IGCSE Biology Paper 4 (Theory / Extended)

Topic 4: Biological Molecules -- 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 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 — Is This Milk Powder Genuine?

Total: 12 marks

Genuine skimmed milk powder contains protein, a very little fat, and lactose, which is a reducing sugar. It contains no starch. A laboratory in Nairobi suspects that a batch has been bulked out with cornflour, which is almost pure starch.

(a) [4]

(i) Name the test that would detect this adulteration and describe how to carry it out. [2]

(ii) State the result expected for genuine milk powder and for adulterated milk powder. [2]

Model Answer — 1a

(i) the iodine test for starch [1]

(i) shake a sample of the powder with distilled water and add a few drops of iodine solution [1]

(ii) genuine powder: the iodine stays orange-brown [1]

(ii) adulterated powder: the iodine turns blue-black, showing starch is present [1]

⚠ If you missed marks here: Both results must be given, because the whole value of this test is that the honest sample and the adulterated one look different. Writing only “it goes blue-black if there is starch” leaves out the evidence that clears a genuine batch. Note also that the powder has to be shaken with water first — iodine dropped onto a dry powder gives an unreliable colour.

1 mark — iodine test for starch named 1 mark — sample shaken with distilled water then a few drops of iodine solution added 1 mark — genuine powder leaves the iodine orange brown 1 mark — adulterated powder turns the iodine blue black

(b) [4]

(i) Explain why the Benedict’s test and the biuret test would not reveal the adulteration. [2]

(ii) Describe a control the laboratory should include and explain what it shows. [2]

Model Answer — 1b

(i) genuine milk powder already contains lactose, a reducing sugar, so Benedict’s gives a positive result whether or not cornflour has been added [1]

(i) milk powder already contains protein, so biuret gives a purple colour either way; neither test can distinguish the two samples [1]

(ii) test a sample of milk powder known to be genuine, prepared and tested in exactly the same way [1]

(ii) it shows the colour expected when no starch is present, so any blue-black in the suspect sample can be attributed to added starch [1]

⚠ If you missed marks here: The reason these tests fail is not that they are unreliable but that both samples are positive — a test can only detect an adulterant that changes the result. The control here is a known-genuine sample, not distilled water: the comparison has to be against the real thing, otherwise you learn nothing about what normal milk powder looks like.

1 mark — milk already contains lactose so Benedict is positive either way 1 mark — milk already contains protein so biuret is purple either way and neither test distinguishes them 1 mark — control is a sample of known genuine milk powder treated identically 1 mark — the control shows the expected negative colour so blue black indicates added starch

The laboratory prepares mixtures of genuine powder and cornflour and tests each with the same volume of iodine solution.

cornflour added / %	0	5	10	20	40
colour with iodine	orange-brown	pale blue-grey	blue	dark blue-black	dark blue-black

The suspect batch gave a blue colour.

(i) Estimate the percentage of cornflour in the suspect batch. [1]

(ii) Give two reasons why this estimate should be treated with caution. [2]

(iii) Suggest one way of obtaining a more trustworthy figure. [1]

Model Answer — 1c

(i) about 10% cornflour [1]

(ii) the colour is judged by eye, so the comparison is subjective and depends on the lighting [1]

(ii) above about 20% the colour stops changing, so the scale cannot distinguish high percentages, and the steps between the known mixtures are wide [1]

(iii) measure the colour with a colorimeter and read the percentage from a calibration graph, or repeat the comparison with mixtures at smaller intervals [1]

⚠ If you missed marks here: Notice the ceiling in the data: 20% and 40% look identical, so this scale is only usable at the low end — spotting that saturation is the mark that separates strong answers. The other caution is the eye: a colour chart is a semi-quantitative tool, and semi-quantitative means an estimate, not a measurement.

