RESPIRATION IN PLANTS MCQs for NEET UG BIOLOGY — Medium

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ATP is a

View answer and explanation

Correct option: B

Nucleotide

Option B (Nucleotide) is correct because ATP (adenosine triphosphate) is a nucleotide, consisting of a nitrogenous base (adenine), a sugar (ribose), and three phosphate groups.

Solution approach

Correct Answer + Core Reason: Nucleotide. ATP is a nucleotide.

Step-by-Step Solution:

Step 1: Define ATP's structure.

Step 2: ATP consists of adenine, ribose, and three phosphate groups. This is the structure of a nucleotide.

Why other options are wrong:
- Option A: Purine is a type of nitrogenous base (adenine is a purine), but ATP is more than just a purine.
- Option C: Nucleoside consists of a nitrogenous base and a sugar, but lacks the phosphate groups.
- Option D: Nucleosome is a structural unit of DNA packaging in eukaryotes.

Key Concepts:
- Nucleotide
- ATP

Memory Aids:
- ATP = Nucleotide

Quick Tips:
- Remember the components of a nucleotide: base, sugar, and phosphate.

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Which of the following process takes place in anaerobic bacteria ?

View answer and explanation

Correct option: C

Glycolysis

Option C (Glycolysis) is correct because glycolysis is the only metabolic pathway listed that can occur in the absence of oxygen, making it essential for anaerobic bacteria.

Solution approach

Correct Answer + Core Reason: Glycolysis. Glycolysis does not require oxygen and occurs in anaerobic bacteria.

Step-by-Step Solution:

Step 1: Recall the requirements of each process.
- ETS (Electron Transport System) requires oxygen.
- Krebs cycle requires oxygen.
- Glycolysis does not require oxygen.
- Link reaction requires oxygen (indirectly).

Step 2: Identify the process that can occur in anaerobic bacteria.\nGlycolysis is the only process listed that does not require oxygen.

Step 3: Final Answer\nGlycolysis

Why other options are wrong:
- Option A (ETS): ETS requires oxygen as the final electron acceptor.
- Option B (Krebs cycle): The Krebs cycle requires oxygen (indirectly through the ETC).
- Option D (Link reaction): The link reaction requires oxygen (indirectly through the ETC).

Key Concepts:
- Anaerobic respiration
- Glycolysis
- Krebs cycle
- Electron transport system

Memory Aids:
- Remember that glycolysis is the first step in both aerobic and anaerobic respiration.

Quick Tips:
- Know which processes require oxygen and which do not.

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Preparatory phase before fermentation is

View answer and explanation

Correct option: A

Upstream process

Option A (Upstream process) is correct because the preparatory phase before fermentation involves processes like media preparation and sterilization, which are upstream activities.

Solution approach

Correct Answer + Core Reason: Upstream process. Preparatory phases like media preparation are considered upstream.

Step-by-Step Solution:

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

Step 2: Define upstream process: Initial steps before the main fermentation process.

Step 3: Examples of upstream processes: Media preparation, sterilization, and inoculation.

Step 4: Conclude: Preparatory phase is an upstream process.

Why other options are wrong:
- Option B: Downstream process involves product recovery and purification after fermentation.
- Option C: Inoculation is part of the upstream process, not the entire preparatory phase.
- Option D: Filtration can be part of both upstream (sterilization) and downstream processes.

Key Concepts:
- Fermentation process
- Upstream and downstream processing

Memory Aids:
- Upstream = Before, Downstream = After

Quick Tips:
- Distinguish between steps before and after the main fermentation.

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Anaerobic respiration takes place in

View answer and explanation

Correct option: C

Cytoplasm

Option C (Cytoplasm) is correct because anaerobic respiration, which does not require oxygen, takes place in the cytoplasm of the cell.

Solution approach

Correct Answer + Core Reason: Cytoplasm. Anaerobic respiration takes place in the cytoplasm.

Step-by-Step Solution:

Step 1: Recall the location of anaerobic respiration.

Step 2: Glycolysis, the first step in both aerobic and anaerobic respiration, occurs in the cytoplasm. In anaerobic respiration, the subsequent steps (e.g., fermentation) also occur in the cytoplasm.

Why other options are wrong:
- Option A: Mitochondria is the site of aerobic respiration.
- Option B: Nucleus contains the genetic material.
- Option D: Vacuole is involved in storage and waste disposal.

Key Concepts:
- Anaerobic Respiration
- Cytoplasm

Memory Aids:
- Anaerobic = Cytoplasm

Quick Tips:
- Anaerobic respiration occurs in the absence of oxygen.

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2{C_{51}}{H_{98}}{O_6} + 146{O_2} \to 102\,C{O_2} + 98\,{H_2}O

, the RQ of above equation is?

View answer and explanation

Correct option: A

0.7

Option A (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 102/146 ≈ 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, 146 O2 molecules are consumed.

