Cambridge International AS & A Level Biology 9700: Testing for biological molecules – Practice Questions

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 sub-topic 2.1, Testing for biological molecules, from topic 2 of the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027. It is AS Level content, tested on Paper 1 and Paper 2 and assumed for Paper 4. Questions 4 and 8 to 10 use the Paper 3 skills of identifying unknown solutions and estimating concentrations, and question 11 is a Paper 5 style plan.

Revise first with the study guide and the revision notes. For all of topic 2, use the biological molecules practice questions. The 9700 course hub and the printable checklist show where this sub-topic sits, and the free 9700 AS diagnostic checks your AS topics together.

Questions

1. State the colour of Benedict’s solution before it is heated with a sample, and describe the sequence of colours seen as the concentration of reducing sugar increases. [2]

2. Describe how to carry out the biuret test on a solution, using potassium hydroxide solution and copper sulfate solution, and state the result if protein is present. [2]

3. A student crushes a peanut, adds water, shakes the tube, then adds ethanol. The mixture stays clear. Explain why this result does not show that the peanut contains no lipid, and state how the procedure should be changed. [3]

4. Five tubes, V to Z, each contain one of: maltose solution, a mixture of starch and albumen, vegetable oil, sucrose solution, distilled water. The results are:

Tube	Benedict’s (heated)	Iodine	Biuret	Emulsion	Benedict’s after acid hydrolysis and neutralisation
V	orange	orange-brown	blue	clear	orange
W	blue	blue-black	purple	clear	not tested
X	blue	orange-brown	blue	cloudy white	not tested
Y	blue	orange-brown	blue	clear	brick-red
Z	blue	orange-brown	blue	clear	blue

Identify the contents of each tube. [5]

5. A solution is known to contain no reducing sugar.

(a) Describe how you would test this solution for a non-reducing sugar. [5] (b) Explain why sucrose gives a positive result in this test but a negative result in a Benedict’s test alone. [2]

6. A student needs 20 cm³ each of 0.2, 0.4, 0.6, 0.8 and 1.0 % glucose solution, made from a 1.0 % stock by proportional dilution. Calculate the volumes of stock and distilled water needed for each. [3]

7.

(a) A 2.0 % glucose solution is used to make a serial dilution in which each solution is half the concentration of the one before. State the concentrations of the first five solutions, including the stock. [2] (b) A glucose solution has a concentration of 0.050 mol dm⁻³. Express this as % (w/v). (Mr of glucose = 180) [2]

8. Two plant extracts were tested with Benedict’s solution before and after acid hydrolysis and neutralisation. Volumes, temperature and heating time were the same in every test.

Extract	Before hydrolysis	After hydrolysis
A	yellow	brick-red
C	brick-red	brick-red

Explain what these results show about the sugars in each extract. [4]

9. A student made Benedict’s colour standards from glucose solutions of 0.2, 0.4, 0.6, 0.8 and 1.0 %, heating each for 4 minutes. A sample of fruit juice, treated the same way, gave a brick-red colour that matched the 1.0 % standard.

(a) Explain why colour standards give an estimate as a range rather than an exact value. [1] (b) Suggest how the student could estimate the concentration of reducing sugar in the juice. [2] (c) Suggest two improvements to make the colour comparison more accurate. [2]

10. A student timed how long Benedict’s solution took to show its first colour change with five glucose solutions. Each was repeated three times.

Glucose / %	Time 1 / s	Time 2 / s	Time 3 / s	Mean time / s
0.25	212	208	216	212
0.50	118	122	120	120
0.75	84	86	85	85
1.00	66	64	92
1.25	52	54	53	53

(a) Identify the anomalous result and calculate the mean time for 1.00 % glucose. [2] (b) Calculate the rate of the reaction, 1/t, for 0.50 % glucose. Give your answer to 3 significant figures. [1] (c) A solution of unknown concentration, tested the same way, took a mean of 100 s. State the range within which its concentration lies and explain your answer. [2] (d) State three variables that must be standardised in this investigation, and how each would be kept constant. [3]

11. Plan an investigation to compare the concentration of reducing sugar in the juice of three varieties of apple, using the time to first colour change in a Benedict’s test. Your plan should include how you would estimate concentrations, controls, and one safety precaution. [8]

Answers

1. Blue before heating [1]. With more reducing sugar: green → yellow → orange → brick-red [1]. Examiner insight: Both ends of the series are needed for the second mark; “it goes red” alone does not describe a sequence.

2. Add potassium hydroxide solution to the sample, then a few drops of dilute copper sulfate solution, and mix (no heating) [1]. Protein present: the blue solution turns purple / lilac [1]. Examiner insight: The reagent order and the final colour gain the marks; adding a heating step can contradict an otherwise correct method.

3. Lipids are insoluble in water, so adding water first does not dissolve the lipid [1]. With no lipid dissolved in ethanol, no droplets come out of solution, so no emulsion forms: a false negative [1]. Change: shake the crushed peanut with ethanol first, then pour the ethanol into water and look for a cloudy white emulsion [1]. Examiner insight: “The order is wrong” without a reason about solubility scores only the procedural mark.

4. V: maltose (reducing sugar) [1]. W: starch and albumen (iodine and biuret both positive) [1]. X: vegetable oil [1]. Y: sucrose (negative Benedict’s, positive after hydrolysis) [1]. Z: distilled water (negative in every test) [1]. Examiner insight: One mark per correct identification; if the command word is “explain”, evidence from the table is also required.

