IGCSE Biology Paper 4 (Theory / Extended)

Topic 7: Human Nutrition -- Challenge Exam 2

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 — Four Diets Compared

Total: 12 marks

The table shows the mean daily intake of four groups of adults, each group living in a different region. All four groups contained adults of similar age and body mass.

group	energy / kJ	protein / g	fat / g	vitamin C / mg	iron / mg	fibre / g
A	9800	62	74	55	15.1	24
B	6100	28	19	48	5.9	31
C	13400	88	141	19	17.0	7
D	9600	58	70	9	14.6	22

(a) [2]

Calculate how many times greater the fat intake of group C is than that of group B. Show your working and give your answer to one decimal place.

Model Answer — 1(a)

working: 141 ÷ 19 [1]

= 7.4 times [1]

⚠ If you missed marks here: A “how many times” question is a division, not a subtraction. Answering 122 gives the difference in grams, which is a real quantity but not the one asked for — read the command words before reaching for the numbers.

1 mark — working shown: 141 divided by 19 1 mark — answer 7.4 (accept 7.42)

(b) [3]

Which group is at greatest risk of scurvy? Justify your choice using the data, and describe two symptoms.

Model Answer — 1(b)

group D, with a vitamin C intake of only 9 mg per day, the lowest in the table [1]

vitamin C is needed to form the connective tissue that holds cells together [1]

two symptoms from: bleeding gums, wounds that heal slowly or reopen, loose teeth, easy bruising [1]

⚠ If you missed marks here: Group C is the tempting answer because its diet looks unhealthy overall, but scurvy is caused by one specific shortage and C has twice D’s vitamin C. Justify a choice with the column that matters, not with a general impression of the row.

1 mark — group D identified 1 mark — justified with the vitamin C figure of 9 mg 1 mark — two correct symptoms of scurvy

Group B has the highest fibre intake and the lowest intake of energy, protein and iron. Describe two health problems this group is most likely to face, and explain each.

Model Answer — 1(c)

anaemia [1]

iron intake of 5.9 mg is well below the others, and iron is needed to make haemoglobin, so less oxygen is carried [1]

poor growth and repair of tissues, or loss of body mass [1]

because protein supplies the amino acids for new cell material and enzymes, and the energy intake is too low so body stores are used [1]

⚠ If you missed marks here: High fibre is a strength of this diet, not a weakness, so an answer that criticises it has misread the table. Each problem needs the nutrient and what that nutrient is used for — naming the disease alone is worth half.

1 mark — anaemia identified 1 mark — linked to low iron and to haemoglobin 1 mark — poor growth or repair, or loss of body mass, identified 1 mark — linked to low protein and amino acids, or low energy intake

(d) [3]

Group C has an energy intake 40 per cent above group A and a fibre intake less than a third of A. Predict two consequences for the alimentary canal and for body mass, and explain the fibre effect.

Model Answer — 1(d)

excess energy intake leads to gain in body mass as the surplus is stored as fat [1]

the low fibre intake makes constipation likely [1]

because there is too little bulk for the gut muscles to grip, so material moves slowly and more water is absorbed from it [1]

⚠ If you missed marks here: Predictions must follow from the data given. Adding untestable claims — heart disease, diabetes — may be true of such a diet but earns nothing here, because nothing in the table measures them.

1 mark — gain in body mass from surplus energy stored as fat 1 mark — constipation predicted from low fibre 1 mark — explained by lack of bulk, slow movement, more water absorbed

Question 2 — Two Proteases, Two Optima

Total: 12 marks

The activity of two human proteases was measured at a range of pH values, at 37 °C. Activity is given as a percentage of each enzyme’s own maximum.

pH	enzyme P / %	enzyme Q / %
1.0	74	0
2.0	100	0
3.0	58	0
5.0	9	6
7.0	0	61
8.0	0	100
9.0	0	47
11.0	0	0

(a) [3]

State the optimum pH of each enzyme and identify each enzyme, giving the region of the alimentary canal in which it acts.

