Many proteins are enzymes - AQA A-Level Biology (Paper 1) Exam Questions

Practise 20 exam-style questions on Many proteins are enzymes for AQA A-Level Biology (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Pectinase is an enzyme that hydrolyses pectin in plant cell walls. A student added pectinase solution to apple pulp and measured the volume of juice that dripped through a filter funnel in 15 minutes.

Describe a suitable control for this investigation and explain why it is needed.

Model answer

Add the same volume of boiled (denatured) pectinase solution to the same mass of apple pulp, at the same temperature, and measure the volume of juice collected in 15 minutes. This shows how much juice drips through without active pectinase, so any extra juice in the test can be shown to be due to the action of the enzyme.

Explanation

A control experiment is identical except for the factor being tested - here, active enzyme. Boiled enzyme is often best because it keeps everything else in the enzyme solution the same while destroying the active site. Students often just write 'no enzyme' without saying what replaces it or why the control is needed.

Mark scheme

1 mark
The same volume of boiled/denatured pectinase added to the same mass of apple pulp (with all other conditions the same)
Accept 'the same volume of water/buffer instead of pectinase' or 'apple pulp with no pectinase, replaced by the same volume of water'. Insufficient: 'no enzyme' with nothing to replace its volume and no reference to the same conditions.
1 mark
To show that the extra juice is due to (the action of) pectinase / to show how much juice is released without (active) pectinase
Accept 'to show that any difference is due to the enzyme'. Insufficient: 'to compare' or 'to make it a fair test' alone.

Question 2

3 marks

Enzymes catalyse a wide range of intracellular and extracellular reactions.

Explain the difference between an intracellular enzyme and an extracellular enzyme.

Give one named example of each type of enzyme and the reaction it catalyses.

Model answer

An intracellular enzyme catalyses reactions inside the cell that made it, whereas an extracellular enzyme is secreted from the cell and catalyses reactions outside it.
Intracellular: DNA polymerase, which joins DNA nucleotides by condensation reactions during DNA replication.
Extracellular: amylase, which is secreted by the salivary glands and the pancreas and hydrolyses starch to maltose in the gut.

Explanation

Most enzymes work inside cells, catalysing the reactions of metabolism such as DNA replication and the hydrolysis of ATP. Extracellular enzymes are secreted and act outside the cells that made them - in the gut, in tears, or, for fungi and bacteria, on the material around them. Each example needs both the enzyme's name and what it does; 'amylase - digestion' is too vague.

Mark scheme

1 mark
Intracellular enzymes catalyse reactions inside the cell that made them AND extracellular enzymes are secreted/released from cells and catalyse reactions outside cells
Both parts are needed. Accept 'intracellular enzymes work inside cells but extracellular enzymes work outside cells'. Insufficient: 'extracellular enzymes are found outside cells' with no idea of secretion or of where they act.
1 mark
Intracellular example with its reaction, e.g. DNA polymerase joins DNA nucleotides / DNA helicase breaks hydrogen bonds between bases / ATP hydrolase hydrolyses ATP / ATP synthase makes ATP from ADP and phosphate / RNA polymerase joins RNA nucleotides / catalase breaks down hydrogen peroxide
The enzyme and its substrate or reaction are both needed. Accept another correctly named intracellular enzyme with its reaction, e.g. lysosomal enzymes hydrolysing engulfed bacteria. List rule: if more than one example of a type is given, a wrong example cancels a correct one. Ignore membrane-bound disaccharidases and dipeptidases (maltase, sucrase, lactase) for either type.
1 mark
Extracellular example with its reaction, e.g. amylase hydrolyses starch (to maltose) / lipase hydrolyses lipids or triglycerides / an endopeptidase, exopeptidase or other protease hydrolyses proteins or peptide bonds / lysozyme hydrolyses bacterial cell walls
The enzyme and its substrate or reaction are both needed. Insufficient: 'amylase - digestion' or 'protease - breaks down food'.

Question 3

4 marks

Pitcher plants trap insects in a pitcher-shaped leaf containing fluid. Cells in the wall of the pitcher secrete a protease into the fluid, which hydrolyses the proteins of trapped insects.

After an insect has been trapped, the hydrogen ion concentration of the fluid is 1.6 × 10⁻³ mol dm⁻³.

pH = -log₁₀[H⁺]

Calculate the pH of the fluid. Give your answer to 1 decimal place.

The protease has an optimum pH of about 2.5. After heavy rain, the pH of the fluid rises to about 5 and the activity of the protease is very low. Suggest why.

