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NCERT Exemplar Class 11 Biology Solutions Chapter 14 Respiration in Plants

NCERT Exemplar Class 11 Biology Solutions Chapter 14 Respiration in Plants

Edited By Irshad Anwar | Updated on Apr 22, 2025 07:19 PM IST

The NCERT Exemplar Class 11 Biology Solutions Chapter 12: Respiration in Plants discusses how plants decompose food to release energy, a process which is critical for their growth and development. Plants lack special respiratory organs as animals do; rather, they exchange gases through stomata (leaves), lenticels (stems), and root hairs (roots). The chapter discusses aerobic and anaerobic respiration, with aerobic respiration being in the presence of oxygen and resulting in the production of ATP, carbon dioxide, and water, whereas anaerobic respiration is done without oxygen and results in the production of ethanol or lactic acid.

This Story also Contains
  1. Detailed Answers to the NCERT Exemplar for Class 11 Science Chapter 12 Respiration in plants (Multiple Choice Questions)
  2. Answers to NCERT Exemplar Class 11 Science Chapter 12 Respiration in Plants (Very Short Answer)
  3. Elaborated Answers to NCERT Exemplar Class 11 Science Chapter 12 Respiration in Plants (Short Answer)
  4. Access Answers to NCERT Exemplar Class 11 Science Chapter 12 Respiration in Plants (Long Answer)
  5. Important Questions for Class 11 Biology Chapter 12
  6. Extra Concepts for NEET for Biology Class 11 Chapter 12 Respiration in Plants
  7. NCERT Exemplar Class 11 Biology Solutions Chapter Wise
  8. Approach to solve Questions for Class 11 Biology Chapter 12
NCERT Exemplar Class 11 Biology Solutions Chapter 14 Respiration in Plants
NCERT Exemplar Class 11 Biology Solutions Chapter 14 Respiration in Plants

Respiration involves three main steps: glycolysis (in the cytoplasm), the Krebs cycle (in the mitochondria), and the Electron Transport Chain (ETC), which produces most of the ATP. Moreover, the chapter describes fermentation, a form of anaerobic respiration in certain plants that yields alcohol and carbon dioxide. NCERT Exemplar Chapter 12 also talks about other important basic factors in biology. All in all, this chapter provides a basis for understanding biodiversity, taxonomy, and the scientific classification of organisms. Also, check NCERT Solutions for Class 11 Biology for other chapters. Generally, plant respiration is vital as it supplies the energy required for different biological processes, such as growth, repair, and metabolism. All of these are detailed in the NCERT Exemplar for Class 11 Biology Chapter 12.

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Detailed Answers to the NCERT Exemplar for Class 11 Science Chapter 12 Respiration in plants (Multiple Choice Questions)

Question:1

The ultimate electron acceptor of respiration in an aerobic organism is
(a) cytochrome
(b) Oxygen
(c) hydrogen
(d) Glucose.

Answer:

The answer is the option b) oxygen
Explanation: The electrons accepted by the oxygen molecules are the ones that are removed from hydrogen, which makes oxygen an electron acceptor.

Question:2

Phosphorylation of glucose during glycolysis is catalysed by
(a) phosphoglucomutase
(b) phosphoglucoisomerase
(c) hexokinase
(d) phosphorylase

Answer:

The answer is the option c) hexokinase
Explanation: hexokinase is responsible for the phosphorylation of glucose which leads to the production of glucose-6-phosphate.

Question:3

Pyruvic acid, the key product of glycolysis, can have many metabolic fates. Under aerobic conditions, it forms
(a) lactic acid
(b) CO2 + H2O
(c) Acetyl CoA + CO2
(d) ethanol+ CO2

Answer:

The answer is the option c) Acetyl CoA + CO2
Explanation: Acetyl CoA, CO2 and NADH are the products obtained when Pyruvic acid undergoes a dehydrogenation reaction.


Question:4

Electron Transport System (ETS) is located in the mitochondrial
(a) outer membrane
(b) intermembrane space
(c) inner membrane
(d) matrix.

Answer:

The answer is option (c) Inner membrane
Explanation: ETS is the electron transport system which as carrier molecules that can act as electron accepters as well as donors, and it is present in the inner membrane of mitochondria.