Model Answer — 2(a)

P: optimum about pH 2; Q: optimum about pH 8 [1]

P is pepsin, acting in the stomach [1]

Q is trypsin, acting in the small intestine (duodenum) [1]

⚠ If you missed marks here: The identification comes from matching each optimum to the pH of a real region, not from remembering names in isolation. If you get them the wrong way round you have an enzyme that is inactive in the only organ it is ever found in, which is the check that catches the swap.

1 mark — optima given as about pH 2 and about pH 8 1 mark — P identified as pepsin acting in the stomach 1 mark — Q identified as trypsin acting in the small intestine

(b) [3]

Explain fully why enzyme P shows zero activity at pH 8.

Model Answer — 2(b)

pH 8 is far from the optimum of enzyme P, so the enzyme is denatured [1]

the shape of the active site changes and is no longer complementary to the substrate [1]

protein molecules can no longer bind, so no enzyme–substrate complexes form and no amino acids are produced [1]

⚠ If you missed marks here: Two words fail here every year. Denatured, never “killed” — an enzyme is a protein molecule and was never alive. And active site, because “the enzyme changes shape” states the event without explaining why the reaction stops.

1 mark — enzyme is denatured at a pH far from its optimum 1 mark — shape of the active site changes, no longer complementary 1 mark — no enzyme-substrate complexes form, so no product

Both enzymes show low but measurable activity at pH 5, yet neither is found working at pH 5 in the body. Suggest why the body uses two proteases in two regions rather than one protease working at pH 5 throughout.

Model Answer — 2(c)

at pH 5 each enzyme works at less than 10 % of its maximum, so digestion would be very slow [1]

each region has a pH set for other reasons — the stomach is acidic to kill microorganisms, the duodenum is alkaline because bile neutralises the acid [1]

using an enzyme whose optimum matches each region gives the fastest possible digestion in both [1]

⚠ If you missed marks here: “Suggest” means build an argument from the data plus what you know, not recall a fact. The strongest answers notice that the pH of each region is fixed by a different requirement, so the enzyme has to fit the region and not the other way round.

1 mark — at pH 5 both enzymes are near-inactive, so digestion would be slow 1 mark — each region has its pH set for another reason (acid kills microorganisms / bile neutralises) 1 mark — matching each enzyme optimum to its region maximises the rate

(d) [3]

A student writes: “The data show that enzyme Q is a better enzyme than enzyme P.” Evaluate this statement.

Model Answer — 2(d)

the statement cannot be supported: activity is expressed as a percentage of each enzyme’s own maximum, so the two columns cannot be compared directly [1]

the actual rates in units of product per minute are not given, so nothing here shows which enzyme is faster [1]

each is best in its own conditions; “better” has no meaning without stating the pH [1]

⚠ If you missed marks here: The single most valuable habit in data questions is reading the units before the numbers. A percentage of a maximum hides the size of that maximum, so 100 % for one enzyme and 100 % for another may be very different real rates.

1 mark — notes that both columns are percentages of each enzyme's own maximum 1 mark — notes that absolute rates are not given 1 mark — concludes that better is meaningless without stating the conditions

Question 3 — Surface Area, Bile and the Rate of Fat Digestion

Total: 12 marks

Cream was mixed with pancreatic lipase in four tubes at 37 °C. The mass of fatty acid produced after 15 minutes was measured. The mean droplet diameter in each tube was measured at the start.

tube	contents	mean droplet diameter / µm	fatty acid produced / mg
1	cream + lipase	46	11
2	cream + lipase + bile	4	58
3	cream + bile	4	0
4	cream + boiled lipase + bile	4	0

(a) [3]

Using tubes 1 and 2, explain the effect of bile on the rate of fat digestion.

Model Answer — 3(a)

bile has emulsified the fat, reducing the mean droplet diameter from 46 to 4 µm [1]

many small droplets have a much larger total surface area than a few large ones [1]

more fat is exposed to lipase, so more enzyme–substrate complexes form per second and over five times as much fatty acid is produced [1]

⚠ If you missed marks here: Quote the numbers. An answer that says “bile increases the surface area so digestion is faster” is correct but does not use the data it was given, and data questions reserve a mark for exactly that.

