Chapter 12: Respiration in Plants Case Study Questions | biology Class 11 CBSE

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A group of students performed an experiment to study anaerobic respiration in yeast. They prepared a sugar solution, added yeast, and sealed the flask with a delivery tube dipped in lime water. Within a few hours, the lime water turned milky and a distinct smell of alcohol was detected. When they measured the sugar content of the solution after 24 hours, it had decreased significantly. One student noted that the yeast population stopped growing after the alcohol concentration reached a certain level. The teacher explained that the yeast was carrying out a specific type of fermentation, converting glucose to simpler products under oxygen-free conditions. The students were asked to identify the enzymes involved, calculate the ATP yield, and explain why yeast growth eventually stopped.

Question 1: Name the two enzymes involved in alcoholic fermentation and state the immediate product formed by each enzyme.

Pyruvic acid decarboxylase catalyses the conversion of pyruvic acid to acetaldehyde and CO2; the milky lime water confirms CO2 production.
Alcohol dehydrogenase then catalyses the reduction of acetaldehyde to ethanol, using NADH+H+ as the reducing agent.
The overall products of alcoholic fermentation are ethanol and CO2, both of which were detected in the experiment — alcohol by smell and CO2 by the lime water test.

Question 2: Calculate the net ATP synthesised per molecule of glucose during this fermentation and explain the significance of NADH reoxidation in the process.

Glycolysis consumes 2 ATP and produces 4 ATP by substrate-level phosphorylation, giving a net gain of 2 ATP per glucose; no additional ATP is produced during the fermentation steps themselves.
NADH+H+ produced during glycolysis (when PGAL is oxidised to BPGA) must be reoxidised to NAD+; this occurs during the reduction of acetaldehyde to ethanol by alcohol dehydrogenase.
Reoxidation of NADH to NAD+ is essential because NAD+ is required as an electron acceptor at the PGAL oxidation step in glycolysis; without it, glycolysis would halt and no ATP at all could be produced anaerobically.

Question 3: Explain why yeast growth stopped after the alcohol concentration reached approximately 13%. How would a distiller obtain a beverage with 40% alcohol content from this fermented solution?

Yeast cells are poisoned by the ethanol they produce; when ethanol concentration in the medium reaches approximately 13%, the concentration becomes lethal to the yeast cells themselves, causing them to die.
Once yeast cells die, fermentation ceases entirely regardless of available sugar; the biological ceiling on natural fermentation is therefore set by yeast's own ethanol toxicity threshold.
To obtain a beverage with 40% alcohol (e.g., whisky or vodka), a distiller uses distillation — a physical process that exploits the lower boiling point of ethanol (78°C) compared to water (100°C) to concentrate and separate ethanol from the fermented mixture, far exceeding what biological fermentation alone can achieve.

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During a practical session, students were asked to measure the respiratory quotient (RQ) of three different seeds: wheat seeds (carbohydrate-rich), groundnut seeds (fat-rich), and soybean seeds (protein-rich). They used a respirometer to measure the volumes of CO2 evolved and O2 consumed over one hour. The results showed that the three seed types gave significantly different RQ values. The teacher explained that the RQ value directly reflects the chemical composition of the respiratory substrate being oxidised and that in living organisms, the measured RQ is rarely a whole number because multiple substrates are often oxidised simultaneously. Students were asked to predict the RQ values and explain the biochemical basis for the differences observed.

Question 1: Define respiratory quotient and state the expected RQ value for wheat seeds respiring carbohydrates.

The respiratory quotient (RQ) is the ratio of the volume of CO2 evolved to the volume of O2 consumed during aerobic respiration: RQ = Volume of CO2 evolved / Volume of O2 consumed.
When carbohydrates such as glucose are the sole substrate, equal volumes of CO2 and O2 are produced and consumed respectively, as shown by the equation C6H12O6 + 6O2 → 6CO2 + 6H2O.
Therefore, wheat seeds respiring carbohydrates would give an RQ of 6CO2/6O2 = 1.0.

Question 2: Explain why groundnut seeds (fat-rich) would give a lower RQ than wheat seeds, and calculate the approximate RQ of tripalmitin.

