mitochondria where two glycine molecules (4 carbon) give rise to one molecule of serine (3 carbon) and one CO2 (one carbon). The serine is taken up by peroxisome and converted into glycerate. The glycerate enters the chloroplast where it is phosphorylated to form PGA. PGA molecules enter the Calvin cycle to make carbohydrates releasing one molecule of CO2 In mitochondria photorespiration is also called the photosynthetic carbon oxidation cycle.

Increased O2 level increases photorespiration whereas increased CO2 level increases photorespiration ( and increases C2 photosynthesis).

In C3 plants photosynthesis occurs only in one cell type i.e. mesophyll cells. Both light reactions and carbon reactions occur in mesophyll cells in C3 plants. In C4 plant photosynthesis requires the presence of two types of photosynthesis cells that is mesophyll cells and bundle sheath cells. The C4 plants contain dimorphic chloroplasts, which means chloroplasts in mesophyll cells are granular. Therefore C2 pathway does not operate in the C4 pathway.

All the important changes can be summarised as
Photosynthesis in Higher Plants Class 11 Important Extra Questions Biology 5

Question 6.
Describe carbon reactions of the C3 pathway. Does this pathway operate also in C4 plants?

Answer:
The reactions catalyzing the assimilation of CO2 to carbohydrates take place in the stroma where all the necessary enzymes are localized. These reactions are referred to as ‘carbon reactions’ (also called dark reactions) leading to the photosynthetic reduction of carbon to carbohydrates.

In the first phase of carbon reaction, C02 enters the leaf through the stroma. This CO2 is accepted by a 5-carbon molecule, ribulose-1-5 bisphosphate (RuBP) already present in the leaf. It forms two molecules of 3-carbon, compound, 3- phosphoglycerate (PGA). This 3-carbon molecule is the first stable product of this pathway and hence it is called C, PATHWAY.

The formation of (PGA) with CO2 combining with RuBP is called carboxylation. This reaction is catalyzed by an enzyme called ribulose bisphosphate carboxylase (Rubisco). This enzyme also possesses oxygenase activity and hence abbreviated as Rubisco. This activity allows O, to compete with C02 for combining with RuBP.

After the carboxylation reduction of PGA occurs and ATP and NADPH, formed during photochemical reactions with the reduction of PGA, glyceraldehyde-3 phosphate-a carbohydrate is formed. These 3-carbon molecules, also called triose phosphates act as precursors for the synthesis of sucrose and starch. To complete the cycle, and to continue it, regeneration of the 5-carbon acceptor molecule, that is RuBP takes place.

The C3 type of carbon reaction occurs in the stroma of the chloroplast. This pathway is called the Calvin cycle.

The CO2concentrating mechanism is called the C pathway. Operation of the C4 pathway requires the cooperation of both cell-type mesophyll and bundle sheath cells. The objective of the C. pathway is to build up a high concentration of CO2 which suppresses photorespiration. This C. pathway is more efficient than the C3 pathway. Hence C? pathway does not operate C4 plants. (See the table)
Photosynthesis in Higher Plants Class 11 Important Extra Questions Biology 6
Schematic representation of C3 pathway Calvin cycle.

Question 7.
Describe briefly the experiment conducted by the scientist, T.W. Englemann.

Answer:
T.W. Englemann plotted the action spectrum of photosynthesis.

Photosynthesis can occur in visible light of wavelength varying between 390 to 763 nm. The rate of photosynthesis is not uniform in light of all wavelengths.

It varies depending upon their relative absorption by chlorophyll pigments. The graph showing the relative yield or rate of photosynthesis in plants exposed to monochromic light of different wavelengths is termed as ACTION SPECTRUM. The rate of photosynthesis, as shown in the action spectrum is maximum in the blue region of light.
Photosynthesis in Higher Plants Class 11 Important Extra Questions Biology 7
Curves showing a comparison of absorption and action spectra of chlorophyll pigments during photosynthesis

Question 8.
What is a photosystem? Which is the pigment that acts as a reaction center? Describe the interaction of photosystem 1 and photosystem II.

Answer:
The light is entrapped by a group of chlorophyll molecules which together constitute a photosystem. Each pigment system has a trap or reaction center, which is either P700 or P680 ao. In this ‘P’ stands for pigment and figures 680 and 700 for the wavelength of light. Chlorophyll molecule acts as a trap center with the transfer of high energy electron to electron transport system (ETS).

The high-energy electrons return rapidly to their normal low energy orbitals in the absence of light and the excited chlorophyll molecule reverts to its original stable condition. These two photosystems: photosystem-I and photosystem-11 exist with different forms of chlorophyll ‘a’ as the reaction center. The PS-II is located in the appressed regions of grana thylakoids and the PS-I in the stroma thylakoids and non-appressed regions of grana.

The function of two photosystems that interact with each other is to trap light energy and convert it to chemical energy (ATP). This chemical energy stored in the form of ATP is used by living cells.
Photosynthesis in Higher Plants Class 11 Important Extra Questions Biology 8
Distribution of pigment in photosystem I and Photosystem II.

Question 9.
What led to the evolution of the C4 pathway of photosynthesis? Describe in detail.

Answer:
Kortschak (1965) observed that in sugarcane, the compound in which CO, got incorporated was oxaloacetic acid or oxaloacetate (OAA), a 4-carbon compound instead of phosphoglyceric acid, a 3-carbon com¬pound.

Hatch and Slack (1965-1967) found it a regular mode of COfIxation in a number of monocots such as sugar cane, maize, sorghum, and Pennisetum. They found that the initial acceptor of CO2 in such plants is Phosphoenalpymvic acid instead of RuBP and the first stable compound is oxaloacetate acid, a 4-carbon compound. These plants are termed C4 plants as the first stable compound is a 4-carbon compound and other plants are termed C3 plants.

Hatch and Slack observed that these plants have another pathway of CO2 a fixation that precedes the Calvin cycle occurring in C3 plants. This cycle is known as the C4 pathway.

Question 10.
Describe in detail how ATP and NADPH2 are formed during photochemical reactions?

Answer:
Photosynthesis at present is thus considered basically an oxidation-reduction process during which water is oxidized to release oxygen and CO2 is reduced to carbohydrates. Photosynthesis involves two steps- the first step is light-dependent called Light reaction or Hill reaction or photochemical phase. The second step is the Dark reaction or Blankman’s reaction or the Biosynthetic phase, which does require light.

Light Reaction or Hill Reaction: During this process, solar energy is converted into chemical energy, light is trapped by chlorophyll and carotenoid pigments and is converted into chemical energy which is stored in the form of ATP energy-rich molecule.

Photolysis of water occurs that leads to the evolution of oxygen and formation of H+ ions, the latter combining with NADP to form NADPH2 often termed as reducing power. ATP and NADPH together termed assimilatory power as CO2 fixation during the dark reaction.

Photolysis of water takes place in presence of light and water oxidizing enzyme as follow:
Photosynthesis in Higher Plants Class 11 Important Extra Questions Biology 9
The unstable OH” combines to form water and molecular oxygen after losing the electrons which are accepted by oxidized chlorophyll molecule. (P680 of PS11) through an unknown electron acceptor compound “Z”. This step requires the presence of Mn++ and Cl’ ions.
Photosynthesis in Higher Plants Class 11 Important Extra Questions Biology 10

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NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | BIOLOGY IMPORTANT QUESTIONS | CHAPTER -11 | TRANSPORT IN PLANTS | EDUGROWN |

In This Post we are  providing Chapter-11 TRANSPORT IN PLANTS NCERT MOST IMPORTANT QUESTIONS for Class 11 BIOLOGY 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 MOST IMPORTANT QUESTIONS ON TRANSPORT IN PLANTS

Question 1.
What are the advantages of transpiration?

Answer:

  1. Ascent of sap: Transpiration pull created in leaves is responsible for ascent of sap.
  2. Absorption of water: Transpiration pull is also responsible for passive absorption of water.
  3. The distribution of minerals in different parts of the plant is done by transpiration.
  4. Cooling effect: Transpiration lowers the temperature of the leaf and causes a cooling effect.
  5. The increased rate of transpiration favours the development of tissue, which provides strength to the plant.
  6. Excessive transpiration induces hardness which imparts resistance of plants to drought.