Step 4: Calculate the RQ. $RQ = \frac{{102}}{{146}} \approx 0.7$

Why other options are wrong:
- Option B (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 not the correct ratio of CO2 produced to O2 consumed.
- 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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What is the net ATP molecules gain, when 4 molecules of glucose undergo anaerobic respiration in plant?

View answer and explanation

Correct option: A

8 ATP

Option A (8 ATP) is correct because anaerobic respiration (fermentation) yields a net gain of 2 ATP molecules per glucose molecule. Therefore, 4 glucose molecules will yield 4 * 2 = 8 ATP molecules.

Solution approach

Correct Answer + Core Reason: 8 ATP. Anaerobic respiration yields 2 ATP per glucose molecule.

Step-by-Step Solution:

Step 1: ATP yield per glucose in anaerobic respiration: 2 ATP.

Step 2: Number of glucose molecules: 4.

Step 3: Total ATP gain: 4 * 2 = 8 ATP.

Step 4: Conclude: The net ATP gain is 8 ATP.

Why other options are wrong:
- Option B: 20 ATP is incorrect and would be closer to aerobic respiration.
- Option C: 144 ATP is incorrect and far too high for anaerobic respiration.
- Option D: 16 ATP is incorrect; the correct calculation is 4*2 = 8.

Key Concepts:
- ATP yield in anaerobic respiration

Memory Aids:
- Anaerobic = 2 ATP per glucose

Quick Tips:
- Remember the ATP yield for anaerobic respiration is much lower than aerobic.

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Gaseous exchange in woody plants occurs through

View answer and explanation

Correct option: C

Lenticels – Stomata

Option C (Lenticels – Stomata) is correct because lenticels facilitate gas exchange in woody stems, while stomata are present in leaves for gas exchange.

Solution approach

Correct Answer + Core Reason: Lenticels – Stomata. Woody plants use lenticels and stomata for gaseous exchange.

Step-by-Step Solution:

Step 1: Identify the question: How do woody plants exchange gases?

Step 2: Recall the function of lenticels: Lenticels are pores in the bark of woody stems that allow gas exchange.

Step 3: Recall the function of stomata: Stomata are pores in leaves that allow gas exchange.

Step 4: Consider other options: Hydathodes and cuticle are not the primary structures for gas exchange in woody plants.

Step 5: Conclusion: Gaseous exchange in woody plants occurs through lenticels and stomata.

Why other options are wrong:
- Option A: Hydathodes are involved in guttation, not primary gas exchange.
- Option B: Hydathodes and cuticle are not the primary structures for gas exchange.
- Option D: Cuticle is a waxy layer that reduces water loss but does not facilitate significant gas exchange.

Key Concepts:
- Gas exchange in plants
- Lenticels
- Stomata
- Woody plants

Memory Aids:
- Lenticels = Woody stems = Gas exchange
- Stomata = Leaves = Gas exchange

Quick Tips:
- Remember that lenticels are specific to woody plants.

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Which one of the following is the terminal electron acceptor?

View answer and explanation

Correct option: B

Molecular

{O_2}

Option B (Molecular

{O_2}

) is correct because molecular oxygen is the final electron acceptor in the electron transport chain during aerobic respiration.

Solution approach

Correct Answer + Core Reason: Molecular ${O_2}$ . Oxygen acts as the terminal electron acceptor in the electron transport chain.

Step-by-Step Solution:

Step 1: Understand the electron transport chain (ETC).

Step 2: Recall the role of the terminal electron acceptor in the ETC.

Step 3: Identify the terminal electron acceptor as molecular oxygen ( ${O_2}$ ).

Step 4: Conclusion: Molecular oxygen is the terminal electron acceptor.

Why other options are wrong:
- Option A: Molecular $C{O_2}$ : Carbon dioxide is a product of respiration, not an electron acceptor.
- Option C: Molecular ${H_2}$ : Hydrogen is involved in the process, but not as the terminal acceptor.
- Option D: NADP ${H_2}$ : NADPH is an electron *donor*, not an acceptor.

Key Concepts:
- Electron Transport Chain (ETC)
- Terminal Electron Acceptor
- Aerobic Respiration

Memory Aids:
- Oxygen is the ultimate electron "sink".

Quick Tips:
- Remember that oxygen's high electronegativity makes it a good electron acceptor.

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In respiration, respiratory substances can be used

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because carbohydrates, proteins, and organic acids can all be used as respiratory substances.

Solution approach

Correct Answer + Core Reason: All of these. Carbohydrates, proteins, and organic acids can all serve as respiratory substrates.

Step-by-Step Solution:

Step 1: Understand the concept of respiratory substances.

Step 2: Recall the different types of molecules that can be oxidized to release energy.

Step 3: Recognize that carbohydrates, proteins, and organic acids can all be used in respiration.

Step 4: Conclusion: All of these substances can be used in respiration.