1 mark — bile emulsifies, droplet diameter falls from 46 to 4 micrometres 1 mark — small droplets give a much larger total surface area 1 mark — more fat exposed to lipase, so much more fatty acid produced

(b) [3]

Explain what tubes 3 and 4 show, and why both were needed.

Model Answer — 3(b)

tube 3 produced no fatty acid, showing that bile alone cannot digest fat because it contains no enzyme [1]

tube 4 produced no fatty acid, showing that the fatty acid in tube 2 was produced by active lipase and not by anything else in the mixture [1]

each tube removes a different possible explanation, so together they show the effect in tube 2 needs both bile and working lipase [1]

⚠ If you missed marks here: This experiment is the cleanest available proof that bile does not digest fat. If you have ever written “bile digests fat”, tube 3 is the result to remember: same emulsified droplets as tube 2, and nothing whatever produced.

1 mark — tube 3 shows bile alone produces nothing because it contains no enzyme 1 mark — tube 4 shows the product in tube 2 required active lipase 1 mark — together they eliminate alternative explanations for tube 2

A cube of solid fat of side 12 mm is cut into cubes of side 3 mm. Calculate the surface area before and after cutting, and state the factor by which it has increased.

Model Answer — 3(c)

before: 6 × 12² = 864 mm² [1]

after: 64 cubes × 6 × 3² = 3456 mm² [1]

increase by a factor of 4 (which equals 12 ÷ 3) [1]

⚠ If you missed marks here: The volume has not changed, so the factor is simply the ratio of the side lengths. Students who answer 64 have given the number of pieces, not the increase in area — and the same confusion turns up whenever villi and microvilli are being discussed.

1 mark — surface area before = 864 mm squared 1 mark — surface area after = 3456 mm squared 1 mark — factor of 4

(d) [3]

Explain how physical digestion in the mouth and in the stomach produces the same kind of advantage as bile does in the duodenum, and state clearly what all three have in common.

Model Answer — 3(d)

in the mouth the teeth cut and grind the food into smaller pieces [1]

in the stomach the muscular wall churns the contents, breaking them up further and mixing them [1]

all three increase the surface area for enzyme action without changing the food molecules — all are physical, not chemical [1]

⚠ If you missed marks here: The final mark is for the general principle rather than for another example. Bile is the odd one out only in appearance: it works on droplets rather than on lumps, but it changes no molecules, so it belongs with chewing and churning and not with the enzymes.

1 mark — teeth cut and grind food in the mouth 1 mark — muscular wall of the stomach churns the contents 1 mark — all three increase surface area without chemical change to the molecules

Question 4 — Where the Nutrients Actually Cross

Total: 12 marks

The concentration of glucose was measured in the contents of the alimentary canal at points along its length, two hours after a starchy meal. The internal surface area of each region was also measured.

region	glucose in contents / arbitrary units	internal surface area / m²
stomach	31	0.1
duodenum	44	2.0
upper ileum	26	110.0
lower ileum	4	95.0
colon	1	0.4

(a) [3]

Describe the pattern in the glucose figures and explain what causes the rise and then the fall.

Model Answer — 4(a)

glucose rises from 31 in the stomach to a peak of 44 in the duodenum, then falls steadily to 1 in the colon [1]

the rise is caused by continued chemical digestion of starch by pancreatic amylase and by maltase, releasing more glucose [1]

the fall is caused by absorption of glucose through the wall of the ileum into the blood [1]

⚠ If you missed marks here: Describe first, explain second — and quote figures when you describe. The commonest loss here is explaining the fall as “the glucose is used up”: nothing in the lumen of the gut respires, so glucose leaves by crossing the wall, not by disappearing.

1 mark — pattern described with figures: rise to 44 in the duodenum, fall to 1 in the colon 1 mark — rise explained by continued digestion of starch 1 mark — fall explained by absorption into the blood

(b) [3]

Use the surface area column to explain why most absorption happens where it does.