Question:5

Which of the following exhibits the highest rate of respiration?
(a) Growing shoot apex
(b) Germinating seed
(c) Root tip
(d) Leaf bud

Answer:

The answer is the option (b) Germinating seed

Explanation: germination is the process which has the maximum growth rate as compared to other options and has the highest respiration rate, hence germinating seed is the correct answer.



Question:6

Mitochondria are called powerhouses of the cell. Which of the following observations support this statement?
(a) Mitochondria synthesise ATP.
(b) Mitochondria have a double membrane.
(c) The enzymes of the Krebs cycle and the cytochromes are found in mitochondria.
(d) Mitochondria are found in almost all plant and animal cells.

Answer:

The answer is the option (a) Mitochondria synthesise ATP
Explanation: Mitochondria which synthesises ATP (Adenosine triphosphate) is called the powerhouse of the cell as ATP provides the energy to fuel many processes of the cell.


Question:7

The end product of oxidative phosphorylation is
(a) NADH
(b) Oxygen
(c) ADP
(d) ATP + H 2 O.

Answer:

The answer is the option (d) ATP + H2O
Explanation: this process is mainly used for oxidisation of nutrients by the use of enzymes in order to release energy and molecular oxygen.

Question:8

Match the following and choose the correct option from those given below.

Column A

Column B

A.Molecular oxygen

i. a-ketoglutaric acid

B.Electron acceptor

ii. hydrogen acceptor

C.Pyruvate dehydrogenase

iii. cytochrome c

D.Decarboxylation

iv. acetyl Co A


(a) A-ii, B-iii, C-iv, D-i
(b) A-iii, B-iv, C-ii, D-i
(c) A-ii, B-i, C-iii, D-iv
(d) A-iv, B-iii, C-i, D-ii

Answer:

The answer is the option (a) A – (ii), B – (iii), C – (iv), D – (i)

Answers to NCERT Exemplar Class 11 Science Chapter 12 Respiration in Plants (Very Short Answer)

Question:1

Energy is released during the oxidation of compounds in respiration. How is this energy stored and released as and when it is needed?

Answer:

During the process of respiration, energy is released. This is the energy which gets stored inside the Mitochondria in the form of ATP molecules. Whenever our body needs energy, it is released from the Mitochondria in the form of ATP molecules which are then broken down to release energy.

Question:2

Explain the term “Energy Currency”. Which substance acts as energy currency in plants and animals?

Answer:

Energy currency means a reserve of energy which can be used as a currency to obtain something at the required times. ATP is called the energy currency as it acts as an energy source which is released as and when our body requires it. It is present in plants as well as animals as an energy source.

Question:3

Different substrates get oxidised during respiration. How does Respiratory Quotient (RQ) indicate which type of substrate, i.e., carbohydrate, fat or protein is getting oxidized?
R.Q.=A/B
What do A and B stand for?
What type of substrates have R.Q. of 1,< 1 or >1?

Answer:

a) The volume of carbon dioxide is indicated by A. The volume of oxygen consumed is indicated by B.
b) Respiratory quotient
=1 (equal to 1), for carbohydrates
<1 (Less than 1), for fats and proteins
>1 (greater than 1), No substance


Question:4

F0 -F1 particles participate in the synthesis of

Answer:

ATP synthesis is the process in which F0−F1particles participate.


Question:5

When does anaerobic respiration occur in man and yeast?

Answer:

In the case of yeast, when there is an unavailability of oxygen, anaerobic respiration takes place. However, in the case of man, the anaerobic process takes place in special cases. These include the situations of strenuous physical exercise in which muscle cells require a lot of extra energy.


Question:6

Which of the following will release more energy on oxidation? Arrange them in ascending order,
(a) Igmoffat
(b) 1 gm of protein
(c) 1 gm of glucose
(d) 0.5 g of protein + 0.5g glucose

Answer:

(a)<(b)<(d)<(c)
As compared to all the elements in the options, 1 gm of glucose will release the highest amount of energy during the process of oxidation.


Question:7

The product of aerobic glycolysis in skeletal muscle and anaerobic fermentation in yeast are respectively __________and__________

Answer:

Carbon-di-oxide and ethanol are the products when the process of aerobic glycolysis in muscles and anaerobic fermentation in yeast takes place.

