Samacheer Kalvi Class 11 Bio Botany Solutions Chapter 14 Respiration

Part-I.

 

Question 1. The number of ATP molecules formed by complete oxidation of one molecule of pyruvic acid is:
(a) 12
(b) 13
(c) 14
(d) 15
Answer: (a) 12
In simple words: When one pyruvic acid molecule is fully broken down, 12 ATP molecules are made. This process creates energy for the cell.

🎯 Exam Tip: Remember the total ATP yield from the complete oxidation of glucose and individual steps like pyruvic acid oxidation to score well.

 

Question 2. During oxidation of two molecules of cytosolic NADH + H+, number of ATP molecules produced in plants are
a) 3
b) 4
c) 6
d) 8
Answer: (b) 4
In simple words: In plants, when two molecules of NADH + H+ from the cytoplasm are oxidized, they produce 4 ATP molecules. This happens during cellular respiration to generate energy.

🎯 Exam Tip: Be mindful of the slight differences in ATP yield between plant and animal cells, especially concerning NADH + H+ oxidation.

 

Question 3. The compound which links glycolysis and Krebs cycle is:
(a) succinic acid
(b) pyruvic acid
(c) acetyl CoA
(d) citric acid
Answer: (c) acetyl CoA
In simple words: Acetyl CoA acts like a bridge, connecting the glycolysis pathway to the Krebs cycle. It's an important molecule in energy production within cells.

🎯 Exam Tip: Understanding the role of acetyl CoA as the entry point for the Krebs cycle is key to explaining cellular respiration pathways.

 

Question 4. Assertion (A): Oxidative phosphorylation takes place during the electron transport chain in mitochondria. Reason (R): One molecule of \(\alpha\)-ketoglutaric acid is phosphorylated into succinic acid by substrate phosphorylation.
a) A and R is correct. R is correct explanation of A
b) A and R is correct but R is not the correct explanation of A,
c) A is correct but R is wrong
d) A and R is wrong
Answer: (c) A is correct but R is wrong
In simple words: Assertion (A) is true because oxidative phosphorylation happens during the electron transport chain in mitochondria, which is how most ATP is made. Reason (R) is false because \( \alpha \)-ketoglutaric acid is not directly phosphorylated into succinic acid by substrate phosphorylation.

🎯 Exam Tip: Clearly differentiate between oxidative phosphorylation (electron transport chain) and substrate-level phosphorylation to avoid common mistakes in assertion-reason questions.

 

Question 5. Which of the following reaction is not involved in Krebs cycle.
(a) Shifting of phosphate from 3C to 2C
(b) Splitting of Fructose 1,6 bisphosphate of into two molecules 3C compounds.
(c) Dephosphorylation from the substrates
(d) All of these
Answer: (d) All of these
In simple words: None of these reactions happen in the Krebs cycle. These types of phosphate shifts, splitting of fructose, and dephosphorylation from substrates are part of other metabolic pathways like glycolysis. The Krebs cycle mainly focuses on oxidizing acetyl CoA.

🎯 Exam Tip: A good understanding of the specific steps and molecules involved in glycolysis versus the Krebs cycle is essential to correctly answer questions about their distinct reactions.

 

Question 6. What are enzymes involved in phosphorylation and dephosphorylation reactions in EMP pathway?
Answer:
(i) Enzymes involved in phosphorylation are
a) Hexokinase and phosphofructokinase.
(ii) Enzymes involved in dephosphorylation are
a) Phosphoglycerate Kinase
b) Pyruvate Kinase
In simple words: In the EMP pathway, enzymes like Hexokinase and Phosphofructokinase add phosphate groups (phosphorylation). Enzymes such as Phosphoglycerate Kinase and Pyruvate Kinase remove phosphate groups (dephosphorylation). These enzymes help control the flow of energy in the pathway.

🎯 Exam Tip: Listing specific enzymes for each type of reaction (phosphorylation and dephosphorylation) shows a detailed understanding of the EMP pathway.

 

Question 7. Why does the RQ value of some succulent plants become zero?
Answer: In some succulent plants, like Opuntia and Bryophyllum, carbohydrates are partly broken down into organic acids, especially malic acid. This process uses oxygen but does not release carbon dioxide. Since the Respiratory Quotient (RQ) is calculated as the ratio of CO2 released to O2 consumed, if no CO2 is released, the RQ value becomes zero. These plants have a special way of breathing called CAM photosynthesis.
In simple words: Succulent plants break down food in a special way that uses oxygen but makes no carbon dioxide. Because RQ is CO2 divided by O2, their RQ becomes zero.

🎯 Exam Tip: When discussing RQ, always explain why CO2 release or O2 consumption might be altered, especially in special plant types like succulents.

 

Question 8. Explain the reactions taking place in mitochondrial inner membrane.
Answer: The mitochondrial inner membrane is where the electron and hydrogen (proton) transport occurs through four main multiprotein complexes (I-IV).
1. Complex-I (NADH dehydrogenase): This complex has a flavoprotein (FMN) and iron-sulphur protein (Fe-S). It transfers electrons and protons from mitochondrial NADH (internal) to Ubiquinone (UQ).
NADH + H\(^+\) \( \implies \) NAD\(^-\) + UQH\(_2\)
2. In plants, an extra NADH dehydrogenase (external) complex is found on the outer surface of the inner mitochondrial membrane. This complex can oxidize cytosolic NADH + H\(^+\). Ubiquinone (UQ) or Coenzyme Q (CoQ) is a small, fat-soluble electron and proton carrier located within the inner membrane of mitochondria. This molecule helps shuttle electrons.
3. Complex-II (succinic dehydrogenase): This complex contains FAD flavoprotein and an iron-sulphur protein (Fe-S). It takes electrons and protons from succinate in the Krebs cycle, turning it into fumarate, and then passes them to ubiquinone. Succinate changes to fumarate in this step.
Succinate + UQ \( \implies \) Fumaraic + UQH\(_2\)
4. Complex-III (Cytochrome bc1 complex): This complex oxidizes reduced ubiquinone (ubiquinol) and moves electrons through Cytochrome bc1 Complex (Iron Sulphur center bcl complex) to cytochrome c.
UQH\(_2\) + 2Cyt c oxidized \( \implies \) UQ + 2Cyt c reduced + 2H\(^+\)
5. Complex IV (Cytochrome c oxidase): This is the final oxidase complex. It reduces oxygen (1/2 O\(_2\)) to water (H\(_2\)O). Two protons are needed to form one molecule of H\(_2\)O, which is called terminal oxidation.
2Cyt c oxidized + 2H\(^+\) + 1/2 O\(_2\) \( \implies \) 2Cyt c reduced - H\(_2\)O
In simple words: The inner part of the mitochondria has special protein groups called complexes (I to IV). These complexes work together to move electrons and hydrogen, making energy. Each complex has a specific job, like passing electrons from NADH or FADH2 to other molecules, and finally, oxygen takes the electrons to make water. Ubiquinone acts as a small shuttle to carry electrons between these complexes.

🎯 Exam Tip: When explaining electron transport, remember to mention all four complexes and their specific functions, along with the key electron carriers like NADH, FADH2, and ubiquinone.

 

Question 9. What is the name of alternate way to glucose breakdown? Explain the process involved in it?
Answer: The Pentose phosphate pathway (PPP) is an alternate way for glucose breakdown. It is also known as the Hexose Monophosphate Shunt (HMP shunt) or Direct Oxidative Phase.
- Pentose phosphate pathway was first described by Warburg, Dickens, and Lipmann in 1938.
- It involves two main phases: an oxidative phase and a non-oxidative phase.
- In the oxidative phase, six molecules of six-carbon Glucose 6 phosphate are converted into six molecules of five-carbon sugar Ribulose – 5 Phosphate. During this conversion, 6CO2 is lost, and 12 NADPH + H\(^+\) are generated. NADPH is important for reductive biosynthesis.
- The non-oxidative pathway then converts Ribulose – 5 – phosphate molecules into various other intermediate sugars like Ribose – 5 – phosphate (5C), Xylulose – 5 - phosphate (5C), Glyceraldehyde – 3 – phosphate (3C), Sedoheptulose – 7 – phosphate (7C), and Erythrose – 4 – phosphate (4C). These intermediate sugars are precursors for making other important molecules.
- Finally, five molecules of glucose 6 – phosphate are regenerated from these intermediates.
The overall reaction is:
6 x Glucose – 6 – phosphate + 12NADP\(^+\) + 6H\(_2\)O
\( \implies \)
5 x glucose - 6 - phosphate + 6CO\(_2\) + Pi + 12 NADPH + 2H\(^+\)
The net result of complete oxidation of one glucose – 6 – phosphate yields 6CO\(_2\) and 12 NADPH + H\(^+\). This pathway is controlled by the glucose – 6 – phosphate dehydrogenase enzyme, which is inhibited by a high ratio of NADPH to NADP\(^+\).
In simple words: The pentose phosphate pathway is another way cells break down glucose. It helps create NADPH, which cells need for building things, and also makes sugars with 5 carbons. It has two parts: one that uses oxygen and one that changes sugars around.

