AQA A-Level Biology Paper 2, June 2025: Question 6

✅ Mark Scheme Answers

• A: Cell body (or soma / dendron / dendrite(s))
• B: Axon (or axoplasm)
• C: Node of Ranvier
• D: Schwann cell

🧠 Exam Technique & Recognition

• Dendron/Cell Body (A): Look for the large central region containing the nucleus and broad cytoplasmic projections.
• Axon (B): Points directly to the elongated conductive core inside the myelin sheath (do not say "cytoplasm").
• Node of Ranvier (C): Points to the unprotected periodic gap between adjacent Schwann cells.
• Schwann Cell (D): Points to the cell wrapped around the axon that produces the myelin sheath.

❌ Common Misconceptions

• Confusing Node of Ranvier with synaptic cleft or axon terminal.
• Spelling: Phonetic spellings of Schwann cell are accepted (e.g., shwan , schwan ), but swan is strictly rejected!
• Writing "cytoplasm" for B is ignored—it must specifically refer to the axon or axoplasm.

✅ Mark Scheme Criteria (3 Marks Max)

• Point 1: No / less saltatory conduction
  OR Action potentials/depolarisation cannot jump from node to node
  OR Depolarisation must occur along the whole length of the axon
• Point 2: (Nerve) impulses / action potentials are slowed down (take longer to travel)
• Point 3: Affects/involves relay neurones (in brain)
  OR Affects transmission to motor neurone(s)
  OR Involves effector / muscle / neuromuscular junction

💡 Biological Mechanism

• Myelin acts as an electrical insulator composed of lipid membranes.
• In myelinated fibres, ion exchange occurs exclusively at the Nodes of Ranvier, allowing the action potential to "jump" via local circuits (saltatory conduction).
• Loss of myelin exposes the whole axon membrane, forcing continuous ion movement along the entire axon, substantially slowing transmission speed.
• Reflex arcs rely on fast relaying from sensory to relay neurones (in the CNS/brain) to motor neurones.

❌ Examiner Watch-Outs

• Never use vague terminology like "messages" or "signals". You must state nerve impulses or action potentials.
• Do not stop at "impulses are slower"—you must explain why (loss of saltatory conduction) and link it directly to the context (relay neurones in brain or transfer to motor neurones).

✅ Marking Guidance (3 Marks Max)

• 1: Less / no acetylcholine (ACh) binds to postsynaptic receptors;
• 2: Acetylcholine normally slows/decreases heart rate (causes inhibitory effect at the SAN)
  OR Fewer / no action potentials/depolarisation sent via parasympathetic neurones to SAN
  OR Less/no inhibition of SAN;
• 3: Explicitly refers to the sinoatrial node / SAN (pacemaker).

🧠 Exam Technique: Structuring the Explanation

Follow a clear 3-step pharmacological reasoning chain:

1. Competitive binding: Atropine binds to cholinergic receptors, blocking ACh from attaching.
2. Normal function of PNS: State clearly that the parasympathetic nervous system releases ACh to inhibit the SAN and reduce heart rate.
3. Net physiological effect: Because the SAN is no longer inhibited by ACh, heart rate increases.

❌ Pitfalls & Guidance

• Do NOT talk about enzymes or active sites: Atropine binds to receptors on the postsynaptic membrane or SAN cell membrane, not an enzyme active site. Mentioning active sites will lose marks!
• Name the pacemaker: Many students write "heart" or "cardiac muscle", missing the vital mark for naming the Sinoatrial Node (SAN).

Step 1: Write down the given values & formula
Formula: LBW = (9.27 × 10³ × TBW) / [(6.68 × 10³) + (216 × BMI)]
Given: TBW = 115.2 kg, BMI = 41, Dosage rate = 0.01 mg kg⁻¹ of LBW

Step 2: Calculate the Numerator
Numerator = 9.27 × 10³ × 115.2 = 9270 × 115.2 = 1,067,904

Step 3: Calculate the Denominator
(6.68 × 10³) = 6680
(216 × BMI) = 216 × 41 = 8856
Denominator = 6680 + 8856 = 15,536

Step 4: Calculate LBW (Lean Body Weight)
LBW = 1,067,904 / 15,536 ≈ 68.737384... kg

Step 5: Calculate Required Dosage & Round to 3 sig figs
Dosage = 0.01 mg kg⁻¹ × 68.737384... kg = 0.68737384... mg
Rounding to 3 significant figures gives: 0.687 mg

Full Marks: Correct answer of 0.687 alone scores 2 marks.

❌ Common Calculation Traps (1 Mark Partial Credit)

• Significant Figures Error: Leaving answer unrounded (e.g. 0.6874 ) or over-rounding to 2 s.f. ( 0.69 ) loses 1 mark.
• Rounding Error: Rounding incorrectly up to 0.688 scores only 1 mark.
• Order of Operations (BIDMAS):
  • Adding before multiplying: giving an answer of 1.54 (1 mark).
  • Multiplying by (216 × BMI) instead of adding: giving an answer of 1.81 × 10⁻⁴ (1 mark).
• Forgetting Dosage Factor: Calculating LBW as 68.7 kg but forgetting to multiply by 0.01 mg kg⁻¹.

🧠 Examiner Checklist for Calculations

• Keep full calculator precision in your working; do not round values mid-calculation.
• Always re-read the final line of the question prompt: "Give your answer to 3 significant figures."
• Check unit consistency: TBW is given in kg, dosage is per kg, final answer required in mg. No unit conversion between grams and milligrams was needed here.