Question 2.
What is the diffusion pressure? What are the factors which affect the rate of diffusion?

Answer:
The pressure exerted by the particles due to their tendency to diffuse from the region of higher concentration to the region of lower concentration.

Various factors affect the rate of diffusion:

  • Diffusion pressure gradient
  • Temperature
  • The density of diffusing substance
  • The density of the medium.

Question 3.
Define water holding capacity or field capacity of the soil.

Answer:
After heavy rainfall or irrigation, the amount of water actually retained by soil even against the force of gravity is termed as water holding capacity or field capacity of the soil. It is expressed in terms of the percentage of water present per unit dry weight of soil.

Question 4.
Describe osmosis as a special case of diffusion.

Answer:
Uptake and distribution of water, solutes, and gases occur in a plant as a result of diffusion. The diffusion of water through a semipermeable membrane is known as osmosis. The diffusion of water molecules continues across the membrane until an equilibrium is attained. Osmosis can be demonstrated by a simple experiment as follows

A thistle funnel is taken and tied with a semipermeable membrane (parchment paper) to the wide mouth of the thistle funnel and made tight.

The thistle funnel is filled with concentrated sugar solution and its wide mouth is dipped into water contained in a beaker. The membrane allows water molecules to pass through and not the sugar molecules. The level of sugar solution will rise in the funnel from ‘A’ to ‘B’. This demonstrates osmosis.
Transport in Plants Class 11 Important Extra Questions Biology 1
A demonstration of osmosis. A thistle funnel is filled with sucrose solution and kept inverted in a beaker containing water,
(a) Water will diffuse across the membrane (as shown by arrows) to raise the level of the solution in the funnel
(b) Pressure can be applied as shown to stop the water movement into the funnel.

Question 5.
What is the factor which affects the rate of transpiration?

Answer:
There are two factors.
A. External (Environmental) factors.
B. Internal (Living factors).

A. External factors:

  1. Light: Causes stomatal opening
  2. Temperature: High temperature decreases relative humidity increasing transpiration.
  3. Humidity: It directly affects the rate of transpiration that is related to the vapour pressure of the atmosphere.
  4. Wind: High velocity of wind causes closure of stomata.
  5. Soil moisture: The rate of transpiration is directly proportional to the quantity of available moisture in the soil.

B. Internal factor:

  1. Root Shoot Ratio: Roots absorbs water, should transpire water, hence their ratio affects transpiration.
  2. Leaf area: Smaller plants tend to transpire more rapidly per unit area than larger plants.
  3. Leaf Anatomy: Modification of leaves affects transpiration.

Question 6.
Describe the role of osmotic potential in regulating the water potential of plant cells.

Answer:
Osmotic potential is the amount by which the water potential of pure water is reduced by the presence of the solute. The osmotic potential has a negative value.

If we apply additional pressure, the water can be flown out of the solution.

Osmosis is driven by two factors:

  1. The concentration of dissolved solutes in solution,
  2. Pressure difference.

Water potential is the driving force for water movement in plants. Water potential represents the free energy associated with water. Osmotic potential regulates the flow of water molecules through the membrane.

Question 7.
Describe the theories related to the translocation of water.

Answer:
There are three most important theories related to the translocation of water.

  1. Root pressure theory
  2. Capillarity
  3. Cohesion theory.

1. Root pressure theory: Water flows from higher water potential to low water potential. Water from the soil is absorbed by root hairs and conducted through xylem vessels. Mineral ions from the soil are taken up by roots and get deposited in the xylem vessels.

When the stem of a plant is cut transversely above the soil surface, a drop of the xylem sap will exude from the cut surface. This indicates the presence of positive pressure in the xylem. This pressure is known as Root Pressure.

2. Capillarity: Capillarity means a rise in water in tubes of small diameter kept in a water vessel. The uptake of water through xylem vessel is possible in small size plants through capillarity. This is due to the forces of adhesion and cohesion.

Adhesive forces attract molecules of different kinds whereas cohesive forces attract molecules of the same kind to each other. According to this theory, water is taken due to the force of adhesion and flows upward due to the force of cohesion.

3. Cohesion: This is the most important theory of water movement through plants. It is based on the force of cohesion between water molecules. This sets up a continuous water column from the top to the root tip of the plant. According to this theory water evaporates from the leaf to the atmosphere, results in a decrease in the water potential of epidermal cells.

This loss of water is balanced by water moving from adjacent cells along a water potential gradient. The movement of the water occurs from the soil to the root. Uptake of water is termed as cohesion theory and also known as transpirational pull.

Question 8.
Is there a general mechanism to explain the opening and closing of stomata? Justify your answer.

Answer:
There is no general mechanism to explain the opening and closing of stomata. Because opening and closing of stomata are regulated by the accumulation of solute in the guard cells. Solutes are taken in the guard cells, as a result, osmotic potential and water potential of guard cells are lowered, the guard cells become turgid and swell size, resulting in the stomatal opening. With a decline in guard cell solutes, water moves out, resulting in the stomatal opening.

There are two theories to explain the mechanism of opening and closing of stomata.

  1. Classical starch sugar conversion theory: According to this theory, the change in osmotic concentration is brought about due to the conversion of starch into glucose and vice-versa.
  2. K+ Influx and Efflux theory: According to this theory when the leaf is exposed to light, the pH of the guard ceils rises due to the active transfer of H+ ions from the cytoplasm into chloroplast’s utilization of CO. in photosynthesis. In the majority of plants, stomata remain open during the day and close at night.

Hence, there is no general mechanism to explain the stomatal opening and closing.

Question 9.
Mention some factors that influence stomatal opening and closing. How are these factors involved in regulating stomatal behaviour?

Answer:
Factors affecting stomatal movements:

  1. Light: In most of the plant’s stomata open during the day. The effect of light causes the opening of stomata or it may be either due to the hydrolysis of starch into glucose.
  2. The water content of leaves: A decrease of water content in stomatal cells results in an increase in their D.P.D. Water from guard cells moves into these cells and stomata close.
  3. CO2 concentration: Low CO2 concentration in guard cells causes the opening of stomata.
  4. pH: High pH stimulates the opening of stomata and low pH causes closure of stomata and high concentration of CO2 causes closure of stomata.
  5. Temperature: High temperature stimulates the opening of the stomata.
  6. Atmospheric Humidity: Humid environment favours opening and dryness causes closure of stomata.
  7. Minerals: Minerals like P, Mg, Ca etc. affect the stomatal opening. A high concentration of K+ ions causes the opening of stomata.
  8. Growth Hormones: Cytokinins stimulates the opening of stomata. Abscisic acid induces the closure of stomata.

Question 10.
Write short notes on:
(i) Cohesion-Tension and Transpiration pull theory.

Answer:
Transpiration pull theory: Ascent of sap has been explained satisfactorily by Dixon with the help of a theory called Transpiration pull theory. According to this theory water continuously evaporates from the turgid and moist cell walls of mesophyll cells in the leaves.

It makes the mesophyll air saturated. The air outside the leaf is dry. So a gradient is set up which allows the water vapours to go out from the interior of the leaf to the outside through the stomata. The mesophyll cells draw water from the deeper tissue, which in turn take water from the xylem of the leaf. It creates a kind of pull in the leaf called transpiration pull.

The xylem of the leaf is connected to the xylem of the stem and further to the xylem of the roots. Since there is a continuous column of water in the plant, water is virtually lifted up due to transpiration pull a situation similar to one like drawing a bucket of water from a well. The column of water does not break because of the great force of cohesion among the water molecules. This theory is also called the cohesion of water molecules theory.

(ii) Mass flow hypothesis.
Answer:
Mass flow hypothesis: The carbohydrates prepared in the leaves are translocated to other parts of the plant in the form of sucrose through phloem at the expense of metabolic energy. Munch’s mass flow hypothesis is the most accepted theory for the translocation of organic food.

According to this hypothesis, organic substances move from the region of high osmotic pressure to the region of low osmotic pressure due to the development of a gradient of turgor pressure. This can be proved by taking two interconnected osmometers. One of the osmometers has a high solute cone than the other. The whole apparatus is placed in water.

