Respiration in Plants Class 11 Biology: Glycolysis, Krebs Cycle, ETS and Respiratory Quotient

What Respiration in Plants is about

Weightage: This chapter is part of Unit IV, Plant Physiology, which carries 12 of the 70 theory marks in the school annual exam together with Photosynthesis in Higher Plants and Plant Growth and Development.Source: CBSE Biology (044) syllabus 2026-27, Class XI course structure, Unit IV Plant Physiology

Key concepts

Do plants breathe? Gas exchange and why respiration is stepwise

NCERT section 12.1 Do Plants Breathe?

Plants need O₂ for respiration and give out CO₂, but they have no specialised respiratory organs. Gas exchange happens through stomata (leaves) and lenticels (stems), by diffusion.

Why plants manage without respiratory organs:
1. Each part looks after itself: roots, stems and leaves handle their own gas exchange; very little gas is transported between parts.
2. Low demand: plant parts respire much more slowly than animals. Large gas volumes are exchanged only in photosynthesis, and then O₂ is produced inside the cells anyway.
3. Short diffusion distances: almost every living cell lies close to the surface. In woody stems and roots, the living cells form thin layers under the bark, which has lenticels; the interior cells are dead and only give support. Loosely packed parenchyma gives a connected network of air spaces.

Combustion vs respiration:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy

• Burning releases all the energy at once, mostly as heat.
• In respiration glucose is oxidised in many small steps, so energy is released in small packets, some of which are coupled to ATP synthesis. Each step is catalysed by an enzyme.

Aerobic and anaerobic life: the first cells probably lived without oxygen. Today some organisms are facultative anaerobes and some obligate anaerobes. All organisms keep the enzymes to partially oxidise glucose without O₂: this is glycolysis.

Worked example

Give three reasons why plants can survive without special respiratory organs, and name the openings used for gas exchange in leaves and woody stems. (3 marks)

Try it first, then show the solutionSolution

1. Each organ handles its own gas exchange with little transport between parts; plants respire slowly, so demand is low. (1)
2. Living cells are close to the surface (thin layers under the bark), and loose parenchyma provides connected air spaces, so diffusion distances are short. (1)
3. Stomata in leaves; lenticels in woody stems. (1)

Answer: Self-sufficient parts, low demand, short diffusion paths; stomata and lenticels.

Glycolysis (EMP pathway)

NCERT section 12.2 Glycolysis

Glycolysis (Greek *glycos* = sugar, *lysis* = splitting) is the partial oxidation of glucose to pyruvic acid. Its scheme was worked out by Gustav Embden, Otto Meyerhof and J. Parnas, so it is called the EMP pathway.

• Occurs in the cytoplasm of all living organisms.
• In anaerobic organisms it is the only process of respiration.
• Glucose → 2 pyruvic acid through ten enzyme-controlled reactions.
• In plants, glucose comes from sucrose (end product of photosynthesis) or from storage carbohydrates. Sucrose is converted into glucose and fructose by invertase, and both enter glycolysis.

Main steps (learn in this order):
1. Glucose → glucose-6-phosphate by hexokinase (uses 1 ATP). Fructose is also phosphorylated.
2. Glucose-6-phosphate → fructose-6-phosphate (isomerisation).
3. Fructose-6-phosphate → fructose-1,6-bisphosphate (uses 1 ATP).
4. Fructose-1,6-bisphosphate splits into two 3-carbon molecules: dihydroxyacetone phosphate (DHAP) and 3-phosphoglyceraldehyde (PGAL); DHAP is converted into PGAL.
5. PGAL → 1,3-bisphosphoglycerate (BPGA): PGAL is oxidised and phosphorylated with inorganic phosphate; NAD⁺ → NADH + H⁺ (one per PGAL, so 2 NADH per glucose).
6. BPGA → 3-phosphoglyceric acid (PGA): energy-yielding step making ATP (×2 per glucose).
7. PGA → 2-PGA → phosphoenolpyruvate (PEP).
8. PEP → pyruvic acid: makes another ATP (×2 per glucose).

Energy accounting per glucose:

• ATP used: 2 (steps 1 and 3).
• ATP made directly: 4 (2 at BPGA → PGA, 2 at PEP → pyruvate).
• Net ATP = 4 − 2 = 2, plus 2 NADH + H⁺.

