⚡ Challenge Paper Preparation

Challenge Prep: Biological Molecules

IGCSE Biology 0610 — Topic 4

Every exam paper in this topic is a challenge paper, so this page does the ramping. Topic 4 is small, which means examiners cannot hide behind obscure content — they test precision instead. The marks disappear in the same dozen places every year: DCPIP that “went clear”, a blue Benedict’s tube read as “no carbohydrate”, a missing water bath, ethanol and water poured in the wrong order, a green Benedict’s result dismissed as negative, a vitamin C comparison read upside down, and a base-pairing sum where nobody remembered to halve. All of them are below — hunted down, explained, and drilled until the food tests feel like free marks.

📋 The Complete Food-Test Reference Table

Learn every row as a sentence: reagent, how it is added, the condition, the colour BEFORE and the colour AFTER, and the negative result. Cover the right-hand columns and recite them.

nutrient	reagent	how it is done	POSITIVE (colour change)	NEGATIVE
starch	iodine solution	add a few drops to the sample; no heating	orange-brown → blue-black	stays orange-brown
reducing sugar	Benedict’s solution	add an excess, then heat in a water bath at about 80 °C for a few minutes	blue → green → yellow → orange → brick-red (semi-quantitative)	stays blue
protein	biuret solution	add and shake gently; no heating	blue → purple (violet / lilac)	stays blue
fat or oil	ethanol, then distilled water	ethanol FIRST — shake, let the solid settle, then decant the ethanol into a second tube of water	cloudy white emulsion forms	stays clear and colourless
vitamin C	DCPIP solution	fixed volume of DCPIP; add juice drop by drop from a syringe, shaking, until the blue just disappears	the blue DCPIP is decolourised (blue → colourless)	stays blue

Three facts that sit outside the table and are worth marks on their own: Benedict’s is the only test that is heated; the emulsion test is the only one with no colour change; and DCPIP is the only one where a bigger measured volume means less of the nutrient.

⚠️ Common Traps & Misconceptions

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Twelve traps that cost marks on Topic 4 challenge papers. Most of them are wording, not knowledge — which is exactly why they are so easy to fix.

⚠️ TRAP

Trap 1: “The DCPIP went clear”

The TrapIt sounds like a perfectly good observation and it is written in thousands of scripts every session. Blue DCPIP goes “clear”, the candidate concludes vitamin C is present, and the mark is refused.

The TruthClear means transparent — you can see through it. The DCPIP was already clear when it was blue, so saying it went clear describes no change at all. Colourless means having no colour. Decolourised means a colour that was there has been removed. The accepted answers are “the DCPIP is decolourised” or “it changes from blue to colourless”.

Why It MattersMark schemes print “colourless — not clear” in bold, and the same distinction is tested in Benedict’s and biuret questions too. One imprecise word deletes an observation you actually made correctly at the bench.

Example Question“State what is seen when orange juice is added to blue DCPIP solution. [1]”

⚠️ TRAP

Trap 2: Reading a blue Benedict’s tube as “no carbohydrate”

The TrapThe tube stays blue and the candidate writes “so the food contains no carbohydrate”. It feels like a safe, general conclusion. It is wrong, and it is probably the single most expensive sentence in Topic 4.

The TruthBenedict’s solution detects reducing sugars only. Starch and cellulose are both carbohydrates and both leave Benedict’s solution completely blue. A blue tube licenses exactly one statement: no reducing sugar is present. Every food test is specific to one named substance, and a negative result rules out that substance and nothing else.

Why It MattersThis appears in nearly every table-interpretation question, usually as the most tempting distractor in a multiple choice, and as an easy two-mark explain question on Paper 4.

Example Question“A student finds that Benedict’s solution stays blue with a sample of pasta. Explain why this does not show that pasta contains no carbohydrate. [2]”

⚠️ TRAP

Trap 3: Heating Benedict’s solution in a Bunsen flame instead of a water bath

The Trap“Heat the tube until the colour changes.” Vague, and usually accompanied by a Bunsen burner in the diagram. Both the method mark and, in a practical, the safety mark go with it.

The TruthBenedict’s must be heated in a water bath at about 80 °C for a few minutes. Two reasons are creditable: direct heating makes the liquid boil suddenly and spit out of the tube, which is dangerous; and a water bath brings every tube to the same temperature, so different foods can be compared fairly.

Why It Matters“Describe how to test a food for reducing sugar” is worth three or four marks and the condition is always one of them. It is also the only heating step anywhere in the five tests, so it is the one detail that distinguishes the tests from each other.

Example Question“Describe how you would test a sample of apple juice for reducing sugar. [3]”

⚠️ TRAP

Trap 4: Adding water before ethanol in the emulsion test

The Trap“Put the food in a tube with water, shake it, then add ethanol.” It reads like a sensible order — dissolve the food first — and it destroys the test.

The TruthEthanol first. Fat is insoluble in water but soluble in ethanol, so it must be taken into ethanol before anything else happens. Only then is the ethanol poured into water, where the fat can no longer stay dissolved and comes out as tiny droplets. Reverse the order and the fat never dissolves, so the test cannot work as designed.

Why It MattersMethod questions on this test are worth three or four marks and the order carries at least one of them. It is also the fault most often planted in “what did the student do wrong?” questions.

Example Question“A student adds water to the food, shakes it, then adds ethanol. Identify the error and explain its effect. [2]”

⚠️ TRAP

Trap 5: Not decanting — the false positive that fools everyone

The TrapThe whole food-and-ethanol mixture is tipped into the water, the tube goes cloudy, and the candidate reports fat. The same thing happens with a fat-free food, and nobody notices.

