0610/51 Oct/Nov 2025 Worked Answers (IGCSE Biology 0610 Core)

17 questions · 40 marks · 75 minutes

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1. Question 1(a)(i)

   1 marksRecording temperatures in a results table
   Step 1: To check whether the temperature stayed steady, each waterbath needs both a starting and a final temperature. Step 2: There are two waterbaths, the cold one (C) and the warm one (W). Step 3: Two readings for each of the two waterbaths gives four temperatures in total.
   Method:

   Count one start plus one final reading for each of the two waterbaths.
   Examiner tips

   • Record a start and a final temperature for every waterbath used.

2. Question 1(a)(ii)

   5 marksDesigning a results table
   Step 1: A results table needs ruled columns with a header line and headings that give the units, here the distance in mm. Step 2: For each stalk it must record the number of sections cut and the total distance the dye moved, for both beaker C and beaker W. Step 3: Because warmth speeds up water movement, the dye moves further in the warm beaker W than in the cold beaker C.
   Method:

   Check the table has headed columns with units for both beakers, then apply the warm-is-faster trend.
   Examiner tips

   • Ruled columns, a header line, units in the headings, and both beakers recorded.

3. Question 1(a)(iii)

   2 marksCalculating the rate of movement
   Step 1: Rate of movement = total distance moved ÷ time taken. Step 2: = 30 mm ÷ 3 minutes = 10. Step 3: The distance is in mm and the time is in minutes, so the rate is 10 mm per minute.
   Method:

   Divide the total distance by the time and state the unit as mm per minute.
   Examiner tips

   • Rate = distance ÷ time; always give the unit that matches the quantities used.

4. Question 1(a)(iv)

   1 marksConclusion: temperature and water movement
   Step 1: The dye marks how far the water has moved up the xylem of the stalk. Step 2: In the same time the dye moved further in the warm waterbath than in the cold one. Step 3: So a higher temperature gives a greater rate of movement of water up the stalk.
   Method:

   Relate the greater distance in the warm beaker to a faster rate at higher temperature.
   Examiner tips

   • A conclusion links the variable changed (temperature) to the result (rate of movement).

5. Question 1(a)(v)

   1 marksReasons for repeating an investigation
   Step 1: A single set of readings could contain a mistake that goes unnoticed. Step 2: Repeating the investigation lets anomalous results, ones that do not fit the trend, be spotted. Step 3: Removing or checking these makes the overall results more reliable.
   Method:

   Explain repeats in terms of spotting anomalies and improving reliability.
   Examiner tips

   • Repeats let you find anomalies and improve reliability.

6. Question 1(b)

   2 marksVariables kept constant
   Step 1: A fair test changes only the variable being investigated, here the temperature, and keeps everything else the same. Step 2: The length of the stalk and the time it spends in the dye must be the same for both beakers. Step 3: Temperature is the variable being changed, and the distances and sections are what is measured, so they are not control variables.
   Method:

   Pick the quantities kept the same, excluding the temperature changed and the results measured.
   Examiner tips

   • Keep the stalk length, dye and timing the same; only the temperature should change.

7. Question 1(c)(i)

   2 marksIdentifying a source of error and an improvement
   Step 1: The changing start and final temperatures show the water temperature did not stay constant, which is the source of error. Step 2: An improvement must fix that error directly. Step 3: A thermostatically controlled waterbath keeps the temperature steady, so it reduces this error.
   Method:

   Read the temperature change as the error, then pick the improvement that controls temperature.
   Examiner tips

   • Match the improvement to the actual error: drifting temperature needs a controlled waterbath.

8. Question 1(c)(ii)

   1 marksImproving the precision of a measurement
   Step 1: Each 5 mm section can only locate the dye to the nearest 5 mm, which is imprecise. Step 2: Cutting thinner sections narrows down where the dye stops. Step 3: So thinner sections give a more precise final position for the dye.
   Method:

   Choose the change that reduces the step size used to find the dye end-point.
   Examiner tips

   • Finer divisions give a more precise measurement of where the dye reaches.

9. Question 1(d)(i)

   3 marksCalculating actual size from magnification
   Step 1: Rearrange magnification = picture length ÷ actual length to give actual length = picture length ÷ magnification. Step 2: Actual length = 60 ÷ 35 = 1.71… mm. Step 3: To one significant figure this is 2 mm.
   Method:

   Divide 60 by 35 and round the answer to one significant figure.
   Examiner tips

   • actual size = picture size ÷ magnification; then round as asked.

10. Question 1(f)

    1 marksTesting for carbon dioxide
    Step 1: Limewater is the usual indicator for carbon dioxide. Step 2: When carbon dioxide is bubbled through it, limewater turns milky (cloudy). Step 3: Iodine tests for starch, biuret for protein and DCPIP for vitamin C, so none of those tests for carbon dioxide.
    Method:

    Recall the standard test for carbon dioxide gas.
    Examiner tips

    • Limewater turns milky with carbon dioxide.

11. Question 2(a)

    1 marksTest for protein
    Step 1: Protein is tested for using biuret solution. Step 2: With protein present, blue biuret solution turns purple (violet). Step 3: Benedict's tests for reducing sugars, iodine for starch and ethanol for fats, so none of those tests for protein.
    Method:

    Recall the reagent used in the protein food test.
    Examiner tips

    • Biuret solution plus protein gives a purple colour.

12. Question 2(b)(i)

    4 marksPlotting a line graph
    Step 1: Both axes are labelled with their units, here height in cm and vital capacity in dm³, and a linear scale is chosen so the points fill more than half of the grid. Step 2: Every point is plotted accurately. Step 3: A single suitable line is then drawn between all the plotted points.
    Method:

    Recall the rules for a good line graph and choose the approach that meets all of them.
    Examiner tips

    • Linear scales, labelled axes with units, all points plotted and a single line.

13. Question 2(b)(ii)

    1 marksDescribing a relationship from a graph
    Step 1: At the smallest height the vital capacity is lowest, and at the greatest height it is highest. Step 2: The values rise steadily all the way through. Step 3: So as the height increases, the vital capacity increases.
    Method:

    State the overall direction of change of vital capacity with height.
    Examiner tips

    • Describe whether the measured quantity rises or falls as the other increases.

14. Question 2(b)(iii)

    2 marksEstimating a value from a graph
    Step 1: A height of 175 cm is halfway between 170 cm and 180 cm. Step 2: The vital capacity halfway between 4.1 dm³ and 5.1 dm³ is about 4.6 dm³. Step 3: So the estimate for a 175 cm person is about 4.6 dm³.
    Method:

    Find 175 cm between the two nearest heights and read off the matching vital capacity.
    Examiner tips

    • Interpolate between the two nearest data points.

15. Question 3

    6 marksPlanning an investigation into pH and enzyme activity
    Step 1: To test pH, several pH values are set with buffers while everything else is kept the same: the temperature and the volumes and concentrations of amylase and starch. Step 2: The breakdown of starch is followed by removing samples at regular times and adding iodine, timing how long the starch takes to disappear. Step 3: Repeating at each pH makes the results reliable, which only the first method does.
    Method:

    Check the plan changes only pH, follows starch breakdown with iodine, controls the rest and repeats.
    Examiner tips

    • A valid plan changes only pH, controls the rest, uses iodine to follow starch breakdown, and repeats.