CBSE Class 11 – Page 28

NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption

June 18, 2021

NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption.

Question 1.
Choose the correct answer among the following:
(a) Gastric juice contains
(i) pepsin, lipase, and rennin
(ii) trypsin, lipase, and rennin
(iii) trypsin, pepsin, and lipase
(iv) trypsin, pepsin, and rennin
(b) Succuss enterics is the name given to:
(i) a junction between the ileum and large intestine
(ii) intestinal juice
(iii) swelling in the gut
Solution:
(a) (i) Pepsin, lipase, and rennin
(b) (ii) Intestinal juice

Question 2.
Match column I with column II
Column I Column II
(a) Bilirubin and biliverdin (i) Parotid
(b) Hydrolysis of starch (ii) Bile
(c) Digestion of fat (iii) Lipases
(d) Salivary gland (iv) Amylases
Solution:
Column I Column II
(a) Bilirubin and biliverdin (ii) Bile
(b) Hydrolysis of starch (iv) Amylases
(c) Digestion of fat (iii) Lipases
(d) Salivary gland (i) Parotid

Question 3.
Answer briefly:
(a) Why are villi present in the intestine and not in the stomach?
(b) How does pepsinogen change into its active form?
(c) What are the basic layers of the wall of the alimentary canal?
(d) How does bile help in the digestion of fats
Solution:
(a) Villi increases surface area for absorption and maximum absorption takes place in the intestine.
(b) Coming in contact with hydrochloric acid in stomach proenzyme pepsinogen convert to its active form pepsin, the proteolytic enzyme of the stomach.
(c) There are four basic layers in the wall of alimentary canal i.e. serosa, muscularis, submucosa and mucosa.
(d) Bile helps in the emulsification of fats i.e. breakdown the fats into very small micelles. Bile also activates lipases.

Question 4.
State the role of pancreatic juice in the digestion of proteins.
Solution:
Pancreatic juice contains inactive enzymes like trypsinogen, chymotrypsinogen, procarboxypeptidases. Trypsinogen is activated by an enzyme enterokinase secreted by the intestinal mucosa into active trypsin which in turn activates other enzymes in the pancreatic juice. Proteins, proteases, and peptones in the chyme are digested by the proteolytic enzymes of pancreatic juice.

Question 5.
Describe the process of digestion of protein in the stomach.
Solution:
The mucosa of the stomach has gastric glands that secrete mucus, proenzyme pepsinogen, HCl, and intrinsic factor. Intrinsic factor is essential for the absorption of vitamin B12. The food mixes thoroughly with acidic gastric juice of the stomach and called the chyme. The proenzyme pepsinogen on exposure to HC1 gets converted to active enzyme pepsin. Pepsin converts proteins into proteases and peptones. Renin found in the gastric juice of infants also helps in the digestion of milk protein.

Question 6.
Give the dental formula of human beings.
Solution:
Dental formula of human beings is 2123/2123.

Question 7.
Bile juice contains no digestive enzymes, yet it is important for digestion. Why?
Solution:
The bile juice released into the duodenum contains bile pigments (bilirubin and biliverdin), bile salts, cholesterol, and phospholipids but no enzymes. Bile helps in the emulsification of fats i.e. break down the fats into very small micelles. Bile also activates lipases. Bile is alkaline (pH about 8). Bile pigments are excreted in faeces. In absence of HCl, the above-mentioned functions will not occur and digestion of food will be affected.

HCl of the gastric juice may cause gastric or duodenal ulcers and damage, the underlying blood vessels. These cause hemorrhage inside. Strong HCl is produced in stress conditions also. Parental or oxyntic cells which secrete HCl and intrinsic factor (Intrinsic factor is essential for the absorption of vitamin B₂)

Question 8.
Describe the digestive role of chymotrypsin. Which two other digestive enzymes of the same category are secreted by their source gland?
Solution:
Chymotrypsin changes proteins into peptides and also milk protein, changes into paracasein (curd). 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.

Trypsin changes proteases and peptones into peptides and amino acids. Procarboxypetidases change peptides into small peptides and amino acids. Amylases change starch into maltose. Lipases change emulsified fat into fatty acids and glycerol. Nucleases changes nucleotides into phosphate, sugar and nitrogen bases.

Question 9.
How are polysaccharides and disaccharides digested?
Solution:
(a) Digestion of carbohydrates starts in the mouth cavity with the help of enzymes salivary amylase.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 1

Question 10.
What would happen if HCl were not secreted in the stomach?
Solution:
HCl is secreted by oxyntic cells in the stomach wall and performs five functions:

Kill bacteria and germs
Loosens fibrous material of food.
Activates proenzyme pepsinogen to its active form pepsin.
Gives an acidic medium for action by pepsin.
Curdles milk. Pepsin changes proteins into proteases and peptones.

Question 11.
How does butter in your food get digested and absorbed in the body?
Solution:
Butter is a kind of fat. Fats are broken down by lipases with the help of bile into di- and monoglycerides:
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 2

Question 12.
Discuss the main steps in the digestion of proteins as the food passes through different parts of the alimentary canal.
Solution:
(a) In the stomach the proenzyme pepsinogen, on exposure to HC1 converted into active pepsin that converts proteins into proteases and peptones.

(b) Proteins, proteases and peptides in the chyme reaching the intestine are acted upon by proteolytic enzymes of pancreatic juice and converted to dipeptides.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 4
(c) The enzymes in succuss entericus act on the end product to form amino acids.

Question 13.
Explain the terms thecodont and diphyodont.
Solution:
Thecodont: In human beings, teeth are
embedded in pits, the sockets of the jawbones. Such teeth are called the thecodont.
Diphyodont: The teeth that appear in two sets, i. e., milk-teeth which are later replaced by permanent teeth. This condition is called diphyodont.

Question 14.
Name different types of teeth and their number in an adult human.
Solution:
Incisors – 8
Canines – 4
Premolars – 8
Molars – 12

Question 15.
What are the functions of the liver?
Solution:

Secretion of bile
Synthesis of blood clotting factors
Regulating carbohydrate, protein, and lipid metabolism.
Synthesis of amino acids, plasma proteins, cholesterol, etc.
Stores glycogen, fats, fat-soluble vitamins, etc.
Detoxifies toxic substances
Elimination of foreign bodies
Produces heat through metabolism
Produces anticoagulant heparin
Synthesis of urea from ammonia.

VERY SHORT ANSWER QUESTIONS

Question 1.
What is digestion?
Solution:
Digestion is the mechanical, enzymatic, and biochemical transformation of complex (polymers) food molecules into simple molecules (monomers) which are suitable for absorption.

Question 2.
Which is the food constituent that bile helps to digest and absorb?
Solution:
Fats.
Question 3.
What is the function of enterokinase?
Solution:
Enterokinase of intestinal juice activates the inactive trypsinogen into trypsin which digests protein in the duodenum.

Question 4.
Which is a nondigestive activating enzyme?
Solution:
Enterokinase.

Question 5.
Mention the role of bile salt in the digestion of fats.
Solution:
Bile salts emulsify fat particles and reduce the surface tension of fat droplets to increase the action of enzyme lipase.5. Bile salts emulsify fat particles and reduce the surface tension of fat droplets to increase the action of the enzyme lipase.

Question 6.
What is the role of HCl in protein digestion?
Solution:
Role of HCl:

It activates pepsinogen into active pepsin.
It provides a suitable acidic medium for the action of proteases in the stomach.

Question 7.
Name the hardest substance in the body.
Solution:
Enamel.

Question 8.
What is a cystic duct?
Solution:
Duct of gall bladder.

Question 9.
Mention two functions of mucus.
Solution:
Role of mucus:
(a) Acts as a lubricant.
(b) Protects the epithelial surface of the stomach from the corrosive effect of hydrochloric acid and digestion by pepsin.

Question 10.
Name the secretion of goblet cells in the human stomach.
Solution:
Goblet cells secrete mucus.

Question 11.
Where the taste buds located?
Solution:
Taste buds are located in the papillae on the upper surface of the tongue.

Question 12.
What is the ‘Pancreatic Enzyme’ which acts on starch? (Oct. 83, 01)
Solution:
The pancreatic amylase(Amylopsin).

Question 13.
What is the function of epiglottis?
Solution:
Epiglottis prevents the entry of food into the trachea, by closing its opening called the glottis.

Question 14.
Which part of the stomach continues into the duodenum?
Solution:
Pyloric region

Question 15.
What name is given to the major lymph vessel present in the intestinal villi?
Solution:
Lacteal

Question 16.
Where are the crypts of Leiberkuhn located?
Solution:
Crypts of Lieberkuhn are located in between the bases of the villi in the intestine.

Question 17.
Name the structural and functional unit of the liver.
Solution:
Hepatic lobules

Question 18.
Mention the secretion of Goblet cells.
Solution:
Mucin (Mucus) (April 93)

Question 19.
What is a bolus?
Solution:
When the thoroughly masticated food mixes with the saliva, the food particles become adhered together by the mucus known as bolus.

Question 20.
What is chyme?
Solution:
After partial digestion in the stomach, the form of food is called chyme.

Question 21.
What is the meaning of deglutition?
Solution:
The act of swallowing is called deglutition.

Question 22.
Name the enzyme involved in the breakdown of nucleotides into sugars and bases.
Solution:

SHORT ANSWER QUESTIONS

Question 1.
Mention the juice secreted by the liver. State its function indigestion. (April 1985)
Solution:
The liver cells secrete a juice called the ‘Bile juice’. ‘Bile salts’ are one of the components of Bile juice. These are very important in digesting lipids or fats. Specifically ‘Bile salts’ aid in digestion than bringing about digestion of fat by emulsifying fats and making the fat molecules accessible to the action of the pancreatic Lipase (or Lipid digesting) enzyme. Apart from this the bile juice also creates an alkaline medium in the intestine so that food particles can be actively acted upon by the pancreatic enzymes.

Question 2.
Name the organs which secret carboxypeptidases and aminopeptidases respectively. Give the function performed by these enzymes
Solution:
Carboxypeptidases are secreted by the pancreas. Aminopeptidases are secreted by the intestine. Both the enzymes act on the terminal peptide bonds and release the terminal/last amino acids of the peptide chain.

Question 3.
State the Role of HCI indigestion. (Oct. 88, April 93, 98)
Solution:
The main role of HCI is to convert the inactive enzyme pepsinogen to pepsin which in turn helps in the digestion of proteins. It also creates an acidic medium in the stomach for the activity of pepsin.

Question 4.
What is succus entericus? Mention any two carbohydrate digesting enzymes present in it? (April 2006)
Solution:
Succus entericus intestinal juice is the secretion of intestinal glands in the ileum of the small intestine. The carbohydrate digesting enzymes are maltase, lactase, and sucrase.

Question 5.
Differentiate between micelles and chylomicrons.
Solution:
The differences between micelles and chylomicrons are
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 7

Question 6.
How is our gut lining protected from its own secretion of proteases?
Solution:
(i) Protease is secreted in an inactive form and poses no threat to the gut lining.
(ii) The mucus provides protection to the epithelial lining.

Question 7.
Name the organs which secrete carboxypeptidases and aminopeptidases respectively. Give the functions performed by these enzymes.
Solution:
Carboxypeptidases are secreted by the exocrine part of the pancreas. Aminopeptidases are secreted by the intestinal mucosa. These enzymes act on the terminal peptide bonds and release the last amino acid from the polypeptide chain, thereby progressively shortening the peptide chain.

Question 8.
Where is the ileocaecal valve present? What is its function?
Solution:
The ileo-caecal valve is present at the junction of the ileum of the small intestine and the caecum of the large intestine.
It prevents the backflow of the matter from the caecum into the ileum.

Question 9.
How does the nervous system control the activities of the gastro-intestinal tract?
Solution:
The sight, smell and presence of food in the oral cavity can stimulate the secretion of saliva. Gastric and intestinal secretions are also stimulated by similar neural signals. Muscular activities of the alimentary canal are coordinated by both local and CNS neural mechanisms. Hormonal control of secretion of digestive enzymes is carried out by local hormones.

Question 10.
Name the final products of the digestion of proteins. How and where are they absorbed from the alimentary canal?
Solution:
The final products of digestion of proteins are amino acids They are absorbed by an active process utilising energy against the concentration gradient in the ileum.

Question 11.
What is the pancreas? Mention the major secretions of the pancreas that are helpful in digestion.
Solution:
The pancreas is a carrot-shaped soft greyish pink gland that lies transversely below the stomach between the duodenum and spleen, which secretes digestive enzymes from its exocrine parts and hormones from its endocrine parts.
The pancreas secretes three enzymes in inactive proenzyme or zymogen state and three in active enzyme state.
Proenzymes. Trypsinogen, chymotrypsinogen, and procarboxypeptidases.
Active Enzymes. Amylase, lipase, and nucleases.

LONG ANSWER QUESTIONS

Question 1.
What is meant by vernalization? Explain the significance of vernalization.
Solution:

Question 2.
Write short notes on
(a) Liver
(b) Layers of the alimentary canal
Solution:
(a) Liver: The liver is the largest gland of the body. It is situated in the abdominal cavity, just below the diaphragm and has two lobes. The hepatic lobules are the structural and functional units of the liver containing hepatic cells arranged in the form of cords. Each lobule is covered by a thin connective tissue sheath called the Glisson’s capsule. The bile secreted by the hepatic cells passes through the hepatic ducts and is stored and concentrated in a thin muscular sac called the gall bladder. The duct of gall bladder (Cystic duct) along with the peptic duct from the liver forms the common bile duct. The bile contains pigments like bilirubin and biliverdin, bile salts, cholesterol and phospholipids but no enzymes. Bile helps in the emulsification of fats and activates lipases.

(b) Layers of the alimentary canal:
The wall of the alimentary canal possesses four layers namely serosa, muscularis, sub-mucosa, and mucosa. The serosa is the outermost layer and is made up of a thin mesothelium with some connective tissues. Muscularis is formed by smooth muscles usually arranged into an inner circular and an outer longitudinal layer. An oblique muscle layer may be present in some regions.

The sub-mucosal layer is formed of loose connective tissues containing nerves, blood, and lymph vessels. In the duodenum, glands are also present in the sub-mucosa. The innermost layer lining the lumen of the alimentary canal is the mucosa. This layer forms irregular folds in the stomach and small finger-like foldings called villi in the small intestine. The cells lining the villi produce numerous microscopic projections called microvilli giving a brush border appearance.

Villi are supplied with a network of capillaries and a large lymph vessel called the lacteal. The mucosal epithelium has goblet cells which secrete mucus that help in lubrication. Mucosa also forms glands in the stomach and crypts in between the bases of villi in the intestine (crypts of lieberkuhn).

Question 3.
Describe the major disorders of the human digestive system.
Solution:
Disorders of the digestive system:
(i) Indigestion:

It is the condition in which the food is not properly digested leading to a feeling of fullness.
It is caused by inadequate secretion of digestive enzymes, food poisoning, overeating or spicy food.

(ii) Constipation

It refers to the condition where the faeces are retained in the rectum for longer periods as the bowel movements occur irregularly.

(iii) Diarrhoea:

It refers to the abnormal frequency of bowel movement and increased liquidity of the faecal discharge; absorption of food is impaired.

(iv) Vomiting:

It is the ejection of stomach contents through the mouth; this reflex action is controlled by the vomit centre in the medulla.
It is due to viral infection, where liver is affected and digestion of fats is impaired.
The eyes and skin turn yellow due to the deposit of bile pigments.

Question 4.
How is the DNA content in our diet digested in the body?
Solution:
DNA content is digested in the intestinal part of our alimentary canal by the enzymes present in pancreatic juice and succus entericus.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 9
DNAase is found in pancreatic juice while nucleotidase and nucleosidase occur in succus entericus and hydrolyse the DNA content in our diet.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 16 Digestion and Absorption, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 16 Digestion and Absorption, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 English Hornbill Chapter 7 The Adventure

June 18, 2021

Here we are providing NCERT Solutions for Class 11 English Hornbill Chapter 7 The Adventure. Students can get Class 11 English The Adventure NCERT Solutions, Questions and Answers designed by subject expert teachers.

The Adventure NCERT Solutions for Class 11 English Hornbill Chapter 7

The Adventure NCERT Text Book Questions and Answers

The Adventure Understanding the text

I. Tick the statements that are true.
Answer:
The true statements are:

The story’ hinges on a particular historical event.
The story’ tries to relate history to science.

II. Briefly explain the following statements from the text.

Question 1.
“You neither travelled to the past nor the future. You were in the present experiencing a different world. ”
Answer:
This statement is made by Rajendra Deshpande to Professor Gangadharpant Gaitonde. Here he tries to give a possible explanation to Prof Gaitonde’s unique experience in the two days following his collision with a truck. According to the speaker, Professor Gaitonde did not travel in time. In fact, he was in the present but was experiencing a different world. This statement emphasises on the idea that reality is not unique. There are many worlds and many realities. A person’s reality is based on his/her experience at a particular time at a particular place.

Question 2.
“You have passed through a fantastic experience: or more correctly, a catastrophic experience. ”
Answer:
In this statement Rajendra Deshpande tries to rationalise Professor Gaitonde’s experience of making a transition to another world. According to him, Professor Gaitonde has had a fantastic experience. Here, the word fantastic means both extraordinary and fanciful. However, he immediately finds a better word for the experience and calls it catastrophic. Now, this word is a reference to Catastrophe Theory which is a branch of mathematics and is concerned with abrupt changes. Professor Gaitonde’s experience of travelling to a parallel world without moving back or forth in time seems to be an apt example of a catastrophic experience to Rajendra Deshpande.

Question 3.
Gangadharpant could not help comparing the country he knew with what he was witnessing around him.
Answer:
This statement is made by the third person narrator of the story. Professor Gaitonde, who travels to another world in the present, witnesses a different India which is completely different from the India he has known from this childhood. The country he witnesses in the alternative world refused to surrender to the colonising powers and stood up against them.

The people of this country are more honest, more efficient, well organised and quite modem. Everything that he sees in this country poses a stark contrast to the country he lives in. This compels him to compare the country he knows so well with the country he witnesses around him.

Question 4.
“The lack of determinism in quantum theory! ”
Answer:
This statement is made by Prof Gaitonde in response to Rajendra Deshpane’s explanation of how indeterminable the behaviour of electrons is. Here he maintains that even a person like him, who is not an expert of science, knows about the lack of determinism in quantum theory. Quantum theory believes in the randomness or uncertainty of the universe. This is quite succinctly explained by Rajendra Deshpande with the example of electrons orbitting the nucleus of an atom. The electrons could move in any of the large number of specified states and could even jump from one state to another. There is no certainty of an electron’s presence in a particular state.

Question 5.
“You need some interaction to cause a transition. ”
Answer:
This statement is made by Rajendra Deshpande in response to Professor Gaintonde’s question of why he made the transition to the other world. Rajendra Deshpande says that in order to make a transition like this, one must have an interaction. By interaction, he means something that gives the transition a particular direction. In this case, the focused thought of Professor Gaitonde about the probable course of history in case the third battle of Panipat had had a different result functioned as the interaction and determined the transition for him.

The Adventure Talking about the text

Question 1.
Discuss the following statements in groups of two pairs, each pair in a group taking opposite points of view.

