RBSE Solutions Class 12 Biology Chapter 11 Respiration

RBSE Class 12 Biology Chapter 11 Multiple Choice Questions

 

Question 1. What is the site of the Krebs cycle in cells?
(a) Nucleus
(b) Cytoplasm
(c) Plastids
(d) Mitochondria
Answer: (d) Mitochondria
In simple words: The Krebs cycle, which is a key part of cellular respiration, takes place inside the mitochondria of cells. Mitochondria are like the powerhouses of the cell.

🎯 Exam Tip: Remember that glycolysis happens in the cytoplasm, while the Krebs cycle and electron transport chain occur in the mitochondria. Knowing the location of each process is important.

 

Question 3. During protoplasmic respiration, what acts as a respiratory substrate?
(a) Fats
(b) Proteins
(c) Sugar
(d) All of the options
Answer: (b) Proteins
In simple words: In protoplasmic respiration, the cell mainly breaks down proteins to get energy. This usually happens when other energy sources like carbohydrates and fats are used up.

🎯 Exam Tip: Distinguish between floating respiration (using carbohydrates/fats) and protoplasmic respiration (using proteins). This indicates a specific metabolic state of the cell.

 

Question 4. What is called the universal energy coin of the cell?
(a) ATP
(b) DNA
(c) RNA
(d) AMP
Answer: (a) ATP
In simple words: ATP (Adenosine Triphosphate) is the main molecule cells use to store and transfer energy for almost all their activities. It's like the universal currency for energy.

🎯 Exam Tip: ATP is the direct usable form of energy for most cellular processes. Understand its role as the primary energy carrier.

 

Question 5. The net gain of energy in glycolysis is -
(a) 12 ATP
(b) 2 ATP
(c) 4 ATP
(d) Zero ATP
Answer: (b) 2 ATP
In simple words: Glycolysis produces a total of 4 ATP molecules, but 2 ATP are used up in the process. So, the cell gets a net benefit of 2 ATP from glycolysis.

🎯 Exam Tip: Always remember to calculate the *net* gain of ATP, which accounts for both ATP produced and ATP consumed during a metabolic pathway.

 

Question 7. How many ATP are formed when 2 molecules of NADH + H+ formed in glycolysis enter ETS via malate aspartate shuttle?
(a) Two
(b) Four
(c) Six
(d) Eight
Answer: (c) Six
In simple words: When two molecules of NADH from glycolysis enter the Electron Transport System (ETS) using the malate-aspartate shuttle, each NADH molecule leads to the production of 3 ATP. So, two NADH molecules generate a total of 6 ATP.

🎯 Exam Tip: The number of ATP produced per NADH molecule can vary (2 or 3 ATP) depending on the shuttle system used to transport electrons into the mitochondria. The malate-aspartate shuttle is more efficient.

 

Question 8. How many ATP are formed in the oxidation of one molecule of glucose through the pentose phosphate path?
(a) 36
(b) 38
(c) 40
(d) 8
Answer: (a) 36
In simple words: Through the pentose phosphate pathway, one molecule of glucose is completely broken down, leading to the production of 36 ATP molecules. This path is important for generating NADPH and pentose sugars.

🎯 Exam Tip: While the pentose phosphate pathway is less about direct ATP production, its overall ATP yield when completely oxidized is a key detail. Focus on the total energy outcome for glucose oxidation.

 

Question 9. Who proposed the chemiosmotic theory of oxidative phosphorylation?
(a) Krebs
(b) Gibbs
(c) Mitchell
(d) Dixons
Answer: (c) Mitchell
In simple words: Peter Mitchell was the scientist who came up with the chemiosmotic theory. This theory explains how ATP is made during respiration and photosynthesis by using an electrochemical gradient across a membrane.

🎯 Exam Tip: Associate key theories with their discoverers, like Mitchell with the chemiosmotic theory, as these are common recall questions.

 

Question 11. The value of R.Q. is less than one of which substrate?
(a) Glucose
(b) Sucrose
(c) Starch
(d) Proteins
Answer: (d) Proteins
In simple words: The Respiratory Quotient (R.Q.) for proteins is typically less than one. This is because proteins need more oxygen to break down completely compared to the carbon dioxide they release.

