Respiration in Plants ISC Class 11: Glycolysis, Krebs Cycle, ETS and RQ

What Respiration in Plants is about

Weightage: Respiration in Plants belongs to the unit Plant Physiology, which CISCE weights at 17 of the 70 marks in the ISC Class XI Biology Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Biology syllabus, examination year 2028, Class XI unit weightage

Key concepts

Gas exchange in plants and types of respiration

Do plants breathe? Plants need O₂ for respiration and give out CO₂, but they have no special respiratory organs. Reasons: (1) each part (root, stem, leaf) takes care of its own gas exchange, so little gas is transported from one part to another; (2) plants have low respiratory demand compared with animals, except during germination and active growth; (3) gases diffuse only short distances: almost every living cell is close to the surface, and loosely packed parenchyma has intercellular air spaces connected to stomata (leaves, young stems) and lenticels (woody stems and roots); in large stems most living cells lie just under the bark. In leaves, the O₂ produced in photosynthesis is also available to cells. Types of respiration: (1) aerobic, using O₂, completely oxidising the substrate to CO₂ and water and releasing much energy; (2) anaerobic, without O₂, incompletely oxidising the substrate to ethanol or lactic acid; in microbes it is called fermentation. Sometimes respiration is also classed by substrate: floating respiration uses carbohydrates or fats; protoplasmic respiration uses proteins (in starving cells).

Worked example

A large tree trunk has no breathing organ. How do its living cells get oxygen?

Try it first, then show the solutionSolution

1. Most of the trunk's interior is dead wood that does not respire
2. The living cells lie in a thin layer just under the bark and get O₂ by diffusion through lenticels and air spaces

Answer: Diffusion through lenticels to living cells near the surface

Glycolysis (EMP pathway): flowchart

Glycolysis (Embden–Meyerhof–Parnas pathway) occurs in the cytoplasm of all living cells and needs no oxygen. In plants the glucose comes from sucrose or from stored starch. Flowchart: glucose (6C) → *(ATP → ADP; hexokinase)* glucose-6-phosphate → fructose-6-phosphate → *(ATP → ADP; phosphofructokinase)* fructose-1,6-bisphosphate (6C) → splits into dihydroxyacetone phosphate and glyceraldehyde-3-phosphate (PGAL), two interconvertible 3C triose phosphates → each PGAL *(NAD⁺ → NADH + H⁺; inorganic phosphate added)* → 1,3-bisphosphoglycerate → *(ADP → ATP)* 3-phosphoglycerate → 2-phosphoglycerate → *(−H₂O)* phosphoenolpyruvate (PEP) → *(ADP → ATP)* pyruvic acid (3C). Balance per glucose: ATP used 2, ATP formed 4 (by substrate-level phosphorylation), so net 2 ATP; 2 NADH formed; 2 pyruvate formed. Pyruvic acid is the key product: its fate depends on whether O₂ is available.

Worked example

In glycolysis, at which two steps is ATP used and at which steps is it formed?

Try it first, then show the solutionSolution

1. Used: glucose → glucose-6-phosphate, and fructose-6-phosphate → fructose-1,6-bisphosphate
2. Formed: 1,3-bisphosphoglycerate → 3-phosphoglycerate, and PEP → pyruvate (each twice per glucose)

Answer: 2 ATP used; 4 ATP formed; net gain 2 ATP

Fermentation (anaerobic respiration): brief idea

Under anaerobic conditions pyruvate is not oxidised further; instead the NADH from glycolysis is used to reduce it, regenerating NAD⁺ so that glycolysis can continue. (1) Alcoholic fermentation (yeast, also plant roots in waterlogged soil): pyruvic acid → *(pyruvic acid decarboxylase; CO₂ released)* acetaldehyde → *(alcohol dehydrogenase; NADH → NAD⁺)* ethanol. (2) Lactic acid fermentation (some bacteria; animal muscle during intense exercise when O₂ is short): pyruvic acid → *(lactate dehydrogenase; NADH → NAD⁺)* lactic acid. In both, the net yield is only 2 ATP per glucose (from glycolysis), and most of the energy remains in ethanol or lactic acid. These products are toxic; yeasts die when ethanol reaches about 13% in the medium.