Fate of pyruvate: pyruvic acid is the key product. Its metabolic fate depends on the cell's needs: lactic acid fermentation, alcoholic fermentation, or aerobic respiration (Krebs cycle).

Worked example

In glycolysis, name the two steps where ATP is consumed and the two steps where ATP is produced. What is the net yield of ATP and NADH per glucose? (3 marks)

Try it first, then show the solutionSolution

1. ATP used: glucose → glucose-6-phosphate (hexokinase) and fructose-6-phosphate → fructose-1,6-bisphosphate. (1)
2. ATP made: 1,3-BPGA → 3-PGA and PEP → pyruvic acid (each twice per glucose). (1)
3. Net: 4 − 2 = 2 ATP, plus 2 NADH + H⁺. (1)

Answer: Net 2 ATP and 2 NADH per glucose.

Fermentation (anaerobic respiration)

NCERT section 12.3 Fermentation

When O₂ is absent or scarce, pyruvic acid is handled without the mitochondria.

Alcoholic fermentation (yeast):

• Pyruvic acid → CO₂ + ethanol.
• Enzymes: pyruvic acid decarboxylase (removes CO₂, giving acetaldehyde) and alcohol dehydrogenase (reduces acetaldehyde to ethanol).

Lactic acid fermentation:

• Some bacteria and animal muscle cells during vigorous exercise (when O₂ is inadequate) convert pyruvic acid → lactic acid, using lactate dehydrogenase.

In both, the reducing agent is NADH + H⁺, which is reoxidised to NAD⁺. This regenerates NAD⁺ so glycolysis can continue.

Limitations:

• Not much energy is released: less than 7% of the energy in glucose, and not all of it is trapped as ATP.
• Net gain is only 2 ATP per glucose (from glycolysis); the fermentation steps make no ATP.
• The products are hazardous: acid or alcohol.
• Yeasts poison themselves when the alcohol concentration reaches about 13%. So naturally fermented drinks cannot exceed this; stronger drinks are made by distillation.

Fermentation vs aerobic respiration:

• Fermentation: partial breakdown of glucose; aerobic: complete breakdown to CO₂ and H₂O.
• Fermentation: net 2 ATP per glucose; aerobic: many more ATP.
• NADH is reoxidised to NAD⁺ slowly in fermentation but vigorously in aerobic respiration.

Worked example

Why must NADH be reoxidised during fermentation even though no ATP is made in the fermentation steps? Name the enzyme that does this in muscle. (2 marks)

Try it first, then show the solutionSolution

1. Glycolysis needs a continuous supply of NAD⁺ to oxidise PGAL; reducing pyruvate (or acetaldehyde) with NADH regenerates NAD⁺, so glycolysis and its 2 ATP can continue. (1)
2. Lactate dehydrogenase (pyruvic acid → lactic acid). (1)

Answer: To regenerate NAD⁺ for glycolysis; lactate dehydrogenase.

Aerobic respiration: link reaction and the Krebs (TCA) cycle

NCERT section 12.4 Aerobic Respiration (12.4.1 Tricarboxylic Acid Cycle)

Aerobic respiration: complete oxidation of organic substances in the presence of O₂, releasing CO₂, water and a large amount of energy. In eukaryotes it takes place in mitochondria.

Two crucial events:
1. Complete oxidation of pyruvate by stepwise removal of all hydrogen atoms, leaving three CO₂ molecules: in the matrix.
2. Passing the electrons from those hydrogen atoms to O₂, with ATP synthesis: on the inner mitochondrial membrane.

Link reaction (oxidative decarboxylation): pyruvate enters the matrix and is converted by the pyruvic dehydrogenase complex, which needs coenzymes NAD⁺ and coenzyme A (and Mg²⁺):
Pyruvic acid + CoA + NAD⁺ → Acetyl CoA + CO₂ + NADH + H⁺
Two pyruvates per glucose give 2 NADH here.

