IGCSE Biology Paper 4 (Theory / Extended)

Topic 7: Human Nutrition -- Challenge Exam 3

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0610

Instructions

Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.

Question Navigation

This paper covers the whole of Topic 7. Like a real Cambridge paper, the seven questions range across every sub-topic — diet, the digestive system, physical and chemical digestion, and absorption — and they are deliberately mixed rather than grouped. All three Topic 7 papers do; they differ in the angle they come at it from, not in what they cover.

Question 1 — Designing a Fair Test on Pancreatic Lipase

Total: 12 marks

A student wants to find the effect of temperature on the digestion of milk fat by lipase. She plans to add lipase to milk containing an alkaline indicator, and to time how long the mixture takes to change colour.

(a) [3]

Explain why the indicator changes colour, and state which product of the reaction is responsible.

Model Answer — 1(a)

lipase breaks fats and oils down into fatty acids and glycerol [1]

the fatty acids are acidic, so the pH of the mixture falls [1]

the alkaline indicator changes colour as the mixture becomes acidic, so the colour change marks the point at which enough fatty acid has been produced [1]

⚠ If you missed marks here: Only one of the two products can move a pH, and the name gives it away. Answers that credit glycerol have listed both products without asking which one the method actually detects — a good habit is to check every measurement against the molecule it depends on.

1 mark — lipase produces fatty acids and glycerol 1 mark — fatty acids are acidic and lower the pH 1 mark — the indicator changes colour as the mixture becomes acidic

(b) [4]

State the independent variable, the dependent variable, and two variables that must be controlled, explaining why one of them matters.

Model Answer — 1(b)

independent variable: temperature [1]

dependent variable: the time taken for the indicator to change colour [1]

two controlled variables from: volume and concentration of lipase, volume of milk, starting pH, volume of indicator, fat content of the milk [1]

explanation, e.g. a higher lipase concentration would produce fatty acids faster whatever the temperature, so the effect of temperature could not be separated from it [1]

⚠ If you missed marks here: The dependent variable is what you measure, not what you conclude — “the rate of digestion” is a calculation made afterwards, and the raw measurement is a time. Losing this mark suggests the method has not been read as a sequence of actions.

1 mark — independent variable: temperature 1 mark — dependent variable: time for the indicator to change colour 1 mark — two sensible controlled variables named 1 mark — explanation of why one of them would confound the result

Describe the control experiment she should set up, and state the result she should expect.

Model Answer — 1(c)

an identical tube in which the lipase has been boiled and cooled before use, everything else being the same [1]

no colour change, showing that the change in the other tubes was caused by active lipase and not by the milk, the indicator or the temperature alone [1]

⚠ If you missed marks here: A control changes exactly one thing. Replacing the lipase with water changes two — the enzyme and the total volume — whereas boiled enzyme keeps every other component of the mixture identical. That is why boiled enzyme is the standard control in every digestion experiment.

1 mark — boiled and cooled lipase used, everything else identical 1 mark — no colour change expected, showing active enzyme was responsible

(d) [3]

Her results are shown below. Explain the result at 70 °C, and explain why the value at 10 °C does not have the same cause.

temperature / °C	time for colour change / s
10	640
20	300
37	110
50	260
70	no change in 900 s

Model Answer — 1(d)

at 70 °C the lipase has been denatured: the shape of the active site has changed and is no longer complementary to the fat, so no complexes form [1]

at 10 °C the enzyme is not damaged, only slow: the molecules have little kinetic energy [1]

so there are few effective collisions between substrate and active site; warming the tube would restore the rate, whereas cooling the 70 °C tube would not [1]

⚠ If you missed marks here: Zero activity at a low temperature and zero activity at a high temperature look identical in a results table and have completely different causes. The word denatured belongs only to the hot tube, and using it for the cold one is the most heavily penalised error in enzyme questions.