The TruthUndissolved food particles scatter light exactly as fat droplets do. The mixture must be left to settle and only the clear ethanol decanted into a separate tube of distilled water. If solid is carried across, the cloudiness is a false positive and proves nothing. Note the wording too: it is an emulsion, a suspension of droplets, not a precipitate — nothing has reacted.

Why It MattersChallenge papers love this one because it looks like a result and is actually an artefact. Recognising a false positive is a genuine evaluation skill and is worth two marks on its own.

Example Question“Both the groundnut paste and the fat-free spread gave a cloudy tube. Explain how this could happen. [2]”

⚠️ TRAP

Trap 6: Heating the biuret test

The TrapBecause Benedict’s solution is blue and biuret solution is blue, the two blur together in revision and the water bath migrates from one to the other.

The TruthThe biuret test needs no heating at all — the purple colour develops at room temperature. Benedict’s is the only one of the five tests that is heated. Learn the tests as five separate procedures with five separate condition lines, and this cannot happen.

Why It MattersIn a “describe the test” question an unnecessary heating step is not neutral; it contradicts the method mark and can cost it. It also signals to the examiner that the tests have been learned as a blur.

Example Question“Describe how you would test a solution of egg white for protein. [2]”

⚠️ TRAP

Trap 7: Giving the final colour and not the change

The Trap“It goes blue-black.” “It goes purple.” “It goes brick-red.” Each of these is half of an observation and usually scores half of nothing.

The TruthObservation marks are written as changes: iodine orange-brown → blue-black; biuret blue → purple; Benedict’s blue → brick-red; DCPIP blue → colourless. Get into the habit of writing an arrow between two colours every single time, and state the negative result as the reagent staying its original colour.

Why It MattersThis is the cheapest mark in the topic to gain and the cheapest to lose, and it appears in Papers 2, 4 and 6 alike. It also protects you against mixing up which reagent starts blue.

Example Question“State the colour change seen when iodine solution is added to a food containing starch. [1]”

⚠️ TRAP

Trap 8: Calling a green Benedict’s result negative

The Trap“It only went green, not brick-red, so there is no sugar.” The candidate has learned one colour instead of a scale.

The TruthAny colour other than blue is a positive result. The ladder runs blue → green → yellow → orange → brick-red as the amount of reducing sugar increases. Green means a small amount, brick-red means a lot. That is precisely what makes the test semi-quantitative: the colour estimates the quantity, judged by eye, without giving an exact number.

Why It MattersData questions frequently give a table of colours across several samples or several days, and the whole interpretation collapses if green is scored as zero.

Example Question“Sample D turned green and sample B turned brick-red. Compare the amounts of reducing sugar in the two samples. [2]”

⚠️ TRAP

Trap 9: Reading the DCPIP comparison upside down

The TrapJuice X needs 2.0 cm³ and juice Y needs 0.5 cm³. Candidates reach for “X used more, so X has more vitamin C” — and reverse the conclusion of the whole question.

The TruthThe relationship is inverse. A fixed volume of DCPIP always needs the same mass of vitamin C to decolourise it, so the juice that supplies that mass in a smaller volume must be more concentrated. Here Y is four times as concentrated as X. Concentration calculations follow the same logic: concentration = (mass that decolourises the dye) ÷ (volume needed).

Why It MattersEvery DCPIP question ends in a comparison or a calculation, and this inversion is the difference between full marks and none. Percentage-change questions compound it: work out the concentration ratio first, then the percentage.

Example Question“Fresh juice needed 0.50 cm³ and stored juice needed 1.25 cm³. Calculate the percentage decrease in vitamin C concentration. [2]”

⚠️ TRAP

Trap 10: Forgetting to halve in a base-pairing calculation

The Trap“18% is adenine, so 82% is guanine.” Or, one step better but still wrong, “64% is guanine.” Both answers come from stopping too early.

The TruthTwo equalities, applied in order. First, %A = %T, so 18% adenine means 18% thymine and 36% between them. Second, the remaining 64% belongs to cytosine and guanine together, and since %C = %G they share it equally: guanine = 32%. Write the four bases as four lines before touching the calculator and the halving step cannot be forgotten.

Why It MattersBase arithmetic appears on almost every Extended paper that touches DNA, either as a percentage question or as a number-of-bases question. The same two-step method solves all of them.

Example Question“A DNA sample contains 4000 bases, of which 1200 are cytosine. Calculate the number of adenine bases. [2]”

⚠️ TRAP

Trap 11: Explaining a difference between two glucose polymers by inventing different elements

The Trap“Cellulose cannot be digested because it contains a different element.” It sounds chemical and confident, and it is simply untrue.

The TruthStarch, glycogen and cellulose all contain only carbon, hydrogen and oxygen, and all three are built from glucose. Every difference between them comes from how the glucose units are joined and how the chains are arranged — coiled with a few branches in starch, heavily branched in glycogen, straight and packed side by side in cellulose. The substance that unpicks the links in starch does not fit those in cellulose.

Why It MattersThis is the standard four-mark explain question in the topic, and it is also the reasoning behind “why is cellulose good for cell walls?” and “why is glycogen branched?”.

Example Question“Cotton and potato starch are both made only of glucose. Explain why humans can digest one but not the other. [2]”

⚠️ TRAP

Trap 12: Treating a fat as a polymer, or forgetting the one-to-three ratio

The Trap“A fat is made of many fatty acids joined together.” The candidate has assumed that because fat is a store like starch, it must be built like starch.

The TruthA fat is built from one glycerol molecule and three fatty acid molecules — four molecules in total. It is not a long chain of repeating units, which is why the phrase “built from many small molecules” belongs to starch, glycogen, cellulose and proteins but not to fats. Both components must be named and the ratio stated for full marks.

Why It Matters“Name the products from which a fat is made” is a routine two-marker, and the second mark is almost always the one-to-three ratio. It also appears as a discriminating idea in comparison questions.

