IGCSE Biology Paper 4 (Theory / Extended)

Topic 4: Biological Molecules -- Challenge Exam 1

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 — Four Unlabelled Food Samples

Total: 12 marks

A food technologist in Kathmandu is given four unlabelled food samples, A, B, C and D. Each sample is ground up and shaken with distilled water before testing. Her results are shown below.

test	A	B	C	D
add iodine solution	blue-black	stays orange-brown	stays orange-brown	blue-black
add Benedict’s solution, heat in a water bath	stays blue	brick-red	stays blue	green
add biuret solution	stays blue	stays blue	purple	purple
ethanol emulsion test	stays clear	stays clear	cloudy white emulsion	cloudy white emulsion
add to DCPIP solution	stays blue	decolourised by 1.0 cm³	stays blue	stays blue

(a) [4]

State the food molecules present in each of samples A, B, C and D. In each case give the evidence from the table.

Model Answer — 1a

A contains starch only — iodine turns blue-black and every other test is negative [1]

B contains a reducing sugar and vitamin C — Benedict’s gives brick-red and the DCPIP is decolourised [1]

C contains protein and fat — biuret gives purple and the ethanol emulsion test gives a cloudy white emulsion [1]

D contains starch, a reducing sugar, protein and fat — it is positive for iodine, Benedict’s, biuret and the emulsion test [1]

⚠ If you missed marks here: Every identification has to be pinned to a named test result, not guessed from what the food might be. Two results are easy to skim past: sample D gives green with Benedict’s, which is still a positive result for reducing sugar, and sample A is negative with DCPIP, so no vitamin C may be claimed for it.

1 mark — A contains starch only from the blue-black iodine result 1 mark — B contains reducing sugar and vitamin C from brick-red Benedict and decolourised DCPIP 1 mark — C contains protein and fat from purple biuret and cloudy white emulsion 1 mark — D contains starch reducing sugar protein and fat

(b) [4]

(i) Sample B decolourised the DCPIP after 1.0 cm³ had been added. A fifth sample, E, needed 2.5 cm³ to decolourise the same volume of DCPIP. State which of B and E contains more vitamin C per cm³ and explain your answer. [2]

(ii) Explain why the negative Benedict’s result for sample A does not prove that sample A contains no carbohydrate. [2]

Model Answer — 1b

(i) sample B contains more vitamin C per cm³ [1]

(i) because a smaller volume of B was needed to decolourise the same volume of DCPIP, so the vitamin C in B is more concentrated [1]

(ii) Benedict’s solution detects reducing sugars only [1]

(ii) starch is a carbohydrate but is not a reducing sugar, and sample A did give a positive iodine test for starch [1]

⚠ If you missed marks here: The DCPIP comparison is an inverse one and it catches people out every year: the juice needing the smaller volume is the richer one, because less of it carries enough vitamin C to decolourise the dye. In part two, “Benedict’s tests for sugar” is too loose — the mark is for the word reducing, and for naming starch as the carbohydrate that slips through.

1 mark — B contains more vitamin C per cm3 1 mark — smaller volume needed to decolourise the same volume of DCPIP so more concentrated 1 mark — Benedict solution detects reducing sugars only 1 mark — starch is a carbohydrate but not a reducing sugar and A gave positive iodine

(i) Sample D gave a green colour with Benedict’s solution while sample B gave brick-red. State what this shows, and explain what is meant by calling the Benedict’s test semi-quantitative. [2]

(ii) In her notebook the technologist wrote “with sample B the DCPIP went clear”. Explain why this wording would score no marks, and give wording an examiner would accept. [2]

Model Answer — 1c

(i) sample D contains less reducing sugar than sample B [1]

(i) semi-quantitative means the final colour (blue → green → yellow → orange → brick-red) estimates how much reducing sugar is present, but does not give an exact value [1]

(ii) “clear” means transparent, not colourless — a blue solution is perfectly clear, so the word describes no change at all [1]

(ii) the accepted wording is that the DCPIP is decolourised, or that it changes from blue to colourless [1]

⚠ If you missed marks here: Green is a positive Benedict’s result, not a failed one — the ladder of colours is the whole point of the test. The vocabulary mark is a pure precision mark: clear means see-through, colourless means having no colour, and decolourised means a colour that was there has been removed. Mark schemes print “colourless — not clear” in bold for exactly this reason.