1 mark — estimate of about 10 per cent cornflour 1 mark — colour is judged by eye so the comparison is subjective 1 mark — colour stops changing above about 20 per cent so high values cannot be distinguished 1 mark — use a colorimeter with a calibration graph or use mixtures at smaller intervals

Question 2 — Why Store Glucose as Glycogen?

Total: 12 marks

A fish stores carbohydrate in its liver as glycogen. A potato stores carbohydrate in its tuber as starch. Both are built from glucose, and both are insoluble in water.

You are told that water moves into a cell when the solution inside the cell is more concentrated than the solution outside it.

(a) [4]

Complete the account below by giving the missing information.

(i) The small molecules from which glycogen is built. [1]

(ii) The small molecules from which a protein is built. [1]

(iii) The elements present in glycogen. [1]

(iv) The additional element present in every protein but in no carbohydrate. [1]

Model Answer — 2a

(i) glucose [1]

(ii) amino acids [1]

(iii) carbon, hydrogen and oxygen [1]

(iv) nitrogen [1]

⚠ If you missed marks here: These are four one-word marks and they should all be automatic, but two errors recur: writing “glycogen is made of starch” (it is not — both are made of glucose) and offering sulfur as the element in every protein. Sulfur is in some proteins; nitrogen is in all of them, and that is what makes nitrogen the diagnostic element.

1 mark — glucose named as the subunit of glycogen 1 mark — amino acids named as the subunits of protein 1 mark — carbon hydrogen and oxygen present in glycogen 1 mark — nitrogen named as the extra element in every protein

(b) [4]

Glycogen molecules are very highly branched. Starch molecules are coiled with only a few branches. Cellulose molecules are long, straight and unbranched.

(i) Suggest why a highly branched storage molecule is useful to an active animal that must release glucose quickly. [2]

(ii) Explain how the shape of cellulose molecules suits them to a completely different job. [2]

Model Answer — 2b

(i) a branched molecule has many more ends (chain tips) exposed at once [1]

(i) so glucose can be released from many points at the same time, allowing a faster supply than from an unbranched chain [1]

(ii) straight unbranched chains can lie close together side by side and be held in bundles or fibres [1]

(ii) this makes cellulose strong and rigid, which suits it to forming plant cell walls rather than to storage [1]

⚠ If you missed marks here: “Branching makes it break down faster” is the conclusion, not the reason — the mark is for saying why: more exposed ends means more places to release glucose simultaneously. In the second part, do not stop at “cellulose is strong”; link the straightness to chains packing side by side, because that is the structural claim the examiner is checking.

1 mark — a branched molecule has many more exposed chain ends 1 mark — glucose can be released from many points at once giving a faster supply 1 mark — straight unbranched chains lie side by side in bundles or fibres 1 mark — this makes cellulose strong and rigid suiting it to cell walls not storage

Explain why animals and plants store carbohydrate as glycogen or starch rather than simply keeping a large amount of glucose dissolved in their cells. [4]

Model Answer — 2c

glucose is soluble, so storing a large amount of it would make the solution inside the cell much more concentrated [1]

water would then move into the cell, which could damage or burst it [1]

glycogen and starch are insoluble, so they have no such effect on the concentration inside the cell [1]

many glucose molecules are joined into one large molecule, so a great deal of carbohydrate is packed compactly into a small space and cannot leak out of the cell [1]

⚠ If you missed marks here: This question rewards a chain of reasoning, not a single fact. Start from solubility, move to concentration, then to water entering the cell, then to the fix: an insoluble polymer. Answers that simply state “starch is insoluble” collect one mark out of four. The compactness point — thousands of units in one molecule that cannot diffuse away — is the one most often missed.