Water enters the osmometer with a high solute cone. It creates high turgor pressure in it. High turgor pressure forces the solution to move through the tube to the other osmometer. It is called mass flow. If somehow, the solute is continuously added to the donor osmometer and converted into the osmotically inactive compound in the other osmometer, this system can work indefinitely.
Transport in Plants Class 11 Important Extra Questions Biology 2
Munch’s mass flow apparatus.

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NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | BIOLOGY IMPORTANT QUESTIONS | CHAPTER -12 | MINERAL NUTRITION | EDUGROWN |

In This Post we are  providing Chapter-12 MINERAL NUTRITION NCERT MOST IMPORTANT QUESTIONS for Class 11 BIOLOGY 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 MOST IMPORTANT QUESTIONS ON MINERAL NUTRITION

Question 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. However, a nutrient is inessential for a plant if it is not involved in the plant’s physiology and metabolism. For example, plants growing near radioactive sites tend to accumulate radioactive metals. Similarly, gold and selenium get accumulated in plants growing near mining sites. However, this does not mean that radioactive metals, gold, or selenium are essential nutrients for the survival of these plants.

Question 2:

Why is purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics?

ANSWER:

Hydroponics is the art of growing plants 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.

Question 3:

Explain with examples: macronutrients, micronutrients, beneficial nutrients, toxic elements and essential elements.

ANSWER:

Macronutrients: They are the nutrients required by plants in large amounts. They are present in plant tissues in amounts more than 10 mmole kg–1 of dry matter. Examples include hydrogen, oxygen, and nitrogen.

Micronutrients: They are also called trace elements and are present in plant bodies in very small amounts, i.e., amounts less than 10 mmole kg– 1 of dry matter. Examples include cobalt, manganese, zinc, etc.

Beneficial nutrients: They are plant nutrients that may not be essential, but are beneficial to plants. Sodium, silicon, cobalt and selenium are beneficial to higher plants.

Toxic elements: Micronutrients are required by plants in small quantities. An excess of these nutrients may induce toxicity in plants. For example, when manganese is present in large amounts, it induces deficiencies of iron, magnesium, and calcium by interfering with their metabolism.

Essential elements: These elements are absolutely necessary for plant growth and reproduction. The requirement of these elements is specific and non-replaceable. They are further classified as macro and micro-nutrients.

Question 4:

Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency.

ANSWER:

The five main deficiency symptoms arising in plants are:

  • Chlorosis
  • Necrosis
  • Inhibition of cell division
  • Delayed flowering
  • Stunted plant growth

Chlorosis or 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.

Question 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:

In plants, the deficiency of a nutrient can cause multiple symptoms. For example, the deficiency of nitrogen causes chlorosis and delayed flowering.

In a similar way, the deficiency of a nutrient can cause the same symptom as that caused by the deficiency of another nutrient. For example, necrosis is caused by the deficiency of calcium, magnesium, copper, and potassium.

Another point to be considered is that different plants respond in different ways to the deficiency of the same nutrient.

Hence, to identify the nutrient deficient in a plant, all the symptoms developed in its different parts must be studied and compared with the available standard tables.

Question 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:

Deficiency symptoms are morphological changes in plants, indicating nutrient deficiency. Deficiency symptoms vary from one element to another. The plant part in which a deficiency symptom occurs depends on the mobility of the deficient element in the plant. 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.

Question 7:

How are the minerals absorbed by the plants?

ANSWER:

The absorption of soil nutrients by the roots of plants occurs in two main phases – apoplast and symplast.

During the initial phase or apoplast, 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 or symplast, 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.

Question 8:

What are the conditions necessary for fixation of atmospheric nitrogen by Rhizobium. What is their role in N2 -fixation?

ANSWER:

Rhizobium is a symbiotic bacteria present in the root nodules of leguminous plants. The basic requirements for Rhizobium to carry out nitrogen fixation are as follows:

(a) Presence of the enzyme nitrogenase

(b) Presence of leg-haemoglobin

(c) Non-haem iron protein, ferrodoxin as the electron-carrier

(d) Constant supply of ATP

(e) Mg2+ions as co-factors

Rhizobium contains the enzyme nitrogenase – a Mo-Fe protein – that helps in the conversion of atmospheric free nitrogen into ammonia.

The reaction is as follows:

N2 + 8e + 8H+ + 16 ATP→ 2 NH3 + H2 + 16ADP + 16Pi

The Rhizobium bacteria live as aerobes under free-living conditions, but require anaerobic conditions during nitrogen fixation. This is because the enzyme nitrogenase is highly sensitive to molecular oxygen. The nodules contain leg-haemoglobin, which protects nitrogenase from oxygen.

Question 9:

What are the steps involved in formation of a root nodule?

ANSWER:

Multiple interactions are involved in the formation of root nodules. The Rhizobium bacteria divide and form colonies. These get attached to the root hairs and epidermal cells. 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. As a result, the cells in the cortex and pericycle undergo division, leading to the formation of root nodules. The nodules finally get connected with the vascular tissues of the roots for nutrient exchange.

Question 10:

Which of the following statements are true? If false, correct them:

(a) Boron deficiency leads to stout axis.

(b) Every mineral element that is present in a cell is needed by the cell.

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

(d) It is very easy to establish the essentiality of micronutrients because they

are required only in trace quantities.

ANSWER:

(a) True

(b) All the mineral elements present in a cell are not needed by the cell. For example, plants growing near radioactive mining sites tend to accumulate large amounts of radioactive compounds. These compounds are not essential for the plants.

(c) Nitrogen as a nutrient element is highly mobile in plants. It can be mobilised from the old and mature parts of a plant to its younger parts.

(d) True

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NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | BIOLOGY IMPORTANT QUESTIONS | CHAPTER -10 | CELL CYCLE AND CELL DIVISION | EDUGROWN |

In This Post we are  providing Chapter-10 CELL CYCLE AND CELL DIVISION NCERT MOST IMPORTANT QUESTIONS for Class 11 BIOLOGY 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 MOST IMPORTANT QUESTIONS ON CELL CYCLE AND CELL DIVISION

Page No 171:

Question 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.

Question 2:

Distinguish cytokinesis from karyokinesis.

ANSWER:

Cytokinesis Karyokinesis 
(i)Cytokinesis is the biological process involving the division of a cell’s cytoplasm during mitosis or meiosis.(i)Karyokinesis is the biological process involving the division of a cell’s nucleus during mitosis or meiosis.
(ii)Stages such as prophase, metaphase, anaphase, and telophase are not present in cytokinesis.(ii)It is divided into four stages –prophase, metaphase, anaphase, and telophase

Question 3:

Describe the events taking place during interphase.

ANSWER:

Interphase involves a series of changes that prepare a cell for division. It is the period during which the cell experiences growth and DNA replication in an orderly manner. Interphase is divided into three phases.

(i) G1 phase

(ii) S phase

(iii) G2 phase

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.

Question 4:

What is G0 (quiescent phase) of cell cycle?

ANSWER:

Gor quiescent phase is the stage wherein cells remain metabolically active, but do not proliferate unless called to do so. Such cells are used for replacing the cells lost during injury.

Question 5:

Why is mitosis called equational division?

ANSWER:

Mitosis is the process of cell division wherein the chromosomes replicate and get equally distributed into two daughter cells. The chromosome number in each daughter cell is equal to that in the parent cell, i.e., diploid. Hence, mitosis is known as equational division.

Question 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

Question 7:

Describe the following:

(a) synapsis (b) bivalent (c) chiasmata

Draw a diagram to illustrate your answer.

ANSWER:

(a) Synapsis

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

(b) Bivalent

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

(c) Chiasmata

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.

Question 8:

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

ANSWER:

Cytokinesis in plant cellsCytokinesis is animal cells
(i)The division of the cytoplasm takes place by cell plate formation.(i)The division of the cytoplasm takes place by cleavage.
(ii)Cell plate formation starts at the centre of the cell and grows outward, toward the lateral walls.(ii)Cleavage starts at the periphery and then moves inward, dividing the cell into two parts.

Question 9:

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

ANSWER:

(a) Spermatogenesis or the formation of sperms in human beings occurs by the process of meiosis. It results in the formation of four equal-sized daughter cells.

(b) Oogenesis or the formation of ovum in human beings occurs by the process of meiosis. It results in the formation of four daughter cells which are unequal in size.

Question 10:

Distinguish anaphase of mitosis from anaphase I of meiosis.

