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This paper covers the whole of Topic 4. Like a real Cambridge paper, the seven questions range across every sub-topic — biological molecules, the food tests, and the structure of DNA — and they are deliberately mixed rather than grouped. All three Topic 4 papers do; they differ in the angle they come at it from, not in what they cover.
Question 1 — Two Polymers, One Building Block
Total: 12 marks
Cotton fibre is almost pure cellulose. A potato tuber stores its carbohydrate as starch. Both molecules are built from exactly the same small molecule, yet they do completely different jobs.
You are told that the human gut produces a substance that breaks the links in starch, but no substance that breaks the links in cellulose.
(a)[4]
(i) Name the small molecule from which both cellulose and starch are built. [1]
(ii) State the three chemical elements present in both molecules. [1]
(iii) Explain what is meant by saying that starch is a large molecule built from many small ones. [2]
Model Answer — 1a
(i) glucose [1]
(ii) carbon, hydrogen and oxygen [1]
(iii) very many glucose molecules are joined together end to end in long chains [1]
(iii) so one starch molecule is enormous compared with the glucose units it is made from, and it is the joining together, not any change of element, that makes it different [1]
⚠ If you missed marks here: “Starch is a carbohydrate” restates the question. The mark is for the process: many glucose molecules joined together into long chains. Note also that building a large molecule from small ones adds no new elements — carbon, hydrogen and oxygen are all that is present before and after.
(b)[4]
Cotton is not digested by humans, but potato starch is.
(i) Explain what this shows about the two molecules, given that both are made only of glucose. [2]
(ii) Suggest why cellulose is suited to forming the walls of plant cells while starch is suited to being a store. [2]
Model Answer — 1b
(i) the glucose units must be joined together in a different way in the two molecules [1]
(i) so a substance that fits and breaks the links in starch does not fit the links in cellulose — the subunit being identical is not enough to make the molecules the same [1]
(ii) cellulose chains are long, straight and unbranched, and lie side by side in bundles, which makes them strong and good for support [1]
(ii) starch chains are coiled and branched, making a compact molecule that is insoluble, so it can be packed away in large amounts without affecting the cell [1]
⚠ If you missed marks here: The trap is answering “because cellulose is tougher” — true, but it is a restatement rather than an explanation. The examiner wants the linking arrangement: same brick, different mortar. For the second part, keep the two properties separate: straight chains side by side gives strength, and coiled, branched and insoluble gives compact storage.
(c)[4]
A student adds iodine solution to a piece of cotton wool and to a slice of raw potato.
(i) State and explain the result in each case. [2]
(ii) The student concludes “cotton contains no carbohydrate”. Explain why this conclusion is wrong and state what the result actually shows. [2]
Model Answer — 1c
(i) the potato turns blue-black because it contains starch [1]
(i) the cotton stays orange-brown because it contains cellulose and no starch [1]
(ii) iodine solution tests for starch specifically, not for carbohydrates in general [1]
(ii) the result shows only that cotton contains no starch; cellulose is still a carbohydrate made of glucose [1]
⚠ If you missed marks here: This is the single most useful habit in the whole topic: a food test detects one named substance, and a negative result rules out that substance only. Iodine detects starch, not carbohydrate; Benedict’s detects reducing sugars, not carbohydrate. Also give the negative result properly — the iodine stays orange-brown, it does not “do nothing”.
Question 2 — Testing an Unfamiliar Food
Total: 12 marks
Baobab fruit powder is sold with the claim that it is “rich in vitamin C and a good source of protein”. A technician in Dar es Salaam is asked to test the claim. She has iodine solution, Benedict’s solution, biuret solution, ethanol, DCPIP solution, distilled water, a water bath, a balance and standard laboratory glassware.
(a)[4]
For each of the four nutrients below, name the reagent used and state the positive result. Give the colour change, not just the final colour.
reducing sugar: Benedict’s solution, heated in a water bath, blue → green/yellow/orange/brick-red [1]
protein: biuret solution, blue → purple (violet) [1]
fat: ethanol emulsion test, clear → cloudy white emulsion [1]
⚠ If you missed marks here: Half of these marks are thrown away by giving only the end colour. “Blue-black” alone does not show you know iodine starts orange-brown, and “purple” alone does not show biuret starts blue. The Benedict’s answer must also carry its condition — heated in a water bath — because without heating the test simply does not happen.