Example Question“Name the small molecules from which a fat is built and state how many of each combine. [2]”

🧩 Multi-Step Reasoning Walkthroughs

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Six challenge-level questions broken into steps. Try each step yourself before revealing the next one.

Walkthrough 1 — Four Unlabelled Samples, One Results TableFour food samples A–D are tested. A: iodine blue-black, all other tests negative. B: Benedict’s brick-red, DCPIP decolourised, all others negative. C: biuret purple and a cloudy white emulsion, all others negative. D: positive for iodine, Benedict’s (green), biuret and the emulsion test. (a) Identify the molecules in each sample. [4] (b) Explain why the negative Benedict’s result for A does not prove A contains no carbohydrate. [2] (c) Explain what D’s green Benedict’s result adds to the picture. [2]

A negative result is data

Go down the column for each sample and turn every cell into a sentence. Blue-black iodine means starch present. Benedict’s staying blue means no reducing sugar — not “no carbohydrate”. Biuret staying blue means no protein. A clear emulsion tube means no fat. DCPIP staying blue means no vitamin C. Only when all five cells are translated should you name the sample.

One positive, four negatives

A = starch only. B = a reducing sugar and vitamin C. C = protein and fat. D = starch, a reducing sugar, protein and fat. Each identification must quote the result that supports it — naming the nutrient without the evidence is only half an answer in a Paper 4 mark scheme.

Specific tests, specific conclusions

Benedict’s detects reducing sugars only. Starch is a carbohydrate that it cannot see, and sample A gave a strongly positive iodine result, so A is in fact rich in carbohydrate. The blue tube licenses one statement and one only: no reducing sugar.

Green is positive, and it is a quantity

Green is a positive Benedict’s result showing a small amount of reducing sugar, whereas B’s brick-red shows a large amount. That comparison is possible because the test is semi-quantitative: the colour estimates how much is present, judged by eye, without giving a number.

Full Mark-Scheme Answer(a) A starch only, from blue-black iodine with all other tests negative [1]; B a reducing sugar and vitamin C, from brick-red Benedict’s and decolourised DCPIP [1]; C protein and fat, from purple biuret and a cloudy white emulsion [1]; D starch, a reducing sugar, protein and fat [1]. (b) Benedict’s detects reducing sugars only [1], and starch is a carbohydrate that gives no result with it — A tested positive for starch with iodine [1]. (c) D contains less reducing sugar than B [1]; the colour ladder estimates the amount, which is what semi-quantitative means [1].

Walkthrough 2 — A Vitamin C Comparison, End to EndA class compares guava and lime juice using DCPIP. A standard vitamin C solution of 1.0 mg/cm³ needs 1.00 cm³ to decolourise 1.0 cm³ of DCPIP. Guava needs 0.80 cm³ and lime needs 2.00 cm³. (a) Describe the method, including the end-point. [4] (b) Calculate the concentration of vitamin C in each juice. [2] (c) State the conclusion and one improvement. [2]

The DCPIP is the fixed quantity

Measure a fixed volume of DCPIP into a test tube. Add the juice from a syringe, drop by drop, shaking after each addition. Record the volume of juice needed for the blue colour to disappear completely. Repeat with each liquid and take a mean. Adding DCPIP to the juice is the standard reversal and loses the method marks.

Every tube is decolourised by the same mass

The standard tells you that 1.00 cm³ × 1.0 mg/cm³ = 1.0 mg of vitamin C decolourises this volume of DCPIP. Because the DCPIP volume never changes, every juice volume in the table also carries 1.0 mg. Writing that sentence down is what makes the calculation trivial.

Concentration = mass ÷ volume

Guava: 1.0 ÷ 0.80 = 1.25 mg/cm³. Lime: 1.0 ÷ 2.00 = 0.50 mg/cm³. Sanity-check the direction: guava needed the smaller volume, so guava must come out the stronger juice. It does — 2.5 times stronger.

Say the number, then improve the method

Conclusion: guava juice contains about 2.5 times as much vitamin C per cm³ as lime juice. Improvements that earn credit: more repeats and a mean; smaller measured drops near the end-point; filtering the juice so that pulp does not obscure the colour change.

Full Mark-Scheme Answer(a) Fixed volume of DCPIP [1]; juice added drop by drop from a syringe with shaking [1]; volume of juice needed to decolourise recorded [1]; repeated and a mean taken [1]. (b) Each decolourising volume contains 1.0 mg of vitamin C [1]; guava 1.25 mg/cm³ and lime 0.50 mg/cm³ [1]. (c) Guava contains about 2.5 times as much vitamin C per cm³ [1]; one valid improvement [1].

Walkthrough 3 — Base Pairing: From a Percentage to a Number of BasesA DNA sample contains 3 200 000 bases, of which 18% are adenine. (a) State the percentage of thymine and explain. [2] (b) Calculate the number of guanine bases. [2] (c) A second sample is reported as 30% adenine and 25% thymine. Explain why this must be wrong. [2]

A, T, C, G — four lines

Before any arithmetic, write the four bases as four lines and fill in what you are given. This makes the two equalities visible: %A = %T and %C = %G. Almost every error in this calculation comes from doing it in your head.

Adenine and thymine are locked together

Adenine pairs only with thymine, so thymine = 18% as well, and A + T = 36%. The explanation mark is for the pairing rule, not for the number — say “because adenine always pairs with thymine, so they are present in equal numbers”.

The step everyone forgets

100 − 36 = 64% for cytosine and guanine together. Since %C = %G, guanine = 64 ÷ 2 = 32%. Then convert to a number: 0.32 × 3 200 000 = 1 024 000 guanine bases. Answers of 82% or 64% are the two standard stopping-too-early errors.