Why other options are wrong:
- Option A: Carbohydrate: While carbohydrates are a common respiratory substrate, they are not the *only* one.
- Option B: Protein: Proteins can be used as a respiratory substrate, but not exclusively.
- Option C: Organic acid: Organic acids can also be used, but again, not exclusively.

Key Concepts:
- Respiratory Substrates
- Carbohydrates, Proteins, and Fats as energy sources

Memory Aids:
- Think of the food groups that provide energy: carbs, proteins, fats (organic acids are derived from these).

Quick Tips:
- Consider the different metabolic pathways that can feed into respiration.

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Conversion of glucose to glucose 6-phosphate in EMP pathway is catalysed by

View answer and explanation

Correct option: D

Hexokinase

Option D is correct because hexokinase catalyzes the phosphorylation of glucose to glucose-6-phosphate in the first step of glycolysis.

Solution approach

Correct Answer + Core Reason: Hexokinase. Hexokinase catalyzes the conversion of glucose to glucose-6-phosphate.

Step-by-Step Solution:
Step 1: Recall the steps of glycolysis.
Step 2: Identify the enzyme that catalyzes the first step: Hexokinase.
Step 3: Understand the reaction: Hexokinase phosphorylates glucose to form glucose-6-phosphate.

Why other options are wrong:
- Option A: Aldolase catalyzes the cleavage of fructose-1,6-bisphosphate into two 3-carbon molecules.
- Option B: Phosphofructokinase catalyzes the phosphorylation of fructose-6-phosphate to fructose-1,6-bisphosphate.
- Option C: Enolase catalyzes the dehydration of 2-phosphoglycerate to phosphoenolpyruvate (PEP).

Key Concepts:
- Glycolysis
- Enzymes
- Hexokinase
- Phosphorylation

Memory Aids:
- Hexokinase = first step of glycolysis.

Quick Tips:
- Memorize the key enzymes in glycolysis and their respective reactions.

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Aerobic respiratory pathway is appropriately termed as

View answer and explanation

Correct option: C

Amphibolic

Option C (Amphibolic) is correct because the aerobic respiratory pathway involves both catabolic (breakdown) and anabolic (synthesis) processes.

Solution approach

Correct Answer + Core Reason: Amphibolic. The pathway involves both catabolic and anabolic processes.

Step-by-Step Solution:

Step 1: Define catabolic pathways: These break down complex molecules into simpler ones, releasing energy.

Step 2: Define anabolic pathways: These build complex molecules from simpler ones, requiring energy.

Step 3: Explain the respiratory pathway: It breaks down glucose (catabolic) but also provides intermediates for synthesis of other molecules (anabolic).

Step 4: Conclude: The aerobic respiratory pathway is appropriately termed as amphibolic.

Why other options are wrong:
- Option A (Catabolic): While it is catabolic, it also has anabolic aspects.
- Option B (Parabolic): This term is not relevant in this biological context.
- Option D (Anabolic): While it has anabolic aspects, it is primarily catabolic.

Key Concepts:
- Catabolism
- Anabolism
- Amphibolic Pathways
- Cellular Respiration

Memory Aids:
- Amphibolic pathways are both catabolic and anabolic.

Quick Tips:
- Consider the dual nature of metabolic pathways.

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How many ATP molecules will be generated in a plant system during complete oxidation of 40 molecules of glucose?

View answer and explanation

Correct option: C

1520

Option C (1520) is correct because each glucose molecule yields approximately 38 ATP molecules during complete oxidation, and 40 x 38 = 1520.

Solution approach

Correct Answer + Core Reason: 1520. Complete oxidation of one glucose molecule yields approximately 38 ATP molecules.

Step-by-Step Solution:

Step 1: Identify the ATP yield per glucose molecule during complete oxidation.

Step 2: The complete oxidation of one glucose molecule typically yields around 38 ATP molecules (net gain).

Step 3: Multiply the ATP yield per glucose molecule by the number of glucose molecules.

Step 4: Calculation: 40 glucose molecules * 38 ATP/glucose molecule = 1520 ATP molecules.

Step 5: Final Answer: 1520 ATP molecules

Why other options are wrong:
- Option A (180): This is incorrect. It might be a result of a miscalculation or misunderstanding of the ATP yield per glucose molecule.
- Option B (360): This is incorrect. It might be a result of a miscalculation or misunderstanding of the ATP yield per glucose molecule.
- Option D (3040): This is incorrect. It might be a result of multiplying by 76 instead of 38, or another miscalculation.

Key Concepts:
- Cellular Respiration
- ATP Production
- Glucose Oxidation

Memory Aids:
- Remember the approximate ATP yield per glucose molecule (around 38 ATP).

Quick Tips:
- Always consider the net ATP gain during complete oxidation, not just a single step.

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How many compounds in the list given below are common products of glycolysis, fermentation and Krebs cycle?