🎯 Exam Tip: When explaining alternative pathways, ensure to highlight their unique products (like NADPH and specific sugars) and their importance beyond just energy production.

 

Question 10. How will you calculate net products of one sucrose molecule upon complete oxidation during aerobic respiration as per recent view?
Answer: To calculate the net products from one sucrose molecule during aerobic respiration, we consider the latest understanding of ATP transport costs.
When ATPs are transported from the mitochondrial matrix into the cytosol, the actual yield is less. For each NADH + H\(^+\) oxidized during the electron transport system, 2.5 ATPs are formed, and for each FADH2, 1.5 ATPs are formed. A sucrose molecule produces two molecules of glucose. Therefore, to calculate the total ATP yield for one sucrose molecule, we first find the yield per glucose molecule and then multiply by two.
In plant cells, the net yield is 30 ATP molecules for complete aerobic oxidation of one glucose molecule. So, for one sucrose molecule, it will be \(30 \times 2 = 60\) ATP.
In animal cells (especially those using the malate shuttle mechanism), the net yield is 32 ATP molecules per glucose. So, for one sucrose molecule, it will be \(32 \times 2 = 64\) ATP.
In simple words: One sucrose molecule breaks down into two glucose molecules. For plants, one glucose makes 30 ATP, so sucrose makes 60 ATP. For animals, one glucose makes 32 ATP, so sucrose makes 64 ATP. These numbers consider the energy used to move ATP around the cell.

🎯 Exam Tip: Remember that sucrose yields two glucose molecules. Always factor in the cell type (plant vs. animal) when calculating ATP yield, as the shuttle systems can cause differences in the net ATP produced.

Part-II.

11th Bio Botany Guide Respiration Additional Important Questions and Answers

I. Choose The Correct Answer

 

Question 1. The term respiration was coined by:
(a) Lamark
(b) Kerb
(c) Pepys
(d) Blackman
Answer: (c) Pepys
In simple words: The word "respiration" was first used by a person named Pepys. He named this process where living things take in oxygen and release carbon dioxide to get energy.

🎯 Exam Tip: Knowing the historical figures associated with scientific terms like "respiration" can sometimes be useful in general knowledge sections.

 

Question 2. Black man divided respiration into floating respiration and protoplasmic respiration based on respiratory ...........
a) Quotient
b) respiratory reaction
c) respiratory pathway
d) substrate
Answer: (d) substrate
In simple words: Blackman sorted respiration into "floating" and "protoplasmic" based on what kind of food (substrate) the living thing was breaking down for energy. Floating respiration uses carbohydrates or fats, while protoplasmic respiration uses proteins.

🎯 Exam Tip: Understanding that the substrate used for respiration can vary helps categorize different types of respiratory processes.

 

Question 3. The discovery of ATP was made by:
(a) Lipman
(b) F. A. Belt
(c) Warburg
(d) Karl Lohman
Answer: (d) Karl Lohman
In simple words: Karl Lohman was the person who discovered ATP. ATP is like the main energy currency that cells use for all their work.

🎯 Exam Tip: ATP is central to all cellular energy processes, so knowing its discoverer is a good fact for biology exams.

 

Question 4. The type of respiration which is rare and liberates toxic ammonia
a) Protoplasmic respiration
b) floating respiration
c) Aerobic respiration
d) Anaerobic respiration
Answer: (a) Protoplasmic respiration
In simple words: Protoplasmic respiration is a less common type of breathing where cells break down proteins. When proteins are broken down, they release ammonia, which can be harmful to the cell.

🎯 Exam Tip: Highlight the key characteristic of protoplasmic respiration, which is the use of proteins as a substrate and the release of toxic ammonia.

 

Question 5. On hydrolysis, one molecule of ATP releases energy of:
(a) 8.2 Kcal
(b) 32.3 kJ
(c) 7.3 Kcal
(d) 7.8 Kcal
Answer: (c) 7.3 Kcal
In simple words: When a molecule of ATP breaks down (hydrolysis), it gives off about 7.3 kilocalories of energy. This energy is then used by the cell to do work.

🎯 Exam Tip: Knowing the approximate energy yield from ATP hydrolysis is fundamental for understanding cellular energy dynamics.

 

Question 6. To convert Kcal to KJ multiply by 4.18 (100 Kcal=418 KJ) calculate the amount KJ energy for 7.3 Kcal
a) 30.6 KJ
b) 32.06 KJ
c) 30.55 KJ
d) 5.01 KJ
Answer: (a) 30.6 KJ
In simple words: To change energy from Kcal to KJ, you multiply by 4.18. So, 7.3 Kcal becomes 30.514 KJ, which is rounded to 30.6 KJ.

🎯 Exam Tip: Always pay attention to units (Kcal vs. KJ) and the conversion factor provided in physics and biology calculations.

 

Question 7. Identify the link reaction:
(a) conversion of glucose into pyruvic acid
(b) conversion of glucose into ethanol
(c) conversion of acetyl CoA into CO2 and water
(d) conversion of pyruvic acid into acetyl coenzyme – A
Answer: (d) conversion of pyruvic acid into acetyl coenzyme – A
In simple words: The link reaction is when pyruvic acid changes into acetyl coenzyme A. This step connects glycolysis to the Krebs cycle, preparing the fuel for the next stage of energy production.

🎯 Exam Tip: Clearly defining the link reaction as the transition between glycolysis and the Krebs cycle is crucial for explaining overall respiration.

 

Question 8. Food materials like carbohydrate, fat and proteins are completely oxidised into CO2, H2O and energy in ................ respiration
a) anaerobic
b) aerobic
c) Bacterial respiration
d) Facultative
Answer: (b) aerobic
In simple words: When food like carbohydrates, fats, and proteins are fully broken down to carbon dioxide, water, and energy, it happens through aerobic respiration. This type of respiration uses oxygen.

🎯 Exam Tip: The key differentiator for aerobic respiration is the complete oxidation of food substances, which produces a large amount of energy using oxygen.

 

Question 9. Kreb's cycle is a:
(a) catabolic pathway
(b) anabolic pathway
(c) amphibolic pathway
(d) hydrolytic pathway
Answer: (c) amphibolic pathway
In simple words: The Krebs cycle is called an amphibolic pathway because it does two things: it breaks down molecules (catabolic) and it also provides building blocks for new molecules (anabolic). It's a central hub in metabolism.

🎯 Exam Tip: An "amphibolic" pathway means it serves both catabolic (breaking down) and anabolic (building up) roles in the cell's metabolism. This is a common and important concept.

 

Question 10. Which process is occur in yeast and some bacteria and 2 ATP molecules are produced during this process.
a) Anaerobic respiration
b) aerobic respiration
c) mixed fermentation
d) CAC cycle
Answer: (a) Anaerobic respiration
In simple words: Anaerobic respiration happens in yeast and some bacteria without oxygen. In this process, only 2 ATP molecules are made, which is much less than with oxygen. Fermentation is a type of anaerobic respiration.

🎯 Exam Tip: Note that anaerobic respiration is less efficient than aerobic respiration, producing significantly fewer ATP molecules.

 

Question 11. The oxidation of one molecule of NADH + H+ gives rise to:
(a) 2 ATP
(b) 3 ATP
(c) 4 ATP
(d) 2.5 ATP
Answer: (b) 3 ATP
In simple words: When one molecule of NADH + H+ is oxidized in the electron transport chain, it generates 3 ATP molecules. This is a key energy yield in cellular respiration.

🎯 Exam Tip: Memorize the ATP yield from NADH + H+ and FADH2 as these are frequently tested values in bioenergetics.

 

Question 12. Net product in Glycolysis are
a) 4ATP and 2NADH + H+
b) 2 ATP and 2NADH + H+
c) 6АТР
d) 24 ATP
Answer: (b) 2ATP and 2NADH + H+
In simple words: After glycolysis, the cell gets a net of 2 ATP molecules and 2 NADH + H+ molecules. While 4 ATP are made, 2 ATP are used up, leading to a net gain of 2 ATP.

🎯 Exam Tip: Distinguish between the gross production and net production of ATP in glycolysis, as ATP is both consumed and produced during the pathway.

 

Question 13. Cyanide acts as electron transport chain inhibitor by preventing:
(a) synthesis of ATP from ADP
(b) flow of electrons from NADH + H+
(c) flow of electrons from cytochrome a3 to O2
(d) oxidative phosphorylation
Answer: (c) flow of electrons from cytochrome a3 to O2
In simple words: Cyanide stops the electron transport chain by blocking the movement of electrons from cytochrome a3 to oxygen. This prevents cells from making ATP and can be deadly.

🎯 Exam Tip: Understand the specific targets of different inhibitors in the electron transport chain to explain their impact on ATP synthesis.

 

Question 14. Which is the raw material for the formation of chlorophylls, cytochrome, phytochrome and pyrrole substance
a) acetyl COA
b) Pyruvic acid
c) Malic acid
d) Succinyl COA
Answer: (d) Succinyl COA
In simple words: Succinyl CoA is a basic building block needed to make important molecules like chlorophyll (for photosynthesis), cytochrome (for electron transport), phytochrome (a light-sensing pigment), and pyrrole. These are all vital for plant life.