Model answer

pH = -log₁₀(1.6 × 10⁻³) = -(-2.796) = 2.8

After heavy rain the pH rises to about 5, so the hydrogen ion concentration is about 160 times lower than at pH 2.8 and far from the protease's optimum of 2.5. The change in H⁺ concentration alters the ionic and hydrogen bonds that hold the tertiary structure of the protease, so the active site changes shape, the protein substrate is no longer complementary and fewer enzyme-substrate complexes form. The rain also dilutes the enzyme.

Explanation

pH = -log₁₀[H⁺], so a low pH means a high hydrogen ion concentration; each pH unit is a ten-fold change in [H⁺]. Enzymes work fastest at their optimum pH because H⁺ ions affect the charges on R groups and so the ionic and hydrogen bonds in the tertiary structure. A common error is to forget the minus sign and give -2.8. This protease is an extracellular enzyme: it is secreted and acts outside the cells that made it.

Mark scheme

1 mark
Correct method: -log₁₀(1.6 × 10⁻³); or a correct final answer of 2.8 with no working
Max 3 if the answer does not refer to pH or hydrogen ions when explaining the low activity. A correct answer of 2.8 with no working scores 2 marks (points 1 and 2). Accept for 1 mark: 2.80 or 2.796 (not given to 1 decimal place) OR -2.8 (minus sign of the formula not applied).
1 mark
2.8
Do not accept 2.80 or -2.8 for this mark.
1 mark
At pH 5 the H⁺ concentration is much lower (more than 100 times lower than at pH 2.8)
Points 3 to 6: max 2. Accept 'the fluid is much less acidic'. Insufficient: 'the pH is not the optimum' alone (this repeats the question).
1 mark
The change in H⁺ concentration changes/breaks ionic and/or hydrogen bonds in the tertiary structure (of the protease)
Accept 'R groups gain or lose charge, so the bonds between R groups change'. Reject 'peptide bonds break'.
1 mark
The active site changes shape, so the substrate is no longer complementary / fewer enzyme-substrate complexes form
Accept 'denatured' for a change of shape of the active site. Reject 'the enzyme is killed'. Accept 'E-S complex' or 'ES complex'. Ignore 'ESC'.
1 mark
Rain also dilutes the protease (and the substrate), so there are fewer collisions / fewer enzyme-substrate complexes form

Question 4

6 marks

Explain how the tertiary structure of an enzyme allows it to catalyse a specific reaction.

Explain how competitive and non-competitive inhibitors reduce the rate of an enzyme-controlled reaction.

Model answer

The sequence of amino acids determines how the polypeptide folds into its tertiary structure, which gives the active site a specific shape. Only a substrate with a complementary shape can bind, forming an enzyme-substrate complex; as it binds, the active site changes shape slightly to mould around the substrate (induced fit), straining bonds in the substrate and lowering the activation energy.

A competitive inhibitor has a shape similar to the substrate and binds to the active site, so fewer substrate molecules can bind and fewer enzyme-substrate complexes form; increasing the substrate concentration reduces its effect. A non-competitive inhibitor binds at a site other than the active site and changes the tertiary structure, so the active site changes shape and the substrate is no longer complementary; increasing the substrate concentration does not reduce its effect.

Explanation

Specificity comes from the tertiary structure: the folding of the polypeptide gives the active site a shape that only a complementary substrate can bind to. Inhibitors reduce the number of enzyme-substrate complexes in different ways - competitive ones occupy the active site, non-competitive ones change its shape. A strong answer covers both parts of the question; using 'complementary' and 'enzyme-substrate complex' correctly earns marks that vague words like 'fits the enzyme' do not.

Mark scheme

1 mark
The sequence of amino acids/primary structure determines the folding/tertiary structure, which gives the active site a specific shape
Max 6. Max 3 from points 1 to 5 and max 4 from points 6 to 11.
1 mark
Only a substrate with a complementary shape can bind to the active site
Accept 'fit' for bind.
1 mark
An enzyme-substrate complex forms
Accept 'E-S complex' or 'ES complex'. Ignore 'ESC'.
1 mark
The active site changes shape slightly to mould around the substrate (induced fit)
1 mark
Puts strain on/distorts bonds in the substrate, lowering the activation energy
1 mark
A competitive inhibitor has a similar shape to the substrate / is complementary to the active site
1 mark
A competitive inhibitor binds to the active site, so (fewer) substrate molecules can bind / fewer enzyme-substrate complexes form
1 mark
The effect of a competitive inhibitor is reduced by increasing substrate concentration
Accept 'substrate out-competes the inhibitor at high concentration'.
1 mark
A non-competitive inhibitor binds at a site other than the active site
Accept 'allosteric site'.
1 mark
A non-competitive inhibitor changes the tertiary structure / shape of the active site, so the substrate is no longer complementary / cannot bind
Reject 'the active site is on the substrate'.
1 mark
The effect of a non-competitive inhibitor is not reduced by increasing substrate concentration

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