1 mark — D contains less reducing sugar than B 1 mark — semi quantitative means colour estimates the amount of reducing sugar not an exact value 1 mark — clear means transparent not colourless so it describes no colour change 1 mark — accepted wording is decolourised or blue to colourless

Question 2 — Comparing Vitamin C in Two Juices

Total: 12 marks

A class in Bangalore compares the vitamin C content of fresh guava juice and fresh lime juice. On the bench they have blue DCPIP solution, syringes, test tubes, distilled water, and a standard vitamin C solution of concentration 1.0 mg/cm³.

(a) [4]

Describe how the class should use DCPIP solution to compare the vitamin C content of the two juices. Include the apparatus used, what is measured, and how the end-point is judged.

Model Answer — 2a

measure a fixed volume of DCPIP solution (for example 1.0 cm³) into a test tube [1]

add the juice from a syringe, drop by drop, shaking the tube after each addition [1]

record the volume of juice needed for the blue colour to be completely decolourised [1]

repeat with the second juice and with the standard vitamin C solution, and take a mean of repeat readings [1]

⚠ If you missed marks here: This is a titration in everything but name, so the marks sit on the measured quantity and the end-point. “Add DCPIP to the juice” is the wrong way round — the DCPIP is the fixed volume and the juice is the variable one being measured. The end-point is the moment the blue colour just disappears, and it must be reached by drop-by-drop addition with shaking, or the end-point is overshot.

1 mark — fixed volume of DCPIP measured into a test tube 1 mark — juice added drop by drop from a syringe with shaking 1 mark — volume of juice needed to decolourise the blue DCPIP is recorded 1 mark — procedure repeated for each juice and a mean taken

(b) [4]

(i) State three variables that must be kept the same if the comparison is to be a fair one. [3]

(ii) Blackcurrant juice is deep purple. Explain the difficulty of using this method with blackcurrant juice. [1]

Model Answer — 2b

(i) the same volume of DCPIP solution in every tube [1]

(i) the same concentration of DCPIP solution [1]

(i) the same temperature, and drops of the same size delivered from the same syringe [1]

(ii) the strong purple colour of the juice masks the blue colour of the DCPIP, so the end-point cannot be seen [1]

⚠ If you missed marks here: “Keep everything the same” earns nothing — each control variable must be named separately, and the two that matter most here are the volume and the concentration of the DCPIP, because both change how much vitamin C is needed to decolourise it. The blackcurrant point is about seeing the end-point, not about the chemistry: the vitamin C still works, but you cannot tell when the blue has gone.

1 mark — same volume of DCPIP in every tube 1 mark — same concentration of DCPIP solution 1 mark — same temperature and same drop size from the same syringe 1 mark — deep purple juice masks the blue DCPIP so the end point cannot be seen

The class found that 1.0 cm³ of the standard 1.0 mg/cm³ vitamin C solution decolourised the DCPIP. Their mean results for the juices are shown.

liquid	mean volume needed to decolourise the DCPIP / cm³
standard vitamin C, 1.0 mg/cm³	1.00
guava juice	0.80
lime juice	2.00

(i) Calculate the concentration of vitamin C in each juice, in mg/cm³. Show your working. [2]

(ii) State the conclusion the class can draw. [1]

(iii) Suggest one improvement to the investigation. [1]

Model Answer — 2c

(i) every volume that just decolourises the DCPIP contains the same mass of vitamin C = 1.00 × 1.0 = 1.0 mg [1]

(i) guava = 1.0 ÷ 0.80 = 1.25 mg/cm³ and lime = 1.0 ÷ 2.00 = 0.50 mg/cm³ [1]

(ii) guava juice contains about 2.5 times as much vitamin C per cm³ as lime juice [1]

(iii) any one improvement, such as more repeats and a mean, adding the juice in smaller measured drops near the end-point, or filtering the juice first so pulp does not obscure the colour [1]

⚠ If you missed marks here: The calculation only works once you see that every tube is decolourised by the same mass of vitamin C — the fixed volume of DCPIP is the constant, and the juice volume is the variable. Dividing the wrong way round (0.80 ÷ 1.0) gives 0.8 mg/cm³ and reverses the conclusion. Always sanity-check: the juice that needed less volume must come out with the higher concentration.