Fats such as tripalmitin have a much lower proportion of oxygen relative to carbon and hydrogen compared to carbohydrates; they require proportionally more O2 for complete oxidation than they produce CO2.
For tripalmitin: 2(C51H98O6) + 145O2 → 102CO2 + 98H2O + energy; the RQ = 102/145 = approximately 0.7.
Since more O2 is consumed than CO2 is evolved during fat oxidation, the RQ is less than 1, explaining why groundnut seeds give a lower RQ value than carbohydrate-rich wheat seeds.

Question 3: Soybean seeds have an intermediate RQ of approximately 0.9. Explain why pure proteins or fats are never used as the sole respiratory substrate in living organisms, and how this affects the measured RQ.

The chapter explicitly states that in living organisms, respiratory substrates are often more than one; pure proteins or pure fats are never exclusively used as respiratory substrates because cells simultaneously oxidise a mixture of carbohydrates, fats, and proteins.
The measured RQ of any organism therefore reflects the weighted average of the RQ values of all substrates being oxidised simultaneously; a value between 0.7 and 1.0 indicates mixed substrate use.
For soybean seeds, the protein-rich composition gives an RQ near 0.9, but in practice the seeds also contain some carbohydrates and fats; the actual measured value would depend on the precise proportions of each substrate being oxidised, making it practically impossible to measure a theoretical RQ for a single pure substrate in a living system.

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Case Set 3

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A biochemistry researcher was studying the TCA cycle in isolated plant mitochondria. She added a labelled acetyl CoA (with radioactive carbon) to the mitochondrial preparation along with all necessary coenzymes and found that radioactively labelled CO2 was released and labelled intermediates appeared throughout the cycle. She also observed that when oxaloacetic acid was not replenished, the cycle quickly stopped even though acetyl CoA was still available. In a separate experiment, she found that inhibiting succinate dehydrogenase (the enzyme that oxidises succinate to fumarate) caused accumulation of succinic acid and cessation of FADH2 production. Students were asked to analyse these observations in the context of the TCA cycle's structure and function.

Question 1: Name the first product of the TCA cycle and the enzyme that catalyses its formation. State the carbon count of the starting materials and the product.

At the start of the TCA cycle, acetyl CoA (2C) condenses with oxaloacetic acid (OAA, 4C) and water to form citric acid (6C).
The reaction is catalysed by the enzyme citrate synthase, and a molecule of CoA is released in the process.
Citric acid, a tricarboxylic acid, gives the cycle its alternate names — the citric acid cycle or tricarboxylic acid (TCA) cycle.

Question 2: Explain why the TCA cycle stopped when OAA was not replenished, even though acetyl CoA was still available.

OAA is the first member of the TCA cycle and is regenerated at the end of each turn when malate is oxidised to OAA by malate dehydrogenase; it is not consumed permanently but must be continuously recycled.
If OAA is not regenerated or replenished, there is no acceptor molecule available to condense with incoming acetyl CoA via citrate synthase, and the cycle cannot begin a new turn.
The continued oxidation of acetyl CoA via the TCA cycle requires the continued replenishment of oxaloacetic acid; without OAA, acetyl CoA accumulates unutilised, and the entire TCA cycle — along with its NADH, FADH2, and ATP outputs — ceases.

Question 3: Analyse the consequences of inhibiting succinate dehydrogenase (Complex II) on the ETS and overall ATP production. Explain why FADH2 production from the Krebs' cycle would cease but NADH-linked ATP production might partially continue.

Succinate dehydrogenase (Complex II) catalyses the oxidation of succinic acid to fumarate with reduction of FAD+ to FADH2; inhibiting it would cause succinate accumulation and halt FADH2 production at this step of the Krebs' cycle.
In the ETS, Complex II is the entry point for electrons from FADH2 into the electron transport chain via ubiquinone; with no FADH2 from this step, electron flow through this branch to ubiquinone would cease, reducing ATP yield from FADH2 (normally 2 ATP per FADH2).
However, NADH-linked electron flow through Complex I would be unaffected by this specific inhibition; the three NADH+H+ producing steps in the Krebs' cycle (isocitrate → alpha-ketoglutarate, alpha-ketoglutarate → succinyl-CoA, malate → OAA) would continue generating NADH, allowing Complex I-driven proton pumping and partial ATP synthesis via oxidative phosphorylation to continue, though total ATP yield per glucose would be significantly reduced.

