RESPIRATION IN PLANTS MCQs for NEET UG BIOLOGY — Medium, Page 2

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Application +4 | -1

In the given equation below, identify the compound which is being oxidized and calculate the R.Q. of the same.

\left({C}_{15} {H}_{98} {O}_6\right)+145 {O}_2 \rightarrow 102 {CO}_2+98 {H}_2 {O}+\text {energy }

View answer and explanation

Correct option: B

Tripalmitin, R.Q. = 0.7

Option B (Tripalmitin, R.Q. = 0.7) is correct because tripalmitin is a fat, and the RQ for fats is approximately 0.7, calculated from the balanced equation.

Solution approach

Correct Answer + Core Reason: Tripalmitin, R.Q. = 0.7. Tripalmitin is oxidized, and its RQ is approximately 0.7.

Step-by-Step Solution:

Step 1: Identify the compound being oxidized. In the given equation, ${C_{15}}{H_{98}}{O_6}$ (Tripalmitin) is being oxidized.

Step 2: Define Respiratory Quotient (RQ). RQ is the ratio of carbon dioxide produced to oxygen consumed: $RQ = \frac{{CO_2 \\ produced}}{{O_2 \\ consumed}}$

Step 3: Identify the number of CO2 molecules produced. From the equation, 102 CO2 molecules are produced (note: there seems to be a typo in the question, it should be C51H98O6, not C15H98O6. Assuming it's C51H98O6, the balanced equation is 2C51H98O6 + 145O2 -> 102CO2 + 98H2O). However, using the numbers given in the question, we will assume that the number of CO2 molecules produced is 102.

Step 4: Identify the number of O2 molecules consumed. From the equation, 145 O2 molecules are consumed.

Step 5: Calculate the RQ. $RQ = \frac{{102}}{{145}} \approx 0.7$

Why other options are wrong:
- Option A (Palmitin, R.Q. = 0.9): This is incorrect because the compound is tripalmitin, not palmitin, and the RQ is not 0.9.
- Option C (Palmitin, R.Q. = 1): This is incorrect because the compound is tripalmitin, not palmitin, and the RQ is not 1.
- Option D (Tripalmitin, R.Q. = 0.2): This is incorrect because the RQ is approximately 0.7, not 0.2.

Key Concepts:
- Respiratory Quotient (RQ)
- Oxidation of fats
- Stoichiometry of respiration

Memory Aids:
- RQ = CO2 out / O2 in

Quick Tips:
- Always check the balanced equation to get the correct stoichiometric coefficients.

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RQ value of 4 may be expected for the complete oxidation of which one of the following?

View answer and explanation

Correct option: C

Oxalic acid

Option C (Oxalic acid) is correct because oxalic acid has the highest carbon to oxygen ratio among the given options, leading to a higher RQ value of 4.

Solution approach

Correct Answer + Core Reason: Oxalic acid. Oxalic acid has a high carbon-to-oxygen ratio, resulting in an RQ of 4.

Step-by-Step Solution:

Step 1: Define Respiratory Quotient (RQ). RQ is the ratio of carbon dioxide produced to oxygen consumed: $RQ = \frac{{CO_2 \\ produced}}{{O_2 \\ consumed}}$

Step 2: Consider the general RQ values. Carbohydrates have an RQ of 1, fats around 0.7, and organic acids can have RQ values greater than 1.

Step 3: Analyze the chemical formulas of the given options. Oxalic acid (C2H2O4) has a high carbon-to-oxygen ratio compared to the other options.

Step 4: Determine the RQ for oxalic acid. The balanced equation for the oxidation of oxalic acid is: $C_2H_2O_4 + O_2 \rightarrow 2CO_2 + 2H_2O$ . Therefore, RQ = 2/0.5 = 4 (2 moles of CO2 produced for every 0.5 moles of O2 consumed). Note: the balanced equation requires 0.5 moles of O2, not 1 mole. The equation should be $C_2H_2O_4 + 0.5O_2 \rightarrow 2CO_2 + H_2O$ .

Why other options are wrong:
- Option A (Glucose): This is incorrect because glucose is a carbohydrate with an RQ of 1.
- Option B (Malic acid): This is incorrect because malic acid has an RQ greater than 1, but not as high as 4.
- Option D (Tartaric acid): This is incorrect because tartaric acid has an RQ greater than 1, but not as high as 4.

Key Concepts:
- Respiratory Quotient (RQ)
- Oxidation of organic acids

Memory Aids:
- High carbon-to-oxygen ratio = High RQ

Quick Tips:
- Remember that organic acids can have RQ values greater than 1.

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Fate of pyruvic acid during aerobic respiration is

View answer and explanation

Correct option: C

Oxidative decarboxylation

Option C (Oxidative decarboxylation) is correct because it describes the process where pyruvic acid is converted to Acetyl CoA, releasing carbon dioxide.

Solution approach

Correct Answer + Core Reason: Oxidative decarboxylation. Pyruvic acid is converted to Acetyl CoA through oxidative decarboxylation.

Step-by-Step Solution:

Step 1: Identify the process: Aerobic respiration involves the breakdown of pyruvic acid.

Step 2: Recall the fate of pyruvic acid: Pyruvic acid undergoes oxidative decarboxylation to form Acetyl CoA.

Step 3: Understand oxidative decarboxylation: This process involves the removal of a carbon atom from pyruvic acid in the form of carbon dioxide and the attachment of the remaining two-carbon fragment to coenzyme A.

Step 4: Conclude: The fate of pyruvic acid during aerobic respiration is oxidative decarboxylation.

Why other options are wrong:
- Option A (Lactic acid fermentation): Lactic acid fermentation occurs under anaerobic conditions, not aerobic respiration.
- Option B (Alcoholic acid fermentation): Alcoholic fermentation also occurs under anaerobic conditions.
- Option D (Oxidative phosphorylation): Oxidative phosphorylation is the process of ATP synthesis using the electron transport chain, which occurs after the Krebs cycle.