(b)[4]
(i) Describe how the technician should prepare the powder before testing, so that the tests work and the four samples are comparable. [2]
(ii) Explain why she should also carry out every test on distilled water alone. [2]
Model Answer — 2b
(i) weigh out the same mass of powder for each test and mix it with the same measured volume of distilled water [1]
(i) stir or shake to make a suspension and filter it, so that the solid does not itself make the tubes cloudy or hide a colour change [1]
(ii) this is a control, in which the food is the only thing left out [1]
(ii) it shows the colour each reagent gives when the nutrient is absent, so any change seen with the powder can be attributed to the food and not to the reagent or the water [1]
⚠ If you missed marks here: A control is not “doing the test again” — it is running the identical procedure with the one factor under test removed. Say what it shows, not just that it is a control. The preparation marks are about comparability: same mass, same volume, filtered. Unfiltered suspensions ruin the emulsion test in particular, because the food itself looks cloudy.
(c)[4]
Her results are shown.
test
result with baobab powder
result with distilled water
iodine solution
stays orange-brown
stays orange-brown
Benedict’s solution, heated
orange
stays blue
biuret solution
very pale purple
stays blue
DCPIP
decolourised by 0.4 cm³
not decolourised by 10 cm³
Evaluate the claim that baobab powder is “rich in vitamin C and a good source of protein”, using all the evidence. [4]
Model Answer — 2c
the vitamin C claim is supported: only 0.4 cm³ was needed to decolourise the DCPIP, a very small volume, while distilled water did not decolourise it at all [1]
the protein claim is weakly supported at best: biuret gave only a very pale purple, so protein is present but in a small amount [1]
the powder also contains a reducing sugar (orange with Benedict’s) but no starch (iodine unchanged) [1]
the water controls were all negative, so the colour changes were genuinely caused by the food and the conclusions are trustworthy [1]
⚠ If you missed marks here: “Evaluate” means weigh the evidence both ways, so a one-sided “the claim is true” caps the mark. A pale purple is a positive biuret result but a weak one, and the biuret test is not designed to measure quantity — so the honest verdict is “protein present, but the evidence does not support good source”. Referring back to the control is the mark most candidates never think to write.
Question 3 — Oils, Fats and a False Positive
Total: 12 marks
A technician sets up the ethanol emulsion test on four liquids: olive oil (a plant oil), liquid paraffin (a mineral oil obtained from crude oil, which is not a food fat), glucose solution and distilled water.
liquid tested
appearance after the test
olive oil
dense cloudy white emulsion
liquid paraffin
cloudy white emulsion
glucose solution
stays clear and colourless
distilled water
stays clear and colourless
(a)[4]
(i) Name the two kinds of small molecule that combine to form a fat or an oil, and state how many of each are used. [2]
(ii) State the elements present in a fat. [1]
(iii) Starch is described as being built from many small molecules, but a fat is not described in the same way. Explain why. [1]
Model Answer — 3a
(i) glycerol and fatty acids [1]
(i) one molecule of glycerol combines with three fatty acid molecules [1]
(ii) carbon, hydrogen and oxygen [1]
(iii) a starch molecule contains thousands of repeated glucose units, whereas a fat is made from only four molecules joined together, so it is not a long repeating chain [1]
⚠ If you missed marks here: The one-to-three ratio is worth its own mark and is regularly given as “two fatty acids” or left out altogether. The last part is a genuinely discriminating idea: starch, glycogen and cellulose are long chains of one repeating unit, while a fat is a small assembly of four molecules — which is why the huge-molecule language belongs to the polysaccharides and proteins, not to fats.
(b)[4]
(i) Explain what the glucose solution and distilled water results tell the technician about the reliability of her method. [2]
(ii) The liquid paraffin also gave a cloudy white emulsion, although it is not a food fat. Explain what this result shows about what the ethanol emulsion test actually detects. [2]
Model Answer — 3b
(i) these two liquids act as negative controls [1]
(i) both stayed clear, so a cloudy result really does indicate something dissolved in the ethanol and not a fault in the apparatus or the ethanol [1]
(ii) the test detects any substance that dissolves in ethanol but is insoluble in water, not specifically a food fat [1]
(ii) so a positive emulsion test on its own does not prove the substance is a fat or oil that can be used as food [1]
⚠ If you missed marks here: The word to reach for is specificity. A positive result means “something behaving like a lipid is present”, not “a nutritious fat is present”. Answers claiming the paraffin result is simply a mistake miss the point entirely — the test worked exactly as designed, and the limitation is in what the design can distinguish.