Elaborated Answers to NCERT Exemplar Class 11 Science Chapter 12 Respiration in Plants (Short Answer)

Question:1

If a person is feeling dizzy, glucose or fruit juice is given immediately but not a cheese sandwich, which might have more energy. Explain.

Answer:

Glucose and fats both provide energy. However, glucose is a quicker and more instant source of energy than fats as aerobic cellular respiration releases instant energy from glucose. This process creates CO2 and water as its by-products. Also, the absorption rate of glucose in our body is faster than fats. Hence when a person is feeling dizzy, he or she should be given glucose at the earliest and a food product containing fats.


Question:2

What is meant by the statement “aerobic respiration is more efficient”?

Answer:

Unlike anaerobic respiration, the process of aerobic respiration leads to complete oxidation of substances. This reaction releases energy and carbon dioxide. So, in the case of aerobic respiration, a larger amount of energy is produced as compared to the anaerobic respiration process as complete oxidation takes place. Hence, aerobic respiration is considered as an efficient process.

Question:4

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?

Answer:

Anaerobic respiration generally does not lead to the production of a higher amount of energy as compared to the aerobic respiration process. But still, in certain circumstances, organisms do undergo anaerobic respiration in the absence of enough supply of oxygen.

C6H12O6+2ADP+2Pi→2C3H6O3+2ATP+2H2O
Humans undergo anaerobic respiration in muscle cells when they are involved in intense physical activities, and the muscles are exhausted. Also, yeast undergoes anaerobic respiration when there is a dearth of atmospheric oxygen in the surroundings.
C6H12O6→2C2H5OH+2CO2+Energy

Question:5

Oxygen is an essential requirement for aerobic respiration, but it enters the respiratory process at the end. Discuss.

Answer:

The role of oxygen in the process of aerobic respiration is only at the end of the terminal of the process. The vital importance of oxygen is that it drives the hydrogen out of the body by driving the whole process and acting as a final hydrogen acceptor.

Question:6

Respiration is an energy releasing and enzymatically controlled catabolic process which involves a step-wise oxidative breakdown of organic substances inside living cells.

In this statement about respiration explain the meaning of
(i) Step-wise oxidative breakdown
(ii)Organic substances (used as substrates)

Answer:

(i) Step-wise oxidative breakdown
The oxidation of an element does not happen in one single go. It happens in a step by step process known as a step-wise oxidative breakdown. The cells of our body also require some energy to perform other functions. This step by step release of energy would ensure proper utilisation of energy and make it large enough to be associated with the synthesis of ATP molecules.
ii) Organic Substances
Organic substances are the ones which are present in the living organisms. These substances are oxidised in the process of respiration to release energy. Examples of organic substances are carbohydrates, proteins and fats.


Question:7

Comment on the statement – Respiration is an energy-producing process, but ATP is being used in some steps of the process.

Answer:

ATP molecules are also used in the process of respiration which also involves energy production. It is used to fuel energy into the formation of some intermediaries amidst the respiration process. It is used two times in the process of respiration, which includes: the conversion of glucose into glucose 6-phosphate and the conversion of fructose 6 - 6-phosphate into fructose 1, 6 - biphosphate. There is a gain of 36 ATP molecules when the oxidation of one molecule of glucose takes place. Hence the net gain is higher than the consumption, and that is why we can label the respiration process as an energy-yielding process.


Question:8

The figure given below shows the steps in glycolysis. Fill in the missing steps A, B, C, and D and also indicate whether ATP is being used up or released at step E.

Answer:
Step A: fructose 6 – phosphate
Step B: fructose 1, 6 biphosphate
Step C: triose phosphate
Step D: triose biphosphate
And finally, in Step E, we can conclude that the energy is being used up.


Question:9

Why is the respiratory pathway referred to as an amphibolic pathway? Explain.

Answer:

Generally, the pathway of respiration is also called the catabolic pathway. This happens as it involves the release of energy by breaking down substances. Most of the time, organic substances like carbohydrates, proteins and fats are converted into energy by breaking them down but sometimes an intermediate product of this process like Acetyl CoA will be used by the body to synthesise proteins or fats again. Hence the process of respiration is both a catabolic as well as an anabolic pathway.


Question:10

We commonly call ATP the energy currency of the cell. Can you think of some other energy carriers present in a cell? Name any two.