\mathrm{NA}{\mathrm{D}^ + }\mathrm{,}\,\,\mathrm{C}{\mathrm{O}_\mathrm{2}},\,

fructose – 6 – phosphate, GTP, Lactic acid, Malic acid,

\mathrm{NADH + }{\mathrm{H}^ + }\mathrm{,}\,\,\mathrm{FAD}{\mathrm{H}_\mathrm{2}}

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because there are no compounds listed that are common products of all three processes: glycolysis, fermentation, and the Krebs cycle.

Solution approach

Correct Answer + Core Reason: Zero. No compound in the list is a common product of glycolysis, fermentation, and the Krebs cycle.

Step-by-Step Solution:

Step 1: Analyze the products of glycolysis. Glycolysis produces pyruvate, ATP, NADH, etc.

Step 2: Analyze the products of fermentation. Fermentation produces ethanol or lactic acid, $CO_2$ , and a small amount of ATP.

Step 3: Analyze the products of the Krebs cycle. The Krebs cycle produces $CO_2$ , NADH, FADH2, GTP, etc.

Step 4: Compare the products of all three processes with the list provided.

Step 5: $NAD^+$ is a reactant, not a product.

Step 6: $CO_2$ is produced in fermentation and Krebs cycle, but not directly in glycolysis.

Step 7: Fructose-6-phosphate is an intermediate in glycolysis, not a product.

Step 8: GTP is produced in the Krebs cycle only.

Step 9: Lactic acid is produced in lactic acid fermentation only.

Step 10: Malic acid is an intermediate in the Krebs cycle, not a product common to all three.

Step 11: $NADH + H^+$ is produced in glycolysis and Krebs cycle, but not directly in fermentation (it's regenerated).

Step 12: $FADH_2$ is produced in the Krebs cycle only.

Step 13: Final Answer: Zero compounds are common products of all three processes.

Why other options are wrong:
- Option B (Four): Incorrect, there are no four compounds common to all three.
- Option C (Two): Incorrect, there are no two compounds common to all three.
- Option D (One): Incorrect, there is no one compound common to all three.

Key Concepts:
- Glycolysis
- Fermentation
- Krebs Cycle (Citric Acid Cycle)
- Metabolic Pathways

Memory Aids:
- Remember the key products of each metabolic pathway.

Quick Tips:
- Review the steps and products of glycolysis, fermentation, and the Krebs cycle.

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Which of the following is true regarding glycolysis?
I. Takes place in cytosol
II. Produces no ATP
III. Has no connection with electron transport chain
IV. Reduces two molecules of NAD

+

for every glucose molecule processed
Choose the correct option

View answer and explanation

Correct option: A

Only I

Option A (Only I) is correct because glycolysis takes place in the cytosol. It does produce ATP, has a connection with the electron transport chain (via NADH), and reduces two molecules of NAD+.

Solution approach

Correct Answer + Core Reason: Only I. Glycolysis takes place in the cytosol.

Step-by-Step Solution:

Step 1: Evaluate statement I.\nTakes place in cytosol. This statement is true.

Step 2: Evaluate statement II.\nProduces no ATP. This statement is false; glycolysis produces ATP.

Step 3: Evaluate statement III.\nHas no connection with the electron transport chain. This statement is false; NADH produced in glycolysis is used in the electron transport chain.

Step 4: Evaluate statement IV.\nReduces two molecules of NAD $+$ for every glucose molecule processed. This statement is true.

Step 5: Identify the correct statement(s).\nOnly statement I is correct.

Step 6: Final Answer\nOnly I

Why other options are wrong:
- Option B (I, II and III): Statements II and III are incorrect.
- Option C (I and II): Statement II is incorrect.
- Option D (None of these): Statement I is correct.

Key Concepts:
- Glycolysis
- ATP production
- Electron transport chain
- NADH

Memory Aids:
- Remember that glycolysis occurs in the cytosol and produces ATP and NADH.

Quick Tips:
- Carefully evaluate each statement before selecting the correct option.

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In yeast during anaerobic respiration, how many glucose molecules are required for production of 38 ATP molecules?

View answer and explanation

Correct option: C

19

Option C (19) is correct because anaerobic respiration yields only 2 ATP per glucose molecule, so 19 glucose molecules are needed to produce 38 ATP.

Solution approach

Correct Answer + Core Reason: 19. Anaerobic respiration produces 2 ATP per glucose molecule, so 38 ATP requires 19 glucose molecules.

Step-by-Step Solution:

Step 1: Understand the ATP yield of anaerobic respiration in yeast.

Step 2: Anaerobic respiration (alcoholic fermentation) in yeast produces approximately 2 ATP molecules per glucose molecule.

Step 3: Calculate the number of glucose molecules required to produce 38 ATP molecules: $\text{{Number of glucose molecules}} = \frac{\text{{Total ATP required}}}{\text{{ATP per glucose molecule}}} = \frac{{38}}{{2}} = 19$

Step 4: Therefore, 19 glucose molecules are required.