1 mark — 70 degrees: enzyme denatured, active site changed, no complexes form 1 mark — 10 degrees: enzyme inactive but undamaged, low kinetic energy 1 mark — few effective collisions; warming restores activity but cooling does not reverse denaturation

Question 2 — A Model Gut Made of Visking Tubing

Total: 12 marks

A length of visking tubing was filled with a mixture of starch solution and amylase, tied at both ends, rinsed on the outside and suspended in a beaker of water at 37 °C. Samples of the water in the beaker were tested at intervals.

(a) [3]

State what the visking tubing, the mixture inside it and the water in the beaker each represent in the human body.

Model Answer — 2(a)

the tubing represents the wall of the small intestine, which is partially permeable [1]

the mixture inside represents the contents of the lumen of the small intestine [1]

the water in the beaker represents the blood in the capillaries of the villi [1]

⚠ If you missed marks here: A model question is asking what each part stands for, not what it is made of. Note also why the tubing must be rinsed on the outside first: any starch or sugar left on the surface would appear in the beaker and imitate the very result being looked for.

1 mark — tubing represents the partially permeable wall of the small intestine 1 mark — contents represent the lumen of the small intestine 1 mark — beaker water represents the blood in the capillaries

(b) [4]

Describe the results you would expect from testing the beaker water with iodine solution and with Benedict’s solution at the start and after one hour, and explain what these results show.

Model Answer — 2(b)

at the start: iodine stays orange-brown and Benedict’s stays blue — neither starch nor sugar is present [1]

after an hour: iodine still orange-brown, showing starch has not passed through the tubing [1]

after an hour: Benedict’s gives an orange or brick-red precipitate, showing reducing sugar has passed through [1]

so the starch has been digested to a small soluble molecule which can then be absorbed through a partially permeable membrane, while the large insoluble molecule cannot [1]

⚠ If you missed marks here: Both tests are needed, and the negative one carries as much information as the positive one. If starch did appear in the beaker the tubing would be leaking, and the experiment would show nothing about digestion at all.

1 mark — both tests negative at the start 1 mark — iodine still negative after an hour: starch has not passed through 1 mark — Benedict's positive after an hour: reducing sugar has passed through 1 mark — conclusion: only small soluble molecules can cross, so digestion must precede absorption

Describe two ways in which this model is a poor representation of the real small intestine, and explain the consequence of each.

Model Answer — 2(c)

the tubing is smooth, with no folds, villi or microvilli, so its surface area is minute compared with the real intestine and absorption is far slower [1]

the water in the beaker is not flowing, unlike blood, so the concentration gradient falls as sugar accumulates instead of being maintained [1]

any further valid point with its consequence: no active transport in tubing so only diffusion occurs; no muscular movement to mix the contents; only one enzyme present [1]

⚠ If you missed marks here: Evaluating a model means naming the difference and saying what it costs. “It is not alive” is true of every model and scores nothing on its own, because it does not identify a specific process the model fails to reproduce.

1 mark — no folds, villi or microvilli, so far less surface area 1 mark — the beaker water does not flow, so the concentration gradient is not maintained 1 mark — a third valid limitation with its consequence

(d) [2]

Name the process by which glucose leaves the tubing in this model, and name a process used in the real ileum that the model cannot show.

Model Answer — 2(d)

in the model glucose leaves by diffusion, down a concentration gradient through the partially permeable membrane [1]

the ileum also uses active transport, which moves glucose against a concentration gradient using energy from respiration and requires living cells [1]

⚠ If you missed marks here: Visking tubing is not alive, so it cannot carry out any process that needs energy — and that single fact explains the limitation. Answers naming osmosis for glucose confuse the movement of water with the movement of a solute.

1 mark — diffusion named for the model 1 mark — active transport named as the process the model cannot show

Question 3 — Testing Foods and Testing Claims

Total: 12 marks

A food manufacturer claims that its new fruit drink “contains more vitamin C than fresh orange juice and provides protein for growth”. A student sets out to test both claims in the laboratory.

(a) [3]

Describe how the student could compare the vitamin C content of the drink and of fresh orange juice, and state how the results would be judged.