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A physiology student observed that marathon runners breathe heavily and produce a large amount of CO2 during the race but also experience muscle cramps and burning sensations in their legs during the final sprint. The student recalled that muscle cells can switch between aerobic and anaerobic modes of respiration. She also noted from her biology textbook that the first cells on Earth lived in an oxygen-free atmosphere and that all organisms retain the enzymatic machinery to partially oxidise glucose without oxygen. She connected these observations to the concept of glycolysis and fermentation discussed in her chapter on plant respiration, recognising that these pathways are universal in living organisms.

Question 1: Why do marathon runners breathe heavily during a race? Name the pathway responsible for the bulk of ATP production during sustained aerobic exercise.

During sustained aerobic exercise, muscles require large amounts of ATP continuously; aerobic respiration consumes O2 and releases CO2, necessitating increased breathing to supply O2 and remove CO2.
The electron transport system (ETS) coupled with oxidative phosphorylation is responsible for the bulk of ATP production (up to 34 of the theoretical 38 ATP per glucose) during aerobic exercise.
The Krebs' cycle generates the NADH+H+ and FADH2 that feed the ETS; their oxidation through the respiratory chain drives proton pumping and ATP synthesis via Complex V (ATP synthase).

Question 2: Explain the biochemical basis for muscle cramps and burning sensation during the final sprint, linking it to lactic acid fermentation.

During a final sprint, the rate of O2 demand exceeds supply; muscle cells switch to anaerobic conditions and pyruvate is reduced to lactic acid by lactate dehydrogenase, using NADH+H+ as the reducing agent.
Lactic acid accumulates in muscle tissue, lowering the local pH; this acidification impairs enzyme activity, disrupts ion gradients, and causes the burning sensation and cramping experienced during intense exercise.
The reoxidation of NADH to NAD+ via lactic acid fermentation allows glycolysis to continue briefly, generating 2 ATP per glucose anaerobically; however, lactic acid must later be transported to the liver for reconversion to pyruvate or glucose when O2 is restored.

Question 3: The student noted that glycolysis is present in all living organisms. Explain the evolutionary and biochemical significance of glycolysis being a universal pathway, and discuss what this implies about the conditions on early Earth.

Glycolysis is present in all living organisms — from the simplest prokaryotes to complex eukaryotes — because it evolved before oxygen became available in Earth's atmosphere; the first cells required an anaerobic ATP-generating pathway, and glycolysis fulfilled this requirement using only cytoplasmic enzymes without any membrane-bound organelle.
The universality of glycolysis implies that it was established in the common ancestor of all life; all subsequent respiratory adaptations (Krebs' cycle, ETS) were additions built upon this ancestral anaerobic framework rather than replacements of it.
The presence of glycolysis in all organisms, including strict aerobes that could survive without it alone, reflects the ancient anaerobic conditions of early Earth; even modern aerobic organisms retain the complete enzymatic machinery to partially oxidise glucose without oxygen — a biochemical inheritance from the oxygen-free primordial environment.

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Case Set 5

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A botanist was studying gas exchange in a large, 50-year-old banyan tree with a thick woody trunk. She was puzzled about how the living cells deep within the bark obtained oxygen for respiration, given that the tree had no lungs or gill-like structures. Upon careful examination, she found small pore-like structures scattered across the grey bark surface. She also noted that when she peeled a section of bark, the living cells were confined to a very thin layer just beneath the outer layers, while the woody interior consisted entirely of dead cells. She concluded that the tree had evolved a highly efficient structural solution to gaseous exchange that made specialised respiratory organs unnecessary.

Question 1: Identify the structures the botanist found on the bark and explain their function in plant respiration.

