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Cambridge IGCSE Biology 0610: Biological molecules – Practice Questions

Original practice questions with marked answers on Cambridge IGCSE Biology 0610 biological molecules: elements, food tests and DNA base pairing.

Subject
Biology
Level
IGCSE
Topic
Biological molecules
Updated

Aligned to Cambridge IGCSE Biology (0610), For examination in 2026, 2027 and 2028. Official specification .

Syllabus page (what it covers and how it is assessed): Cambridge IGCSE Biology.

Syllabus points this page covers, with Core and Extended

0610

  • 4 Biological molecules (whole topic)
  • 4.1 Biological molecules · Core and Extended

"Core and Extended" means part of that syllabus point is Extended only. The page's own tier notes say which part.

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These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover topic 4, Biological molecules (sub-topic 4.1), of the Cambridge IGCSE Biology 0610 syllabus for examination in 2026, 2027 and 2028. Outcomes 1 to 3 (elements, building blocks and food tests) are Core. Outcome 4, the structure of DNA, is Supplement content: questions and parts on it are labelled (Extended), and a Core candidate can skip them. Several questions use the practical skills tested on Paper 5 and Paper 6.

Revise first with the study guide and the revision notes. The 0610 course hub and printable checklist show where this topic sits in the course.

Questions

1. State the chemical elements present in:

(a) starch [1] (b) protein. [1]

2. State the smaller molecules from which each of these is made:

(a) cellulose [1] (b) protein [1] (c) fats. [1]

3. A student wants to find out whether a slice of banana contains starch. Describe how she should test the banana, and state the result she would see if starch is present. [3]

4. Five sugar solutions, P, Q, R, S and T, are each tested with Benedict’s solution under the same conditions. The final colours are: P green, Q blue, R brick-red, S yellow, T orange.

(a) Put the solutions in order of reducing sugar concentration, highest first. [2] (b) State which solution contained no detectable reducing sugar. [1] (c) State two variables that must be kept the same for this comparison to be valid. [2]

5. Describe how you would carry out the ethanol emulsion test on a peanut, and state one safety precaution you would take. [4]

6. A student places 1.0 cm³ of DCPIP solution in each of three test-tubes. She adds three different fruit juices, X, Y and Z, drop by drop until the DCPIP is decolourised. The volumes needed are: X 1.5 cm³, Y 4.5 cm³, Z 3.0 cm³.

(a) State the colour change at the end point. [1] (b) State which juice contains the most vitamin C. Explain your answer. [2] (c) Calculate how many times more concentrated the vitamin C is in juice X than in juice Y. [2] (d) Suggest one way to make the results more reliable. [1]

7. (Extended) Describe the structure of a DNA molecule. [4]

8. (Extended) In a sample of DNA, 22% of the bases are T. Calculate the percentage of each of the other three bases. [3]

9. (Extended)

(a) One strand of a DNA molecule has the base sequence C G G A T T C A. Write the sequence of bases on the other strand. [1] (b) A short section of DNA is 40 base pairs long. 13 of the base pairs are A–T pairs. Calculate the number of C bases in this section. [3]

10. A technician has a white powder labelled “sports drink mix”. She dissolves it in distilled water and tests the solution. Her results are shown below.

Test Final colour or appearance
Iodine solution orange-brown
Benedict’s solution, heated orange
Biuret solution purple
Ethanol emulsion clear

(a) State what each of the four results shows about the powder. [4] (b) Describe a control she should set up for the biuret test, and explain its purpose. [2] (c) Protein in the drink is digested in the body. Name the smaller molecules produced, and name the element they contain that glucose does not contain. [2]

11. A student makes an extract from dry bean seeds and another from bean seeds that have been germinating for five days. She tests both extracts. Iodine solution turns blue-black with both extracts. Benedict’s solution stays blue with the dry-seed extract but turns orange with the germinating-seed extract.

(a) Describe how she should carry out the Benedict’s test on an extract. [3] (b) Starch is made from glucose. Suggest an explanation for the Benedict’s results. [2] (c) (Extended) In a sample of bean DNA, 18% of the bases are C. Calculate the percentage of bases that are A. [2] (d) (Extended) State how the two strands of DNA are held together, and explain why the bases on one strand decide the bases on the other. [2]

Answers

1. (a) Carbon, hydrogen and oxygen (C, H, O) [1] (b) Carbon, hydrogen, oxygen and nitrogen (C, H, O, N) [1]; sulfur may be added but is not needed. Examiner insight: Each part needs the full list; a list for protein without nitrogen scores zero for that part, and adding nitrogen to starch also loses the mark.

2. (a) Glucose [1] (b) Amino acids [1] (c) Fatty acids and glycerol [1] Examiner insight: In (c) both molecules are needed for the single mark; “fatty acids” alone is incomplete, and “sugars” for cellulose is too vague where the syllabus names glucose.