Why other options are wrong:
- Option A: 1 glucose molecule produces only 2 ATP, not 38.
- Option B: 2 glucose molecules produce only 4 ATP, not 38.
- Option D: 38 glucose molecules would produce 76 ATP, not 38.

Key Concepts:
- Anaerobic respiration
- ATP production
- Alcoholic fermentation

Memory Aids:
- Remember that anaerobic respiration is much less efficient than aerobic respiration in terms of ATP production.

Quick Tips:
- Know the ATP yield of different metabolic pathways.

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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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The mechanism of breakdown of food materials within the cells to release energy is called?

View answer and explanation

Correct option: C

Cellular respiration

Option C (Cellular respiration) is correct because cellular respiration is the process by which cells break down food materials to release energy.

Solution approach

Correct Answer + Core Reason: Cellular respiration. Cellular respiration is the breakdown of food materials to release energy.

Step-by-Step Solution:

Step 1: Define cellular respiration.

Step 2: Cellular respiration is the process by which cells break down glucose and other organic molecules to release energy in the form of ATP.

Why other options are wrong:
- Option A: Cellular digestion is the breakdown of food into smaller molecules, but not specifically for energy release.
- Option B: Cellular photosynthesis is the process by which plants convert light energy into chemical energy.
- Option D: Cell division is the process by which cells reproduce.

Key Concepts:
- Cellular Respiration

Memory Aids:
- Respiration = Release Energy

Quick Tips:
- Cellular respiration is essential for all living organisms.

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Assertion: Plants do not have specialised
respiratory organs.

Reason: There is very little transport of gases
from one plant part to another.

View answer and explanation

Correct option: A

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

Option A (If both Assertion and Reason are true and Reason is the correct explanation of Assertion) is correct because plants do not have specialized respiratory organs due to the limited transport of gases between plant parts; each part handles its own gas exchange.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true and Reason is the correct explanation of Assertion. Plants lack specialized respiratory organs because gas transport is limited.

Step-by-Step Solution:

Step 1: Analyze the assertion: Plants do not have specialized respiratory organs.

Step 2: Analyze the reason: There is very little transport of gases from one plant part to another.

Step 3: Determine the relationship: The reason explains why plants do not need specialized organs.

Step 4: Conclusion: Both are true, and the reason correctly explains the assertion.

Why other options are wrong:
- Option B: The reason correctly explains the assertion.
- Option C: Both the assertion and reason are true.
- Option D: Both the assertion and reason are true.

Key Concepts:
- Plant respiration
- Gas transport
- Specialized organs

Memory Aids:
- Limited gas transport = No specialized organs

Quick Tips:
- Remember that plants rely on diffusion for gas exchange.

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Correct sequence of electron acceptor of ATP synthesis is

View answer and explanation

Correct option: B

cyt-b, c ,a,

{a_3}

Option B (cyt-b, c ,a,

{a_3}

) is correct because it represents the correct sequence of electron carriers in the electron transport chain for ATP synthesis.

Solution approach

Correct Answer + Core Reason: cyt-b, c ,a, ${a_3}$ . This is the correct sequence of cytochrome electron carriers in the electron transport chain.

Step-by-Step Solution:

Step 1: Understand the electron transport chain (ETC).

Step 2: Recall the sequence of electron carriers in the ETC.

Step 3: Identify the correct order of cytochromes: cyt-b, cyt-c, cyt-a, cyt- ${a_3}$ .

Step 4: Conclusion: The correct sequence is cyt-b, c, a, ${a_3}$ .

Why other options are wrong:
- Option A: cyt-a, ${a_3}$ , b, c: This is an incorrect sequence.
- Option C: cyt-b, c , ${a_3}$ , a: This is an incorrect sequence.
- Option D: cyt-c, b, a, ${a_3}$ : This is an incorrect sequence.

Key Concepts:
- Electron Transport Chain (ETC)
- Cytochromes
- Electron Flow

Memory Aids:
- Remember the order based on their redox potentials.

Quick Tips:
- Memorize the sequence of electron carriers in the ETC.

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Application +4 | -1

If 5 molecule of water are produced in ETS of mitochondria then find the number of ATP produced and

NAD{H_2}

are oxidised:-

View answer and explanation

Correct option: C

15ATP,\mathrm{ }5NAD{H_2}

Option C (

15ATP,\mathrm{ }5NAD{H_2}

) is correct because each NADH2 molecule yields 3 ATP molecules in ETS, so 5 NADH2 will produce 15 ATP.

Solution approach

Correct Answer + Core Reason: $15ATP,\mathrm{ }5NAD{H_2}$ . Each $NAD{H_2}$ produces 3 ATP in the electron transport system (ETS).

Step-by-Step Solution:

Step 1: Determine the ATP yield per $NAD{H_2}$ molecule in ETS.

Step 2: Each $NAD{H_2}$ molecule yields approximately 3 ATP molecules.

Step 3: Calculate the total ATP produced from 5 $NAD{H_2}$ molecules.