Model Answer — 3(a)

add each liquid drop by drop from a syringe or burette to a fixed volume of DCPIP solution of fixed concentration, shaking after each drop [1]

record the volume needed to decolourise the DCPIP for each liquid [1]

the liquid needing the smaller volume contains more vitamin C, since less of it is required to reduce the same amount of DCPIP [1]

⚠ If you missed marks here: The judgement is the reverse of what most people expect: a smaller volume means a higher concentration. Fixing the volume and concentration of DCPIP is what makes the two results comparable at all, so it is a mark in its own right.

1 mark — DCPIP method described with each liquid added drop by drop 1 mark — volume needed to decolourise the DCPIP recorded 1 mark — smaller volume indicates a higher vitamin C concentration

(b) [3]

Describe the test the student should use for protein, including the expected positive and negative results.

Model Answer — 3(b)

the biuret test: add sodium hydroxide solution then a few drops of copper(II) sulfate solution, and shake [1]

positive result: the solution turns purple or violet [1]

negative result: the solution stays blue [1]

⚠ If you missed marks here: Naming a test is only part of the answer; the mark scheme wants the colour change in both directions, because a colour that is described only as “it changes” could be any test in the syllabus. Note that biuret needs no heating, unlike Benedict’s.

1 mark — biuret test named and reagents described 1 mark — positive result: purple or violet 1 mark — negative result: stays blue

The biuret test on the drink stays blue. Explain what this shows about the manufacturer’s claim, and explain why protein is important in the diet.

Model Answer — 3(c)

there is no detectable protein in the drink, so that part of the claim is not supported [1]

protein supplies the amino acids used for the growth and repair of tissues [1]

and for making enzymes and antibodies; a shortage cannot be made up from carbohydrate or fat, which contain no nitrogen [1]

⚠ If you missed marks here: Be careful with the wording of a negative result: it shows no detectable protein, not that protein is impossible in any quantity. Precision about what a test can and cannot prove is exactly what a challenge paper is looking for.

1 mark — no detectable protein, so the claim is not supported 1 mark — protein supplies amino acids for growth and repair 1 mark — protein needed for enzymes and antibodies; carbohydrate and fat cannot substitute

(d) [3]

The manufacturer replies that the drink is heat-treated for safety. Suggest how this could affect the vitamin C result, and describe how the student could investigate it.

Model Answer — 3(d)

heating destroys vitamin C, so the treated drink may contain less than the fresh juice it was made from [1]

investigation: heat samples of fresh juice to different temperatures, or for different times, keeping the volume and everything else the same [1]

then measure the volume needed to decolourise DCPIP for each sample and compare with an unheated control [1]

⚠ If you missed marks here: The investigation must have one independent variable and a control. A common answer describes heating the juice but forgets the unheated sample, and without it there is nothing to compare the results against.

1 mark — heating destroys vitamin C, lowering the content 1 mark — method: heat samples to different temperatures or for different times, other variables fixed 1 mark — DCPIP volumes compared with an unheated control

Question 4 — Chewing, Surface Area and an Experiment With Four Faults

Total: 12 marks

A student wrote this method. “To show that chewing helps digestion, put a whole piece of bread into one tube of protease solution and a chewed piece of bread into another tube. Leave them on the bench overnight. Whichever tube looks emptier in the morning shows that chewing helps.”

(a) [4]

Identify four faults in this method and explain the effect of each.

Model Answer — 4(a)

the enzyme is wrong: bread is mainly starch, so amylase is needed and protease will produce almost no change in either tube [1]

the masses of bread are not stated or matched, so any difference could be caused by the quantity rather than by the surface area [1]

the temperature is not controlled and no time is measured; “overnight” on a bench varies from day to day [1]

the measurement is subjective: “looks emptier” produces no numbers and cannot be repeated or compared [1]

⚠ If you missed marks here: Work through the method in order — substrate, enzyme, quantities, conditions, measurement — rather than writing down whatever you notice first. The wrong enzyme is the fault most often missed, because candidates read “digestion” and stop checking.