(c)[4]
The technician now wants to compare how much fat is present in two brands of crisps. The emulsion test cannot give her a number.
Describe a method that would let her compare the fat content of the two brands, and state one variable she must control. [4]
Model Answer — 3c
weigh out the same mass of each brand of crisps and crush each one [1]
shake each with the same volume of ethanol so that the fat dissolves, then filter [1]
evaporate the ethanol from the filtrate and weigh the fat left behind (or reweigh the dried crisp residue and find the loss in mass) [1]
control variable: the same mass of crisps, the same volume of ethanol, the same shaking time or the same drying time [1]
⚠ If you missed marks here: The question asks for a quantitative method, so any answer that ends in a colour or a cloudiness cannot score. The measured quantity has to be a mass, and the fat has to be separated first — dissolve it in ethanol, filter, then evaporate the solvent. Saying “use a balance” without saying what is weighed and when earns nothing.
Question 4 — Base Composition in Four Species
Total: 12 marks
The table shows the percentages of the four bases in DNA samples from four species. Some values have been left out, and one row contains an impossible set of results.
species
adenine / %
thymine / %
cytosine / %
guanine / %
W
29
21
X
31
19
19
Y
15
15
35
35
Z
28
22
25
25
You are also told that the more closely two species are related, the more similar their DNA base sequences are.
(a)[4]
(i) Give the missing values for species W and species X. [2]
(ii) Identify the row that cannot be correct and explain why. [2]
Model Answer — 4a
(i) species W: thymine = 29% and guanine = 21% [1]
(i) species X: adenine = 31% [1]
(ii) species Z cannot be correct [1]
(ii) because adenine pairs only with thymine, so the two percentages must be equal, but 28% and 22% are not [1]
⚠ If you missed marks here: Fill the table by rule, not by making the row add to 100 — both routes reach the same answer for W and X, but only the pairing rule spots the faulty row. Check every row against two equalities: %A = %T and %C = %G. Species Z passes the second and fails the first, which is exactly the kind of half-right data an examiner plants.
(b)[4]
(i) Explain, in terms of the structure of DNA, why the percentage of adenine always equals the percentage of thymine. [2]
(ii) Describe how the two strands of a DNA molecule are held together. [2]
Model Answer — 4b
(i) DNA has two strands, and every base on one strand is paired with a base on the other strand [1]
(i) adenine can only pair with thymine, so each adenine is matched by one thymine and their numbers must be equal [1]
(ii) there are bonds between the pairs of bases [1]
(ii) these bonds run between the two strands along the whole length of the coiled double helix, holding the strands together [1]
⚠ If you missed marks here: “They are equal because they pair” is only half the argument — the answer must first establish that every base is paired, which is a consequence of DNA having two strands running the full length of the molecule. For the second part, the bonds are specifically between the pairs of bases; answers saying the strands are “wrapped around each other” describe the shape, not what holds them.
(c)[4]
The base sequences of a short section of the same gene were compared. Species W and species Y differ at 4 bases in every 100. Species W and species X differ at 27 bases in every 100.
(i) State which two species are more closely related and justify your answer. [2]
(ii) Explain why comparing base sequences is far more useful than comparing the base percentages in the first table. [2]
Model Answer — 4c
(i) species W and species Y are more closely related [1]
(i) because their base sequences differ at far fewer positions (4 in 100 compared with 27 in 100), and more closely related species have more similar base sequences [1]
(ii) the percentages only say how much of each base is present, not the order in which the bases are arranged [1]
(ii) two completely unrelated species could have identical percentages while having entirely different sequences, so percentages cannot show relatedness [1]
⚠ If you missed marks here: The justification mark needs the numbers quoted and the principle stated — “because they are more similar” is circular. The second part turns on the difference between composition and order: the same four letters in the same proportions can spell utterly different messages, and it is the message, not the letter count, that carries the information.
Question 5 — Putting a Number on Benedict
Total: 10 marks
A laboratory wants a numerical version of the Benedict’s test. Glucose solutions of known concentration are each tested in the same way, and the colour of the final mixture is measured with a colorimeter, which gives a reading from 0 (blue, no reducing sugar) to 100 (deep brick-red).
glucose concentration / g per dm³
0
2
4
6
8
10
colorimeter reading
0
17
34
50
66
80
A sample of diluted honey tested in exactly the same way gave a reading of 42.