Key Concepts:
- Aerobic Respiration
- Oxidative Decarboxylation
- Krebs Cycle

Memory Aids:
- Remember that aerobic respiration uses oxygen and breaks down pyruvic acid to Acetyl CoA.

Quick Tips:
- Differentiate between aerobic and anaerobic processes.

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Respiration is completely absent in

View answer and explanation

Correct option: A

Bacteriophage

Option A (Bacteriophage) is correct because bacteriophages are viruses, and viruses are not capable of independent respiration; they rely on host cells.

Solution approach

Correct Answer + Core Reason: Bacteriophage. Bacteriophages are viruses and do not respire.

Step-by-Step Solution:

Step 1: Identify the question: Which organism does not respire?

Step 2: Recall the characteristics of viruses: Viruses are non-cellular and require a host to replicate and perform metabolic processes.

Step 3: Consider other options: Cyanobacteria, Chlorella, and Spirulina are capable of respiration.

Step 4: Conclusion: Respiration is completely absent in bacteriophages.

Why other options are wrong:
- Option B: Cyanobacteria are photosynthetic bacteria that also respire.
- Option C: Chlorella is a type of algae that respires.
- Option D: Spirulina is a type of cyanobacteria that respires.

Key Concepts:
- Respiration
- Viruses
- Bacteria
- Algae

Memory Aids:
- Viruses are non-living and cannot respire.

Quick Tips:
- Remember that viruses are obligate intracellular parasites.

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What will happen, when glucose is administered orally?

View answer and explanation

Correct option: D

Respiration

Option C (Circulation) is the most direct answer. Glucose, when administered orally, is absorbed into the bloodstream and then circulated throughout the body for use in cellular respiration.

Solution approach

Correct Answer + Core Reason: Circulation. Orally administered glucose enters the bloodstream and is circulated throughout the body.

Step-by-Step Solution:

Step 1: Consider the fate of orally administered glucose. Glucose is a simple sugar that is readily absorbed into the bloodstream from the digestive system.

Step 2: Trace the path of glucose after absorption. Once in the bloodstream, glucose is transported to various cells and tissues throughout the body via the circulatory system.

Step 3: Relate circulation to cellular processes. The circulatory system delivers glucose to cells where it can be used for respiration or stored as glycogen.

Why other options are wrong:
- Option A: Excretion: Glucose is normally not excreted unless there is a medical condition like diabetes where the kidneys cannot reabsorb all the glucose.
- Option B: Digestion: Glucose is already a simple sugar and does not need to be digested further before absorption.
- Option D: Respiration: While glucose is used in respiration, the immediate effect of oral administration is its entry into the circulation, which then allows it to be used in respiration.

Key Concepts:
- Glucose absorption
- Circulatory system
- Cellular respiration

Memory Aids:
- Oral glucose -> Bloodstream -> Circulation -> Cellular use

Quick Tips:
- Consider the immediate fate of the substance after administration.

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How many ATP molecules can be produced through oxidative phosphorylation of

2NAD{{H}_{2}}\,\,\text{and}\,\,3FAD{{H}_{2}}

?

View answer and explanation

Correct option: D

12

Option D (12) is correct because each NADH2 produces 3 ATP and each FADH2 produces 2 ATP through oxidative phosphorylation.

Solution approach

Correct Answer + Core Reason: 12. Calculate the ATP produced by $NAD{{H}_{2}}$ and $FAD{{H}_{2}}$ during oxidative phosphorylation.

Step-by-Step Solution:

Step 1: Determine the ATP yield per $NAD{{H}_{2}}$ molecule during oxidative phosphorylation.

Step 2: Each $NAD{{H}_{2}}$ molecule yields approximately 3 ATP molecules.

Step 3: Determine the ATP yield per $FAD{{H}_{2}}$ molecule during oxidative phosphorylation.

Step 4: Each $FAD{{H}_{2}}$ molecule yields approximately 2 ATP molecules.

Step 5: Calculate the total ATP produced from 2 $NAD{{H}_{2}}$ molecules: 2 * 3 = 6 ATP.

Step 6: Calculate the total ATP produced from 3 $FAD{{H}_{2}}$ molecules: 3 * 2 = 6 ATP.

Step 7: Calculate the total ATP produced: 6 ATP + 6 ATP = 12 ATP.

Step 8: Final Answer: 12 ATP molecules

Why other options are wrong:
- Option A (15): This is incorrect. It might be a result of incorrect ATP yield calculation for $NAD{{H}_{2}}$ and $FAD{{H}_{2}}$ .
- Option B (24): This is incorrect. It might be a result of incorrect ATP yield calculation for $NAD{{H}_{2}}$ and $FAD{{H}_{2}}$ .
- Option C (6): This is incorrect. It only considers the ATP produced by either $NAD{{H}_{2}}$ or $FAD{{H}_{2}}$ , not both.

Key Concepts:
- Oxidative Phosphorylation
- ATP Production
- $NAD{{H}_{2}}$ and $FAD{{H}_{2}}$ Oxidation

Memory Aids:
- Remember that each $NAD{{H}_{2}}$ yields approximately 3 ATP and each $FAD{{H}_{2}}$ yields approximately 2 ATP.

Quick Tips:
- Always remember the ATP yield per $NAD{{H}_{2}}$ and $FAD{{H}_{2}}$ molecule in oxidative phosphorylation.

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Four respiratory enzymes are given below. Arrange them in increasing order of the carbon number of the substrates on which they act.
I. Enolase
II. Aconitase
III. Fumarase
IV. Alcohol Dehydrogenase

View answer and explanation

Correct option: D

IV, I, III, II

Option D (IV, I, III, II) is correct because it arranges the enzymes based on the increasing carbon number of their substrates: Alcohol Dehydrogenase (2C), Enolase (3C), Fumarase (4C), Aconitase (6C).

Solution approach

Correct Answer + Core Reason: IV, I, III, II. The order is based on the increasing carbon number of the substrates.

Step-by-Step Solution:

Step 1: Identify the substrates of each enzyme.

Step 2: Alcohol dehydrogenase acts on ethanol (2 carbons).