Model Answer — 4(b)

the ileum has by far the largest internal surface area, 110 and 95 m² against 2.0 m² for the duodenum [1]

this is produced by folding of the lining, by villi and by microvilli on the epithelial cells [1]

a larger surface allows more molecules to be absorbed per second, which is why the glucose figure falls most steeply there [1]

⚠ If you missed marks here: The two columns have to be read together — that is why both were given. An answer that describes villi without connecting them to the falling glucose has answered a recall question rather than the data question in front of it.

1 mark — ileum has much the largest surface area, quoting figures 1 mark — surface area produced by folds, villi and microvilli 1 mark — larger surface allows more absorption per second, matching the steep fall

In the lower ileum the glucose concentration in the contents is 4 units while in the blood of the villus capillaries it is 9 units. Explain how glucose can still be absorbed.

Model Answer — 4(c)

glucose is moving against the concentration gradient, so diffusion cannot account for it [1]

it is moved by active transport, using protein carriers in the cell membrane [1]

this requires energy from respiration, which is why the epithelial cells contain many mitochondria [1]

⚠ If you missed marks here: The word osmosis appears here in a large minority of answers and scores nothing — osmosis moves water, not solutes. The clue that active transport is wanted is always the direction: uphill movement cannot be diffusion, whatever else is true.

1 mark — identifies movement against the concentration gradient 1 mark — names active transport using protein carriers 1 mark — requires energy from respiration / many mitochondria

(d) [3]

A patient has an illness that flattens the villi of the ileum. Predict the effect on the figures in this table and on the patient, and explain why.

Model Answer — 4(d)

the internal surface area of the ileum would fall sharply [1]

less glucose would be absorbed, so the glucose figure in the lower ileum and colon would be higher than shown [1]

the patient would lose mass and become weak or anaemic, because nutrients pass out in the faeces instead of entering the blood [1]

⚠ If you missed marks here: A prediction has to say which way each number moves. Note the direction carefully: damaged absorption leaves more in the gut, not less — getting that backwards is the commonest error in questions about coeliac disease and similar conditions.

1 mark — internal surface area of the ileum falls 1 mark — glucose remaining in the lower ileum and colon would be higher 1 mark — patient loses mass or becomes anaemic as nutrients are lost in the faeces

Question 5 — What Happens to the Mass of a Meal

Total: 10 marks

A researcher followed a measured meal of dry mass 320 g through the alimentary canal of a volunteer and recorded the dry mass of material remaining at four points.

point	dry mass remaining / g
leaving the stomach	318
leaving the duodenum	301
leaving the ileum	64
leaving the colon (faeces)	58

(a) [2]

Calculate the percentage of the original dry mass that was absorbed by the time the material left the ileum. Show your working.

Model Answer — 5(a)

mass absorbed = 320 − 64 = 256 g; working 256 ÷ 320 × 100 [1]

= 80 % [1]

⚠ If you missed marks here: Work out what has gone, not what is left, before dividing. Answering 20 % gives the fraction still in the gut — the right calculation attached to the wrong question, which is a very expensive way to lose two marks.

1 mark — working: (320 - 64) divided by 320, times 100 1 mark — answer 80 %

(b) [3]

Explain why the mass falls only slightly between the stomach and the end of the duodenum, but very steeply along the ileum.

Model Answer — 5(b)

the stomach and duodenum are mainly regions of digestion, which breaks molecules up but does not remove them from the canal [1]

very little absorption occurs there, so the mass in the canal barely changes [1]

the ileum is the main region of absorption, where nutrients cross into the blood and so leave the canal [1]

⚠ If you missed marks here: Digestion and absorption are different events and this table separates them beautifully: breaking a molecule in half changes nothing about the mass present. The mass only falls when something crosses the wall.

1 mark — stomach and duodenum are mainly regions of digestion 1 mark — digestion changes molecules but does not remove mass from the canal 1 mark — ileum is the main absorbing region, so mass leaves there

Only 6 g of dry mass is lost along the colon. Explain what the colon does, and why the dry mass changes so little there.

