AQA AS Level Biology Paper 2, November 2021: Question 7

✅ Correct Answer / Mark Scheme

• Component 1 & 2 (2 marks): Adenine and ribose (or pentose sugar), plus three phosphate groups. (Note: ADP + Pi is also accepted for 2 marks).
• Component 3 (1 mark): Condensation reaction.
• Component 4 (1 mark): Catalysed by the enzyme ATP synthase.

💡 Key Knowledge

• ATP stands for Adenosine Triphosphate. It is a nucleotide derivative.
• Energy is required to join ADP and inorganic phosphate ( P_i ) together via a condensation reaction.
• ATP synthase is the specific enzyme embedded in membranes (like mitochondrial or bacterial cell-surface membranes) that facilitates this phosphorylation.

🧠 Exam Technique

• Read command words carefully: "Describe" requires a detailed account of the structure and the reaction process, not just a list of words.
• Ensure you name all three distinct chemical components clearly to secure both initial marking points.

❌ Common Errors

• Saying "adenosine" instead of "adenine" (adenosine is the combination of adenine and ribose, so naming it as a component is chemically imprecise).
• Writing "ATPase" instead of "ATP synthase". ATPase breaks down ATP; ATP synthase builds it. Examiners heavily penalise this confusion.
• Stating "phosphodiester bond" instead of recognising it is a condensation reaction forming phosphoanhydride bonds.

✅ Correct Answer

57% or 57.1% (Accept 57 to 1 sig fig or 57.1 to 3 sig figs).

📐 Step-by-Step Calculation

1. Step 1: Find the initial rate of amino acid uptake for H cells (at 0 mmol dm⁻³ cyanide) from Figure 8 = 2.8 (or accept 2.8 - 3.0 depending on grid resolution, though standard mark scheme uses initial value 2.8). Let's use start = 2.8. At 30 mmol dm⁻³, the plateau value for H cells = 1.2 .
2. Step 2: Calculate the decrease: Initial (2.8) - Final (1.2) = 1.6 .
3. Step 3: Apply the percentage formula: (Decrease ÷ Initial) × 100
4. Step 4: (1.6 ÷ 2.8) × 100 = 57.14% . Round appropriately.

❌ Common Calculation Traps

• Dividing the final value by the initial value instead of calculating the change first.
• Using the wrong baseline starting value by misreading the y-axis intercept for curve H.

✅ Correct Answer / Mark Scheme (Max 3 marks)

• Point 1: Amino acid uptake occurs by active transport.
• Point 2: Cyanide reduces or stops amino acid uptake in both cell types (though G drops to zero, H plateaus).
• Point 3: ATP production stops on cell-surface membranes (or enzymes not working on membranes) in both cells.
• Point 4: ATP production continues in the cytoplasm (or enzymes active in cytoplasm) for H cells only, explaining why H cells maintain residual uptake.

💡 Key Knowledge

• Cyanide is a metabolic inhibitor that typically halts ATP production (inhibits respiration/electron transport chain).
• Active transport requires metabolic energy in the form of ATP. If uptake drops to zero with an inhibitor, it indicates active transport.
• Cellular compartmentalisation: G cells only have ATP-producing enzymes on their surface membrane (blocked entirely by external cyanide), whereas H cells have them in the cytoplasm as well (allowing internal glycolysis/cytoplasmic ATP generation to sustain partial uptake).

🧠 Exam Technique

• Link the shape of the graph directly to the stem information provided about enzyme locations in G vs H cells.
• Ensure you explicitly mention active transport to gain the mechanism mark. Top-level responses compare both cell lines rather than describing them in isolation.