AQA A-Level Biology Paper 1, June 2025: Question 3

✅ Correct Answer (Table 1)

💡 Key Knowledge

• Glycogen: A branched polysaccharide made exclusively of α-glucose monomers joined by 1,4- and 1,6-glycosidic bonds.
• Sucrose: A disaccharide formed from one α-glucose and one fructose unit. Because it consists of only two units, it is not a polymer.
• DNA: A polymer of deoxyribonucleotides. Each nucleotide contains a deoxyribose sugar, a nitrogenous base, and a phosphate group. Monomers join via phosphodiester bonds.
• Condensation: Both glycosidic and phosphodiester bonds form via condensation (releasing H₂O). Hence, all three molecules are formed by condensation.

❌ Common Errors

• Calling sucrose a polymer: Disaccharides contain two monomer units, but polymers must consist of many repeating units.
• Thinking DNA does not form by condensation: Phosphodiester bond formation between the 3'-OH and 5'-phosphate releases water molecules.
• Ticking galactose for sucrose: Sucrose = glucose + fructose. Galactose is found in lactose (glucose + galactose).

🧠 Exam Technique

Read column-by-column rather than row-by-row. Check off the chemical identity first:

• Does it contain sugars, phosphates, or nitrogenous bases?
• Count monomer units: Disaccharide = 2 (not a polymer); Polynucleotide/Polysaccharide = many (polymer).

📐 Step-by-Step Calculation

Given:

• Ratio = Lactose : Galactose : Lactase = 15 : 3 : 1
• Total mixture volume = 90 cm³

1. Find total number of parts:
   15 + 3 + 1 = 19 parts
2. Find volume of 1 part:
   90 cm³ ÷ 19 = 4.7368... cm³
3. Calculate galactose volume (3 parts):
   3 × (90 ÷ 19) = 270 ÷ 19 = 14.2105... cm³
4. Round appropriately:
   Acceptable answers: 14 or 14.2 cm³

❌ Common Calculation Traps

• Forgetting the third component: Adding only 15 + 3 = 18 parts, giving (3 ÷ 18) × 90 = 15 cm³. This is incorrect because lactase represents 1 part of the final mixture.
• Calculating the wrong substance: Working out lactose (15 parts = 71.1 cm³) instead of galactose (3 parts).
• Incorrect rounding: Writing 14.21 without rounding or truncation errors (e.g. 14.1).

✅ Model Answers by Tube

Tube X (100 mg dm⁻³):

• Galactose binds to the active site of lactase as a competitive inhibitor; [1 mark]
• This slows / reduces / inhibits lactose hydrolysis (fewer enzyme-substrate complexes form). [1 mark]

Tube Y (180 mg dm⁻³):

• Without inhibitor, all / more lactose is hydrolysed (or hydrolysed faster), producing more glucose (>100 mg dm⁻³). [1 mark]

Tube Z (0 mg dm⁻³):

• No lactose / substrate is present (so no glucose can be formed) OR galactose cannot be hydrolysed by lactase / galactose is not a substrate. [1 mark]

💡 Scientific Principles

• Lactase reaction: Lactose + H₂O → Glucose + Galactose. Glucose is the measured product.
• Tube Y is the positive control: It shows the uninhibited rate/yield of lactose hydrolysis under identical conditions.
• Tube Z is the negative control: It proves that lactase cannot break down galactose to produce glucose, confirming that lactase is specific to lactose and that galactose alone does not release glucose.
• Tube X shows competitive inhibition: Galactose is complementary in shape to lactase's active site, competing directly with lactose.

🧠 Exam Technique: Structuring 4-Mark Comparisons

When the question divides the answer lines by tube ( Tube X , Tube Y , Tube Z ), address each tube explicitly:

• State the mechanism: name the active site and the process (hydrolysis).
• Compare yields using the figures provided: Tube Y produced 180 vs Tube X's 100 mg dm⁻³.
• Explain the zero result: never just say "it did not work"—state why (absence of substrate).

❌ Examiner Pitfalls

• Saying competitive inhibitors "stop" the reaction: Competitive inhibitors only slow down the rate of hydrolysis; they do not permanently halt it. Mark schemes explicitly state: Reject 'stops' hydrolysis .
• Failing to mention the active site: Stating galactose "blocks the enzyme" without naming the active site misses Mark 1.
• Vague comments on Tube Z: Saying "no reaction happened" is insufficient; you must state that lactose was absent or galactose is not the substrate.

✅ Correct Answer

• Change: Increase the concentration of lactose / substrate (or add more lactose). [1 mark]
• Result: More product / glucose formed OR increased rate of reaction (glucose concentration exceeds 100 mg dm⁻³ / approaches 180 mg dm⁻³). [1 mark]

💡 The Biochemical Mechanism

How to distinguish between competitive and non-competitive inhibitors experimentally:

• Competitive inhibition is reversible by substrate: As [S] increases, substrate molecules vastly outnumber inhibitor molecules. The probability of an enzyme-substrate (E-S) collision increases, overcoming the inhibition and restoring maximum rate (V_max).
• Non-competitive inhibition cannot be overcome: Non-competitive inhibitors bind to an allosteric site and alter the active site tertiary structure. Increasing [S] has no restorative effect.

❌ Common Errors

• Suggesting decreasing the inhibitor: While reducing galactose would increase glucose, it does not distinguish competitive from non-competitive inhibition. The diagnostic test for competitive inhibition is always increasing substrate concentration.
• Changing temperature or pH: Altering environmental variables affects enzyme kinetic energy or denatures proteins, invalidating the test.
• Vague result descriptions: Stating "it works better" or "reaction goes back to normal" instead of specific biological terms like "more glucose produced" or "rate of reaction increases".

🧠 Exam Golden Rule

Whenever an AQA question asks how to prove an inhibitor is competitive, your reflex answer must be: increase substrate concentration. The expected result is always that the rate increases / maximum rate is achieved because the substrate outcompetes the inhibitor for the active site.