1 mark — recognises that each decolourising volume contains 1.0 mg of vitamin C 1 mark — guava 1.25 mg per cm3 and lime 0.50 mg per cm3 1 mark — conclusion that guava contains about 2.5 times more vitamin C per cm3 1 mark — sensible improvement such as more repeats smaller drops or filtering the juice

Question 3 — Burning Three Purified Food Substances

Total: 12 marks

A researcher burns 1.00 g of each of three purified food substances, X, Y and Z, in an excess of oxygen and analyses all the gases produced.

substance	carbon dioxide produced / g	water produced / g	oxides of nitrogen	sulfur dioxide
X	1.47	0.60	none detected	none detected
Y	1.31	0.51	detected	trace detected
Z	2.85	1.15	none detected	none detected

One substance is a carbohydrate, one is a protein and one is a fat.

(a) [4]

(i) State the three chemical elements present in all three of X, Y and Z. [1]

(ii) Identify Y as a carbohydrate, a protein or a fat, and give two pieces of evidence from the table. [2]

(iii) Explain which of X and Z is the fat. [1]

Model Answer — 3a

(i) carbon, hydrogen and oxygen [1]

(ii) Y is a protein [1]

(ii) the oxides of nitrogen show that Y contains nitrogen, and the trace of sulfur dioxide shows sulfur, which is found in some proteins but never in carbohydrates or fats [1]

(iii) Z is the fat, because 1.00 g of Z produced far more carbon dioxide and water than 1.00 g of X, showing a higher proportion of carbon and hydrogen (and correspondingly less oxygen) [1]

⚠ If you missed marks here: Nitrogen is the element that identifies a protein — sulfur is a bonus clue because only some proteins contain it, so an absence of sulfur dioxide would not rule protein out. For the fat, the argument must be comparative and per gram: fats are richer in carbon and hydrogen than carbohydrates, which is why the same mass releases far more carbon dioxide and water when burned.

1 mark — carbon hydrogen and oxygen named 1 mark — Y identified as a protein 1 mark — nitrogen shown by oxides of nitrogen and sulfur by sulfur dioxide 1 mark — Z is the fat because more carbon dioxide and water per gram means more carbon and hydrogen

(b) [4]

(i) Name the small molecules from which starch, glycogen and cellulose are all built. [1]

(ii) Name the small molecules from which proteins are built. [1]

(iii) Name the two kinds of small molecule from which a fat is built and state how many of each combine to make one fat molecule. [2]

Model Answer — 3b

(i) glucose [1]

(ii) amino acids [1]

(iii) glycerol and fatty acids [1]

(iii) one molecule of glycerol combines with three fatty acid molecules [1]

⚠ If you missed marks here: “Sugar” is not precise enough for the first mark — the subunit of all three polysaccharides is glucose, which is exactly why the three molecules can be so different while being built from the same brick. For fats, both components must be named and the ratio stated: one glycerol to three fatty acids. Writing “fatty acids and glucose” is a very common slip.

1 mark — glucose named as the subunit of starch glycogen and cellulose 1 mark — amino acids named as the subunits of proteins 1 mark — glycerol and fatty acids named 1 mark — one glycerol combines with three fatty acids

A student writes: “Carbohydrates and fats contain exactly the same three elements, so they must be the same kind of molecule and must do the same job in the body.”

Discuss this statement. [4]

Model Answer — 3c

it is true that both contain only carbon, hydrogen and oxygen [1]

but the three elements are present in different proportions: a fat contains a much higher proportion of hydrogen and carbon and much less oxygen than a carbohydrate [1]

and the atoms are arranged differently because the molecules are built from different subunits — carbohydrates from glucose, fats from glycerol and fatty acids [1]

so a list of elements can never identify a molecule; a food test is needed, such as iodine or Benedict’s for carbohydrate and the ethanol emulsion test for fat [1]

⚠ If you missed marks here: This is a “discuss” question, so the answer has to concede the true part before dismantling it — a flat “the student is wrong” throws away the first mark. The two reasons the examiner wants are different proportions of the same elements and different subunits and arrangement. Finish with the practical consequence: elemental analysis alone cannot tell you what a molecule is, which is precisely why food tests exist.

1 mark — agrees that both contain only carbon hydrogen and oxygen 1 mark — elements present in different proportions with fat richer in hydrogen and poorer in oxygen 1 mark — different subunits and arrangement glucose versus glycerol and fatty acids 1 mark — concludes that a food test is needed because elements alone cannot identify a molecule

Question 4 — The Structure of DNA

Total: 12 marks

A sample of DNA extracted from the bacterium Deinococcus radiodurans is found to contain 3 200 000 bases in total. Chemical analysis shows that 18% of these bases are adenine.

(a) [4]

Describe the structure of a DNA molecule.