1 mark — wrong enzyme: bread is starch, so amylase is needed 1 mark — masses of bread not matched, so quantity is confounded with surface area 1 mark — temperature and time not controlled 1 mark — measurement is subjective and not quantitative

(b) [3]

Rewrite the method as a valid investigation, giving three specific improvements.

Model Answer — 4(b)

use equal measured masses of bread, one left whole and one broken up, with the same volume and concentration of amylase [1]

keep both tubes in a water bath at 37 °C and start timing at the same moment [1]

measure the time taken for iodine solution to stop giving a blue-black colour, and repeat each tube three times to calculate a mean [1]

⚠ If you missed marks here: An improvement must be specific enough to follow. “Control the variables” and “be more accurate” are instructions to somebody else, not a method — name the variable, the value and the instrument.

1 mark — equal measured masses, same volume and concentration of amylase 1 mark — water bath at a fixed temperature, timing started together 1 mark — quantitative measurement using iodine, repeated to give a mean

State the prediction for this improved investigation and explain it in terms of physical digestion.

Model Answer — 4(c)

the broken-up bread loses its blue-black iodine colour in a shorter time [1]

breaking it up is physical digestion — smaller pieces with no chemical change to the starch molecules [1]

the greater surface area exposes more starch to the amylase, so more enzyme–substrate complexes form per second and the reaction is faster [1]

⚠ If you missed marks here: The middle mark is the one that separates a full answer from a partial one: the pieces get smaller, the molecules do not. An answer that says chewing “breaks the starch down” has described chemical digestion and contradicts the definition it is being tested on.

1 mark — prediction: broken-up bread digested in a shorter time 1 mark — identifies physical digestion: smaller pieces, molecules unchanged 1 mark — greater surface area gives more enzyme-substrate complexes per second

(d) [2]

Explain how the structure of the human jaw and teeth together achieve the same effect as breaking up the bread by hand.

Model Answer — 4(d)

incisors and canines bite and tear pieces off the food while the premolars and molars grind them between broad ridged surfaces [1]

the teeth are embedded in the bone of the jaw and the gums, and cement anchors each root, so large forces can be applied without the tooth moving [1]

⚠ If you missed marks here: The second mark rewards the detail that makes chewing possible at all. A tooth sitting loosely in soft tissue could not grind anything, which is why cement, bone and gum appear in the syllabus statement alongside enamel and dentine.

1 mark — different tooth types described biting, tearing and grinding 1 mark — teeth anchored by cement in the jaw bone and gums so force can be applied

Question 5 — Measuring Absorption Directly

Total: 10 marks

A researcher tied off a short length of living small intestine from a mammal, filled it with a glucose solution of known concentration and bathed it in an oxygenated salt solution. The glucose concentration in the bathing solution was measured every 10 minutes.

(a) [3]

Explain why the bathing solution was oxygenated, and predict what would happen to the results if oxygen were removed.

Model Answer — 5(a)

the epithelial cells need oxygen for aerobic respiration, which releases the energy used in active transport [1]

without oxygen much less energy is available, so active transport slows or stops [1]

glucose would still appear in the bathing solution while a concentration gradient favoured it, but far more slowly, and absorption would stop once the gradient was gone [1]

⚠ If you missed marks here: The prediction has to distinguish the two mechanisms working side by side. Saying “absorption would stop completely” ignores diffusion, which needs no energy at all and continues for as long as the gradient lasts.

1 mark — oxygen needed for aerobic respiration to supply energy 1 mark — active transport slows or stops without it 1 mark — diffusion continues while a gradient exists, so absorption is slower rather than absent

(b) [3]

Explain how the structure of the intestinal lining makes this piece of tissue an efficient absorbing surface.

Model Answer — 5(b)

the lining is folded and carries millions of villi, each covered in microvilli, giving a very large surface area [1]

the wall of each villus is only one epithelial cell thick, giving a very short diffusion distance [1]

each villus contains a dense network of capillaries and a lacteal to carry absorbed material away [1]

⚠ If you missed marks here: Three separate adaptations, three separate marks — area, distance and removal. Answers that describe only the surface area lose two thirds of the question, and it is the removal of absorbed material that keeps the process going rather than stalling.