Krebs cycle (tricarboxylic acid / citric acid cycle), worked out by Hans Krebs:
1. Acetyl group (2C) + oxaloacetic acid (OAA, 4C) + water → citric acid (6C), catalysed by citrate synthase; CoA is released.
2. Citrate is isomerised to isocitrate.
3. Two successive decarboxylations give α-ketoglutaric acid (5C) and then succinyl CoA (4C).
4. The rest of the cycle converts succinyl CoA → OAA, allowing the cycle to continue.
5. During succinyl CoA → succinic acid, one GTP is made (substrate-level phosphorylation); GTP passes its phosphate to ADP, giving ATP.
6. At three points, NAD⁺ is reduced to NADH + H⁺; at one point, FAD is reduced to FADH₂.

For the cycle to continue: OAA must be regenerated, and NAD⁺ and FAD must be regenerated from NADH and FADH₂ (by the ETS).

Per pyruvate (one turn):
Pyruvic acid + 4NAD⁺ + FAD + 2H₂O + ADP + Pi → 3CO₂ + 4NADH + 4H⁺ + FADH₂ + ATP
(The 4 NADH include the one from the link reaction.)

Per glucose (two turns + link): 6 CO₂, 8 NADH, 2 FADH₂, 2 ATP.

Worked example

Starting from one pyruvate, list the products of the link reaction and one turn of the Krebs cycle. Where does each process occur? (3 marks)

Try it first, then show the solutionSolution

1. Link reaction (matrix): 1 acetyl CoA, 1 CO₂, 1 NADH + H⁺. (1)
2. Krebs cycle (matrix): 2 CO₂, 3 NADH + H⁺, 1 FADH₂, 1 ATP (via GTP). (1)
3. Total per pyruvate: 3 CO₂, 4 NADH, 1 FADH₂, 1 ATP; both processes occur in the mitochondrial matrix. (1)

Answer: 3 CO₂, 4 NADH, 1 FADH₂, 1 ATP per pyruvate, in the matrix.

Electron transport system and oxidative phosphorylation

NCERT section 12.4.2 Electron Transport System (ETS) and Oxidative Phosphorylation

The energy stored in NADH + H⁺ and FADH₂ is released when their electrons pass along the electron transport system (ETS), in the inner mitochondrial membrane, and is used to make ATP.

Path of electrons:

• Complex I (NADH dehydrogenase): oxidises NADH from the matrix; electrons pass to ubiquinone.
• Complex II: FADH₂ (from succinate oxidation in the Krebs cycle) also passes electrons to ubiquinone.
• Reduced ubiquinone (ubiquinol) is oxidised by complex III (cytochrome bc₁ complex), which passes electrons to cytochrome c.
• Cytochrome c, a small protein sitting on the outer face of the inner membrane, acts as a mobile carrier between complex III and complex IV.
• Complex IV (cytochrome c oxidase): contains cytochromes a and a₃ and two copper centres; it passes electrons to O₂, which combines with protons to form water.
• As electrons move from one carrier to the next (I → IV), energy is used to pump protons; this is coupled to ATP synthase (complex V).

ATP yield: oxidation of 1 NADH → 3 ATP; 1 FADH₂ → 2 ATP (FADH₂ enters later, at complex II, so fewer protons are pumped).

Role of oxygen: O₂ acts only at the final step, as the final hydrogen (electron) acceptor. Yet it is vital: by removing hydrogen it drives the whole chain. Without O₂, NADH cannot be reoxidised and the Krebs cycle stops.

Oxidative phosphorylation: ATP made using the energy of oxidation-reduction (electron flow) to build the proton gradient. (In photophosphorylation, light energy builds the gradient.) The mechanism is chemiosmosis.

ATP synthase (complex V): two parts:

• F₀: an integral membrane protein complex forming the channel through which protons cross the inner membrane.
• F₁: a peripheral headpiece with the site of ATP synthesis from ADP + Pi.
• Proton flow through F₀ is coupled to F₁'s catalytic site. For each ATP, 4 H⁺ pass through F₀ from the intermembrane space to the matrix, down the electrochemical gradient.

Worked example

Why is oxygen called the final hydrogen acceptor, and why does respiration stop without it even though O₂ acts only at the last step? (3 marks)

Try it first, then show the solutionSolution

1. At complex IV, O₂ accepts electrons and combines with protons to form water, so it is the last acceptor of hydrogen. (1)
2. Without O₂, electrons cannot leave the chain; all carriers stay reduced, so NADH and FADH₂ cannot be oxidised. (1)
3. NAD⁺ and FAD are not regenerated, so the Krebs cycle and link reaction stop, and oxidative phosphorylation ceases. (1)

Answer: O₂ accepts electrons at complex IV; without it the chain jams and NAD⁺/FAD are not regenerated.