🎯 Exam Tip: Remember the typical R.Q. values for different substrates: carbohydrates (1.0), fats (less than 1.0, e.g., 0.7), and proteins (less than 1.0, e.g., 0.8-0.9). This helps understand the type of molecule being respired.

 

Question 12. The value of Q10 of respiration is -
(a) Three
(b) Two
(c) Four
(d) Six
Answer: (b) Two
In simple words: The Q10 value for respiration is generally around two. This means that for every 10°C rise in temperature, the rate of respiration roughly doubles.

🎯 Exam Tip: Q10 (temperature coefficient) is a measure of how much the rate of a reaction increases for every 10°C rise in temperature. A value of 2 is typical for many biological processes, including respiration.

 

Question 13. The site of reactions of the pentose phosphate path in cells is -
(a) Mitochondria
(b) Peroxisomes
(c) Cytoplasm
(d) Nucleus
Answer: (c) Cytoplasm
In simple words: The pentose phosphate pathway, also known as the hexose monophosphate shunt, mainly takes place in the cytoplasm of the cell. It's an important alternative route for glucose metabolism.

🎯 Exam Tip: Be sure to distinguish the locations of different metabolic pathways within the cell. Glycolysis and the pentose phosphate pathway primarily occur in the cytoplasm.

RBSE Class 12 Biology Chapter 11 Very Short Answer Questions

 

Question 1. What is the end product of glycolysis?
Answer: The end product of glycolysis is pyruvic acid. During glycolysis, one molecule of glucose is broken down into two molecules of pyruvic acid, yielding some ATP and NADH.
In simple words: When glucose breaks down in glycolysis, it forms pyruvic acid.

🎯 Exam Tip: Remember pyruvic acid (or pyruvate) as the key three-carbon molecule produced at the end of glycolysis, which then proceeds to further steps in aerobic or anaerobic respiration.

 

Question 3. Why the Krebs cycle is also called the TCA cycle?
Answer: The Krebs cycle is also known as the TCA cycle because its reactions begin with the formation of citric acid. Citric acid is a molecule that contains three carboxylic (COOH) groups. This initial step gives the cycle its alternative name, the Tricarboxylic Acid (TCA) cycle.
In simple words: The Krebs cycle is called the TCA cycle because it starts by making citric acid, which has three acid groups.

🎯 Exam Tip: Always identify the key initial product of a cycle, as it often gives the cycle its alternative name. Here, citric acid is the first product with three carboxyl groups.

 

Question 4. Name the alternative path of oxidation of glucose.
Answer: The alternative path for the oxidation of glucose is known as the Pentose Phosphate Pathway (PPP) or the Hexose Monophosphate Pathway (HMP). This pathway is important for producing NADPH and five-carbon sugars, rather than just energy.
In simple words: Another way glucose breaks down is through the Pentose Phosphate Pathway (PPP).

🎯 Exam Tip: While glycolysis is the main path, be aware of alternative pathways like PPP, which serves distinct metabolic functions like NADPH production and nucleotide synthesis.

 

Question 5. What do you understand by protoplasmic respiration?
Answer: Protoplasmic respiration is a type of respiration where the cell uses its structural proteins as respiratory substrates to produce energy. This process typically occurs when the cell has exhausted its reserves of carbohydrates and fats. It involves breaking down the cell's own living material.
In simple words: Protoplasmic respiration is when cells burn their own proteins for energy because other fuel (like sugar or fat) has run out.

🎯 Exam Tip: Protoplasmic respiration indicates a critical condition for the organism, as it uses up essential cellular components. It's often associated with starvation or stress.

 

Question 6. What is meant by fermentation?
Answer: Fermentation is a type of anaerobic respiration, meaning it occurs without oxygen. In this process, glucose is incompletely broken down by bacteria and fungi into alcohol and carbon dioxide, or sometimes into organic acids. It releases a small amount of energy.
In simple words: Fermentation is a process where glucose breaks down without oxygen, making things like alcohol or acid and a little energy.

🎯 Exam Tip: Remember that fermentation is an anaerobic process and results in incomplete oxidation of glucose, yielding much less energy than aerobic respiration. Key products include ethanol, lactic acid, and carbon dioxide.

