Paper 62, worked through. Cambridge IGCSE Biology, May/June 2026: worked solutions

Forty marks in an hour. This paper is about doing an experiment on paper: reading scales, drawing a results table, plotting a graph, drawing a specimen, and planning an investigation. Bring a ruler and a sharp pencil. Most marks are for doing a routine thing carefully, so learn the routines: table headings with units, labelled axes, one clean outline.

1. 1(a)(i)Working out the concentration of a diluted solution.[1]
   Show how to do itHide the working

   1. Beaker V3 holds 4 cm³ of the 0.6% solution and 8 cm³ of water: 12 cm³ in all.
   2. The vitamin C solution is 4 parts out of 12, which is one third, so the concentration is one third of 0.6%.
   3. 0.6 × 4 ÷ 12 = 0.2%.

   Answer0.2

   Check it against the row above: 8 cm³ out of 12 gave 0.4%, two thirds of 0.6. The pattern 0.6, 0.4, 0.2 confirms the answer.

   Revise this: Biological molecules and food tests
2. 1(a)(ii)Drawing a results table and filling it in from three syringe diagrams.[3]
   Show how to do itHide the working

   1. Draw a table with ruled lines and a heading row. You need two columns: "percentage concentration of vitamin C solution / %" and "volume of iodine solution remaining / cm³".
   2. Read each syringe. There are five divisions between each number, so each one is 0.2 cm³. The iodine solution remaining is 1.4 cm³ in V1, 3.0 cm³ in V2 and 4.6 cm³ in V3.
   3. Enter the concentrations 0.6, 0.4 and 0.2 with the three volumes beside them.

   AnswerA ruled table headed concentration / % and volume remaining / cm³, with the rows 0.6 and 1.4, 0.4 and 3.0, 0.2 and 4.6.

   Three marks: a table with a header line, headings that include units, and correct data. Put the units in the headings only. Writing "cm³" after each number in the cells loses the heading mark.

   Revise this: Biological molecules and food tests
3. 1(a)(iii)Finding how much iodine solution was added to each tube.[1]
   Show how to do itHide the working

   1. Each syringe started with 5.0 cm³. Volume added = 5.0 − volume remaining.
   2. V1: 5.0 − 1.4 = 3.6 cm³. V2: 5.0 − 3.0 = 2.0 cm³. V3: 5.0 − 4.6 = 0.4 cm³.

   AnswerV1 3.6 cm³, V2 2.0 cm³, V3 0.4 cm³

   All three must be right for the one mark. They fit the idea in the question: the stronger the vitamin C solution, the more iodine solution it takes.

   Revise this: Biological molecules and food tests
4. 1(a)(iv)Naming the dependent variable.[1]
   Show how to do itHide the working

   1. The independent variable is what the student changes: the concentration of vitamin C.
   2. The dependent variable is what is measured as a result: the volume of iodine solution.

   AnswerThe volume of iodine solution added (or remaining in the syringe).

   Independent is what you change. Dependent is what you measure. The colour change is the end-point you watch for, not the variable.

   Revise this: Biological molecules and food tests
5. 1(a)(v)Why the mixture turns blue-black.[1]
   Show how to do itHide the working

   1. Step 4 put starch suspension into every test-tube.
   2. Iodine solution turns blue-black when starch is present.

   AnswerThe tubes contain starch, and iodine solution turns blue-black with starch.

   While vitamin C is left it reacts with the iodine, so no colour appears. Once the vitamin C is used up, the next drop of iodine meets the starch. That is why the colour "remains".

   Revise this: Biological molecules and food tests
6. 1(a)(vi)The effect of using one syringe for all three vitamin C solutions.[1]
   Show how to do itHide the working

   1. The syringe was used for the strongest solution first. Some of that solution stays in the syringe and is carried into the weaker ones.
   2. Test-tube V3 would then contain more vitamin C than it should, so more iodine solution would be needed.

   AnswerA greater volume of iodine solution would be required.

   This is contamination. The fix is a clean syringe for each solution, or rinsing it between uses.

   Revise this: Biological molecules and food tests
7. 1(a)(vii)Why a volume is a better measurement than a count of drops.[1]
   Show how to do itHide the working

   1. Drops are not all the same size, so the same number of drops can be different volumes.
   2. It is also easy to lose count.

   AnswerDrops vary in volume (or: it is easy to miscount them).