Respiratory balance sheet: 38 ATP

NCERT section 12.5 The Respiratory Balance Sheet

The net ATP from one glucose can be calculated only theoretically, using these assumptions:
1. The pathway is sequential and orderly: glycolysis → TCA → ETS, one after another, each substrate forming the next.
2. NADH from glycolysis enters the mitochondria and undergoes oxidative phosphorylation.
3. No intermediate is withdrawn to make other compounds.
4. Only glucose is respired; no other substrate enters midway.

In a real cell these are not valid: all pathways run simultaneously, substrates enter and leave as needed, ATP is used as made, and enzyme rates are controlled in many ways. Still, the calculation shows how efficiently cells capture energy.

Calculation per glucose:

• Glycolysis: 2 ATP (net) + 2 NADH (× 3 = 6 ATP) = 8 ATP.
• Link reaction: 2 NADH (× 3) = 6 ATP.
• Krebs cycle (2 turns): 2 ATP + 6 NADH (× 3 = 18) + 2 FADH₂ (× 2 = 4) = 24 ATP.
• Total = 8 + 6 + 24 = 38 ATP.

Check by counting carriers: total NADH = 2 + 2 + 6 = 10 → 30 ATP; FADH₂ = 2 → 4 ATP; substrate-level ATP = 2 (glycolysis) + 2 (Krebs) = 4. 30 + 4 + 4 = 38.

So aerobic respiration gives 19 times the net ATP of fermentation (38 ÷ 2).

Worked example

Suppose a cell respires glucose but uses half of its acetyl CoA to make fatty acids instead of sending it into the Krebs cycle. Which assumption of the balance sheet is broken, and roughly how many ATP per glucose would the cell actually gain? (3 marks)

Try it first, then show the solutionSolution

1. The assumption that no intermediate is withdrawn for synthesis of other compounds is broken. (1)
2. Glycolysis and link reaction still give 8 + 6 = 14 ATP. (1)
3. Only one acetyl CoA enters the Krebs cycle: 1 ATP + 3 NADH (9) + 1 FADH₂ (2) = 12 ATP; total about 14 + 12 = 26 ATP. (1)

Answer: The 'no withdrawal' assumption fails; about 26 ATP.

Amphibolic pathway and respiratory quotient

NCERT section 12.6 Amphibolic Pathway; 12.7 Respiratory Quotient

Other substrates and their entry points:

• Carbohydrates: usually converted to glucose first (glucose is the favoured substrate).
• Fats: broken into glycerol and fatty acids. Fatty acids → acetyl CoA; glycerol → PGAL.
• Proteins: broken by proteases into amino acids, which after deamination enter as pyruvate, acetyl CoA or at some step within the Krebs cycle, depending on structure.

Amphibolic pathway: respiration was traditionally called catabolic (breaking down). But the same intermediates are withdrawn for synthesis: e.g. acetyl CoA is taken out to make fatty acids; Krebs-cycle intermediates are used to make amino acids/proteins. Since the pathway serves both catabolism and anabolism, it is better called amphibolic.

Respiratory quotient (RQ), also called the respiratory ratio, compares the gases exchanged:
RQ = CO₂ given out (volume) ÷ O₂ taken in (volume)

• Carbohydrates: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O; RQ = 6/6 = 1.
• Fats: RQ less than 1. Tripalmitin: 2C₅₁H₉₈O₆ + 145O₂ → 102CO₂ + 98H₂O; RQ = 102/145 ≈ 0.7. Fats are poor in oxygen, so more O₂ is needed to oxidise them.
• Proteins: about 0.9.
• In real organisms the substrate is usually a mixture; pure fats or proteins are never the only substrate.

Worked example

Why is the RQ of fats lower than that of carbohydrates? Show using the RQ of glucose and tripalmitin. (3 marks)

Try it first, then show the solutionSolution

1. Glucose: 6 CO₂ released for 6 O₂ used, RQ = 1. (1)
2. Tripalmitin: 102 CO₂ for 145 O₂, RQ = 102 ÷ 145 ≈ 0.70. (1)
3. Fats contain much less oxygen relative to carbon and hydrogen, so extra O₂ from air is needed to oxidise them; O₂ used exceeds CO₂ released. (1)

Answer: Fats are oxygen-poor and need more O₂, so RQ ≈ 0.7 vs 1 for glucose.

