AQA A-Level Biology Paper 2, June 2025: Question 1

✅ Correct Answers

• 1: Cytoplasm (or cytosol)
• 2: Phosphorylation
• 3: Pyruvate (accept pyruvic acid)
• 4: NAD (accept NADH, NADH₂, reduced NAD)

💡 Key Knowledge

• Glycolysis is strictly anaerobic and occurs in the cytoplasm, not inside the mitochondria.
• Glucose (6C) is activated by phosphorylation using 2 molecules of ATP to form hexose bisphosphate before splitting into two triose phosphate (3C) molecules.
• Triose phosphate is oxidised by losing hydrogen to the coenzyme NAD, forming pyruvate and generating a net gain of 2 ATP per glucose.

🧠 Exam Technique

Be precise with terminology. When the passage already says "reduced _____", filling in simply NAD is standard because together it reads "reduced NAD". However, AQA generously accepts written equivalents like NADH or NADH₂.

❌ Common Errors

• Writing NADP instead of NAD (NADP is used exclusively in photosynthesis; NAD is used in respiration).
• Writing "substrate-level phosphorylation" or "oxidative phosphorylation" for blank (2) — this describes ATP synthesis, not the phosphorylation of glucose.
• Confusing the cytoplasm with the mitochondrial matrix.

✅ Correct Answers (Awarded in Pairs)

Option Pair A (Per-turn basis):

• 1. 1 CO₂ is produced from the link reaction and 2 CO₂ are produced from the Krebs cycle [1 mark]
• 2. The link reaction and Krebs cycle occur twice per glucose molecule (or 2 pyruvate are produced per glucose) [1 mark]

Option Pair B (Total glucose basis):

• 1. 2 CO₂ molecules are produced in total by the link reaction [1 mark]
• 2. 4 CO₂ molecules are produced in total by the Krebs cycle [1 mark]

📐 Stoichiometric Breakdown

Trace the 6 carbons of glucose (C₆H₁₂O₆):

🧠 Exam Technique

The prompt specifically instructs you to use knowledge of both the link reaction and the Krebs cycle. You must state the exact numbers of CO₂ produced at both stages to access both marks.

❌ Common Errors

• Forgetting that 1 glucose splits into 2 pyruvates, leaving the final tally as 1 + 2 = 3 CO₂.
• Vaguely stating "carbon dioxide is released" without specifying which process produces how many molecules.

✅ Correct Answers (1 mark per sub-heading)

Oxygen uptake decreased:

• Less/no oxygen is required as the final electron acceptor (at the end of the electron transport chain) [Accept: less oxygen needed to bind with protons/H⁺ and electrons to form water].

ATP production decreased:

• Less/no oxidative phosphorylation (or: less/no proton gradient formed, reduced chemiosmosis / flow through ATP synthase).

Ethanol was produced:

• Anaerobic respiration occurs / pyruvate is reduced to ethanol (by reduced NAD).

💡 Key Knowledge

• Oxygen's role: Oxygen is the terminal electron acceptor in oxidative phosphorylation. It binds electrons from the ETC and protons from the matrix:
  ½O₂ + 2H⁺ + 2e⁻ → H₂O
• Coupled proton pumping: Without electron flow along the ETC carriers, protons (H⁺) cannot be actively pumped into the intermembrane space, collapsing the proton gradient needed for ATP synthase.
• Fermentation in yeast: With the ETC blocked, NADH cannot be re-oxidised aerobically. To regenerate free NAD for glycolysis to continue, yeast reduces pyruvate to ethanal and then to ethanol.

🧠 Exam Technique

• Notice the organism: yeast. Yeast carries out ethanolic fermentation, producing ethanol + CO₂, not lactate!
• Marks for points 1 and 2 can technically be awarded under either of the first two headings, but always write each answer under its specific provided sub-heading to guarantee full credit.

❌ Common Errors

• Rejecting lactate/lactic acid: Saying yeast produces lactic acid is an immediate disqualifier for mark point 3.
• Vaguely stating "oxygen isn't needed anymore" without explaining why (it acts as the terminal/final electron acceptor).
• Writing "hydrogen" instead of specifying protons / H⁺ ions when describing the electrochemical gradient.