Step 3: Enolase acts on 2-phosphoglycerate (3 carbons).

Step 4: Fumarase acts on fumarate (4 carbons).

Step 5: Aconitase acts on citrate/isocitrate (6 carbons).

Step 6: Arrange the enzymes in increasing order of carbon number of their substrates: Alcohol Dehydrogenase (IV), Enolase (I), Fumarase (III), Aconitase (II).

Step 7: Final Answer: IV, I, III, II

Why other options are wrong:
- Option A (II, IV, III, I): Incorrect order.
- Option B (IV, I, II, III): Incorrect order.
- Option C (I, IV, III, II): Incorrect order.

Key Concepts:
- Enzyme Substrates
- Carbon Number
- Metabolic Pathways

Memory Aids:
- Remember the substrates of key respiratory enzymes.

Quick Tips:
- Review the metabolic pathways and the substrates of the enzymes involved.

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Select the wrong statement.

View answer and explanation

Correct option: B

The intermediate compound which links glycolysis with Krebs’ cycle is malic acid

Option B (The intermediate compound which links glycolysis with Krebs’ cycle is malic acid) is wrong because Acetyl-CoA, not malic acid, links glycolysis to the Krebs cycle.

Solution approach

Correct Answer + Core Reason: The intermediate compound which links glycolysis with Krebs’ cycle is malic acid. Acetyl-CoA is the link, not malic acid.

Step-by-Step Solution:

Step 1: Analyze the statement about tripalmitin. Tripalmitin is a fat, and fats have an RQ of less than 1 (around 0.7). So, this statement is correct.

Step 2: Analyze the statement about the intermediate compound linking glycolysis and Krebs' cycle. Glycolysis produces pyruvate, which is then converted to acetyl-CoA. Acetyl-CoA enters the Krebs' cycle. Therefore, acetyl-CoA is the linking compound, not malic acid. This statement is incorrect.

Step 3: Analyze the statement about ATP production during aerobic respiration. One glucose molecule yields a net gain of approximately 36-38 ATP molecules during aerobic respiration. So, this statement is correct.

Step 4: Analyze the statement about ATP production during fermentation. One glucose molecule yields a net gain of 2 ATP molecules during fermentation. So, this statement is correct.

Step 5: Final Answer: The wrong statement is that malic acid links glycolysis to the Krebs cycle.

Why other options are wrong:
- Option A (When tripalmitin is used as a substrate in respiration, the RQ is 0.7): This statement is correct.
- Option C (One glucose molecule yields a net gain of 36 ATP molecules during aerobic fermentation): This statement is correct (approximately).
- Option D (One glucose molecule yields a net gain of 2 ATP molecules during fermentation): This statement is correct.

Key Concepts:
- Respiratory Quotient (RQ)
- Glycolysis
- Krebs Cycle (Citric Acid Cycle)
- Fermentation
- ATP Production

Memory Aids:
- Remember that Acetyl-CoA is the link between glycolysis and the Krebs cycle.

Quick Tips:
- Review the steps of glycolysis, the link reaction, and the Krebs cycle.

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Refer the given equation?

2\left( {{C_{51}}{H_{98}}{O_6}} \right) + 145\,\,{O_2} \to 102C{O_2} + 98{H_2}O +

energy the RQ in this case is

View answer and explanation

Correct option: B

0.7

Option B (0.7) is correct because the respiratory quotient (RQ) is calculated as the ratio of CO2 produced to O2 consumed, which in this case is 102/145 ≈ 0.7.

Solution approach

Correct Answer + Core Reason: 0.7. The respiratory quotient is calculated as the ratio of CO2 produced to O2 consumed.

Step-by-Step Solution:

Step 1: Define Respiratory Quotient (RQ). RQ is the ratio of carbon dioxide produced to oxygen consumed: $RQ = \frac{{CO_2 \\ produced}}{{O_2 \\ consumed}}$

Step 2: Identify the number of CO2 molecules produced. From the equation, 102 CO2 molecules are produced.

Step 3: Identify the number of O2 molecules consumed. From the equation, 145 O2 molecules are consumed.

Step 4: Calculate the RQ. $RQ = \frac{{102}}{{145}} \approx 0.7$

Why other options are wrong:
- Option A (1): This is incorrect because the ratio of CO2 to O2 is not equal to 1.
- Option C (1.45): This is incorrect because it is the inverse of the correct ratio (145/102).
- Option D (1.62): This is incorrect as it is not the correct ratio of CO2 produced to O2 consumed.

Key Concepts:
- Respiratory Quotient (RQ)
- Stoichiometry of respiration

Memory Aids:
- RQ = CO2 out / O2 in

Quick Tips:
- Always check the balanced equation to get the correct stoichiometric coefficients.

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The energy-releasing metabolic process in which substrate is oxidized without an external electron acceptor, is called

View answer and explanation

Correct option: B

Fermentation

Option B (Fermentation) is correct because it is an energy-releasing metabolic process where a substrate is oxidized without an external electron acceptor.

Solution approach

Correct Answer + Core Reason: Fermentation. Fermentation is an energy-releasing process where a substrate is oxidized without an external electron acceptor.

Step-by-Step Solution:

Step 1: Define the Process: The process involves oxidation of a substrate without an external electron acceptor.

Step 2: Consider Glycolysis: Glycolysis involves oxidation but uses NAD+ as an electron acceptor.

Step 3: Consider Fermentation: Fermentation regenerates NAD+ without an external electron acceptor.

Step 4: Consider Aerobic Respiration: Aerobic respiration uses oxygen as the final electron acceptor.

Step 5: Consider Photorespiration: Photorespiration is related to photosynthesis, not energy release from a substrate.

Why other options are wrong:
- Option A (glycolysis): Glycolysis uses NAD+ as an electron acceptor.
- Option C (Aerobic respiration): Aerobic respiration uses oxygen as the final electron acceptor.
- Option D (Photorespiration): Photorespiration is related to photosynthesis, not energy release from a substrate.

