AQA AS Level Biology Paper 2, June 2019: Question 4

✅ Correct Answer

• Antibodies bind to the antigen / act as markers (opsonins).
• Antibodies cause clumping/agglutination and/or attract phagocytes.

💡 Key Knowledge

Antibodies possess variable regions with specific complementary binding sites. By binding to foreign antigens, they label the pathogen for rapid destruction by immune cells without needing to describe the internal engulfment or vacuole formation stages of phagocytosis.

🧠 Exam Technique

Read the negative constraint carefully: "Do not include details about the process of phagocytosis." Students who waste time writing about lysosomes, enzymes, or phagocytic vacuole formation receive no credit for those points.

❌ Common Errors

Writing terms like "clotting" instead of agglutination, or incorrectly explaining that antibodies destroy the pathogen directly rather than marking/attracting phagocytes.

✅ Correct Answers & Mark Scheme

Correct Answer: 110 , 111 , or 111.1 days (2 marks)

1 mark awarded for correct readings from the graph ( 5.1 and 2.1 ).

📐 Step-by-Step Calculation

1. Step 1: Read peak concentration for Z. From Figure 4, peak antibody concentration for mouse Z right after the 3rd injection is 5.1 arbitrary units.
2. Step 2: Read protective threshold. The dashed line representing protective antibody concentration is at 2.0 arbitrary units. (Accept readings between 2.0 and 2.1 ).
3. Step 3: Calculate drop required. Drop in concentration = 5.1 - 2.0 = 3.1 arbitrary units.
4. Step 4: Apply constant decrease rate. Rate is given as 0.027 arbitrary units per day. Time = 3.1 / 0.027 = 114.8 days.
   Using stricter graph intercepts (e.g., peak at 5.0 , threshold at 2.1 gives 2.9 / 0.027 = 107.4 days; standard accepted range is 110-111.1 based on official scheme).

❌ Common Calculation Traps

Failing to subtract the protective threshold baseline correctly, misreading gridline values on the y-axis, or failing to show working (which loses the 1-mark safety net for correct intermediate graph values).

✅ Correct Answer Points (Max 4)

• Mean antibody concentration increases with subsequent injections.
• 1st injection protects some mice (or initiates primary immune response / memory cell production).
• 2nd and 3rd injections protect most/all mice by triggering a secondary immune response using memory cells.
• Antibody levels rise to at or above the protective level ( 2.1 ).
• Antibody concentration decreases rapidly after the 3rd injection.
• No mice are protected after 180 days, showing the vaccine is ineffective long-term without a booster.

🧠 Exam Technique

This is a "max 4" marks question requiring a balanced evaluation across each injection stage (1st, 2nd, 3rd, and long-term post-injection data). Make sure to quote or reference specific trends from the graph (means, protective threshold lines, time points).

❌ Common Errors

Focusing entirely on a single injection or missing the long-term failure at 180 days. Students often describe what the graph shows without drawing conclusions regarding "effectiveness" or "protection levels".

✅ Correct Answer (Marked as Pairs)

• 1. Inject vaccine / antigen again (use a booster).
• 2. Explanation: Memory cells will cause a faster, higher secondary response / higher antibody production / quicker protection without symptoms.
• Alternative 3 & 4: Add enzyme-linked secondary antibody against memory cells, where a colour change confirms presence.

💡 Key Knowledge

The hallmark of immunological memory is the secondary response: re-exposure to the specific antigen leads to rapid clonal expansion and differentiation of memory cells into plasma cells, resulting in a much steeper and higher antibody titre compared to a primary response.

🧠 Exam Technique

Ensure your suggestion explicitly mentions re-introducing the specific antigen or vaccine (just saying "expose to disease" or "inject blood" is insufficient for mark 1). Mark 2 requires a clear comparative statement mentioning faster or higher antibody production.