(a)[4]
Describe how each of the known glucose solutions should be tested so that the readings can be fairly compared with one another and with the honey sample.
Model Answer — 5a
take the same volume of each glucose solution and add the same volume of Benedict’s solution [1]
heat all the tubes in a water bath at the same temperature [1]
heat every tube for the same length of time, then remove them together [1]
measure each mixture in the colorimeter in the same way, and repeat each concentration and take a mean [1]
⚠ If you missed marks here: Every mark here is a control variable, and the two that are most often forgotten are time and volume of Benedict’s. Benedict’s colour deepens the longer it is heated, so tubes pulled out at different moments cannot be compared even if everything else was identical. “Do it the same way each time” without naming variables scores nothing.
(b)[3]
(i) Describe the relationship between glucose concentration and colorimeter reading. [1]
(ii) Use the data to estimate the glucose concentration of the diluted honey sample. Show how you obtained your answer. [2]
Model Answer — 5b
(i) the colorimeter reading increases as the glucose concentration increases, in an almost directly proportional way [1]
(ii) a reading of 42 lies between the readings for 4 g/dm³ (34) and 6 g/dm³ (50) [1]
(ii) 42 is halfway between 34 and 50, so the concentration is about 5 g/dm³ [1]
⚠ If you missed marks here: Reading between known points is called interpolation, and the working must be shown — an unsupported “5” risks losing the method mark. Say which two data points the value falls between and how far along it lies. Describing the relationship needs a direction word (increases) linked to both variables, not just “they go up”.
(c)[3]
(i) Explain why using a colorimeter makes the Benedict’s test quantitative, while using the eye makes it only semi-quantitative. [2]
(ii) Honey is a pale yellow liquid. Suggest one difficulty this creates for the method. [1]
Model Answer — 5c
(i) the colorimeter gives an actual numerical measurement that can be compared with a calibration series [1]
(i) judging by eye gives only a description such as green or orange, which is subjective and varies between people and lighting conditions [1]
(ii) the honey is already yellow, so its own colour adds to the colour produced by the test and would make the reading too high unless a honey-only blank is used [1]
⚠ If you missed marks here: The distinction is measurement versus description. A colour chart still relies on a human judgement, so it stays semi-quantitative however many bands it has. The honey point is about interference: any coloured sample needs its own blank, otherwise the pigment in the food is being measured as though it were reducing sugar.
Question 6 — Proteins and the Nitrogen Clue
Total: 12 marks
Proteins are built from amino acids. A food laboratory estimates the protein content of a food by measuring how much nitrogen it contains, then using the fact that protein contains about 16% nitrogen by mass.
A 100 g sample of dried lentils was found to contain 4.0 g of nitrogen.
(a)[4]
(i) Name the small molecules from which proteins are built. [1]
(ii) State the elements always present in a protein, and name one element found in some proteins but not all. [2]
(iii) Explain why two proteins built from the same set of small molecules can be completely different from each other. [1]
Model Answer — 6a
(i) amino acids [1]
(ii) carbon, hydrogen, oxygen and nitrogen are always present [1]
(ii) sulfur is present in some proteins [1]
(iii) because the amino acids are joined in a different order, so the chain folds into a different shape and the protein does a different job [1]
⚠ If you missed marks here: Nitrogen is the element that separates protein from carbohydrate and fat, and it must appear in the “always” list; sulfur belongs in the “sometimes” list and swapping them costs both marks. For the last part, the key word is order (sequence) — the same twenty building blocks in a different order give a completely different molecule, exactly as the same letters give different words.
(b)[4]
(i) Calculate the mass of protein in 100 g of the dried lentils. Show your working. [2]
(ii) The same method is applied to a sample of gelatine dessert powder that also contains added ammonium salts. Explain why the protein figure obtained would be too high. [2]
Model Answer — 6b
(i) mass of protein = 4.0 ÷ 0.16 [1]
(i) = 25 g of protein per 100 g [1]
(ii) the method assumes that all of the nitrogen in the sample comes from protein [1]
(ii) ammonium salts also contain nitrogen, so this extra nitrogen is counted as though it were protein and the result is an overestimate [1]
⚠ If you missed marks here: Multiplying by 0.16 instead of dividing gives 0.64 g and is the standard error — sanity-check the direction, because protein must be a larger mass than the nitrogen inside it. The second part is an assumption question: state the assumption first (all nitrogen comes from protein), then show how the added salts break it.
