Digestive system, enzymes and food tests (RP4, RP5) – GCSE Biology exam questions with mark scheme

1Grade 1–33 marks

Complete the table about digestive enzymes.

Enzyme	Substrate	Product(s)
amylase	starch	(i) ______
protease	(ii) ______	amino acids
lipase	lipids	(iii) ______

(i) (simple) sugars; accept maltose; allow glucose(1)
(ii) proteins(1)
(iii) fatty acids and glycerol (both needed)(1)
Answer: (i) sugars (maltose) (ii) protein (iii) fatty acids and glycerol
Examiner tip: Lipids break down into TWO products — fatty acids and glycerol — and both are needed for the mark.

2Grade 4–54 marks

A student wants to find out whether a food sample contains starch and protein.

(a) Describe the test for starch, including the positive result. [2]

(b) Describe the test for protein, including the positive result. [2]

(a) add (a few drops of) iodine solution (to the food sample / solution); allow iodine(1)
(a) (positive result) turns blue-black; accept black / dark blue; ignore starting colour (orange-brown); do not accept blue alone or purple(1)
(b) add Biuret reagent / sodium hydroxide solution then (dilute) copper sulfate solution; do not accept Benedict's; ignore any reference to heating (not needed, but does not lose the mark)(1)
(b) (positive result) turns purple / lilac / mauve / violet; ignore starting colour (blue); do not accept pink alone(1)
Answer: (a) iodine: orange → blue-black (b) Biuret: blue → purple
Examiner tip: Name the reagent and the exact final colour: starch gives blue-black (not just 'blue') and protein gives purple with Biuret; no heating is needed for either test.

3Grade 6–76 marks

Amylase is an enzyme that breaks down starch.

Plan an investigation to find out how pH affects the rate of reaction of amylase.

You should include how you would make the investigation a fair test. [6]

place one drop of iodine solution in each well of a spotting tile(1)
use buffer solutions to set pH values, e.g. pH 4, 5, 6, 7, 8(1)
mix fixed volume of amylase, buffer and starch solution (e.g. 2 cm³ each) and start a stopwatch(1)
keep temperature constant using a water bath (e.g. at 30–37 °C) and allow solutions to reach temperature before mixing(1)
every 10 / 30 seconds add a drop of the mixture to a fresh well of iodine(1)
record the time when iodine stays orange-brown (no longer turns blue-black) = starch has all been broken down(1)
repeat at each pH and calculate a mean(1)
control variables: same concentration and volume of starch and of amylase, same temperature(1)
rate = 1000 ÷ time (or 1 ÷ time) to compare pH values(1)

Level	Marks	What it looks like
Level 3	5–6	A coherent, logically ordered method that could be followed by someone else to produce valid results: pH varied with buffers, starch end-point detected with iodine and timed, temperature and other variables controlled with named methods, repeats.
Level 2	3–4	A method with most key steps in a sensible order, but some details or controls are missing so results may not be fully valid.
Level 1	1–2	Some relevant steps identified (e.g. use iodine, change pH) but the method is unclear or out of order and could not be followed.
No relevant content	0	Nothing worthy of credit.

Answer: Method using iodine on spotting tile, buffers for pH, water bath for temperature, time until iodine stays orange, repeats and controls.
Examiner tip: Saying 'keep it a fair test' earns nothing — name the control variables and say how each is controlled (e.g. water bath for temperature).

4Grade 6–74 marks

The graph shows the effect of temperature on the rate of reaction of a human enzyme.

(a) Use the graph to give the optimum temperature of the enzyme. [1]

(b) Explain why the rate of reaction decreases above the optimum temperature. [3]

(a) 37 °C (accept 36–38 °C); unit not required(1)
(b) (high temperature changes the shape of the enzyme so) the active site changes shape(1)
(b) so the substrate no longer fits (into) / is no longer complementary to the active site; accept fewer enzyme–substrate complexes form(1)
(b) the enzyme is denatured (as temperature rises, more enzyme molecules are denatured); do not accept enzyme is killed / dies(1)
Answer: (a) 37 °C (b) active site changes shape, substrate no longer fits, enzyme denatured
Examiner tip: Enzymes are denatured, never 'killed' — and say what happens to the ACTIVE SITE, not just the enzyme.

5Grade 1–3Foundation5 marks

The digestive system is an organ system.

(a) Write these four levels of organisation in order, from the smallest to the largest.

organ cell organ system tissue [1]

(b) Amylase is released into the mouth in saliva.

Name the glands that make this amylase. [1]

Name the other organ. [1]

(d) Amylase can break down starch but it cannot break down protein.