1 mark — villi and microvilli give a large surface area 1 mark — wall one epithelial cell thick, short diffusion distance 1 mark — capillaries and a lacteal remove absorbed material

Suggest why the same experiment gives almost no absorption if the tissue is first boiled and cooled.

Model Answer — 5(c)

boiling kills the cells and denatures their membrane carrier proteins and enzymes, so no active transport is possible [1]

only simple diffusion could then occur through the damaged tissue, and no energy from respiration is available [1]

⚠ If you missed marks here: Note the word used here. The cells are killed, because cells are alive; the proteins are denatured, because proteins are not. Using the wrong verb for either one is the mistake this part is designed to expose.

1 mark — boiling kills the cells and denatures carrier proteins 1 mark — no energy from respiration, so no active transport, only diffusion

(d) [2]

Explain why a result from this experiment cannot by itself tell you anything about assimilation.

Model Answer — 5(d)

the experiment measures only the movement of glucose from the lumen into the bathing solution, which is absorption [1]

assimilation is the uptake and use of a nutrient by a cell, which would need a different measurement such as the rate at which glucose is respired or stored in a tissue [1]

⚠ If you missed marks here: Two words, two different events, and the experiment can only see one of them. Knowing what a piece of apparatus cannot measure is as valuable in an exam as knowing what it can.

1 mark — identifies that the measurement is of absorption only 1 mark — assimilation defined as uptake and use by cells, needing a different measurement

Question 6 — Five Processes, One Cheese Sandwich

Total: 12 marks

A cheese sandwich contains starch, protein and fat. Follow it through the body.

(a) [3]

Name the five processes carried out by the digestive system, in the order in which they affect the sandwich, and define the two that are most often confused with each other.

Model Answer — 6(a)

ingestion, digestion, absorption, assimilation, egestion, in that order [1]

absorption: the movement of nutrients from the intestines into the blood [1]

assimilation: the uptake and use of nutrients by cells [1]

⚠ If you missed marks here: The order is forced by the biology: nothing can be absorbed before it is small enough, and nothing can be assimilated before it has reached a cell. Writing assimilation before absorption describes a cell using a nutrient that is still in the gut.

1 mark — five processes named in the correct order 1 mark — absorption correctly defined 1 mark — assimilation correctly defined

(b) [4]

For the protein in the cheese, describe the complete route of chemical digestion, naming both enzymes, both organs and the conditions in each.

Model Answer — 6(b)

pepsin is secreted by the stomach wall and begins protein digestion in the acidic conditions of the stomach, at about pH 2 [1]

the hydrochloric acid provides that optimum pH and also kills harmful microorganisms [1]

trypsin is secreted by the pancreas and continues digestion in the alkaline conditions of the small intestine [1]

the products are amino acids, which are small and soluble and can therefore be absorbed [1]

⚠ If you missed marks here: Both proteases are needed for full marks and each must be tied to its own organ and pH. The commonest error is to give trypsin the stomach, which places an enzyme with an optimum near pH 8 into a solution of pH 2 where it would be denatured at once.

1 mark — pepsin from the stomach wall, acting in acidic conditions about pH 2 1 mark — hydrochloric acid provides that pH and kills microorganisms 1 mark — trypsin from the pancreas, acting in alkaline conditions in the small intestine 1 mark — products are amino acids, small and soluble, so absorbable

Describe what happens to the starch in the bread from the mouth to the moment glucose enters the blood, naming all three enzymes involved.

Model Answer — 6(c)

salivary amylase begins the breakdown of starch to maltose in the mouth [1]

pancreatic amylase continues it in the small intestine after the salivary enzyme has been denatured by stomach acid [1]

maltase, on the membranes of the epithelium lining the small intestine, breaks maltose down to glucose, which is then absorbed into the capillaries of a villus [1]

⚠ If you missed marks here: Three enzymes, not one. Candidates who name only salivary amylase cannot explain how starch is fully digested, and those who have amylase producing glucose directly have skipped the enzyme that Cambridge names explicitly in the Supplement.