Respiration in Plants in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• Glycolysis, Fermentation and Link Reaction
• Krebs Cycle, ETS, ATP Balance and RQ

Respiration in Plants: key facts

Aerobic respiration of glucose

${\text{C}}_6{\text{H}}_{12}{\text{O}}_6+6{\text{O}}_2\to 6{\text{CO}}_2+6{\text{H}}_2\text{O}+\text{energy}$

Net 38 ATP (theoretical).

Link reaction

$\text{pyruvic acid}+\text{CoA}+{\text{NAD}}^{+}\stackrel{{\text{pyruvate dehydrogenase, Mg}}^{2+}}{\to }\text{acetyl CoA}+{\text{CO}}_2+\text{NADH}+{\text{H}}^{+}$

In the mitochondrial matrix.

ATP from carriers

$1\text{NADH}=3\text{ATP};1{\text{FADH}}_2=2\text{ATP}$

4 H⁺ through F₀ per ATP.

Respiratory quotient

$\text{RQ}=\frac{{\text{volume of CO}}_2\text{evolved}}{{\text{volume of O}}_2\text{consumed}}$

Carbohydrate 1; fat about 0.7; protein about 0.9.

Tripalmitin oxidation

$2{\text{C}}_{51}{\text{H}}_{98}{\text{O}}_6+145{\text{O}}_2\to 102{\text{CO}}_2+98{\text{H}}_2\text{O}$

RQ = 102/145 ≈ 0.7.

Glycolysis net

$\text{glucose}+2{\text{NAD}}^{+}+2\text{ADP}+2{\text{P}}_i\to 2\text{pyruvate}+2\text{NADH}+2{\text{H}}^{+}+2\text{ATP}$

4 ATP made − 2 used = 2 net.

Respiration in Plants: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Respiration in Plants

The question patterns to expect, and the method that works for each.

Tricks to remember Respiration in Plants

Short memory aids. Say each one aloud once, then use it on a question the same day.

• EMP = Every Morning Porridge

  Embden, Meyerhof, Parnas gave glycolysis.

  Use it when: Naming scientists.

• Krebs: Citrate Is Krebs' Starting Substrate For Making OAA

  Citrate → Isocitrate → α-Ketoglutarate → Succinyl CoA → Succinate → Fumarate → Malate → OAA.

  Use it when: Order of Krebs intermediates.

• 3-2-1 from NADH, FADH₂, GTP

  NADH gives 3 ATP, FADH₂ gives 2, one GTP gives 1 ATP per turn.

  Use it when: ATP counting.

• 8 + 6 + 24 = 38

  Glycolysis 8, link 6, Krebs 24.

  Use it when: Balance sheet.

• RQ: Carbs 1, Proteins 0.9, Fats 0.7

  'One Nice Seven': the more oxygen-poor the fuel, the lower the RQ.

  Use it when: RQ values.

• I-Q-III-c-IV-O₂

  Complex I → ubiQuinone → complex III → cytochrome c → complex IV → O₂.

  Use it when: Order of the ETS.

Respiration in Plants: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 2 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Respiration in Plants

The slips examiners see most often, and what to write instead.

• ✗Mistake: Writing that glycolysis occurs in mitochondria

  All respiration is linked with mitochondria.

  ✓Do this: Glycolysis is in the cytoplasm; link reaction and Krebs cycle in the matrix; ETS on the inner membrane.

  Usually costs: 1 mark.

• ✗Mistake: Giving 4 ATP as the net yield of glycolysis

  Forgetting ATP used in the first phase.

  ✓Do this: 4 ATP made − 2 ATP used = net 2 ATP (plus 2 NADH).

  Usually costs: 1 mark.

• ✗Mistake: Counting 36 or 30 ATP without explanation

  Other books use different assumptions.

  ✓Do this: For NCERT-based answers use 38, showing NADH = 3 ATP and FADH₂ = 2 ATP and the assumptions.

  Usually costs: 1-2 marks.

• ✗Mistake: Writing RQ as O₂/CO₂

  Ratio inverted.

  ✓Do this: RQ = CO₂ evolved ÷ O₂ consumed.