ANSWER:

Anaphase of mitosisAnaphase I of meiosis
Anaphase is the stage during which the centromere splits and the chromatids separate. The chromosomes move apart, toward the opposite poles. These chromosomes are genetically identical.During anaphase I, the homologous chromosomes separate, while the chromatids remain attached at their centromeres.Hence, in anaphase I, the chromosomes of each bivalent pair separate, while the sister chromatids remain together.

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NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | BIOLOGY IMPORTANT QUESTIONS | CHAPTER -9| BIOMOLECULES | EDUGROWN |

In This Post we are  providing Chapter-9 BIOMOLECULES NCERT MOST IMPORTANT QUESTIONS for Class 11 BIOLOGY 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 MOST IMPORTANT QUESTIONS ON BIOMOLECULES

Question 1.
Enlist the functions of small carbohydrates?

Answer:

  1. Monosaccharides are formed during the photosynthetic pathway. They are stored in plants and are utilized by other living organisms depending on them.
  2. Glucose is the blood sugar of many animals and on oxidation, it provides energy for all vital activities.
  3. Nucleotides and nucleosides contain pentose sugar in the form of ribose and deoxyribose sugars. They form a part of nucleic acids.
  4. Lactose of milk is formed from glucose and galactose and mammary glands of mammals.
  5. Glucose is used for the synthesis of fats and amino acids.
  6. Structural polysaccharides like cellulose and oligosaccharides are derived from mono-saccharides.
  7. Food storage polysaccharides like starch and glycogen are derived from monosaccharides.

Question 2.
Enumerate the functions of Lipids.

Answer:

  1. Lipids are storage products in plants as well as animals.
    (a) In plants, fats are stored in cotyledons or endosperm to provide nourishment to the developing embryo.
    (b) In animals fats are stored in adipocytes to be used whenever required by the body.
  2. In animals, subcutaneous fats act as an insulation layer and shock \ absorber.
  3. They form structural components of membranes, phospholipids, glycolipids, and sterols.
  4. They take part in the synthesis of steroid hormones, vitamin D, and bile salts.
  5. Act as a solvent for fat-soluble vitamins i.e., vitamin A, D, E, and K.
  6. The neutral fats form a concentrated fuel producing more than twice as much energy per gram as do the carbohydrates. They thus, represent an economical food reserve in the body.
  7. The wax lipids form a waterproof protective coating on animal furs, plant stem, leaves, and fruits.

Question 3.
How does water help in maintaining the constancy of the internal environment of an organism?

Answer:
Some substances, capable of neutralizing acids or bases, remain in solution in the cytoplasm as extracellular fluids, e.g., bicarbonate (HCO3), carbonic acid, dibasic phosphate (HPO4-2). Acids and bases mix in the body fluids with these substances and are neutralized by them. Because of its solvent action water aids in keeping a constant pH.

Water also helps in maintaining constant body temperature by eliminating excess heat through the evaporation of sweat. Elimination of waste products through urine also helps in maintaining the constancy of the internal environment of an organism.

Question 4.
What are peroxisomes and phagosomes?

Answer:
Peroxisomes: These were for the first time observed in the kidney of rodents. They are found both in plants and animals. Their size varies from 0.5 to lp in diameter. They are delimited by a single membrane and contain a finely granular matrix. They often possess a central core called nucleoid which may consist of parallel tubules or twisted with strands. Peroxisomes are generally observed in close association with the endoplas¬mic reticulum.

Peroxisomes in different plant and animal cells differ con¬siderably in their enzymatic make-up, but they contain some peroxide-producing enzymes like urate, oxidase, D-amino acid oxidase, B-hydroxy acid oxidase, and catalase. Peroxisomes are somehow associated with some metabolic processes like photorespiration and lipid metabolism in animal cells.

Sphaerosomes: There are cell organelles bounded by a single membrane. They contain enzymes and are visible under the light microscope. These show some affinities for fat stains, including Sudan stain and sodium tetroxide.

These organelles originate from E.R. by budding. They contain enzymatic proteins which help in synthesizing oils and fats. Further devel¬opment of phagosomes takes place through an increase in the lipid content with a concomitant decrease in protein.

Question 5.
Enumerate the importance of Energy carriers.

Answer:
Energy carriers consist of nucleotides having one or two additional phosphate groups linked up at their phosphate end forming diphosphates and triphosphates. Linkage of additional phosphate groups occurs at the cost of a large amount of energy. This energy is provided by the oxidation of food mainly glucose and by photosynthesis.

Separation of the additional phosphate groups from the nucleotides by enzymatic hydrolysis releases a correspondingly large amount of energy.

Thus, ADP and ATP provide ready energy for biological activities.

The bonds joining the additional phosphate groups to the nucleotides are called high energy or energy-rich bonds, as they carry a great deal of energy. The nucleotides having more than one phosphate group are called higher nucleotides.

The energy of energy carriers, when set free is utilized for driving energy-dependent reactions in the cell and is biologically useful energy. ATP is the most common energy carrier in cells and is often called the energy currency of the cell.

Question 6.
Explain the functions of amino acids.

Answer:

  1. Amino acids are the building blocks for proteins.
  2. The amino acid Tyrosine takes part in the formation of the skin pigment melanin as well as hormones thyroxine and adrenaline.
  3. Glycine is important for the formation of heme.
  4. Tryptophan takes part in the formation of the vitamin nicotinamide.
  5. In plants, tryptophan forms the growth hormone indole-3- acetic acid.
  6. Amino acids are converted into glucose by deamination.
  7. Histamine and other biogenic amines are formed by the removal of carboxyl groups from amino acids.

Question 7.
Give reasons for following
(i) Salts dissolve in water but oil does not

Answer:
Water molecules are hydrogen-bonded to form short-lived macromolecular aggregates. To dissolve in water, a solute molecule must form hydrogen bonds with water molecules. Salts are polar compounds, their hydrophilic polar groups form hydrogen bonds with water molecules. So they dissolve oils having hydrophobic non-polar groups that cannot join the lattice structure of water. Thus non-polar molecules of oil do not dissolve in water.

(ii) Amino acid can be basic
Answer:
A free amino group is basic and a free carboxyl group is acidic. Amino acids can be basic because they may carry two amino groups and one carboxyl group e.g., Arginine. One free amino group causes amino acids to be basic.

(iii) Phospholipids form a thin layer on the surface of an aqueous medium.
Answer:
Phospholipids form a thin layer on the surface of an aqueous medium due to the simultaneous presence of both polar and non-polar groups in the molecule. As a result, the phospholipid molecules may arrange themselves in a double-layered membrane in aqueous media.

Question 8.
Illustrate lock and key hypothesis of enzyme action?

Answer:
Mechanism of Enzyme action: The working of enzymes is a complex one. All enzymes first of all combine with the reactions they catalyze. In other words, enzymes with substrates form an intermediate complex before decomposition of the substrate can occur.

This two-way reaction can be represented as follows.
1st step: Enzyme substrate complex = Enzyme + Product.
Formation of the enzyme-substrate complex during enzyme action.

From the above, it is clear that the enzymes must combine first with substrate molecules in order to act. In order to explain the mode of action of an enzyme. Fischer proposed a lock and key theory. According to him if the right key fits in the right lock. The lock can be opened, otherwise not.
Biomolecules Class 11 Important Extra Questions Biology 11
Model of enzyme activity

To explain the above in context with the enzyme action it is believed that molecules have specific configurations into which other molecules can fit. The molecules which are acted upon by the enzymes are called substrates of the enzymes. Under the above assumption, only those substrate molecules with the proper geometric shape can fit into the active site of the enzymes.

If this happens, the above molecules may compete with the substrate, and the reaction may either slow down or stop. Substances are called competitive inhibitors because they act to prevent the production of a substance.

An induced-fit model of enzyme action was given by Koshland (1959). Buttressing and catalytic are two groups of the active site of the enzyme. Their site when the substrate attaches to its bonds is broken.

Question 9.
What is the structure of DNA?

Answer:
The nucleic acids are among the largest of all molecules found in living beings. They contain three types of molecules (a) 5 carbon sugar, (b) Phosphoric acid (usually called phosphates when in chemical combi¬nation), and nitrogen-containing bases (Purines and Pyrimidines). The three join together to form a nucleotide i.e., sugar+ base + phosphate = Nucleotide. Only a few nucleotides are possible. They differ only in the kind of purines or pyrimidine (nitrogen-containing bases).