1 mark — salivary amylase produces maltose in the mouth 1 mark — pancreatic amylase continues in the small intestine 1 mark — maltase on the epithelial membranes produces glucose, absorbed into a capillary

(d) [2]

Some of the sandwich leaves the body in the faeces. Name the process, and explain why this material is treated differently from urea leaving in the urine.

Model Answer — 6(d)

egestion, the removal of undigested food such as cellulose from the body as faeces [1]

urea is a waste product of metabolism made inside the body’s own cells, so its removal is excretion; the undigested food was never absorbed and never took part in metabolism [1]

⚠ If you missed marks here: The distinction is about origin, not about state or route. The material in faeces has been inside the tube but never inside a cell — which is why the alimentary canal is sometimes described as a tunnel running through the body rather than as part of it.

1 mark — egestion named and described 1 mark — distinguished from excretion: urea is a metabolic waste product, undigested food is not

Question 7 — Two Claims to Evaluate

Total: 10 marks

A magazine article makes two claims. Claim 1: “Take a bile supplement with fatty meals — bile is the enzyme that digests fat.” Claim 2: “Chewing food thoroughly digests it before it reaches the stomach, so people with no teeth cannot digest starch.”

(a) [3]

Evaluate claim 1. State what is wrong with it and describe accurately what bile does.

Model Answer — 7(a)

bile is not an enzyme and contains none, so it cannot catalyse the breakdown of fat [1]

bile emulsifies fats and oils, breaking large droplets into many small ones and so increasing the surface area for chemical digestion [1]

bile is also alkaline and neutralises the acidic mixture arriving from the stomach, giving a suitable pH for the pancreatic enzymes; the fat is actually digested by lipase [1]

⚠ If you missed marks here: “Bile digests fat” is the single most commonly refused sentence in Topic 7. Emulsification changes the size of the droplets; digestion changes the molecules. Only lipase does the second.

1 mark — bile is not an enzyme and cannot digest fat 1 mark — bile emulsifies fat, increasing surface area 1 mark — bile is alkaline and provides a suitable pH; lipase digests the fat

(b) [4]

Evaluate claim 2. Identify the part that is right, the part that is wrong, and explain both.

Model Answer — 7(b)

right: chewing does begin the process, because salivary amylase starts converting starch to maltose in the mouth [1]

wrong: chewing itself is physical digestion, which changes the size of the pieces but makes no chemical change to the molecules [1]

wrong: only a small fraction of the starch is digested in the short time food is in the mouth; most is broken down by pancreatic amylase in the small intestine [1]

so a person with no teeth still digests starch, though physical digestion is poorer and the reduced surface area makes chemical digestion slower [1]

⚠ If you missed marks here: A good evaluation separates the true part from the false part instead of rejecting the whole claim. The last mark is the one that answers the question as asked — whether the conclusion about people with no teeth actually follows.

1 mark — identifies that salivary amylase does begin starch digestion in the mouth 1 mark — chewing is physical digestion, no chemical change to the molecules 1 mark — most starch is digested by pancreatic amylase in the small intestine 1 mark — concludes starch is still digested, though more slowly

A person with no teeth eats only soft, low-fibre foods. Describe two nutritional risks and explain the cause of one of them.

Model Answer — 7(c)

risk of constipation from the low fibre intake [1]

risk of deficiency of vitamin C, iron or calcium, since raw fruit, vegetables and meat are difficult to eat [1]

explanation, e.g. fibre provides bulk for the gut muscles to grip so the contents move along; without it material moves slowly and more water is absorbed, making the faeces hard [1]

⚠ If you missed marks here: Both risks must come from the description given — soft and low-fibre. A general answer about “an unhealthy diet” earns nothing, because nothing in the stem says the diet is short of energy or protein.

1 mark — constipation identified as a risk of low fibre 1 mark — a named vitamin or mineral deficiency risk 1 mark — mechanism of one risk explained fully

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