Adding to 100 is not enough

Adenine can only pair with thymine, so their percentages must be equal. 30% and 25% are not, so the analysis contains an error or the sample is contaminated. Notice that the figures could still add to 100 with the other two bases — validity here is about the pairing rule, not the total.

Full Mark-Scheme Answer(a) Thymine = 18% [1], because adenine always pairs with thymine so the two are present in equal numbers [1]. (b) C + G = 64%, so guanine = 32% [1]; 0.32 × 3 200 000 = 1 024 000 [1]. (c) Adenine pairs only with thymine so their percentages must be equal [1]; 30 and 25 are unequal, so there is an error or contamination [1].

Walkthrough 4 — Marking a Flawed Emulsion TestA student writes: “I put a little food into a test tube, added 2 cm³ of distilled water and shook it hard. Then I added 2 cm³ of ethanol to the same tube and looked for a cloudy layer.” Both his groundnut paste and his “fat free” spread went cloudy. (a) Give the correct method and the result in both directions. [4] (b) Identify two errors and their effects. [4] (c) Explain why the emulsion forms and suggest a control. [3]

Ethanol, settle, decant, water

Add the food to ethanol and shake so any fat dissolves. Let the solid settle. Decant the ethanol into a second tube of distilled water. Positive: a cloudy white emulsion. Negative: the mixture stays clear and colourless.

Water first means the fat never dissolves

Water was added before ethanol. Fat is insoluble in water, so it is never taken into solution and the test cannot work as designed. Naming the error alone is half a mark pair — the consequence has to be stated.

No decanting means a false positive

The ethanol went into the same tube instead of being decanted into clean water. Undissolved food particles stay in the mixture and cloud it on their own, so a fat-free food looks positive. That is exactly why both of his tubes went cloudy — the result is an artefact, not a measurement.

Physical, not chemical

Fat dissolves in ethanol but is insoluble in water, so when the two mix the fat comes out of solution as tiny droplets suspended in the water, which scatter light and look milky white. Nothing has reacted, so “precipitate” is the wrong word. Control: run the whole procedure with ethanol and water but no food; it should stay clear and colourless.

Full Mark-Scheme Answer(a) Food added to ethanol and shaken [1]; ethanol decanted into a second tube of water [1]; positive = cloudy white emulsion [1]; negative = stays clear and colourless [1]. (b) Water added before ethanol [1] so the fat never dissolves [1]; ethanol not decanted [1] so food particles give a false positive [1]. (c) Fat is soluble in ethanol but insoluble in water [1] so it separates as tiny suspended droplets that scatter light [1]; control of ethanol and water with no food [1].

Walkthrough 5 — Identifying a Molecule From Combustion Data1.00 g samples of three purified substances are burned in excess oxygen. X gives 1.47 g CO₂ and 0.60 g H₂O only. Y gives 1.31 g CO₂, 0.51 g H₂O, oxides of nitrogen and a trace of SO₂. Z gives 2.85 g CO₂ and 1.15 g H₂O only. Identify each and justify. [4]

Nitrogen means protein

Only Y produces nitrogen-containing gases, and nitrogen is present in every protein and in no carbohydrate or fat. So Y is the protein. The trace of sulfur dioxide confirms it, because sulfur is found in some proteins and nothing else on this syllabus — though its absence would prove nothing.

Same three elements, both

X and Z both give only carbon dioxide and water, so both contain only carbon, hydrogen and oxygen. At this point the elements have run out of information: carbohydrates and fats share exactly the same three. The next clue has to be quantitative.

Fats are richer in carbon and hydrogen

Per gram, Z produces roughly twice as much carbon dioxide and water as X. Carbon dioxide comes from the carbon and water from the hydrogen, so Z contains a much higher proportion of both — and correspondingly less oxygen. That is the composition of a fat, so Z is the fat and X is the carbohydrate.

Every identification needs its evidence

The mark scheme awards the identification and the reason separately. Say “Y is a protein because oxides of nitrogen show nitrogen is present” and “Z is a fat because the same mass releases far more carbon dioxide and water, showing a higher proportion of carbon and hydrogen”.

Full Mark-Scheme AnswerX is a carbohydrate [1]; Y is a protein, shown by the oxides of nitrogen (and confirmed by the sulfur dioxide) [1]; Z is a fat [1], because 1.00 g of it releases far more carbon dioxide and water than the carbohydrate, showing a higher proportion of carbon and hydrogen and less oxygen [1].

Walkthrough 6 — Designing a Fair Benedict ComparisonA student wants to compare the reducing sugar content of a banana on five consecutive days. Design a valid, reliable method and explain how she should make her comparison as quantitative as possible. [8]

Same mass, same volume, filtered

Crush the same mass of banana with the same volume of distilled water each day, then filter. If the mass or the dilution varies, the colour reflects the preparation rather than the fruit, and the comparison is invalid before the reagent is even added.

Same volume, same bath, same time

Add the same volume of Benedict’s solution to every tube, heat them all in the same water bath at the same temperature, and remove them all after the same length of time. Benedict’s deepens with heating time, so tubes pulled out at different moments cannot be compared even if everything else was identical.

The observer is a variable

Match the colours against a printed colour chart in consistent lighting, with the same observer each day. This is what keeps the test semi-quantitative: it gives an estimate, not a measurement, because a human eye is doing the deciding.

Standards plus an instrument

Prepare a series of glucose solutions of known concentration, test them identically, measure each final colour with a colorimeter, and plot a calibration graph. The banana readings can then be read off the line as actual concentrations. Repeats and a mean, with anomalies discarded, complete the reliability side.