In 1953 J.D. Watson and F.H.C. Crick working in Cambridge Uni¬versity, England prepared a model of DNA molecule elucidating the struc¬ture of DNA molecule. They were awarded the Nobel Prize for this outstanding work.
Biomolecules Class 11 Important Extra Questions Biology 12
Structure of DNA

Watson and Crick model of DNA: According to Watson and Crick, the DNA molecule consisted of two strands twisted around each other in the form of a helix. Each strand is made of polynucleotides, each polynucleotide consisting of many nucleotides which remain united with its complimentary’ chain with the help of bases.

Adenine always unites with thymine and cytosine with guanine. It means that one polynucleotide chain of DNA molecule is complementary to the other.

The distance between two chains of the helix is about 20 A and the helix turns over every 34 A. Each mm of the chain consists of about 10 nucleotides.
Biomolecules Class 11 Important Extra Questions Biology 13
Structure of DNA

Question 10.
How does the substrate concentration affect the velocity of enzyme reaction?

Answer:
Michaelis constant or more appropriately Michaelis-Menten constant (Km) is a mathematical derivation given by Leonor Michaelis and Monde Menten in 1913 with the help of which velocity of reaction can be calculated for any substrate concentration.
Biomolecules Class 11 Important Extra Questions Biology 14
Effect of substrate concentration on enzyme action

Km or Michaelis constant is the substrate concentration at which the chemical reaction attains half its maximum velocity. The constant is an inverse measure of the affinity of an enzyme for its substrate, that is the smaller the Km the greater the substrate affinity and vice versa. The value usually lies between 104 – 105 M

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NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | BIOLOGY IMPORTANT QUESTIONS | CHAPTER -8 | CELL : THE UNIT OF LIFE | EDUGROWN |

In This Post we are  providing Chapter-8 CELL : THE UNIT OF LIFE NCERT MOST IMPORTANT QUESTIONS for Class 11 BIOLOGY 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 MOST IMPORTANT QUESTIONS ON CELL : THE UNIT OF LIFE

1. Define Cell. Explain its Structure.

Meaning and Types of Cell- Basic Unit of Life:

Robert Hooke (1665) observed the so called cells for the first time in a thin slice of cork under a very primitive microscope invented by him. He coined the term “cell”. Living cells were seen for the first time by Anton Van Leeuwenhoek (1632-1723), with his improved microscope. Much later (1838-39) cell-theory was proposed by two German biologists separately – viz., M.J. Schleiden for plants and Theodor Schwann for animals. According to them, “Cells are the structural and functional units of living organisms.” Later, Rudolph Virchow (1855) extended the cell theory and suggested that all living cells arise from pre-existing cells (Omnis cellula e cellula).

Viruses are the most notable exception to the cell theory because they lack internal organisation and protoplasm. Other exceptions include protozoans, fungi and algae because their entire organisation is represented by just one cell.

Size of the Cell:

Usually the cells are microscopic and their size varies between 10 µm and 100 µm. The smallest cells are those of PPLO (Pleuropneumonia like organisms) whose size may vary between 0.1 to 0.4 µm. The largest cell is the egg of ostrich measuring about 15 cm in its outer diameter. The longest animal cell is the nerve cell which may be approximately one metre long, while the longest plant cell is the sclerenchymatous fibre of Boehmeria nevia (about 55 cm long).

The factors governing the size of the cell are:

(i) The ratio between the volume of the nucleus and that of the cytoplasm.

(ii) The ratio of the cell surface to the cell volume.

(iii) The rate of metabolism.

(iv) The size and the number of chromosomes.

Shape of the Cell:

There is a great variation in the shape of cell. Some cells, e.g., Amoeba, slime moulds and WBCs have a constantly changing shape while others (e.g., neurons, muscle cells, RBCs, etc.) have a characteristic shape. The shape is governed by the plasma membrane and the cell wall, if present.

Numbers of the Cells:

The body of protozoans, bacteria, certain fungi and algae is represented by one cell only. They are called as unicellular or acellular forms. Most of the animals and plants are made up of several cells. They are called multicellular.

On the basis of the nuclear organisation, cells are of two types:

1. Prokaryotic:

Cells in which mitochondria, chloroplast and nuclear membrane are absent are called prokaryotic cells. For example bacteria, blue green algae (cyanobacteria), and mycoplasma.

2. Eukaryotic:

Cells in which nucleus and membrane bound organelles are present are called eukaryotic cells. They are found in all plants and animals.

2. What is a Cell Wall? What are the Functions of it?

The presence of a cell wall is a characteristic feature of plant cells. It is always formed by the activity of the protoplasm.

The adjacent cell walls are cemented together by middle lamella composed of calcium or magnesium pectate cell wall is differentiated into three layers, viz.:

1. Primary Cell Wall:

It is the outermost layer of the cell wall present on both the sides of middle lamella. It is usually thin (1-3 µm), and elastic. The chief constituents of the primary cell wall are cellulose and hemicellulose, some amount of pectin and a structural protein extension (rich in proline and hydroxyproline) may also be present. In thin walled cells (like meristematic cells, parenchyma, collenchyma and mesophyll cell) primary cell wall remains as the only layer.

2. Secondary Cell Wall:

It is much thicker (5-10 µm), rigid and inelastic It is formed only when the growth in the surface area of the primary cell wall ceases. Its position is in between the primary wall and the protoplast. The secondary cell wall may be thickened on account of the deposition of substances like cutin, suberin, lignin and pectin.

3. Tertiary Cell Wall:

It is of rare occurrence (in tracheids of gymnosperms). It is deposited on the inner side of the secondary cell wall. It is relatively richer in xylem (a polymer of pentose sugar D-xylose) than cellulose.

Functions of Cell Wall:

Functions of cell wall are largely mechanical. It acts like a skeleton of the plant by providing rigidity strength and flexibility. It maintains the shape and structure of the cells and the tissues, and protects the protoplasm against mechanical injuries. By impregnation of cutin and suberin it also reduces loss of water by transpiration. It, being freely permeable, helps in the absorption and transportation of water and solutes in the different parts of the plant.

3. What is Protoplasm? Explain the Nature and Properties of it.

Felix Dujardin (1835) described protoplasm in protozoa. He called it “Sarcode”. The term protoplasm was coined by Johannes E. Purkinje (1839) and Hugo Von Mohl (1846) independently. Huxley called protoplasm as the physical basis of life, for life cannot exist apart from it.

Protoplasm refers to the living substance of the cell and includes all parts of the cell. It is the set of all metabolic functions. The protoplasm can be divided into cytoplasm and nucleus.

Cytoplasm:

It is the part of cell occurring between plasma membrane and nucleus. This term was introduced by Strassburger (1882).

It is composed of two distinct types of structures, viz.:

(i) A continuous fluid like substance called cytosol.

(ii) A number of organelles which are having definite function.

Physical Nature of Protoplasm:

(i) It is a thick, greyish, viscous jelly-like translucent fluid of colloidal nature.

(ii) The colloid particles exhibit Brownian movement.

(iii) It shows Tyndall’s effect, i.e., when a beam of strong light is passed through it in a dark room, the path of light appears like a cone.

(iv) It is a reversible colloidal system. It can be watery (Sol) at one time and jelly like at another (Gel). The sol and gel states are reversible.

Chemical Nature of Protoplasm:

Generally the elements of protoplasm are grouped in the following three categories according to their abundance in the protoplasmic matrix:

(i) Major Constituents:

These include Oxygen (62%), Carbon (20%), Hydrogen (10%) and Nitrogen (3%).

(ii) Trace Elements:

These occur in very low quantities or in traces.

The trace elements are- calcium, potassium, phosphorus, sodium, chlorine, magnesium, sulphur, iodine and iron.

(iii) Ultrastructure Elements:

These are required by the cell as co-factors for various metabolic reactions, e.g., copper, cobalt, manganese, zinc, molybdenum, boron, silicon, etc.

Biological Properties of Protoplasm:

(i) Irritability:

Protoplasm shows the ability to respond to stimuli.

(ii) Conductivity:

Protoplasm of neurons is specially adapted to conduct the impulses.