(i) A single event may change the course of the history of a nation.
For:

Events like the defeat of important leaders like Napoleon, Hitler and so on have helped construct today’s version of historic events for different nations.
Had one of these events not happened, chances are that history may have been entirely different.
Similarly, this text tries to present what would have happened if the Marathas had not been defeated by the British in India.
The India that is constructed in the text, is heavily influenced by the Marathi kingdom, unlike the British India that we’ve known through historic events. Therefore, altering historic events can change the history of a nation.

Against:

It is impossible to predict if altering a single event can in reality change the course of history.
While this text attempts to show that, had the Marathas won the war, the British could have still taken over and created the British India that we know of.
All that is portrayed here is speculation, as it is impossible to go back and change historic events.

(ii) Reality is what is directly experienced through the senses.
For:

Gaitonde’s construction of the alternate version of history is so real that he believes he actually witnessed a different version of reality than the one we are familiar with.
Similarly, he is able to convince Rajendra of this reality because he has a piece of paper which contains notes of all these alternate events.
Both these characters believe that what they experience through the senses is real.
They show us that the senses construct reality for us.
The school of thought that believes reality is constructed through our experiences is called empiricism.

Against:

The lack of clarity or established proof that any of the events Gaitonde believed actually happened indicates that all of this could have just been his imagination.
Similarly, the piece of paper he presents as proof may have just been a work of fiction constructed by him.
This may also be the impact of the accident where Gaitonde’s memory may have been affected or damaged, leading him to believe in alternate realities.
All of the events in this story can be explained through rational thought as well.

(iii) The methods of inquiry of history, science and philosophy are similar.
For:

In the story we see Gaitonde, a history academic experiencing an alternate version of historical events closely connected to time, which Rajendra compares to the quantum theory in physics.
The method that Rajendra uses to explain the science behind this to Gaitonde is very close to how philosophers explained things, attempting to simplify complicated concepts using examples.
Thus we see all three disciplines closely connected in the text through their understanding of time and reality.

Against:

If all three disciplines were so closely connected in their methods and fields of inquiry, they wouldn’t be studied separately.
But because the concepts they deal with are radically different, they are studied separately.
Further, while the inquiry may be similar in the text, this does not imply that-all methods of inquiry used by these disciplines are similar, as the text only picks up one particular incident.
And this may broadly differ in these fields on other instances.

Question 2.
(i) The story is called‘The Adventure’. Compare it with the adventure described in ‘We’re Not Afraid to Die…’
Answer:
“We’re Not Afraid to Die…” narrates a family’s adventures on a boat at sea facing a terrible storm. It describes various efforts made by the parents to guide their boat to safety and presents the courage of their children while the parents panicked. “The Adventure” constructs an alternate version of reality, which is what would have happened had the Marathas defeated the British. How the Indian nation would have appeared under the influence of Indian rulers instead of the British. While the events of “We’re Not Afraid to Die.. .’’are realistic and depict encounters that may have happened around the world, the events in “The Adventure” are fictional and speculative.

(ii) Why do you think Professor Gaitonde decided never to preside over meetings again?
Answer:
In the alternate reality of India, Professor Gaitonde attempted to take the empty chair on stage that he was so accustomed to take, only to be met with hostility from the crowd, who threw eggs and tomatoes at him and ran him off stage. The impact of this incident on him is so traumatic, at what he considers his thousandth address as chair, that when he returns to his actual reality he refuses to preside over any more meetings again.

The Adventure Thinking about language

Question 1.
In which language do you think Gangadharpant and Khan Sahib talked to each other? Which language did Gangadharpant use to talk to the English receptionist?
Answer:
Since Khan Sahib is from Peshawar, he may have spoken to Gangadharpant in Hindi or Urdu. However, since Gangadharpant was speaking to the English receptionist in territory controlled by the British, one can assume he spoke to her in the English language.

Question 2.
In which language do you think Bhausahebanchi Bakhar was written?
Answer:
Since the Bhausahebanchi Bakhar describes the victory of the Marathas over the British, and appears to be a seminal book in a library in Maharashtra, the language used may have been Marathi.

Question 3.
There is mention of three communities in the story: the Marathas, the Mughals, the Anglo-Indians. Which language do you think they used within their communities and while speaking to the other groups?
Answer:
Within the community, the Marathas would have used Marathi, the Mughals Urdu, and the Anglo-Indians would’ve used English. While speaking to other groups, considering that the British had little influence over India in the alternate version of history, Marathi may have been the common language because of the influence that Marathi rulers had on the outcome of India. Urdu could also have been the common language since the Mughal rule continued in Delhi or Hindi because of the historical association of the nation with the language.
(Answers will vary)

Question 4.
Do you think that the ruled always adopt the language of the ruler?
Answer:
While the ruled may be inclined to learn the language of the ruler, history has shown us through the diversity of languages that exists across the nation that not all people solely adopt a single language being promoted. While people across the nation use a common language to speak to one another, which is the ruler or administration’s language, they also retain association with local languages used in smaller spaces like the household.
(Answers will vary)

The Adventure Working with words

I. Tick the item that is closest in meaning to the following phrases.

to take issue with—iii) to disagree .
to give vent to—(i) to express
to stand on one’s feet—(ii) to be independent
to be wound up—(ii) to stop operating
to meet one’s match—(iii) to meet someone who is equally able as oneself

II. Distinguish between the following pairs of sentences.

Question 1.
(i) He was visibly moved.
Answer:
The features on his face showed how he was emotionally affected,

(ii) He was visually impaired.
Answer:
He had trouble seeing things because of loss of vision.

Question 2.
(i) Green and black stripes were used alternately.
Answer:
Green stripes and black stripes were both used, one after the other.

(ii) Green stripes could be used or alternatively black ones.
Answer:
Either green stripes could be used everywhere or black ones in its place.

Question 3.
(i) The team played the two matches successfully.
Answer:
The team performed well in two specific matches they played

(ii) The team played two matches successively.
Answer:
The team played two consecutive matches, one followed by the other.

Question 4.
(i) The librarian spoke respectfully to the learned scholar.
Answer:
The librarian showed a sense of respect for the scholar while speaking to him.

(ii) You will find the historian and the scientist in the archaeology and natural science sections of the museum respectively.
Answer:
You will find the historian in the archaeology section and the scientist in the science section of the museum.

The Adventure Things to do

II. Look up the Internet or an encyclopedia for information on the following theories.

(i) Quantum theory
Quantum theory in physics is the study of the smallest particles in the universe found in atomic and subatomic particles. It observes the energy produced by particles like protons, neutrons and electrons where one unit is referred to as quanta, which led to the name Quantum theory. This theory was put together by the physicist Max Planck in 1900, who was later awarded the Nobel Prize.

(ii) Theory of relativity
This was established by the famous scientist Albert Einstein first in 1905. Einstein drew connections between space and time, arguing that these concepts were not fixed as previously understood by physics but fluid with possibilities of change. In the absence of a gravitational field, it was possible to convert matter into huge quantities of energy, and different bodies could experience time differently.

(iii) Big Bang theory
The Big Bang theory is a scientific explanation to the beginning of the world. According to this, the universe that we know began with a small singularity, a small and dense source of energy, which gradually expanded over time. This expanded into the stars and planets that we know of in the universe today.

(iv) Theory of evolution
The theory of evolution was given by Charles Darwin in his book On the Origin of Species in 1859. Darwin believed that all life on earth was connected to each other, and that humans evolved through a process of natural selection. Natural selection states that through a process of evolution from previous animal forms in different centuries, the human form has evolved by learning to adapt to natural habitat on the planet.

NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom

June 18, 2021

NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom.

Question 1.
What is the basis of the classification of algae?
Solution:
Classification of Algae: The algae are divided into three main classes: Chlorophyceae, Phaeophyceae, and Rhodophyceae.

Chlorophyceae: The members of Chlorophyceae are commonly called green algae.
Phaeophyceae: The members of Phaeophyceae or brown algae, are found primarily in marine habitats.
Rhodophyta is commonly called red algae. (See Table on the opposite page)

Question 2.
When and where does reduction division take place in the life cycle of a liverwort, a moss, a fern, a gymnosperm and an angiosperm?
Solution:
I. Liverwort – The main part of the body of liverwort is thalloid. Haploid gametes are produced from the male and female sex organs garnets fuses to form zygote. Zygote develops in the form of sporophytes. These sporophytes are further differentiated into foot, seta and capsule. As a result of reduction division many haploid spores are produced in capsule.
II. Moss – In the first stage in moss primary protonema develops into secondary ’ protonema. Both these stages are haploid. Zygote formed by the fusion of gametes further produce sporophytes.
III. Fern – Leaves of sporophyte bear sporangia in which spores are produced by reduction division in meiosis.
IV. Gymnosperm – In microsporophylls and megasporophylls that bear microsporangia and megasporangia respectively, reduction division occurs to produce microspores (pollen grains) and megaspore.
V. Angiosperm – Main part of the body is sporophytic and bears flowers. Reduction division takes place in anthers of stamen i.e. haploid pollen grains and in the ovary of pistil producing eggs.

Question 3.
Name three groups of plants that bear archegonia. Briefly describe the life cycle of any one of them.
Solution:

Archegonia is the female reproductive organ of

Bryophytes
Pteridophytes and
Gymnosperms groups of plants.

The life cycle of gymnosperms: The gymnosperms are heterosporous, they produce haploid microspores and megaspores. The two kinds of spores are produced within sporangia that are borne oil sporophylls which are arranged spirally along an axis to form lax or compact strobili or cones.
The strobili bearing microsporophylls and microsporangia are called microsporangia or male strobili. The microspores develop into a male gametophytic generation which is highly reduced and is confined to only a limited number of cells. This reduced gametophyte is called pollen-grain. The development of pollen- grains takes place within the microsporangia.

The cones Rearing megasporophylls with ovules or megasporangia are called microsporangia or female strobili. The male or female cones or strobili may be borne on the same tree or on different trees. The megaspore mother cell is differentiated from one of the cells of the nucellus. The nucellus is protected by envelopes and the composite structure is called an ovule.

The ovules are borne on megasporophylls which may be clustered to form the female cones. The female cones are borne on the main plant body of the sporophyte. The megaspore mother cell divides meiotically to form four megaspores. One of the megaspores enclosed within the megasporangium (nucleus) develops into a multicellular female gametophyte that bears two or more archegonia or female sex organs. The multicellular female gametophytes are also retained within megasporangium.

Three groups of plant that bear archegonia are
1. Bryophyta,
2. Pteridophyta and
3. Gymnosperm.

Life cycle of gymnosperms: The gymnosperms are heterosporous, they produce haploid microspores and megaspores.
The two kinds of spores are produced within sporangia that are borne on sporophyll which are arranged spirally along an axis to form strobili or cones.
The strobili bearing microsporophylls and microsporongia are called microsporangiate or male strobili.
The microspores develop into a male gametophytic generation which is highly reduced, and confined to only limited number of cells. The reduced gametophyte is called pollen grain.
The development of pollen grain takes place within microsporangia.
The cones bearing megasporophylls with ovules or megasporangia are called macrosporangiate or female strobili.
The male or female cones or strobili may be borne on the same tree or different trees. The megaspore mother cell is differentiated from one of the cells of the nucellus.
The nucellus is protected by envelopes and the composite structure is called an ovule. The ovules are borne on megasporophylls which may be clustered to form the female cones.
The females cones are borne on the main plant body of the sporophyte. The megaspore mother cell divides meiotically to form four megaspore.
One of the megaspore enclosed within the megasporangium develops into a multicellular female gametophyte that bears 2 or more archegonia or female sex organs.
The multicellular female gametophyte is also remaining within megasporangium.

Question 4.
Mention the ploidy of the following: protonemal cells of a moss; primary endosperm nucleus in dicot, leaf cell of a moss; prothallus cell of a fern; gemma cell in Marchantia; Meristem cells of monocot, ovum of liverwort and zygote of a fern.
Solution:
Protonemal cells of a moss – Haploid (n) Primary endosperm nucleus in dicot – Triploid (3n)
Leaf cell of a moss – Haploid (n)
Prothallus cell of a fem – Haploid (n)
Gemma cell in Marchantia – Haploid (n) Meristem cells of monocot – Diploid (2n) Ovum of liverwort – Haploid (n)
Zygote of a fem – Diploid (2n)

Question 5.
Write a note on economic importance of algae and gymnosperms.
Solution:

Algae like Chlorella, Gelidium (produce agar- agar) are used as food. Many algae like diatoms (used in manufacture of glass, polish, etc), algin (used in vulcanisation, artificial fibres, etc.), are used in industry.
Nos toe, Anabaena, etc., are useful in increasing fertility of soil. Antibiotic chlorellin is extracted from Chlorella.
Many algae have harmful effect also, for example, Microcystis, Chlrococcus, Oscillatoria cause water blooms and Cephaleuros species of algae are parasitic on tea leaves and cause harm to tea industry.
v Alginic acid are extracted from the members Phaeophycea such as laminaria, Macrocystis and carrageenin is extracted from red algae chondris Crispos.

Economic importance of gymnosperms :

Gymnosperms helps in checking soil erosion.
Seeds of Pinns gerardiana, Gnetnm gnemon and Ginkgo biloba are eaten.
Conifers like Pinus longifolia, Cedrus deodara, Picea, Tsngo, etc., produce soft wood. Bark of Tsugo yields tannins for making inks, seeds and bark of Cycas are used as poultica for wounds and sores.
Ephedra, Gnetum, Taxus baccata, Cycles rumphii are used for medicinal purposes.

Question 6.
Both gymnosperms and angiosperms bear seeds, then why are they classified separately?
Solution:
Gymnosperms are ‘naked seeded’ plants because their seeds are not enclosed in fruit wall whereas angiosperm are ‘enclosed seeds’ as seeds (ovules) are found enclosed in the ovary wall.

Question 7.
What is heterospory? Briefly comment on its significance. Give two examples.
Solution:

Genera like Selaginella and Salvinia which produce two kinds of spores macro and microspores are known as heterosporous.
The megaspore and microspores germinate and give rise to female and male gametophytes, respectively.
The female gametophytes in these plants are retained on the parent sporophyte for variable periods.
The development of the zygotes into young embryo takes place within the female gametophytes.
This event is considered as an important step in evolution leading to seed habit. Heterospory is considered as the first step towards seed habit. Selaginella and Marsilea, show seed habit.

Question 8.
Explain briefly
(i) Protonema
(ii) Antheridium
(iii) Archegonium
(iv) Diplontic
(v) Sporophyll
(vi) Isogamy
Solution:
(i) Protonema: The predominant stage of moss gametophyte which directly develops from spore is known as protonema.
(ii) Antheridium: The male sex organ in bryophytes, pteridophytes and gymnosperms is called antheridium. It bears male gamete.
(iii) Archegonium : It is the female sex organ found in bryophytes, pteridophytes and gymnosperms. It bears a female gamete.
(iv) Diplontic: In the life cycle of plants when their diploid stage is dominant for long time then this is called diplontic.
(v) Sporophyll: The sporophyte bears sporangia that are subtended by leaf-like appendages called sporophylls.
(vi) Isogamy: When the gametes involved in sexual reproduction are morphologically similar then this is called isogamy. These gametes are physiologically different.

Question 9.
Differentiate between the following:-
• Red algae and brown algae
• Liverworts and moss
• Homosporous and heterosporous pteridophyte
• Syngamy and triple fusion
Solution:

Differences between red algae and brown algae are as follows
NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom 1

Question 10.
How would you distinguish monocots from dicots?
Solution:
Dicots are characterized by having two cotyledons in their seeds while the monocotyledons have only one.

Question 11.
Match the following content of column I with column II
Column I Column II
(a) Chlamydomonas (i) Moss
(b) Cycas (ii) Pteridophyta
(c) Selaginella (iii) Algae
(d) Sphagnum (iv) Gymnosperm
Solution:
(a) Chlamydomonas (iii) Algae
(b) Cycas (iv) Gymnosperm
(c) Selaginella (ii) Pteridophyta
(d) Sphagnum (i) Moss

Question 12.
Describe the important characteristics of gymnosperms.
Solution:
The seed forming vascular plants which produce seeds but no fruits are called gymnosperms. General characters of gymnosperm are as follows:

Fertilization does not require water.
Leaves may be of two kinds: foliage leaves and scale leaves.
The ovules are orthotropus.
Most primitive seed-bearing plants.
Mostly these plants are evergreen.
Have no ovary wall, seeds are naked.
Exhibit polyembryony. Sexual reproduction oogamous type.
Large, tall and woody trees.

VERY SHORT ANSWER QUESTIONS

Question 1.
Which group of plants are commonly called “Amphibians of Plant Kingdom”?
Solution:
Bryophytes

Question 2.
Which pigment does provide red colour to red algae?
Solution:
Phycoerythrin.

Question 3.
Which alga does reproduce sexually by conjugation?
Solution:
Spiro gym

Question 4.
Name the organ that fixes the Plant body of Riccia to the soil.
Solution:
Hepaticopsida.

Question 5.
Which plant group is called vascular cryptograms?
Solution:
Pteridophytes.

Question 6.
Which plant is commonly called as walking fern?
Solution:
Adiantum

Question 7.
Which group of plants produces seed but not fruits?
Solution:
Gymnosperm.

Question 8.
Which part of ovule is haploid in gymnosperm?
Solution:
Endosperm

Question 9.
What is name of megasporophyll bearing ovules of angiosperm?
Solution:
Carpel.

Question 10.
Why do gymnosperms fail to produce fruits?
Solution:
Fruits are formed from ovaries. Since the gymnosperm ovules are not enclosed inside the ovaries, they do not produce fruit.

Question 11.
Why life cycle of angiosperm is called diplontic?
Solution:
Life cycle of angiosperm is called diplontic because the diploid (sporophytic) phase is more prominent and long-lived whereas the haploid (gametophytic) phase is short-lived.

Question 12.
Which algae is known as Rolling alga?
Solution:
Volvox

Question 13.
Which are the specialized structures in selaginella which bear adventitious roots?
Solution:
Rhizophores

Question 14.
What is the number of cells and nuclei present in the embryo sac of angiosperm?
Solution:
7 cells and 8 nuclei

Question 15.
Which plant is known as a living fossil?
Solution:
Cycas

SHORT ANSWER QUESTIONS

Question 1.
What are the salient features of pteridophyte?
Solution:
Pteridophytes are vascular cryptogams i.e., plants of this group possess vascular tissue (i. e., xylem, and phloem) for the conduction of water and minerals and for the translocation of foods. They are flowerless and seedless plants.
The general characters of pteridophytes are as following.
I. Primary root is short-lived. It is replaced by adventitious roots.
II. All vegetative parts possess vascular tissues. A cambium is altogether absent. In xylem trachea are absent and in phloem companion cells are absent.
III. Pteridophytes show origin and evolution of stele {i.e., vascular tissue, pericycle, and if present the pith).
IV. Pteridophytes are characterized by having only tracheids in their xylem and only sieve tube in their phloem.
V. The main plant body is the sporophyte (diploid), usually differentiated into true roots, true stems, and true leaves. A stem is usually an underground rhizome or an erect trunk as in tree ferns. Leaves are large (megaphyllous) and variously shaped.

Question 2.
What is the Alternation of Generation?
Solution:
gametophytic and sporophytic generations alternate in the life cycle, it is called the Alter-nation of generation. Or vice versa gametophyte produces gametes and their fusion product zygote or Oospore. Oospore produces sporophyte. The sporophyte produces spores. Spores germinate and produce gametophyte.