Explain why. [2]

(a) cell → tissue → organ → organ system (all four in the correct order for the mark)(1)
(b) salivary glands (allow saliva glands; ignore mouth / tongue / saliva)(1)
(c) stomach (do not accept pancreas, small intestine, salivary glands or mouth)(1)
(d) (amylase has an) active site with a specific shape (allow amylase is specific to starch)(1)
(d) protein does not fit the active site / protein is not complementary to the active site (allow lock-and-key idea: protein is the wrong ‘key’ for the amylase ‘lock’; do not accept ‘the active site is the same shape as protein’)(1)
Answer: (a) cell, tissue, organ, organ system (b) salivary glands (c) stomach (d) the active site of amylase has a specific shape; protein does not fit it (is not complementary)
Examiner tip: In (d), saying ‘amylase only works on starch’ just repeats the question — you must mention the active site and that protein does not fit it.

6Grade 4–56 marks

A student tested four foods, P, Q, R and S, for different nutrients.

(a) Describe how the student should test a food solution for sugars using Benedict’s solution.

Include the colour of a positive result. [3]

The table shows the student’s results.

Food	Benedict’s test (after heating)	Iodine test	Biuret test	Ethanol (emulsion) test
P	blue	blue-black	blue	clear
Q	brick red	orange-brown	blue	clear
R	blue	orange-brown	purple	cloudy white
S	orange	orange-brown	purple	clear

(b) Give the letter of the food that contains both protein and lipid. [1]

(d) The student used the same volume of each food solution and of Benedict’s solution.

Which food, Q or S, contained more sugar? Give a reason for your answer. [1]

(a) add (a few drops / equal volume of) Benedict’s solution to the food solution (in a test tube)(1)
(a) heat in a (hot / boiling) water bath, e.g. about 75–100 °C, for a few minutes (allow ‘heat’ / ‘warm’ only if a water bath or kettle water is used; do not accept heating directly over a Bunsen flame)(1)
(a) positive result: colour changes from blue to green / yellow / orange / brick red
allow red / orange-red for brick red
do not accept colours from other tests (e.g. purple, blue-black)(1)
(b) R(1)
(c) P(1)
(d) Q and brick red is further along the colour sequence than orange / brick red shows a higher concentration of sugar than orange (no mark for Q without a reason)(1)
Answer: (a) add Benedict’s; heat in a water bath (~75 °C) for a few minutes; blue → green/yellow/orange/brick red if sugar present (b) R (c) P (d) Q – brick red indicates more sugar than orange
Examiner tip: Benedict’s is the only food test that needs heating — forgetting the water bath is the most common lost mark.

7Grade 4–56 marks

A student investigated the effect of pH on the breakdown of starch by amylase.

The student mixed amylase, starch solution and a buffer solution in a water bath at 35 °C.
Every 30 seconds the student put a drop of the mixture into iodine solution on a spotting tile.
The student recorded the time when the iodine first stayed orange-brown.

The table shows the results.

pH	Time for starch to be broken down in seconds			Mean time in seconds
pH	Trial 1	Trial 2	Trial 3	Mean time in seconds
4	300	330	300	310
5	150	180	150	160
6	90	90	120	100
7	120	150	120	130
8	240	270	450	X

(a) Identify the anomalous result in the table. [1]

(b) Calculate the mean time X at pH 8. Do not include the anomalous result. [1]

$$\text{rate} = \frac{1000}{\text{mean time}}$$

Calculate the rate of reaction at pH 8. Give your answer to 2 significant figures. [2]

(d) The student concluded that the optimum pH of amylase is exactly pH 6.

Suggest how the student could find the optimum pH more precisely. [1]

(e) Give one reason why testing a drop only every 30 seconds makes the times less accurate. [1]

(a) 450 (s) / pH 8 trial 3 (allow the result circled or described)(1)
(b) 255 (s) — $(240+270)\div 2$ (allow 255.0; 320 scores 0 as it includes the anomaly)(1)
(c) $1000 \div 255$ (allow ecf from their answer to (b))(1)
(c) 3.9 (allow ecf, e.g. 1000 ÷ 320 = 3.1; 3.92 / 3.921… not to 2 s.f. scores 1 mark only; ignore units)(1)
(d) repeat using smaller pH intervals between pH 5 and pH 7 (allow e.g. pH 5.5, 6.0, 6.5 / every 0.2 pH units around pH 6; ‘use more pH values’ alone = 0 unless they are between 5 and 7 or at smaller intervals)(1)
(e) the starch could have been fully broken down at any time during the 30 seconds before the test / the actual end time could be up to 30 s earlier than recorded (allow ‘sample more often, e.g. every 10 s’ only if linked to the end point being missed; ignore ‘human error’)(1)
Answer: (a) 450 s (pH 8, trial 3) (b) 255 s (c) 3.9 (d) test more pH values at smaller intervals between pH 5 and 7 (e) the end point could be anywhere in the 30 s interval before it was detected
Examiner tip: Leave the anomaly out before working out the mean — including 450 s gives 320 s and loses the mark.

8Grade 6–76 marks

A scientist added lipase to two flasks of milk at 37 °C. Bile was added to one flask. The pH in both flasks was kept the same using a buffer.

The graph shows the percentage of the fat in the milk that was digested over 20 minutes.