🎯 Exam Tip: Recognize the dual role of Krebs cycle intermediates, not just for energy, but also as precursors for biosynthesis of other vital compounds.

 

Question 15. End products of fermentation in yeast is:
(a) pyruvic acid and CO2
(b) lactic acid and CO2
(c) ethyl alcohol and CO2
(d) mixed acid and CO2
Answer: (c) ethyl alcohol and CO2
In simple words: When yeast undergoes fermentation, it produces ethyl alcohol and carbon dioxide as the final products. This is how alcoholic drinks and bread are made.

🎯 Exam Tip: Differentiate between alcoholic fermentation (yeast) and lactic acid fermentation (muscles/bacteria) by their respective end products.

 

Question 26. Match the Column I with the enzyme responsible for its production in column II

(a) A-4, B-1, C-2, D-3
(b) A-4, B-2, C-3, D-1
(c) A-3, D-2, C-1, B-4
(d) A-1, B-2, C-3, C-4
Answer: (a) A-4, B-1, C-2, D-3
In simple words: This question matches key substances in metabolic pathways with the specific enzymes that create them. Understanding these enzyme-substrate relationships is important for understanding cell energy.

🎯 Exam Tip: When matching, look for key terms in both columns. For instance, "Citric acid" strongly points to "Citric acid Synthetase".

 

Question 27. Which one is wrongly matched

Answer: (B) Glycolysis – Twenty four ATP
In simple words: Glycolysis produces a net of only 2 ATP directly, not 24 ATP. The 24 ATP value is incorrect for this process.

🎯 Exam Tip: Memorize the net ATP yield for each stage of respiration (Glycolysis, Krebs cycle, ETC) to avoid confusion in such questions.

II. 2 Marks Questions

 

Question 1. Define respiration?
Answer: Respiration is a natural process in living things where food, like carbohydrates, proteins, and fats, is broken down by oxidation inside cells. This process creates energy and releases carbon dioxide while taking in oxygen. It is how living organisms get energy to function.
In simple words: Respiration is when cells break down food using oxygen to get energy, giving off carbon dioxide.

🎯 Exam Tip: Include the key elements: "biological process," "oxidation of food substances," "energy production," "O2 intake," and "CO2 liberation."

 

Question 2. Name some High energy compounds present in a cell
Answer: Some high energy compounds found in a cell are:
• GTP \( \rightarrow \) Guanosine Tri Phosphate
• UTP \( \rightarrow \) Uridine Tri Phosphate
These molecules store and transfer energy for various cellular activities.
In simple words: Cells use special molecules like GTP and UTP to store and move energy around.

🎯 Exam Tip: Remember common high-energy compounds like ATP, GTP, and UTP, as they are crucial for cellular metabolism.

 

Question 3. What do you understand by compensation of point?
Answer: The compensation point is a specific moment when the amount of carbon dioxide \( (CO_2) \) released by a plant during respiration is exactly balanced by the amount of \( CO_2 \) absorbed for photosynthesis. At this point, there is no net exchange of gases with the environment, meaning the plant is neither taking in nor giving out \( CO_2 \). This balance is critical for plant survival, especially in low light conditions.
In simple words: Compensation point is when a plant's breathing out carbon dioxide matches the amount it uses for making food, so no gas goes in or out.

🎯 Exam Tip: Clearly state that \( CO_2 \) released by respiration equals \( CO_2 \) fixed by photosynthesis, leading to "no net gaseous exchange."

 

Question 4. What is Anaerobic respiration? What are its steps?
Answer: Anaerobic respiration is a process where glucose is broken down to produce energy without the presence of oxygen. This type of respiration is less efficient than aerobic respiration but allows organisms to generate ATP when oxygen is scarce. It happens in two main steps:
• In the absence of molecular oxygen, glucose is incompletely broken down into either ethyl alcohol or lactic acid.
• It includes two steps:
(i) Glycolysis
(ii) Fermentation
In simple words: Anaerobic respiration makes energy from sugar without oxygen. It has two steps: glycolysis and fermentation, ending in alcohol or lactic acid.

🎯 Exam Tip: Key points are "absence of oxygen," "incomplete degradation," and the two steps: glycolysis and fermentation, mentioning the end products.

 

Question 5. What is anaerobic respiration?
Answer: Anaerobic respiration is the process where glucose is partially broken down in the absence of oxygen, producing either ethyl alcohol or lactic acid. This process allows cells to generate a small amount of energy when oxygen is not available. It occurs in two stages: glycolysis and fermentation.
In simple words: Anaerobic respiration is when sugar breaks down without oxygen, making alcohol or lactic acid, to get a little energy.

🎯 Exam Tip: Define anaerobic respiration by emphasizing "absence of molecular oxygen," "incomplete degradation," and the end products (ethyl alcohol/lactic acid).

 

Question 6. What is Link reaction?
Answer: The link reaction, also known as the transition reaction, happens in the mitochondrial matrix during aerobic respiration. It converts pyruvic acid into acetyl coenzyme-A. This process also produces two molecules of NADH + H\( ^+ \) and two molecules of \( CO_2 \). This step connects glycolysis to the Krebs cycle.
In simple words: The link reaction turns pyruvic acid into acetyl coenzyme-A, making NADH and carbon dioxide. It connects two main energy-making steps in cells.

🎯 Exam Tip: Highlight the conversion of "pyruvic acid to acetyl coenzyme-A" and its role as a "link" between glycolysis and the Krebs cycle.

 

Question 7. Who is Sir Hans Adolf Krebs?
Answer: Sir Hans Adolf Krebs was a German biochemist born on August 25th, 1900. He was awarded the Nobel Prize in Physiology in 1953 for his discovery of the citric acid cycle, a key metabolic pathway in all aerobic organisms. His work explained how cells generate energy.
In simple words: Sir Hans Adolf Krebs was a scientist who won a Nobel Prize for discovering the citric acid cycle, which is how cells make energy.

🎯 Exam Tip: Focus on his key contribution: "discovery of the Citric acid cycle" and the "Nobel Prize."

 

Question 8. Why Kreb cycle is called as citric acid cycle (or) Tri Carboxylic acid Cycle?
Answer: The Krebs cycle is known as the citric acid cycle or Tri Carboxylic acid Cycle for two main reasons:
• The cycle begins with the condensation of acetyl CoA with oxaloacetate in the presence of water to produce citric acid (or citrate). Citric acid is the first product formed in the cycle. This step is crucial for the cycle's initiation.
• Because citric acid has three carboxyl groups, it is also called a tricarboxylic acid. The cycle is named after this important molecule which plays a central role in the entire process.
In simple words: The Krebs cycle is called the citric acid cycle because citric acid is the first molecule made. It's also called tricarboxylic acid cycle because citric acid has three acid parts.

🎯 Exam Tip: Emphasize that "citric acid is the first product" and that it contains "three carboxyl groups," hence "tricarboxylic acid."

 

Question 9. Mention the role of NADH dehydrogenase enzyme in the electron transport system.
Answer: NADH dehydrogenase is an enzyme that contains flavoprotein (FMN) and is associated with non-heme iron Sulphur protein (Fe-S). This complex is responsible for passing electrons and protons from mitochondrial NADH (internal) to ubiquinone (UQ). This is the first step in the electron transport chain, crucial for generating a proton gradient.
In simple words: NADH dehydrogenase moves electrons and protons from NADH to ubiquinone in the electron transport chain, helping to start energy production.

🎯 Exam Tip: Clearly state its function: "passing electrons and protons from NADH to ubiquinone" at the start of the electron transport chain.

 

Question 10. Which cycle is amphibolic pathway? Why? The Krebs cycle is called an amphibolic pathway.
Answer: The Krebs cycle is considered an amphibolic pathway because it plays a dual role in metabolism:
• The Krebs cycle is primarily a catabolic pathway, meaning it breaks down molecules (like acetyl CoA) to release energy.
• Later, it also acts as an anabolic pathway by providing intermediate compounds that are used to build larger molecules, such as amino acids and nucleic acids. This dual nature of both breakdown and synthesis is why it's called amphibolic.
In simple words: The Krebs cycle is amphibolic because it both breaks down molecules for energy and provides building blocks for new molecules.

🎯 Exam Tip: Define "amphibolic" as having both "catabolic" (breakdown) and "anabolic" (synthesis) functions, and link it to the Krebs cycle's role in providing intermediates.

 

Question 11. Mention any two electron transport chain inhibitors.
Answer: Two common electron transport chain inhibitors are:
1. 2, 4 DNP (Dinitrophenol) – This chemical prevents the making of ATP from ADP. It does this by directing electrons from CoQ to \( O_2 \) without generating a proton gradient, effectively uncoupling electron transport from ATP synthesis.
2. Cyanide – Cyanide stops the flow of electrons from Cytochrome \( a_3 \) to \( O_2 \). This blocks the entire electron transport chain, preventing oxygen from being the final electron acceptor.
In simple words: 2,4 DNP stops ATP from being made by letting electrons flow without making energy. Cyanide blocks electrons from going to oxygen, stopping the whole energy chain.