Key Concepts:
- Fermentation
- Oxidation-Reduction Reactions
- Electron Acceptors

Memory Aids:
- Fermentation = no external electron acceptor.

Quick Tips:
- Remember the role of electron acceptors in different metabolic processes.

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Most of the biological energy is supplied by mitochondria through

View answer and explanation

Correct option: D

Oxidising TCA (tricarboxylic acid) substrate

Option D (Oxidising TCA (tricarboxylic acid) substrate) is correct because the electron transport chain, powered by the oxidation of TCA cycle intermediates, generates the majority of ATP.

Solution approach

Correct Answer + Core Reason: Oxidising TCA (tricarboxylic acid) substrate. Oxidation of TCA cycle intermediates powers ATP production.

Step-by-Step Solution:

Step 1: Recall the role of mitochondria: They are the powerhouses of the cell, producing ATP.

Step 2: Understand the process: The TCA cycle (Krebs cycle) generates electron carriers (NADH and FADH2).

Step 3: Explain the electron transport chain: These carriers donate electrons to the electron transport chain, which drives ATP synthesis through oxidative phosphorylation.

Step 4: Conclude: Most biological energy is supplied by mitochondria through oxidizing TCA substrates.

Why other options are wrong:
- Option A (Breaking of proteins): Proteins are not the primary energy source; carbohydrates and fats are.
- **Option B (Reduction of NADP $^ +$ ):** NADP $^+$ reduction is more important in photosynthesis than in mitochondrial energy production.
- Option C (Breaking of sugars): Sugar breakdown (glycolysis) yields some ATP, but most comes from the TCA cycle and oxidative phosphorylation.

Key Concepts:
- Mitochondria
- TCA Cycle (Krebs Cycle)
- Oxidative Phosphorylation
- Electron Transport Chain

Memory Aids:
- Remember that the TCA cycle and electron transport chain are the main ATP-producing pathways in mitochondria.

Quick Tips:
- Understand the flow of electrons in cellular respiration.

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The reactions of Pentose Phosphate Pathway (PPP) take place in

View answer and explanation

Correct option: B

Cytoplasm

Option B (Cytoplasm) is correct because the enzymes of the pentose phosphate pathway are located in the cytoplasm.

Solution approach

Correct Answer + Core Reason: Cytoplasm. The enzymes of the Pentose Phosphate Pathway (PPP) are located in the cytoplasm.

Step-by-Step Solution:

Step 1: Define the PPP: The Pentose Phosphate Pathway (PPP) is a metabolic pathway that produces NADPH and pentoses.

Step 2: Identify the Location: The PPP occurs in the cytoplasm of cells.

Why other options are wrong:
- Option A (Mitochondrion): The PPP does not occur in the mitochondrion.
- Option C (Chloroplast, peroxisome and mitochondrion): The PPP does not occur in the chloroplast, peroxisome or mitochondrion.
- Option D (Chloroplast, glyoxysome and mitochondrion): The PPP does not occur in the chloroplast, glyoxysome or mitochondrion.

Key Concepts:
- Pentose Phosphate Pathway
- Cytoplasm

Memory Aids:
- PPP = Cytoplasm.

Quick Tips:
- Remember the location of major metabolic pathways.

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In citric acid cycle first step is

View answer and explanation

Correct option: A

Acetyl Co-A combines with oxalo acetic acid

Option A (Acetyl Co-A combines with oxalo acetic acid) is correct because the first step of the citric acid cycle involves the combination of acetyl-CoA with oxaloacetic acid to form citric acid.

Solution approach

Correct Answer + Core Reason: Acetyl Co-A combines with oxalo acetic acid. This is the initial condensation reaction of the citric acid cycle.

Step-by-Step Solution:

Step 1: Understand the citric acid cycle (Krebs cycle).

Step 2: Recall the first step of the cycle.

Step 3: Identify the reactants: Acetyl Co-A and oxaloacetic acid.

Step 4: Recognize that they combine to form citric acid.

Step 5: Conclusion: Acetyl Co-A combines with oxaloacetic acid in the first step.

Why other options are wrong:
- Option B: Acetyl Co-A combines with citric acid: This is incorrect; citric acid is the *product* of the first step.
- Option C: Citric acid combines with oxaloacetic acid: This is incorrect; they don't combine directly.
- Option D: Citric acid combines with malic acid: This is incorrect; these are intermediates in the cycle, but not the first step.

Key Concepts:
- Citric Acid Cycle (Krebs Cycle)
- Acetyl Co-A
- Oxaloacetic Acid
- Citric Acid

Memory Aids:
- "Citrate Is Krebs' Starting Substrate For Making Oxaloacetate". (Citrate is formed from Acetyl-CoA and Oxaloacetate)

Quick Tips:
- Memorize the key steps and intermediates of the citric acid cycle.

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In plants, glucose is derived from which of the following?

View answer and explanation

Correct option: D

Sucrose

Option D (Sucrose) is correct because sucrose is a disaccharide that is broken down into glucose and fructose, providing glucose for plant metabolism.

Solution approach

Correct Answer + Core Reason: Sucrose. Sucrose is a disaccharide that breaks down into glucose and fructose.

Step-by-Step Solution:

Step 1: Recall the composition of sucrose. Sucrose is a disaccharide composed of glucose and fructose.

Step 2: Explain the breakdown of sucrose. During metabolism, sucrose is hydrolyzed into its constituent monosaccharides, glucose and fructose.

Step 3: Relate glucose to plant metabolism. Glucose is a primary substrate for respiration and other metabolic processes in plants.

Why other options are wrong:
- Option A: Protein: Proteins are primarily used for structural and enzymatic functions, not as a direct source of glucose.
- Option B: Fat: Fats can be converted to glucose through gluconeogenesis, but this is not the primary source of glucose in plants.
- Option C: Oxalic acid: Oxalic acid is a waste product and not a source of glucose.

Key Concepts:
- Carbohydrate metabolism
- Sucrose structure
- Glucose as a metabolic substrate

Memory Aids:
- Sucrose = Glucose + Fructose

Quick Tips:
- Consider the breakdown products of complex carbohydrates.