   Revise this: Biological molecules and food tests
8. 1(b)(i)Calculating a mean of three readings.[1]
   Show how to do itHide the working

   1. Add the three volumes: 2.8 + 2.9 + 2.7 = 8.4.
   2. Divide by the number of readings: 8.4 ÷ 3 = 2.8.

   Answer2.8 cm³

   Give a mean to the same number of decimal places as the readings it came from.

   Revise this: Biological molecules and food tests
9. 1(b)(ii)Estimating the vitamin C concentration of the drink by comparison.[1]
   Show how to do itHide the working

   1. Compare the drink's 2.8 cm³ with the known solutions: 0.4% needed 2.0 cm³ and 0.6% needed 3.6 cm³.
   2. 2.8 is between 2.0 and 3.6, so the concentration is between 0.4% and 0.6%. It is exactly halfway, which suggests about 0.5%.

   AnswerAbout 0.5% (any value between 0.4% and 0.6% is accepted).

   An estimate needs a number, not just "between 0.4 and 0.6". Say where between the two your value falls.

   Revise this: Biological molecules and food tests
10. 1(b)(iii)A change that would make the estimate more trustworthy.[1]
    Show how to do itHide the working

    1. The estimate lies in the gap between two known solutions, 0.4% and 0.6%. More known solutions inside that gap would narrow it down.

    AnswerTest more concentrations of vitamin C between 0.4% and 0.6% (use smaller intervals).

    The question asks about the procedure in part (a), which made the known solutions. Repeating the drink test is already done in step 19.

    Revise this: Biological molecules and food tests
11. 1(b)(iv)Describing the test for reducing sugars.[2]
    Show how to do itHide the working

    1. Add Benedict's solution to a sample of the drink.
    2. Heat the mixture, in a hot water-bath. If a reducing sugar is present, the blue solution changes colour, through green and yellow to orange or brick-red.

    AnswerAdd Benedict's solution and heat. A change from blue to green, orange or brick-red shows that reducing sugar is present.

    Two marks: the reagent and the heating. Forgetting to heat is the usual way the second mark is lost.

    Revise this: Biological molecules and food tests
12. 1(b)(v)Another test for vitamin C.[1]
    Show how to do itHide the working

    1. Vitamin C decolourises DCPIP, a blue solution.

    AnswerThe DCPIP test.

    The food tests to know: iodine for starch, Benedict's for reducing sugars, biuret for protein, ethanol emulsion for fats and oils, DCPIP for vitamin C.

    Revise this: Biological molecules and food tests
13. 1(c)(i)Plotting a line graph of vitamin C concentration against drying time.[4]
    Show how to do itHide the working

    1. Axes: drying time / hours along the x-axis (the variable that was changed), and vitamin C concentration / mg per 100 g of kiwi fruit up the y-axis.
    2. Scales: even steps that spread the points over more than half the grid, for example 0 to 16 hours across and 0 to 250 up.
    3. Plot (0, 90), (4, 125), (8, 185), (12, 230) and (16, 230) with small neat crosses.
    4. Join the points with ruled straight lines from one to the next. Do not extend the line beyond the first or last point.

    AnswerLabelled axes with units, even scales, five accurate points, joined point to point with a ruler.

    One mark each for labels with units, scales, plotting and the line. In biology a line graph of results is joined dot to dot with a ruler, or drawn as a smooth line that touches every point. Keep the line thin.

    Revise this: A balanced diet
14. 1(c)(ii)Describing and explaining how the vitamin C concentration changes as the fruit dries.[2]
    Show how to do itHide the working

    1. Describe: the concentration rises from 90 to 230 mg per 100 g over the first 12 hours, then stays at 230.
    2. Explain: drying removes water, so each 100 g of fruit contains more of everything else, including vitamin C. After 12 hours no more water can be lost, so the concentration stops rising.

    AnswerThe concentration increases and then levels off after 12 hours. It increases because water is lost from the fruit, and levels off when no more water can be lost.

    "Describe and explain" is two jobs for two marks: what happens, and why. No vitamin C is being made. The same amount is in less fruit.

    Revise this: A balanced diet
15. 1(c)(iii)Reading a value from the graph for a time that was not tested.[2]
    Show how to do itHide the working

    1. Find 10 hours on the x-axis and rule a line up to your graph line.
    2. From there rule a line across to the y-axis and read the value.
    3. Ten hours is halfway between 8 hours (185) and 12 hours (230), so on a point-to-point graph the answer is about 208.