1 mark — glucose is soluble so storing a lot of it makes the cell contents much more concentrated 1 mark — water would move into the cell and could damage or burst it 1 mark — glycogen and starch are insoluble so they do not change the concentration inside the cell 1 mark — many glucose units in one large molecule pack compactly and cannot leak out of the cell

Question 3 — Five Unlabelled Liquids

Total: 12 marks

A technician finds five bottles whose labels have fallen off. She knows that they contain, in some order: glucose solution, starch suspension, egg-white solution, cooking oil and vitamin C solution. She labels them P, Q, R, S and T and tests each one with all five food-test reagents.

liquid	iodine solution	Benedict’s solution, heated	biuret solution	ethanol emulsion test	DCPIP
P	orange-brown	brick-red	blue	clear	blue
Q	blue-black	blue	blue	clear	blue
R	orange-brown	blue	purple	clear	blue
S	orange-brown	blue	blue	cloudy white	blue
T	orange-brown	blue	blue	clear	decolourised

(a) [4]

Identify liquids P, Q, R, S and T, giving the evidence from the table for each.

Model Answer — 3a

P is the glucose solution — the only liquid giving a brick-red colour with heated Benedict’s solution [1]

Q is the starch suspension — the only liquid turning iodine blue-black [1]

R is the egg-white solution — the only liquid turning biuret purple, showing protein [1]

S is the cooking oil (cloudy white emulsion) and T is the vitamin C solution (decolourises DCPIP) [1]

⚠ If you missed marks here: Each identification must quote the one result that is uniquely positive; naming the liquids in the order they appear in the question is not evidence. Watch Q in particular: it is positive with iodine only, which tells you it is starch and simultaneously tells you it is not a reducing sugar, since Benedict’s stayed blue.

1 mark — P is glucose solution shown by the brick red Benedict result 1 mark — Q is starch suspension shown by the blue black iodine result 1 mark — R is egg white solution shown by the purple biuret result 1 mark — S is cooking oil from the cloudy white emulsion and T is vitamin C from the decolourised DCPIP

(b) [4]

(i) Explain why the technician needed to carry out more than one test on each liquid, even though each liquid gave one clearly positive result. [2]

(ii) Explain why she must take a fresh sample of a liquid for each test rather than adding the next reagent to the same tube. [2]

Model Answer — 3b

(i) a single positive result only shows that one named substance is present, not that the others are absent [1]

(i) the negative results are needed as evidence too — they rule out the other four possibilities and make each identification certain [1]

(ii) reagents already in the tube would mix with the next one and change the colour seen [1]

(ii) for example iodine is orange-brown and would mask the blue-to-purple change of the biuret test, giving a false reading [1]

⚠ If you missed marks here: The idea being tested is that negative results are data. A liquid identified from one positive result alone could be a mixture; the four negatives are what make the answer safe. For the second part, give a concrete example of interference rather than the vague “it would contaminate it”.

1 mark — one positive result shows only that one named substance is present 1 mark — negative results rule out the other possibilities and confirm the identification 1 mark — reagents left in the tube would mix with the next reagent and change the colour 1 mark — example of interference such as orange brown iodine masking the biuret colour change

The technician now receives a sixth bottle, U, and finds that it gives a brick-red colour with Benedict’s solution and a purple colour with biuret solution and a cloudy white emulsion.

(i) State what U contains. [1]

(ii) Suggest what kind of substance U might be, and explain your suggestion. [1]

(iii) Explain why the results for U show that the five food tests can be used on mixtures as well as on pure substances. [2]

Model Answer — 3c

(i) U contains a reducing sugar, protein and fat [1]

(ii) it could be a natural food such as milk, since milk contains all three of these nutrients [1]

(iii) each reagent reacts only with the one substance it detects, and takes no notice of the others present [1]

(iii) so several positive results in the same sample simply mean several nutrients are present together, which is what happens with almost every real food [1]

⚠ If you missed marks here: Multiple positive results are not a contradiction and not a sign of contamination — they are the normal picture for real food. The reasoning mark is for stating that each test is specific to one substance, so results add rather than interfere. Do not claim U contains starch: the iodine result is not given, and inventing data loses more than it gains.