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In the electron transport system present in the inner mitochondrial membrane, complexes I and IV are respectively

View answer and explanation

Correct option: C

NADH Dehydrogenase and cytochrome-c oxidase complex

Option C (NADH Dehydrogenase and cytochrome-c oxidase complex) is correct because Complex I is NADH dehydrogenase and Complex IV is cytochrome-c oxidase complex in the electron transport system.

Solution approach

Correct Answer + Core Reason: NADH Dehydrogenase and cytochrome-c oxidase complex. Complex I is NADH dehydrogenase and Complex IV is cytochrome-c oxidase complex in the electron transport system.

Step-by-Step Solution:

Step 1: Recall the components of the electron transport system (ETS).

Step 2: Complex I is NADH dehydrogenase, which accepts electrons from NADH.

Step 3: Complex IV is cytochrome-c oxidase complex, which transfers electrons to oxygen.

Why other options are wrong:
- Option A: $FADH_2$ is associated with Complex II, not Complex IV.
- Option B: The order is incorrect; Complex I is NADH dehydrogenase.
- Option D: ATP synthase is Complex V, not Complex IV.

Key Concepts:
- Electron transport system (ETS)
- NADH dehydrogenase
- Cytochrome-c oxidase complex

Memory Aids:
- Remember the order of the complexes in the ETS.

Quick Tips:
- Know the function of each complex in the electron transport system.

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Seeds respire in

View answer and explanation

Correct option: D

both 1 and 3

Option D (both 1 and 3) is correct because seeds can respire both aerobically (in the presence of oxygen) and anaerobically (in the absence of oxygen), although aerobic respiration is more efficient.

Solution approach

Correct Answer + Core Reason: both 1 and 3. Seeds can respire in the presence or absence of oxygen.

Step-by-Step Solution:

Step 1: Identify the question: Under what conditions do seeds respire?

Step 2: Recall seed respiration: Seeds can respire aerobically (with oxygen) and anaerobically (without oxygen).

Step 3: Conclusion: Seeds respire in the presence and absence of oxygen.

Why other options are wrong:
- Option A: Seeds can respire in the presence of oxygen, but it's not the only condition.
- Option B: Carbon dioxide is a product of respiration, not a requirement.
- Option C: Seeds can respire in the absence of oxygen, but it's not the only condition.

Key Concepts:
- Seed respiration
- Aerobic respiration
- Anaerobic respiration

Memory Aids:
- Seeds can respire with or without oxygen.

Quick Tips:
- Remember that anaerobic respiration is less efficient than aerobic respiration.

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Assertion: Plants can get along without respiratory organs.

Reason: Roots, stems and leaves respire at rates per lower than animals do.

View answer and explanation

Correct option: A

If both Assertion & Reason are true and the reason is the correct explanation of the assertion

Option A is correct because plants don't have specialized respiratory organs, and their respiration rate is lower than animals, making the assertion and reason true and related.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are true and the reason is the correct explanation of the assertion. Plants do not possess specialized respiratory organs, and their lower respiration rates support this.

Step-by-Step Solution:
Step 1: Analyze the Assertion: Plants do not have specialized respiratory organs like lungs or gills.
Step 2: Analyze the Reason: Roots, stems, and leaves respire at lower rates compared to animals.
Step 3: Determine the relationship: The lower respiration rate in plants is a reason why they can function without specialized respiratory organs.
Step 4: Conclude: Both the assertion and reason are true, and the reason correctly explains the assertion.

Why other options are wrong:
- Option B: While both statements are true, the reason directly explains the assertion, making option A more accurate.
- Option C: The assertion is true, so this option is incorrect.
- Option D: Both the assertion and reason are true, so this option is incorrect.

Key Concepts:
- Plant respiration
- Absence of specialized respiratory organs in plants
- Respiration rates in plants and animals

Memory Aids:
- Plants don't "breathe" like animals; they respire at a slower pace.

Quick Tips:
- Consider the overall physiology of plants and animals when comparing respiration.

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Which of the given organic molecules have RQ less than unity?
A. Fat B. Oxalic acid C. Protein D. Carbohydrate

View answer and explanation

Correct option: D

A and C

Option D is correct because fats and proteins have RQ values less than unity, while oxalic acid and carbohydrates have RQ values greater than or equal to unity.

Solution approach

Correct Answer + Core Reason: A and C. Fats and proteins have RQ values less than unity.

Step-by-Step Solution:

Step 1: Recall RQ values for different organic molecules.

Step 2: RQ value for Fat: Fats have an RQ value of approximately 0.7, which is less than unity.

Step 3: RQ value for Oxalic acid: Oxalic acid has an RQ value greater than 1.

Step 4: RQ value for Protein: Proteins have an RQ value of approximately 0.8-0.9, which is less than unity.

Step 5: RQ value for Carbohydrate: Carbohydrates have an RQ value of 1.

Step 6: Identify molecules with RQ less than unity: Fat and Protein.

Why other options are wrong:
- Option A: Oxalic acid has RQ > 1, so this is incorrect.
- Option B: Oxalic acid has RQ > 1 and Carbohydrate has RQ = 1, so this is incorrect.
- Option C: Carbohydrate has RQ = 1, so this is incorrect.

Key Concepts:
- Respiratory Quotient (RQ)
- RQ values for different substrates

Memory Aids:
- Fats and Proteins < 1, Carbohydrates = 1

Quick Tips:
- Memorize the approximate RQ values for common respiratory substrates.

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In photosynthesis, NADPH

_2

is formed but in respiration it forms during

View answer and explanation

Correct option: A

HMP

Option A (HMP) is correct because NADPH

_2

is formed during the Hexose Monophosphate Pathway (HMP), also known as the Pentose Phosphate Pathway, which can operate alongside respiration.

Solution approach

Correct Answer + Core Reason: HMP. NADPH2 is formed during the Hexose Monophosphate Pathway.

Step-by-Step Solution:

Step 1: Photosynthesis: NADPH2 is formed during the light-dependent reactions.