Question 3.
What are gemmae? Name two plants that produce gemmae.
Solution:
Gemmae are green, multicellular asexual buds, which develop in small receptacles, called gemmae cups, on the thallus.
The gemmae become detached from the parent thallus and germinate to form new individuals. e.g., Marchantia, Riccia.

Question 4.
Differentiate between cytotaxonomy and chemotaxonomy.
Solution:
Cytotaxonomy that is based on cytological information like chromosome number, structure, behaviour and chemotaxonomy that uses the chemical constituents of the plant to resolve confusions, are also used by taxonomists these days.

Question 5.
Differentiate between monocotyledon and dicotyledons.
Solution:

Question 6.
Mention four types of Placentation with suitable examples.
Solution:

Marginal placentation: Example members of the family fabaceae (Crotalaria)
Axile placentation: Example members of the family Malvaceae (Hibiscus)
Parietal placentation. Example: Cucumber.
Basal placentation Example: Sunflower.

Question 7.
What are the features of Gemmae found in bryophytes?
Solution:
Gemmae are the means of asexual reproduction found in many bryophytes (ex-Liverworts). They are one to many-celled, specially produced clonal plant fragments. They are green multicellular, asexual buds which develop in small receptacles called “gemma cups” located on the thalli. Gemmae become detached from the parent body and germinate to form a new individual

Question 8.
What is the basis of the classification of the phylogenetic system?
Solution:
It indicates evolutionary as well as the genetic relationship among organism, it is based on the fossil record, biochemical, anatomical, morphological, embryological, physiological, genetics, Karyotype, and other studies.

Question 9.
What are the main features of Anthocerotopsida?
Solution:
These are also known as hornworts because typical horn-like appearances are present of their sporophyte. These contain thalloid gametophyte, distinctly dorsiventral, rhizoids are present, Thalloid do not possess air chambers and scales. Each cell of thallus has a single large chloroplast with a pyrenoid.

LONG ANSWER QUESTIONS

Question 1.
Enumerate distinguishing characters of Bryophytes.
Solution:

Bryophytes are called “Amphibians of the plant kingdom” They grow in moist shady places. They require water for their sperms to swim towards eggs, hence the term ‘amphibians’.
The plant body is gametophyte so that it produces garnets.
The gametophyte is nonvascular so that xylem and phloem tissues are absent.
The gametophyte is thalloid in structure or it may be differentiated into stem-like, root-like and leaf-like organs.
Gametophyte reproduces sexually. The male sex organs are antheridia and the female sex organs are archegonia. Antheridia produce flagellated motile sperms. Archegonia produce female gamete or egg.
Fertilization results in the formation of diploid oospore. It produces diploid sporophyte Sporophyte reproduces asexually and produces haploid spores. They germinate and produce haploid gametophyte. The sporophyte is dependant on the gametophyte.
In the life cycle, there is an alternation of generation between haploid gametophyte and diploid sporophyte.
Bryophytes also show vegetative reproduction by fragmentation, gemmae formation etc.

Question 2.
Write short notes on the following :
(a) Peristomial teeth of moss
(b) Protonema of moss
(c) Archegonia of moss
Solution:
(a) Peristomal teeth are located just below the operculum. It helps in the dispersal of spores by the hygroscopic movement of its outer ring while the inner ring does not show the hygroscopic movements.
(b) Each spore produces a filamentous juvenile stage called protonema. Protonema has two types of branches, subterranean nongreen rhizoidal, and green Mediterranean branches. Buds develop on green prostrate branches which grow to form new moss plants.
(c) Female reproductive organ of moss is called archegonium.
NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom 4

Question 3.
Explain the general characters of the Gymnosperm plant body. (Tumkur 2008)
Solution:

Gymnosperm plant body is sporophyte.
Sporophyte consists of stem, root and leaves. Stem is unbranched in cycas. the stem bears crown of leaves at the tip.
The foliage leaves are pinnately compound in cycas and needle-like in conifers. Brown small non-photosynthetic leaves called scale leaves/cataphylls are present in cycas.
In cycas, the cataphylls are thickly covered with brown hairs called ramenta.
The foliage leaves are green. The pinnae are tough, leathery and has only midrib without veins.
Young leaves show circinate vernation.
The roots are inhabited by blue-green algae like Nostoc help the plant in nitrogen fixation
The sporophyte produces two types of spores so it is heterosporous.

Question 4.
Give a comparative account of gymnosperms and angiosperms.
Solution:
Comparison of gymnosperms and angiosperms
NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom 5

Question 5.
Describe major features of Plant kingdom.
Solution:
Features of Plant Kingdom:

Plants are autotrophic, except for some carnivorous, plants. They trap photo energy from sunlight and convert it to chemical energy through I photosynthesis. Because of this plants are the main channel for supplying energy in the food chain on earth.
Reproduction in plants can be by any of the following modes: Vegetative or Asexual, and Sexual Reproduction.
Plant cell is unique because of presence of cell wall and large vacuoles. Green parts of plant contain chlorophyll, which helps them in trapping the photo energy.
Sizes of plants can vary from microscopic to a very large tree. Plants are mainly divided into Algae, Bryophytes, Pteridophytes, Gymnosperms and Angiosperms.
Lower plants, like algae and bryophytes, have thalloid structures, while higher plants, like gymnosperms and angiosperms, have clearly defined roots and stems.
In higher plants root gives a means to anchor in the soil and helps the plant in taking minerals and water from the soil. Green leaves on the stem help them in photosynthesis.
Most of the plant growth as a result of photosynthesis. After photosynthesis, extra food is utilized to facilitate growth.
Usually in higher plants growth is unlimited and some taller trees can live a life of more than 1000 years.
Being the main carbon fixation agent, plants are very important for the whole ecology.
The whole food basket for humans is being filled by the plant kingdom. Even animal products, like milk and poultry, are indirect results of a plant’s carbon fixation.
Plants supply raw materials for a majority of economic activities. Wood for furniture and building materials comes from plants. The whole paper industry is dependent on the plant kingdom. Think of life if there was no paper and you may understand the larger impact on human civilization.
Angiosperms have special organs, called flowers, to bear sexual parts. Flowers are a helpful tool in facilitating variations and further evolution of the plant kingdom.

NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom 6

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NCERT Solutions for Class 11 Biology Chapter 8 Cell: The Unit of Life

June 18, 2021

NCERT Solutions for Class 11 Biology Chapter 8 Cell: The Unit of Life

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 8 Cell: The Unit of Life.

Question 1.
Which of the following is not correct?
(a) Robert brown discovered the cell.
(b) Schleiden and Schwann formulated the cell theory.
(c) Virchow explained that cells are from pre¬existing cells.
(d) A unicellular organism carries out its life activities within single cell.
Solution:
(a) Robert brown discovered the cell.

Question 2.
New cells generate from
(a) bacterial fermentation
(b) regeneration of old cells
(c) pre-existing cells
(d) abiotic material
Solution:
(c) pre-existing cells

Question 3.
Match the following
Column A Column B
(a) Cristae (i) Flat membranous sac in stroma
(b) Cisternae (ii) Infoldings in mitochondria
(c) Thylakoids (iii) Disc-shaped sacs in Golgi apparatus
Solution:
(a) Cristae (ii) Infoldings in mitochondria
(b) Cisternae (iii) Disc-shaped sacs in Golgi apparatus
(c) Thylakoids (i) Flat membranous sac in stroma

Question 4.
Which of the following is correct
(a) Cells of living organisms have a nucleus.
(b) Both animals and plant cells have a well defined cell wall.
(c) In prokaryotes, there are no membrane bound organelles.
(d) Cells are formed de novo from abiotic materials
Solution:
(c) In prokaryotes, there are no membrane-bound organelles.

Question 5.
What are mesosomes in a prokaryotic cells? Mention the function that it performs.
Solution:
A special membranous structure is a mesosome that is formed by the extensions of the plasma membrane into the cell. These extensions are in the form of vesicles, tubules, and lamellae. They help in cell wall formation, DNA replication, and distribution to daughter cells. They also help in respiration, in the secretion process increase plasma membrane surface and enzymatic content.

Question 6.
How do neutral solutes move across the plasma membrane? Can the polar molecules also move across it in the same way? If not, then how are these transported across the membrane?
Solution:
Neutral solutes may move across the membrane by the process of simple diffusion along the concentration gradient i.e. from higher concentration to the lower. Water may also move across this membrane from higher to lower concentrations. The movement of water by diffusion is called osmosis. As the polar molecules cannot pass through the non-polar lipid bilayer, they require a carrier protein of the membrane to facilitate their transport across the membrane.

A few ions or molecules are transported across the membrane against their concentration gradient i.e. from lower to the higher concentration. Such transport is an energy-dependent process, in which ATP is utilized and is called active transport. e,g., Na’/K* Pump.

Question 7.
Name two cell-organelles that are double membrane-bound. What are the characteristics of these two organelles? State their functions and draw labelled diagrams of both.
Solution:
Two double membrane-bound cell organelles:
(a) Mitochondria: It has finger-like folds in the inner membrane called cristae. Mitochondria is the place for aerobic respiration.
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1
(b) Chloroplast: Chloroplast is responsible for converting light energy into chemical energy. Chloroplast contains stacked thylakoid in its matrix.

Functions of mitochondria: The double membrane mitochondria are actively associated with aerobic respiration and the release of energy for cellular activity. The biological oxidation of the fats and carbohydrate release much amount of energy which is utilised by mitochondria for ATP synthesis. The required energy is released from ATP molecules for various cell processes in cells so they are termed as “the powerhouse of the cell”
Functions of chloroplast :
(i) Their main function of the chloroplast is to trap the sun’s energy and to convert it into chemical energy of food by photosynthesis.
(ii) Storage of starch.
(iii) Chloroplast in fruits and flowers changes into chromoplasts.

Question 8.
What are the characteristics of prokaryotic cells?
Solution:
The prokaryotic cells are represented by bacteria, blue-green algae, mycoplasma, and PPLO (Pleuro Pneumonia-like organisms). They are generally smaller and multiply more rapidly than the eukaryotic cells. They may vary greatly in shape and size. The four basic shapes of bacteria are bacillus (rod-like), coccus (spherical), vibrio (comma-shaped), and spirillum (spiral).

The organization of the prokaryotic cell is fundamentally similar even though prokaryotes exhibit a wide variety of shapes and functions. All prokaryotes have a cell wall surrounding the cell membrane. The fluid matrix filling the cell is the cytoplasm. There is no well-defined nucleus. The genetic material is basically naked being not enveloped by a nuclear membrane.

In addition to the genomic DNA (the single chromosome/ DNA circle), many bacteria have small DNA circles outside the genomic DNA. These smaller DNA circles are called plasmids, The plasmid DNA confers certain unique phenotypic to antibiotics. Prokaryotes have something unique in the form of inclusions.

A specialised differentiated form of the cell membrane called mesosome is the characteristic of prokaryotes. They are essentially infoldings of the cell membrane.

Question 9.
Multicellular organisms have a division of labour. Explain.
Solution:
In unicellular organisms, there is no division of labour.

The single cell of the organisms is capable of performing all the vital activities of life i.e., respiration, movement, digestion and reproduction, etc. Respiration, nutrition, and excretion in most of these unicellular organisms takes place through the general body surface.
No special organs for these are present in them because they are too small to need them. Most of these unicellular organisms reproduce by simple binary division, to maintain their continuity.
However, in some, sexual reproduction has also been observed.

Question 10.
Cell is the basic unit of life. Discuss in brief.
Solution:
Cell: The Basic Unit of life: All living organisms are composed of small, tiny structures or compartments called cells. These cells are called the ‘building blocks’ of life.

The cells in true sense are considered as the basic unit of life because all the life processes i.e., metabolism, responsiveness, reproduction are carried out by the cells.
Respiration, nutrition, release of energy for the body are carried out within the cells only.
Even the animals and plants reproduce because the cells reproduce individually.
Growth occurs because cell grow and multiply.
In Amoeba all the life processes are performed within the boundaries of the single cell.
This is true of all other multicellular organisms. The only difference in the multicellular organisms is that the body of these organisms is made up of many cells.
In these organisms, the cell do not behave independently but get organized into tissues. Each tissue is specialized to perform specific functions. Different tissues then get organised into tissues.
Each tissue is specialized to perform specific functions. Different tissues then get organised into organs which perform certain specific functions.
Different organs are finally organised to form organ systems. Now it must be very clear that the basic structure to tissues, organs and organ system are the cells only.
These tissues, organs and organ system of the organisms work because the cells work.
Thus “the cells are structural and functional unit of the living beings” hence it is the basic unit of life.

Question 11.
What are nuclear pores? State their function.
Solution:
At a number of places, the nuclear envelope is intercepted by minute pores which are called nuclear pores. These are formed by the fusion of two nuclear membranes. These nuclear pores are the passages through which the movement of RNA and protein molecules takes place in both directions between the nucleus and the cytoplasm.

Question 12.
Both lysosomes and vacuoles are * endomembrane structures, yet they differ in
terms of their functions. Comment.
Solution:
Lysosomes are filled with hydrolytic enzymes that are capable of digesting carbohydrates, proteins, lipids and nucleic acids whereas vacuoles contain water, sap, excretory product and other materials not useful for cell.

Question 13.
Describe the structure of the following with the help of labelled diagrams.
(i) Nucleus
(ii) Centrosome
Solution:

(i) Nucleus:

1. The nucleus is a large organelle controlling all the activities of the eukaryotic cells. Some cells have more than one nucleus.
2. Binucleate cells have 2 nuclei per cell eg. Paramoecium. Multinucleate cells have many nuclei e.g. Ascaris.
3. Some cells lack nucleus (anucleate) at maturity. Examples: mammalian RBCs and sieve tube cells in vascular plants.
4. The nucleus is bounded by two membranes, which make the nuclear envelope.
5. The outer and inner membranes are separated by a narrow space, perinuclear space.
6. The outer membrane remains in continuation with endoplasmic reticulum (ER) and the inner one surrounds the nuclear contents.
7. At some points, the nuclear evelope is interrupted by the presence of small structures called nuclear pores.
8. These pores help in exchange of materials between nucleoplasm and cytoplasm. Nuclear membrane dissappears during cell division. It reappears during nuclear reorganization in stage.
9. The nucleoplasm contains chromatin and nucleolus. The nucleolus is a rounded structure. It is not separated from the rest of the nucleoplasm by membrane.
10. It is associated with a specific nucleolar organizing region (NOR) of some chromosomes. Nucleolus is the “site for ribosomal RNA synthesis”.
11. The cells which remain engaged in protein synthesis have larger and more numerous nuclei in their nucleoplasm.

NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 4

(ii) Centrosome:

Under the electron microscope, each centriole is seen to be formed of nine sets of tubular structures arranged in a circular fashion.
Each of these sets is a triplet composed of three microtubules. Each microtubule has a diameter of about 250A. The triplets are found in the matrix.
Sometimes delicate strands appear to connect sets of the triplet to each other.
Also can be seen radiating from the central core of the cylinder, delicate strands which connect sets of the triplets to each other giving a cartwheel appearance.
Basal bodies are structures similar to the centrioles. They produce cilia and flagella.

Question 14.
What is a centromere? How does the position of the centromere form the basis of the classification of chromosomes? Support your answer with a diagram showing the position of the centromere on different types of chromosomes.
Solution:
Eukaryotic chromosomes: The chromosomes are uncoiled in a loose, indistinct network called the chromatin that contains DNA, RNA and protein in interphase.

The types of proteins present and associated with DNA are histone and non-histone proteins.

Chromosomes are thread-like structures. They become visible (under light microscope) during cell division.

In higher organisms, the well-developed nucleus contains a definite number of chromosomes of definite size and shape.
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 5
The shape of a chromosome is usually observable at metaphase and anaphase when the position of primary constriction {centromere) is clearly seen. Based on the position of the centromere, chromosomes are of 3 types:

telocentric – with terminal centromere,
the acrocentric – terminal centromere is capped by a telomere
submetacentric – the centromere is subterminal in position
metacentric – these have median centromere.

VERY SHORT ANSWER QUESTIONS

Question 1.
Who discovered the Golgi body?
Solution:
Camillo Golgi (1898).

Question 2.
Give the location of 70S ribosomes.
Solution:
Prokaryotic cells, plastids, and mitochondria

Question 3.
Name the cell organelle rich in acid hydrolases.
Solution:
Lysosomes

Question 4.
Who proposed the cell theory?
Solution:
Schleiden and Schwann.

Question 5.
Expand PPLO.
Solution:
PPLO (Pleuro Pneumonia Like Organisms)

Question 6.
Name the organelle responsible for protein synthesis in a cell.
Solution:
Ribosome

Question 7.
Give the full form of SER and RER.
Solution:

SER – Smooth endoplasmic reticulum.
RER – Rough endoplasmic reticulum.

Question 8.
Name the membrane which surrounds the vacuole in the cell.
Solution:
Tonoplast

Question 9.
Name two types of constituents of the plasma membrane.
Solution:
Proteins and lipids.

Question 10.
Name two processes of passive transport.
Solution:

Osmosis
Diffusion

Question 11.
What is plasmodesmata? What is its function?
Solution:
Plasmodesmata: Adjoining the cells and the linking gap of cytoplasmic protoplasmic presence is called Plasmodesmata. It links the neighbouring cells together.

SHORT ANSWER QUESTIONS

Question 1.
Why fluid-mosaic model is more accepted than other models of the plasma membrane?
Solution:
The fluid mosaic model explains
(i) quasifluid state of the plasma membrane,
(ii) It differentiates two 6. types of proteins
(iii) It explains functional specificity and variability in two surfaces of PM.

Question 2.
Name three types of elements in the Golgi body. List two major functions of the Golgi body.
Solution:
Three types of elements in golgi body are cistemae, vesicles and vacuoles. The main function of golgi bodies are cellular secretion and acrosome formation.

Question 3.
What is peculiar about mitochondrial DNA?
Solution:
Mitochondrial DNA is circular double-stranded and not associated with histone proteins.

Question 4.
How does cytokinesis take place in plant and animal cells?
Solution:
In plant cell cytokinesis take place by cell plate formation and in animal cells it occurs by constriction

Question 5.
Differentiate between prokaryotic and eukaryotic cells.
Solution:
The main differences between prokaryotic cell and eukaryotic cell are
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 6

Question 6.
Differentiate between active and passive transport across the membrane.
Solution:
The main differences between active transport and passive transport are
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 7

Question 7.
Describe the different methods of transport of nutrients in cell.
Solution:
Different methods of transport of nutrients in the cell are:
(i) Simple diffusion : It is the movement of ions/ molecules of any substance from a region of higher concentration to ‘the region of lower concentration, until equilibrium is reached. Many neutral solutes move by diffusion.
(ii) Osmosis : It is the movement of solvent molecules across a semipermeable membrane, from the region of higher concentration to the region of lower concentration, until equilibrium is reached. Water moves by osmosis from one cell to the other.
(iii) Facilitated diffusion : It refers to the movement of ions/molecules across the membrane with the help of transmembrane proteins.

Question 8.
Why are ribosomes of prokaryotes different from eukaryotes?
Solution:
The type of ribosomes of prokaryotes is different from eukaryotes because the prokaryotes were primitive, simpler and have remained intact during evolution while at the base level eukaryotes have adapted with the environment and are retaining their kind of entities for the complex structure. Prokaryotes have 70S ribosomes with 30S and 50S subunit and eukaryotes have 80S ribosome with 40S subunit and 60S sub unit.