Full Mark-Scheme AnswerSame mass of banana with the same volume of water, filtered [1]; same volume of Benedict’s solution [1]; same water bath temperature [1]; same heating time [1]; colours matched against a chart by the same observer in consistent lighting [1]; repeats with a mean and anomalies discarded [1]; series of known glucose standards tested identically [1]; colorimeter readings plotted as a calibration graph so concentrations can be read off [1].

🔍 Spot the Difference

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Six pairs that look almost identical and have different answers. The distinction is where the marks live.

Question A

A food leaves Benedict’s solution blue. What can you conclude?

That the food contains no reducing sugar. Nothing else.

Question B

A food leaves iodine solution orange-brown. What can you conclude?

That the food contains no starch. Nothing else.

Key DifferenceBoth are negative carbohydrate results and neither means “no carbohydrate”. Starch is invisible to Benedict’s; reducing sugars and cellulose are invisible to iodine. A negative result rules out one named substance, never a whole class.

Question A

Which reagent is blue before the test and needs heating?

Benedict’s solution. Heat in a water bath at about 80 °C; positive result blue → brick-red.

Question B

Which reagent is blue before the test and needs no heating?

Biuret solution. Add and shake at room temperature; positive result blue → purple.

Key DifferenceTwo blue reagents, one water bath. Benedict’s is the only test of the five that is heated, and adding a heating step to biuret contradicts the method mark. Anchor them by their end colours: Benedict’s ends warm (brick-red), biuret ends cool (purple).

Question A

What does it mean to say a solution is clear?

It is transparent — you can see through it. A blue copper solution is perfectly clear.

Question B

What does it mean to say a solution has been decolourised?

A colour that was present has been removed, leaving it colourless.

Key Difference“Clear” is about transparency; “colourless” is about colour; “decolourised” is about a change. Blue DCPIP is clear before the test, so “it went clear” reports nothing at all and is refused by mark schemes.

Question A

How do you improve the reliability of a food test comparison?

Repeat each measurement several times, take a mean, and discard anomalous values.

Question B

How do you improve the validity of a food test comparison?

Control the variables — same mass of food, same volume of water and reagent, same temperature and time — so that the thing you are measuring is the thing you meant to measure.

Key DifferenceReliability is about repeatability; validity is about measuring the right quantity. A perfectly repeatable measurement of the wrong thing is reliable and invalid, and questions frequently ask you to name which one an improvement addresses.

Question A

Are the two strands of a DNA molecule identical?

No. Identical would mean the same sequence on both strands, which the pairing rule forbids.

Question B

Are the two strands of a DNA molecule complementary?

Yes. Every base is matched by its fixed partner — A with T, C with G — so A C C T G faces T G G A C.

Key DifferenceComplementary means matched, not the same. This is also the reason one strand carries enough information to rebuild the other: each base has exactly one possible partner, so nothing is left to guess.

Question A

Why is cellulose good at forming plant cell walls?

Its chains are long, straight and unbranched, so they lie side by side in bundles and make strong fibres.

Question B

Why is glycogen good as an animal energy store?

It is heavily branched, giving many free ends, so glucose can be released quickly from many points at once — and it is insoluble and compact.

Key DifferenceIdentical building block, opposite jobs. Whenever a question compares two glucose polymers, the answer is the linking and the arrangement of the chains — never a difference in elements and never a difference in subunit.

🔗 Biological Molecules Concept Map

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Click each node to see how the pieces of Topic 4 connect.

⭐ CORE FRAMEWORK 1

Elements → building blocks → large molecules

The Elements — and What They Give Away ▶

Carbohydrates and fats: carbon, hydrogen, oxygen. Proteins: those three plus nitrogen, and sulfur in some.

Nitrogen present → protein. No other food molecule contains it, so this is the fastest identification in the topic.

Nitrogen absent → carbohydrate or fat, and elements alone cannot choose. Use the proportions: fats are rich in C and H, poor in O, so burning a gram of fat gives far more CO₂ and water.

The Building Blocks ▶

Starch, glycogen and cellulose → glucose. Proteins → amino acids. Fats and oils → one glycerol + three fatty acids.

Joining small molecules into big ones adds no new elements — a starch molecule is thousands of times more massive than glucose but still contains only C, H and O.

A fat is the odd one out: only four molecules join, so it is not a long chain of repeating units and the phrase “built from many small molecules” does not apply to it.

Three Glucose Polymers, Three Jobs ▶

Starch — coiled chains, a few branches; the plant store; insoluble and compact.

Glycogen — the same glucose units, heavily branched; the animal store; many free ends means fast glucose release.

Cellulose — long straight unbranched chains lying side by side; strong fibres for plant cell walls; gives a negative iodine test although it is a carbohydrate.

Why Insoluble Stores? ▶

Storing a lot of soluble glucose would make the solution inside the cell much more concentrated.

Water would then move into the cell, which could damage or burst it.

Starch and glycogen are insoluble, so they do not affect the concentration, they pack compactly, and they cannot leak out of the cell.

⭐ CORE FRAMEWORK 2

The five tests: reagent, condition, colour before, colour after

Iodine — Starch ▶

A few drops of iodine solution; no heating.

Positive: orange-brown → blue-black. Negative: stays orange-brown.

Tests for starch specifically — cellulose and reducing sugars both give negative results despite being carbohydrates.

Benedict’s — Reducing Sugars ▶

Add an excess of Benedict’s solution, then heat in a water bath at about 80 °C for a few minutes. The only test that is heated.

Positive: blue → green → yellow → orange → brick-red as the amount increases. Negative: stays blue.

Semi-quantitative: the colour estimates the amount by eye. A colorimeter and a calibration graph make it quantitative.

Biuret — Protein ▶

Add biuret solution and shake; no heating.

Positive: blue → purple (violet). Negative: stays blue.

Protein is the nitrogen-containing class, so biuret is the direct test for it — more trustworthy than estimating protein from a nitrogen measurement, which counts any nitrogen from any source.