The structures observed on the bark surface are lenticels — small pore-like openings present in the bark of woody stems.
Lenticels allow gaseous exchange by diffusion: O2 from the atmosphere diffuses into the living cells beneath the bark, and CO2 produced by cellular respiration diffuses out.
Lenticels serve the same gas-exchange function in woody stems as stomata do in leaves, ensuring that all living cells in the stem have access to atmospheric O2 for aerobic respiration.

Question 2: Explain why dead cells in the woody interior do not pose a problem for the tree's respiratory needs.

The dead cells in the interior of woody stems (xylem) have no metabolic activity — they do not respire, do not require O2, and do not produce CO2; they serve purely as structural support (mechanical strength).
Because these dead cells have no respiratory demands, the tree only needs to supply O2 to the thin layer of living cells just beneath the bark; the diffusion distance from lenticels to these cells is very short.
This structural arrangement — living cells at the periphery and dead cells in the interior — dramatically reduces the volume of tissue requiring gas exchange, making specialised respiratory organs entirely unnecessary.

Question 3: A student argues that large trees must have a circulatory system for gas transport because simple diffusion cannot be sufficient across their large volume. Using evidence from the chapter, refute this argument.

The argument incorrectly assumes that gas must travel through the entire volume of the tree; in reality, gas exchange only needs to reach the thin peripheral layer of living cells, not the dead interior — diffusion distances are therefore short even in large trees.
The loose packing of parenchyma cells throughout the plant creates an interconnected network of air spaces that allows gases to diffuse rapidly; each living cell is located close to a surface or air space exposed to the atmosphere, making long-distance gas transport unnecessary.
Plants also respire at rates far lower than animals, meaning the absolute demand for O2 delivery and CO2 removal is minimal; combined with lenticels for direct atmospheric access and the near-surface location of all living cells, simple diffusion is entirely sufficient — no active circulatory system is needed.

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Chapter 12: Respiration in Plants | Case Study Questions

Passage

Question 1: Name the two enzymes involved in alcoholic fermentation and state the immediate product formed by each enzyme.

Question 2: Calculate the net ATP synthesised per molecule of glucose during this fermentation and explain the significance of NADH reoxidation in the process.

Question 3: Explain why yeast growth stopped after the alcohol concentration reached approximately 13%. How would a distiller obtain a beverage with 40% alcohol content from this fermented solution?

Passage

Question 1: Define respiratory quotient and state the expected RQ value for wheat seeds respiring carbohydrates.

Question 2: Explain why groundnut seeds (fat-rich) would give a lower RQ than wheat seeds, and calculate the approximate RQ of tripalmitin.

Question 3: Soybean seeds have an intermediate RQ of approximately 0.9. Explain why pure proteins or fats are never used as the sole respiratory substrate in living organisms, and how this affects the measured RQ.

Passage

Question 1: Name the first product of the TCA cycle and the enzyme that catalyses its formation. State the carbon count of the starting materials and the product.

Question 2: Explain why the TCA cycle stopped when OAA was not replenished, even though acetyl CoA was still available.

Question 3: Analyse the consequences of inhibiting succinate dehydrogenase (Complex II) on the ETS and overall ATP production. Explain why FADH2 production from the Krebs' cycle would cease but NADH-linked ATP production might partially continue.

Passage

Question 1: Why do marathon runners breathe heavily during a race? Name the pathway responsible for the bulk of ATP production during sustained aerobic exercise.

Question 2: Explain the biochemical basis for muscle cramps and burning sensation during the final sprint, linking it to lactic acid fermentation.

Question 3: The student noted that glycolysis is present in all living organisms. Explain the evolutionary and biochemical significance of glycolysis being a universal pathway, and discuss what this implies about the conditions on early Earth.

Passage

Question 1: Identify the structures the botanist found on the bark and explain their function in plant respiration.

Question 2: Explain why dead cells in the woody interior do not pose a problem for the tree's respiratory needs.

Question 3: A student argues that large trees must have a circulatory system for gas transport because simple diffusion cannot be sufficient across their large volume. Using evidence from the chapter, refute this argument.

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