Worked example

Why must yeast convert pyruvate to ethanol rather than simply stopping at pyruvate when O₂ is absent?

Try it first, then show the solutionSolution

1. Glycolysis needs NAD⁺ at the PGAL oxidation step; without O₂ the NADH cannot be oxidised by the ETS
2. Reducing acetaldehyde to ethanol oxidises NADH back to NAD⁺, letting glycolysis continue and make ATP

Answer: To regenerate NAD⁺ so that glycolysis keeps making ATP

Aerobic respiration: link reaction and Krebs cycle (flowchart)

In the presence of O₂, pyruvate enters the mitochondrial matrix. Link reaction (oxidative decarboxylation): pyruvic acid (3C) + coenzyme A + NAD⁺ → *(pyruvate dehydrogenase complex; Mg²⁺)* acetyl CoA (2C) + CO₂ + NADH + H⁺. Krebs cycle (citric acid / TCA cycle, described by Hans Krebs) in the matrix: acetyl CoA (2C) + oxaloacetic acid (4C) + H₂O → *(citrate synthase; CoA released)* citric acid (6C) → isocitrate (6C) → *(NAD⁺ → NADH; CO₂ out)* α-ketoglutaric acid (5C) → *(NAD⁺ → NADH; CO₂ out)* succinyl CoA (4C) → *(GDP → GTP → ATP; substrate-level phosphorylation)* succinic acid (4C) → *(FAD → FADH₂)* fumaric acid → *(+H₂O)* malic acid → *(NAD⁺ → NADH)* oxaloacetic acid (4C), which accepts the next acetyl CoA. Per turn (per acetyl CoA): 2 CO₂, 3 NADH, 1 FADH₂, 1 ATP (via GTP). Each glucose gives two pyruvates, so the cycle turns twice. Up to the Krebs cycle the glucose has been fully broken to CO₂, but most energy is still held in NADH and FADH₂.

Worked example

How many CO₂, NADH and FADH₂ are produced from one pyruvate entering the mitochondrion (link reaction plus one Krebs turn)?

Try it first, then show the solutionSolution

1. Link reaction: 1 CO₂ and 1 NADH
2. Krebs turn: 2 CO₂, 3 NADH, 1 FADH₂ (and 1 ATP)

Answer: 3 CO₂, 4 NADH and 1 FADH₂ (plus 1 ATP)

Electron transport system and oxidative phosphorylation

The electron transport system (ETS) lies in the inner mitochondrial membrane. Flowchart: NADH (from the matrix) → complex I (NADH dehydrogenase) → ubiquinone (coenzyme Q) → complex III (cytochrome bc₁) → cytochrome c (a small mobile protein on the outer face of the inner membrane) → complex IV (cytochrome c oxidase, with cytochromes a and a₃ and copper) → O₂, which accepts the electrons and protons to form water. FADH₂ enters through complex II (succinate dehydrogenase) to ubiquinone, by-passing complex I. As electrons flow, energy is used to pump protons from the matrix into the intermembrane space, building a proton gradient. Protons flow back through complex V (ATP synthase: an F₀ channel in the membrane and an F₁ head facing the matrix), which makes ATP. Oxidative phosphorylation (definition): the synthesis of ATP from ADP and inorganic phosphate using the energy released when electrons from NADH and FADH₂ are passed along the ETS to oxygen as the final acceptor. It differs from photophosphorylation, where light energy drives the proton gradient. Oxygen is needed only at the end of the chain, but without it the whole chain stops, because it removes the electrons.

Worked example

Why does cyanide, which blocks cytochrome c oxidase, stop almost all ATP production in a cell, even though glycolysis does not need oxygen?