1 mark — U contains a reducing sugar protein and fat 1 mark — suggestion of a real food such as milk containing all three nutrients 1 mark — each reagent detects only the substance it is specific for 1 mark — so several positive results simply mean several nutrients are present together

Question 4 — Rebuilding a Damaged DNA Strand

Total: 12 marks

A DNA molecule from a plant cell contains 4000 bases in total. Of these, 1200 are cytosine.

Part of a DNA molecule is shown below. Some bases on the second strand have been lost.

strand 1	A	C	G	T	T	G	C	A
strand 2	T		C			C		T

(a) [4]

(i) Write out the complete base sequence of strand 2. [2]

(ii) State the rule you used. [1]

(iii) Explain why the lost bases could be worked out with certainty, whereas a message written in a language you do not know could not be reconstructed. [1]

Model Answer — 4a

(i) the four missing bases are G, A, A and G in that order [1]

(i) the complete strand 2 reads T G C A A C G T [1]

(ii) adenine always pairs with thymine and cytosine always pairs with guanine [1]

(iii) because each base has only one possible partner, so the second strand is completely determined by the first — there is no choice to guess at [1]

⚠ If you missed marks here: Work along the strand one base at a time and write the partner directly beneath, rather than trying to hold the sequence in your head. The most common error is pairing A with C or G with T. The final mark is a reasoning mark: the pairing rule is one-to-one, so the information on one strand is enough to rebuild the other exactly.

1 mark — missing bases identified as G A A and G 1 mark — complete strand 2 written as TGCAACGT 1 mark — rule stated that adenine pairs with thymine and cytosine pairs with guanine 1 mark — each base has only one possible partner so the second strand is fully determined

(b) [4]

Use the information about the 4000-base molecule.

(i) State the number of guanine bases and explain your answer. [2]

(ii) Calculate the number of adenine bases. Show your working. [2]

Model Answer — 4b

(i) guanine = 1200 [1]

(i) because cytosine always pairs with guanine, so the two are present in equal numbers [1]

(ii) cytosine + guanine = 2400, so adenine + thymine = 4000 − 2400 = 1600 [1]

(ii) adenine and thymine are equal in number, so adenine = 1600 ÷ 2 = 800 [1]

⚠ If you missed marks here: Two divisions are needed and skipping either one is fatal: first subtract both C and G from the total, then halve what is left. Answering 2800 (4000 − 1200) forgets guanine; answering 1600 forgets that adenine and thymine share the remainder equally. Setting the work out as four lines — C, G, A + T, A — makes the slip impossible.

1 mark — guanine is 1200 1 mark — because cytosine always pairs with guanine so they are equal in number 1 mark — cytosine plus guanine is 2400 so adenine plus thymine is 1600 1 mark — adenine is 800

(i) Describe how the two strands of a DNA molecule are arranged and what holds them together. [3]

(ii) A student says “the two strands are joined by bonds between the two long strands themselves”. Explain why this is not correct. [1]

Model Answer — 4c

(i) the two strands are coiled around each other to form a double helix [1]

(i) each strand carries a sequence of bases along its length [1]

(i) bonds between the pairs of bases hold the two strands together [1]

(ii) the bonds form specifically between a base on one strand and its partner base on the other, not between the strands themselves [1]

⚠ If you missed marks here: The syllabus wording is worth learning exactly: two strands, coiled into a double helix, bases along each strand, bonds between pairs of bases holding the strands together. The student’s version fails on the last point, and that is the detail examiners test, because it is what makes the pairing rule structurally meaningful.

1 mark — the two strands are coiled around each other into a double helix 1 mark — each strand carries a sequence of bases along its length 1 mark — bonds between the pairs of bases hold the two strands together 1 mark — the bonds are between paired bases not between the strands themselves

Question 5 — What Chewing Does to Bread

Total: 10 marks

A student chews identical pieces of bread for different lengths of time and spits each sample into a labelled tube. She is told that saliva contains a substance that breaks starch down into a reducing sugar. Her results are shown.

time chewed / s	result with iodine solution	result with Benedict’s solution
0	blue-black	stays blue
30	blue-black	green
60	blue	orange
120	orange-brown	brick-red

(a) [4]

Describe how the student should test each sample so that the two sets of results are valid and comparable.