Step 4: Calculation: 5 $NAD{H_2}$ * 3 ATP/ $NAD{H_2}$ = 15 ATP.

Step 5: Final Answer: 15 ATP and 5 $NAD{H_2}$ oxidized.

Why other options are wrong:
- **Option A ( $30ATP,\mathrm{ }10NAD{H_2}$ ): This is incorrect. It assumes that each NADH2 produces 6 ATP, which is wrong.
- Option B ( $10ATP,\mathrm{ }5NAD{H_2}$ ): This is incorrect. It assumes that each NADH2 produces 2 ATP, which is wrong.
- Option D ( $6ATP,\mathrm{ }6NAD{H_2}$ ):** This is incorrect. The number of $NAD{H_2}$ oxidized is 5, not 6, and the ATP yield is also incorrect.

Key Concepts:
- Electron Transport System (ETS)
- ATP Production
- $NAD{H_2}$ Oxidation

Memory Aids:
- Remember that each $NAD{H_2}$ yields approximately 3 ATP in ETS.

Quick Tips:
- Always remember the ATP yield per $NAD{H_2}$ and $FAD{H_2}$ molecule in ETS.

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How many times ATP is utilised in glycolysis?

View answer and explanation

Correct option: A

2

Option A is correct because 2 ATP molecules are utilized in the initial steps of glycolysis: one for glucose to glucose-6-phosphate and another for fructose-6-phosphate to fructose-1,6-bisphosphate.

Solution approach

Correct Answer + Core Reason: 2. Two ATP molecules are consumed in the initial steps of glycolysis.

Step-by-Step Solution:
Step 1: Recall the steps of glycolysis where ATP is utilized.
Step 2: Identify the phosphorylation of glucose to glucose-6-phosphate as one ATP-consuming step.
Step 3: Identify the phosphorylation of fructose-6-phosphate to fructose-1,6-bisphosphate as another ATP-consuming step.
Step 4: Count the total number of ATP molecules utilized: 2.

Why other options are wrong:
- Option B: Three ATP molecules are not utilized in glycolysis.
- Option C: Four ATP molecules are not utilized in glycolysis. While 4 ATP are produced, 2 are initially used.
- Option D: Five ATP molecules are not utilized in glycolysis.

Key Concepts:
- Glycolysis
- ATP utilization
- Phosphorylation

Memory Aids:
- 2 ATP used in glycolysis.

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

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Application +4 | -1

The respiratory quotient of

{C_{51}}{H_{98}}{O_6}

(tripalmitin) is

View answer and explanation

Correct option: D

0.7

Option D (0.7) is correct because the respiratory quotient (RQ) for fats like tripalmitin is typically around 0.7 due to the higher proportion of carbon and hydrogen compared to oxygen.

Solution approach

Correct Answer + Core Reason: 0.7. The respiratory quotient of tripalmitin is 0.7 because it is a fat.

Step-by-Step Solution:

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

Step 2: General RQ values. Carbohydrates have an RQ of 1, proteins around 0.9, and fats around 0.7.

Step 3: Tripalmitin is a fat. Therefore, its RQ is approximately 0.7.

Why other options are wrong:
- Option A (1.4): This is incorrect because an RQ of 1.4 is not typical for any common respiratory substrate.
- Option B (1): This is incorrect because an RQ of 1 is characteristic of carbohydrates.
- Option C (0.9): This is incorrect because an RQ of 0.9 is characteristic of proteins.

Key Concepts:
- Respiratory Quotient (RQ)
- RQ values for different substrates (carbohydrates, fats, proteins)

Memory Aids:
- Carbohydrates: RQ = 1 (think "C" for carbohydrate and "1")
- Fats: RQ ~ 0.7 (lower because they require more oxygen to oxidize)
- Proteins: RQ ~ 0.9

Quick Tips:
- Remember the general RQ values for the main respiratory substrates.

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Understanding +4 | -1

Lowest rate of respiration will be in

View answer and explanation

Correct option: C

Dry seeds

Option C (Dry seeds) is correct because dry seeds have very low metabolic activity and therefore a minimal rate of respiration compared to metabolically active tissues.

Solution approach

Correct Answer + Core Reason: Dry seeds. Dry seeds have the lowest rate of respiration due to their dormancy.

Step-by-Step Solution:

Step 1: Identify the question: Which has the lowest rate of respiration?

Step 2: Recall the respiration rate in different tissues: Dry seeds are dormant and have a very low metabolic rate.

Step 3: Conclusion: Dry seeds have the lowest rate of respiration.

Why other options are wrong:
- Option A: Collenchyma is a living tissue and has a higher respiration rate than dry seeds.
- Option B: Leaves are actively photosynthesizing and respiring.
- Option D: Germinating seeds have a high metabolic rate and respiration rate.