Step 2: Respiration: NADPH2 is formed during the HMP pathway.

Step 3: HMP pathway: Also known as the Pentose Phosphate Pathway.

Step 4: Conclude: NADPH2 forms during HMP in respiration.

Why other options are wrong:
- Option B: ETS (Electron Transport System) uses NADH and FADH2, not directly forming NADPH2.
- Option C: Krebs' cycle produces NADH and FADH2, not directly forming NADPH2.
- Option D: HMP is the correct answer.

Key Concepts:
- HMP pathway
- NADPH2 formation

Memory Aids:
- HMP = NADPH2

Quick Tips:
- Remember that HMP is a source of NADPH2 in respiration.

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In oxidative decarboxylation, only a carbon molecule of pyruvic acid is get oxidised, other two carbon molecule goes to form

View answer and explanation

Correct option: A

Acetyl Co-A

Option A (Acetyl Co-A) is correct because the two-carbon molecule resulting from oxidative decarboxylation binds to Coenzyme A to form Acetyl Co-A.

Solution approach

Correct Answer + Core Reason: Acetyl Co-A. The remaining two-carbon molecule forms Acetyl Co-A.

Step-by-Step Solution:

Step 1: Recall the process of oxidative decarboxylation.

Step 2: Understand the products: Pyruvic acid (3 carbons) loses one carbon as $CO_2$ , leaving a two-carbon molecule.

Step 3: Identify the fate of the two-carbon molecule: This two-carbon molecule combines with Coenzyme A to form Acetyl Co-A.

Step 4: Conclude: The other two carbon molecules go to form Acetyl Co-A.

Why other options are wrong:
- **Option B ( $C{O_2}$ ):** $CO_2$ is the carbon molecule that is oxidized and released, not the destination of the other two carbons.
- Option C (Citric acid): Citric acid is formed later in the Krebs cycle, after Acetyl Co-A combines with oxaloacetate.
- Option D (Both (a) and (b)): Only Acetyl Co-A is formed from the remaining two carbons.

Key Concepts:
- Oxidative Decarboxylation
- Acetyl Co-A Formation
- Krebs Cycle

Memory Aids:
- Remember that Acetyl Co-A is the entry molecule for the Krebs cycle.

Quick Tips:
- Focus on the carbon balance during the reactions.

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Anaerobic respiration is also called as

View answer and explanation

Correct option: B

Fermentation

Option B (Fermentation) is correct because anaerobic respiration is the process of energy production in the absence of oxygen, which is also known as fermentation.

Solution approach

Correct Answer + Core Reason: Fermentation. Anaerobic respiration is also known as fermentation.

Step-by-Step Solution:

Step 1: Define Anaerobic Respiration: Anaerobic respiration is respiration without oxygen.

Step 2: Define Fermentation: Fermentation is a type of anaerobic respiration.

Step 3: Relate the Terms: Anaerobic respiration and fermentation are often used interchangeably, although fermentation is a more specific term.

Why other options are wrong:
- Option A (β-oxidation): β-oxidation is the breakdown of fatty acids, which requires oxygen.
- Option C (Oxidation): Oxidation can occur in both aerobic and anaerobic respiration.
- Option D (None of these): Fermentation is the correct answer.

Key Concepts:
- Anaerobic Respiration
- Fermentation

Memory Aids:
- Anaerobic = no air = fermentation.

Quick Tips:
- Recognize the relationship between anaerobic respiration and fermentation.

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Fermentation is

View answer and explanation

Correct option: D

both 1 and 2

Option D (both 1 and 2) is correct because fermentation is a type of anaerobic respiration and involves the incomplete oxidation of carbohydrates.

Solution approach

Correct Answer + Core Reason: both 1 and 2. Fermentation is anaerobic and involves incomplete oxidation.

Step-by-Step Solution:

Step 1: Define fermentation: Metabolic process in the absence of oxygen.

Step 2: Anaerobic respiration: Respiration without oxygen.

Step 3: Incomplete oxidation: Partial breakdown of carbohydrates.

Step 4: Conclude: Fermentation is both anaerobic respiration and involves incomplete oxidation.

Why other options are wrong:
- Option A: Fermentation is anaerobic respiration, but this is not the complete answer.
- Option B: Fermentation involves incomplete oxidation, but this is not the complete answer.
- Option C: Fermentation involves incomplete, not complete, oxidation.

Key Concepts:
- Fermentation
- Anaerobic respiration
- Incomplete oxidation

Memory Aids:
- Fermentation = Anaerobic + Incomplete

Quick Tips:
- Remember that fermentation is both anaerobic and involves incomplete oxidation.

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Which one is correct sequence in glycolysis?

View answer and explanation

Correct option: B

G-6-P→3-PGAL → 3-PGA → PEP

Option B is correct because the correct sequence is Glucose-6-Phosphate -> 3-Phosphoglyceraldehyde -> 3-Phosphoglycerate -> Phosphoenolpyruvate.

Solution approach

Correct Answer + Core Reason: G-6-P→3-PGAL → 3-PGA → PEP. This is the correct sequence of intermediates in glycolysis.

Step-by-Step Solution:
Step 1: Recall the sequence of intermediates in glycolysis.
Step 2: Start with Glucose-6-Phosphate (G-6-P).
Step 3: Follow with 3-Phosphoglyceraldehyde (3-PGAL).
Step 4: Then 3-Phosphoglycerate (3-PGA).
Step 5: End with Phosphoenolpyruvate (PEP).

Why other options are wrong:
- Option A: PEP does not come before 3-PGAL and 3-PGA in glycolysis.
- Option C: PEP does not come before 3-PGA and 3-PGAL in glycolysis.
- Option D: 3-PGA does not come before 3-PGAL in glycolysis.

Key Concepts:
- Glycolysis
- Intermediates in glycolysis
- Biochemical pathways

Memory Aids:
- Remember the order of key intermediates in glycolysis.

Quick Tips:
- Draw out the glycolysis pathway to memorize the sequence.