(a) Determine how many more percentage points of the fat had been digested after 10 minutes in the flask with bile than in the flask without bile. [1]

(b) Explain how bile caused the difference shown in the graph. [3]

Bile is alkaline. Explain why this is important in the small intestine. [2]

(a) 45 (75 − 30) (allow 44–46)(1)
(b) bile emulsifies fat / breaks (large) fat drops into (many) small droplets (do not accept ‘bile breaks down / digests fat’ or ‘bile dissolves fat’)(1)
(b) this increases the surface area of the fat(1)
(b) so more fat molecules can reach / come into contact with lipase active sites at once / more enzyme–substrate complexes formed (per second), so fat is digested faster (allow lipase breaks fat into fatty acids and glycerol faster)(1)
(c) neutralises (hydrochloric) acid from the stomach(1)
(c) (so the pH is suitable for) enzymes in the small intestine, which work best at alkaline / neutral pH / optimum pH (allow so enzymes in the small intestine are not denatured by acid)(1)
Answer: (a) 45 (b) bile emulsifies fat into small droplets → larger surface area → lipase digests fat faster (c) it neutralises stomach acid so small-intestine enzymes are at their optimum (alkaline) pH
Examiner tip: Bile is not an enzyme and does not digest fat — it emulsifies it, and you must say this increases surface area for lipase.

9Grade 6–76 marks

A person eats a slice of bread with butter.

Bread contains mainly starch. Butter contains mainly lipid (fat).

Explain how the starch and the lipid in this food are digested, and how the body uses the products of digestion. [6]

starch is digested by amylase / a carbohydrase(1)
amylase is produced by the salivary glands, pancreas and small intestine; starch digestion begins in the mouth and is completed in the small intestine(1)
starch is broken down into (simple) sugars, e.g. maltose / glucose(1)
bile (made in the liver, stored in the gall bladder) emulsifies the lipid into small droplets, increasing the surface area for lipase (do not credit ‘bile digests / breaks down fat’ as digestion)(1)
lipase, produced by the pancreas (and small intestine), breaks lipids into glycerol and fatty acids in the small intestine(1)
bile neutralises stomach acid so enzymes in the small intestine work at their optimum pH(1)
enzymes break large insoluble molecules into small soluble molecules that can be absorbed into the blood (from the small intestine)(1)
glucose is used in respiration to release energy(1)
products are used to build new carbohydrates (e.g. glycogen / starch), new lipids (from glycerol and fatty acids) and new proteins(1)

Level	Marks	What it looks like
Level 3	5–6	Digestion of both starch and lipid is explained accurately with the correct enzymes, where they are made or act, and the correct products; the role of bile is included; and at least one way the products are used is linked clearly to the body (respiration or building new molecules). Logically sequenced.
Level 2	3–4	Digestion of both starch and lipid is described with mostly correct enzymes and products, but with gaps (e.g. no sites of production, no role for bile, or use of products only vaguely stated).
Level 1	1–2	Simple relevant statements, e.g. names an enzyme or a product, or states that food is broken down so it can be absorbed. Links are weak or there are errors.
No relevant content	0	Nothing worthy of credit.

Answer: Amylase (salivary glands, pancreas, small intestine) breaks starch into sugars; bile emulsifies lipid, then lipase (pancreas, small intestine) breaks it into glycerol + fatty acids; products absorbed into blood; glucose used in respiration; products used to build new carbohydrates, lipids and proteins.
Examiner tip: Top-level answers cover all three parts — starch, lipid and the use of the products; many students stop after digestion and cap themselves at Level 2.

10Grade 8–9Higher6 marks

A student mixed amylase with starch solution at 37 °C and measured the mass of sugar produced over 10 minutes.

The graph shows the results. A tangent has been drawn to the curve at 2 minutes.

(a) Calculate the mean rate of sugar production over the first 10 minutes.

Give the unit. [2]

(b) Use the tangent to calculate the rate of sugar production at 2 minutes. [2]

Explain why the rate of reaction decreased between 2 minutes and 10 minutes. [2]

(a) $35.4 \div 10$ / reading of 35–36 mg divided by 10 = 3.5 (allow 3.5–3.6)(1)
(a) unit: mg/min (allow mg per minute, mg min$^{-1}$)(1)
(b) correct readings from the tangent, e.g. $\dfrac{40-4}{6-0}$ (allow any two points on the tangent, e.g. $\dfrac{16-4}{2-0}$)(1)
(b) 6 (mg/min) (allow 5.5–6.5)(1)
(c) the starch / substrate is being used up / starch concentration decreases (allow less starch left to break down)(1)
(c) so fewer starch molecules fit into / collide with the active sites (per minute) / fewer enzyme–substrate complexes formed (per minute) (do not accept ‘enzyme is denatured’ or ‘enzyme is used up’)(1)
Answer: (a) 3.5 mg/min (b) 6 mg/min (c) starch is used up, so fewer enzyme–substrate complexes form per minute
Examiner tip: The question tells you the temperature was constant — answers about the enzyme denaturing score zero; the rate falls because the substrate is running out.

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