May/June 2025 Paper 51 Worked Answers (IGCSE Biology 0610 Core)

17 questions · 40 marks · 75 minutes

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1. Question 1(a)(i)

   4 marksRecording diffusion results in a table
   Step 1: A results table needs ruled columns with clear headings and the correct units, here concentration in mol per dm³ and time in s. Step 2: Every concentration must be recorded against its measured time. Step 3: A higher acid concentration diffuses to the centre faster, so the 1.0 mol per dm³ cube turns completely red in the shortest time.
   Method:

   Recall the features of a good results table, then use the diffusion rule to predict which cube changes fastest.
   Examiner tips

   • Tables need ruled columns, headings and units; include every measured value.

2. Question 1(a)(ii)

   1 marksConclusion on concentration and diffusion
   Step 1: A shorter time to turn red means the acid reached the centre of the cube faster, so diffusion was faster. Step 2: The cubes in the more concentrated acid turned red fastest. Step 3: So as the concentration of acid increases, the rate of diffusion increases.
   Method:

   Translate the shorter times into a faster rate and state how it changes with concentration.
   Examiner tips

   • Shorter time = faster rate; relate the trend directly to the variable changed.

3. Question 1(a)(iii)

   2 marksIndependent and dependent variables
   Step 1: The independent variable is the one the student deliberately changes, here the concentration of the hydrochloric acid. Step 2: The dependent variable is the one measured in response, here the time for the cube to turn red. Step 3: So concentration is the independent variable and time is the dependent variable.
   Method:

   Decide which quantity is deliberately changed and which is measured in response.
   Examiner tips

   • Independent = what you change; dependent = what you measure.

4. Question 1(a)(iv)

   2 marksSources of error in the method
   Step 1: A source of error is a feature of the method that makes the results less reliable. Step 2: If the cubes are not all the same size, the acid has different distances to diffuse, so the times cannot be fairly compared. Step 3: The other options are normal, correct parts of the method, not errors.
   Method:

   Test each option for whether it reduces the reliability of the comparison.
   Examiner tips

   • Errors make results less reliable; correct standard steps are not errors.

5. Question 1(a)(v)

   1 marksIdentifying hazards
   Step 1: A hazard is something that could cause harm. Step 2: A sharp knife can cut the skin and hydrochloric acid is corrosive and can irritate skin and eyes, so both are hazards. Step 3: The stop-clock, white card, test-tube rack and distilled water are not harmful.
   Method:

   Pick out the items that could physically harm the student.
   Examiner tips

   • Hazards are usually sharp, hot, corrosive or toxic items.

6. Question 1(b)

   6 marksPlanning an osmosis investigation
   Step 1: To test temperature, temperature must be the only variable changed, so the solution concentration, piece size and time are all kept the same. Step 2: Osmosis is detected by the change in mass, so each piece is weighed before and after. Step 3: Using several temperatures and repeating the readings gives reliable results, which only the first method does.
   Method:

   Check each plan changes only temperature, measures mass change and is repeated.
   Examiner tips

   • A valid plan changes one variable, controls the rest, measures a relevant quantity and repeats.

7. Question 2(a)(i)

   1 marksColour of DCPIP
   Step 1: DCPIP is a blue dye. Step 2: Vitamin C decolourises DCPIP, turning it colourless. Step 3: So before any vitamin C is added the DCPIP solution is blue.
   Method:

   Recall the starting colour of DCPIP before the reaction.
   Examiner tips

   • Remember DCPIP starts blue and is decolourised by vitamin C.

8. Question 2(a)(ii)

   2 marksVariables kept constant
   Step 1: A fair test keeps everything constant except the variable being investigated, which here is the vitamin C concentration. Step 2: Each test must start with the same volume and the same concentration of DCPIP. Step 3: So the volume of DCPIP and the concentration of DCPIP are the variables kept constant.
   Method:

   Identify which quantities must stay the same so only vitamin C concentration is tested.
   Examiner tips

   • Control everything about the DCPIP so only the vitamin C concentration varies.