Model Answer — 5(c)

the colon absorbs mainly water, together with some mineral ions [1]

water contributes nothing to dry mass, so removing it leaves the dry figure almost unchanged [1]

most of the remaining 58 g is undigested material such as fibre, which cannot be absorbed at all [1]

⚠ If you missed marks here: The word dry in the table is the key to this part, and it is regularly skipped. The colon is doing a great deal of work; it simply happens to be work that a dry mass measurement cannot detect.

1 mark — colon absorbs mainly water and some mineral ions 1 mark — water does not contribute to dry mass 1 mark — remaining material is largely undigested fibre, which cannot be absorbed

(d) [2]

State which of the five processes of the digestive system removes the final 58 g from the body, and justify your choice.

Model Answer — 5(d)

egestion [1]

the 58 g was never absorbed into the blood or into any cell, so it is not a waste product of metabolism and its removal is not excretion [1]

⚠ If you missed marks here: Naming the process alone gets one mark; the justification requires you to say what egestion is not. This distinction is tested in almost every Topic 7 paper because the two words sound so similar.

1 mark — egestion named 1 mark — justified: the material was never absorbed, so it is not a product of metabolism

Question 6 — Following Starch With Iodine and Benedict’s

Total: 12 marks

Starch solution was mixed with saliva at 37 °C. Every two minutes a sample was tested with iodine solution and a second sample was tested with Benedict’s solution.

time / min	colour with iodine solution	colour with Benedict’s solution
0	blue-black	blue
2	blue-black	blue
4	dark brown	green
6	orange-brown	orange
8	orange-brown	brick-red
10	orange-brown	brick-red

(a) [3]

Explain what each of the two tests is showing, and state what has happened by 8 minutes.

Model Answer — 6(a)

iodine tests for starch: the loss of the blue-black colour shows starch is disappearing [1]

Benedict’s tests for reducing sugar: blue to green to orange to brick-red shows the sugar concentration is rising [1]

by 8 minutes essentially all the starch has been converted to maltose by salivary amylase [1]

⚠ If you missed marks here: Each test answers one question only. A frequent error is to read the Benedict’s colours backwards — blue is the negative result, and brick-red is the strongest positive, not the other way round.

1 mark — iodine tests for starch; loss of blue-black shows starch disappearing 1 mark — Benedict's tests for reducing sugar; colour sequence shows sugar increasing 1 mark — by 8 minutes starch has been converted to maltose by salivary amylase

(b) [3]

The student concludes that saliva converts starch to glucose. Explain why this conclusion goes beyond the evidence, and state what would actually complete the conversion, and where.

Model Answer — 6(b)

Benedict’s solution tests for reducing sugars in general and cannot distinguish maltose from glucose [1]

amylase produces maltose, not glucose [1]

maltase, on the membranes of the epithelium lining the small intestine, breaks maltose down to glucose [1]

⚠ If you missed marks here: Two ideas are being tested together: the limits of a food test, and the two-step nature of starch digestion. The location of maltase is a Supplement point and is worth memorising as a phrase — it is on the membranes, not free in the fluid.

1 mark — Benedict's cannot distinguish maltose from glucose 1 mark — amylase produces maltose, not glucose 1 mark — maltase acts on the membranes of the epithelium lining the small intestine

The experiment was repeated with the saliva at pH 2 instead of pH 7. Predict the results of both tests at 10 minutes and explain your prediction.

Model Answer — 6(c)

iodine would still give a blue-black colour and Benedict’s would remain blue [1]

pH 2 is far from the optimum of salivary amylase, so the enzyme is denatured and the shape of its active site is changed [1]

starch is no longer complementary to the active site, so no complexes form and no maltose is produced [1]

⚠ If you missed marks here: Predict both tests, not one. This is also the biological reason that starch digestion pauses in the stomach and has to be restarted by pancreatic amylase in the duodenum — a useful synoptic link to make in one extra sentence.

1 mark — predicts blue-black with iodine and blue with Benedict's 1 mark — amylase denatured at pH 2, active site changed 1 mark — no complexes form, so no maltose produced

(d) [3]

Name the two places in the alimentary canal where amylase acts, name the organ that secretes it in each case, and explain why digestion of starch has to begin again after the stomach.