3. Add a few drops of iodine solution to the banana slice [1]. If starch is present, the colour changes from orange-brown [1] to blue-black [1]. Examiner insight: The colour change is worth two separate marks, one for each colour; “goes black” or a final colour with no starting colour limits you to fewer marks.

4. (a) R (highest) and Q (lowest) at the two ends [1]; T, S, P in that order between them [1]: R, T, S, P, Q. (b) Q [1], because it stayed blue. (c) Any two of: volume of sugar solution; volume of Benedict’s solution; temperature of the water bath; heating time [1] [1]. Examiner insight: For “state two variables”, each correct variable scores separately; vague answers such as “the amount” or “same conditions” do not score unless you name what is measured.

5. Crush the peanut and shake it with ethanol [1]. Pour the ethanol into a test-tube of water [1]. A cloudy white emulsion shows that fat or oil is present [1]. Ethanol is flammable, so keep it away from naked flames [1]. Examiner insight: The order is marked: ethanol first, then water; a description that adds water first loses the method mark, and a safety point needs the reason (flammable), not just “be careful”.

6. (a) From blue to colourless [1]. (b) Juice X [1]. It needed the smallest volume to decolourise the same volume of DCPIP [1]. (c) 4.5 ÷ 1.5 [1] = 3 times [1]. (d) Repeat each juice several times and calculate a mean / add the juice in small measured volumes from a syringe [1]. Examiner insight: In (b) the choice of X earns its mark only with the reason about volume; answers that pick Y because it “used the most juice” have the relationship backwards and score zero.

7. (Extended) Two strands coiled together to form a double helix [1]. Each strand contains chemicals called bases [1]. Bonds between pairs of bases hold the two strands together [1]. Bases always pair the same way: A with T and C with G [1]. Examiner insight: Four separate structural points are needed for four marks; “a double helix” alone is one mark, and the letters A, T, C and G are accepted because full base names are not required.

8. (Extended) A pairs with T, so A = 22% [1]. C + G = 100 − (22 + 22) = 56% [1]. C and G are equal, so C = 28% and G = 28% [1]. Examiner insight: Show the subtraction; a wrong final answer can still earn the method mark if the working shows A = T and C = G, but a bare wrong number earns nothing.

9. (a) (Extended) G C C T A A G T [1] (b) 40 base pairs, of which 13 are A–T pairs [1]. So 40 − 13 = 27 are C–G pairs [1]. Each C–G pair contains one C, so there are 27 C bases [1]. Examiner insight: In (b) the common slip is to count 80 bases and then divide wrongly; setting out the pairs in words earns the method marks even if the arithmetic slips.

10. (a) No starch detected (iodine stayed orange-brown) [1]. Reducing sugar present (Benedict’s turned orange) [1]. Protein present (biuret turned purple) [1]. No fat detected (the emulsion test stayed clear) [1]. (b) Test distilled water with biuret solution in the same way [1]. It shows that the reagent alone does not turn purple, so the purple colour came from the powder [1]. (c) Amino acids [1]; the element is nitrogen [1]. Examiner insight: Each conclusion in (a) is a separate mark and must name the nutrient; “positive” or “negative” without saying which molecule does not score.

11. (a) Add an equal volume of Benedict’s solution to the extract [1]. Heat in a water bath for a few minutes [1]. A colour change from blue to green, yellow, orange or brick-red shows a reducing sugar [1]. (b) In the germinating seeds, some starch has been broken down [1] into reducing sugar / glucose, which Benedict’s detects [1]. (c) (Extended) G = 18%, so A + T = 100 − 36 = 64% [1]. A = 32% [1]. (d) (Extended) By bonds between pairs of bases [1]. Bases always pair A with T and C with G, so each base on one strand fixes its partner on the other [1]. Examiner insight: A “suggest” question like (b) rewards a reasoned link to the data; stating that the germinating seeds “have more sugar” repeats the result and gains no credit without the idea that starch was broken down.

Where marks are usually lost

  • Listing C, H and O for protein and forgetting nitrogen.
  • Giving “fatty acids” without glycerol as the building blocks of fats and oils.
  • Describing a colour change with only the end colour, or writing “black” for starch.
  • Forgetting to say “heat” or “water bath” in the Benedict’s method.
  • Getting the emulsion-test order wrong: ethanol first, then water.
  • Reading DCPIP volumes backwards: the smallest volume shows the most vitamin C.
  • Naming vague control variables such as “same conditions” instead of a measurable quantity.
  • (Extended) Pairing A with C or G with T, or giving only “double helix” for a four-mark structure question.
  • (Extended) Forgetting that A = T and C = G before splitting the remaining percentage.

Next steps

Official syllabus

Cambridge IGCSE Biology 0610 syllabus for examination in 2026, 2027 and 2028 (Version 3), Cambridge International. Topic 4, Biological molecules, sub-topic 4.1.

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