  Usually costs: 1 mark.

• ✗Mistake: Saying fermentation produces ATP in the fermentation step itself

  All 2 ATP are assumed to come from fermentation.

  ✓Do this: The 2 ATP come from glycolysis; the fermentation step only regenerates NAD⁺.

  Usually costs: ½-1 mark.

• ✗Mistake: Calling respiration a purely catabolic pathway

  Old textbook idea.

  ✓Do this: It is amphibolic: intermediates are withdrawn for synthesis of fatty acids and amino acids.

  Usually costs: 1 mark.

NCERT exercise guide: Respiration in Plants

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Respiration in Plants

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The enzyme that joins acetyl CoA with oxaloacetic acid is:
   (a) pyruvic dehydrogenase (b) citrate synthase (c) hexokinase (d) alcohol dehydrogenase

   Hint

   Product is citrate.

   Answer

   (b) citrate synthase

2. Q2 · 1 mark · MCQExam level

   RQ of a pure protein substrate is about:
   (a) 1.0 (b) 0.7 (c) 0.9 (d) 4.0

   Hint

   Between fat and carbohydrate.

   Answer

   (c) 0.9

3. Q3 · 2 marks · Very short answerExam level

   Name the two parts of ATP synthase in mitochondria and state the function of each. (2 marks)

   Hint

   Channel and headpiece.

   Answer

   F₀: integral membrane protein forming the proton channel (1); F₁: peripheral headpiece with the site of ATP synthesis from ADP and Pi (1).

4. Q4 · 2 marks · Very short answerHarder exam level

   How do fats and proteins enter the respiratory pathway? (2 marks)

   Hint

   Entry points.

   Answer

   Fats: glycerol → PGAL; fatty acids → acetyl CoA (1). Proteins: amino acids after deamination enter as pyruvate, acetyl CoA or Krebs cycle intermediates (1).

5. Q5 · 2 marks · Very short answerHarder exam level

   Why does FADH₂ yield fewer ATP than NADH? (2 marks)

   Hint

   Entry point in ETS.

   Answer

   FADH₂ passes electrons to ubiquinone via complex II, skipping complex I (1); fewer protons are pumped, so it supports 2 ATP instead of 3 (1).

6. Q6 · 2 marks · Very short answerHarder exam level

   Distinguish between substrate-level phosphorylation and oxidative phosphorylation with one example each from respiration. (2 marks)

   Hint

   Direct vs via ETS.

   Answer

   Substrate-level: ATP (or GTP) made directly in a reaction of a pathway, e.g. PEP → pyruvate or succinyl CoA → succinate (1). Oxidative: ATP made by ATP synthase using the proton gradient built by electron transport from NADH/FADH₂ to O₂ (1).

7. Q7 · 3 marks · Short answerHarder exam level

   Compare alcoholic and lactic acid fermentation in terms of organisms, enzymes and products. (3 marks)

   Hint

   Yeast vs muscle/bacteria.

   Answer

   Alcoholic: yeast; pyruvic acid decarboxylase and alcohol dehydrogenase; ethanol + CO₂ (1½). Lactic acid: some bacteria and exercising muscle; lactate dehydrogenase; lactic acid, no CO₂ (1½). Both reoxidise NADH and give net 2 ATP.

8. Q8 · 3 marks · Short answerHarder exam level

   In a tissue respiring glucose aerobically, the NADH formed in glycolysis is not passed into the mitochondria; it is used instead to reduce other compounds in the cytoplasm. (a) Which assumption of the respiratory balance sheet does this break? (b) Recalculate the net ATP per glucose. (c) Give one other reason why the real yield in a living cell differs from the textbook figure. (3 marks)

   Hint

   Take 3 ATP per NADH away for each glycolytic NADH.

   Answer

   (a) The assumption that the NADH made in glycolysis enters the mitochondria and undergoes oxidative phosphorylation (1). (b) The 2 glycolytic NADH would have given 2 × 3 = 6 ATP, so the yield is 38 − 6 = 32 ATP per glucose (1). (c) Any one: intermediates are withdrawn to make other compounds; other substrates enter the pathway midway; all pathways run at the same time and ATP is used as soon as it is made (1).