Answer:

The other energy carriers present in the cell include NADP (Nicotinamide adenine dinucleotide phosphate) and NADPH (Nicotinamide Adenine Dinucleotide Phosphate Hydrogen).


Question:11

ATP produced during glycolysis is a result of substrate-level phosphorylation. Explain.

Answer:

Substrate-level phosphorylation is the stage where the ATP formation happens by the transferring of the phosphoryl (PO3) group from a phosphorylated reactive intermediate to ADP. In the process of glycolysis, ATP formation happens at two main stages which are as follows:
1) Conversion of BPGA (biphosphoglyceric acid) into PGA (phosphoglyceric acid).
2) Conversion of PEP (phosphoenolpyruvate) into pyruvic acid.


Question:12

Do you know any step in the TCA cycle where there is substrate-level phosphorylation Which one?

Answer:

There is one stage in the citrus cycle where there is substrate-level Phosphorylation. It is the one where succinic acid is produced from succinyl CoA. So, it leads to a conversion of the GTP (Guanosine triphosphate) molecule to GDP molecule.


Question:13

ln, a way green plants and cyanobacteria have synthesised all the food on the Earth. Comment.

Answer:

The statement green plants and cyanobacteria have synthesised all the food on the Earth is quite true and can be justified as follows. Since the major producers in the aquatic environment are cyanobacteria and the major producers for terrestrial organisms are plants, all the other organisms are dependent in a direct or indirect manner on these producers.


Question:14

When a substrate is being metabolised, why does not all the energy that is produced get released in one step? It is released in multiple steps. What is the advantage of step-wise release?

Answer:

The advantages of stepwise energy release in oxidation are:
- Prevention of all the energy being used up in one go
- Stored energy reserves in the form of ATP can also be used in future
- The making of intermediary products also requires energy
- The stored energy can also be used in other processes in the body


Question:15

Respiration requires O2 . How did the first cells on the Earth manage to survive in an atmosphere that lacked O2 ?

Answer:

According to the scientists, the category of the cells which were found on the Earth was anaerobes. These cells carried out anaerobic respiration and did not require a continuous supply of oxygen for this process. Also, under special situations, all organisms can respire in an anaerobic manner and partially oxidise the glucose. Owing to this ability, the cells may have survived the atmosphere without oxygen.


Question:16

lt is known that red muscle fibres in animals can work for longer periods of time continuously. How is this possible?

Answer:

The reasons that red muscle fibres in animals can work for longer periods Continuously are:
- They are thin muscle fibres
- They get plenty of oxygen supply due to the high amount of myoglobin
- Lactic acid is not formed here as they resort to aerobic respiration
- The number of mitochondria is high in number and hence has a good supply of energy through ATP molecules


Question:17

The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Explain.

Answer:

In the case of aerobic respiration, the oxidation of glucose happens completely, which leads to a net gain of 36 ATP molecules for one glucose molecule. While, in anaerobic respiration, the oxidation of glucose does not happen in a complete manner, and the number of ATP molecules produced is just 2 for one molecule of glucose.


Question:18

RuBP carboxylase, PEPcase, Pyruvate dehydrogenase, ATPase, Cytochrome oxidase, Hexokinase, Lactate dehydrogenase. Select/ choose enzymes from the list above which are involved in
(a) Photosynthesis
(b) Respiration
(c) Both in photosynthesis and respiration

Answer:

(a) enzymes involved in photosynthesis are: RuBP carboxylase, PEPcase
(b) enzymes involved in respiration are: Pyruvate dehydrogenase, Hexokinase, Lactate dehydrogenase
(c) enzymes involved in photosynthesis as well as respiration are ATPase, cytochrome oxidase


Question:19

How does a tree trunk exchange gases with the environment, although it lacks stomata?

Answer:

The tree trunk does not have stomata like in the green leaves. But it is covered with a wooden bark which contains lenticels. These lenticels facilitate the exchange of gases in the case of tree trunks.


Question:20

Write two energy-yielding reactions of glycolysis.

Answer:

Glycolysis
The two reactions in the glycolysis which yield energy are:
(a) Formation of PGA (phosphoglyceric acid) from BPGA (biphosphoglyceric acid)
(b) Formation of Pyruvic acid (phosphoenolpyruvate) from PEP


Question:21

Name the site(s) of pyruvate synthesis. Also, write the chemical reaction wherein pyruvic acid dehydrogenase acts as a catalyst.