Try it first, then show the solutionSolution

1. With complex IV blocked, electrons cannot pass to O₂, so the whole chain backs up and no proton gradient is made
2. NADH and FADH₂ cannot be reoxidised, so the Krebs cycle stops; only glycolysis (2 ATP per glucose) remains

Answer: Oxidative phosphorylation, which makes most ATP, stops completely

Energy relations: ATP count per glucose

The classical balance sheet assumes each NADH yields 3 ATP and each FADH₂ yields 2 ATP in the ETS, and that NADH made in the cytoplasm enters the mitochondrion without loss. Glycolysis: 2 ATP (net) + 2 NADH (= 6 ATP) → 8 ATP. Link reaction (×2): 2 NADH → 6 ATP. Krebs cycle (×2): 2 ATP + 6 NADH (18 ATP) + 2 FADH₂ (4 ATP) → 24 ATP. Total = 8 + 6 + 24 = 38 ATP per glucose, of which 4 are made by substrate-level phosphorylation and 34 by oxidative phosphorylation. This is a theoretical maximum: it assumes a strictly sequential, orderly pathway with one substrate, no intermediates withdrawn for synthesis, and full efficiency. In living cells some intermediates are used to build other molecules, transporting cytoplasmic NADH into mitochondria can cost energy, and the membrane is somewhat leaky, so the real yield is lower (often given as about 30–32 ATP). In fermentation the yield is only 2 ATP per glucose. Taking roughly 7.3 kcal per mole of ATP and 686 kcal per mole of glucose, aerobic respiration traps about 40% of the energy as ATP; the rest is lost as heat.

Worked example

How many ATP are produced by the complete aerobic oxidation of 5 molecules of glucose, and how many glucose molecules would yeast need to ferment to produce the same amount?

Try it first, then show the solutionSolution

1. Aerobic: 5 × 38 = 190 ATP
2. Fermentation gives 2 ATP per glucose: 190 ÷ 2 = 95 glucose molecules

Answer: 190 ATP; 95 glucose molecules

Amphibolic pathway: brief idea

The respiratory pathway is not only for breakdown (catabolism). Its intermediates are also used to build molecules (anabolism), so it is called amphibolic. Entry points for breakdown: carbohydrates enter as glucose; fats are broken into glycerol, which enters glycolysis as PGAL, and fatty acids, which are broken down to acetyl CoA; proteins are broken into amino acids, which after removal of the amino group enter as pyruvate, acetyl CoA or Krebs-cycle acids such as α-ketoglutarate and oxaloacetate. Withdrawals for synthesis: when a cell needs to make fatty acids, acetyl CoA is withdrawn from the pathway; Krebs-cycle acids such as α-ketoglutarate and oxaloacetate are used to make amino acids. Thus the same pathway can run materials towards breakdown or away for synthesis.

Worked example

Why is the respiratory pathway called amphibolic and not just catabolic?

Try it first, then show the solutionSolution

1. Fats and proteins are broken down and fed into it, which is catabolic
2. Acetyl CoA and Krebs acids are also withdrawn to make fatty acids and amino acids, which is anabolic

Answer: Because it serves both breakdown and synthesis

Respiratory quotient (RQ)

The respiratory quotient is the ratio of the volume of CO₂ evolved to the volume of O₂ consumed in respiration: RQ = CO₂ ÷ O₂. It depends on the substrate. (1) Carbohydrates: $C_6{H}_{12}{O}_6+6O_2\to 6C{O}_2+6H_{2}O$, so RQ = 6/6 = 1. (2) Fats contain relatively little oxygen, so they need more O₂ to oxidise; for tripalmitin, $2C_{51}{H}_{98}{O}_6+145O_2\to 102C{O}_2+98H_{2}O$, RQ = 102/145 = 0.7 (less than 1). (3) Proteins: about 0.9. (4) Organic acids, which are rich in oxygen, give RQ more than 1: malic acid, $C_4{H}_6{O}_5+3O_2\to 4C{O}_2+3H_{2}O$, RQ = 4/3 = 1.33. (5) Anaerobic respiration: CO₂ is released but no O₂ is used, so RQ is infinite. In living organisms the substrate is usually a mixture, so measured RQ values are intermediate; e.g. an RQ near 0.7 in germinating castor seeds shows that fat is being used.

Worked example

Germinating seeds use 14 mL of O₂ and give out 10 mL of CO₂ in an hour. Calculate the RQ and identify the likely substrate.