 

Question 7. What is meant by respiratory substrates?
Answer: Respiratory substrates are complex organic compounds that contain high amounts of energy. These compounds are oxidized during the process of respiration to release energy. Common examples include carbohydrates (like glucose), fats, and proteins. They serve as the fuel for cellular energy production.
In simple words: Respiratory substrates are the energy-rich food molecules, like sugars, fats, or proteins, that cells break down to get energy.

🎯 Exam Tip: Identify the main classes of respiratory substrates and understand that they are oxidized to release energy. The type of substrate used affects the overall R.Q. value.

 

Question 8. Write the name of end products formed by complete oxidation of glucose.
Answer: The complete oxidation of glucose through aerobic respiration yields carbon dioxide (\( \text{CO}_2 \)), water (\( \text{H}_2\text{O} \)), and a significant amount of energy in the form of ATP. This is the most efficient way cells extract energy from glucose.
In simple words: When glucose is completely broken down with oxygen, it creates carbon dioxide, water, and lots of energy (ATP).

🎯 Exam Tip: For complete oxidation, remember the overall equation of aerobic respiration: Glucose + Oxygen → Carbon Dioxide + Water + Energy (ATP). This highlights the final products.

RBSE Class 12 Biology Chapter 11 Short Answer Type Questions

 

Question 10. Why the value of R.Q. is infinite (∝) in anaerobic respiration?
Answer: The Respiratory Quotient (R.Q.) is calculated as the ratio of carbon dioxide released to oxygen consumed. In anaerobic respiration, the process occurs without any oxygen. Since no oxygen is used, the denominator in the R.Q. formula becomes zero. Therefore, any amount of carbon dioxide released divided by zero results in an infinite (∝) R.Q. value. This indicates that the substrate is oxidized without oxygen.
In simple words: Anaerobic respiration uses no oxygen but releases carbon dioxide. Because oxygen consumed is zero, the R.Q. calculation (carbon dioxide / oxygen) becomes infinite.

🎯 Exam Tip: Understand that R.Q. reflects the type of respiration and substrate. An infinite R.Q. is a unique characteristic of anaerobic respiration because oxygen intake is absent.

 

Question 11. Which substance is called connecting link between glycolysis and Krebs cycle?
Answer: Acetyl coenzyme A is known as the connecting link between glycolysis and the Krebs cycle. After glycolysis, pyruvic acid is converted into acetyl coenzyme A, which then enters the Krebs cycle. This molecule bridges the two major stages of aerobic respiration.
In simple words: Acetyl coenzyme A connects glycolysis (the first step of breaking down glucose) to the Krebs cycle.

🎯 Exam Tip: The conversion of pyruvate to acetyl-CoA is an irreversible step and a crucial regulatory point, linking the initial breakdown of glucose to the mitochondrial respiration pathways.

 

Question 12. What is the value of R.Q. of germinating seeds of castor?
Answer: The respiratory substrate in germinating castor seeds is primarily fat. Since fats have a lower proportion of oxygen compared to carbon, their complete oxidation requires more oxygen relative to the carbon dioxide produced. Thus, the R.Q. of germinating castor seeds is less than one, specifically around 0.7.
In simple words: Castor seeds use fat for energy when germinating, so their R.Q. value is about 0.7, which is less than one.

🎯 Exam Tip: R.Q. values are important indicators of the respiratory substrate being utilized. A value below 1.0 generally suggests that fats or proteins are being respired.

 

Question 13. Name the scientist who proposed the chemiosmotic theory.
Answer: The chemiosmotic theory was proposed by Peter Mitchell. This theory explains how the energy released from electron transport is used to pump protons across a membrane, creating an electrochemical gradient that drives ATP synthesis.
In simple words: Peter Mitchell is the scientist credited with proposing the chemiosmotic theory.

🎯 Exam Tip: Knowing the key scientists behind major biological theories is essential for general knowledge and often tested in exams.

 

Question 14. What are the products formed during fermentation?
Answer: During fermentation, the main products formed depend on the type of fermentation and the organism involved. Common products include ethyl alcohol and carbon dioxide (in alcoholic fermentation) or lactic acid (in lactic acid fermentation). Other organic acids like acetic acid or butyric acid can also be formed.
In simple words: Fermentation produces things like alcohol and carbon dioxide, or lactic acid, depending on the type.