Model Answer — 4a

a DNA molecule is made of two strands [1]

the two strands are coiled together to form a double helix [1]

each strand contains a sequence of bases [1]

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

⚠ If you missed marks here: All four marks are for exact wording, and three of them are routinely thrown away. “DNA is a helix” misses that there are two strands; “the strands are stuck together” misses that the bonds are specifically between pairs of bases; and describing DNA as a “spiral” or a “ladder” without the words double helix is not credited. Learn the four-line description as four separate sentences.

1 mark — DNA is made of two strands 1 mark — the strands are coiled to form a double helix 1 mark — each strand contains a sequence of bases 1 mark — bonds between pairs of bases hold the two strands together

(b) [4]

(i) State the percentage of the bases in this sample that are thymine and explain your answer. [2]

(ii) Calculate the number of guanine bases in this DNA sample. Show your working. [2]

Model Answer — 4b

(i) thymine = 18% [1]

(i) because adenine always pairs with thymine, so the two bases must be present in equal numbers [1]

(ii) adenine + thymine = 36%, so cytosine + guanine = 64%; cytosine pairs with guanine, so guanine = 32% [1]

(ii) 32% of 3 200 000 = 0.32 × 3 200 000 = 1 024 000 guanine bases [1]

⚠ If you missed marks here: The whole calculation hangs on one sentence: A pairs with T and C pairs with G, so %A = %T and %C = %G. The classic error is to subtract 18% from 100% and call the answer guanine (82%), forgetting that thymine has to be removed as well and that the remaining 64% is then shared equally between cytosine and guanine. Write the pairing rule down before touching the numbers.

1 mark — thymine is 18 per cent 1 mark — because adenine always pairs with thymine so they are equal in number 1 mark — cytosine plus guanine is 64 per cent so guanine is 32 per cent 1 mark — guanine number is 1 024 000

(i) Part of one strand of a DNA molecule has the base sequence T A C G G A T. Write the base sequence of the matching part of the other strand. [1]

(ii) Explain why the two strands are described as complementary rather than identical. [1]

(iii) A technician reports that a different DNA sample contains 30% adenine and 25% thymine. Explain why this result cannot be correct. [2]

Model Answer — 4c

(i) A T G C C T A [1]

(ii) every base is paired with a fixed partner on the other strand (A with T, C with G), so the two sequences match up but are not the same sequence [1]

(iii) adenine can only pair with thymine, so adenine and thymine must be present in equal percentages [1]

(iii) 30% and 25% are not equal, so the analysis contains an error or the sample is contaminated [1]

⚠ If you missed marks here: Reading the complementary strand slowly, one base at a time, is the only way to avoid the classic mistakes — writing the same sequence back, or pairing A with C. In the last part it is not enough to say “the numbers are different”: the mark is for stating the rule (A pairs only with T, so the percentages must be equal) and only then applying it to the figures given.

1 mark — complementary strand written as ATGCCTA 1 mark — each base has a fixed partner so the strands match but are not the same 1 mark — adenine pairs only with thymine so their percentages must be equal 1 mark — thirty and twenty five are unequal so the result shows an error or contamination

Question 5 — Benedict Testing a Ripening Banana

Total: 10 marks

A student in Colombo follows a single banana as it ripens. Every two days she crushes a 5 g piece with 10 cm³ of distilled water, filters the mixture and tests the filtrate with Benedict’s solution. She matches the final colour against a printed colour chart scored from 0 to 4.

day	0	2	4	6	8
colour after the test	blue	green	yellow	orange	brick-red
score from colour chart	0	1	2	3	4

(a) [4]

Describe how the student carries out the Benedict’s test on a sample of the banana filtrate. Give both the positive and the negative result.

Model Answer — 5a

add an excess of Benedict’s solution to the filtrate in a test tube [1]

heat the tube in a water bath at about 80 °C for a few minutes [1]

a positive result is a colour change from blue through green, yellow and orange to brick-red [1]

if no reducing sugar is present the solution stays blue [1]

⚠ If you missed marks here: The heating step is the mark most often lost, and “warm it” is not enough — say heat in a water bath. The colour change must be given as a change (blue to brick-red), not just an end colour, and the negative result must be stated as stays blue. Writing “it turns red” without mentioning the starting blue leaves the examiner unable to award the observation mark.