(c)[4]
A student is given two colourless solutions, one of egg white in water and one of starch that has been broken down into sugars. She has only biuret solution and Benedict’s solution.
(i) Describe how she should use both reagents to identify which solution is which, and give the results she expects. [3]
(ii) State why the biuret test does not need heating, unlike the Benedict’s test. [1]
Model Answer — 6c
(i) add biuret solution to a sample of each: the egg white solution turns purple, the sugar solution stays blue [1]
(i) add Benedict’s solution to a fresh sample of each and heat in a water bath: the sugar solution turns brick-red, the egg white solution stays blue [1]
(i) testing with both reagents confirms the identification, since each solution is positive for one test and negative for the other [1]
(ii) the biuret test gives its purple colour at room temperature, so no heating step is required [1]
⚠ If you missed marks here: Identification answers must give the result for both solutions in both tests — a negative result is evidence too. Using a fresh sample for the second test matters, because biuret solution contains sodium hydroxide and copper sulfate and would interfere with Benedict’s. And Benedict’s is the only one of the five food tests that requires heating.
Question 7 — Vitamin C on the Shelf
Total: 10 marks
A student measures the vitamin C content of freshly squeezed orange juice, then again after storing it in a clear bottle on a warm windowsill. She also tests a sample that was boiled for 10 minutes. Each time she records the volume of juice needed to decolourise 1.0 cm³ of DCPIP solution.
sample
mean volume of juice needed to decolourise the DCPIP / cm³
fresh
0.50
after 3 days
0.80
after 6 days
1.25
boiled for 10 minutes
2.50
(a)[4]
(i) Describe the trend shown by the stored samples. [1]
(ii) Explain what an increase in the volume of juice needed tells you about the juice. [1]
(iii) Calculate the percentage decrease in the vitamin C concentration of the juice after 6 days. Show your working. [2]
Model Answer — 7a
(i) the volume of juice needed increases the longer the juice is stored [1]
(ii) a larger volume is needed because the juice contains less vitamin C per cm³ — the relationship is inverse [1]
(iii) concentration after 6 days as a fraction of the fresh juice = 0.50 ÷ 1.25 = 0.40 [1]
(iii) so the concentration has fallen to 40% of the original, a decrease of 60% [1]
⚠ If you missed marks here: Everything in a DCPIP experiment is inverted: more volume needed means less vitamin C. The percentage calculation is the place candidates lose marks by computing (1.25 − 0.50) ÷ 0.50 = 150% — that is the increase in volume, not the decrease in concentration. Convert to concentration first by taking the ratio the other way round, then subtract from 100%.
(b)[3]
(i) The student concludes that boiling destroyed 80% of the vitamin C. Show the calculation that supports this figure. [1]
(ii) State one assumption she has made in using DCPIP to reach this conclusion. [1]
(iii) Suggest how the juice should have been stored to keep more of its vitamin C. [1]
Model Answer — 7b
(i) 0.50 ÷ 2.50 = 0.20, so 20% remains and 80% has been lost [1]
(ii) she assumes that only vitamin C in the juice decolourises DCPIP, and that a fixed mass of vitamin C always decolourises the same volume of DCPIP [1]
(iii) store it cold in a refrigerator, in a sealed opaque container, away from light and air [1]
⚠ If you missed marks here: An assumption is something taken for granted, not a source of error — write it as “she assumes that…”. The strongest one here is specificity: DCPIP is decolourised by vitamin C, but the conclusion only holds if nothing else in orange juice does the same. For storage, name at least two of cold, dark and sealed; one word rarely convinces.
(c)[3]
Give three ways the student could improve the reliability of her results. [3]
Model Answer — 7c
repeat each measurement several times and calculate a mean, discarding anomalous values [1]
use the same volume and concentration of DCPIP for every test, and add the juice from the same syringe in equal small drops [1]
keep all samples at the same temperature when tested, and filter the juice so that pulp does not obscure the colour change [1]
⚠ If you missed marks here: Reliability improvements must be things that reduce random variation — repeats and means, standardised volumes, consistent drop size. “Be more careful” and “use better equipment” are not creditable. Note the difference from validity: filtering the juice improves how clearly you can see the end-point, which is a fair criticism, but the mark-earning core is always repeats and controlled volumes.
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