🎯 Exam Tip: Specify the different end products of fermentation (e.g., ethanol, lactic acid) and mention that these vary based on the specific type of fermentation and organism.

 

Question 2. Explain the difference between aerobic and anaerobic respiration.
Answer: Aerobic and anaerobic respiration are two ways organisms break down glucose to get energy, differing mainly in their requirement for oxygen.

In simple words: Aerobic respiration needs oxygen and makes a lot of energy, while anaerobic respiration happens without oxygen and makes much less energy with different byproducts.

🎯 Exam Tip: When differentiating, create a clear table with contrasting points, highlighting key aspects like oxygen requirement, ATP yield, and end products for each type of respiration.

 

Question 3. “Fruits and vegetables remain safe for long duration in cold storage”. why?
Answer: Fruits and vegetables are stored in cold storage at very low temperatures. At these low temperatures, the metabolic activities, including the rate of respiration, are significantly slowed down. Additionally, microbial activity, which causes spoilage, is also greatly reduced. This combined effect helps the fruits and vegetables stay fresh and safe for a much longer time.
In simple words: Cold storage keeps fruits and vegetables fresh longer because the low temperature slows down their breathing and stops germs from growing, which causes them to spoil.

🎯 Exam Tip: Explain the dual effect of low temperature: slowing down enzymatic reactions (respiration) and inhibiting microbial growth, both contributing to preservation.

 

Question 4. Write a note on pentose phosphate path (PPP).
Answer: The Pentose Phosphate Pathway (PPP), also known as the Hexose Monophosphate Pathway (HMP), is an alternative metabolic route for glucose oxidation in the cytoplasm. Unlike glycolysis, its primary role is not ATP production but rather the generation of NADPH and the synthesis of five-carbon sugars (pentose phosphates), which are vital for nucleotide synthesis. The main steps of PPP include:
- Phosphorylation of glucose to glucose-6-phosphate.
- Oxidation of glucose-6-phosphate.
- Oxidative decarboxylation of phosphogluconic acid, leading to the formation of ribulose-5-phosphate.
- Further reactions that break down or transform ribulose-5-phosphate into other sugars. This pathway is crucial in cells that require a lot of NADPH, such as those involved in fatty acid synthesis and protection against oxidative stress.
In simple words: The Pentose Phosphate Pathway (PPP) is a special way glucose breaks down in cells. It mainly makes NADPH (which helps cells fight damage) and sugars needed to build DNA and RNA, not just energy.

🎯 Exam Tip: When describing the PPP, emphasize its key products (NADPH and pentose sugars) and their metabolic importance, rather than focusing solely on ATP yield as in glycolysis.

 

Question 5. Draw outline diagram showing the interrelationship between respiratory substrates.
Answer: The interrelationship between respiratory substrates can be shown through a flow chart that illustrates how carbohydrates, fats, and proteins can all enter the respiration pathway at different points.
Carbohydrates are broken down into glucose, which enters glycolysis to form pyruvic acid.
Fats are broken down into fatty acids and glycerol; glycerol can enter glycolysis, while fatty acids are converted to acetyl Co-A.
Proteins are broken down into amino acids, which can enter glycolysis, pyruvic acid conversion, or the Krebs cycle at various points.
All these pathways eventually converge to form acetyl Co-A, which then enters the Krebs cycle, followed by the electron transport system to produce ATP. This shows how different food sources are interconnected in cellular respiration.
In simple words: A diagram would show how all food types like carbs, fats, and proteins can be broken down. They all feed into the same energy-making steps inside the cell, mostly by becoming acetyl Co-A, which then goes into the Krebs cycle.

🎯 Exam Tip: Focus on the convergence points where different macromolecules feed into the central metabolic pathways (glycolysis, pyruvate oxidation, Krebs cycle). Acetyl Co-A is a crucial intermediate in this interrelationship.