Key Concepts:
- Respiration rate
- Dormancy
- Metabolic activity

Memory Aids:
- Dry seeds = Dormant = Low respiration

Quick Tips:
- Remember that metabolic activity is directly related to respiration rate.

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Anaerobic respiration can occur

View answer and explanation

Correct option: C

Both (a) and (b)

Option C (Both (a) and (b)) is correct because anaerobic respiration can occur in both lower organisms like bacteria and fungi, as well as in higher plants and animals under certain conditions (e.g., muscle cells during intense exercise).

Solution approach

Correct Answer + Core Reason: Both (a) and (b). Anaerobic respiration occurs in both lower and higher organisms.

Step-by-Step Solution:

Step 1: Anaerobic respiration in lower organisms: Common in bacteria and fungi.

Step 2: Anaerobic respiration in higher organisms: Occurs in muscle cells during intense exercise.

Step 3: Conclude: Anaerobic respiration can occur in both lower and higher organisms.

Why other options are wrong:
- Option A: Anaerobic respiration also occurs in higher organisms.
- Option B: Anaerobic respiration also occurs in lower organisms.
- Option D: Anaerobic respiration does occur in both lower and higher organisms.

Key Concepts:
- Anaerobic respiration
- Occurrence in different organisms

Memory Aids:
- Anaerobic = Universal (to some extent)

Quick Tips:
- Remember that both lower and higher organisms can perform anaerobic respiration.

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Beer and butter milk are products of fermentation by

View answer and explanation

Correct option: D

Saccharomyces cerevisiae

Option D (Saccharomyces cerevisiae) is correct because this yeast species is widely used in the fermentation of beer and other alcoholic beverages, as well as in dairy fermentation.

Solution approach

Correct Answer + Core Reason: Saccharomyces cerevisiae. This yeast is used in alcoholic fermentation (beer) and also in some dairy fermentations (butter milk).

Step-by-Step Solution:

Step 1: Consider Beer Production: Beer is produced through alcoholic fermentation.

Step 2: Consider Butter Milk Production: Butter milk is produced through lactic acid fermentation.

Step 3: Identify the Microorganism: Saccharomyces cerevisiae is a yeast used in alcoholic fermentation. Some bacteria are used in dairy fermentation, but *Saccharomyces cerevisiae* can also contribute to dairy fermentation.

Why other options are wrong:
- Option A (Rhizopus stolonifer): Rhizopus is a mold, not typically used in beer or butter milk production.
- Option B (Caedobacter taeniospiralis): Caedobacter is a bacterium, but not commonly associated with beer or butter milk production.
- Option C (Bacillus subtilis): Bacillus subtilis is a bacterium, but not typically used in beer or butter milk production. It is used in natto fermentation.

Key Concepts:
- Fermentation
- Alcoholic Fermentation
- Lactic Acid Fermentation
- Yeast
- Bacteria

Memory Aids:
- Saccharomyces = sugar fungus = fermentation.

Quick Tips:
- Remember common microorganisms used in food production.

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Net gain of ATP molecules per hexose during aerobic respiration is

View answer and explanation

Correct option: C

36

Option C (36) is correct because the net gain of ATP molecules per hexose during aerobic respiration is approximately 36.

Solution approach

Correct Answer + Core Reason: 36. Aerobic respiration yields approximately 36 ATP molecules per hexose.

Step-by-Step Solution:

Step 1: Understand the process of aerobic respiration.

Step 2: Recall the ATP production from glycolysis, Krebs cycle, and electron transport chain.

Step 3: Sum the ATP molecules produced: Glycolysis (2 ATP net), Krebs cycle (2 ATP), and ETC (32 ATP).

Step 4: Total ATP = 2 + 2 + 32 = 36 ATP.

Step 5: Conclusion: Net gain of ATP is 36.

Why other options are wrong:
- Option A: 12: This is a significantly lower number than the actual ATP yield.
- Option B: 18: This is also a lower number than the actual ATP yield.
- Option D: 30: While some older estimates suggest 30-32, the generally accepted value is closer to 36.

Key Concepts:
- Aerobic Respiration
- ATP Production
- Glycolysis, Krebs Cycle, ETC

Memory Aids:
- Remember the approximate ATP yield from each stage.

Quick Tips:
- Be aware of different estimates for ATP yield, but 36 is a commonly accepted value.

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Application +4 | -1

How many PGAL are produced by glycolysis of 3 molecules of glucose? How many ATP are released by respiration of these PGAL till formation of

C{O_2}

and

{H_2}O

?

View answer and explanation

Correct option: D

6 PGAL-120ATP

Option D (6 PGAL-120ATP) is correct because 3 glucose molecules produce 6 PGAL molecules, and each PGAL molecule yields approximately 20 ATP molecules during respiration.

Solution approach

Correct Answer + Core Reason: 6 PGAL-120ATP. 3 glucose molecules yield 6 PGAL, and each PGAL yields ~20 ATP.