Answer:

The main site for the formation of pyruvate is the cytoplasm of a cell. Pyruvic acid dehydrogenase acts as a catalyst in the decarboxylation of pyruvate, as shown below:

The reaction of decarboxylation of pyruvate produces Acetyl CoA as a by-product.


Question:22

Mention the important series of events of aerobic respiration that occurs in the matrix of the mitochondrion as well as one that takes place in the inner membrane of the mitochondrion.

Answer:

The important series of events of aerobic respiration that occur in the are as follows:
a) The matrix of mitochondria

  1. Complete oxidation of pyruvate. This oxidation is a step-by-step procedure. It involves the elimination of all the atoms of hydrogen, leaving behind three molecules of CO2.

  2. The inner membrane of the mitochondrion.

Oxygen acts as an electron accepter here. As the hydrogen atoms are passed to oxygen atoms, removal of electrons take place along with the formation of ATP molecules.


Question:23

Respiratory pathway is believed to be a catabolic pathway. However, nature of the TCA cycle is amphibolic. Explain.

Answer:

The citrus cycle involves breaking down of the glucose molecules depicting catabolic behaviour. And it also involves the formation of FADH2 and ATP molecules which depict its anabolic behaviour. Hence both these process takes place in the TCA cycle, and this is the reason because of why it can be labelled as amphibolic.

Access Answers to NCERT Exemplar Class 11 Science Chapter 12 Respiration in Plants (Long Answer)

Question:1

In the following flow chart, replace the symbols a,b,c and d with appropriate terms. Briefly explain the process and give any two applications of it.
Glucose Pathway

Answer:

Glucose Pathway
The major pathways or the sequence of the steps are shown in the image above. The left flowchart depicts the process of glycolysis. The right flow chart depicts the processing of pyruvic acid, which then leads to the formation of lactic acid. In some organisms to produce CO2 and ethanol, pyruvic acid is processed to release some amount of energy. The applications of anaerobic respiration are:
- To bring the rise in bread and cakes and make them fluffy, the yeast undergoes anaerobic reproduction.
- In the formation of curd, when inoculum is lactobacillus


Question:3

Oxygen is critical for aerobic respiration. Explain its role with respect to ETS.

Answer:

NADH+, H+ and FADH2 are the carriers of energy in the electron transport system. These carriers are utilised to produce ATP during this process. Oxygen molecules accept electrons which are transferred to them through a series of reactions. Oxygen molecules accept the hydrogen molecules and lead to the formation of water molecules. This process is important for the aerobic respiration process.

The process of Oxidative phosphorylation involves the transfer of electrons which can be related to the flow of water through a tap. Just as in that case, the droplets cannot flow down until the first drop is allowed to pass through, after the first electron is allowed, others follow when the oxygen starts accepting electrons as a hydrogen acceptor. That is the reason due to which the presence of oxygen is necessary for the electron transfer to be made possible by the creation of a gradient.


Question:4

Enumerate the assumptions that we undertake in making the respiratory balance sheet. Are these assumptions valid for a living system? Compare fermentation and aerobic respiration in this context.

Answer:

The assumptions that are made during the calculation of a respiratory balance sheet are:
- An orderly and sequential pathway is followed in the process of respiration.
- During the process of glycolysis, NADH is produced. This undergoes oxidative phosphorylation inside the mitochondria.
- No intermediate products are utilised in the formation of other products
- Glucose is oxidised only, and no other substance is utilised at any in-between stages The above assumptions are invalid for a living organism. The processes need to take place simultaneously and not follow a strict sequential path.

Fermentation

Aerobic respiration

Glucose gets converted to ethanol during its partial breakdown.

Glucose breaks down completely, which leads to the formation of carbon dioxide.

Gain of ATP molecules: 2

Gain of ATP molecules: 36

3slow oxidation process of NADH to NAD+

Fast oxidation process of NADH to NAD+



Question:5

Give an account of glycolysis. Where does it occur? What are the end products? Trace the fate of these products in both aerobic and anaerobic respiration.

