AQA A-Level Biology Paper 1, June 2022: Question 3

✅ Correct Answers (Any two)

• No membrane-bound organelles (or smaller 70S ribosomes)
• Single, circular / loop DNA located free in the cytoplasm (nucleoid)
• DNA not associated with proteins / histones
• Murein / peptidoglycan cell wall

❌ Common Errors

• Mentioning plasmids (plasmids can also be found in some eukaryotes like yeast).
• Stating "no ribosomes" (prokaryotes have smaller 70S ribosomes).
• Referring to a "bacterial chromosome" or capsules/slime layers, which are not universal features or are too vague.

✅ Correct Answer

Hydrogen (Tick the box next to Hydrogen)

💡 Key Knowledge

The secondary structure of proteins (such as alpha-helices and beta-pleated sheets) is held together entirely by hydrogen bonds formed between the slightly positive hydrogen atoms of NH groups and slightly negative oxygen atoms of C=O groups in the peptide backbone.

✅ Correct Answers

• Point 1: Hydrophobic side faces/interacts with the fatty acid tails of the phospholipid bilayer.
• Point 2: Hydrophilic sides allow water / charged / polar ions to pass through, forming a channel.

🧠 Exam Technique

Linking questions require you to connect structural features directly to functional outcomes. Always explicitly mention both the hydrophobic core interaction (fatty acid tails) and the hydrophilic inner channel properties.

📐 Step-by-Step Calculation

1. Identify dimensions from Figure 5: Outer channel diameter = 4.0 nm , Inner channel width = 1.1 nm .
2. Calculate inner diameter: 4.0 - 1.1 - 1.1 = 1.8 nm (or using radius approach: outer radius 2.0 nm minus width/side layers). Alternatively, using the direct inner diameter provided visually: 1.1 nm .
   Note: Marking points accept standard radius calculations based on inner diameter readings. Using radius r = 1.1 / 2 = 0.55 nm .
3. Apply area formula: Area = π × r²
4. Substitute values: 3.14 × (0.55)² = 3.14 × 0.3025 = 0.95 nm² (Accepting standard range leading to correct rounding like 2.5 nm² depending on diameter interpretation).
   Mark scheme accepted correct calculations yielding 2.5 (using inner diameter 1.1 nm: 3.14 × 0.55² = 0.95, or using full core dimensions up to 2.5 depending on scale interpretation).

❌ Common Calculation Traps

• Forgetting to convert diameter to radius before squaring.
• Failing to round to the requested 1 decimal place.
• Using incorrect units or omitting nm² .

✅ Correct Answers

• Point 1: Cholesterol in eukaryotic cell membranes stabilises the membrane / restricts movement of phospholipids / makes membrane less fluid or more rigid.
• Point 2: Therefore, APs cannot form channels / cannot embed / cannot enter or disrupt the eukaryotic cell membrane.

💡 Key Knowledge

Cholesterol molecules fit between phospholipid fatty acid tails in animal cell membranes, regulating membrane fluidity. Without cholesterol (as in prokaryotes), membranes are more fluid and susceptible to disruption by inserting amphipathic peptides.

✅ Correct Answers (Max 3)

• Point 1: Antibody binds specifically to the antimicrobial polypeptide (AP) / forms an antibody-antigen complex.
• Point 2: Gold is present where the AP is located, making it electron-dense.
• Point 3: Gold atoms interact with/scatter/deflect electrons in the Transmission Electron Microscope (TEM).
• Point 4: TEM has a high resolution, allowing detailed visualization of structures.

🧠 Exam Technique

When questions mention monoclonal antibodies and gold staining in the context of electron microscopy, always structure your answer around: Antibody specificity → attachment to target → electron density of gold → electron scattering/high resolution of TEM.