Step-by-Step Solution:

Step 1: Calculate PGAL production: 1 glucose yields 2 PGAL during glycolysis. Therefore, 3 glucose molecules yield 3 * 2 = 6 PGAL molecules.

Step 2: Calculate ATP production per PGAL: Each PGAL molecule, when completely oxidized, yields approximately 20 ATP molecules (this includes glycolysis, pyruvate oxidation, Krebs cycle, and oxidative phosphorylation).

Step 3: Calculate total ATP production: 6 PGAL * 20 ATP/PGAL = 120 ATP.

Step 4: Conclude: 6 PGAL are produced, and their respiration yields 120 ATP.

Why other options are wrong:
- Option A (4 PGAL- 80 ATP): Incorrect PGAL and ATP count.
- Option B (6 PGAL-160ATP): Incorrect ATP count. Assumes higher ATP yield per PGAL.
- Option C (4 PGAL-40ATP): Incorrect PGAL and ATP count.

Key Concepts:
- Glycolysis
- PGAL (Glyceraldehyde-3-phosphate)
- ATP Production
- Cellular Respiration

Memory Aids:
- Remember that 1 glucose yields 2 PGAL.
- Remember the approximate ATP yield per PGAL molecule.

Quick Tips:
- Know the stoichiometry of glycolysis and the ATP yield of respiration.

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The tissue of highest respiratory activity is?

View answer and explanation

Correct option: A

Meristems

Option A (Meristems) is correct because meristematic tissues, which are actively dividing cells in plants, have the highest respiratory activity due to their high metabolic rate.

Solution approach

Correct Answer + Core Reason: Meristems. Meristems have the highest respiratory activity.

Step-by-Step Solution:

Step 1: Recall the characteristics of meristematic tissue.

Step 2: Meristems are actively dividing cells with high metabolic activity, requiring more energy and thus higher respiration rates.

Why other options are wrong:
- Option B: Ground tissue performs various functions, but its respiratory activity is lower than meristems.
- Option C: Phloem is involved in transport, and its respiratory activity is lower than meristems.
- Option D: Mechanical tissue provides support, and its respiratory activity is lower than meristems.

Key Concepts:
- Meristems
- Respiratory Activity

Memory Aids:
- Meristems = High Activity

Quick Tips:
- Actively dividing cells have higher metabolic rates and thus higher respiratory activity.

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Understanding +4 | -1

In succulent plants like Opuntia, the RQ value will be

View answer and explanation

Correct option: D

Zero

Option D is correct because succulent plants like Opuntia fix carbon dioxide at night and have minimal respiration, resulting in an RQ value close to zero.

Solution approach

Correct Answer + Core Reason: Zero. Succulent plants like Opuntia perform CAM photosynthesis, leading to minimal respiration and an RQ value close to zero.

Step-by-Step Solution:

Step 1: Understand CAM photosynthesis. Succulent plants perform Crassulacean Acid Metabolism (CAM) photosynthesis.

Step 2: Explain $CO_2$ fixation in CAM plants. CAM plants fix $CO_2$ at night and store it as organic acids.

Step 3: Explain respiration in CAM plants. During the day, these plants use the stored $CO_2$ for photosynthesis, minimizing respiration.

Step 4: Determine the RQ value. Due to minimal respiration, the RQ value is close to zero.

Why other options are wrong:
- Option A: RQ less than one is associated with fats and proteins, but not as low as zero in this context.
- Option B: RQ greater than one is associated with organic acids.
- Option C: Infinite RQ is associated with anaerobic respiration.

Key Concepts:
- Respiratory Quotient (RQ)
- CAM photosynthesis
- Xerophytes

Memory Aids:
- CAM plants = zero RQ

Quick Tips:
- Remember that CAM plants have a unique photosynthetic pathway that minimizes respiration.

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Remembering +4 | -1

During photorespiration, loss of

C{O_2}

occurs in which of the given cell organelles?

View answer and explanation

Correct option: C

Mitochondria

Option C (Mitochondria) is correct because the release of

CO_2

during photorespiration occurs in the mitochondria.

Solution approach

Correct Answer + Core Reason: Mitochondria. The release of $CO_2$ during photorespiration occurs in the mitochondria.

Step-by-Step Solution:

Step 1: Understand the process of photorespiration.

Step 2: Photorespiration involves the interaction of chloroplasts, peroxisomes, and mitochondria.

Step 3: The release of $CO_2$ occurs in the mitochondria during the conversion of glycine to serine.

Why other options are wrong:
- Option A: Chloroplast is where the initial oxygenation of RuBP occurs.
- Option B: Peroxisome is involved in the conversion of glycolate to glycine.
- Option D: Sphaerosomes are involved in lipid storage.

Key Concepts:
- Photorespiration
- Chloroplast
- Peroxisome
- Mitochondria

Memory Aids:
- Remember that photorespiration is a wasteful process that involves multiple organelles.

Quick Tips:
- Know the roles of different organelles in photorespiration.

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