Ethanol Emulsion — Fat and Oil ▶

Ethanol first: shake the food with ethanol so fat dissolves, let the solid settle, then decant the ethanol into a second tube of distilled water.

Positive: cloudy white emulsion. Negative: stays clear and colourless. The only test with no colour change.

Two traps: adding water first (the fat never dissolves) and failing to decant (food particles give a false positive). Safety: ethanol is highly flammable.

DCPIP — Vitamin C ▶

Fixed volume of blue DCPIP; add juice drop by drop from a syringe, shaking, until the blue just disappears; record the volume of juice.

Positive: the DCPIP is decolourised. Never write “went clear”.

Inverse: less juice needed means more vitamin C. Concentration = (mass that decolourises the dye) ÷ (volume needed).

⭐ CORE FRAMEWORK 3

DNA: structure, pairing rule, and the arithmetic that follows

The Four-Sentence Structure ▶

DNA is made of two strands.

The strands are coiled together to form a double helix.

Each strand contains a sequence of bases, and bonds between pairs of bases hold the two strands together. The full names of the bases are not required.

The Pairing Rule and Its Consequences ▶

A pairs with T; C pairs with G. Every base has exactly one possible partner.

Consequence 1: %A = %T and %C = %G — the two equalities behind every calculation.

Consequence 2: one strand completely determines the other, so a damaged strand can be rebuilt exactly. Consequence 3: the strands are complementary, not identical.

The Two-Step Calculation ▶

Step 1: use one equality to fill in the partner of the base you are given, and add the pair together.

Step 2: subtract from 100% (or from the total number of bases) and halve what is left between the other pair. The halving is the step everyone forgets.

Validity check: any data set where %A ≠ %T, or %C ≠ %G, is impossible — even if the four figures add neatly to 100.

Sequences and Classification ▶

Base sequences carry information; base percentages do not, because percentages ignore the order.

More closely related species have more similar base sequences, which is why sequence comparison is used in classification.

Two unrelated species could share identical base percentages while spelling out completely different sequences — a favourite distractor.

❌ “Why Is This Wrong?” Exercises

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Six real student answers. Find the fault before you reveal it.

Exercise 1: “Benedict’s solution stayed blue with the pasta sample. State what this shows. [1]”

Student’s Answer“It shows that pasta contains no carbohydrate.”

The FlawThe conclusion is far too broad. Benedict’s solution does not test for carbohydrate — it tests for reducing sugars, and pasta is loaded with starch, which is a carbohydrate the test cannot see.

Correct Answer“It shows that the pasta contains no reducing sugar [1].” An iodine test on the same sample would give a strong blue-black colour.

Key RuleEvery food test detects one named substance. A negative result rules out that substance and nothing wider.

Exercise 2: “Describe what is seen when lemon juice is added to blue DCPIP solution. [1]”

Student’s Answer“The DCPIP goes clear.”

The Flaw“Clear” means transparent, and the DCPIP was transparent to begin with — it was clear and blue. The answer therefore describes no change at all, and mark schemes explicitly refuse it.

Correct Answer“The blue DCPIP is decolourised / changes from blue to colourless [1].”

Key RuleClear = transparent. Colourless = no colour. Decolourised = a colour has been removed. Three words, three meanings, and only two of them describe what happened here.

Exercise 3: “Describe how to test a food sample for fat. [3]”

Student’s Answer“Put the food in a test tube with water and shake it, then add ethanol. If it goes white, fat is present.”

The FlawTwo faults. The order is reversed — fat is insoluble in water, so it never dissolves. And there is no decanting, so undissolved food particles cloud the tube and produce a false positive even for a fat-free food.

Correct Answer“Add the food to ethanol and shake so that any fat dissolves [1]. Let the solid settle, then decant the ethanol into a second tube of distilled water [1]. A cloudy white emulsion shows fat is present; with no fat the mixture stays clear and colourless [1].”

Key RuleEthanol first, settle, decant, water. And describe the result as a cloudy white emulsion — not a precipitate, because nothing has reacted.

Exercise 4: “A DNA sample contains 30% adenine. Calculate the percentage of guanine. [2]”

Student’s Answer“100 − 30 = 70, so guanine is 70%.”

The FlawTwo bases have been ignored. Thymine has not been subtracted, and the remainder has not been shared between cytosine and guanine. The answer treats a four-base molecule as though it had two.

Correct Answer“Adenine 30% means thymine 30%, so A + T = 60% [1]. C + G = 40%, and since C = G, guanine = 20% [1].”

Key RuleTwo steps, always: use %A = %T to remove the first pair, then subtract and halve. Stopping after the subtraction gives 40%, which is the other standard wrong answer.

Exercise 5: “Cotton and potato starch are both made of glucose. Explain why humans can digest one but not the other. [2]”

Student’s Answer“Because cellulose contains a different element that humans cannot break down.”

The FlawInvented chemistry. Cellulose contains exactly the same three elements as starch — carbon, hydrogen and oxygen — and exactly the same building block, glucose. The explanation has reached for the one thing that definitely is not different.

Correct Answer“The glucose units are joined together in a different way in cellulose [1], so the substance that breaks the links in starch does not fit the links in cellulose [1].”

Key RuleWhen two molecules share a subunit, the difference is always the linking and the arrangement of the chains — never the elements.

Exercise 6: “Describe how to test apple juice for reducing sugar. [3]”

Student’s Answer“Add Benedict’s solution and warm it gently over a Bunsen burner until it turns red.”

The FlawThe heating method is wrong and the result is under-described. Direct heating makes the liquid boil suddenly and spit out of the tube, and it heats tubes unevenly so results cannot be compared. Saying only “turns red” gives neither the starting colour nor the negative result.