Model Answer — 6(d)

mouth, secreted by the salivary glands [1]

small intestine or duodenum, secreted by the pancreas [1]

the salivary amylase carried into the stomach is denatured by the acid, so fresh enzyme is needed once the food has passed on [1]

⚠ If you missed marks here: Where an enzyme is secreted and where it acts are separate syllabus statements, and amylase is the enzyme for which the answer is two places rather than one. Candidates who learn only “amylase is in saliva” cannot explain how a starchy meal is ever fully digested.

1 mark — mouth, secreted by the salivary glands 1 mark — small intestine, secreted by the pancreas 1 mark — salivary amylase denatured by stomach acid, so pancreatic amylase is needed

Question 7 — Vitamin C and the Uptake of Iron

Total: 10 marks

Volunteers were each given a meal containing exactly 5.0 mg of iron. The mass of iron absorbed into the blood was measured. Each volunteer took the meal four times, with a different drink each time.

drink taken with the meal	iron absorbed / mg	percentage of the iron absorbed
water	0.10	2.0
orange juice (rich in vitamin C)	0.42	8.4
tea	0.05	1.0
orange juice from which vitamin C had been removed	0.11	2.2

(a) [3]

Explain what the fourth row adds to the investigation, and state the conclusion the whole table supports.

Model Answer — 7(a)

the fourth row is a control: it keeps everything about the orange juice the same except the vitamin C [1]

its result, 0.11 mg, is almost identical to water, showing the effect was not caused by any other component of the juice [1]

conclusion: vitamin C increases the absorption of iron, roughly fourfold in these conditions [1]

⚠ If you missed marks here: Without the fourth row the experiment shows only that orange juice helps, which would leave the sugar, the acid or anything else in the juice as possible causes. Recognising the row that removes the alternative explanations is the whole skill being tested.

1 mark — fourth row identified as a control isolating the vitamin C 1 mark — its result matches water, ruling out other components of the juice 1 mark — conclusion: vitamin C increases iron absorption

(b) [2]

Calculate how many times more iron is absorbed with orange juice than with tea, and comment on the size of the difference.

Model Answer — 7(b)

0.42 ÷ 0.05 = 8.4 times [1]

the choice of drink changes iron uptake by nearly an order of magnitude, so it matters as much as the iron content of the meal itself [1]

⚠ If you missed marks here: The comment mark is not decoration. A number on its own does not show that you understand what it means, and challenge papers routinely reserve a mark for saying whether a difference is large enough to matter.

1 mark — factor of 8.4 calculated 1 mark — comment that the effect is large, comparable in importance to the iron content

A person eating a diet rich in iron still develops anaemia. Using the table, suggest an explanation, and describe two symptoms of anaemia.

Model Answer — 7(c)

the iron in the diet may not be absorbed, for example if the diet is low in vitamin C or if the person drinks tea with meals [1]

iron that is not absorbed is egested in the faeces and never reaches the blood [1]

two symptoms from: tiredness, pale skin, breathlessness on exertion, poor concentration [1]

⚠ If you missed marks here: Eating a nutrient and having it is not the same thing, and this is the clearest example in the syllabus. The mark for egestion checks that you know where the unabsorbed iron went — it was never inside the body, so it cannot have been excreted.

1 mark — suggests the iron is not absorbed, linked to low vitamin C or to tea 1 mark — unabsorbed iron is egested in the faeces 1 mark — two symptoms of anaemia

(d) [2]

Suggest two improvements to the design of this investigation.

Model Answer — 7(d)

use a larger number of volunteers and calculate a mean, since individuals absorb iron at different rates [1]

control other variables: the same meal composition, the same volume of drink, the same time of day, and the same interval since the previous meal [1]

⚠ If you missed marks here: Improvements must be things that would change the reliability or the fairness of the result. “Repeat the experiment” on its own is too vague to score; say what would be repeated and what would be done with the extra readings.

1 mark — more volunteers and a mean calculated 1 mark — a named further variable controlled, e.g. same meal, same volume of drink

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