Question 9.
What is the function of
(1) Nuclear Pores
(2) Slimy Capsule in Bacteria
(3) Golgi Bodies:
(4) Centrosome with Centrioles
Solution:
(1) Nuclear Pores: There is exchange of RNA and proteins through the nuclear pores
(2) Slimy Capsule in Bacteria : A slimy capsule is the outer covering of cell wall of bacteria and is an additional protection for the bacteria.
(3) Golgi Bodies :
(i) It takes part in packaging materials delivered either to the intra-cellular targets or secteted outside the cell.
(ii) It is also a important site of formation of glycoproteins and glycolipids.
(4) Centrosome with Centrioles
(i) Centrioles help in organising the spindle fibres and astral rays during cell division.
(ii) It also provides basal bodies which give rise to cilia and flagella.

LONG ANSWER QUESTIONS

Question 1.
Describe the structure of cell wall.
Solution:

A non-living rigid structure called the cell wall forms an outer covering for the plasma membrane of fungi and plants.
Cell wall does not only give shape to the cell and but protect the cell from mechanical damage and infection.
It also helps in cell-to-cell interaction and provides barrier to undesirable macromolecules.
Algae have cell wall, made of cellulose, galactans, mannans and minerals like calcium carbonate, while in other plants it consists of cehulose, hemicellulose, pectins and proteins.
The cell wall of a young plant cell, the primary wall is capable of growth, which gradually diminishes as the cell matures and the secondary wall is formed on the inner side of the cell.
The middle lamella is a layer mainly of calcium pectate which holds or glues the different neighbouring cells together.
The cell wall and middle lamellae may be transferred by plasmodesmata which connect the cytoplasm of neighbouring cells.

Question 2.
Give an account of prokaryotic cells.
Solution:

The prokaryotic cells are represented by bacteria, blue green algae, mycoplasma and PPLO (Pleuro Pneumonia Like Organisms). They are generally smaller and multiply more rapidly than the eukaryotic cells.
They may vary greatly in shape and size. The four basic shapes of bacteria are bacillus (rod like), coccus (spherical), vibrio (comma shaped) and spirillum (Spiral).
The organisation of the prokaryotic cell is fundamentally similar even though prokaryotes exhibit a wide variety of shapes and functions. All prokaryotes have a cell wall surrounding the cell membrane.
The fluid matrix filling the cell is the cytoplasm. There is no well defined nucleus.
The genetic material is basically naked, not enveloped by a nuclear membrane.
In addition to the genomic DNA (the single chromosome/circular DNA), many bacteria have small circular DNA outside the genomic DNA. These smaller DNA is called plasmids.
The plasmid DNA confers certain unique phenotypic characters to such bacteria. One such character is resistance to antibiotics. Nuclear membrane is found in eukaryotes.
No organelles, like the ones in eukaryotes, are found in prokaryotic cells except ribosomes.
Prokaryotes have something unique in the form of inclusions. A specialised differentiated form of cell membrane called mesosome is the characteristic of prokaryotes which helps in respiration process.
They are essentially infoldings of cell membrane.

Question 3.
Give an ultrastructure of mitochondria.
Solution:

Mitochondria, unless specifically stained, are not easily visible under the microscope.
The number of mitochondria per cell is variable depending on the physiological activity of the cells.
In terms of shape and size also, considerable degree of variability is observed.
Typically it is sausage shaped or cylindrical having a diameter of 0.2-1.0 ft m (average 0.5 film) and length (1.0 -4.1 ft).
Each mitochondrion is a double membrane-bound structure with the outer membrane and the inner membrane dividing its lumen distinctly into two aqueous compartments, i.e. the outer compartment and the inner compartment.
The inner compartment is called the matrix. The outer membrane forms the continuous limiting boundary of the organehe.
The inner membrane forms a number of infoldings called the cristae. The cristae increase the surface area.
The two membranes have their own specific enzymes associated with the mitochondrial function. Mitochondria are the sites of aerobic respiration.
They produce cellular energy in the form of ATP, hence they are called. “Power houses” of the cell.
The matrix also possesses single circular DNA molecule, a few RNA molecules, ribosomes (70s) and the components required for the synthesis of proteins. The Mitochondria divide by fission.

Question 4.
Describe the fluid mosaic model of membrane.
Solution:
The characteristic features of fluid mosaic model
• This model was proposed by Singer and Nicholson.
• According to this model, there is a central bilipid layer (of phospholipids) with their polar head group toward the outside and the non-polar tails pointing inwards.
• Some proteins which are embedded in the lipid layer are called integral proteins and they cannot be separated from the membrane easily.
• Some large globular integral proteins which project beyond the lipid layer on both the sides are believed to have channels through which water soluble materials can pass across.
• Those proteins which are superficially attached are called peripheral (extrinsic) proteins and they can be easily removed.
• Some membrane lipids and integral proteins remain bound to oligosaccharides; such oligosaccharides project into the extracellular fluid and they influence the manner in which cells interact with the other cell.
• There are also certain specific proteins called membrane receptors, which mediate the flow of materials and information into the cell.

Question 5.
What are plastids? How are they classified on the basis of the type of pigments? Name them and their pigments and mention their functions.
Solution:
Plastids are double-membrane bound organelles
of different shapes, that are found only in plant
cells and contain pigments and storage products.
They are of three types :
(i) Leucoplasts
These are the oval, spherical, rod-like or filamentous colourless plastids which are found in storage organs. Their main function is to store reserve materials like starch (amyloplasts), proteins (aleuroplasts) and fats (elaioplasts).
(ii) Chromoplasts
• These are coloured plastids containing mainly the yellow, red and orange pigments (carotene and xanthophyll).
• These are found in petals of flowers and skin of fruits.
• They attracts agents for pollination and dispersal of fruits/seeds.
(iii) Chloroplasts
• These are the green plastids containing mainly chlorophylls and very little carotene and xanthophyll.
• Their main function is photosynthesis and the formation of starch.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 8 Cell: The Unit of Life, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 8 Cell: The Unit of Life, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division

June 18, 2021

NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division.

Question 1.
What is the average cell cycle span for a mammalian cell?
Solution:
It is significant to note that in the 24-hour average duration of the cell cycle of a human cell or mammalian cells, cell division proper lasts for only about an hour.

Question 2.
Distinguish cytokinesis from karyokinesis.
Solution:

Karyokinesis is a division of the nucleus (mitosis or meiosis) while cytokinesis is a division of the cytoplasm.
Cytokinesis in an animal cells is achieved by the appearance of a furrow in the plasma membrane.
The furrow gradually deepens and ultimately joins in the center dividing the cell cytoplasm into two.
In-plant cell cell wall formation starts in the center of the cell and grows outward to meet the existing lateral walls.
The formation of a new cell wall begins with the formation of a simple precursor, called the cell plate that represents the middle lamella between the walls of two adjacent cells.
It is the time of cytoplasmic division, organelles like mitochondria and plastids get distributed between the two daughter cells.
In some organisms, karyokinesis is not followed by ”cytokinesis as a result of which multinucleate condition arises leading to the formation of syncytium (liquid endosperm of coconut).

Question 3.
Describe the events taking place during the interphase.
Solution:
The interphase though called the resting phase is the time during which the cell is preparing for division by undergoing both cell growth and DNA replication in an orderly manner. The interphase is divided into three further phases :

G₁ phase (Gap)
S phase (Synthesis)
G₂ phase (Gap2)

G₁ phase corresponds to the interval between mitosis and initiation of DNA replication. During G₁ phase the cell is metabolically active and continuously grows but does not replicate its DNA. S or synthesis phase marks the period during which DNA synthesis or replication takes place. During this time the amount of DNA per cell doubles. If the initial amount of DNA is denoted as 2C then it increases to 4C. However, there is no increase in the chromosome number; if the cell had diploid or 2n number of chromosomes at G₁ even after the S phase the number of chromosomes remains the same i.e. 2n.

In animal cells, during the S phase, as DNA replication begins in the nucleus, the centrioles, initiate replication in the cytoplasm. Dumping the G₂ phase proteins are synthesized in preparation for mitosis while cell growth continues.

Question 4.
What is the G₀ (quiescent phase) of the cell cycle?
Solution:
Some cells in adult animals do not appear to exhibit division (e.g., heart cell, nerve cell). These cells become inactive and become specialized by differentiating and do not further exit the G₁ phase, and enter a stage but divide occasionally called the quiescent stage (G₀) of the cell cycle.
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1
A cell in this stage remains metabolically active but no longer proliferate unless called to do so depending on the requirement of the organism.

Question 5.
Why is mitosis called equational division?
Solution:
M phase is the most dramatic period of the cell cycle involving a major reorganization of virtually all components of the cell. Since the number of chromosomes in the parent and progeny cells is the same it is also called equational division. Though for convenience mitosis has been divided into four stages of nuclear division, it is very essential to understand that cell division is a progressive process and very clear-cut lines cannot be drawn between various stages.
Mitosis is divided into the following four stages:

Prophase
Meta phase
Anaphase
Telophase

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

Question 7.
Describe the following:
(a) synapsis
(b) bivalent
(c) chiasmata
Draw a diagram to illustrate your answer.
Solution:
(a) During zygotene of prophase I of meiosis homologous chromosomes pair together. This pairing is called synapsis.
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 2
(b) Bivalent: The complex formed by homologous chromosomes during zygotene is called a bivalent.
(c) Chiasmata: During diplotene, the paired chromosomes make an X-shaped structure. This is called chiasmata.

Question 8.
How does cytokinesis in plant cells differ from that in animal cells?
Solution:
In animal cells, cytokinesis is achieved by the appearance of a furrow in the plasma membrane. The furrow gradually deepens and ultimately joins in the centre dividing the cell cytoplasm into two.

Due to the presence of cell walls, cytokinesis is different in plants. In plants, wall formation starts in the centre of the cell and grows outward to meet the existing lateral walls. The formation of the new cell wall begins with the formation of a simple precursor, called the cell plate that represents the middle lamella between the walls of two adjacent cells. Organelles like mitochondria and plastids get distributed between the two daughter cells during this process.

Question 9.
Find examples where the four daughter cells from meiosis are equal in size and where they are found unequal in size.
Solution:
The four daughter cells produced may be equal in size in the sperm of animals. They may be unequal in size as gametes in plants-pollen grains and egg in ovules.

Question 10.
Distinguish anaphase of mitosis from anaphase I of meiosis.
Solution:
The main difference between anaphase of mitosis and anaphase I of meiosis are as follows:
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 3

Question 11.
List the main differences between mitosis and meiosis.
Solution:
The main differences between mitosis and meiosis are as follows :
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 4

Question 12.
What is the significance of meiosis?
Solution:
1. Meiosis ensures the production of the haploid phase in the life cycle of sexually reproducing organisms whereas fertilization restores the diploid phase. We come across meiosis during gametogenesis in plants and animals. This lead to the formation of haploid gemotes.

2. Crossing over takes place during the Pachytene stage of meiosis, in which an exchange of genetic matter occurs. This produces variation, the raw material for evolution.

3. Meiosis has an impact on the genetic consequences due to pairing, crossing over, recombination, and segregation of homologous chromosomes.

Question 13.
Discuss with your teacher about

haploid insects and lower plants where cell-division occurs and
some haploid cells in higher plants where cell-division does not occur.

Solution:

Some insects like honey bee drones are haploid. They are not fertile.
In lower plant, main plant body is haploid produces haploid microspores by mitosis. The gametes of Chlamydomonas fuse to form diploid zygotes. Meiosis take place at this stage in lower plants forming haploid spous which give rise to the new plant.

Question 14.
Can there be mitosis without DNA replication in S-phase?
Solution:
S or synthesis phase marks the period during which DNA synthesis or replication takes place. During this time the amount of DNA per cell doubles. If the initial amount of DNA is denoted as 2C then it increases to 4C. However, there is no increase in the chromosome number; if the cell had diploid or 2n number of chromosomes at G, even after the S phase the number of chromosomes remains the same i.e. 2n.

Question 15.
Can there be DNA replication without cell division?
Solution:
DNA replication takes place in order to prepare cells for division. Cell division is the next logical step after DNA replication.

Question 16.
Analyze the events during every stage of the cell cycle and notice how the following two parameters change
i. Number of chromosomes (N) per cell
ii. Amount of DNA content (C) per cell
Solution:
Cell division is a very important process in all living organisms. During the division of a cell, DNA replication and cell growth also take place. Although cell growth is a continuous process, DNA synthesis occurs only during one specific stage in the cell cycle. The replicated chromosomes (DNA) are then distributed to daughter nuclei by a complex series of events during cell division. These events are themselves under genetic central.

The cell cycle is divided .nto two basic phases: The M phase starts with the nuclear division, corresponding to the separation of daughter chromosomes (Kaiyokinesis), and usually ends with the division of cytoplasm (Cytokinesis’). The interphase though called the resting phase is the time during which the cell is preparing for division by undergoing both cell growth and DNA replication in an orderly manner. The interphase is divided into three further phases:

G₁ phase (Gap)
S phase (Synthesis)
G₂ phase (Gap2)

VERY SHORT ANSWER QUESTIONS

Question 1.
What is karyokinesis?
Solution:
Division of nucleus is Known as karyokinesis.

Question 2.
Name the phase in which chromatids move apart.
Solution:
Anaphase.

Question 3.
Name the synthetic phase of interphase.
Solution:
S phase

Question 4.
Name the cell divisions which help recombination of genes.
Solution:
Meiosis.

Question 5.
Which type of cell division occurs in the meristematic cell of the root apex?
Solution:
Mitosis

Question 6.
Name the stage during which astral and spindle fibres disappear and nuclear membrane and nucleoli reappear.
Solution:
Telophase.

Question 7.
Name the sub-phases of prophase-I of Meiosis.
Solution:
Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

Question 8.
In which stage, the actual reduction of chromosome number occurs in meiosis?
Solution:
Anaphase I

Question 9.
What is Synapsis? (Bijapur, Belgaum, Shimoga 2004)
Solution:
The process of pairing homologous chromosomes is called Synapsis.

Question 10.
What is the peculiarity of zygotene?
Solution:
In the zygotene phase pairing of homologous chromosomes or synapsis takes place.

Question 11.
Define crossing over. Give its significance.
Solution:
It is an exchange of genes between non-sister chromatids of homologous chromosomes. It produces new combination of genes and variation.
Question 12.
Why mitosis is an equational division called?
Solution:
Mitosis is a process of cell division, in which chromosomes are equally distributed into two daughter cells so it is equational division.

SHORT ANSWER QUESTIONS

Question 1.
Differentiate between:
(a) S-phase and G2 phase.
(b) G1 and G2 phase
Solution:
(a) Difference between S-phase and G2 phase:
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 5
(b) Difference between Gj and G2 phase

Question 2.
Write the significance of Mitosis / Meiosis. (Udupi 2006, D.Kannada 2010)
Solution:
Significance of Mitosis:

The distribution of an equal number of chromosomes to the daughter cells maintains a constant chromosome number.
Mitosis increases the number of cells so it contributes to growth.

Significance of Meiosis:

Meiosis brings genetic crossing over and random distribution of paternal and maternal chromosomes to daughter cells.
Recombination produces variations and variations are the sources of organic evolution.

Question 3.
Mention the significance of mitosis.
Solution:

In multicellular organisms, the growth of the body is due to the mitotic division of cells.
Replacement of worn-out cells and repair of the damaged cells is by mitosis.
In unicellular organisms, mitosis results in the asexual reproduction of cells.
In plants, vegetative propagation involves only mitotic divisions.

Question 4.
When and why does a reduction in the number of chromosomes take place in meiosis?
Solution:
The actual reduction in the number of chromosomes takes place in anaphase I.
This is because in anaphase I, one member from each homologous pair moves to one pole; the two chromatids of the chromosomes do not separate as the centromeres do not divide at this stage.

Question 5.
Differentiate between prophase I and prophase of mitosis.
Solution:
Difference between prophase I and prophase of mitosis are :
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 7

Question 6.
Imagine a situation if there was no meiosis. Then what would have happened to the next generation?
Solution:
In the absence of meiosis, the next generation would have double the number of chromosomes after the fusion of gametes. This would have resulted in the birth of an altogether new species. The maintenance of characters’ sets would have been possible only through asexual reproduction.

Question 7.
Which stage in meiosis is marked with genetic recombination? How
Solution:
In the pachytene stage of meiosis I the chromosomes appear as tetrads of homologous chromosomes and cross as the genetic material exchange takes place. Thus this stage is marked with genetic recombination.

Question 8.
What is Synapsis and Synaptonemal Complex? In which stage do these occur?
Solution:
The pairing of chromosomes is called Synapsis.
In the chromosomal synapsis, a complex structure is made known as the Synaptonemal complex These occur in the zygotene stage of meiosis.

Question 9.
Describe the events in the prophase of animal cells.
Solution:

Chromosomal material condenses to form compact mitotic chromosomes.
Chromosomes are seen to be composed of two chromatids attached together at the centromere.
The centrioles formed in interphase start moving to the opposite poles of the cell.
Initiation of the assembly of the mitotic spindle, the microtubules, the proteinaceous components of the cell cytoplasm help in the process.
By the end of the prophase, the nuclear membrane and nucleolus disappear.

Question 10.
Diagrammatically shows the cell cycle and answers the following questions.

(a) Which is the resting stage.
(b) Which is the stage where replication takes palce.
(c) Which is the stage where mitosis takes place.
(d) Which is the post cell synthetic but pre-cell division stage.
Solution:
(a) Go Phase
(b) S Phase
(c) M Phase
(d) G2 Phase

Question 11.
Define cytokinesis. How is it accomplished in animal and plant cells?
Solution:
The process by which the cytoplasm of the cell divides resulting in the formation of two daughter nuclei is called cytokinesis.

In an animal cell, this is achieved by the appearance of a furrow in the plasma membrane. The furrow gradually deepens and ultimately joins in the centre dividing the cell cytoplasm into two.

In-plant cells, wall formation starts in the centre of the cell and grows outward to meet the existing lateral walls. The formation of a new cell wall begins with the formation of a simple precursor, called the cell plate that represents the middle lamella between the walls to two adjacent cells.

LONG ANSWER QUESTIONS

Question 1.
Describe briefly cytokinesis in animal cells and plant cells.
Solution:
(a) Cytokinesis in animal cells: In the animal cells, the cytoplasm divides by constriction. It appears on the equator and slowly deepens. The constriction converges on all the sides and pinches off the parent cell into 2 daughter cells. Constriction is the result of a peripheral band of microfilaments. This constriction divide the cytoplasm finally,

(b) Cytokinesis in plant cells: Plant cells have a rigid cell wall and this cannot undergo cytokinesis by invaginating cleavage furrow. Therefore, in them, the cytokinesis is accomplished by the formation of phragmoplast from carbohydrate and lipid-containing vesicles of Golgi apparatus and endoplasmic reticulum vesicles. A cell plate at the equator of the dividing cell is formed and divides the cytoplasm.

Question 2.
Mention the significance of mitosis.
Solution:
Significance of mitosis: Mitosis or the equational division is usually restricted to the diploid cells only.