    AnswerAbout 208 mg per 100 g (read from your own graph).

    One mark is for the construction lines on the graph, so draw them. The other is for a value that matches your line.

    Revise this: A balanced diet
16. 2(a)(i)Making a large drawing of three beans in a pod.[4]
    Show how to do itHide the working

    1. Draw it at least as big as the photograph, in the space given, without running into the text.
    2. Use one continuous, clean pencil line for the pod and for each bean. No shading and no sketchy double lines.
    3. Get the proportions and positions right: the top bean and the middle bean are close together, with a bigger gap between the middle bean and the bottom one.
    4. Show the top bean reaching the edge of the pod.

    AnswerA large, unshaded outline drawing of the pod and its three beans, with the top two beans close together and the top bean touching the edge of the pod.

    The four marks are for the outline, the size and two details of what is actually in the photograph. Draw what you see, not what you think a bean pod looks like. Never use a ruler for the outline, and never shade.

    Revise this: Magnification
17. 2(a)(ii)Measuring a line and calculating a magnification to two significant figures.[3]
    Show how to do itHide the working

    1. Measure line PQ with a ruler: 33 mm.
    2. Magnification = length of line PQ ÷ actual length = 33 ÷ 19 = 1.7368…
    3. To two significant figures that is 1.7.

    AnswerPQ = 33 mm; magnification ×1.7

    Both lengths must be in the same unit, here millimetres. A magnification has no unit, so do not write "mm" after it. Round only at the end.

    Revise this: Magnification
18. 2(b)(i)The independent variable in the germination experiment.[1]
    Show how to do itHide the working

    1. It is the thing the student deliberately changed: the temperature, set at 10, 20, 30, 40 and 50 °C.

    AnswerTemperature.

    Revise this: Reproduction in flowering plants
19. 2(b)(ii)Something that was kept the same.[1]
    Show how to do itHide the working

    1. Read the method for what did not change between temperatures: 200 beans each time, and 45 cm³ of water each day.

    AnswerThe volume of water added each day (or the number of beans).

    Take your answer from the method given. Anything not mentioned there, such as light, cannot be said to have been kept constant.

    Revise this: Reproduction in flowering plants
20. 2(b)(iii)Why so many beans were planted.[1]
    Show how to do itHide the working

    1. Some beans may never germinate, and individual beans vary. With 200, one odd bean hardly changes the mean, and unusual results can be spotted.

    AnswerSo that anomalous results can be identified and the mean is reliable (some beans may not germinate).

    A large sample does not prevent anomalies. It lets you see them and reduces their effect, and the mark scheme rejects "to prevent anomalous results".

    Revise this: Reproduction in flowering plants
21. 2(b)(iv)The apparatus for measuring 45 cm³ of water.[1]
    Show how to do itHide the working

    1. A measuring cylinder measures volumes of liquid of this size. A 50 cm³ one would suit.

    AnswerA measuring cylinder.

    Revise this: Reproduction in flowering plants
22. 2(c)Planning an investigation into whether the age of beans affects how much oxygen their catalase releases.[6]
    Show how to do itHide the working

    1. What to change: use beans of at least two different ages, for example fresh beans and beans stored for a year.
    2. Prepare them: crush or blend the same mass of each kind of bean with a little water, to release the catalase.
    3. Start the reaction: add the crushed beans to a measured volume of hydrogen peroxide solution in a flask.
    4. Measure: collect the oxygen in a gas syringe, or over water in an upturned measuring cylinder, and record the volume collected in a fixed time, such as two minutes.
    5. Keep the same: the type and mass of bean, the volume and concentration of hydrogen peroxide, and the temperature.
    6. Repeat each age at least three times and work out a mean. Wear eye protection and gloves, because hydrogen peroxide is an irritant.

    AnswerUse beans of different ages. Crush the same mass of each, add to the same volume and concentration of hydrogen peroxide at the same temperature, and collect the oxygen in a gas syringe for a fixed time. Repeat three times, and wear goggles.

    Six marks from nine possible points. Three of them are for naming variables to keep constant, so list at least three. State how the gas is collected and measured: "see how much gas is made" is not a method.

    Revise this: Enzymes

These explanations are Papermunch’s own, written to teach the method. The answers have been checked against the exam board’s mark scheme, which has the final say: open it above. The question paper and mark scheme belong toCambridge University Press & Assessment: each question shown here is drawn from the paper itself as you read, and is not kept on this site.