Answer:

Pyruvic acid is formed in the process of glycolysis due to the breakdown of glucose which takes place in the cytoplasm of the cell. This process also involves the partial oxidation of glucose molecules known as the EMP pathway. The important steps involved in the process of glycolysis are as follows:

- Phosphorylation of glucose for the production of glucose-6-phosphate
- Formation of PGAL (Phosphoglyceraldehyde) through the conversion of Fructose-6-phosphate
- The molecules of PGAL are further processed for the production of Pyruvic acid.
- Two molecules of ATP are gained in the process of glycolysis of one glucose molecule
Aerobic Respiration of Pyruvate:
Complete oxidation of Pyruvic acid takes place and carbon dioxide and Energy are produced as a result of this reaction.

Anaerobic Respiration Pyruvate:
In the absence of oxygen Anaerobic Respiration takes place, and ethanol and CO2 are produced from the conversion of pyruvic acid. In some organisms who face a complete absence of oxygen, the pyruvic acid can also be converted to lactic acid.

Also, Check NCERT Books and NCERT Syllabus here

Important Questions for Class 11 Biology Chapter 12

Below are a few solved important questions:-

Q1. What is the end product of oxidative phosphorylation?

Answer:

A metabolic process called oxidative phosphorylation releases chemical energy and oxidises foods to create adenosine triphosphate (ATP) in cells.

Q2. Match the following

Column A

Column B

A.Molecular oxygen

i. a-ketoglutaric acid

B.Electron acceptor

ii. hydrogen acceptor

C.Pyruvate dehydrogenase

iii. cytochrome c

D.Decarboxylation

iv. Acetyl-Coa

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Answer:

A. Molecular oxygen: ii. hydrogen acceptor

B.Electron acceptor: iii. cytochrome c

C. Pyruvate dehydrogenase: iv. acetyl Co A

D. Decarboxylation: i. a-ketoglutaric acid


Q3. What is the fate of pyruvic acid, the key product of glycolysis under aerobic condition?

Answer:

Acetyl CoA, CO2 and NADH are the products obtained when Pyruvic acid undergoes a dehydrogenation reaction. An organic molecule with a backbone of three carbons, pyruvate is essential to an organism's processes of synthesis and degradation. It is used in bodybuilding supplements, the food business, and cancer screening, and when it is present in abnormal amounts, it is linked to a number of illnesses.

Q4. Where is the Electron Transport System (ETS) located in the mitochondria? Write few lines about it.

Answer:

ETS is the electron transport system, which is a carrier molecule that can act as electron acceptors as well as donors, and it is present in the inner membrane of mitochondria. The electron transport chain is present in multiple copies in the inner mitochondrial membrane of eukaryotes and the plasma membrane of prokaryotes. It results in oxidative phosphorylation. The ETC is a series of proteins that receive the high-energy electrons from NADH and FADH2 and move them to the final acceptor, molecular oxygen.

Q5. Which enzyme catalyses the phosphorylation of glucose during glycolysis?

Answer:

An enzyme called hexokinase catalyses the initial stage of glycolysis, which converts glucose into energy for cellular metabolism.
Glucose is phosphorylated by this enzyme, which turns it into glucose-6-phosphate. ATP serves as both a phosphate donor and an energy source for hexokinase.

Also, read the NCERT Solution subject-wise

Extra Concepts for NEET for Biology Class 11 Chapter 12 Respiration in Plants

The following extra concepts can be studied from the perspective of the NEET exam:

  1. Respiratory quotient variations: NEET often ask questions on RQ values on different substrates. NCERT introduces the concept, while NEET requires a deeper understanding of the values of all the biomolecules.
  2. Details steps of glycolysis and Krebs' cycle: The enzymes, intermediates involved in glycolysis and Krebs' cycle, and the ATP yield should be memorised properly to answer those tricky match the column or sequence-based questions in NEET.
  3. Types of anaerobic respiration: One should know the differences between lactic acid fermentation (in muscle) and alcoholic fermentation (yeast), including the byproducts, ATP yield and organisms involved.
  4. ATP yield in different pathways: NEET frequently asks about the total ATP generation in all the pathways, such as aerobic respiration and anaerobic respiration and also the exact breakdown per stage of any pathway.