However, in some lower plants and in some social insects haploid cells also divide by mitosis.
It is very essential to understand the significance of this division in the life of an organism.
Mitosis results in the production of diploid daughter cells with identical genetic complement usually.
The growth of multicellular organisms is due to mitosis. Cell growth results in disturbing the ratio between the nucleus and the cytoplasm.
It, therefore, becomes essential for the cell to divide to restore the nucleo- cytoplasmic ratio. A very significant contribution to mitosis is cell repair.
The cells of the upper layer of the epidermis, cells of the lining of the gut, and blood cells are being constantly replaced.
Mitotic divisions in the meristematic tissues – the apical and the lateral cambium, result in the continuous growth of plants throughout their life.

Question 3.
Describe meiosis II with the help of suitable diagrams.
Solution:
Meiosis II is divided into four phases.

Prophase II. Meiosis II is initiated immediately after cytokinesis, usually, before the chromosomes have fully elongated. In contrast to meiosis I, meiosis II resembles a normal mitosis. The nuclear membrane disappears by the end of prophase II. The chromosomes again become compact.

Metaphase II- At this stage, the chromosomes align at the equator and the microtubules from opposite poles of the spindle get attached to the kinetochores of sister chromatids. Anaphase II- It begins with the simultaneous splitting of the centromere of each chromosome (which was holding the sister chromatids together), allowing them to move toward opposite poles of the cell.

Telophase II- Meiosis ends with telophase II, in which the two groups of chromosomes once again get enclosed by a nuclear envelope; cytokinesis follows resulting in the formation of a tetrad of cells i.e., four haploid daughter cells.
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 9

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 10 Cell Cycle and Cell Division, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 10 Cell Cycle and Cell Division, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 English Hornbill Poem 1 A Photograph

June 18, 2021

Here we are providing NCERT Solutions for Class 11 English Hornbill Poem 1 A Photograph. Students can get Class 11 English A Photograph NCERT Solutions, Questions and Answers designed by subject expert teachers.

A Photograph NCERT Solutions for Class 11 English Hornbill Poem 1

A Photograph NCERT Text Book Questions and Answers

Question 1.
What does the word ‘cardboard’ denote in the poem? Why has this word been used?
Answer:
The word “cardboard” in the poem refers to the photograph. It is just a piece of paper a memory as the poet’s mother is dead and gone. It signifies the transience of life and what remains is an insignificant piece of paper.

Question 2.
What has the camera captured?
Answer:
The camera has captured the scene when the poet’s mother and her two female cousins, Betty and Dolly, went paddling. They were holding the poet’s mother’s hands. The elder of the three was about twelve years. The three of them stood smiling as the wind tousled their hair.

Question 3.
What has not changed over the years? Does this suggest something to you?
Answer:
The sea has not changed over the years. It remains the same through generations. On the contrary, life is transient. The mother has now been dead for years. Human life is transitory and this transience is contrasted with the permanence of nature.

Question 4.
The poet’s mother laughed at the snapshot. What did this laugh indicate?
Answer:
The mother laughed at the fleeting moments that had long passed. She relived the memories when they were dressed as children and taken out. She laughed as she recalled the happy memories.

Question 5.
What is the meaning of the line “Both wry with the laboured ease of loss.”
Answer:
The poet’s mother had been out on a beach holiday, years back and felt nostalgic about it, similar to what the poet felt when she relived the memories of her dead mother. The memories, in each case, were beautiful, but painful to recall as time slipped away, so easily.

Question 6.
What does “this circumstance” refer to?
Answer:
“This circumstance” is the death of the mother. This fact is as true and as real as the one that her mother had experienced, on the beach. Both the situations are now a memory of the past. The first is a memory of the mother’s past and the second of the poet’s past.

Question 7.
The three stanzas depict three different phases. What are they?
Answer:
The first stanza is the poet’s description of the photograph that had been captured from her mother’s childhood. The second stanza deals with recollections. The mother’s recollection of her childhood just as the poet recalls her mother who is now dead. The third stanza philosophises death and the transience of life.

NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals

June 18, 2021

NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals.

Question 1.
Answer in one word or one line.
(i) Give the common name of Periplaneta americana.
(ii) How many spermathecae are found in earthworms?
(iii) What is the position of ovaries in cockroach?
(iv) How many segments are present in the abdomen of the cockroach?
(v) Where do you find malpighian tubules?
Solution:
(i) Cockroach
(ii) Four pairs of spermathecae are found in the 6“ to 9th segments (one pair in each segment).
(iii) Cockroach includes a pair of ovaries, that lies laterally in the 2nd to 6th abdominal segments of the abdomen.
(iv) Ten
(v) At the junction of midgut and hindgut. 100-150 yellow coloured thin filamentous ring is present in earthworm, which is called malpighian tubules.

Question 2.
Answer the following :
(i) What is the function of nephridia?
(ii) How many types of nephridia are found in earthworms based on their location?
Solution:
(i) Nephridia is the excretory organ of the earthworm or pheretima.
(ii) There are three types of nephridia –
(i) Septal nephridia – Present on both the sides of intersegmental septa of segment 15 to the last that open into the intestine.
(ii) Integumentary nephridia – Attached to the lining of the body wall of segment 3 to the last that opens on the body surface.
(iii) Pharyngeal nephridia – Present as three paired tufts in the 4th, 5th, and 6th segments.
These three different types of nephridia are almost similar in structure.

Question 3.
Draw a labelled diagram of the reproductive organs of an earthworm.
Solution:

Question 4.
Draw a labelled diagram of the alimentary canal of a cockroach
Solution:

Question 5.
Distinguish between the followings :
(a) Prostomium and peristomium
(b) Septal nephridium and pharyngeal nephridium
Solution:
(a) The main difference between prostomium and peristomium are :
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 3
(b) The main difference between septal nephridium and pharyngeal nephridium are:

Question 6.
What are the cellular components of blood?
Solution:
Red blood cells and white blood cells are the cellular components of blood.

Question 7.
What are the following and where do you find them in an animal body.
1. Chondrocytes
2. Axons
3. Ciliated epithelium
Solution:
1. Chondriocytes: The intercellular material of cartilage is solid and pliable and resists compression. Cells of this tissue (Chondriocytes) are enclosed in small cavities within the matrix secreted by them. Most of the cartilages in vertebrate embryos are replaced by bones in adults. Cartilage is present in the tip of the nose, outer ear joints, between adjacent bones of the. vertebral column, limbs, and hands in adults.

2. Axons: It is found in the nervous system neuron. It is a long fiber, the distal end of which is branched. The main function of axons is the transmission of impulses by means of neurotransmitters.

3. Ciliated epithelium: If the columnar or cuboidal cells bear cilia on their free surface they are called the ciliated epithelium. They are found in the lining of the stomach and intestine and help in secretion and absorption, their function is to move particles or mucus in a specific direction over the epithelium. They are mainly present in the inner surface of hollow organs like bronchioles and fallopian tubes.

Question 8.
Describe various types of epithelial tissues with the help of labelled diagrams.
Solution:
Epithelial tissues
Epithelial tissues provide covering to the inner and outer lining of various organs. The cells of epithelial tissues are compactly packed with a little intercellular matrix.
There are two types of epithelial tissues:
(i) Simple epithelium
(ii) Compound epithelium
(i) Simple epithelium :
Composed of a single layer of cells and functions as a lining for body cavities ducts and tubes. It is further divided into three types on the basis of structure modifications-
(a) Squamous epithelium – It is made of a single layer of flattened cells with irregular boundaries.
It is found as a lining for body cavities, ducts, and tubes such as in the walls of blood vessels and air sacs of lungs.
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 5
Functions – It helps in forming a diffusion boundary.
(b) Cuboidal epithelium – It is made of a single, layer of cube-like cells. It is commonly found in the ducts of glands and tubular parts of nephrons in kidneys. Specialized cuboidal cells are capable of producing gametes found in gonads called the germinal epithelium.
Functions- It helps in secretion and absorption and also in moving particles or mucus in a specific direction over the epithelium.
(c) Columnar epithelium – It is composed of a single layer of tall and slender cells. Nuclei are located at the base.
Its free surface may have microvilli.
It is found inlining of the stomach and intestine.
Functions – It helps in secretion and absorption.
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 6

Question 9.
Distinguish between
(a) Simple epithelium and compound epithelium
(b) Cardiac muscle and striated muscle
(c) Dense regular and dense irregular connective tissues
(d) Adipose and blood tissue
(e) Simple gland and compound gland
Solution:
(a) The main difference between simple epithelium and compound epithelium are as following.
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 7
The main difference between Cardiac muscles and striated muscle are as following.

The main difference between dense regular connective tissues and dense irregular connective tissues are as following.

The main difference between adipose tissue and blood tissue are as following.
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 10
The main difference between simple gland and compound gland are as following

Question 10.
Mark the odd one in each series:
(a) Areolar tissue; blood; neuron; tendon
(b) RBC; WBC; platelets; cartilage
(c) Exocrine: endocrine; salivary gland;ligament
(d) Maxilla; mandible; labrum; antennae
(e) Protonema; mesothorax; metathorax; coxa
Solution:
(a) Neuron
(b) Cartilage
(c) Ligament
(d) Antennae
(e) Protonema

Question 11.
Match the terms in column I with those in column II:
Column I Column II
(a) Compound epithelium – (i) Alimentary canal
(b) Compound eye – (ii) Cockroach
(c) Septal nephridia – ( iii) Skin
(d) Open circulatorysystem – (iv) Mosaic vision
(e) Typhlosole – (v) Earthworm
(f) Osteocytes – (vi) Phallomere
(g) Genitalia – (vii) Bone
Solution:
(a) Compound epithelium ( iii) Skin
(b) Compound eye – (iv) Mosaic vision
(c) Septal nephridia – (v) Earthworm
(d) Open circulatory system – (ii) Cockroach
(e) Typhlosole – (i) Alimentarycanal
(f) Osteocytes – (vii) Bone
(g) Genitalia – (vi) Phallomere

Question 12.
Mention briefly the circulatory system of earthworms.
Solution:
Pheretima exhibits a closed type of blood vascular system, consisting of blood vessels capillaries, and heart. Due to the closed circulatory system, blood is confined to the heart and blood vessels. Contractions keep blood circulating in one direction. Smaller blood vessels supply the gut, nerve cord, and body wall. Blood glands are present on the 4th, 5th, and 6th segments. They produce blood cells and hemoglobin which is dissolved in blood plasma. Blood cells are phagocytic in nature. Earthworms lack specialized breathing devices. Respiratory exchange occurs through moist body surfaces into their bloodstream.

Question 13.
Draw a neat diagram of the digestive system of the frog.
Solution:
The digestive system of frog:

Question 14.
Mention the function of the following:
(a) Ureters in frog
(b) Malpighian tubules
(c) Body wall in earthworm
Solution:
Functions:
(1) Ureters in frog:

They carry the urine from the kidneys to the cloaca.
In males, it also conducts the sperm as it is the urinogenital duct.
In females, the ureters and oviduct open- separately in the cloaca.

(2) Malpighian tubules:

They are the excretory organs of a cockroach.
They collect the nitrogenous wastes from the haeomolymph and send them into the intestine.
Each tubule is lined by glandular and ciliated cells. They absorb nitrogenous waste products and convert them into uric acid which is excreted out through the hindgut.

(3) Body wall of earthworm:
The body wall of the earthworm is covered externally by a thin non-cellular cuticle

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the protein found in white fibres.
Solution:
Collagen.

Question 2.
State the function of setae.
Solution:
It helps in locomotion by gripping the earth.

Question 3.
What is the functional unit of the cockroach eye?
Solution:
Ommatidium.

Question 4.
Name the unit of neural or nervous system.
Solution:
Neurons.

Question 5.
What is worm casting ?
Solution:
It is the insoluble and undigested food that is given out along with soil through anus.

Question 6.
Which cell covers the exposed surfaces of the body (skin) and internal passage ways (digestive tract and glands) ?
Solution:
Squamous epithelium

Question 7.
Name the type of epithelium that lines the inner surface of stomach.
Solution:
Columnar epithelium

Question 8.
Name the type of epithelium that lines the buccal cavity.
Solution:
Stratified squamous epithelium.

Question 9.
N ame of type of tissue that is the most abundant in animal body.
Solution:
Connective tissue.

Question 10.
What is the other name given to the gizzard of cockroach?
Solution:
Proventriculus.

Question 11.
Name the larva of a frog.
Solution:
Tadpole.

Question 12.
What is the scientific name of Indian (bull) frog?
Solution:
Rana tigrina.

SHORT ANSWER QUESTIONS

Question 1.
What are the neuroglia cells?
Solution:
Cells which holds the neuron together are known as neuroglia cells.

Question 2.
What is vermicompositing?
Solution:
The process of increasing soil fertility by earthworms is known as vermicomposting.

Question 3.
What are exocrine glands? Name any two secretions of them.
Solution:
Exocrine glands : Those glands which have ducts to pour their secretion(s) into the respective site of action, are called exocrine.
(i) Salivary glands secrete saliva into the buccal cavity.
(ii) Liver secretes bile into the duodenum.

Question 4.
Write four functions of bones.
Solution:
Functions of bones:
(i) They provide place for attachment of muscles and help in movement and locomotion.
(ii) Bone marrow is the site of manufacture of blood cells.
(iii) Bones provide protection to the internal organs.
(iv) The long bones of the limbs serve the weight-bearing function.
(v) They act as the depot of calcium and phosphorus.

Question 5.
Describe the three types of cell junctions present in the epithelium and other tissues.
Solution:
Cell junctions
(i) Tight junctions – They check leaking of substances across a tissue.
(ii) Adhering junctions – They help in cementing the neighbouring cells together.
(iii) Gap junctions – They facilitate the cells to communicate with each other by cytoplasmic connections, for rapid transfer of ions, small molecules, etc.

Question 6.
How is the gizzard in the alimentary canal of a cockroach suitable for grinding the food?
Solution:
Gizzard of cockroach has following characteristics:
(i) The gizzard has an outer layer of thick circular muscles.
(ii) The inner thick layer of cuticle forms six plate like teeth.
(iii) The movement with the help of muscles and the teeth-like structures help in grinding the food.

Question 7.
What is a typhlosole in an earthworm? Where is it found? What is its function?
Solution:
Typhlosole: It is an internal median fold of the dorsal wall of the intestine. It is found in the intestine between 26th and 35th segments of the body.
It increases the effective area of absorption.

LONG ANSWER QUESTIONS

Question 1.
Describe the female reproductive organs of frog.
Solution:
The female reproductive organs include a pair of ovaries. The ovaries are situated near kidneys and there is no functional connection with kidneys. A pair of oviduct arising from the ovaries opens into the cloaca separately. A mature female can lay 2500 to 3000 ova at a time.
Fertilisation is external and takes place in water. Development involves a larval stage called tadpole. Tadpole undergoes metamorphosis to form the adult.

NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 13

Question 2.
Describe with examples, various types of connective tissues.
Solution:
Connective tissues are most abundant and widely distributed in the body of complex animals. They are named connective tissues because of their special function of linking and supporting other tissues/C ••gans of the body. Connective tissues are classified into three types:
(i) Loose connective tissue,
(ii) Dense connective tissue and
(iii) Specialised connective tissue.
(i) Loose connective tissue:

It has cells and fibres loosely arranged in a semi-fluid ground substance, for example, areolar tissue present beneath the skin.
Often it serves as a support framework for epithelium. It contains fibroblasts (cells that produce and secrete fibres), macrophages and mast cells.
Adipose tissue is another type of loose connective tissue located mainly beneath the skin. The cells of this tissue are specialised to store fats.

(ii) Dense connective tissue :

Fibres and fibroblasts are compactly packed in the dense connective tissues.
Orientation of fibres show a regular or irregular pattern and are called dense regular and dense irregular tissues.
In the dense regular connective tissues, the collagen fibres are present in rows between many parallel bundles of fibres.
Tendons, which attach skeletal muscles to bones and ligaments which attach one bone to another are examples of this tissue.
Dense irregular connective tissue has fibroblasts and many fibres (mostly collagen) that are oriented differently. This tissue is present in the skin.

(iii) Specialised connective tissue :

Cartilage, bones and blood are various types of specialised connective tissues.
The intercellular material of cartilage is solid and pliable and resists compression.
Cells of this tissue (chondrocytes) are enclosed in small cavities within the matrix secreted by them.
Cartilage is present in the tip of nose, outer ear joints, between adjacent bones of the vertebral column, limbs and hands in adults.

Bones : It has a hard and non-pliable ground substance rich in calcium salts and collagen fibres which give bone its strength. It is the main tissue that provides structural frame to the body.
Blood : It is a fluid connective tissue containing plasma, red blood cells (RBC), white bloo4 cells (WBC) and platelets. It is the main circulating fluid that helps in the transport of various substances.

Question 3.
Describe the alimentary canal of earthworm.
Solution:
The alimentary canal is a straight tube and mns between first to last segment of the boay. It has following parts.
Mouth : A terminal mouth opens into the buccal cavity (1-3 segments) which leads into muscular pharynx.
oesophagus: A small narrow tube, oesophagus (5-7 segments), continues into a muscular gizzard (8-9 segments). It helps in grinding the soil particles and decaying leaves, etc.
Stomach : The stomach extends from 9-14 segments. The food of the earthworm is decaying leaves and organic matter mixed with soil. Calciferous glands, present in the stomach, neutralise the humic acid present in humus.
• Intestine starts from the 15th segment onwards and continues till the last segment. A pair of short and conical intestinal caecae project from the intestine on the 26th segment.
Typhosole : The characteristic feature of the intestine between 26-35 segments is the presence of internal median fold of dorsal wall called typhlosole. This increases the effective area of absorption in the intestine.
Anus : The alimentary canal opens to the exterior by a small rounded aperture called anus. The ingested organic rich soil passes through the digestive tract where digestive enzymes breakdown complex food into smaller absorbable units. These simpler molecules are absorbed through intestinal membranes and are utilised.

Question 4.
Draw a labelled diagram of external features of cockroach.
Solution:
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 14

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NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

June 18, 2021

NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants.

Question 1.
“All elements that are present in a plant need not be essential to its survival”. Comment.
Solution:
Plants obtain their inorganic nutrients from the air, water, and soil. Plants absorb a wide variety of mineral elements. Not all the mineral elements that they absorb are required by plants. Out of the more than 105 elements discovered so far, less than 21 are essential and beneficial for normal plant growth and development. The elements required in large quantities are called macronutrients. While those required in fewer quantities or in the trace are termed micronutrients. These elements are either essential constituents of proteins, carbohydrates, fats, nucleic acid, etc. and/or take part in various metabolic processes.

Question 2.
Why is the purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics?
Solution:
The technique of growing plants in a nutrient solution is known as hydroponics. Since a number of improvised methods have been employed to try and determine the mineral nutrients essential for plants. The essence on all these methods involves the culture of plants in a soil-free, defined mineral solution. These method require purified water and mineral nutrients salts. Purification of water and nutrient salt is important to find out other influencing factors

Question 3.
Explain with examples:
Macronutrients, micronutrients, beneficial nutrients, toxic elements, and essential elements.
Solution:
Based upon the criteria only a few elements have been found to be absolutely essential for plant growth and metabolism. These elements are further divided into two broad categories based on their quantitative requirements,

Macronutrients
Micronutrients

Macronutrients must generally be present in plant tissues in the concentration of 1 to 10 mg/L of dry matter. The macronutrients include carbon, hydrogen, oxygen, nitrogen, phosphorous, sulfur, potassium, calcium, and magnesium. Of these, carbon, hydrogen, and oxygen are mainly obtained from CO₂, and H₂0, while the others are absorbed from the soil as mineral nutrition.