Try it first, then show the solutionSolution

1. RQ = CO₂ ÷ O₂ = 10 ÷ 14 = 0.71
2. An RQ near 0.7 indicates fats

Answer: RQ ≈ 0.71; the seeds are respiring fat (oily seeds)

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

• Electron Transport, ATP Yield and Respiratory Quotient
• Glycolysis, Fermentation and the Krebs Cycle

Respiration in Plants: key facts

Aerobic respiration of glucose

$C_6{H}_{12}{O}_6+6O_2\to 6C{O}_2+6H_{2}O+\text{energy}$

About 686 kcal per mole released.

Respiratory quotient

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

Carbohydrate 1; fat about 0.7; protein about 0.9; organic acids above 1; anaerobic infinite.

ATP balance per glucose (classical)

$\underset{\text{glycolysis}}{\underbrace{2+2\times 3}}+\underset{\text{link}}{\underbrace{2\times 3}}+\underset{\text{Krebs}}{\underbrace{2+6\times 3+2\times 2}}=8+6+24=38$

NADH = 3 ATP, FADH₂ = 2 ATP; actual yield lower.

Per pyruvate and per acetyl CoA

$\text{pyruvate}\to 4NADH+1FAD{H}_2+1ATP=15ATP;\text{acetyl CoA}\to 12ATP$

Useful for quick checks.

Fermentation

$\text{pyruvate}\to \text{acetaldehyde}+C{O}_2\to \text{ethanol};\text{pyruvate}\to \text{lactic acid}$

NADH reoxidised; net 2 ATP per glucose.

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.

• Citrate Is Krebs' Starting Substrate For Making Oxaloacetate

  Citric acid, Isocitric, α-Ketoglutaric, Succinyl CoA, Succinic, Fumaric, Malic, Oxaloacetic.

  Use it when: Order of Krebs cycle acids.

• 2-2-6-2 for one glucose through Krebs

  2 ATP, 2 FADH₂, 6 NADH, 4 CO₂ (two turns).

  Use it when: Krebs cycle totals.

• 8 + 6 + 24 = 38

  Glycolysis 8, link 6, Krebs 24.

  Use it when: ATP count.

• One, point-seven, point-nine

  RQ: carbohydrate 1, fat 0.7, protein 0.9; acids above 1; anaerobic infinite.

  Use it when: RQ values.

• I-Q-III-c-IV-O

  NADH → complex I → ubiquinone → complex III → cytochrome c → complex IV → oxygen.

  Use it when: ETS order.

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 D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 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: Counting glycolysis as 4 ATP

  The investment steps are forgotten.

  Wrong working

  Glycolysis gives 4 ATP.

  Right working

  4 ATP are formed but 2 are used: net 2 ATP (plus 2 NADH).

  ✓Do this: Always write 'net'.

  Usually costs: 1 mark.

• ✗Mistake: Counting the Krebs cycle once per glucose

  One turn is learned, then not doubled.

  Wrong working

  Krebs gives 1 ATP, 3 NADH and 1 FADH₂ per glucose.

  Right working

  One glucose gives two acetyl CoA, so the cycle turns twice: 2 ATP, 6 NADH, 2 FADH₂.

  ✓Do this: Multiply post-glycolysis steps by two.

  Usually costs: 1-2 marks.

• ✗Mistake: Inverting the RQ ratio

  O₂ is written on top.

  Wrong working

  RQ of fat = 145/102 = 1.42.

  Right working

  RQ = CO₂/O₂ = 102/145 = 0.7.

  ✓Do this: Quotient of what comes out over what goes in: CO₂ over O₂.

  Usually costs: 1-2 marks.

• ✗Mistake: Saying oxygen is used in the Krebs cycle

  Krebs is linked to aerobic respiration.

  Wrong working

  O₂ combines with acetyl CoA in the Krebs cycle.

  Right working

  O₂ is used only at the end of the ETS as the final electron acceptor, forming water.

  ✓Do this: Oxygen waits at complex IV.

  Usually costs: 1 mark.

• ✗Mistake: Writing full mechanisms instead of flowcharts

  Over-preparation.

  Wrong working

  Every enzyme name and structural formula in glycolysis.

  Right working

  The syllabus asks for flowcharts only: intermediates in order with key inputs and outputs.

  ✓Do this: Practise a neat one-page flowchart for each.

  Usually costs: Time.