(iii) Metabolism:

Protoplasmic matrix is the seat of various metabolic processes of the cell.

(iv) Growth:

A successful metabolism always results into synthesis of new protoplasm thereby causing growth of the cell.

(v) Reproduction:

Protoplasm is a self-perpetuating substance.

4. Define Plasma Membrane. What are the Functions of it? Explain its Structure.

Cytoplasm of all the living cells is enclosed by a living membrane called as cell membrane, plasma membrane or plasmalemma. The term cell membrane was given by C. Nageli and C. Cramer (1855) while the term plasmalemma was coined by J.Q. Plower (1931).

To explain the structure of plasma membrane several models have been proposed. The most accepted model is fluid mosaic model.

The Fluid Mosaic Model:

Proposed by Singer and Nicholson (1972), this model is now widely accepted. According to this model, there is a continuous bilayer of phospholipid molecules and globular proteins are embedded in it. The membrane is, thus, considered to be a semifluid structure in which lipids as well as intrinsic proteins are able to make movements within the bilayer. The concept of fluidity implies that lipids and proteins are held in their position by non-covalent bonds.

The proteins in the membrane are of two types:

(i) Extrinsic (Peripheral) Proteins:

These are superficially attached to outer and inner surfaces of lipid bilayer. They are soluble and can readily dissociate from the membrane, e.g., spectrin of RBC membrane.

(ii) Intrinsic (Integral) Proteins:

They penetrate partially or even completely through the lipid bilayer, e.g., ATPase, cytochrome oxidase, rhodopsin, etc. These are amphipathic like the phospholipids. Their hydrophilic head protrudes from the surface of the membrane while hydrophobic end is embedded in the membrane. These are capable of lateral diffusion in the lipid bilayer.

Besides lipids and proteins, carbohydrates also occur at the outer surface of the membrane. These are covalently linked to polar heads of phospholipids or proteins forming glycolipids, or glycoproteins. The glycoproteins form the glycocalyx of the animal-cell surface, which is helpful in cell adhesion and cell recognition.

Functions of Plasma Membrane:

1. It forms a limiting boundary of the cell.

2. Being selectively permeable it allows only useful substances to enter the cell and thus maintains the homeostasis of the cell.

3. Through its receptors it helps in binding hormones, drugs, neurotransmitters, growth factors, etc.

4. The glycocalyx of the membrane helps in cell recognition, adhesion and in exchange of materials or information.

5. It helps in bulk transport by phagocytosis, pinocytosis and exocytosis.

5. What is Mitochondria? Explain its Structure and Functions.

Historical Background:

Mitochondria were first observed in flight muscles of insects by Kolliker (1850). W. Fleming (1882) called them as fila. R. Altman (1892) called them bioplast. The term mitochondria were given by Benda. In plant cells, mitochondria of a cell are collectively called as “chondriome”, whereas those of muscle cells are called as sarcosomes.

Shape:

Mitochondria vary in shape but are generally rod shaped, filamentous or granular.

Size:

The average length of mitochondria is between 3-4 µm and the average diameter 0.5-2.0 µm. In the oocytes of Rana pipiens 40 µm long mitochondria have been reported.

Number:

On an average 200-300 mitochondria are present in a cell. But variations are also reported. For example in the algae Micromonas and Microsterias, only one mitochondria is seen in a cell. Their maximum number has been reported in a protozoan Chaos where it is estimated to be approximately 5,00,000.

Ultrastructure:

A mitchondrion is a double membrane bound structure, each membrane being about 60 Å thick. The two membranes are separated by a perimitochondrial space of about 60-80 Å.

The outer membrane is smooth, tightly stretched and elastic. The inner membrane is rough and selectively permeable. It encloses an inner chamber filled with a matrix which contains most of the enzymes of Krebs’ cycle, 70 S ribosomes, two to six circular DNA molecules, divalent cations, e.g., Ca++, Mg++, etc.

The inner membrane is rich in enzymes like succinate dehydrogenase, cytochrome oxidase, ATPase, etc. The side of the inner membrane facing the matrix is called the M-face while the side facing the outer chamber or cytoplasm is called the C-face. The inner membrane is thrown into several fingers like folds projecting into the matrix. These folds are called as crests or cristae. The cavity of the cristae is called the intracristae space and is continuous with the perimitochondrial space.

Attached to M-face of inner membrane are several elementary particles, or oxysomes. Each particle is made up of three parts-viz.; a polyhedral head, a stalk, and a cuboidal base.

These particles are placed at regular intervals of 100 Å. In a mitochondrion their number may vary from 104 – 105.

Functions of Mitochondria:

(i) Pyruvic acid produced during glycolysis enters mitochondria where it is subjected to Krebs cycle and electron transport system to produce ATP by oxidative phosphorylation. Almost total usable energy of a cell is produced by mitochondria. Hence, it is called as the “Power­house” of the cell.

(ii) It accumulates certain ions (e.g., Ca++ and Fe+++), ferritin, phospholipids, and bile pigments.

(iii) It stores neutral fats and lipids, vitamin C, vitamin A, carotenoids and carcinogenic hydrocarbons.

(iv) It is involved in the elongation of fatty acids and in the synthesis of lipids.

(v) It has an important role in the synthesis of structural proteins, yolk and glycogen. Mitochondrion is a semiautonomous organelle. It is so, because it contains DNA as well as ribosomes, and is therefore able to synthesize some of the proteins required by it.

Mitochondrion is a symbiotic prokaryote. Altman believed mitochondria (and chloroplast also) to be a prokaryotic organism which had entered the cytoplasm of a eukaryotic cell during early days of evolutionary history. The reasons in favour of this hypothesis are the resemblances between their ribosomes, DNA and structure of membranes.

6. What are Plastids? What are the Types of Plastids? Explain the Structure and Function of Chloroplast.

The term plastid was coined by A.F.W. Schimper (1885). Schimper and Meyer (1883-85) is covered these organelles. These are the second largest organelle of the cell and are scattered in the cytoplasm of all green plants. They are not found in blue-green algae, fungi and bacteria.

The plastids are of three types, viz.:

1. Leucoplasts

2. Chromoplasts

3. Chloroplasts

One form of plastid can change into the other.

1. Leucoplasts:

These are the plastids without any pigment and are chiefly concerned with food storage. These are found in embryonic cells, gametes, meristematic regions, seeds and in underground parts.

They are of the following three types:

(i) Amyloplasts:

These store starch. They are found in tubers, cotyledons and endosperm.

(ii) Elaioplasts:

These store fats and oils; for example, in seeds.

(iii) Proteinoplasts or Aleuronolasts:

They store proteins. These are abundant in cotyledons of pulses.

2. Chromoplasts:

These contain pigments of various colours, e.g., carotenoids and xanthophylls.

3. Chloroplasts:

These are the most important plastids found in almost all plants except parasitic plants.

Shape:

Their shape is usually ovoid, discoid or ellipsoid in higher plants. In lower plants the chloroplasts have very unusual shape and size. In Spirogyra chloroplast is ribbon shaped, in Oedogonium it forms a net-work, in Desmids and Zygnema the chloroplasts are like radiating platelets, in Chlamydomonas it is cup shaped, in Ulothrix girdle shaped and in Anthoceros spindle shaped.

Size:

The size of chloroplast is vari­able but on an average its diameter ranges between 4-6 µm and the length between 90-100 µm. These are relatively larger in polyploid plants and sciophytes.

Number:

Normally, there are 20-50 chloroplasts in a plant cell but in algae just one chloroplast may be present in a cell.

Ultrastructure:

Chloroplast is a double membrane bound organelle. Each membrane is 60 Å thick. The space between outer and inner membranes is called as periplastidial space (100-300 Å). The inner space is filled with a granular and transparent fluid known as stroma or matrix. The stroma contains fat globules, starch grains, osmiophilic granules, pyrenoids, and enzymes of dark reaction. The matrix also contains RNA, DNA and 70S ribosomes.

In the stroma, a characteristic system of lamellae is present. These are made up of unit mem­brane bound structures called thylakoids (100-300 Å wide) which are stacked over one another. One such stack is called as a granum. A granum may comprise 50-100 thylakoids. In a chloroplast usually 40-60 grana are found. The grana are interconnected by intergrana lamellae (also called as stroma lamellae or the frets).