🎯 Exam Tip: For each inhibitor, name it and briefly explain its mechanism of action within the electron transport chain.

 

Question 12. How many ATP molecules are produced in aerobic respiration present in plants?
Answer: In aerobic respiration, the net gain of ATP molecules produced from the complete oxidation of one glucose molecule in plants is typically 36 ATP molecules. This includes ATP generated through substrate-level phosphorylation and oxidative phosphorylation.
In simple words: Aerobic respiration in plants makes about 36 ATP molecules from one glucose molecule.

🎯 Exam Tip: Remember the total net ATP yield for aerobic respiration in plants, which is often around 36 ATP.

 

Question 13. What are the significances of Respiratory Quotient?
Answer: The Respiratory Quotient (RQ) is important for several reasons:
1. RQ value indicates which type of respiration occurs in living cells, either aerobic or anaerobic. Different substrates lead to different RQ values.
2. It also helps to identify which type of respiratory substrate (like carbohydrates, fats, or proteins) is being used in the process. For example, carbohydrates have an RQ of 1, while fats have an RQ less than 1.
In simple words: Respiratory Quotient shows if breathing uses oxygen or not, and what kind of food (like sugar or fat) is being burned for energy.

🎯 Exam Tip: Focus on the two main significances: identifying the "type of respiration" and the "respiratory substrate."

 

Question 14. Who was awarded the Nobel Prize for coupling of oxidation and phosphorylation in mitochondria?
Answer: Peter Mitchell, a British Biochemist, received the Nobel Prize for Chemistry in 1978. He was awarded for his chemiosmotic theory, which explained how ATP synthesis is coupled with electron transport and proton movement across the mitochondrial membrane. This was a groundbreaking discovery in bioenergetics.
In simple words: Peter Mitchell won the Nobel Prize for explaining how cells use a special process in mitochondria to make energy (ATP).

🎯 Exam Tip: Associate Peter Mitchell with the "chemiosmotic theory" and the mechanism of "ATP synthesis in mitochondria."

 

Question 15. Mention any two industrial uses of alcoholic fermentation.
Answer: Two industrial uses of alcoholic fermentation are:
1. In bakeries, it is used for preparing bread, cakes, and biscuits. The yeast in the dough ferments sugars, producing \( CO_2 \) gas which makes the dough rise, and alcohol which evaporates during baking.
2. In beverage industries, it is used for preparing wine and alcoholic drinks. Yeast ferments sugars in fruits or grains to produce ethanol, which is the alcohol in these drinks.
In simple words: Alcoholic fermentation helps make bread rise and is used to produce wine and other alcoholic beverages.

🎯 Exam Tip: Remember the two main industries: "bakeries" (for leavening) and "beverage industries" (for alcohol production).

 

Question 16. Define mixed acid fermentation.
Answer: Mixed acid fermentation is a process where pyruvic acid is converted into several different organic acids, such as lactic acid, ethanol, and formic acid, along with gases like \( CO_2 \) and \( H_2 \). This type of fermentation is common in bacteria like Enterobacteriaceae. It's a complex process that produces a variety of end products.
• Formation of Lactic acid, ethanol, formic acid and gases like \( CO_2 \) and \( H_2 \) from pyruvic acid.
• For example, it is carried out by bacteria such as Enterobacteriaceae.
In simple words: Mixed acid fermentation turns pyruvic acid into many different acids and gases like lactic acid and ethanol, commonly done by certain bacteria.

🎯 Exam Tip: The key characteristic is the production of "multiple organic acids" (lactic, ethanol, formic) and "gases" from pyruvic acid.

 

Question 17. Mention any two internal factors, that affect the rate of respiration in plants.
Answer: Two internal factors that influence the rate of respiration in plants are:
1. The amount of protoplasm and its state of activity significantly affect the rate of respiration. More active protoplasm means higher respiration.
2. The concentration of respiratory substrate (like glucose) is directly proportional to the rate of respiration. More substrate means more material to break down for energy.
In simple words: The amount of living material (protoplasm) and how much food (substrate) is available inside a plant cell both affect how fast it breathes.

🎯 Exam Tip: Focus on "protoplasm (amount and activity)" and "respiratory substrate concentration" as internal factors.

 

Question 18. Why microorganisms respire an anaerobically?
Answer: Microorganisms often respire anaerobically for specific reasons:
• Bacteria are prokaryotes, which means they do not have membrane-bound organelles like mitochondria. Mitochondria are where aerobic respiration mainly occurs. This structural difference limits their ability for full aerobic respiration.
• So, many of them are forced to respire anaerobically, especially in environments where oxygen is scarce or absent. They adapt to generate energy through fermentation or other anaerobic pathways.
In simple words: Many tiny living things breathe without oxygen because they don't have mitochondria, the cell parts needed for breathing with oxygen.

🎯 Exam Tip: The core reason is the "absence of membrane-bound organelles like mitochondria" in prokaryotic microorganisms.

 

Question 19. Write down any two significance of the pentose phosphate pathway.
Answer: Two significant roles of the pentose phosphate pathway (PPP) are:
• The HMP shunt (another name for PPP) is important for producing two key products: NADPH and pentose sugars. NADPH is vital for reductive biosynthesis reactions and protecting against harmful oxygen-free radicals.
• Coenzyme NADPH, which is made in this pathway, is used for building new molecules (reductive biosynthesis) and helps protect cells by neutralizing damaging oxygen-free radicals. Pentose sugars are used to make DNA and RNA.
In simple words: This pathway makes NADPH, which helps build new cell parts and protects against harm. It also makes special sugars used for DNA and RNA.

🎯 Exam Tip: Remember the two main products and their functions: "NADPH" (reductive biosynthesis, antioxidant) and "pentose sugars" (nucleotide synthesis).

 

Question 20. Complete the following Picture.
A diagram showing a graph titled "Rate of Photosynthesis" and "Carbohydrate balance" over "Time in a day (hours)" with two points labeled (A)-? and (B)-?.
Answer:
A. Compensation point
B. Rate of Respiration
In simple words: On the graph, 'A' marks the point where photosynthesis and respiration are equal, and 'B' represents the rate at which a plant breathes.

🎯 Exam Tip: Understand that the "compensation point" is where the rates of photosynthesis and respiration balance out, often occurring during dawn and dusk.

 

Question 21. Write the missing A and B.
A diagram showing the structure of an ATP molecule with two parts labeled A-? and B-?.
Answer:
A. Ribose
B. Adenine
In simple words: In the ATP molecule, 'A' is the sugar part called ribose, and 'B' is the base part called adenine.

🎯 Exam Tip: Be able to identify the three main components of ATP: adenine (nitrogenous base), ribose (pentose sugar), and three phosphate groups.

III. 3 Mark Questions

 

Question 1. In the biosphere how do plants and animals are complementary systems, which are integrated to sustain life?
Answer: Plants and animals work together as complementary systems in the biosphere to support life. Plants take in carbon dioxide and release oxygen through photosynthesis, producing carbohydrates. Animals, including humans, inhale this oxygen and use it for cellular respiration, which breaks down carbohydrates to produce energy and releases carbon dioxide. This constant exchange forms a vital cycle where each group provides what the other needs, maintaining balance on Earth. Oxygen is transported through the blood to cells, where energy is obtained through cellular respiration.
In simple words: Plants give out oxygen and take in carbon dioxide, and animals do the opposite. This way, they help each other breathe and keep life going on Earth.

🎯 Exam Tip: Explain the key processes: "photosynthesis" in plants (takes \( CO_2 \), releases \( O_2 \)) and "respiration" in animals (takes \( O_2 \), releases \( CO_2 \)), highlighting their interdependence.

 

Question 2. What is Respiration?
Answer: Respiration is a fundamental biological process where complex organic compounds, such as carbohydrates, are broken down through oxidation inside cells. This breakdown involves releasing energy, which is then stored in ATP molecules, along with the release of heat. Respiration is crucial for all living organisms, as it provides the energy needed for various cellular activities. It also occurs in all living cells of organisms.
• Breaking of C-C bonds of complex organic compounds through oxidation within the cells.
• The energy released during respiration is stored in the form of ATP and heat is liberated.
• It occurs in all the living cells of organisms.
In simple words: Respiration is how cells break down food to get energy. It happens in all living cells, making ATP and giving off heat.

🎯 Exam Tip: For a comprehensive answer, define respiration, explain energy storage (ATP), heat liberation, and its universal occurrence in living cells.

 

Question 3. What are the factors associated with the compensation point in respiration?
Answer: The compensation point in respiration is mainly influenced by two common factors: carbon dioxide concentration and light intensity. There are two types of compensation points:
1. The \( CO_2 \) compensation point: This is the \( CO_2 \) concentration at which the rate of photosynthesis equals the rate of respiration. For C3 plants, this range is typically 40–60 ppm (parts per million), while for C4 plants, it is much lower, around 1–5 ppm \( CO_2 \).
2. The light compensation point: This is the light intensity at which the rate of photosynthesis exactly matches the rate of respiration. Above this point, the plant gains biomass; below it, it loses biomass.
In simple words: The compensation point depends on how much carbon dioxide is in the air and how much light there is. Different plant types have different balance points.