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 final acceptor of electrons in the ETS is:
   (a) NAD⁺
   (b) cytochrome c
   (c) oxygen
   (d) FAD

   Hint

   Forms water.

   Answer

   (c)

2. Q2 · 1 mark · MCQExam level

   The RQ of a carbohydrate is:
   (a) 0.7
   (b) 0.9
   (c) 1
   (d) infinite

   Hint

   6/6.

   Answer

   (c)

3. Q3 · 1 mark · Very short answerExam level

   Name the 6-carbon compound formed first in the Krebs cycle.

   Hint

   Citric acid cycle.

   Answer

   Citric acid

4. Q4 · 1 mark · Very short answerExam level

   Name the end product of glycolysis.

   Hint

   3-carbon.

   Answer

   Pyruvic acid (pyruvate)

5. Q5 · 2 marks · Short answerExam level

   Distinguish between aerobic and anaerobic respiration.

   Hint

   O₂, products, ATP.

   Answer

   Aerobic: uses O₂; complete oxidation to CO₂ and H₂O; in cytoplasm and mitochondria; 38 ATP per glucose. Anaerobic: no O₂; incomplete oxidation to ethanol + CO₂ or lactic acid; in cytoplasm only; 2 ATP per glucose.

6. Q6 · 2 marks · Short answerHarder exam level

   What is the link reaction? Where does it occur?

   Hint

   Pyruvate to acetyl CoA.

   Answer

   Oxidative decarboxylation of pyruvic acid to acetyl CoA by the pyruvate dehydrogenase complex, releasing CO₂ and forming NADH; it occurs in the mitochondrial matrix and links glycolysis to the Krebs cycle.

7. Q7 · 2 marks · Short answerHarder exam level

   A seedling is respiring stored fat. At which point of the respiratory pathway does (i) glycerol and (ii) a fatty acid enter? (iii) Name one Krebs-cycle acid that the same cell can withdraw to make amino acids.

   Hint

   Fats split into two parts that enter at different points.

   Answer

   (i) Glycerol is converted to PGAL (glyceraldehyde-3-phosphate) and enters glycolysis. (ii) Fatty acids are broken down to acetyl CoA, which enters the Krebs cycle. (iii) α-Ketoglutaric acid (or oxaloacetic acid). Because the pathway both receives breakdown products and supplies building blocks, it is amphibolic.

8. Q8 · 3 marks · Short answerHarder exam level

   Distinguish between substrate-level phosphorylation and oxidative phosphorylation, naming one step or site where each occurs in aerobic respiration.

   Hint

   Is the phosphate moved directly from a substrate, or is a proton gradient used?

   Answer

   Substrate-level phosphorylation: ATP is formed by direct transfer of a phosphate group from a phosphorylated intermediate to ADP, without the ETS or O₂; e.g. PEP → pyruvic acid in glycolysis (cytoplasm) or succinyl CoA → succinic acid in the Krebs cycle (matrix). Oxidative phosphorylation: ATP is made by ATP synthase (F₀–F₁) in the inner mitochondrial membrane, using the proton gradient set up as electrons from NADH and FADH₂ pass along the ETS to O₂. Classically 4 of the 38 ATP per glucose come from the first and 34 from the second.

9. Q9 · 3 marks · Short answerHarder exam level

   How many ATP are formed from (i) one acetyl CoA (ii) 4 molecules of glucose aerobically (iii) 4 molecules of glucose by fermentation?

   Hint

   12, 38, 2.

   Answer

   (i) 3 NADH × 3 + 1 FADH₂ × 2 + 1 ATP = 12 ATP (ii) 4 × 38 = 152 ATP (iii) 4 × 2 = 8 ATP.

10. Q10 · 2 marks · Short answerHarder exam level

    A plant tissue consumed 25 mL of O₂ and released 20 mL of CO₂. Calculate its RQ and suggest the substrate.

    Hint

    CO₂ ÷ O₂.

    Answer

    RQ = 20 ÷ 25 = 0.8, between the values for fat (0.7) and protein (0.9): probably proteins or a mixture of fat and carbohydrate.