On the inner surface of thylakoid membranes particles of 185 Å lengths, 150 Å widths and 100 Å thicknesses are present. These are called “quantasomes”. These were discovered by Park and Biggins (1963). These are the smallest photosynthetic units capable of carrying out photochemical reaction.

Chloroplast is a semiautonomous organelle, due to presence of DNA, RNA and ribosomes. Chloroplast is capable of synthesising some of its proteins required for integrity of thylakoid membranes. The DNA of chloroplast is responsible for cytoplasmic inheritance and dividing ability of chloroplasts. Due to all these facts, chloroplast, like mitochondria, is said to be semiautonomous organelle.

Functions of Chloroplast:

In the presence of light, carbon dioxide and water chloroplasts manufacture organic food for the plants by the process of photosynthesis. The food prepared by plants is then made available to heterotrophs. The entire process of photosynthesis is completed in two steps- viz.; light reaction and dark reaction. The light reaction takes place in grana which trap the solar energy and store it as chemical energy. During dark reaction, which occurs in stroma, this energy is utilised to combine CO2 and water to build carbohydrates.

Another important function of chloroplasts is to manufacture ATP by the process of photophosphorylation.

7. What is Endoplasmic Reticulum? What are the types and functions of it?

Meaning of Endoplasmic Reticulum:

The endoplasmic reticulum is an extensive network of vesicles and tubules in the cytoplasm. It is more concentrated in the inner region of the cytoplasm than in its peripheral region hence the name endoplasmic reticulum. It was discovered by Porter (1945). It is found in most of plant and animal cells, except mature RBCs, prokaryotes and blue green algae.

The endoplasmic reticulum has the typical unit membrane structure having a thickness of 50 – 60 Å. It exists in three main forms in different cells, depending upon their metabolic state.

These forms are as follows:

1. Cisternae (Lamellae):

These are elongated and unbranched tubules arranged in parallel bundles. They may be interconnected with each other. This form of endoplasmic reticulum is characteristic of cells which are actively involved in protein synthesis.

2. Tubules:

These are small, smooth walled and branched structures of different sizes and shapes. These are characteristic of non-secretory cells, e.g., developing spermatids, muscle cells, etc., and are mainly concerned with storage and transport of steroid hormones, choles­terol, glycerides, etc.

3. Vesicles:

They are large, rounded or irregular structures of smooth membrane. They are abundant in synthetically active cells, e.g., Liver cells, pancreatic cells, developing spermatocytes, etc.

Types of Endoplasmic Reticulum:

There are two distinct morphological types of endoplasmic reticulum, viz.:

1. Rough Endoplasmic Reticulum (RER):

It is called as rough or granular because the membranes are covered with ribosomes giving them a rough appearance. Ribosomes are attached to the membranes through their larger subunit (60S) by a specific glycoprotein called as ‘ribophorin’. The RER is more stable and is predominantly found in those cells which are actively engaged in protein synthesis, e.g., the enzyme secreting cells.

2. Smooth Endoplasmic Reticulum (SER):

In this type, the membranes do not bear ribosomes hence appear smooth. They are usually tubular; cisternae are rare. This form of endoplasmic reticulum is less stable. It is characteristic of cells in which synthesis of non-protein substances, phospholipids, glycolipids and steroid hormones takes place, for example in adipose tissue, adrenal cortex, interstitial cells of testis, etc.

When a cell type has abundant SER, it usually has little RER and vice-versa.

Functions of Endoplasmic Reticulum:

1. The endoplasmic reticulum provides mechanical support for the colloidal structure of the cyto­plasm.

2. It helps in exchange of materials between nucleus and cytoplasm.

3. It separates the cytoplasm into compartments and maintains the ionic gradients and electrical potential across these compartments.

4. The endoplasmic reticulum may help intracellular circulation of various substances.

5. The RER provides a site for protein synthesis by attaching ribosomes on it.

6. Jones and Fawcett (1966) have shown the presence of drug metabolising and detoxifying enzyme-systems in the endoplasmic reticulum.

7. The SER helps in synthesis and storage of lipids, cholesterol and glycogen.

8. In testis, ovary and adrenal cortex it synthesises steroid hormones.

8. What are Lysosomes? Explain the Ultrastructure and Functions of it.

These are smallest membrane bound organelles. They originate directly from the endoplasmic reticulum or from the Golgi complex. These were discovered by Christian de Duve (1955).

It is universally present in animal cells. It is not found in plant cells and prokaryotes.

Ultrastructure:

These are single unit-membrane bound globular structures filled with enzymes. Their diameter varies between 0.2 to 0.8 µm. The lysosomal membrane is impermeable to substrates of enzymes contained in the lysosome.

The enzymes are kept in an inert condition through electrostatic binding of acid groups in the lipoprotein matrix of membrane.

If the enzymes are released, they can digest the cell itself, hence, the lysosomes are also called as “suicide bags” of the cell. Since most of the lysosomal enzymes function better under acidic conditions, they are collectively termed as “acid hydrolases”.

Functions of Lysosome:

1. Extracellular Digestion:

Lysosomal enzymes are released outside the cell where they digest the substrate.

2. Intracellular Digestion:

It may involve autophagy or heterophagy. During autophagy the lysosomes digest the organelles of their own cell while during heterophagy exogenous materials are broken down. Sometimes both autophagy and heterophagy may occur simultaneously in the same lysosomal vesicle. Such vesicles are called as “ambilysosomes.”

3. Hormone Secretion:

Lysosomes modify the secretory products synthesised by the cell before they are released. For example thyroid hormones are released by hydrolysis of thyroglobulin in the secondary lysosomes. Secretion of prolactin from anterior pituitary is controlled by lysosomes.

4. Fertilisation:

The acrosome of sperm is looked upon as a giant lysosome. It helps in dissolving the egg membrane to facilitate the entry of sperm.

5. Developmental Processes:

Resorption of the tadpole tail and regression of insect larval tissues involves lysosomal acid hydrolases. In mammalian females the involution of uterus and mammary glands immediately after the child-birth involves lysosomes.

6. Malfunctioning of Lysosome:

Malfunctioning of lysosome results in tissue damage and may cause several diseases including some cancers.

9. What is Golgi Complex? Explain the Ultrastructure and Functions of it.

It was discovered by Camillo Golgi (1898) in the nerve cells of barn owl.

In plant cells these are also called dictyosomes.

Ultrastructure:

Electron microscope reveals the presence of three membranous components in it, viz.:

1. Cisternae or Lamellae:

They are flattened; parallel sacs piled one upon the other to form stacks.

The cisternae may be flat but are more usually slightly curved. This gives the whole stack convex and concave faces. There are named them as forming or proximal face and maturing or distal face respectively as new lamellae are formed on the forming face and mature lamellae are lost on the maturing face.

2. The Small Vesicles:

They arise from the cisternae by budding.

3. Tubules:

These are like cisternae, but are highly branched.

Functions of Golgi Complex:

1. General Secretion:

These are involved in extra and intra-cellular secretions.

2. Synthesis of Polysaccharides:

The Golgi complex in the goblet cells of the colon produces mucigen. This secretory material contains a large portion of carbohydrate.

3. Glycosylation:

Addition of carbohydrates to the proteins occurs in the Golgi complex as well as in the rough endoplasmic reticulum as both of them contain the enzyme glycosyl trans­ferase. After completion of glycosylation the glycoprotein is released into the lumen of Golgi cisternae.

4. Sulphation:

Golgi complex takes part in sulphate metabolism. Compounds containing active sulphur are formed in two steps. Sulphate is first activated by ATP then the activated sulphur is transferred to acceptor molecule by sulphotransferases.

5. Plasma Membrane Formation:

Secretory granules originating from the Golgi complex fuse with the plasma membrane. The membrane of the granules becomes incorporated in to the plasma membrane and thus contributes to the renewal of the membrane.

6. Cell-Plate Formation:

Substances like pectin and hemicelluloses, which form the matrix of the cell plate, are contributed by the Golgi complex.

7. Lipid Packaging and Secretion:

Golgi complex provides a membrane for envelopment of lipid, so that it can be released from the cell.

8. Acrosome Formation:

Electron microscopic studies have revealed the derivation of acrosomal membrane from the membranes of Golgi derived vesicles.

9. Lysosome Formation:

Primary lysosomes are formed by the Golgi complex.