🎯 Exam Tip: Distinguish between \( CO_2 \) compensation point and light compensation point, and briefly mention the difference between C3 and C4 plants regarding their \( CO_2 \) compensation points.

 

Question 4. Differentiate floating respiration and protoplasmic respiration.
Answer:

In simple words: Floating respiration uses sugars or fats as fuel and is common, while protoplasmic respiration uses proteins, is less common, and can make toxic ammonia.

🎯 Exam Tip: Focus on the type of "respiratory substrate" (carbohydrates/fats vs. proteins) and whether "toxic products" are formed for differentiation.

 

Question 5. What is a redox reaction?
Answer: A redox reaction, or oxidation-reduction reaction, involves the transfer of electrons between molecules. In biological systems, this often means that one molecule gains electrons (is reduced) while another loses electrons (is oxidized). For example:
\( NAD^+ + 2e^- + 2H^+ \rightarrow NADH + H^+ \)
\( FAD + 2e^- + 2H^+ \rightarrow FADH_2 \)
When \( NAD^+ \) (oxidized form) and FAD pick up electrons and protons, they get reduced to NADH + H\( ^+ \) and \( FADH_2 \), respectively. When they release these electrons and hydrogens, they return to their original, oxidized forms. These electron transfers are vital for energy production in cellular respiration.
In simple words: A redox reaction is when molecules swap electrons; one molecule gains electrons (reduction) and another loses them (oxidation). This is important for energy in cells.

🎯 Exam Tip: Define redox reaction as "transfer of electrons," clarifying that "reduction is gain" and "oxidation is loss" of electrons. Use NAD/FAD as examples.

 

Question 6. Define ETS (or) Electron transport chain (or) What is the importance of ETS and oxidative Phosphorylation in respiration.
Answer: The Electron Transport System (ETS) or electron transport chain (ETC) and oxidative phosphorylation are critical parts of respiration. They work together to produce a large amount of ATP, the cell's energy currency. The importance is:
• The electron transport chain and oxidative phosphorylation remove hydrogen atoms (as electrons and protons) from the products of glycolysis, the link reaction, and the Krebs cycle. This is a vital step in cellular energy release.
• This process releases water molecules and a significant amount of energy in the form of ATP molecules. This happens on the inner membrane of the mitochondria, making it the primary site of ATP generation in aerobic respiration.
In simple words: The electron transport chain and oxidative phosphorylation take hydrogen atoms from energy-making steps and use them to create a lot of ATP (energy) and water inside the cell's powerhouses.

🎯 Exam Tip: Emphasize the removal of "hydrogen atoms" from glycolysis/Krebs cycle intermediates and the subsequent "release of ATP and water" on the "inner mitochondrial membrane."

 

Question 7. The significance of Kreb's cycle:
Answer: The Krebs cycle, also known as the citric acid cycle, holds several key significances:
1. The TCA cycle (Krebs cycle) provides energy in the form of ATP, which is essential for the metabolism of plants.
2. It offers carbon skeletons or raw materials that are used in various anabolic processes, helping to build new molecules.
3. Many intermediate compounds of the TCA cycle are further processed to produce important molecules like amino acids, proteins, and nucleic acids.
4. Succinyl CoA, an intermediate, serves as a raw material for making chlorophylls, cytochrome, phytochrome, and other pyrrole substances, which are vital for plant functions.
5. \( \alpha \)-ketoglutarate and oxaloacetate, also intermediates, undergo reductive amination to produce amino acids, contributing to protein synthesis.
6. It acts as a central metabolic hub, or "metabolic sink," playing a crucial role in intermediary metabolism by connecting various biochemical pathways.
In simple words: The Krebs cycle is important because it makes energy (ATP), provides building blocks for new molecules like amino acids and chlorophyll, and links different cell processes.

🎯 Exam Tip: List both the "energy production" (ATP) and "provision of raw materials/intermediates" for synthesis as its main significances.

 

Question 8. Write the differences between ubiquinone and Cytochrome C.
Answer:

In simple words: Ubiquinone is a small fat-like carrier inside the mitochondrial membrane, while Cytochrome C is a small protein on the outer part of the inner membrane. They both move electrons but are different in what they are made of and where they are located.

🎯 Exam Tip: Focus on "solubility (lipid-soluble vs. protein)," "location" (within vs. outer surface of inner membrane), and "associated complex" (Complex I vs. Complex III) for differentiation.

 

Question 9. Write down the characteristic of Anaerobic respiration.
Answer: The characteristics of Anaerobic respiration are:
1. Anaerobic respiration is less efficient in producing energy compared to aerobic respiration. It yields significantly fewer ATP molecules.
2. Only a limited number of ATP molecules are generated per glucose molecule, typically just 2 ATP, which is much lower than aerobic respiration.
3. It is characterized by the production of \( CO_2 \) and is also utilized for carbon fixation in photosynthesis in some organisms. The end products are usually alcohol or lactic acid.
In simple words: Anaerobic respiration makes less energy, produces only a few ATPs per sugar molecule, and can create carbon dioxide or other products like alcohol.

🎯 Exam Tip: Key characteristics include "lower efficiency" (fewer ATP), "limited ATP yield," and the nature of "end products" (alcohol, lactic acid, \( CO_2 \)).

 

Question 10. RQ will be less than one in Red colour Parts Present in Plants? Why?
Answer: In red-colored parts of plants, the Respiratory Quotient (RQ) is typically less than one. This happens because:
• The red color in plants is due to the presence of a pigment called anthocyanin. Anthocyanins are often produced from organic acids, which require more oxygen for their synthesis or metabolism.
• The synthesis of anthocyanin requires more oxygen \( (O_2) \) than the carbon dioxide \( (CO_2) \) evolved. When organic acids are the primary substrate, more oxygen is consumed relative to carbon dioxide produced. This higher oxygen consumption leads to an RQ value below 1.
• So, in these parts, the RQ will be less than one due to the metabolic activities related to anthocyanin and organic acid breakdown.
In simple words: Red plant parts have an RQ less than one because they make a red color pigment called anthocyanin, which uses more oxygen than it releases carbon dioxide.

🎯 Exam Tip: Link "red color" to "anthocyanin synthesis" and explain that this process consumes "more \( O_2 \) than \( CO_2 \) evolved," leading to an RQ less than 1.

 

Question 11. Write down any three external factors, that affect respiration in plants.
Answer: Three external factors that influence the rate of respiration in plants are:
1. Temperature: The optimum temperature for respiration is typically around 30\( ^\circ \)C. Both very low and very high temperatures can decrease the rate of respiration, as enzymes involved in the process are temperature-sensitive.
2. Oxygen concentration: When a sufficient amount of \( O_2 \) is available, the rate of aerobic respiration will be optimal. Conversely, anaerobic respiration is completely stopped under high oxygen levels. The point at which anaerobic respiration stops is called the extinction point.
3. Carbon dioxide concentration: A high concentration of \( CO_2 \) in the atmosphere can reduce the rate of respiration. This is because high \( CO_2 \) levels can inhibit certain respiratory enzymes or shift the equilibrium of metabolic reactions.
In simple words: Three outside factors that change how fast plants breathe are temperature (too hot or cold slows it down), oxygen (enough oxygen makes it fast), and carbon dioxide (too much slows it down).

🎯 Exam Tip: Remember temperature (optimum, extremes), oxygen concentration (sufficiency, extinction point), and carbon dioxide concentration (inhibitory effect).

 

Question 12. Define Lactic acid fermentation.
Answer: Lactic acid fermentation is a metabolic process where pyruvic acid is converted into lactic acid. This process occurs in the absence of oxygen and is a way for organisms to produce ATP when aerobic respiration is not possible. For example, it happens in Bacillus bacteria, some fungi, and in the muscles of vertebrates during intense exercise when oxygen supply is low. The chemical equation is:
\( 2CH_3COCOOH + 2NADH + 2H^+ \xrightarrow{\text{Lactate dehydrogenase}} 2CH_3CHOHCOOH + 2NAD^+ \)
Pyruvic acid Lactic acid
In simple words: Lactic acid fermentation changes pyruvic acid into lactic acid without oxygen. This happens in some bacteria and in our muscles when we exercise hard.

🎯 Exam Tip: Key points are the conversion of "pyruvic acid to lactic acid," "absence of oxygen," and common examples like "muscle cells" or "Bacillus bacteria."

 

Question 13. How alcoholic beverages like beer and wine is made?
Answer: Alcoholic beverages such as beer and wine are produced through a process called fermentation, primarily carried out by yeast. Here's how:
• The conversion of pyruvate to ethanol occurs in malted barley for beer and grapes for wine. This is the core biochemical reaction of fermentation.
• Yeast performs this process under anaerobic conditions (without oxygen). During fermentation, yeast consumes sugars and converts them into ethanol (alcohol) and carbon dioxide.
• As the ethanol concentration increases, it eventually becomes toxic to the yeast cells, which then die. The remaining liquid is collected as beer or wine.
In simple words: Beer and wine are made by yeast, which eats sugars from barley or grapes without oxygen. Yeast turns the sugar into alcohol and carbon dioxide, and when there's too much alcohol, the yeast dies.

