AQA AS Level Biology Paper 1, June 2024: Question 2

10 marks · Medium difficulty · Short Answer

Complete a table on starch and dipeptide digestion, describe and explain a protein test, and calculate the maximum rate of an enzyme-controlled reaction using turnover number.

Practise this question

Question

A multipart biology exam question consisting of Table 2 to complete with enzymes, bonds, and digestion products for starch and dipeptide; a description of a student investigating starch digestion and testing for protein; questions asking to describe the biuret test and explain why it is positive; and a calculation of the maximum rate of a protease-controlled reaction using the kcat formula.
Question text

02.1 Table 2 contains information on the digestion of two biological molecules.

Complete Table 2.

[4 marks]

Table 2

Biological Name of bond

Enzyme Product of digestion

molecule hydrolysed

Starch Maltose

Dipeptide Dipeptidase

A student investigated starch digestion by mixing starch with a solution of the enzyme

used to digest starch.

The student did a biochemical test for protein when starch digestion was completed.

02.2 Describe a biochemical test to show the presence of protein.

[2 marks]

02.3 The student’s test for protein was positive.

Explain why.

[2 marks]

02.4 An enzyme’s turnover number (kcat) is the number of substrate molecules converted

*05* into product molecules by one enzyme molecule in 1 second. It is determined using

this equation.

kcat = Maximum rate of enzyme-controlled reaction / μmol dm–3 s–1

Enzyme concentration / μmol dm–3

A scientist investigated the action of a protease enzyme. The scientist prepared a

reaction mixture with a protease concentration of 0.0118 μmol dm–3. The k for the

cat

protease is 110 substrate molecules per second.

Use this information and the formula to calculate the maximum rate of the

protease-controlled reaction.

Give your answer to 3 significant figures.

Show your working.

[2 marks]

Answer μmol dm–3 s–1

Mark scheme

Show the mark scheme Mark scheme giving the expected answers for each part of question 02: Table answers (amylase, glycosidic, peptide, amino acid), biuret test steps (add biuret, purple color), explanation for positive protein test (enzymes are proteins and not used up), and calculation working giving an answer of 1.30.

Question Marking Guidance Mark Comments

1. Amylase;

2. Glycosidic;

3. Peptide; 3. Reject dipeptide/

4. Amino acid(s); polypeptide bond

02.1 Biological Name of bond Product of (4 x

Enzyme AO1)

molecule hydrolysed digestion

Amylase Glycosidic

Amino

Peptide

acid(s)

1. Add biuret (solution); 1. Reject burette or

Beirut

2. Purple (colour produced);

1. Reject heat

1. Accept a

description of the

2 biuret test: eg copper

02.2 (2 x sulfate and sodium

AO1) hydroxide or CuSO4 +

NaOH or alkaline

copper sulfate

2. Accept

lilac/violet/mauve for

purple

1. (Positive because) enzymes are protein; Accept correct answer

in any order

2. (Because) enzymes not used up (in reactions) 2 2. Accept enzymes

02.3 (2 x reused for enzymes

OR AO2) not used up

2. Presence of

(Because) enzymes still present; enzyme after reaction

must be implied

Correct answer of 1.30 = 2 marks;;

Evidence of correct rearranged equation, for

example 2

02.4 Maximum rate of reaction = kcat x Enzyme (2 x

concentration= 1 mark AO2)

OR

Evidence of 1 or 1.298 or 1.29 or 1.3 (correct

answer but incorrect significant figures) = 1 mark;

How to answer it

Carbohydrate & Protein Digestion & Enzyme Kinetics

What this question tests

This question assesses your knowledge of digestive enzymes (amylase and dipeptidases), specific types of biological bonds broken during digestion, biochemical testing procedures for proteins (the Biuret test), and the application of enzyme kinetics formulas using standard scientific units and significant figures.

Question 02.1 (4 marks)

Completing the Digestion Table

✅ Correct Answers

  • Starch Enzyme: Amylase
  • Starch Bond: Glycosidic
  • Dipeptide Bond: Peptide
  • Dipeptide Product: Amino acid(s)

💡 Key Knowledge

  • Amylase hydrolyses starch into maltose by breaking internal glycosidic bonds.
  • Dipeptidases break the peptide bond located between two amino acids in a dipeptide, releasing individual amino acids.

❌ Common Errors

  • Writing "dipeptide bond" or "polypeptide bond" instead of the correct biological term peptide bond.
  • Confusing enzyme names with substrate names.
Mark allocation: 1 mark per correct table entry (Total: 4 marks).
Question 02.2 (2 marks)

Biochemical Test for Protein

✅ Correct Answers

  1. Add Biuret solution (or sodium hydroxide + copper sulfate).
  2. Observation: Colour changes from blue to purple (or lilac / violet / mauve).

🧠 Exam Technique

When describing tests, be precise with reagents and final observations. Never include "heat" in a Biuret test (heating is only for Benedict's test for reducing sugars).

❌ Common Errors

  • Misspelling Biuret as "burette" or "beirut".
  • Stating that heating is required for the test to work.
Mark allocation: 1 mark for naming the Biuret reagent; 1 mark for the purple/lilac colour observation.
Question 02.3 (2 marks)

Explaining a Positive Protein Test After Starch Digestion

✅ Correct Answers

Any two of the following points:

  • Enzymes are proteins.
  • Enzymes are not used up / are unchanged during the reaction.
  • Enzyme molecules are still present in the mixture at the end.

💡 Key Knowledge

Because the amylase used to digest the starch is itself a functional protein catalyst, and catalysts are not consumed or permanently altered during a chemical reaction, it remains dissolved in the solution at the end, triggering a positive Biuret test.

Mark allocation: 2 marks total (1 mark per valid explanatory point).
Question 02.4 (2 marks)

Enzyme Turnover Calculation

📐 Step-by-Step Calculation

Given equation:

kcat = (Maximum rate of enzyme-controlled reaction) / (Enzyme concentration)

  1. Rearrange the formula:
    Maximum rate = kcat × Enzyme concentration
  2. Substitute the values:
    Maximum rate = 110 × 0.0118
  3. Calculate raw value:
    1.298 µmol dm⁻³ s⁻¹
  4. Apply significant figures (3 s.f.):
    1.30

❌ Common Calculation Traps

  • Failing to round to the requested 3 significant figures (writing 1.298 loses 1 mark).
  • Incorrectly transposing variables when rearranging the division formula.
Mark allocation: 1 mark for correct rearrangement/working out (or unrounded answer 1.298), 1 mark for the final correctly rounded answer 1.30 .

Topics

Biology · Practical skills · 3.1 Biological molecules · 3.3 Organisms exchange substances with their environment · Data analysis

Question and mark scheme from the AQA AS Level Biology examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.