Model Answer — 5a

divide each sample into two equal portions, because a sample tested with iodine cannot then be used for Benedict’s [1]

add a few drops of iodine solution to the first portion and record the colour [1]

add an equal volume of Benedict’s solution to the second portion and heat it in a water bath for the same length of time as all the others [1]

include a control of unchewed bread mixed with distilled water, and keep the mass of bread and the volume of reagents the same each time [1]

⚠ If you missed marks here: The splitting step is the mark that decides this question — the two tests cannot share a tube, because the strong orange-brown of iodine masks the Benedict’s colours completely. After that, the marks are ordinary control variables: same mass of bread, same volume of reagent, same heating time. A control with no saliva is what shows the change is caused by chewing.

1 mark — each sample divided into two equal portions because one sample cannot be used for both tests 1 mark — iodine solution added to the first portion and the colour recorded 1 mark — equal volume of Benedict solution added to the second portion and heated in a water bath for the same time 1 mark — control of unchewed bread with distilled water and same mass and volumes throughout

(b) [3]

(i) Describe what the iodine results show. [1]

(ii) Describe what the Benedict’s results show. [1]

(iii) Explain how the two sets of results together support the statement made about saliva. [1]

Model Answer — 5b

(i) the amount of starch falls the longer the bread is chewed, until after 120 s no starch remains [1]

(ii) the amount of reducing sugar increases the longer the bread is chewed [1]

(iii) starch disappears at the same time as reducing sugar appears, which is exactly what is expected if the starch is being broken down into reducing sugar [1]

⚠ If you missed marks here: Do not simply copy the colours out of the table — translate each colour into an amount of a named substance. The third mark is for the link: it is the mirror-image shape of the two trends, one falling as the other rises, that turns a pair of observations into evidence for a single explanation.

1 mark — amount of starch falls with chewing time until none remains 1 mark — amount of reducing sugar increases with chewing time 1 mark — starch disappears as reducing sugar appears which is what breaking starch down would produce

(i) Explain why a positive Benedict’s result alone would not have been enough to support the statement about saliva. [2]

(ii) The student writes that after 120 s “the iodine did nothing”. Give the correct way to record this observation. [1]

Model Answer — 5c

(i) bread might already have contained some reducing sugar before chewing, so a positive result on its own does not prove any starch was broken down [1]

(i) the falling iodine result is needed as well, to show that starch actually disappeared [1]

(ii) the iodine solution stayed orange-brown, showing that no starch was present [1]

⚠ If you missed marks here: “It did nothing” and “there was no change” are not observations — a negative result has a colour, and the colour is orange-brown. Examiners want the colour stated and then interpreted. The first two marks are about evidence: one measurement showing a substance appearing is much weaker than two showing one substance replacing another.

1 mark — bread might already contain reducing sugar so a positive Benedict result alone proves nothing 1 mark — the falling iodine result is needed to show that starch actually disappeared 1 mark — correct record is that the iodine stayed orange brown showing no starch

Question 6 — Checking a Vitamin C Tablet

Total: 12 marks

A vitamin C tablet is labelled “contains 500 mg of vitamin C”. A student crushes one tablet, dissolves it in distilled water and makes the solution up to exactly 250 cm³.

She finds that 1.00 cm³ of a standard vitamin C solution of concentration 1.0 mg/cm³ is needed to decolourise 1.0 cm³ of DCPIP solution.