🎯 Exam Tip: Highlight the role of "yeast," "anaerobic conditions," and the conversion of "pyruvate to ethanol" from specific raw materials like barley or grapes.

 

Question 14. Tabulate the comparison between glycolysis and fermentation.
Answer:

In simple words: Glycolysis turns glucose into pyruvic acid and makes some ATP, working with or without oxygen. Fermentation takes pyruvic acid and makes alcohol or lactic acid without making more ATP, always without oxygen.

🎯 Exam Tip: Compare the "starting and end products," "oxygen requirement," "ATP yield," "NADH utilization," and "organisms/cells where it occurs" for a complete comparison.

IV. 5 Mark Questions

 

Question 1. Ambulate the differences between aerobic and anaerobic respiration.
Answer:

In simple words: Aerobic respiration uses oxygen, happens in complex living things, makes a lot of ATP, and has many steps. Anaerobic respiration doesn't use oxygen, happens in simpler organisms like yeast, makes very little ATP, and has only two main steps.

🎯 Exam Tip: Focus on the presence/absence of "oxygen," "ATP yield," "stages involved," "end products," and "location" within the cell for clear differentiation.

 

Question 2. Draw the flow chart diagram For Glycolysis (or) EMP pathway.
Answer: The answer requires drawing a detailed flow chart diagram of the Glycolysis (EMP pathway), illustrating all the intermediate steps, enzymes, and energy changes involved.
In simple words: Draw a step-by-step diagram showing how sugar is broken down into pyruvic acid in the glycolysis pathway.

🎯 Exam Tip: When drawing, ensure all intermediates, enzymes, ATP, and NADH consumption/production are correctly labeled in their respective steps.

 

Question 3. Explain the Pay-off phase (oxidative phase or triose phase) of Glycolysis (EMP Pathway).
Answer: The pay-off phase of glycolysis involves several steps to produce energy and pyruvate.

• Two molecules of glyceraldehyde-3-phosphate are changed through oxidation and phosphorylation into two molecules of 1,3-bisphosphoglycerate. This step is important for capturing energy.
• During this reaction, two molecules of NAD+ are converted into two molecules of NADH + H+ by the glyceraldehyde-3-phosphate dehydrogenase enzyme. NADH acts as an electron carrier.
• After this, other enzymes continue the reactions, eventually creating two molecules of pyruvate.
• In this phase, a total of 4 ATP molecules are made at step 7 and step 10.
• The direct transfer of a phosphate group from a substrate molecule to ADP creates ATP; this process is called substrate phosphorylation or transphosphorylation.
• At step 9, two molecules of phosphoglycerate lose water to become phosphoenol pyruvate, with the help of the enolase enzyme.
• This leads to the formation of an enol group within the molecule, a process called enolation.

**Energy Budget of Pay-off Phase:**

• In the payoff phase, a total of 4 ATP and 2 NADH + H+ molecules are produced.
• Since 2 ATP molecules were used up in the preparatory phase, the net output from glycolysis is 2 ATP and 2 NADH + H+.

In simple words: This is the energy-making part of glycolysis where the sugar molecule breaks down further. It produces ATP and NADH, which are like small energy packets for the cell, and the final product is pyruvate.

🎯 Exam Tip: Remember to clearly distinguish between the ATP produced and the net ATP gain, accounting for the ATP consumed in the preparatory phase.

 

Question 4. Explain the preparatory phase of Glycolysis (or) EMP pathway (or) Describe the energonic phase phase of Glycolysis (or) EMP pathway. Describe the hexose phase of Glycolysis (or) EMP pathway.
Answer: The preparatory phase of glycolysis uses energy to prepare glucose for breakdown.

• Glucose, which is the final product of photosynthesis, enters the glycolysis pathway.
• It is first changed into glucose-6-phosphate through a process called phosphorylation, with the help of the enzyme hexokinase. Other enzymes then carry out the next reactions.
• Towards the end of this phase, fructose-1,6-bisphosphate is split into two smaller molecules: glyceraldehyde-3-phosphate and dihydroxyacetone phosphate, a reaction catalyzed by the enzyme aldolase.
• These two molecules are isomers, meaning they have the same chemical formula but different arrangements of atoms.
• Dihydroxyacetone phosphate is then converted into glyceraldehyde-3-phosphate by the enzyme triose phosphate isomerase, so that both molecules can continue in the pathway.
• After these steps, two molecules of glyceraldehyde-3-phosphate are ready to enter the pay-off phase of glycolysis. During this preparatory phase, the cell uses up two ATP molecules.

In simple words: This is the first part of glycolysis where the cell uses a little energy to change glucose into a form that can be broken down later. It's like preparing the fuel before burning it.

🎯 Exam Tip: Focus on the energy investment (ATP consumption) and the splitting of the 6-carbon sugar into two 3-carbon molecules in the preparatory phase.

 

Question 5. Explain pyruvate oxidation (or) Link reaction of Glycolysis.
Answer: Pyruvate oxidation, also called the link reaction, connects glycolysis to the Krebs cycle.

• Two molecules of pyruvate, which are formed during glycolysis in the cytoplasm, then move into the mitochondria.
• In aerobic respiration, each pyruvate molecule combines with coenzyme A. It undergoes oxidative decarboxylation, meaning it loses a carbon atom as \( \text{CO}_2 \) and is oxidized. This reaction forms acetyl CoA. This conversion is handled by a group of enzymes called the pyruvate dehydrogenase complex.
• This process produces two molecules of NADH + H+ and two molecules of \( \text{CO}_2 \) for every glucose molecule (since one glucose makes two pyruvates).
• This reaction is called the transition reaction or link reaction because it links glycolysis, which happens in the cytoplasm, with the Krebs cycle, which happens inside the mitochondria.

In simple words: Pyruvate oxidation is like a bridge. It changes the pyruvate made in glycolysis into acetyl CoA, which is then ready to enter the main energy-making cycle in the cell's powerhouses (mitochondria).

🎯 Exam Tip: Remember that the link reaction produces acetyl CoA, \( \text{CO}_2 \), and NADH, and it's the crucial transition step between glycolysis and the Krebs cycle.

 

Question 6. Draw the flow chart diagram For the Kreb cycle (or) Citric acid cycle.
Answer: The Krebs cycle, also known as the Citric Acid Cycle or TCA cycle, is a series of chemical reactions that play a central role in the energy production of cells. It starts with acetyl CoA combining with oxaloacetate to form citrate. This cycle involves several key steps:

1.  **Condensation:** Acetyl CoA joins with oxaloacetate (4 carbons) to form citrate (6 carbons).
2.  **Isomerization:** Citrate is converted to isocitrate (still 6 carbons) through an intermediate.
3.  **Oxidative Decarboxylation:** Isocitrate is oxidized and decarboxylated (loses \( \text{CO}_2 \)) to form \( \alpha \)-ketoglutarate (5 carbons), producing NADH.
4.  **Another Oxidative Decarboxylation:** \( \alpha \)-ketoglutarate is further oxidized and decarboxylated (loses another \( \text{CO}_2 \)) to form succinyl CoA (4 carbons), producing another NADH.
5.  **Substrate-Level Phosphorylation:** Succinyl CoA is converted to succinate, directly producing ATP (or GTP).
6.  **Oxidation:** Succinate is oxidized to fumarate, producing FADH2 (another electron carrier).
7.  **Hydration:** Fumarate is hydrated (water is added) to form malate.
8.  **Final Oxidation:** Malate is oxidized back to oxaloacetate, producing NADH and regenerating the starting molecule for the cycle to continue.

These steps involve several enzymes like citrate synthase, aconitase, isocitrate dehydrogenase, \( \alpha \)-ketoglutarate dehydrogenase complex, succinyl CoA synthetase, succinate dehydrogenase, fumarase, and malate dehydrogenase. The cycle efficiently breaks down acetyl CoA, releasing energy in the form of ATP, NADH, and FADH2, and producing carbon dioxide as a byproduct.In simple words: The Krebs cycle is a main energy-making process in the cell. It takes acetyl CoA and breaks it down step by step, creating energy carriers like NADH and FADH2, along with some ATP and carbon dioxide. Think of it as a revolving wheel that helps cells get energy from food.

🎯 Exam Tip: While drawing, ensure all intermediates, enzymes, and cofactors like NADH, FADH2, and \( \text{CO}_2 \) release points are correctly placed for full marks.

 

Question 7. What is the significance of Kreb's cycle (or) TCA cycle?
Answer: The Krebs cycle, also called the TCA cycle, is very important for many reasons in the cell.

1.  It directly provides energy in the form of ATP for the plant's metabolism, powering various cellular activities.
2.  It also gives carbon skeletons or raw materials that are needed for various anabolic processes, which are building-up reactions in the cell.
3.  Many intermediate compounds from the TCA cycle are further processed and changed to create important molecules like amino acids, proteins, and nucleic acids.
4.  Succinyl CoA, one of the intermediate products of the cycle, is essential for making chlorophylls (for photosynthesis), cytochrome (for electron transport), phytochrome, and other pyrrole substances.
5.  Compounds like \( \alpha \)-ketoglutarate and oxaloacetate can undergo a process called reductive amination to produce amino acids, which are the building blocks of proteins.
6.  The cycle acts as a central point for many different metabolic pathways, making it a key part of how the cell manages its chemical reactions.