Correct Answer“Add an excess of Benedict’s solution to the juice [1]; heat in a water bath at about 80 °C for a few minutes [1]; a positive result is a change from blue through green and yellow to brick-red, and if no reducing sugar is present it stays blue [1].”

Key RuleWater bath, not flame. And every observation mark is a colour change with the negative result stated alongside it.

✍️ Ultra-Detailed Practice Questions

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Ten Cambridge-style challenge questions. Write your answer first, then reveal the model answer and the examiner’s notes.

Question 1

[6 marks]

A purified substance contains carbon, hydrogen, oxygen, nitrogen and a small amount of sulfur. (a) Identify the class of biological molecule and justify your answer. [2] (b) Name the small molecules from which it is built. [1] (c) Explain why two molecules of this class built from the same set of small molecules can be completely different from one another. [2] (d) Name the reagent you would use to confirm your identification. [1]

Model Answer(a) It is a protein [1], because nitrogen is present in every protein and in no carbohydrate or fat, and sulfur is found in some proteins only [1].
(b) Amino acids [1].
(c) The amino acids are joined in a different order [1], so the chain folds into a different three-dimensional shape and the protein does a different job [1].
(d) Biuret solution [1] (blue → purple).

Examiner’s NotesPart (a) needs the element and the reason; naming the class alone is one mark at best. In (c), “they are different amino acids” misses the point — the question says the same set is used, so only the order can differ. Candidates who reach for sulfur as the always-present element lose the justification mark, since only some proteins contain it.

Question 2

[8 marks]

Describe, in full, how you would test one food sample for starch, reducing sugar, protein and fat. For each test give the reagent, any special condition, and the result you would expect if the nutrient is present and if it is absent. [8]

Model AnswerStarch: add a few drops of iodine solution [1]; positive orange-brown → blue-black, negative stays orange-brown [1].
Reducing sugar: add an excess of Benedict’s solution and heat in a water bath at about 80 °C for a few minutes [1]; positive blue → green/yellow/orange/brick-red, negative stays blue [1].
Protein: add biuret solution, no heating [1]; positive blue → purple, negative stays blue [1].
Fat: shake the food with ethanol, let it settle and decant into distilled water [1]; positive cloudy white emulsion, negative stays clear and colourless [1].

Examiner’s NotesEight marks split neatly into four pairs: method and result for each test. The two most commonly dropped are the water bath (“heat it” is not enough) and the ethanol-before-water order. Take a fresh portion of the sample for each test, and if you have time say so — adding one reagent on top of another makes the colours unreadable.

Question 3

[7 marks]

A student is given four colourless liquids: glucose solution, starch suspension, egg-white solution and vitamin C solution. (a) Describe the tests she should carry out and the results that would identify each liquid. [4] (b) Explain why she must use a fresh sample of each liquid for every test. [2] (c) State one result that would show a liquid contains none of these four substances. [1]

Model Answer(a) Iodine: blue-black identifies the starch suspension, the others stay orange-brown [1]. Benedict’s, heated: brick-red identifies the glucose solution, the others stay blue [1]. Biuret: purple identifies the egg-white solution, the others stay blue [1]. DCPIP: decolourised identifies the vitamin C solution, the others leave it blue [1].
(b) Reagents left in a tube mix with the next one and change the colour seen [1]; for example the orange-brown of iodine would mask the blue-to-purple change of the biuret test [1].
(c) All four tests negative — iodine stays orange-brown, Benedict’s and biuret stay blue and the DCPIP stays blue [1].

Examiner’s NotesIdentification questions want the result for every liquid in every test, because the negatives are what make each identification certain. In (b), a concrete example of interference scores far better than “it would be contaminated”. Part (c) is a check that you can state a set of negatives as a meaningful result in its own right.

Question 4

[8 marks]

A student compares the vitamin C content of two juices using DCPIP. (a) Describe the method, including how the end-point is judged. [4] (b) State three variables that must be kept the same. [3] (c) Explain the difficulty of using this method with a deeply coloured juice such as blackcurrant. [1]

Model Answer(a) Measure a fixed volume of DCPIP into a test tube [1]; add the juice from a syringe drop by drop, shaking after each addition [1]; record the volume of juice needed for the blue colour to disappear completely [1]; repeat for each juice and take a mean [1].
(b) The same volume of DCPIP [1]; the same concentration of DCPIP [1]; the same temperature and the same drop size from the same syringe [1].
(c) The strong colour of the juice masks the blue DCPIP, so the end-point cannot be seen [1].

Examiner’s NotesThe roles of the two liquids must be the right way round: DCPIP fixed, juice measured. Naming control variables individually is essential — “keep everything else the same” earns nothing. The blackcurrant point is about visibility, not chemistry: the vitamin C still works, you simply cannot see the end-point.

Question 5

[6 marks]

A DNA sample contains 4 000 000 bases, of which 1 200 000 are guanine. (a) State the number of cytosine bases and explain your answer. [2] (b) Calculate the number of thymine bases. Show your working. [2] (c) Explain why a reported sample containing 26% cytosine and 21% guanine cannot be correct. [2]

Model Answer(a) 1 200 000 [1], because cytosine always pairs with guanine so the two are present in equal numbers [1].
(b) C + G = 2 400 000, so A + T = 4 000 000 − 2 400 000 = 1 600 000 [1]; adenine and thymine are equal, so thymine = 800 000 [1].
(c) Cytosine pairs only with guanine, so their percentages must be equal [1]; 26% and 21% are unequal, so the analysis contains an error or the sample is contaminated [1].

Examiner’s NotesShow every line of working: partner, sum, subtraction, halving. Answering 2 800 000 for (b) subtracts only the guanine; answering 1 600 000 forgets the final halving. In (c) the mark is for stating the rule before applying it — “the numbers are different” on its own is not an explanation.