10. Neurosecretion:

In many cells, neurosecretory material is synthesised by ribosomes or endoplasmic reticulum, and are packed in Golgi complex.

10. What are Chromosomes? Explain its Structure and Functions.

Meaning of Chromosomes:

All the living organisms have specific characteristics which they transmit to their offspring through successive generations. The characteristics are identified as hereditary traits. These traits are controlled by special units, called as genes, which are borne by the chromosomes. The chromosomes are, thus defined as self-duplicating nuclear filaments having specific organisation and individuality.

Historical Aspects:

W. Fleming (1897) saw deeply stainable thread like material in the nucleus and called it as chromatin. These threads were named chromosomes by Waldeyer (1888). Sutton and Boveri suggested and later proved experimentally that chromosomes were the physical carriers of hereditary characters. Based on this fact they proposed the chromosomal theory of inheritance.

Chromosome Number:

Benden and Boveri (1887) reported that the number of chromosomes is constant for a particular species. The number of chromosomes present in gamete is said to represent one complete set and is called haploid number which is represented by ‘n’. The total number of genes present in a haploid set of chromosomes is known as genome. The somatic cells contain two sets of chromosomes which together represent the diploid number (2n).

Similarly if an individual possesses more than two sets of chromosomes it is said to be polyploid condition (3n, 4n …… and so on). In polyploid individuals the ancestral primitive number is called as base number and is represented as X. For example in the common Triticum aestivum the diploid number (2n) is 42 and haploid number (n) is 21, but its base number (x) is 7 which means that Triticum is a hexaploid (i.e., 2n = 6x).

The minimum number of chromosomes recorded in plants is n = 2 in Haplopappus gracilis (Compositae). The maximum number has been reported from the fern Ophioglossum reticulatum (2n = 1260).

Morphology of Chromosomes:

Size:

The size of chromosomes varies from species to species but is generally specific for a particular species. The chromosome size is generally measured at the mitotic prophase. The chromosomes may be 0.2 to 50 mm in length, for e.g., 3 mm in Drosophila, 5 mm in human beings, 8-12 mm in Zea Mays, 0.25 mm in fungi and 30 mm in Tillium.

Plant chromosomes are usually larger than animal chromosomes and likewise the chromosomes of the monocots are larger than those of the dicots.

Shape:

The shape of the chromosome changes from phase to phase during the continuous process of cell division. During interphase the chromosomes appear as extended fine thread like stainable structures called chromatin threads. However, the shape and structure of the chromosomes can be studied best at the metaphase and anaphase stages of cell division because at these phases the chromosomes contract to the maximum. They
may be rod shaped, J-shaped, L-shaped or V-shaped, depending on position of the primary constriction (centromere) along the length of a chromosome.

Structure:

During interphase, the stained chromosome appears as a thin and coiled filament, composed of chromatin. This filament was named as chromonema by Vejdovsky (1912). Sometimes chromonema and chromatid are used synonymously, but actually these are different. A chromatid refers to one half of the chromosome which is connected at the centromere, while the chromonema represents thread like structures constituting respective chromatids.

Earlier it was thought that chromonema remain embedded in a Paranemic coils amorphous matrix which in turn is covered by a very thin chromosomal sheath or pellicle. However, the electron microscopic studies have not confirmed the presence of matrix and pellicle. The chromonema may be composed of two or more fibres depending on the species.

These fibres remain coiled with each other forming either paranemic or plectonemic coils. In paranemic coiling the coils of the chromonemal fibres are easily separable but in the plectonemic coiling the chromonemal filaments remain so intimately coiled that they cannot be separated easily.

The following parts are distinguished in a condensed chromosome:

1. Primary Constriction:

Each chromosome has a non-stainable region at a specific point along its length. This region is called primary constriction.

2. Centromere:

Within the primary constriction, there is a clear central zone called centromere. This is the point of attachment of the sister chromatids and also the site of attachment of the mitotic spindle fibre. The portion of the chromosome on either side of the centromere is called arm of the chromosome. Functionally the centromere is related to the movement of the chromosomes at anaphase. During this movement, depending upon the relative ratio of the two arms, the chromosomes acquire the shape of I, J, L or V.

The centromere is made up of four very small granules arranged in a square. These granules are called centromeric chromomeres which remain connected to the chromatid fibres. The chromosomes of many organisms contain only one centromere. Such chromosomes are called monocentric, those with two or more centromeres are respectively called dicentric and polycentric.

3. Kinetochore:

The kinetochore is a proteinaceous disc attached to the centromeric chro­momeres. Two kinetochores, one in each chromatid, are observed. These are centres of assembly for the microtubules at the metaphase.

4. Secondary Constriction I:

It is also called as nucleolar organiser. The part of the chromosome beyond the secondary constriction is called as satellite or trabant. The chromosomes having a satellite are called as SAT chromosome. SAT stands for “sine acid thymonucleinico”, means absence of thymonucleic acid in this part. It contains genes for synthesis of ribosomal RNAs.

5. Secondary Constriction II:

One or more additional constrictions called secondary con­striction II may also be present on the chromosome. Their position is fixed hence these are useful in identifying a chromosome in a set.

6. Telomere:

Tips of the chromosomes containing heterochromatic material or repetitive DNA sequences are called telomeres. Each telomere has definite polarity. It does not allow other chromosomes to stick with it or its union with the broken ends.

According to position of centromere, following types of chromosomes are identified:

(i) Telocentric:

Their centromere is situated at one end. At anaphase, it looks like ‘I’.

(ii) Acrocentric:

Their centromere lies almost near the tip of chromosome so that one arm is exceptionally short and the other is long. At anaphase, these chromosomes look like ‘J’.

(iii) Submetacentric:

In this type of chromosome the centromere lays a little away from the centre, dividing the chromosome into two unequal arms. Such chromosome looks like ‘L’ at the anaphase.

(iv) Metacentric:

In this type of chromosome, the centromere lies in the middle of the chromosome, dividing it into two equal arms. The metacentric chromosome becomes V-shaped during anaphase.

Chemical Composition and Models of Chromosomes:

The major constituents of the chromosomes include DNA, RNA, histone and non-histone proteins and metal ions. It carries the genetic information from one generation to other. The RNA is transcribed by DNA and most of it is transported to the cytoplasm. The histone proteins present in the chromosomes are basic proteins that are composed of basic amino acid such as lysine and arginine.

These remain associated with the DNA and act as repressors of gene activity. The non-­histone proteins are mostly acidic and act as enzymes, important among them are DNA polymerase, RNA polymerase and nucleoside triphosphatase. Besides these metallic ions such as Mg++, Ca++, etc. keep them intact and also act as regulators of various enzymes.

Models of Chromosome Structure:

Various models showing the mode of attachment between DNA and proteins in a chromosome have been proposed, among which the nucleosome model is most accepted one.

Nucleosome Model:

A.L. Olins and D. E. Olins (1974) reported the presence of a series of bead like structures in electron micrographs of interphase chromatin fibres. These particles were called as nu (h) bodies. Later Outdet (1975) called them as nucleosome. R.D. Kornberg and Thomas discovered that each spherical unit representing the core is com­posed of 140 base pairs of DNA and a histone octamer having two molecules of each of four different histones- viz., H2A, H2B, H3 and H4.

The histone H1 is loosely associated with the chromatin. Around the core particle of nucleosome, DNA molecule is wrapped 1.75 times. The complete nucleosome is a flattened particle of 55 Å in height and 110 Å in diameter. The core particle made up of histone octamer is 40 Å high and 80 Å wide. The beaded nucleosomes are interconnected by DNA filaments called linker DNA. Their length may vary from 8 to 114 nucleotide base pairs.

The nucleosomal beads are further coiled to form a super-coiled structure called as solenoid.

Functions of Chromosomes:

The chromosomes are most vital component of the cell. These control almost all cellular activities at physiological, molecular and morphological levels.

Besides these, they perform the following main functions:

1. The chromosomes maintain the identity of species.

2. These determine the sex of species of animals and plants.

3. These act as a vehicle of hereditary characters from one generation to other.

4. With the help of their chemical constituents, the DNA and RNA, they synthesize proteins and enzymes.

5. The lampbrush chromosomes synthesize yolk in oocytes of many vertebrates.

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