Micronutrients or trace elements are needed in very small amounts (equal to or less than 0.1 mg/L of dry matter). These include iron, manganese, copper, molybdenum, zinc, boron, chlorine, and nickel. In addition to the 17 essential elements named above, there are some beneficial elements such as sodium, silicon, cobalt, and selenium. They are required by higher plants.

Essential elements can also be grouped into four broad categories on the basis of their diverse functions. These categories are:
(1) Essentia] elements as components of biomolecules and hence structural elements of cells, (eg: carbon, hydrogen, oxygen, and nitrogen).

(2) Essential elements that are components of energy-related chemical compounds in plants, for example, magnesium in chlorophyll and phosphorous in ATP.

(3) Essential elements that activate or inhibit enzymes, for example, Mg²⁺ is an activator for both ribulose bisphosphate carboxylase-oxygenase and phosphoenolpyruvate carboxylase, both of which are critical enzymes in photosynthetic carbon fixation; Zn²⁺ is an activator of alcohol dehydrogenase and Mo of nitrogenase during nitrogen metabolism.

(4) Some essential elements can alter the osmotic potential of a cell. Potassium plays an important role in the opening and closing of stomata. Any mineral ion concentration in tissues that reduces the dry weight of tissues by about 10 percent is considered toxic. Such critical concentrations vary widely among different micronutrients. The toxicity symptom! are difficult to identify.

Toxicity levels for any element also vary for different plants. Many times excess of an element may inhibit the uptake of another element. For example, the prominent symptoms of manganese toxicity are the appearance of brown spots surrounded by chlorotic veins.

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

Chlorosis: Chlorosis is the loss of chlorophyll leading to yellowing in leaves. It is caused by a deficiency of N, K, Mg, S, Fe, Mn, Zn, and Mo.
Necrosis: It is the death of tissue. It occurs due to deficiency of Ca, Mg, Cu, K.
Inhibition of cell division: It occurs due to deficiency of N, K, S, Mo.
Stunted plant growth: It occurs due to deficiency of Ca, N, etc.
Premature fall of leaf and buds: It occurs due to deficiency of calcium, magnesium.

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?
Solution:

The deficiency symptoms can be distinguished on the basis pf the region of occurrence, presence or absence of dead spots, and chlorosis of entire leaf or interveinal chlorosis.
The region of the appearance of deficiency symptoms depends on the mobility of nutrients in plants. The nutrient deficiency symptoms of N, P, K, Mg, and Mo appear in lower leaves.
Zinc is moderately mobile in plants and deficiency symptoms, therefore, appear in middle leaves.
The deficiency symptoms of less mobile elements (S, Fe, Mn, and Cu) appear on new leaves.
Ca and B are immobile in plants, deficiency symptoms appear on terminal buds.
Chlorine deficiency is less common in crops.

Question 6.
Why is it that in certain plants deficiency symptoms appear first in younger parts of the plant while in other they do so in mature organs?
Solution:
The deficiency symptoms tend to appear first in the young tissues whenever the elements are relatively immobile and are not transported out of the mature organs, for example, elements like sulphur and calcium are a part of the structural component of the cell and hence are not easily released.

Question 7.
How are the minerals absorbed by the plants?
Solution:
Uptake of mineral ions, by plants, occurs through
two main phases.
Passive Absorption: It is the process of absorption of minerals through it’s outer space(Intercellular space and cell wall) by physical process. Direct expenditure of metabolic energy is not involved. A substance moves from a region of higher chemical potential to lower chemical potential. It occurs through ion channels (transmembrane protein). The theories to explain the movement of ions:
(a) Ion exchange: Both cation and anion gets absorbed on the surface of cell wall. The absorbed ions are exchanged with ions present in soil solution.
(b) Mass flow hypothesis: According to this hypothesis mass flow of ions occur along with absorption of water as a result of transpirational pull.
Active Absorption: It is the process of movement of ions against a concentration gradient, by utilizing ATP as energy. Both influx and efflux of ions are carried out by carrier mechanism. The activated ions combine with carrier proteins and form ion carrier complex. This complex moves
across all the membrane and reaches inner surface, where it breaks and releases ions into the cytoplasm.

Question 8.
What are the conditions necessary for the fixation of atmospheric nitrogen by Rhizobium? What is their role in nitrogen fixation?
Solution:
Rhizobia are unique because they live in a symbiotic relationship with legumes. Necessary conditions:

Requires a strong reducing agent and energy in the form of ATP.
The enzyme nitrogenase which is very sensitive to oxygen is required.
The processes take place in an anaerobic environment
The energy is provided by the respiration of host cells.

The reduction of nitrogen to ammonia by living organisms is called biological nitrogen fixation. The enzyme, nitrogenase which is capable of nitrogen reduction is present exclusively in prokaryotes. Several types of symbiotic biological nitrogen-fixing associations are known. The most common association on roots is nodules. Their role in N₂– fixation is to supply the plants with nitrogenase that converts nitrogen to amino acids.

Question 9.
What are the steps involved in the formation of root nodule?
Solution:
Nodule formation involves a sequence of multiple interactions between Rhizobium and the roots of the host plant. Stages in the nodule formation are summarised as follows:
Steps in the development of root nodules:
(a) When a root hair of a leguminous plant comes in contact with Rhizobium, it is deformed due to the secretion from the bacterium.
(b) At the site of curling Rhizobia invades the root tissue and proliferate within root hairs.
(c) Some bacteria enlarge to form membrane-bound structures, bacteroids which cannot divide.
(d) The plants form the infection thread, made up of plasma membrane that grows inward, separating the infected tissue from the rest of the plant.
(e) Cell division is stimulated in the infected tissue and more bacteria invade the newly formed tissues.
(f) It is believed that a combination of cytokinin produced by invading bacteria and auxins produced by plant cells, promotes cell division and extension, leading to nodule formation.
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 2

Question 10.
Which of the following statements are true? If false correct them:
(a) Boron deficiency leads to the 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 plants.
(d) It is very easy to establish the essentiality of micronutrients because they are required only in trace quantities.
Solution:
(a) True
(b) False
Correct sentence: Every mineral element that is present in a cell is not needed by the cell.
(c) False
Correct sentence: Nitrogen as a nutrient element is highly mobile in the plants.
(d) False
Correct sentence: It is very difficult to establish the essentiality of micronutrients because they are required only in trace quantities.

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the enzymes that reduce nitrogen in the root nodules of a bean plant.
Solution:
Nitrogenase.

Question 2.
Name the enzymes used in biologically nitrogen fixation. What are the mineral elements needed for the activity of the enzyme?
Solution:
Nitrogenase enzyme.

Question 3.
Name the enzyme that can reduce nitrogen to ammonia.
Solution:
Nitrogenase.

Question 4.
What is the importance of phosphorus for plants?
Solution:
Phosphorus is a constituent of cell membranes, certain proteins, all nucleic acids, and nucleotides, and is required for all phosphorylation reactions.

Question 5.
Name two crops that are commonly produced by hydroponics.
Solution:
Tomato, Lettuce

Question 6.
Name the group of enzymes activated by zinc.
Solution:
Carboxylases.

Question 7.
Define critical concentration of elements with reference to plant nutrition.
Solution:
Critical concentration refers to the concentration of the essential element, below which the plant growth is retarded.

Question 8.
Name the element which is a limiting nutrient for both natural and agricultural ecosystems.
Solution:
Nitrogen.

Question 9.
Name two bacteria that oxidise ammonia into nitrite.
Solution:
Nitrosomonas, Nitrococcus

Question 10.
Name two symbiotic nitrogen-fixing bacteria.
Solution:
Rhizobium, Frankia.

Question 11.
What is hydroponics (Tank farming)?
Solution:
It is plant growth in the liquid culture medium.

Question 12.
What are the framework elements of a plant?
Solution:
Carbon, hydrogen, and oxygen are called framework elements.

Question 13.
What type of condition is created by leghaemoglobin in the root nodules of legumes?
Solution:
Anaerobic condition.

Question 14.
What is meant by active absorption?
Solution:
Active absorption: The uptake of mineral ions against the concentration gradient is called active absorption.

Question 15.
How are amides transported In plants?
Solution:
Amides are transported along with water through the xylem.

Question 16.
What is the function of the enzyme nitrite reductase?
Solution:
It reduces nitrate ions to ammonia.

Question 17.
Name two free-living micro-organisms which can fix nitrogen.
Solution:
Azotobacter, Beijemickia.

SHORT ANSWER QUESTIONS

Question 1.
What is nitrification?
Solution:
Questions It is the conversion of ammonium ion to nitrite and then to nitrate. Nitrosomonas converts ammonium into nitrites, Nitrobacter converts nitrites into nitrates.

Question 2.
Prior to sowing rice, a legume crop was cultivated and ploughed back in this field. Why? Explain.
Solution:
Leguminous plants possess root nodules in which the symbiotic bacteria Rhizobium fixes nitrogen. The fixed nitrogen makes the soil rich in nitrogen fertilizer where the leguminous plant is ploughed back into the field.

Question 3.
Name the organism that fixes nitrogen in symbiotic association with a legume. Where does it live in such plants?
Solution:
Rhizobium. It lives in the root nodules of leguminous plants.

Question 4.
Bring out at similarity and difference between leghaemoglobin and hemoglobin.
Solution:
Both leghaemoglobin and hemoglobin are iron-containing molecules but leghaemoglobin is present in the root nodules in the plants belonging to the family Fabaceae while hemoglobin in human blood pigment.

Question 5.
Bring out at similarity and difference between leghaemoglobin and hemoglobin.
Solution:

Question 6.
In what form is boron absorbed by plants from the soil? Mention its two uses in plants and give two deficiency symptoms of boron in them.
Solution:
Boron is absorbed as \({ H }_{ 2 }{ Po }_{ 4- }\quad and\quad { HPo }_{ 4 }^{ 2- }\)

Translocation of carbohydrates
Pollen germination

Deficiency symptoms

Death of root and shoot tips
Abscission of flowers

Question 7.
List the macronutrients and mention three major function, (any three)
Solution:
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 4

Question 8.
In what form is boron absorbed by plants from the soil? Mention its two uses in the plants and give two deficiency symptoms of boron in them.
Solution:

(i) Translocation of carbohydrates.
(ii) Pollen germination.
(iii) Absorption and utilisation of calcium.
(iv) Cell elongation and differentiation.

Question 9.
Distinguish between micronutrients and macronutrients.
Solution:
Differences between micronutrients and macronutrients are as following:
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 5

LONG ANSWER QUESTIONS

Question 1.
With the help of a suitable diagram describe the nitrogen cycle.
Solution:
Nitrogen cycle: Plants compete with microbes for the limited nitrogen that is available in the soil. Thus, nitrogen is a limiting nutrient for both natural and agricultural ecosystems.

In nature, lightning and ultraviolet radiation provide enough energy to convert nitrogen to nitrogen oxides (NO, NO2, N2O).
Industrial combustions, forest fires, automobile exhausts, and power-generating stations are also sources of atmospheric nitrogen oxides.
The decomposition of organic nitrogen of dead plants and animals into ammonia is called ammonification. Some of this ammonia

volatilises and re-enters the atmosphere but most of it is converted into nitrate by soil bacteria in the following steps
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 6

Question 2.
Name at least five different deficiency symptoms in plants. Describe them and correlate them with concerned mineral deficiency.
Solution:

Chlorosis – yellowing of leaves due to loss of chlorophyll caused by the deficiency of N, S, mg, Fe.
Necrosis – Death of tissues, especially in leaves caused by the deficiency of Ca, Mg, Ca & K.
Delay in flowering caused by the deficiency of molybdenum, nitrogen, and sulphur.
Dieback of roots caused by a deficiency of copper.
Inhibition of cell division caused by a deficiency of potassium, calcium and nitrogen.

Question 3.
Write notes on :
(a) Reductive animation
(b) Transamination
Solution:
(a) Reduction animation – In this process ammonia (formed by nitrogen assimilation) reacts with a ketoglutaric acid to form the amino acid – glutamic acid. Here ∝ – ketoglutaric acid comes from Kreb’s cycle and hydrogen is donated by co-enzyme NADH or NADPH.
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 8

(b) Transamination – Once the glutamic acid is synthesized by reductive amination, other amino acids are synthesized by the transfer of an amino groups to other carbon skeletons. Glutamic acid is the starting material from which 17 other amino acids are formed by the transfer of the amino group of an amino donor compound to the carboxyl position of an amino acceptor compound. Transaminase is the enzyme responsible for such a reaction.

Question 4.
Write an account of the role of mineral elements in a plant.
Solution:
Role of the mineral elements: Plants require mineral elements for various metabolic activities of their body. The following are some of the important functions which the mineral elements perform:

(i) Constituents of the plant body: Elements form the constitution of the plant body. For example carbon, hydrogen and oxygen are essential for the production of carbohydrates hence they are termed framework elements. Nitrogen, sulphur, and phosphorus are required for the synthesis of proteins. Magnesium is an important part of the chlorophyll molecule.

(ii) Influence on the pH of the cell sap: They also influence the pH of the cell sap.

(iii) Maintenance of osmotic pressure in the plant cells: The mineral salts and organic compounds of the cell sap produce necessary osmotic pressure.

(iv) They influence the permeability of cytoplasmic membrane: Different minerals decrease or increase the permeability of plasma membrane.

(v) They have balancing function reactions: Some of the minerals balance the effects of the other.

(vi) They take part in enzymatic reactions: Some elements work as activators while others work as inhibitors in various enzymatic reactions.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 12 Mineral Nutrition, helps you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 12 Mineral Nutrition, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis

June 18, 2021

NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis.

Question 1.
By looking at a plant externally can you tell whether a plant is C₃ or C₄ ? Why and how?
Solution:
Plants that are adapted to dry tropical regions have the C₄ pathway. They have a special type of leaf anatomy, they tolerate higher temperatures, they show a response to highlight intensities. Study vertical sections of leaves, one of a C₃ plant and the other of a C₄ plant.

Question 2.
By looking at which internal structure of a plant can you tell whether a plant is C₃ or C₄? Explain.
Solution:
In C₄ plant internal structure of the leaf possess a special type of anatomy called ‘Kranz’ anatomy. ‘Kranz’ means ‘wreath’ and is a reflection of the arrangement of cells.

The bundle sheath cells may form several layers around the vascular bundles; they are characterised by having large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces.

While in C₃ plants, there is no special type of leaf anatomy. There is only a single type of chloroplast inC₃ i.e. granal, while in C₄ chloroplasts are dimorphic, i.e, granite in the mesophyll cells and agranal in the bundle sheath cells.

Question 3.
Even though very few cells in a C₄ plant carry out the biosynthetic-Calvin pathway, yet they are highly productive, can you discuss why?
Solution:
Though these plants have the C₄ oxalacetic acid as the first CO₂ fixation product they use the C₃ pathway or the Calvin cycle as the main biosynthetic pathway.
In C₄ plants photorespiration does not occur. This is because they have a mechanism that increases the concentration of CO₂ at the enzyme site.
This takes place when the C₄ acid from the mesophyll is broken down in the bundle cells to release CO₂ this results in increasing the intracellular concentration of CO₂ In turn, this ensures that the Rubisco functions as a carboxylase minimizing the oxygenase activity.

Now that you know that the C₄ plants lack photorespiration, you probably can understand why productivity and yields are better in these plants. In addition, these plants show tolerance to higher temperatures.

Question 4.
RuBisCO is an enzyme that acts both as carboxylase and oxygenase. Why do you think RuBisCO carries out more carboxylation in C₄ plants.
Solution:
RuBisCO or Ribulose bisphosphate carboxylase – oxygenase enzyme can bind to both C0₂ and O₂. This binding is competitive. The relative concentration of C0₂ and 0₂ determines which one of the two will bind to the enzyme.

In C₄ plants photorespiration does not occur. This is because they have a mechanism that increases the concentration of C0₂ at the enzyme site.

This takes place when oxaloacetic acid is broken down in the bundle sheath cells to release C0₂.

It results in increased intracellular concentration of C0₂. This ensures that the RuBisCO functions as a carboxylase and minimising the oxygenase activity.

Question 5.
Suppose there were plants that had a high concentration of chlorophyll b, but lacked chlorophyll a, would it carry out photosynthesis? Then why do plants have chlorophyll b and other accessory pigments?
Solution:
Though chlorophyll is the major pigment responsible for trapping light, other thylakoid pigments like chlorophyll b, xanthophylls, and carotenoids, which are called accessory pigments, also absorb light and transfer the energy to ‘chlorophyll a’.

Indeed, they not only enable a wider range of wavelengths of incoming light to be utilized for photosynthesis but also protect ‘chlorophyll a’ from photo-oxidation. Reaction centre chlorophyll-protein complexes are capable of directly absorbing light and performing charge separation events without other chlorophyll pigments but the absorption cross-section is small.

Question 6.
Why is the colour of a leaf kept in the dark frequently yellow, or pale green? Which pigment do you think is more stable?
Solution:
Chlorophyll is unable to absorb energy in the absence of light and loses its stability, giving the leaf a yellowish colour. This shows that xanthophyll is more stable.

Question 7.
Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?
Solution:
Light is a limiting factor for photosynthesis Leaves get lesser light for photosynthesis when they are in shade. Therefore, the leaves or plants in shade perform lesser photosynthesis as compared to the leaves or plants kept in sunlight. In order to increase the rate of photosynthesis, the leaves present in shade have more chlorophyll pigments.

This increase in chlorophyll content increases the amount of light absorbed by the leaves, which in turn increases the rate of photosynthesis. Therefore, the leaves or plants in shade are greener than the leaves or plants kept in the sun.

Question 8.
The figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions.
(a) At which point/s (A, B, or C) in the curve is light a limiting factor?
(b) What could be the Jimiting factor/s in region A?
(c) What do C and D represent on the curve?
Solution:

(a) In the region ‘A’ and half of ‘BTight is limiting factor because rate of photosynthesis is increasing with the intensity of light.
(b) All the other factors except light.
(c) C represents a region where a factor other than light is limiting, e.g., CO₂. D represents the light intensity at which rate of photosynthesis is maximum under existing conditions (e.g., CO₂).

Question 9.
Give a comparison between the following:
(a) C₃ and C₄ pathways
(b) Cyclic and non-cyclic photophosphorylation
(c) Anatomy of leaf in C₃ and C₄.
Solution:
(a) Differences between C₃ and C₄ pathway
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 2
(b) Differences between cyclic and non-cyclic photophosphorylation are

(c) Differences between the anatomy of leaf in C₃ plants and anatomy of leaf in C₄ plants are

VERY SHORT ANSWER QUESTIONS

Question 1.
Write one anatomical feature of C₄ plants.
Solution:
Kranz anatomy in leaf.

Question 2.
Which of the following is not a useful function of the light reaction in photosynthesis?
(a) splitting water
(b) synthesis of NADPH
(c) converting light energy into chemical energy
(d) releasing oxygen for photorespiration
Solution:
(d) Releasing oxygen for photorespiration.

Question 3.
What is the starting substance in the CO₂ fixation cycle? (Apr. 91)
Solution:
RuMP.