(a) [4]

(i) Calculate the concentration of the tablet solution, in mg/cm³, if the label is correct. Show your working. [2]

(ii) Predict the volume of the tablet solution that should decolourise 1.0 cm³ of the same DCPIP solution. Show your working. [2]

Model Answer — 6a

(i) concentration = 500 ÷ 250 [1]

(i) = 2.0 mg/cm³ [1]

(ii) 1.00 cm³ of the 1.0 mg/cm³ standard contains 1.0 mg of vitamin C, so 1.0 mg is the mass needed to decolourise this DCPIP [1]

(ii) volume needed = 1.0 ÷ 2.0 = 0.50 cm³ [1]

⚠ If you missed marks here: The hinge of the whole question is the sentence “1.0 mg of vitamin C decolourises this DCPIP”; without writing it down, the prediction becomes guesswork. Also check the direction: a solution twice as concentrated as the standard needs half the volume, so any predicted answer above 1.00 cm³ must be wrong.

1 mark — working shown as 500 divided by 250 1 mark — concentration 2.0 mg per cm3 1 mark — recognises that 1.0 mg of vitamin C decolourises this volume of DCPIP 1 mark — predicted volume 0.50 cm3

(b) [4]

When she carries out the test, the mean volume of tablet solution needed is 0.80 cm³.

(i) Calculate the actual mass of vitamin C in the tablet. Show your working. [2]

(ii) Comment on the manufacturer’s claim. [1]

(iii) Suggest one explanation for the difference other than the claim being untrue. [1]

Model Answer — 6b

(i) actual concentration = 1.0 ÷ 0.80 = 1.25 mg/cm³ [1]

(i) mass in the tablet = 1.25 × 250 = 312.5 mg (about 310 mg) [1]

(ii) the tablet contains clearly less vitamin C than the 500 mg claimed — only about 63% of it [1]

(iii) any sensible alternative, such as the tablet having lost vitamin C during storage, some of the crushed tablet not dissolving completely, or errors in judging the end-point [1]

⚠ If you missed marks here: The two-step calculation must go concentration first, then mass in the whole 250 cm³ — stopping at 1.25 mg/cm³ answers a question that was not asked. In the last part, an alternative explanation means a reason the measurement could understate the truth, not a criticism of the manufacturer; storage losses and incomplete dissolving are the two strongest.

1 mark — actual concentration calculated as 1.0 divided by 0.80 giving 1.25 mg per cm3 1 mark — mass in the tablet 312.5 mg 1 mark — comment that the tablet contains less vitamin C than the 500 mg claimed 1 mark — alternative explanation such as loss during storage incomplete dissolving or end point error

Her teacher suggests that instead of using a single standard she should prepare a series of vitamin C solutions of known concentration.

(i) Describe how she could prepare solutions of 0.8, 0.6, 0.4 and 0.2 mg/cm³ from the 1.0 mg/cm³ standard. [2]

(ii) Explain the advantage of using a series rather than a single standard. [2]

Model Answer — 6c

(i) mix measured volumes of the standard with measured volumes of distilled water, for example 8 cm³ of standard with 2 cm³ of water to give 0.8 mg/cm³ [1]

(i) keep the total volume the same each time (6 + 4, 4 + 6 and 2 + 8 for the other three) and use a syringe or pipette to measure accurately [1]

(ii) testing several known concentrations allows a calibration graph of concentration against volume of DCPIP decolourised to be drawn [1]

(ii) an unknown can then be read off the line, and any anomalous point can be spotted, which a single standard could never reveal [1]

⚠ If you missed marks here: Dilution answers must specify both volumes and keep the total constant — “add water until it is weaker” earns nothing. For the advantage, the key idea is that a series produces a line, and a line both interpolates unknown values and exposes anomalies. One standard gives you a single point and no way of knowing whether it was a good one.