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In which of the following reactions of glycolysis, oxidation takes place?

View answer and explanation

Correct option: B

Glyceraldehydes 3-phosphate to 1, 3-diphosphoglycerate

Option B (Glyceraldehydes 3-phosphate to 1, 3-diphosphoglycerate) is correct because, in this step, glyceraldehyde 3-phosphate is oxidized, and NAD+ is reduced to NADH.

Solution approach

Correct Answer + Core Reason: Glyceraldehydes 3-phosphate to 1, 3-diphosphoglycerate. This step involves oxidation of glyceraldehyde 3-phosphate and reduction of NAD+.

Step-by-Step Solution:

Step 1: Recall the steps of glycolysis.\nGlycolysis involves several oxidation-reduction reactions.

Step 2: Analyze the given reactions.
- Glucose 6- $P{O_4}$ to fructose 6- $P{O_4}$ : Isomerization, no oxidation.
- Glyceraldehydes 3-phosphate to 1, 3-diphosphoglycerate: Oxidation of glyceraldehyde 3-phosphate and reduction of NAD+ to NADH.
- 1,3-diphosphoglycerate to 3-phosphoglycerate: Substrate-level phosphorylation, no oxidation.
- 2-phosphoglycerate to phosphoglycerate: Dehydration, no oxidation.

Step 3: Identify the reaction where oxidation takes place.\nGlyceraldehydes 3-phosphate to 1, 3-diphosphoglycerate

Step 4: Final Answer\nGlyceraldehydes 3-phosphate to 1, 3-diphosphoglycerate

Why other options are wrong:
- **Option A (Glucose 6- $P{O_4}$ to fructose 6- $P{O_4}$ ): This is an isomerization reaction, not an oxidation-reduction reaction.
- Option C (1,3-diphosphoglycerate to 3-phosphoglycerate): This is a substrate-level phosphorylation reaction, not an oxidation-reduction reaction.
- Option D (2-phosphoglycerate to phosphoglycerate): This is a dehydration reaction, not an oxidation-reduction reaction.

Key Concepts:
- Glycolysis
- Oxidation-reduction reactions

Memory Aids:
- Remember that oxidation involves the loss of electrons or hydrogen atoms.

Quick Tips:**
- Look for changes in oxidation states to identify oxidation-reduction reactions.

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Where is ATP synthesised in glycolysis?

View answer and explanation

Correct option: A

When 1, 3 di PGA is changed into 3PGA

Option A is correct because ATP is synthesized when 1,3-bisphosphoglycerate (1,3 di PGA) is converted to 3-phosphoglycerate (3PGA) by phosphoglycerate kinase.

Solution approach

Correct Answer + Core Reason: When 1, 3 di PGA is changed into 3PGA. ATP is generated during this step in glycolysis.

Step-by-Step Solution:
Step 1: Recall the steps of glycolysis where ATP is produced.
Step 2: Identify the conversion of 1,3-bisphosphoglycerate to 3-phosphoglycerate as one such step.
Step 3: Recognize that this reaction is catalyzed by phosphoglycerate kinase, which generates ATP.

Why other options are wrong:
- Option B: ATP is consumed, not synthesized, when glucose is converted to glucose-6-phosphate.
- Option C: Only option A is correct.
- Option D: No ATP is directly synthesized when 1,6-diphosphate is broken down into triose phosphates.

Key Concepts:
- Glycolysis
- ATP synthesis
- Substrate-level phosphorylation

Memory Aids:
- 1,3 di PGA to 3PGA = ATP production.

Quick Tips:
- Memorize the steps in glycolysis where ATP is produced or consumed.

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Which of the following exhibits the highest rate of respiration?

View answer and explanation

Correct option: B

Germinating seed

Option B (Germinating seed) is correct because germinating seeds have a high metabolic rate due to rapid cell division and growth, requiring a high rate of respiration.

Solution approach

Correct Answer + Core Reason: Germinating seed. Germinating seeds have the highest metabolic activity due to rapid growth and cell division.

Step-by-Step Solution:

Step 1: Consider the metabolic activity of each option.

Step 2: Growing shoot apex: Actively growing, but not as rapid as germination.

Step 3: Germinating seed: Undergoing rapid cell division and growth, requiring a high rate of respiration.

Step 4: Root tip: Actively growing, but generally slower than a germinating seed.

Step 5: Leaf bud: Relatively dormant compared to a germinating seed.

Why other options are wrong:
- Option A: Growing shoot apex: While growing, it doesn't have the same rapid growth rate as a germinating seed.
- Option C: Root tip: Similar to the shoot apex, slower growth than a germinating seed.
- Option D: Leaf bud: Generally more dormant and has a lower respiration rate than a germinating seed.

Key Concepts:
- Respiration rate
- Metabolic activity
- Germination

Memory Aids:
- Germination = Rapid growth = High respiration

Quick Tips:
- Consider the growth rate and metabolic demands of each option.

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How many ATP molecules could maximally be generated from one molecule of glucose, if the complete oxidation of one mole of glucose to carbon dioxide and water yields 686 kcal and the useful chemical energy available in the high energy phosphate bond of one mole of ATP is 12 kcal?

View answer and explanation

Correct option: C

Fifty seven

Option C (Fifty seven) is correct because dividing the total energy released from glucose by the energy per ATP molecule gives the maximum number of ATP molecules that can be generated.

Solution approach

Correct Answer + Core Reason: Fifty seven. Dividing total energy released by energy per ATP gives the maximum ATP molecules.

Step-by-Step Solution:

Step 1: Identify the total energy released: 686 kcal.

Step 2: Identify the energy per ATP molecule: 12 kcal.

Step 3: Calculate the maximum number of ATP molecules: 686 kcal / 12 kcal/ATP = 57.17 ATP.

Step 4: Round down to the nearest whole number: Since you can't have a fraction of an ATP molecule, round down to 57.

Step 5: Conclude: Approximately 57 ATP molecules can be generated.