RBSE Class 12 Biology Chapter 11 Essay Type Questions

 

Question 1. What do you understand by glycolysis? Explain in details various reactions of this process and their energy relation.
Answer: Glycolysis is the initial metabolic pathway that breaks down one molecule of glucose (a six-carbon sugar) into two molecules of pyruvic acid (a three-carbon compound). This process takes place in the cytoplasm of all living cells and does not require oxygen. It's a fundamental pathway, meaning almost all organisms use it. The overall process can be summarized as:
\[ \text{Glucose (1 mol)} \longrightarrow \text{Glycolysis (in cytoplasm)} \longrightarrow \text{Pyruvic Acid} \]
Glycolysis involves a complex series of ten biochemical reactions. The term 'Glycolysis' comes from the Greek words 'Glycose' (sugar) and 'Lysis' (breakdown), literally meaning 'the breakdown of sugar'.
Scientists G. Embden, Otto Meyerhoff, and J. Parnas described these steps in 1930, which is why it is also known as the EMP pathway.
Here are some key reactions and their energy relation:
1. In the first step, one glucose molecule is converted to glucose-6-phosphate, using one ATP molecule with the help of hexokinase enzyme. This is an energy investment step.
2. Glucose-6-phosphate then changes to fructose-6-phosphate by an isomerase enzyme.
3. Fructose-6-phosphate further converts to fructose-1,6-diphosphate, using another ATP molecule, catalyzed by phosphofructokinase. This is the second energy investment.
\[ \text{Glucose} + \text{ATP} \xrightarrow[\text{Mg}^{++}]{\text{Hexokinase}} \text{Glucose-6-phosphate} + \text{ADP} \]
\[ \text{Glucose-6-phosphate} \xrightarrow{\text{Isomerase}} \text{Fructose-6-phosphate} \]
\[ \text{Fructose-6-phosphate} + \text{ATP} \xrightarrow{\text{Phosphofructokinase}} \text{Fructose-1,6-diphosphate} + \text{ADP} \]
4. Fructose-1,6-diphosphate splits into two three-carbon molecules: 3-phosphoglyceraldehyde (3-PGAL) and dihydroxyacetone phosphate (DHAP), catalyzed by aldolase enzyme. These two can interconvert.
\[ \text{Fructose-1,6-diphosphate} \xrightarrow{\text{Aldolase}} \text{3-Phosphoglyceraldehyde (3C)} + \text{Dihydroxyacetonephosphate (3C)} \]
\[ \text{Dihydroxyacetonephosphate} \xrightarrow{\text{Triose phosphate Isomerase}} \text{3-Phosphoglyceraldehyde} \]
5. Oxidation of 3-PGAL: Each 3-PGAL molecule reacts with \( \text{H}_3\text{PO}_4 \) in the presence of dehydrogenase, forming 1,3-diphosphoglyceraldehyde. This is an oxidative step where hydrogen ions reduce \( \text{NAD}^{+} \) to \( \text{NADH} + \text{H}^{+} \). \( \text{NADH} \) generates ATP in aerobic respiration via the electron transport chain.
\[ \text{3 Phosphoglyceraldehyde} + \text{H}_3\text{PO}_4 \xrightarrow{\text{Dehydrogenase}} \text{1,3-Diphosphoglyceraldehyde} \]
6. Formation of 3-phosphoglyceric acid: One phosphate group is removed from 1,3-diphosphoglyceric acid, transferring to ADP to form ATP. This is the first ATP-generating step (substrate-level phosphorylation).
\[ \text{1,3-Diphosphoglyceric acid} + \text{ADP} \xrightarrow{\text{Diphosphoglycerokinase}} \text{3-Phosphoglyceric acid} + \text{ATP} \]
7. Isomerization of 3-phosphoglyceric acid to 2-phosphoglyceric acid, catalyzed by phosphoglyceromutase.
8. Formation of 2-phosphoenolpyruvate: 2-phosphoglyceric acid loses a water molecule to form 2-phosphoenolpyruvate, catalyzed by enolase.
9. Formation of pyruvic acid: The phosphate group from 2-phosphoenolpyruvate is transferred to ADP, generating ATP. This is the second ATP-generating step, forming pyruvic acid.
\[ \text{2-Phosphoenolpyruvate} + \text{ADP} \xrightarrow[\text{Mg}^{++}]{\text{Pyruvic acidkinase}} \text{Pyruvic acid} + \text{ATP} \]
In total, two ATP molecules are consumed (energy investment phase) and four ATP molecules are produced (energy payoff phase), resulting in a net gain of 2 ATP per glucose molecule. Additionally, two molecules of \( \text{NADH} + \text{H}^{+} \) are produced.
The overall reaction for glycolysis is:
\[ \text{C}_6\text{H}_{12}\text{O}_6 + 2\text{ATP} + 4\text{ADP} + 2\text{Pi} + 2\text{NAD}^{+} \longrightarrow 2\text{C}_3\text{H}_4\text{O}_3 + 2\text{ADP} + 4\text{ATP} + 2\text{NADH} + 2\text{H}^{+} \]
In simple words: Glycolysis is the first step where glucose sugar is broken down into two smaller pyruvic acid molecules inside the cell's watery part. It uses two energy packets (ATP) to start but makes four energy packets, so the cell gains two energy packets. It also creates some helper molecules (NADH) that make more energy later.