5. (a) Do a Benedict’s test first and confirm the solution stays blue [1]. Add dilute hydrochloric acid to a fresh sample and heat in a water bath close to boiling [1]. Cool, then neutralise with sodium hydrogencarbonate until fizzing stops / check with pH paper [1]. Add Benedict’s solution and heat in the water bath [1]. A change from blue to green, yellow, orange or brick-red shows a non-reducing sugar is present [1]. (b) Sucrose is a non-reducing sugar, so it cannot reduce the copper(II) ions in Benedict’s solution [1]. Acid hydrolysis breaks the glycosidic bond, releasing glucose and fructose, which are reducing sugars [1]. Examiner insight: The neutralisation step is a separate marking point in (a); a method that goes straight from acid to Benedict’s loses that mark even if every other step is correct.

6. Volume of stock = (concentration / 1.0) × 20 [1]. 0.2 %: 4 cm³ stock + 16 cm³ water; 0.4 %: 8 cm³ + 12 cm³ [1]. 0.6 %: 12 cm³ + 8 cm³; 0.8 %: 16 cm³ + 4 cm³; 1.0 %: 20 cm³ + 0 cm³ [1]. Examiner insight: Giving stock volumes without the water volumes loses the accuracy marks.

7. (a) 2.0, 1.0, 0.50, 0.25, 0.125 %: at least three correct [1]; all five correct [1]. (b) 0.050 × 180 = 9.0 g dm⁻³ [1]; 9.0 / 10 = 0.90 % (w/v) [1]. Examiner insight: In (b) the method mark is for using Mr to get a mass per dm³; forgetting to divide by 10 for 100 cm³ gives 9.0 % and loses the accuracy mark.

8. A: yellow before hydrolysis shows a reducing sugar is present [1]. The colour moves to brick-red after hydrolysis, so more reducing sugar was released, showing a non-reducing sugar is also present [1]. C: brick-red before shows a reducing sugar is present [1]. Nothing can be concluded about a non-reducing sugar in C: the colour was already at the end of the series, so any increase cannot show [1]. Examiner insight: For C, saying “no non-reducing sugar” is not credited; the mark is for recognising that the test cannot show any increase.

9. (a) Colour matching only places the sample between or at standards; the colours are ordinal and cannot give values between them [1]. (b) Dilute the juice by a known factor (for example tenfold) with distilled water and repeat the test [1]. Match the diluted juice to the standards, then multiply the estimate by the dilution factor [1]. (c) Any two [2]: more standards at smaller intervals; view tubes against a white tile in the same light; a second person judges the colours; heat all tubes together in a thermostatically controlled bath. Examiner insight: In (b), multiplying back by the dilution factor is a separate marking point.

10. (a) The anomalous result is 92 s at 1.00 % [1]. Mean of the other two = (66 + 64) / 2 = 65 s [1]. (b) 1/120 = 0.00833 s⁻¹ [1]. (c) Between 0.50 % and 0.75 % [1]. 100 s is shorter than the 0.50 % time (120 s) but longer than the 0.75 % time (85 s), and a shorter time means a higher concentration [1]. (d) Any three, each with a method [3]: volume of juice/glucose solution, with the same syringe volume; volume of Benedict’s, same volume and in excess; temperature, thermostatically controlled water bath; judging the colour change, same observer against a white tile. Examiner insight: In (a), including 92 s gives 74 s and loses the accuracy mark; in (d) a variable named without how it is controlled earns no credit.

11. One mark per point:

• Independent variable: apple variety; dependent variable: time to first colour change from blue; extract juice by crushing and filtering the same mass of each variety [1].
• Make at least five known glucose concentrations by proportional or serial dilution [1].
• Add the same volume of juice or standard (for example 2 cm³) to the same volume of Benedict’s solution, in excess [1].
• Heat in a thermostatically controlled water bath close to boiling; start timing as the tube goes in [1].
• Record the time to the first colour change; three replicates of each and calculate means [1].
• Plot mean time against glucose concentration and read the juice concentrations from the curve [1].
• Control: distilled water with Benedict’s, which should stay blue [1].
• Safety: eye protection (Benedict’s is a moderate hazard) or a holder for hot tubes [1]. Examiner insight: Planning marks need stated volumes, temperatures and methods; “fair test” on its own scores nothing.

Where marks are usually lost

• Writing “Benedict’s turns red” with no heating, or no water bath.
• Missing the neutralisation step after acid hydrolysis.
• Adding water before ethanol in the emulsion test.
• Treating a shorter time to colour change as a lower concentration.
• Including an anomalous replicate in a mean.
• Giving a single exact value from colour standards.
• Concluding “no non-reducing sugar” when Benedict’s is already brick-red before hydrolysis.

Next steps

• Go back over the testing for biological molecules revision notes.
• Re-read any weak section in the testing for biological molecules study guide.
• Revise the rest of topic 2 with the biological molecules study guide.
• See the 9700 course hub and the printable checklist.
• Take the free 10-minute 9700 AS diagnostic.
• Book a free trial class.

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for 2025, 2026 and 2027 (Version 1), Cambridge International (Cambridge University Press & Assessment). Topic 2, Biological molecules, sub-topic 2.1 Testing for biological molecules.