Question 4.
Where is PS II located in a chloroplast?
Solution:
PS II is located in the appressed regions of grana thylakoid

Question 5.
Name the reaction centre of PS I and PS II.
Solution:
P₇₀₀ & P₆₈₀‘

Question 6.
What type of light causes maximum photo-synthesis? (Oct. 1995)
Solution:
Red light

Question 7.
How many molecules of ATP and how many molecules of NADPH are spent to fix three molecules of CO₂ in the Calvin cycle?
Solution:
9 ATP and 6 NADPH

Question 8.
Why do the stomata of CAM plants open during the night?
Solution:
As these plants grow in dry areas, they keep stomata close during the day to conserve water and open their stomata during the night for the diffusion of gases.

Question 9.
Mention one useful role of photorespiration in plants.
Solution:
It protects the plants from photooxidative damage.

Question 10.
Cyanobacteria and some other photosynthetic bacteria don’t have chloroplasts. How do they conduct photosynthesis?
Solution:
Cyanobacteria have bluish pigment phycocyanin, which they use to capture light for photosynthesis. Some green bacteria (cyanobacteria) are red or pink due to pigment phycoerythrin. Whatever the colour of cyanobacteria, they are photosynthetic and so can manufacture food.

Question 11.
What is phosphorylation? (M.Q.P.)
Solution:
Synthesis of ATP either with the help of light (during photosynthesis) or in presence of oxygen (during respiration) is called phosphorylation.

Question 12.
Name the organism Englemann used in his experiment.
Solution:
Cladophora.

Question 13.
Write the currently accepted equation of photosynthesis in plants.
Solution:

Question 14.
What is a pigment?
Solution:
A pigment is a substance that absorbs light of certain wavelength(s).

Question 15.
Write the full form of NADP
Solution:
NADP – Nicotinamide adenine dinucleotide Phosphate.

Question 16.
Expand RuBP
Solution:
Ribulose 1, 5 bisphosphates.

Question 17.
Give a reason for the following:
Some bacteria exhibit photosynthesis but they do not produce oxygen. (July 2006)
Solution:
Some photosynthetic bacteria do not use water as their source of hydrogen, hence do not liberate oxygen.

Question 18.
Mention two conditions where light can become a limiting factor.
Solution:
Conditions in which light can become a limiting factor:
(i) Plants in the shade.
(ii) Plants growing under the canopy in a dense forest.

Question 19.
What are antenna molecules?
Solution:
Antenna molecules are light-harvesting pigment molecules that occur on the outer side of a photosynthetic unit.

Question 20.
What is a quantasome? Where is it present?
Solution:
Quantasome means photosynthetic units. It is equivalent is 230 chlorophyll molecules. These are present in the grana lamellae.

SHORT ANSWER QUESTIONS

Question 1.
Specify how C₄ photosynthetic pathway increases carbon dioxide concentration in bundle sheath cells of sugarcane?
Solution:
In C₄ pathway of sugarcane, C0₂ from atmosphere enters through the stomata in the mesophyll cell and combines with phosphoenol pyruvate to form a 4-C compound oxaloacetic acid. The OAA is then transported to the bundle sheath where it is decarboxylatedto release C0₂ in bundle sheath.

Question 2.
Differentiate between absorption spectrum and action spectrum.
Solution:
The main differences between absorption spectrum and action spectrum are as follows.
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 6

Question 3.
What are quantasomes? (Oct. 94)
Solution:
Quantasome is a functional unit (Photo-synthetic unit) made of a group of pigment molecules required for carrying out a photochemical reaction. The Pigment molecules are embedded in the grana and differentiated as pigment system I (with chi 670, chi 680, P 700) and pigment system ll(with chi 670, chi 680, P 680, and Xanthophylls)

Question 4.
Distinguish between cyclic and non-cyclic photophosphorylation. (M.Q.P., March 2011)
Solution:
Non cyclic photophosphorylation (a) Cyclic photophosphorylation

The path traversed by an electron is non-cyclic.
(a) Path traversed by electron is cyclic.
Both PSI and PSII are active.
(b) Only PSI is active.
It is accompanied by photolysis.
(c) No photolysis.
The major pathway that takes place.
(d) Secondary pathway when additional ATP is needed.

Question 5.
What is Blackman’s law of limiting factors?
Solution:
F.F. Blackman (1905) extended a law to formulate the principle of limiting factors. “When a process is conditioned as to its rapidity by a number of separate factors, the rate of the process is limited by the pace of slowest factors.”

Question 6.
In the condition of water stress why the rate of photosynthesis declines?
Solution:
Due to water stress, stomata remain closed and so there is a decrease in CO₂concentration and the leaf water potential is also reduced, decline the rate of photosynthesis.

Question 7.
What is a reaction centre? Give the reaction centres of PSI and PSII.
Solution:
Reaction centre is a chlorophyll component of the photosystem and it absorbs as well as accepts energy from other pigments and ejects an electron. The reaction centre of PSII is Chla₆₈₀or P₆₈₀ and PSI is Chla₇₀₀ or P₇₀₀

Question 8.
Why is photorespiration considered a wasteful process?
Solution:
Photorespiration considered a wasteful process because
(i) 25% of photosynthetically fixed carbon is lost in the form of C0₂.
(ii) There is no energy-rich useful compound produced during this process.

Question 9.
Give two reasons as to why photosynthesis is important for sustaining life on earth.
Solution:
Photosynthesis is the most important process because;
(i) it is the only natural process by which oxygen is liberated into the atmosphere.
(ii) it is the process by which food is manufactured for all living organisms.

Question 10.
Why does the rate of photosynthesis decrease at higher light intensities? What plays a protective role in such situations?
Solution:
Rate of photosynthesis decreases for two reasons :
(i) Other factors required for photosynthesis become limiting.
(ii) Destruction of chlorophyll by photo-oxidation.
Carotenoids play a protective role by:
(i) absorbing the excess light and
(ii) acting as an antioxidant to detoxify the effect of activated oxygen species.

Question 11.
What is C₄ -pathway? Give an example. (March 2008)
Solution:
CA -pathway is an alternative photosynthetic pathway seen in plants like sugarcane/sorghum/ maize in which the stable compound is oxaloacetate a 4-C compound. It is called the Hatch-slack pathway.

Question 12.
What is kranz anatomy in plants?
Solution:
In Kranz Anatomy vascular bundles are surrounded by a layer of bundle sheath that contains a large number of chloroplasts in mesophyll cells and it is present in C₄ plants e.g, Maize, Sugarcane, etc.

LONG ANSWER QUESTIONS

Question 1.
How is photosystem I different from photosystem II?
Solution:
The main differences between photosystem I and photosystem II are
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 7

Question 2.
Describe the factors that influence the rate of Photosynthesis. (Oct. 1989)
Solution:
The factors that affect photosynthesis may be both internal & External.

Internal factors:

Chlorophyll: it is the light-absorbing pigment and only portions of the plant having chlorophyll can help in photosynthesis.
Protoplasmic factor: young seedlings when transferred from darkness to light show the presence of some factors which is believed to be enzymatic initiates photosynthesis and is called the protoplasmic factor.

External factors:

Light: It is one of the most important factors which affects the process in 3 ways i.e. quantity, quality, and intensity. Quantity of light is otherwise duration and depends upon the photoperiod that is required by the plant quality refers to the wavelength, maximum photosynthesis occurs in red and blue light while minimum in green light. Intensity favours the process and low intensity decreases the rate of photosynthesis. Very high intensity brings about photooxidation of pigments which is called solarization.
CO₂: An increase in CO₂ concentration favours the process provided other factors are not limiting but very high concentrations are toxic and inhibit photosynthesis.
Temperature: Increase in temperature favour photosynthesis but above the optimum range the process decreases due to the denaturation of enzymes.

Question 3.
Explain the process of the biosynthetic phase of photosynthesis occurring in the chloroplasts.
Solution:
The biosynthetic phase of photosynthesis :

It occurs in the stroma of chloroplasts.
These reactions reduce the carbon dioxide into carbohydrates, making use of the ATP and NADPH₂ produced in the photochemical reactions.
The reactions are also called as Calvin cycle.
The three phases of the Calvin cycle are as follows:

(i) Carboxylation
Six molecules of Ribulose 1,5 bisphosphate
react with six molecules of carbon dioxide to form six molecules of a short-lived 6C- compound.
The reaction is catalysed by RuBP carboxylase (RuBisCo).
The six molecules of the 6C-intermediate break into 12 molecules of 3- phosphoglyceric acid (3-PGA), an SC- compound.
It is through this step that carbon dioxide is fixed in the plant.
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 8
(ii) Reduction
12 molecules of 3-phosphoglyceric acid are converted into 12 molecules of 1, 3 diphosphate-glyceric acid, utilising 12 molecules of ATP and then reduced to 3- phosphoglyceraldehyde making use of 12 molecules of NADPH. Two molecules of phosphoglyceraldehyde react to form one molecule of glucose. It is in this step that there is an actual reduction of carbon dioxide leading to sugar formation.

(iii) Regeneration of RuBP
10 molecules of phosphoglyceraldehyde, by a series of complex enzyme-catalyzed reactions, are converted into six molecules of ribulose 1,5-bisphosphate; six molecules of ATP are needed for this step. This step of ‘ regeneration of RuBP is important for the cycle to continue

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, helps you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants

June 18, 2021

NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants.

Question 1.
State the location and function of different types ofmeristems.
Solution:

Question 2.
Cork cambium forms tissues that form the cork. Do you agree with this statement? Explain.
Solution:
Sooner or later, another meristematic tissue called cork cambium or phellogen develops, usually in the cortex region. Cork cambium is a couple of layers thick. It is made of narrow, thin-walled, and nearly rectangular cells. Cork cambium cuts off cells on both sides. The outer cells differentiate into cork or phellem while the inner cells differentiate into secondary cortex or phelloderm. The cork is impervious to water due to suberin deposition in the cell wall. The cells of the secondary cortex are parenchymatous. Phellogen, phellem, and phelloderm are collectively known as periderm.

Question 3.
Explain the process of secondary growth in the stems of woody angiosperms with the help of schematic diagrams. What is its significance?
Solution:

showing secondary growth.
Secondary growth in dicot stem:

It is a “permanent increase in thickness due to the activity of vascular cambium and cork cambium in stellar and extrasolar regions”. In dicot stem intra fascicular cambium is present.
The cells of the medullary ray become meristematic and form interfascicular cambium.
These two cambiums unite and make a complete cambial ring.
The cells of it divide and produce new cells both on its outer and inner sides.
The cells formed on the outer side differentiate into secondary phloem while the cells of the inner side form secondary xylem.
The epidermis is replaced by a secondary protective tissue by an increase in the growth of the stem of the plant. It is made of phellogen (cork cambium).
It arises from the peripheral cells of the cortex. The phellogen forms new cells on the outer side which make phellem (cork) and phelloderm on its inner side also.
Significance: Secondary growth increases the girth or thickness of the plant.
Annual rings of woody angiosperms are very distinct and thus helps in determining the age of the plant.

Question 4.
Draw illustrations to bring out the anatomical difference between
(a) Monocot root and dicot root
(b) Monocot stem and dicot stem
Solution:

NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 4

Question 5.
Cut a transverse section of the young stem of a plant from your school garden and observe it under a microscope. How would you ascertain whether it is a monocot stem or a dicot stem? Give reasons.
Solution:
After observing the transverse section of the stem we can differentiate that stem is monocot or dicot on the basis of the following characters:
NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 5
Question 6.
The transverse section of a plant material shows the following anatomical features –

the vascular bundles are conjoint, scattered, and surrounded by a sclerenchymatous bundle sheath,
phloem parenchyma is absent What will you identify it as?

Solution:
The transverse section of a typical young monocotyledonous stem shows that

The vascular bundles are conjoint, scattered, and surrounded by sclerenchymatous bundle sheaths
Phloem parenchyma is absent, and water containing cavities are present within the vascular bundles.

Question 7.
Why xylem and phloem are called complex tissues?
Solution:
Xylem and phloem a composed of several types of cells and they work as a unit. Hence they are called complex tissues.

Question 8.
What is the stomatal apparatus? Explain the structure of stomata with a labelled diagram.
Solution:

Several minute openings or stomata are found on the epidermis of all the green aerial parts of plants but are abundant on the lower surface on the leaves as they regulate the process of transpiration.
A large number of stomata occur on the upper surface of leaves of aquatic plants.
Each stomata is surrounded by two cells known as the guard cells. In the dicotyledons plants these are bean-shaped, but in sedges and grasses these are dumb-bell-shaped.
The guard cell is living. Their outer walls are thin where as the inner ones surrounding the aperture are highly thickened.
Due to this variation in the thickening, the guard cell may become turgid and flaccid, depending upon the supply of water in them, which makes the opening and closing of stomata possible.
Some times a few neighbouring epidermal cells in the vicinity of guard cells become specialized in their shape and size and contents. These are known as subsidiary cells.
The stomatal aperture, guard cells and the surrounding subsidiary cell are together called stomatal apparatus.

Question 9.
Name the three basic tissue systems in the flowering plants. Give the tissue names under each system.
Solution:
On the basis of their structure and location, there are three types of tissue systems. These are the

Epidermal tissue system,
The ground or fundamental tissue system and
The vascular or conducting tissue system.

1. Epidermal tissue system The epidermal tissue system forms the outer-most covering of the whole plant body and comprises epidermal cells, stomata, and the epidermal appendages the trichomes, and hairs.
2. All tissues except epidermis and vascular bundles constitute the ground tissue. It consists of simple tissues such as parenchyma, collenchyma, and sclerenchyma.
3. The vascular system consists of complex tissues, the phloem, and the xylem. The xylem and phloem together constitute vascular bundles.

Question 10.
How is the plant anatomy useful to us?
Solution:
‘The study of plant anatomy is useful in many ways. First of all the study helps us understand the way a plant functions carrying out its routine activities like transpiration, photosynthesis, and growth and repair. Second, it helps botanists and agriculture scientists to understand the disease and cure for plants. Plants are important to maintain the ecological balance of the earth, so understanding plant anatomy is a way to understand the large system of the ecology on this planet.

Question 11.
What is periderm? How does periderm formation take place in the dicot stems?
Solution:
Phellogen, phellem, and phelloderm are collectively known as periderm. Phellogen develops, usually in the cortex region. Phellogen is a couple of layers thick. It is made of narrow, thin-walled, and nearly rectangular cells. Phellogen cuts off cells on both sides. The outer cells differentiate into cork or phellem while the inner cells differentiate into secondary cortex or phelloderm. The cork is impervious to water due to suberin deposition in the cell wall. The cells of the secondary cortex are parenchymatous.

Question 12.
Describe the internal structure of a dorsiventral leaf with the help of labelled diagrams.
Solution:
Dorsiventral (dicotyledonous) leaf: The vertical section of a dorsiventral leaf through the lamina shows three main parts, namely, epidermis, mesophyll, and vascular system.
Epidermis: The epidermis which covers both the upper surface (adaxial epidermis) and lower surface (abaxial epidermis) of the leaf has a conspicuous cuticle. The abaxial epidermis generally bears more stomata than the adaxial epidermis. The latter may even lack stomata.
Mesophyll:

The tissue between the upper and the lower epidermis is called the mesophyll.
It possesses chloroplasts and carries out photosynthesis, is made up of parenchyma.
It has two types of cells – the palisade parenchyma and the spongy parenchyma.
The adaxially placed palisade parenchyma is made up of elongated cells, which are arranged vertically and parallel to each other.
The oval or round and loosely arranged spongy parenchyma is situated below the palisade cells and extends to the lower epidermis.

6. There are numerous large spaces and air cavities between these cells.

Vascular system:

This includes vascular bundles, which can be seen in the veins and the midrib.
The size of the vascular bundles is dependent on the size of the veins.
The veins vary in thickness in the reticulate venation of the dicot leaves. The vascular bundles are surrounded by a layer of thick-walled bundle sheath cells.

VERY SHORT ANSWER QUESTIONS

Question 1.
Vascular bundles having cambium are known as.
Solution:
Open, Vascular bundle

Question 2.
Name the two types of sclerenchyma.
Solution:
Sclerenchyma fibers and stone cells.

Question 3.
From where do the secondary meristems originate?
Solution:
Permanent tissue.

Question 4.
What does make the root apical meristem subterminal?
Solution:
The presence of the root cap makes the root apical meristem subterminal.

Question 5.
Where are companion cells located in flowering plants? What are their functions?
Solution:
Companion cells are located in phloem cells of vascular tissues, they support the sieve tubes in water conduction.

Question 6.
What is the advantage of lignocellulose in the wall of the xylem?
Solution:
It provides rigidity, thickness, and resistance

Question 7.
A cross-section of a plant material shows the following features under the microscope: vascular bundles are radially arranged. These are found xylem strands showing exarch condition. What type of plant part of is this?
Solution:
Dicot root.

Question 8.
Based on position, classify various types of meristems
Solution:
Apical, intercalary and lateral meristems.

Question 9.
Name the various component cells of xylem. Which of them does not have a nucleus?
Solution:
Tracheids, vessels, xylem parenchyma andxylem fibres. Only xylem parenchyma have nucleus and living.

Question 10.
Give an example of a secondary meristem.
Solution:
Examples of secondary meristem are cork cambium and interfascicular cambium.

Question 11.
Name the tissue involved in linear and lateral growth in plants.
Solution:
Linear growth is caused by apical meristem and lateral growth is caused by lateral meristem.

Question 12.
Heartwood is more durable than springwood. Why?
Solution:
Heartwood is more durable than spring wood due to its little susceptibility to the attack of pathogens and insects.

Question 13.
Where these present:

Hypodermis layer
Mesophyll tissue
Stomata
Cambium

Solution:

Hypodermis layer – is found in stems
Mesophyll tissue – in leaves
Stomata – lower epidermis in leaves
CambiumIn vascular bundles which are open

SHORT ANSWER QUESTIONS

Question 1.
What are the differences between root hairs and stem hairs?
Solution:
The main difference between stem hairs and root hairs are :
. NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 9

Question 2.
Draw well labelled diagrams of the T.S. of dicotyledonous leaf.
Solution:

Question 3.
Why is cambium considered to be a lateral meristem?
Solution:
Cambium is responsible for the increase in the thickness of stems and roots as a result of the addition of secondary tissues (secondary cortex, secondary phloem and secondary xylem). They are located at the lateral position so-known as lateral meristems.

Question 4.
Name the plant part in which the endodermis is absent. Give one basic difference between the endodermis and epidermis.
Solution:
The endodermis is absent in leaves. Cells of endodermis possess Casparian strips or bands in their radial and transverse walls which are not found in the epidermis.

Question 5.
What are Casparian strips?
Solution:
These are thickenings of lignin and suberin formed around the lateral walls of the endodermis to prevent plasmolysis.

Question 6.
Which tissue is most abundantly found in plants? Where all is it present in plants?
Solution:
The tissue most abundantly found in plants is parenchyma. It is found in pith, cortex, and in entire mesophyll of the leaves.

Question 7.
What is present in the phloem of leaves besides sieve elements and is it living or dead? How are these functional & used?
Solution:
Besides sieve elements, in phloem parenchyma, living cells are present. These store food other cells are phloem fibres that are dead and provide mechanical strength. These are also used in making ropes and coarse textiles.