In simple words: The Krebs cycle is vital because it makes energy (ATP) and provides basic ingredients for building all sorts of important molecules in the cell, like proteins and DNA. It's like the central hub for many cell activities.

🎯 Exam Tip: When discussing significance, always mention both energy production (ATP) and its role in providing precursors for biosynthesis (amino acids, chlorophyll, etc.) as these are its dual roles.

 

Question 8. Significance of Kreb Cycle.
Answer: The Krebs cycle has several important roles in cell metabolism.

1.  It helps provide energy in the form of ATP for plant cell activities.
2.  It supplies carbon structures or basic materials needed for various building-up processes in the body.
3.  Many of the substances made during the TCA cycle are further changed to produce important things like amino acids, proteins, and nucleic acids.
4.  Succinyl CoA, which is a key part of this cycle, is a necessary ingredient for making chlorophylls, cytochrome, phytochrome, and other related compounds.
5.  Compounds like \( \alpha \)-ketoglutarate and oxaloacetate can be used to make amino acids through a process called reductive amination.
6.  The Krebs cycle acts like a main junction in metabolism, connecting many different chemical pathways together.

In simple words: The Krebs cycle is crucial because it helps cells create energy (ATP) and provides essential building blocks for making other important cell parts like proteins and DNA. It's a central hub for many life processes.

🎯 Exam Tip: Highlight that the Krebs cycle is an amphibolic pathway, meaning it's involved in both breaking down (catabolic) and building up (anabolic) molecules, making it versatile.

 

Question 9. Write four Electron transport chain inhibitors.
Answer: Here are four substances that stop the electron transport chain:

1.  **2,4 DNP (Dinitrophenol):** This chemical prevents ATP from being made from ADP by changing how electrons flow from coenzyme Q (CoQ) to oxygen (O2). It basically uncouples electron transport from ATP synthesis.
2.  **Cyanide:** It prevents electrons from moving from cytochrome a3 to \( \text{O}_2 \), which effectively stops the entire electron transport chain and blocks respiration.
3.  **Rotenone:** This substance specifically blocks the flow of electrons from NADH + H+ to coenzyme Q (CoQ) in Complex I of the chain.
4.  **Oligomycin:** It directly inhibits the ATP synthase enzyme (Complex V), which is responsible for oxidative phosphorylation, preventing the synthesis of ATP.

In simple words: These chemicals are like roadblocks that stop the cell's energy-making factory (electron transport chain) from working. Each one blocks a different step, preventing ATP from being made.

🎯 Exam Tip: For each inhibitor, specify which complex or specific step of the electron transport chain it affects. This shows a deeper understanding of the mechanism.

 

Question 10. Tabulate Net Products of ATP gained during aerobic respiration per glucose molecule.
Answer: The table below shows the net products of ATP generated during aerobic respiration from one glucose molecule. This summary helps us understand the energy yield at each stage of the process.

In simple words: This table shows all the energy (ATP) that a cell gets from completely breaking down one sugar molecule using oxygen. It helps us see where the energy is made at each part of the process.

🎯 Exam Tip: Ensure that the ATP equivalents for NADH and FADH2 are correctly used (typically 2.5 or 3 ATPs for NADH, and 1.5 or 2 ATPs for FADH2, depending on the specific model taught in the curriculum). The values shown in the table (2x2, 2x3, 6x3, 2x2) correctly reflect a 3 ATP per NADH and 2 ATP per FADH2 model which sums to 36.

 

Question 11. Explain the experiment to demonstrate the production of \( \text{CO}_2 \) in aerobic respiration.
Answer: Here's how to show that carbon dioxide is made during aerobic respiration:

1.  First, take a small quantity of any seeds, like groundnut or bean seeds, and allow them to germinate by soaking them in water. Germinating seeds respire actively.
2.  Once they are germinating, place these seeds inside a conical flask.
3.  Hang a small glass tube inside the flask. This tube should contain 4 ml of freshly prepared Potassium hydroxide (KOH) solution, supported by a thread. Then, close the flask tightly with a cork that has only one hole. KOH is used to absorb carbon dioxide.
4.  Next, take a bent glass tube. Insert its shorter end into the conical flask through the hole in the cork.
5.  Dip the longer end of the bent glass tube into a beaker filled with water.
6.  Note down where the water level initially stands in the bent glass tube.
7.  Keep this whole setup undisturbed for about two hours.
8.  After two hours, you will see the water level in the bent glass tube has risen. This happens because the germinating seeds produce carbon dioxide during aerobic respiration, and the KOH solution inside the flask absorbs this \( \text{CO}_2 \). This absorption reduces the gas pressure inside the flask, causing the water from the beaker to be pushed up the bent tube. The chemical reaction is \( \text{CO}_2 + 2\text{KOH} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \).

In simple words: This experiment uses germinating seeds and KOH to show that living things breathe out carbon dioxide. When the KOH sucks up the carbon dioxide from the seeds, the water level in a tube goes up, proving the gas was there.

🎯 Exam Tip: Clearly state the role of each component: germinating seeds (respiring organism), KOH (absorbs \( \text{CO}_2 \)), and the bent tube/beaker (indicates pressure change and \( \text{CO}_2 \) absorption).

 

Question 12. Define Lactic acid fermentation.
Answer: Lactic acid fermentation is a biological process where pyruvic acid is converted into lactic acid. This process occurs in the absence of oxygen. The reaction for this fermentation can be written as:
\[ \text{2CH}_3\text{COCOOH (Pyruvic acid)} + \text{2NADH} + \text{2H}^+ \xrightarrow{\text{Lactate dehydrogenase}} \text{2CH}_3\text{CHOHCOOH (Lactic acid)} + \text{2NAD}^+ \] This type of fermentation happens in organisms like Bacillus bacteria, fungi, and in the muscles of vertebrates when oxygen supply is low, for example, during intense exercise. It is important for producing foods like yogurt and cheese.In simple words: Lactic acid fermentation is a process where cells change pyruvic acid into lactic acid when there's no oxygen. It happens in some bacteria and in our muscles during hard exercise.

🎯 Exam Tip: Highlight that lactic acid fermentation does not produce \( \text{CO}_2 \) and regenerates \( \text{NAD}^+ \) so glycolysis can continue, providing a small amount of ATP.

 

Question 13. Tabulate the differences between Alcoholic fermentation and Lactic acid fermentation.
Answer: Here is a comparison of alcoholic fermentation and lactic acid fermentation. Both are types of anaerobic respiration, but they produce different end products.

In simple words: These are two ways cells make energy without oxygen, but they create different things. One makes alcohol and gas, while the other makes only lactic acid.

🎯 Exam Tip: Pay attention to the number of steps involved (two for alcoholic, one for lactic acid), the enzymes, and whether \( \text{CO}_2 \) is released. These are key distinguishing features.

 

Question 13. Write the Industrial uses of alcoholic fermentation.
Answer: Alcoholic fermentation has several important uses in industries.

1.  In bakeries, it is used to prepare bread, cakes, and biscuits because the carbon dioxide gas produced during fermentation helps the dough to rise and become light and airy.
2.  In beverage industries, it is crucial for making alcoholic drinks like wine, beer, and spirits. Yeast converts sugars into ethanol and \( \text{CO}_2 \).
3.  It is also used in producing vinegar, which is made from alcoholic solutions, and in tanning and curing leather, where it helps in the preservation process.
4.  Ethanol, a product of this fermentation, is used to make gasohol, a mixture of gasoline and ethanol that serves as a fuel in cars, especially in countries like Brazil.

In simple words: Alcoholic fermentation is very useful! It helps make bread fluffy, creates alcoholic drinks, produces vinegar, and even provides fuel for cars.

🎯 Exam Tip: When listing industrial uses, clearly state the product and the industry it benefits. For example, "Bread (bakeries)" rather than just "bread."

 

Question 14. Tabulate the comparison between glycolysis and fermentation.
Answer: Here's a comparison between glycolysis and fermentation, two key processes in cellular respiration. Glycolysis is the first step, while fermentation happens afterward in the absence of oxygen.

In simple words: Glycolysis is the first stage where sugar breaks down, making some energy and pyruvate. Fermentation happens next when there's no oxygen, changing pyruvate into other products like alcohol or lactic acid without making much more energy.

🎯 Exam Tip: The key difference is that glycolysis is common to both aerobic and anaerobic respiration, while fermentation is specifically an anaerobic process that follows glycolysis to regenerate \( \text{NAD}^+ \).