1 mark — measured volumes of standard mixed with measured volumes of distilled water such as 8 to 2 1 mark — total volume kept the same each time and volumes measured with a syringe or pipette 1 mark — a series allows a calibration graph of concentration against volume to be drawn 1 mark — unknown values can be read from the line and anomalies spotted

Question 7 — Reading a Cereal Packet

Total: 10 marks

The table gives the nutrition information printed on two breakfast cereals, per 100 g.

	cereal A	cereal B
carbohydrate / g	78	62
of which sugars / g	8	34
of which fibre, almost all cellulose / g	10	3
protein / g	10	7
fat / g	2	9
vitamin C added	yes	no

(a) [4]

(i) Predict which cereal gives the stronger Benedict’s result and justify your prediction with data. [2]

(ii) Predict the iodine result for both cereals and explain. [1]

(iii) Predict which cereal gives the more obvious cloudy white emulsion. [1]

Model Answer — 7a

(i) cereal B gives the stronger result [1]

(i) because it contains far more sugars, 34 g compared with 8 g per 100 g [1]

(ii) both turn iodine blue-black, because in both cereals most of the carbohydrate is starch rather than sugar or fibre [1]

(iii) cereal B, which contains 9 g of fat per 100 g compared with 2 g [1]

⚠ If you missed marks here: Predictions must quote figures; naming a cereal without the number is only half an answer. Take care with the iodine prediction: the label does not print a “starch” line, so you have to work it out — carbohydrate minus sugars minus fibre leaves 60 g and 25 g of starch respectively, so both are strongly positive.

1 mark — cereal B predicted to give the stronger Benedict result 1 mark — because it contains 34 g of sugars compared with 8 g 1 mark — both cereals turn iodine blue black because most of their carbohydrate is starch 1 mark — cereal B gives the more obvious emulsion with 9 g of fat compared with 2 g

(b) [3]

A sample of purified cereal fibre is tested on its own. It gives a negative result with iodine solution and a negative result with Benedict’s solution.

Explain how fibre can be counted as a carbohydrate on the label yet give both of these negative results. [3]

Model Answer — 7b

fibre is almost all cellulose, which is a carbohydrate because it is built from glucose and contains only carbon, hydrogen and oxygen [1]

iodine solution tests specifically for starch, and cellulose is not starch, so the iodine stays orange-brown [1]

Benedict’s solution tests for reducing sugars, and the glucose units in cellulose are locked into long chains rather than being free sugar molecules, so it stays blue [1]

⚠ If you missed marks here: This is the clearest example in the topic of why “negative test” never means “absent nutrient”. Cellulose is unmistakably a carbohydrate, yet it fails both carbohydrate tests, because each test is specific to one form. State what each reagent actually detects, then say why cellulose falls outside it.

1 mark — fibre is cellulose which is a carbohydrate built from glucose containing carbon hydrogen and oxygen 1 mark — iodine tests specifically for starch and cellulose is not starch so it stays orange brown 1 mark — Benedict tests for reducing sugars and the glucose in cellulose is locked into chains so it stays blue

A student tries to confirm the added vitamin C by adding DCPIP solution to a bowl of cereal A mixed with milk. The result is impossible to read.

(i) Explain why. [2]

(ii) Describe how she should prepare the sample instead. [1]

Model Answer — 7c

(i) milk is opaque and white, so the blue colour of the DCPIP and its disappearance cannot be seen [1]

(i) the solid pieces of cereal also make the mixture cloudy, hiding any colour change [1]

(ii) shake a weighed sample of the dry cereal with distilled water, filter the mixture, and add the clear filtrate to the DCPIP drop by drop [1]

⚠ If you missed marks here: Every DCPIP question comes back to whether the end-point can be seen. Opaque or strongly coloured samples must be filtered and diluted with distilled water, never milk, because milk both hides the colour and adds its own nutrients to the tube. Naming the filtration step is the mark here.

1 mark — milk is opaque and white so the blue DCPIP and its disappearance cannot be seen 1 mark — solid cereal pieces also make the mixture cloudy and hide the colour change 1 mark — shake weighed dry cereal with distilled water filter it and add the clear filtrate to the DCPIP

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