Why other options are wrong:
- Option A (Two): This is far too low and doesn't account for the energy released from glucose oxidation.
- Option B (Thirty): This is a common estimate for ATP production, but the question asks for the *maximum* possible based on the given energy values.
- Option D (One): This is incorrect as glucose oxidation yields significantly more than one ATP molecule.

Key Concepts:
- ATP Production
- Cellular Respiration
- Energy Yield

Memory Aids:
- Remember that glucose oxidation yields a significant amount of ATP.

Quick Tips:
- Pay attention to the units and what the question is asking for (maximum vs. typical). This question is a theoretical maximum and does not reflect actual ATP yield.

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Read the following statements (a-e) and choose the suitable option w.r.t. respiration
a. A series of slow step-wise reactions controlled by enzymes.
b. Breaking of the

C-C

bonds of respiratory substrates through oxidation within the cells.
c. Liberation of energy.
d. Breakdown of complex molecules to yield energy which takes place in mitochondria only.
e. Carbon skeleton produced during respiration is not used as precursor for cellular biosynthesis of other molecules.

View answer and explanation

Correct option: B

d and e are incorrect

Option B (d and e are incorrect) is correct because respiration occurs in both mitochondria and cytoplasm, and carbon skeletons are used for biosynthesis.

Solution approach

Correct Answer + Core Reason: d and e are incorrect. Statement d is incorrect because glycolysis occurs in the cytoplasm, and statement e is incorrect because carbon skeletons are used as precursors.

Step-by-Step Solution:

Step 1: Analyze statement a: A series of slow step-wise reactions controlled by enzymes - Correct.

Step 2: Analyze statement b: Breaking of the $C-C$ bonds of respiratory substrates through oxidation within the cells - Correct.

Step 3: Analyze statement c: Liberation of energy - Correct.

Step 4: Analyze statement d: Breakdown of complex molecules to yield energy which takes place in mitochondria only - Incorrect. Glycolysis occurs in the cytoplasm.

Step 5: Analyze statement e: Carbon skeleton produced during respiration is not used as precursor for cellular biosynthesis of other molecules - Incorrect. Carbon skeletons are used for biosynthesis.

Why other options are wrong:
- Option A (b, c and e are correct): Statement e is incorrect.
- Option C (Only d is incorrect): Statement e is also incorrect.
- Option D (a and e are correct): Statement e is incorrect.

Key Concepts:
- Respiration
- Glycolysis
- Krebs Cycle
- Oxidative Phosphorylation
- Biosynthesis

Memory Aids:
- Remember that glycolysis occurs in the cytoplasm.

Quick Tips:
- Pay attention to the location of different stages of respiration.

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Find the incorrect match w.r.t. respiratory substrates and their RQ.

View answer and explanation

Correct option: D

Oxalic acid – 0.6

Option D (Oxalic acid – 0.6) is correct because the RQ of oxalic acid is 4, not 0.6. The other RQ values are approximately correct.

Solution approach

Correct Answer + Core Reason: Oxalic acid – 0.6. The RQ of oxalic acid is 4, not 0.6.

Step-by-Step Solution:

Step 1: Review the RQ values for different respiratory substrates.

Step 2: Tripalmitin: RQ ≈ 0.7 (Fat)

Step 3: Proteins: RQ ≈ 0.9

Step 4: Carbohydrates: RQ ≈ 1

Step 5: Oxalic acid: RQ = 4 (Organic acid)

Step 6: Identify the incorrect match. Oxalic acid – 0.6 is the incorrect match because the RQ of oxalic acid is 4.

Why other options are wrong:
- Option A (Tripalmitin – 0.7): This is correct.
- Option B (Proteins – 0.9): This is correct.
- Option C (Carbohydrates – 1): This is correct.

Key Concepts:
- Respiratory Quotient (RQ)
- RQ values for different substrates

Memory Aids:
- Remember the RQ values for common substrates.

Quick Tips:
- Know the RQ values for carbohydrates, fats, proteins, and organic acids.

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During conversion of all the following,

C{O_2}

is released along with synthesis of

NADH\mathrm{ } + \mathrm{ }{H^ + }

, except
a. Pyruvic acid to acetyl CoA b. -ketoglutaric acid to succinyl CoA
c. Malic acid to oxaloacetic acid

View answer and explanation

Correct option: D

c only

Option D (c only) is correct because the conversion of malic acid to oxaloacetic acid produces NADH + H+, but does not release

CO_2

. Decarboxylation does not occur in this step.

Solution approach

Correct Answer + Core Reason: c only. The conversion of malic acid to oxaloacetic acid does not involve the release of $CO_2$ .

Step-by-Step Solution:

Step 1: Analyze the conversion of pyruvic acid to acetyl CoA. This process involves decarboxylation (release of $CO_2$ ) and the reduction of $NAD^+$ to $NADH + H^+$ .

Step 2: Analyze the conversion of $\alpha$ -ketoglutaric acid to succinyl CoA. This process involves decarboxylation (release of $CO_2$ ) and the reduction of $NAD^+$ to $NADH + H^+$ .

Step 3: Analyze the conversion of malic acid to oxaloacetic acid. This process involves the oxidation of malate, reducing $NAD^+$ to $NADH + H^+$ , but does not involve decarboxylation (no release of $CO_2$ ).

Step 4: Final Answer: Only the conversion of malic acid to oxaloacetic acid does not release $CO_2$ .

Why other options are wrong:
- Option A (a only): Incorrect, pyruvic acid to acetyl CoA releases $CO_2$ .
- Option B (b only): Incorrect, $\alpha$ -ketoglutaric acid to succinyl CoA releases $CO_2$ .
- Option C (b and c only): Incorrect, $\alpha$ -ketoglutaric acid to succinyl CoA releases $CO_2$ , but malic acid to oxaloacetic acid does not.

Key Concepts:
- Krebs Cycle (Citric Acid Cycle)
- Decarboxylation
- $NADH$ Production

Memory Aids:
- Remember the steps in Krebs cycle that involve decarboxylation.

Quick Tips:
- Review the Krebs cycle and identify the steps where $CO_2$ is released.

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