 

Question 15. Explain the demonstration of alcoholic fermentation.
Answer: Here's how to demonstrate alcoholic fermentation using a Kuhne's fermentation tube:

1.  First, get a Kuhne's fermentation tube, which is a glass tube with an upright section and a side bulb.
2.  Pour a 10% sugar solution mixed with baker's yeast into the fermentation tube. Fill the side tube, and then plug the mouth with a lid. Yeast contains the zymase enzyme that breaks down sugar.
3.  After some time, the glucose solution will undergo fermentation. You will notice an alcoholic smell coming from the solution.
4.  The level of the solution in the upright glass column will fall. This happens because carbon dioxide gas is produced during fermentation and accumulates in the top part of the tube. This gas pushes the liquid down.
5.  Now, introduce a pellet of KOH into the tube. The KOH will absorb the \( \text{CO}_2 \) gas. As the carbon dioxide is absorbed, the gas pressure inside decreases, causing the liquid level in the upright tube to rise again.
6.  This experiment clearly shows that \( \text{CO}_2 \) gas is produced during fermentation and that yeast converts sugar into alcohol and \( \text{CO}_2 \).

In simple words: This experiment uses a special tube, sugar water, and yeast to show how fermentation works. It proves that yeast makes alcohol and carbon dioxide gas from sugar, which you can see by the liquid level changing and smell the alcohol.

🎯 Exam Tip: Note the role of each component: Kuhne's tube (to observe gas production), sugar solution (substrate), yeast (source of enzymes), and KOH (to confirm \( \text{CO}_2 \) production by absorption and subsequent liquid level change).

 

Question 16. Write the Factors (Internal and External) which affect the process of respiration.
Answer: Many things can affect how fast respiration happens in plants, both from outside and inside the plant. **External Factors:**

1.  **Temperature:** The best temperature for respiration is around 30°C. If the temperature is too low or very high, the rate of respiration will slow down significantly because enzymes work best at optimum temperatures.
2.  **Oxygen Availability:** When there is enough oxygen, aerobic respiration works at its best, and anaerobic respiration stops completely. This specific point is called the "extinction point" for anaerobic respiration.
3.  **Carbon Dioxide Concentration:** A very high amount of carbon dioxide in the air can slow down the rate of respiration. This is because high \( \text{CO}_2 \) concentrations can inhibit some respiratory enzymes.

**Internal Factors:**

1.  **Amount of Protoplasm:** The amount of living material (protoplasm) in a cell and its state of activity influence the rate of respiration. More active protoplasm usually means a higher respiration rate.
2.  **Respiratory Substrate Concentration:** The amount of raw materials (like carbohydrates, fats, or proteins) available for respiration directly affects its rate. If there's more substrate, respiration can happen faster.
3.  **Water Content:** If a plant or tissue is moved from water to a salt solution, its respiration rate can increase. This phenomenon is sometimes referred to as silt respiration.
4.  **Light:** Light doesn't directly affect respiration but plays an indirect role. It influences photosynthesis, which in turn produces the glucose that acts as the primary substrate for respiration.
5.  **Wounding:** When plant organs are wounded or damaged, it often stimulates an increased rate of respiration in that specific area as part of the healing and repair process.

In simple words: Respiration in plants is affected by things outside like how hot it is, how much oxygen and carbon dioxide are around. Inside the plant, things like how much living material, how much food (sugar), and even if it's wounded can change how fast it breathes.

🎯 Exam Tip: When listing factors, specify whether they are internal or external. For each factor, explain *how* it affects the rate (increase, decrease, optimal point) and provide a brief reason if possible.

 

Question 17. Write about the alternate pathway for glucose break down (or) Write about pentose phosphate pathway. (or) Phosphogluconate pathway (or) Warburg-Dickens-Lipmann pathway (or) Hexose Monophosphate pathway (or) HMP Shunt pathway.
Answer: The Pentose Phosphate Pathway (PPP), also known by several other names like Hexose Monophosphate Shunt (HMP Shunt) or Warburg-Dickens-Lipmann pathway, is an alternative way for glucose to be broken down.

1.  This pathway was first described by scientists Warburg, Dickens, and Lipmann in 1938, hence one of its names.
2.  It primarily happens in the cytoplasm of mature plant cells and provides another route for glucose breakdown besides glycolysis.
3.  The pathway has two main parts: an oxidative phase and a non-oxidative phase.
4.  In the **oxidative phase**, six molecules of six-carbon glucose-6-phosphate are converted into six molecules of five-carbon sugar ribulose-5-phosphate. This process also releases 12 NADPH + H+ (which is different from NADH) and 6 \( \text{CO}_2 \). NADPH is crucial for reductive biosynthesis (building complex molecules) and protecting cells from oxidative damage.
5.  The **non-oxidative pathway** then changes ribulose-5-phosphate molecules into various intermediate sugars like ribose-5-phosphate (5C), xylulose-5-phosphate (5C), glyceraldehyde-3-phosphate (3C), sedoheptulose-7-phosphate (7C), and erythrose-4-phosphate (4C). These intermediate sugars are important for other metabolic pathways.
6.  Finally, five molecules of glucose-6-phosphate are remade, allowing the cycle to continue or providing intermediates for glycolysis.
7.  The overall reaction for the complete oxidation of one glucose-6-phosphate molecule through this pathway yields 6 \( \text{CO}_2 \) and 12 NADPH + H+.

In simple words: The Pentose Phosphate Pathway is a special way cells break down sugar to get two main things: NADPH, which helps build new cell parts and fight damage, and different kinds of sugars that are used to make DNA and other important molecules. It's an alternative to the usual sugar-breaking path.

🎯 Exam Tip: Emphasize that the primary outputs of the HMP shunt are NADPH (for biosynthesis and antioxidant defense) and pentose sugars (for nucleotide synthesis), distinguishing it from glycolysis and the Krebs cycle's main role in ATP production.

 

Question 18. Draw the cycle for pentose phosphate pathway (or) Draw the flow chart for HMP Shunt.
Answer: The pentose phosphate pathway (HMP Shunt) is a complex metabolic route that branches off from glycolysis. Instead of drawing the entire detailed flowchart, here is a description of its main features and purpose. This pathway involves a series of sugar transformations that generate NADPH and pentose sugars.

1.  **Glucose-6-Phosphate Dehydrogenation:** The first major step involves the enzyme glucose-6-phosphate dehydrogenase, which oxidizes glucose-6-phosphate to 6-phosphogluconolactone. This crucial reaction produces the first molecule of NADPH.
2.  **Hydrolysis:** The 6-phosphogluconolactone is then hydrolyzed (water is added) to form 6-phosphogluconate.
3.  **Oxidative Decarboxylation:** 6-phosphogluconate is further oxidized and decarboxylated (loses \( \text{CO}_2 \)) to yield ribulose-5-phosphate. This step generates the second molecule of NADPH. This marks the end of the oxidative phase.
4.  **Isomerization:** Ribulose-5-phosphate can be isomerized to either ribose-5-phosphate (which is vital for nucleotide and nucleic acid synthesis) or xylulose-5-phosphate.
5.  **Non-oxidative Interconversions:** In this phase, various enzymes like transketolase and transaldolase perform a series of interconversions. They rearrange the carbon skeletons of these five-carbon sugars into three-carbon, four-carbon, six-carbon, and seven-carbon sugars. These reactions ultimately regenerate glucose-6-phosphate, allowing the cycle to continue, or they provide intermediates that can enter glycolysis.

This cycle is important for providing building blocks for nucleic acids and for generating reducing power in the form of NADPH.In simple words: The Pentose Phosphate Pathway is like a special path that sugar takes in the cell. It's a bit complicated, but its main job is to make special energy packets called NADPH and different types of sugars that are used to build important cell parts like DNA.

🎯 Exam Tip: When asked to "draw" a complex cycle, a detailed description of key steps, inputs, outputs, and major enzymes can be an acceptable alternative if an actual drawing is impractical or highly complex. Always specify the purpose of the cycle's products.

 

Question 19. Write about the Significance of Pentose Phosphate pathway.
Answer: The Pentose Phosphate Pathway (PPP) is very important for several reasons in the cell.

1.  The HMP shunt, as it's also known, creates two crucial products: NADPH and pentose sugars. These play a vital role in processes that build up molecules (anabolic reactions).
2.  The NADPH produced is used for reductive biosynthesis, meaning it helps in creating new molecules like fatty acids and steroids. It also protects cells by counteracting the harmful effects of oxygen-free radicals, thus acting as an antioxidant. This helps maintain cell health and prevents damage.
3.  Pentose sugars like ribose-5-phosphate and its related compounds are essential for making important molecules like DNA (deoxyribonucleic acid), RNA (ribonucleic acid), ATP (adenosine triphosphate), NAD (nicotinamide adenine dinucleotide), FAD (flavin adenine dinucleotide), and Coenzyme A. These are fundamental for genetic information, energy, and metabolism.
4.  Erythrose-4-phosphate, another product of this pathway, is used for the synthesis of anthocyanin (plant pigments), lignin (a component of plant cell walls), and other aromatic compounds.

In simple words: The Pentose Phosphate Pathway is super important because it makes special energy carriers (NADPH) for building cell parts and protecting the cell. It also makes specific sugars that are the building blocks for DNA, RNA, and other key molecules.

🎯 Exam Tip: Focus on the dual significance of the PPP: production of NADPH (for anabolic reactions and antioxidant defense) and the synthesis of pentose sugars (for nucleotide and nucleic acid biosynthesis).