AQA AS Level Biology Paper 1, November 2021: Question 5

✅ Correct Answer / Mark Scheme

• Compact / Branched / Coiled: Fits a large amount of glucose into a small cellular space.
• Polymer of alpha-glucose: Easily hydrolysed to release monosaccharides.
• Branched structure: Provides multiple terminal ends simultaneously for rapid enzyme action (faster hydrolysis).
• Insoluble: Has no osmotic effect on the cell and cannot easily diffuse out of the cell.

💡 Key Knowledge

Glycogen is the main energy storage molecule in animals. Always link a structural feature explicitly to a functional advantage. For instance, never just state "it is branched"—you must explain that branching creates more ends for faster enzyme hydrolysis to release glucose during high respiration demand.

🧠 Exam Technique

This is a classic "structure-function" question worth up to 4 marks. Aim to make 4 distinct, paired points linking feature and function. Avoid vague statements like "it stores energy well"—specify that it provides a respiratory substrate for energy release.

❌ Common Errors

Students frequently confuse glycogen with cellulose or starch, mistakenly attributing plant-specific properties (like microfibrils or rigidity) to glycogen. Another common error is stating glycogen "is insoluble so it doesn't dissolve" without mentioning the vital consequence regarding water potential/osmosis.

✅ Correct Answer

Endopeptidase (accept correct phonetic spelling like endopeptidase).

💡 Key Knowledge

Endopeptidases hydrolyse peptide bonds within the central region of a polypeptide chain, producing smaller peptide fragments. Conversely, exopeptidases hydrolyse peptide bonds at the terminal ends of proteins to remove single amino acids.

✅ Correct Answer

3

💡 Key Knowledge

Figure 5 uses distinct geometric shapes (circles, triangles, squares) to represent different amino acids. Counting the unique shapes reveals how many types of monomer are present, which directly corresponds to the number of different R groups.

📐 Step-by-Step Calculation

Total starting length: 101 amino acids.

Step 1: Calculate total amino acids accounted for by known rows in Table 1.

• Length 5 polypeptides: 2 polypeptides × 5 amino acids = 10 amino acids.
• Length 15 polypeptides: 3 polypeptides × 15 amino acids = 45 amino acids.
• Total accounted for = 10 + 45 = 55 amino acids.

Step 2: Find remaining amino acids for lengths 6 and 20.

101 - 55 = 46 remaining amino acids.

Step 3: Solve using the remaining constraints.

We need a combination of length 6 and length 20 polypeptides that sum to 46 amino acids. Testing possible integer values for the number of length-20 polypeptides:

• If 1 × 20-mer = 20 amino acids. Remaining = 26 (not divisible by 6).
• If 2 × 20-mers = 40 amino acids. Remaining = 6 amino acids. This means exactly 1 × 6-mer.

Final Answers:

• 6 amino acids in length = 1
• 20 amino acids in length = 2

❌ Common Calculation Traps & Partial Marks

Students often forget to multiply the number of amino acids per peptide by the frequency of polypeptides given in the table. If you correctly calculated total amino acids accounted for (55) but couldn't solve the final split, the mark scheme awards 1 mark as a consolation for showing working (2 × 5 + 3 × 15) .