1 mark — add an excess of Benedict solution to the sample 1 mark — heat in a water bath at about eighty degrees for a few minutes 1 mark — positive result blue changing through green yellow orange to brick red 1 mark — negative result the solution stays blue

(b) [3]

(i) Describe the trend shown by these results. [1]

(ii) Explain why the Benedict’s test is described as semi-quantitative rather than quantitative, and state what the student would need in order to make it quantitative. [2]

Model Answer — 5b

(i) the amount of reducing sugar increases steadily as the banana ripens [1]

(ii) the colour only gives an estimate of the amount of reducing sugar, and it is judged by eye against a chart [1]

(ii) to make it quantitative the colour would have to be measured with an instrument such as a colorimeter, giving an actual numerical value [1]

⚠ If you missed marks here: A trend statement needs a direction and a variable: “the colour changes” describes the observation, not the biology. The semi-quantitative marks are about judgement by eye versus measurement by instrument; simply repeating the phrase “semi-quantitative” back at the examiner earns nothing.

1 mark — amount of reducing sugar increases as the banana ripens 1 mark — colour gives only an estimate judged by eye against a chart 1 mark — a colorimeter would give a numerical measurement making it quantitative

(i) Explain why the tubes are heated in a water bath rather than directly in a Bunsen flame. [2]

(ii) Explain why the same volume of Benedict’s solution must be added to every tube. [1]

Model Answer — 5c

(i) heating directly in a flame makes the liquid boil suddenly and spit out of the tube, which is a safety hazard [1]

(i) a water bath heats every tube evenly and to the same temperature, so the colours produced can be compared fairly [1]

(ii) if different volumes were used, the same amount of reducing sugar would give a different depth of colour, so the comparison between days would not be valid [1]

⚠ If you missed marks here: Two different ideas are being tested here and only one of them is safety. The second mark is about fair comparison: a water bath fixes the temperature for every tube, and a fixed volume of Benedict’s fixes the amount of reagent, so any difference in colour must come from the banana rather than from the method.

1 mark — direct flame heating causes sudden boiling and spitting which is dangerous 1 mark — water bath heats all tubes evenly to the same temperature for fair comparison 1 mark — equal volumes of Benedict solution needed or the colour comparison is not valid

Question 6 — A Flawed Emulsion Test

Total: 12 marks

A student in Lagos tests groundnut paste and a spread labelled “fat free”. He writes down what he did:

“I put a little of the 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 of his tubes went cloudy.

(a) [4]

Describe how the ethanol emulsion test should be carried out, in the correct order, and give both the positive and the negative result.

Model Answer — 6a

add the food sample to ethanol in a test tube and shake, so that any fat dissolves in the ethanol [1]

allow the solid to settle, then pour (decant) the ethanol into a second test tube containing distilled water [1]

a cloudy white emulsion forms if fat or oil is present [1]

if there is no fat, the mixture stays clear and colourless [1]

⚠ If you missed marks here: The order is the test. Ethanol first, water second — the fat has to be dissolved before it can be made to reappear. The positive result must be described as a cloudy white emulsion; “it goes milky” is usually accepted but “it goes white” alone is not, and “a precipitate forms” is wrong because nothing has reacted.

1 mark — food added to ethanol and shaken so fat dissolves 1 mark — ethanol decanted into a second tube of distilled water 1 mark — positive result is a cloudy white emulsion 1 mark — negative result stays clear and colourless

(b) [4]

Identify two errors in the student’s method and explain the effect of each one on his results.

Model Answer — 6b

first error: water was added to the food before the ethanol [1]

fat is insoluble in water, so it never dissolves and the test cannot work as intended [1]

second error: the ethanol was added to the same tube instead of being decanted into a separate tube of clean water [1]

undissolved food particles are still in the tube and make it look cloudy, giving a false positive — which is why even the “fat free” spread appeared to contain fat [1]

⚠ If you missed marks here: Naming the error only earns half of each pair of marks — the consequence must be spelled out. The single most valuable idea here is false positive: suspended food particles scatter light exactly like an emulsion does, so a cloudy tube is only evidence of fat when the liquid poured into the water was clear ethanol from which the solid had been allowed to settle.