Question 6

[7 marks]

Genuine skimmed milk powder contains protein, a little fat and lactose (a reducing sugar), but no starch. A laboratory suspects a batch has been bulked out with cornflour. (a) Name the test that would detect this and give the result for genuine and adulterated powder. [3] (b) Explain why the Benedict’s and biuret tests would not detect the adulteration. [2] (c) Describe a suitable control and explain what it shows. [2]

Model Answer(a) The iodine test for starch [1]; genuine powder leaves the iodine orange-brown [1]; adulterated powder turns it blue-black [1].
(b) Genuine milk powder already contains lactose, so Benedict’s is positive either way [1]; it already contains protein, so biuret is purple either way, and neither test can distinguish the samples [1].
(c) Test a sample of known genuine milk powder in exactly the same way [1]; 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].

Examiner’s NotesThe reason the other tests fail is not unreliability — it is that both samples are positive, so the result carries no information. The control here is genuine powder, not distilled water; the comparison has to be against the real product. Give both colours in (a): the negative result is what clears an honest batch.

Question 7

[8 marks]

Explain why plants store carbohydrate as starch rather than as glucose, and why cellulose, built from the same glucose units, is used for cell walls instead. [8]

Model AnswerGlucose is soluble, so storing a large amount would make the solution inside the cell much more concentrated [1]; water would then move into the cell and could damage or burst it [1]. Starch is insoluble, so it has no such effect [1], and joining many glucose units into one large molecule makes the store compact and stops it leaking out of the cell [1]. Starch chains are coiled with a few branches, which suits a compact store [1]. Cellulose is built from the same glucose units but they are joined in a different way [1], giving long straight unbranched chains [1] that lie side by side in bundles and make strong fibres, which is what a cell wall requires [1].

Examiner’s NotesEight marks means a chain of reasoning, not a list of facts. Build it in order: solubility → concentration → water entering the cell → the insoluble solution → compactness → then switch to cellulose and repeat the shape-to-function argument. Answers that simply assert “starch is insoluble and cellulose is strong” collect two marks out of eight.

Question 8

[7 marks]

A student tests a food and records: iodine stays orange-brown; Benedict’s turns green; biuret turns purple; the ethanol emulsion test gives a cloudy white result; DCPIP stays blue. (a) State what the food contains and what it does not contain. [3] (b) Explain what the green Benedict’s result tells you that a brick-red result would not. [2] (c) Give two reasons why the biuret result cannot be used to say how much protein the food contains. [2]

Model Answer(a) Contains a reducing sugar and protein and fat [1]; contains no starch [1]; contains no vitamin C [1].
(b) Green is a positive result showing a small amount of reducing sugar [1], whereas brick-red would show a large amount — the colour ladder makes the test semi-quantitative [1].
(c) The depth of colour is judged by eye, so it is subjective [1]; and the result depends on how much food was used and how much it was diluted, which are not stated [1].

Examiner’s NotesIn (a) the negatives are worth marks too, so list them explicitly. The classic error is writing “no carbohydrate” for the negative iodine result when a reducing sugar has just been detected. Part (c) is an evaluation question: name specific method faults, because “it might be inaccurate” is not creditable.

Question 9

[6 marks]

A vitamin C tablet is labelled “500 mg”. A student dissolves one tablet and makes the solution up to 250 cm³. A standard solution of 1.0 mg/cm³ needs 1.00 cm³ to decolourise 1.0 cm³ of DCPIP. Her tablet solution needs 0.80 cm³. (a) Calculate the actual mass of vitamin C in the tablet. Show your working. [3] (b) Comment on the manufacturer’s claim. [1] (c) Suggest two explanations for the difference other than the claim being untrue. [2]

Model Answer(a) 1.00 cm³ × 1.0 mg/cm³ = 1.0 mg of vitamin C decolourises this DCPIP [1]; concentration of the tablet solution = 1.0 ÷ 0.80 = 1.25 mg/cm³ [1]; mass in the tablet = 1.25 × 250 = 312.5 mg [1].
(b) The tablet contains clearly less than the 500 mg claimed — only about 63% of it [1].
(c) The tablet may have lost vitamin C during storage [1]; some of the crushed tablet may not have dissolved completely, or the end-point may have been overshot [1].

Examiner’s NotesThree steps in (a): the fixed mass, the concentration, then the mass in the whole 250 cm³. Stopping at 1.25 mg/cm³ answers a different question. In (c) an alternative explanation means a reason the measurement could understate the truth — storage losses and incomplete dissolving are the two strongest, and both are worth remembering for any “why does the result differ from the label?” question.

Question 10

[7 marks]

A student writes this method: “Crush the food, add water, then test the same tube with iodine, then Benedict’s, then biuret, then ethanol. Heat every tube in a water bath so the reactions work faster.” (a) Identify three faults and explain the effect of each. [6] (b) State one safety precaution she has omitted. [1]

Model Answer(a) Fault 1: the same tube is used for every test [1] — the orange-brown iodine masks the Benedict’s and biuret colours, so the results cannot be read [1]. Fault 2: every tube is heated [1] — only Benedict’s requires heating, and heating the others is an unnecessary step that contradicts the correct method [1]. Fault 3: the ethanol is added to a tube that already contains water [1] — fat is insoluble in water, and food particles carried into the tube would give a false positive [1].
(b) Ethanol is highly flammable, so it must be kept away from naked flames [1].

Examiner’s NotesFault-finding questions award the fault and its consequence separately, so always write “this is wrong because…”. The heating fault is the one candidates most often miss, because heating feels harmless; the point is that only Benedict’s needs it, and a method that heats everything shows the five tests have been learned as a blur. The safety mark is free if you remember that ethanol appears in the method.