NCERT Exemplar Class 11 Biology Solutions Chapter Wise

The biology chapter-wise exemplar solutions of Class 11 are given below:

NCERT Exemplar Solutions for Class 11 Biology Chapter 1 The Living World
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification
NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants
NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organisation in Animals
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division
NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition
NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis in Higher Plants
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption
NCERT Exemplar Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation
NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination
NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural Control and Coordination
NCERT Exemplar Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration
















Approach to solve Questions for Class 11 Biology Chapter 12

The candidates can begin by thoroughly understanding the key concepts, such as glycolysis, TCA cycle and the electron transport system, focusing on the steps, locations and ATP yield. They can use diagrams and flow charts in their answers to present them clearly. They should include scientific terms such as oxidative phosphorylation and substrate-level phosphorylation to show conceptual accuracy. The answers can be structured in well-organised points or short paragraphs with appropriate headings. They should revise the NCERT content line by line, as questions are mostly based on the textbook wording.

NCERT Exemplar Class 11 Solutions:


NCERT Exemplar Class 11 Biology Solutions Chapter 12 provides detailed information about these three metabolic pathways. Both glycolysis and Kreb's cycle are amphibolic pathways as they provide several intermediates. The amphibolic pathway is used for both breakdown and break-up reactions.

Must Read NCERT Notes subject-wise

Frequently Asked Questions (FAQs)

1. Is this chapter important for the entrance exam?

 Yes, this chapter and the NCERT Exemplar Class, 11 Biology chapter 12 solutions is most crucial if one is preparing for the entrance exam. 

2. Can one score high based on solutions?

Yes, one can understand the answering style and can also revise the topics from the solutions. 

3. Who has prepared these solutions?

These solutions are prepared by our biology teachers who know the CBSE syllabus in detail. 

4. Are these solutions downloadable?

 Yes, one can easily download the solutions using the NCERT Exemplar Class 11 Biology chapter 12 solutions PDF Download function.

5. How does glycolysis occur in respiration in plants?

Glycolysis of plant respiration takes place in the cytoplasm, wherein glucose is disintegrated into two molecules of pyruvate. The process produces ATP and NADH without needing oxygen. The process consists of a series of enzyme-catalyzed steps that transform glucose into energy intermediates. The resulting pyruvate flows into the mitochondria for more respiration in case oxygen is present.

6. Why is the Krebs cycle important in plant respiration?

The Krebs cycle, which takes place in the mitochondria, plays a significant role in plant respiration because it produces ATP, NADH, and FADH₂ for the generation of energy. It also releases CO₂ as a waste product and offers intermediates for biosynthesis. The generated NADH and FADH₂ power the electron transport chain for additional ATP. The Krebs cycle facilitates the effective release of energy from organic molecules.

7. How does electron transport chain function in plant respiration?

The electron transport chain (ETC) in plant respiration takes place in the inner mitochondrial membrane, with electrons from NADH and FADH₂ being passed via protein complexes. This moves protons into the intermembrane space, establishing a proton gradient. The proton return via ATP synthase forces the creation of ATP. Oxygen acts as the terminal electron acceptor to give water.

8. What are the major steps involved in respiration in plants?

The major steps in plant respiration are:

  1. Glycolysis – Glucose is broken down into pyruvate in the cytoplasm, producing ATP and NADH.

  2. Krebs Cycle – Pyruvate is further oxidized in the mitochondria, releasing CO₂, ATP, NADH, and FADH₂.

  3. Electron Transport Chain (ETC) – NADH and FADH₂ donate electrons to generate ATP, with oxygen acting as the final electron acceptor.

9. How does fermentation differ from aerobic respiration?

Fermentation is oxygen-free and results in ATP with byproducts such as ethanol or lactic acid. Aerobic respiration is an oxygen-requiring process and produces much more ATP by completely oxidizing glucose to CO₂ and H₂O. Fermentation takes place in the cytoplasm, whereas aerobic respiration goes on in the mitochondria. Aerobic respiration is more efficient in terms of energy compared to fermentation.

10. What is the role of ATP in cellular respiration of plants?

ATP is the primary energy carrier in plant cellular respiration, storing and transferring energy for cellular processes. It is produced during glycolysis, the Krebs cycle, and oxidative phosphorylation. ATP powers essential functions like nutrient transport, biosynthesis, and growth. It is continuously regenerated to sustain metabolic activities in plant cells.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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