1 mark — error one water added before the ethanol 1 mark — fat is insoluble in water so it never dissolves and the test fails 1 mark — error two ethanol added to the same tube rather than decanted into clean water 1 mark — suspended food particles make the tube cloudy giving a false positive

(i) Explain why a cloudy white emulsion forms when ethanol containing dissolved fat is poured into water. [2]

(ii) Describe a control the student should set up. [1]

(iii) State one safety precaution needed in this test. [1]

Model Answer — 6c

(i) fat dissolves in ethanol but is insoluble in water [1]

(i) so when the ethanol mixes with the water the fat comes out of solution as tiny droplets suspended in the water, which scatter light and look milky white [1]

(ii) repeat the whole procedure with ethanol and water but no food added; it should stay clear and colourless [1]

(iii) ethanol is highly flammable, so it must be kept away from naked flames [1]

⚠ If you missed marks here: An emulsion is a physical suspension of droplets, not a chemical product, so answers that talk about a reaction or a precipitate do not score. The control has to change one thing only — leave out the food and keep everything else identical — and a control that uses a food known to contain no fat is a different (and weaker) answer because the food itself could still cloud the tube.

1 mark — fat is soluble in ethanol but insoluble in water 1 mark — fat comes out as tiny droplets suspended in water that scatter light and look milky 1 mark — control of ethanol and water with no food which stays clear 1 mark — ethanol is highly flammable so keep away from naked flames

Question 7 — Food Stores in Seeds

Total: 10 marks

The table shows the composition of three kinds of seed, per 100 g of dry seed.

seed	carbohydrate (almost all starch) / g	protein / g	fat / g
wheat grain	71	13	2
sunflower seed	20	21	51
kidney bean	60	24	1

A student crushes a sample of each seed and tests it.

(a) [4]

(i) Predict the result of the iodine test on all three seeds and explain your prediction. [1]

(ii) Predict which seed gives the most obvious cloudy white emulsion, and explain why. [2]

(iii) State the result of the biuret test on the sunflower seed. [1]

Model Answer — 7a

(i) all three turn iodine solution blue-black, because all three contain starch [1]

(ii) the sunflower seed [1]

(ii) because it contains by far the most fat per 100 g (51 g, compared with 2 g and 1 g) [1]

(iii) a purple (violet) colour, because the sunflower seed contains protein [1]

⚠ If you missed marks here: Predictions from a data table must quote the numbers that justify them — naming the sunflower seed without the 51 g comparison is an assertion, not an explanation. Also watch the biuret result: it is purple or violet, from an original blue, and candidates who write “it turns pink” or “it turns lilac blue” are unlikely to be credited.

1 mark — all three give blue black with iodine because all contain starch 1 mark — sunflower seed predicted to give the strongest emulsion 1 mark — because it contains the most fat per hundred grams at fifty one grams 1 mark — biuret gives a purple colour showing protein

(b) [3]

Dry wheat grains give a negative Benedict’s result. After three days of germination, crushed wheat grains give a brick-red result. During germination the stored starch is broken down.

Explain these two results. [3]

Model Answer — 7b

in the dry grain the carbohydrate is stored as starch, which is not a reducing sugar, so Benedict’s stays blue [1]

during germination the starch is broken down into much smaller sugar molecules [1]

these sugars are reducing sugars, so Benedict’s now gives a brick-red result [1]

⚠ If you missed marks here: The whole answer turns on the fact that starch and reducing sugar are both carbohydrates but only one of them reacts with Benedict’s. A negative Benedict’s result on the dry grain therefore means “no reducing sugar”, never “no carbohydrate” — the iodine test on the same grain would be strongly positive.

1 mark — dry grain stores starch which is not a reducing sugar so Benedict stays blue 1 mark — during germination starch is broken down into smaller sugar molecules 1 mark — these sugars are reducing sugars so Benedict gives brick red

The biuret test gives a pale purple colour with the wheat grain and a deep purple colour with the kidney bean.

(i) State what this shows. [1]

(ii) Give two reasons why using the depth of purple to compare protein content is unreliable. [2]

Model Answer — 7c

(i) the kidney bean contains more protein than the wheat grain, which agrees with the figures in the table (24 g compared with 13 g) [1]

(ii) the depth of colour is judged by eye, so different people would judge it differently [1]

(ii) the mass of seed crushed and the volume of water used may not have been the same, so the two samples are not directly comparable [1]

⚠ If you missed marks here: Reliability answers need method faults, not vague doubts. “It might be wrong” scores nothing; “the colour is judged by eye rather than measured” and “the samples were not the same mass or dilution” are both specific, testable criticisms. Note that the biuret test is not officially semi-quantitative in the way Benedict’s is, which is exactly why this comparison is shaky.

1 mark — kidney bean contains more protein than the wheat grain agreeing with the table 1 mark — depth of colour is judged by eye so it is subjective 1 mark — mass of seed and volume of water may differ so samples are not comparable

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