LONG ANSWER QUESTIONS

Question 1.
Describe the structure and functions of xylem tissues in angiosperm plants.
Solution:

Xylem is a complex tissue. It forms a part of the vascular bundle.
It is mainly concerned with the conduction of water and minerals. It also provides mechanical support to the plant.
As a conducting strand, xylem forms a continuous channel through the roots, stem, leaves and other aerial parts.
It consists of four different types of cells—xylem vessels, trachieds, xylem fibres and xylem parenchyma.
Xylem vessels and tracheids are concerned with the conduction of water and minerals from roots to aerial parts of the plant.
Xylem fibres provide mechanical strength to the plant body. Xylem parenchyma are the only living components of xylem.
These are concerned with the storage of food and other vital functions.

Question 2.
What is collenchyma? Explain its structure and function in the plant body of a herbaceous angiosperm.
Solution:

The cells of collenchyma are somewhat elongated with cellulose thickening, found as longitudinal strips.
These are usually confined to the comers of the cells.
Collenchyma cells appear circular, oval or angular in the transverse section. Internally, each cell possesses a large 4 central vacuole, peripheral cytoplasm and a nucleus.
Collenchyma is usually found beneath the epidermis in stem, petiole and leaves of herbaceous dicot plants. It is usually absent in monocot stems and monocot roots.

Functions:

It provides tensile strength and rigidity to the plants due to thickening.
Chloroplasts containing collenchyma cells are responsible for photosynthesis.
Collenchyma also provides elasticity to the plant organs.
Collenchyma are alive and also stores food.

Question 3.
Explain sclerenchyma with a well labelled diagram.
Solution:
Sclerenchyma is a simple permanent tissue. It consists of two types of cells. They are sclerenchyma fibres and sclereids.
(a) Sclerenchyma fibres –

These are much elongated fibers with tapering ends.
On ipaturity, they lose their protoplasm and become dead. Their cell wall is made up of cellulose or lignin, or both.
Central cavity of the cell is greatly reduced due to the formation of secondary thickening. Sclerenchyma provides mechanical strength to the plants.
They help in conduction when present in the secondary xylem.

(b) Sclereids –

They develop from ordinary parenchyma cells by the deposition of lignin.
These cells are thick-walled and highly lignified and become dead on maturity.
They are broader as compared to fibers and their cell lumen is veiy narrow.
Sclereids protect the plant from environmental forces like a strong wind.
They provide mechanical strength and rigidity to the plant.

Question 4.
Describe the structure of a monocotyledonous leaf.
Solution:
Anatomy of Monocot/isobilateral leaf: The upper and lower surfaces are covered by a single-layered epidermis.

The upper epidermis has some cells larger than the others; such large cells are known as bulliform/motor cells.
Stomata are found on both upper and lower epidermal layers. The mesophyll is not differentiated into palisade and spongy parenchyma.
Mesophyll cells are isodiametric and are arranged compactly; they contain a number of chloroplasts. Since monocot leaves have parallel veins, a number of vascular bundles can be seen in a row in the section.
Each vascular bundle has sclerenchyma cells (caps) on its upper and lower edges.
The xylem is on the upper side and the phloem on the lower side. There is a parenchymatous bundle sheath, which often contains chloroplasts and performs the function of photosynthesis.

Question 5.
Give two examples & salient features of
(1) Simple Tissue
(2) Complex Tissue
Solution:
(1) Simple Tissue:
(i) Parenchyma
(ii) Collenchyma
(2) Complex Tissue:
(i) Xylem
(ii) Phloem
(1) Simple Tissue:
(i) Parenchyma: These are living, thin-walled cells. It is used for storage of food, induction of substances, provides turgidity to softer parts of the plants
(ii) Collenchyma: These are longer than parenchyma. These are living mechanical tissue, it provides mechanical strength to organs and is present in peripheral position in plants to resist bending my the mind.
(2) Complex Tissue:
(i) Xylem: This is also called Hadrome, which is a water-conducting tissue. It is made up of cells like tracheids, xylem fibers, and xylem parenchyma only xylem parenchyma is living and all others are dead.
(ii) Phloem: This is also called Bast, which is a conducting tissue of food from leaves to all parts of the body. The parts of phloem are sieve elements, companion cells, phloem fibres, and phloem parenchyma. Phloem fibres are dead while parenchyma is living. Together these perform their function.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 6 Anatomy of Flowering Plants, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 6 Anatomy of Flowering Plants, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants

June 18, 2021

NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants.

Question 1.
Differentiate between
(a) Respiration and Combustion
(b) Glycolysis and Krebs’ cycle
(c) Aerobic respiration and Fermentation
Solution:
(a) Differences between respiration and combustion are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 1
(b) Differences between glycolysis and krebs’ cycle are as follows :

(c) Differences between aerobic respiration and fermentation are as follows :

Question 2.
What are respiratory substrates? Name the most common respiratory substrate.
Solution:
The compounds that are oxidized during this process are known as respiratory substrates. Usually, carbohydrates are oxidized to release energy, but proteins, fats, and even organic acids can be used as respiratory substances in some plants, under certain conditions.

Question 3.
Give the schematic representation of glycolysis.
Solution:

Question 4.
What are the main steps in aerobic respiration? Where does it take place?
Solution:
In aerobic respiration which takes place within the mitochondria, the final product of glycolysis, pyruvate is transported from the cytoplasm into the mitochondria.
The crucial events in aerobic respiration are:
• The complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms, leaving three molecules of CO₂.
• The passing on of the electrons removed as part of the hydrogen atoms to molecular O₂ with the simultaneous synthesis of ATP.
• The first process takes place in the matrix of the mitochondria while the second process is located on the inner membrane of the mitochondria.
• Pyruvate, which is formed by the glycolytic catabolism of carbohydrates in the cytosol, after it enters mitochondrial matrix undergoes oxidative decarboxylation by a complex set of reactions catalysed by pyruvic dehydrogenase. The reactions catalysed by pyruvic dehydrogenase require the participation of several coenzymes, including NAD+ and Coenzyme A.
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 5
During this process, two molecules of NADH are produced from the metabolism of two molecules of pyruvic acid (produced from one glucose molecule during glycolysis).
The acetyl CoA then enters a cyclic pathway, the tricarboxylic acid cycle, more commonly called as Krebs’ cycle.

Question 5.
Give the schematic representation of an overall view of Krebs’ cycle.
Solution:

Question 6.
Explain ETS.
Solution:
ETS or electron transport system is located in the inner mitochondrial membrane. It helps in releasing and utilizing the energy stored in NADH + H⁺ and FADH₂ NADH+ H⁺, which is formed during glycolyis and citric acid cycle, gets oxidized by NADH dehydrogenase. The electrons so generated get transferred to ubiquinone through FMN. In a similar manner, FADH₂ generated during citric acid cycle gets transferred to ubiquinone. The electrons from ubiquinone are received by cytochrome bc₁, and further get transferred to cytochrome C. The cytochrome C acts as a mobile carrier between complex III and cytochrome C oxidase complex containing cytochrome a and a₃, along with copper centres.

NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 7
During the transfer of electrons from each complex, the process is accompanied by the production of ATP from ADP and inorganic phosphate by the action ATP synthase. The amount of ATP produced depends on the molecule, which has been oxidized.

Question 7.
Distinguish between the following:
(a) Aerobic respiration and anaerobic respiration.
(b) Glycolysis and fermentation.
(c) Glycolysis and citric acid cycle.
Solution:
Differences between aerobic respiration and anaerobic respiration are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 8
Differences between glycolysis and fermentation are as follows :

Differences between glycolysis and citric acid cycle are as follows :

Question 8.
What are the assumptions made during the calculation of the net gain of ATP?
Solution:

There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle, and ETS pathway following one after another.
The NADH synthesized in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
None of the intermediates in the pathway are utilized to synthesize any other compound.
Only glucose is being respired -no other alternative substrates are entering the pathway at any of the intermediary stages.

Question 9.
Discuss “The respiratory pathway is an amphibolic pathway”.
Solution:
Respiration is generally assumed to be a catabolic process because, during respiration, various substrates are broken down for deriving energy. Carbohydrates are broken down into glucose before entering respiratory pathways. Fats get converted into fatty acids and glycerol whereas fatty acids get converted into acetyl CoA before entering respiration. In a similar manner, proteins are converted into amino acids, which enter respiration after deamination.

During the synthesis of fatty acids, acetyl CoA is withdrawn from the respiratory pathway. Also, in the synthesis of proteins, respiratory substances get withdrawn. Thus, respiration is also involved in anabolism. Therefore, respiration can be termed as. amphibolic pathway as it involves both anabolism and catabolism.

Question 10.
Define RQ. What is its value for fats?
Solution:
The ratio of the volume of CO₂ evolved to the volume of O₂ consumed in respiration is called respiratory quotient (RQ) or respiratory ratio.
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 11

Question 11.
What is oxidative phosphorylation ?
Solution:

Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. Yet, the presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system. Oxygen acts as the final hydrogen acceptor. Unlike photophosphorylation where it is the light energy that is utilized for the production of proton gradient required for phosphorylation, in respiration, it is the energy of oxidation-reduction utilized for the same process. It is for this reason that the process is called oxidative phosphorylation.

Question 12.
What is the significance of the step-wise release of energy in respiration?
Solution:
During oxidation within a cell, all the energy contained in respiratory substrates is not released free into the cell, or in a single step. It is released in a series of slow step-wise reactions controlled by enzymes, and it is trapped as chemical energy in the form of ATP.

Hence, it is important to understand that the energy released by oxidation in respiration is not used directly but is used to synthesise ATP, which is broken down whenever (and wherever) energy needs to be utilised. Hence, ATP acts as the energy currency of the cell.

This energy trapped in ATP is utilised in various energy-requiring processes of the organisms, and the carbon skeleton produced during respiration is used as precursors for the biosynthesis of other molecules in the cell.

VERY SHORT ANSWER QUESTIONS

Question 1.
What is anaerobic respiration? (Oct. 83)
Solution:
Incomplete or partial breakdown of fuel molecules into compounds such as ethyl alcohol, lactic acid in the absence of molecular oxygen.

Question 2.
Name the final acceptor of an electron in ETC.
Solution:
Oxygen is the electron acceptor of ETC.
Question 3.
The function of oxygen in aerobic respiration:
(i) It acts as the final electron acceptor.
(ii) It drives the whole process by removing hydrogen from the system.
Solution:
The function of oxygen in aerobic respiration:
(i) It acts as the final electron acceptor.
(ii) It drives the whole process by removing hydrogen from the system.

Question 4.
What is respiration? (Oct. 86)
Solution:
The oxidative process in which chemically bound energy from complex organic fuel molecules such as carbohydrates, proteins, and fats is captured in the form of ATP.

Question 5.
Where does the electron transport system operate in the mitochondria?
Solution:
Phosphofructokinase catalyses the formation of fructose 1, 6 bisphosphates from fructose 6-phosphate.

Question 6.
Give the function of phosphofructokinase in glycolysis.
Solution:
Hexokinase-helps in the phosphorylation of glucose.

Question 7.
Name the enzyme that catalyses the phosphorylation of glucose.
Solution:
The formation of acetyl CoA takes place in the mitochondrial matrix.

Question 8.
Where does the formation of acetyl CoA take place in a cell?
Solution:
The first step in the Krebs cycle is the condensation of an acetyl group (acetyl CoA) with oxaloacetic acid (OAA) to form citric acid and release the Coenzyme A.

Question 9.
What is the first step of reaction in the TCA cycle?
Solution:
Fatty acids may be converted to acetyl CoA before they from the respiratory substrates.

Question 10.
What is alcoholic fermentation?
Solution:
Alcoholic fermentation is the process by which yeast cells breakdown glucose into ethyl alcohol and carbon-dioxide under anaerobic conditions.

Question 11.
Name the oxidative pathway through which intermediate metabolites of glucose, fatty acids, and amino acids are finally oxidised.
Solution:
36 ATP/38 ATP molecules are obtained in the process of respiration and it is related to the aerobic respiration type.

Question 12.
What is lactic acid fermentation? (Oct. 2001)
Solution:
It is the process of fermentation by which lactose found in milk is converted to lactic acid by the action of lactobacillus.

Question 13.
What are the two molecules obtained by the action of aldolase from fructose -1, -6- biphosphate?
Solution:
ATP is produced.

SHORT ANSWER QUESTIONS

Question 1.
How is the proton gradient established?
Solution:
The proton gradient is established by passing proton (H⁺) from the matrix across the inner mitochondrial membrane into intermembrane space with the energy released during electron transfers in ETC.

Question 2.
Describe the steps in the formation of lactic acid from pyruvic acid.
Solution:
Pyruvic acid is catalysed by the enzyme lactic dehydrogenase. NADH formed in glycolysis is used up for the reduction.
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 12

Question 3.
How is ATP formed by the energy released during the electron transport system in mitochondria?
Solution:
ATP formations require an enzyme called ATP synthase. It has two components F₀– F₁. ATP- synthase becomes active in ATP formation when the concentration of H⁺ on the F_o side is higher than the F₁ side. Fligher proton concentration in the outer chamber causes the proton to pass the inner chamber. F₁ particle induced by the flow of proton through F_o channel. The energy of the proton gradient attaches the phosphate radicle to ADP. This produces ATP.

Question 4.
Give a detailed account of the net gain of ATP at a different stages of respiration.
Solution:

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

Question 5.
Enumerate the functions of ATP.
Solution:
Functions of ATP:-
(i) ATP functions as a universal energy carrier of living systems.
(ii) ATP stores small packets of energy in its molecules.
(iii) It is mobile in the cell. Therefore, it reaches all parts of the cell away from the region of ATP synthesis.
(iv) It activates a number of chemicals by functioning as a phosphorylating agent.
(v) ATP provides energy for muscle contraction.
(vi) It is involved in the transport of substances against a concentration gradient.

Question 6.
Where is cytochrome c located? What is its function?
Solution:
Cytochrome c is located on the outer surface of the inner mitochondrial membrane. It acts as a mobile carrier for the transfer of electrons between complex III and complex IV of the electron transport system.

Question 7.
Define respiratory quotient.
Solution:
The respiratory quotient is defined as the ratio of the volume of carbon dioxide evolved to the volume of oxygen consumed in respiration.

Question 8.
What is oxidative phosphorylation?
Solution:
The whole process by which oxygen effectively allows the production of ATP by phosphorylation of ADP is called oxidative phosphorylation.

Question 9.
The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Why is there anaerobic respiration even in organisms that live in aerobic conditions like human beings and angiosperms?
Solution:
Aerobic organisms do face situations where oxygen availability is little. For example, overworked muscles do not receive enough oxygen during strenuous exercise. Similarly, deep-seated tissues of angiosperms do not receive enough oxygen through diffusion from outside. In such situations, only anaerobic respiration can help in the survival of the tissue.

Question 10.
Comment on the statement- “Respiration is an energy-producing process but ATP is used in some steps of the process”.
Solution:
ATP is required in all those reactions where phosphorylative activation of the substrate is required. Therefore, despite producing energy (as ATP), respiration requires ATP in certain steps, e.g., glucose – glucose 6-phosphate, fructose 6-phosphate —fructose 1, 6- bisphosphate.

LONG ANSWER QUESTIONS

Question 1.
Explain the major steps in Krebs’ cycle. Why is this cycle also called the citric acid cycle?
Solution:
Krebs cycle: This process occurs in the mitochondrial matrix.
Major steps of Krebs cycle are as follows :

Acetyl Co-A, formed by the oxidative decarboxylation of pyruvic acid enters the Krebs’ cycle.
It combines with oxalo acetic acid (OAA), a 4C-compound, to form a 6C-compound, citric acid; the reaction is catalysed by citrate synthase.
Citrate is then isomerised into isocitrate.
Isocitrate is converted into oxalosuccinic acid in the presence of NAD and isocitrate dehydrogenase.
Oxalosuccinic acid is then decarboxylated into a-ketoglutaric acid (KG), in the presence of a decarboxylase enzyme.
a-ketoglutaric acid is converted into succinyl Co-A in the presence of NAD, Co- A, and enzyme a-ketoglutarate dehydrogenase.
When succinyl Co-A is converted into succinic acid, one molecule of GTP is formed and Co-A is released.
In the remaining part of the cycle, succinic acid is converted into OAA, so that the citric acid cycle can continue to operate.
During this cycle, three molecules of NAD and one molecule of FAD are reduced to NADH and FADH respectively.
This cycle is called as a citric acid cycle because the first product is citric acid which is 3-C compound.

Question 2.
Name the end product of glycolysis. Where is it produced in the cell? Discuss oxidative decarboxylation.
Solution:
Glycolysis results in the formation of two molecules of pyruvic acid, NADH, and ATP. It occurs in the cytosol of the cell.
Aerobic oxidation: One of the three carbons of pyruvic acid is oxidised to carbon dioxide in the reaction called oxidative decarboxylation. Pyruvic acid is first decarboxylated and then oxidised by the enzyme pyruvic dehydrogenase. The two-carbon units are readily accepted by coenzyme-A (Co-A) to form acetyl Co-A. The summary of the reaction is given in the following equation :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 15
Thus, pyruvic acid enters the Krebs cycle as acetyl Co-A. Krebs’ cycle occurs in the mitochondrial matrix.
Acetyl Co-A, formed by the oxidative decarboxylation of pyruvic acid enters the Krebs’ cycle.

Question 3.
Represent schematically the interrelationship among metabolic pathways in a plant, showing respiration mediated breakdown of different organic compounds.
Solution:
Schematic representation among metabolic pathways showing respiration mediated breakdown of different organic molecules to CO₂ and H₂O:
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 16
Question 4.
How do plants manage the exchange of gases? Give an overview of respiration in plants.
Solution:
Plants, unlike animals, have no specialized organs for gaseous exchange but they have stomata and lenticels for this purpose. There are several reasons why plants can get along without respiratory organs.

Each plant part takes care of its own gas- exchange needs. There is very little transport of gases from one plant part to another.
Plants do not present great demands for gas exchange. Roots stem and leave respire at a lower rate than animals do.
Only during photosynthesis, large volumes of leases exchanged and, each leaf is well adapted to take care of its own needs during these periods.
When cells perform photosynthesis, the availability of O₂ is not a problem in these cells since 0₂ is released
The distance that gases must diffuse even in large, bulky plants is not great. Each living cell in a plant is located quite close to the surface of the plant.
Even in woody stems, the ‘living’ cells are organised in thin layers inside and beneath the bark. They also have openings called lenticels. The cells in the interior are dead and provide only mechanical support.
Thus, most cells of a plant have attested to a part of their surface in contact with air. This is also facilitated by the loose packing of parenchyma cells in leaves, stems, and roots, which provide an interconnected network of air spaces.
The complete combustion of glucose, which produces C0₂ and H₂0 as end products, yields energy. Most of the energy is given out as heat.
C₆H₁₂O₆ + 6O₂ → 6C0₂+ 6H₂0 + Energy
If this energy is to be useful to the cell, it should be able to utilise it to synthesis other molecules that the cell requires.
The strategy that the plant cell uses is to catabolize the glucose molecule in such a way that not all the liberated energy goes out as heat.
The key is to oxidise glucose not in one step but in several small steps enabling some steps to be just large enough so that the energy released can be coupled to ATP synthesis.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 14 Respiration in Plants, helps you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 14 Respiration in Plants, drop a comment below and we will get back to you at the earliest.

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