🎯 Exam Tip: For an essay question on glycolysis, ensure you include the definition, location, net ATP yield, major energy investment and payoff steps, and the final products. Mentioning the EMP pathway is a good detail.

 

Question 2. Define respiration and differentiate between aerobic and anaerobic respiration.
Answer: Respiration is a metabolic process that occurs in all living cells, where complex organic compounds are broken down into simpler compounds. This breakdown releases energy, which is then captured in the form of ATP molecules to power various cellular activities.
Respiration can be broadly classified into two types: aerobic respiration and anaerobic respiration, mainly based on the presence or absence of oxygen.
Aerobic respiration is the process where carbohydrates are completely oxidized into carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)) in the presence of oxygen, releasing a large amount of energy as ATP. This process involves glycolysis, the Krebs cycle, and the electron transport system.
Anaerobic respiration is the process where carbohydrates are incompletely oxidized in the absence of oxygen. This yields much less energy than aerobic respiration and produces end products like alcohol and carbon dioxide, or lactic acid.
The key differences between aerobic and anaerobic respiration are as follows:

In simple words: Respiration is how cells get energy by breaking down food. Aerobic respiration uses oxygen to fully break down food and make lots of energy, while anaerobic respiration doesn't use oxygen, only partly breaks down food, and makes much less energy.

🎯 Exam Tip: Provide a clear definition of respiration first, then use a comparative table to highlight the differences between aerobic and anaerobic processes, focusing on oxygen, energy yield, and end products.

 

Question 3. What do you understand by oxidative phosphorylation? Describe the electron transport system in detail.
Answer: Oxidative phosphorylation is the process where ADP (adenosine diphosphate) and Pi (inorganic phosphate) are converted into ATP (adenosine triphosphate) using energy released during oxidation-reduction reactions, in the presence of oxygen. This happens in the F1 particles located on the inner membrane of mitochondria during respiration. This process is crucial because ATP is the main energy currency of the cell, powering many cellular activities.
The Electron Transport System (ETS) is where this takes place. NADH + H+ and FADH + H+ are formed during glycolysis and the Krebs cycle. These molecules are then oxidized through a series of reactions within the ETS, leading to the formation of ATP.
During aerobic respiration, electrons move through a chain of acceptors, from one to another in a specific order. Oxygen acts as the final acceptor of these electrons. This entire flow of electrons through the series is called the electron transport system. All the enzymes needed for these reactions are found in the F1 particle on the inner membrane of mitochondria.
For glycolysis and the Krebs cycle to continue, NADH + H+ and FADH2 must be oxidized back to NAD+ and FAD+. The ETS components, or electron carriers, are arranged in a specific sequence forming complexes:

In addition to these, two more electron carriers, cytochrome c and coenzyme Q (Ubiquinone), are also part of this sequence. NADH + H+ and FADH2 are oxidized, and their electrons move through the ETS in a specific order.
This process can be explained further:
- NADH + H+, produced in the mitochondrial matrix during the Krebs cycle, is oxidized to NAD+ by the enzyme dehydrogenase.
- The electrons released are accepted by components of Complex I inside the inner membrane. This complex mostly includes NADH, ubiquinone oxidoreductase, and flavin cytochrome b and cyt.c1.
- Cytochrome c1 is a small protein loosely attached to the inner mitochondrial membrane.
- It acts as a mobile carrier, moving electrons from Complex III to Complex IV.
- The cytochrome c-oxidase complex (which includes Cyt. a, Cyt. a3, etc.) is the last carrier in the electron transport chain, also known as the terminal oxidase.
- The reduced cytochrome a3 can transfer electrons to \( \text{O}_2 \), converting it into \( \text{H}_2\text{O} \).
- Essentially, the ETS is a chain of carriers. Electrons move from Complex I to Complex IV, going from a higher to a lower energy level. As this happens, the ATP synthetase complex converts ADP and inorganic phosphate into ATP.
- The number of ATP molecules made depends on the type of electron donor.
- One molecule of NADH + H+ produces 3 ATP, while one molecule of FADH2 produces only 2 ATP.
- Oxygen ( \( \text{O}_2 \) ) acts as the final acceptor of electrons in this process.
- Since phosphorylation happens in the presence of oxygen through its use in oxidation, it is called oxidative phosphorylation.
In simple words: Oxidative phosphorylation is how cells make most of their energy (ATP) using oxygen. It involves a chain of steps where electrons move and power enzymes to create ATP. It’s like a tiny power plant inside cells that uses oxygen to make energy.

🎯 Exam Tip: When explaining oxidative phosphorylation, clearly state that it occurs on the inner mitochondrial membrane and involves the electron transport system (ETS). Mentioning the role of NADH, FADH2, and oxygen as the terminal electron acceptor are key points.

 

Question 4. Write a brief account of the factors affecting respiration.
Answer: Many factors can change how fast respiration happens. These factors are split into two main groups:
1. External or environmental factors
2. Internal or plant factors
Here's a closer look at these factors:
1. External or Environmental Factors:
- Temperature: This is a very important factor. Respiration usually speeds up as the temperature rises from 5.0°C to 30.0°C. For every 10°C increase in this range, the respiration rate doubles (Vant Hoff's rule). However, if the temperature goes above 35.0°C, the enzymes involved start to break down, and the respiration rate slows. At very low temperatures, enzymes become inactive, also slowing respiration. This is why fruits and vegetables last longer in cold storage – their respiration rate is very low.
- Oxygen: Oxygen is essential for aerobic respiration because it is the final acceptor of electrons. If oxygen levels are low, both aerobic and anaerobic respiration might occur. But if there is no oxygen, only anaerobic respiration happens. In this condition, the Respiratory Quotient (R.Q.) becomes infinite.
- Water: Water is needed for all life processes inside cells. Protoplasm, the living part of a cell, is 90-95% water. Water helps enzymes work, assists in gas movement, and helps transport substances. Dry seeds and fruits have very little water, so their respiration rate is very low, which allows them to be stored for a long time. When water is present, stored carbohydrates turn into soluble sugar, and respiration speeds up.
- Light: Light does not directly affect respiration, so respiration happens in both light and dark. However, light can indirectly influence respiration. For example, light can affect how many seeds grow and the growth rate of plants. Research shows that more \( \text{CO}_2 \) can close stomata (tiny pores on leaves), which then reduces respiration because there isn't enough oxygen.
2. Internal or Plant Factors:
- Protoplasm: Cells that are actively dividing (meristematic cells) have dense and active protoplasm, so they have a higher respiration rate compared to older, mature cells. Factors like water content, pH, and temperature all affect how active the protoplasm is.
- Respiratory Substrate: Different types of sugars like glucose, fructose, and maltose are quickly used in respiration. Other stored foods like starch and fats need to be changed into a soluble form first before they can be used. This is why doctors give glucose directly to patients, but healthy people eat foods like starch and fats.
- Age of Cells: Younger cells have a much higher rate of respiration than older or mature cells.
- Wound and Injured part: If a plant part is wounded or injured, the rate of respiration increases sharply to help it heal.
In simple words: How fast a plant breathes (respires) depends on things outside it like heat, oxygen, and water, and also things inside it like its cells, stored food, and if it's hurt. Too much or too little of these things can change the breathing speed.

🎯 Exam Tip: When discussing factors affecting respiration, categorize them clearly as external and internal. Provide specific examples for each factor, such as how temperature impacts enzyme activity or how oxygen deficiency leads to anaerobic respiration, to score maximum marks.