9. Question 2(b)(i)

   4 marksPlotting a line graph
   Step 1: Both axes are labelled with their units and a linear (evenly spaced) scale is chosen so the plotted points fill more than half of the grid. Step 2: Every one of the five points is plotted accurately. Step 3: A single suitable line is then drawn through the points.
   Method:

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

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

10. Question 2(b)(ii)

    2 marksReading a value from a graph
    Step 1: A larger volume needed means a lower vitamin C concentration, so a volume of 1.2 cm³ lies between the 1.1 cm³ result (500 mg per dm³) and the 1.4 cm³ result (250 mg per dm³). Step 2: A volume of 1.2 cm³ is just above 1.1 cm³, so the concentration is a little below 500 mg per dm³. Step 3: This gives an estimate of about 420 mg per dm³.
    Method:

    Locate 1.2 cm³ between the nearest known volumes and read off the matching concentration.
    Examiner tips

    • Interpolate between the two nearest data points using the trend.

11. Question 2(c)(i)

    1 marksConclusion on storage and vitamin C
    Step 1: The fresh samples are all around 272 to 281 mg per dm³, much higher than the stored samples. Step 2: After storage the values drop to between about 96 and 170 mg per dm³. Step 3: So storing the apple juice decreases its vitamin C concentration.
    Method:

    Compare the two sets of values and describe the overall effect of storage.
    Examiner tips

    • A conclusion states how the dependent variable changes with the independent variable.

12. Question 2(c)(ii)

    1 marksMeaning of an anomalous result
    Step 1: Most results in a reliable set follow a clear pattern or trend. Step 2: An anomalous result is one that stands out and does not fit that pattern. Step 3: So an anomalous result is a result that does not fit the trend of the others.
    Method:

    Define an anomaly as a value that breaks the overall pattern of results.
    Examiner tips

    • An anomalous result is the one that does not fit the trend.

13. Question 2(c)(iii)

    1 marksIdentifying an anomalous result
    Step 1: An anomalous result does not fit the pattern of the others. Step 2: Two of the stored values are close together at 96 and 104 mg per dm³, but 170 mg per dm³ is much higher. Step 3: So 170 mg per dm³ is the anomalous result.
    Method:

    Compare the three values and pick the one that does not fit with the others.
    Examiner tips

    • The anomaly is the value that stands apart from the cluster of the others.

14. Question 2(c)(iv)

    3 marksCalculating a percentage change
    Step 1: Percentage change = (final value − initial value) ÷ initial value × 100. Step 2: = (104 − 272) ÷ 272 × 100 = −168 ÷ 272 × 100. Step 3: = −61.76…, which rounds to −61.8 % to one decimal place.
    Method:

    Find the change, divide by the initial value, multiply by 100 and round to one decimal place.
    Examiner tips

    • Always divide the change by the original (starting) value, not the final value.

15. Question 2(d)(ii)

    3 marksCalculating actual size from magnification
    Step 1: Rearrange magnification = image length ÷ actual length to give actual length = image length ÷ magnification. Step 2: Actual diameter = 61 ÷ 1.8 = 33.888… mm. Step 3: To three significant figures this is 33.9 mm.
    Method:

    Rearrange the magnification formula, divide the measured length by 1.8 and round to 3 significant figures.
    Examiner tips

    • actual size = image size ÷ magnification.

16. Question 2(e)

    2 marksTest for reducing sugars
    Step 1: Reducing sugars are tested for with Benedict's solution. Step 2: The mixture must be heated, usually in a hot water-bath, and a brick-red colour shows reducing sugar is present. Step 3: Iodine tests for starch, biuret tests for protein and the ethanol emulsion test is for fats, so none of those is used here.
    Method:

    Recall the reagent and conditions for the reducing-sugar test.
    Examiner tips

    • Benedict's solution plus heat gives a brick-red colour with reducing sugars.