11. Q11 · 5 marks · Long answerHarder exam level

    (a) Give a flowchart of glycolysis. [3]
    (b) What is the net gain of ATP and NADH per glucose? Where is ATP formed by substrate-level phosphorylation? [2]

    Hint

    Ten steps.

    Answer

    (a) Glucose → glucose-6-P (ATP used) → fructose-6-P → fructose-1,6-bisP (ATP used) → DHAP + PGAL → (×2) 1,3-bisphosphoglycerate (NADH formed) → 3-PGA (ATP formed) → 2-PGA → PEP → pyruvic acid (ATP formed). (b) Net 2 ATP and 2 NADH; substrate-level phosphorylation at 1,3-bisphosphoglycerate → 3-PGA and PEP → pyruvate.

12. Q12 · 5 marks · Long answerHarder exam level

    (a) Describe the electron transport system and explain how ATP is made. [3]
    (b) Why is oxygen essential although it is used only at the last step? [2]

    Hint

    Final acceptor.

    Answer

    (a) In the inner mitochondrial membrane, electrons from NADH pass through complex I, ubiquinone, complex III, cytochrome c and complex IV to O₂; FADH₂ enters at complex II. Energy released pumps protons into the intermembrane space; protons return through ATP synthase (F₀–F₁), which makes ATP (oxidative phosphorylation). (b) O₂ removes electrons from the end of the chain; without it carriers stay reduced, electron flow and proton pumping stop, NADH cannot be reoxidised and the Krebs cycle also halts.

13. Q13 · 3 marks · Short answerHarder exam level

    Why is the efficiency of aerobic respiration much higher than that of fermentation? Use approximate energy values (7.3 kcal per mole ATP; 686 kcal per mole glucose).

    Hint

    38 vs 2.

    Answer

    Aerobic: 38 × 7.3 = 277 kcal trapped out of 686, about 40%. Fermentation: 2 × 7.3 = 14.6 kcal, about 2%; most of the energy stays in ethanol or lactic acid because the substrate is only partly oxidised.

14. Q14 · 3 marks · Short answerHOTS / competency

    During a long fast, a plant tissue shows an RQ of about 0.9 instead of 1. What does this suggest, and what kind of respiration is this?

    Hint

    Substrate change.

    Answer

    An RQ near 0.9 shows proteins are being used as the respiratory substrate after carbohydrate reserves are exhausted. Breakdown of the cell's own proteins is protoplasmic respiration; amino acids are deaminated and enter as pyruvate, acetyl CoA or Krebs acids (amphibolic pathway).

15. Q15 · 3 marks · Short answerHOTS / competency

    A student says, 'In the Krebs cycle all the energy of glucose is released, since all its carbon leaves as CO₂.' Evaluate this statement.

    Hint

    Where is the energy?

    Answer

    Partly correct: by the end of the Krebs cycle all six carbons have left as CO₂. But only 4 ATP have been made directly; most energy is held as high-energy electrons in 10 NADH and 2 FADH₂. It is released only when these are oxidised in the ETS, giving 34 more ATP by oxidative phosphorylation.

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.Net ATP of glycolysis?Show answer

   2 (plus 2 NADH).

2. 2.Product of alcoholic fermentation?Show answer

   Ethanol and CO₂.

3. 3.Enzyme reducing pyruvate in muscle?Show answer

   Lactate dehydrogenase.

4. 4.Acceptor of acetyl CoA in Krebs cycle?Show answer

   Oxaloacetic acid.

5. 5.Site of ETS?Show answer

   Inner mitochondrial membrane.

6. 6.Complex IV?Show answer

   Cytochrome c oxidase.

7. 7.ATP per glucose (classical)?Show answer

   38.

8. 8.RQ in anaerobic respiration?Show answer

   Infinite.

Quick revision checklist

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

• ☐Can you give two reasons why plants lack respiratory organs?
• ☐Can you draw the glycolysis flowchart?
• ☐Can you write both fermentation pathways?
• ☐Can you list the products of one Krebs turn?
• ☐Can you name the ETS complexes in order?
• ☐Can you define oxidative phosphorylation?
• ☐Can you build the 38 ATP balance sheet?
• ☐Can you calculate RQ for glucose, malic acid and tripalmitin?

Respiration in Plants: frequently asked questions