9. Q9 · 5 marks · Long answerHarder exam level

   (a) Describe the link reaction. (b) Explain the role of O₂ in aerobic respiration. (c) In oxidative phosphorylation and in photophosphorylation, what supplies the energy to build the proton gradient, and in which compartment do protons accumulate in each case? (5 marks)

   Hint

   Pyruvate dehydrogenase; final acceptor.

   Answer

   (a) In the matrix, pyruvate is oxidatively decarboxylated by pyruvic dehydrogenase with NAD⁺ and CoA (Mg²⁺) to give acetyl CoA, CO₂ and NADH; 2 NADH per glucose (2). (b) O₂ is the final hydrogen acceptor at complex IV, forming water; it keeps electrons flowing so NAD⁺ and FAD are regenerated (1½). (c) Oxidative phosphorylation: energy from oxidation-reduction along the ETS; protons collect in the mitochondrial intermembrane space and return to the matrix through F₀. Photophosphorylation: light energy; protons collect in the thylakoid lumen and return to the stroma through CF₀ (1½).

10. Q10 · 3 marks · Short answerHOTS / competency

    Suppose a plant cell is supplied only with fatty acids and no sugar. Predict its RQ and explain why glycolysis would contribute little to its ATP. (3 marks)

    Hint

    Entry at acetyl CoA.

    Answer

    RQ would be below 1 (about 0.7), since fats are oxygen-poor and need more O₂ (1). Fatty acids are broken to acetyl CoA, entering directly into the Krebs cycle and bypassing glycolysis (1). So ATP comes mainly from Krebs-cycle NADH/FADH₂ via the ETS, not from glycolysis (1).

11. Q11 · 3 marks · Short answerHOTS / competency

    A student says: 'Since O₂ is used only at the very end, a cell without O₂ should still run the Krebs cycle and make most of its ATP.' Evaluate this claim. (3 marks)

    Hint

    Recycling of NAD⁺ and FAD.

    Answer

    The claim is wrong (½). The Krebs cycle needs a continuous supply of NAD⁺ and FAD; these are regenerated only when NADH and FADH₂ give electrons to the ETS, whose final acceptor is O₂ (1½). Without O₂ the chain stops, NAD⁺ runs out, the cycle halts, and the cell relies on glycolysis plus fermentation for just 2 ATP per glucose (1).

12. Q12 · 1 mark · Assertion-reasonHOTS / competency

    Assertion (A): The respiratory pathway is described as amphibolic.
    Reason (R): Respiration only breaks down substrates to release energy.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

    Hint

    Amphi = both.

    Answer

    (c) A is true; R is false because intermediates are also used for synthesis.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Through which structures do woody stems exchange gases?Show answer

   Lenticels.

2. 2.Which enzyme converts glucose to glucose-6-phosphate?Show answer

   Hexokinase.

3. 3.Which enzyme splits sucrose for glycolysis?Show answer

   Invertase.

4. 4.Name the two enzymes of alcoholic fermentation.Show answer

   Pyruvic acid decarboxylase and alcohol dehydrogenase.

5. 5.At about what alcohol concentration do yeasts die?Show answer

   About 13%.

6. 6.Which enzyme starts the Krebs cycle?Show answer

   Citrate synthase.

7. 7.What is the first product of the Krebs cycle?Show answer

   Citric acid (6C).

8. 8.Which cytochrome is a mobile carrier between complexes III and IV?Show answer

   Cytochrome c.

9. 9.What does complex IV contain?Show answer

   Cytochromes a and a₃ and two copper centres.

10. 10.How many protons pass through F₀ per ATP made?Show answer

    Four.

11. 11.What is the RQ of tripalmitin?Show answer

    About 0.7 (102/145).

12. 12.At which point does glycerol enter the respiratory pathway?Show answer

    As PGAL (in glycolysis).

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you give three reasons why plants need no respiratory organs?
• ☐Can you write glycolysis with the ATP-using and ATP-making steps marked?
• ☐Can you compare alcoholic and lactic acid fermentation?
• ☐Can you list the products of one Krebs cycle turn?
• ☐Can you draw the ETS from NADH to O₂ with complexes I-V?
• ☐Can you calculate 38 ATP in two different ways?
• ☐Can you state the four assumptions of the balance sheet and why they fail in cells?
• ☐Can you calculate RQ from a balanced equation?

Respiration in Plants: frequently asked questions