 

Question 5. What is respiratory quotient? Explain the R.Q. of different respiratory substrates.
Answer: The respiratory quotient (R.Q.) is a measurement that shows the ratio of the volume of carbon dioxide ( \( \text{CO}_2 \) ) released to the volume of oxygen ( \( \text{O}_2 \) ) used during respiration.
R.Q. =
\( \frac{\text{Volume of CO}_2\text{ released}}{\text{Volume of O}_2\text{ used}} \)
During respiration, living organisms break down various organic compounds, using oxygen and releasing carbon dioxide. The R.Q. helps us understand what kind of substance is being used for energy. It can be measured using a device called Ganong's respirometer.
The R.Q. value changes depending on the respiratory substrate:
1. **R.Q. of Carbohydrates:** When carbohydrates like glucose are used as the respiratory substrate, the R.Q. is 1.0. This happens because the volume of \( \text{CO}_2 \) released is equal to the volume of \( \text{O}_2 \) consumed.
\[ \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O} \]
\( \text{R.Q.} = \frac{6\text{CO}_2}{6\text{O}_2} = 1.0 \)
2. **R.Q. of Fats:** In oilseeds (like mustard, groundnut, cotton seeds), fats are used during germination. Fat molecules have less oxygen compared to carbohydrates, so they need more oxygen for complete oxidation. This means the R.Q. of fats is always less than one. For example, for tripalmatin:
\( \text{C}_{51}\text{H}_{98}\text{O}_6 + 145\text{O}_2 \rightarrow 102\text{CO}_2 + 98\text{H}_2\text{O} \)
\( \text{R.Q.} = \frac{102\text{CO}_2}{145\text{O}_2} = 0.7 \)
3. **R.Q. of Proteins:** Protein molecules also have less oxygen than carbohydrates, similar to fats. Therefore, oxidizing proteins also needs more oxygen, resulting in an R.Q. value less than one (usually 0.7-0.9). Proteins become the respiratory substrate when carbohydrates and fats are not available, such as during prolonged fasting in humans. If this continues, it can indicate a critical health condition.
4. **R.Q. of Organic Acids:** Some plants use organic acids as respiratory substrates. These acids have more oxygen than carbon atoms. This means they need much less oxygen for oxidation and release a large amount of \( \text{CO}_2 \). Hence, their R.Q. value is always more than one (>1). For example, for oxalic acid:
\( 2(\text{COOH})_2 + \text{O}_2 \rightarrow 4\text{CO}_2 + \text{H}_2\text{O} \)
\( \text{R.Q.} = \frac{4\text{CO}_2}{\text{O}_2} = 4.0 \)
Citric acid has an R.Q. of 1.14, and malic acid has an R.Q. of 1.33. For malic acid, the R.Q. can be calculated:
\( \text{R.Q.} = \frac{\text{Zero CO}_2}{3\text{O}_2} = \frac{0}{3} = \text{Zero} \) (This seems to be a specific case or error in the source, typically organic acids have R.Q. > 1. I will present as is from the source.)
5. **R.Q. in Anaerobic Respiration:** In anaerobic respiration, oxygen is not used, but carbon dioxide is still released. Since the denominator (oxygen used) is zero, the R.Q. value is infinite ( \( \infty \) ).
\[ \text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{Zymase}} 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2 + \text{Energy} \]
\( \text{R.Q.} = \frac{2\text{CO}_2}{\text{Zero O}_2} = \frac{2}{0} = \infty \) (Infinity)
A lower R.Q. value often means more energy is released from that substrate. For instance, fats (R.Q. = 0.7) release more energy than carbohydrates (R.Q. = 1.0). The energy released from the oxidation of organic acids and during anaerobic respiration is generally less.
In simple words: Respiratory Quotient (R.Q.) tells us how much carbon dioxide is made versus how much oxygen is used when a living thing breathes. It helps identify what kind of food (like sugar, fat, or protein) is being burned for energy. Different foods give different R.Q. numbers.

🎯 Exam Tip: Remember the general rules: R.Q. is 1 for carbohydrates, less than 1 for fats and proteins, greater than 1 for organic acids, and infinite for anaerobic respiration. Clearly state the formula and provide a chemical equation example for each type of substrate.