0610/61 May/June 2025 Worked Answers (IGCSE Biology 0610 Extended)

17 questions · 40 marks · 60 minutes

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1. Question 1(a)(i)

   4 marksDesigning a results table for a diffusion investigation
   Step 1: A results table needs ruled columns with a header line and headings that include units, here the concentration of acid in mol per dm3 and the time to turn red in seconds. Step 2: A converted time must be recorded for all four concentrations so they can be compared. Step 3: A higher acid concentration speeds up diffusion, so the time to turn red falls as the concentration rises.
   Method:

   Recall the features of a good results table, then use diffusion to predict the trend with concentration.
   Examiner tips

   • Tables need ruled columns, a header line, headings with units, and every converted value.

2. Question 1(a)(ii)

   1 marksDrawing a conclusion about concentration and diffusion
   Step 1: A shorter time to turn red means the acid has diffused to the centre of the agar faster. Step 2: The most concentrated acid gave the shortest time, so a higher concentration gave a faster rate of diffusion. Step 3: The conclusion is that increasing the concentration of acid increases the rate of diffusion.
   Method:

   Translate the times into rate of diffusion and relate it to the acid concentration.
   Examiner tips

   • A conclusion links the measured result (time) to the variable changed (concentration).

3. Question 1(a)(iii)

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

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

   • Independent = changed; dependent = measured; controlled = kept the same.

4. Question 1(a)(iv)

   2 marksSuggesting improvements to a practical method
   Step 1: Cutting the agar into equal-sized pieces means each piece has the same surface area and distance to the centre, making the comparison fair. Step 2: Using a white background makes the colour change easier to judge, giving a more accurate end-point. Step 3: Together these improve the reliability and accuracy of the results.
   Method:

   Pick changes that make the test fairer or the measurement more accurate.
   Examiner tips

   • Good improvements keep the test fair or make the end-point easier to judge.

5. Question 1(a)(v)

   1 marksIdentifying hazards in a practical investigation
   Step 1: A hazard is something that could cause harm. Step 2: The knife can cut the skin and hydrochloric acid is corrosive and can irritate or burn, so both are hazards. Step 3: Distilled water, a stop-clock, a rack, a tile and a scale are not harmful, so they are not hazards.
   Method:

   Pick the two items that could actually cause harm.
   Examiner tips

   • Hazards are things that could cause harm, such as sharp tools or corrosive acids.

6. Question 1(b)

   6 marksPlanning an investigation into temperature and osmosis
   Step 1: Temperature is the independent variable, so it must be changed over several values while everything else is held constant. Step 2: Osmosis is measured by the change in mass (or length) of equal-sized potato pieces before and after a fixed time in the same solution. Step 3: The concentration of solution and the size of the pieces are kept constant and the readings repeated; only the first plan does all of this.
   Method:

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

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

7. Question 2(a)(i)

   1 marksColour of DCPIP before vitamin C is added
   Step 1: DCPIP is a blue dye. Step 2: Adding vitamin C decolourises it, so it becomes colourless only at the end-point. Step 3: Before any vitamin C is added the DCPIP is still blue.
   Method:

   Recall the starting colour of DCPIP before it is decolourised.
   Examiner tips

   • DCPIP is blue and turns colourless when decolourised by vitamin C.

8. Question 2(a)(ii)

   2 marksControlled variables in a DCPIP investigation
   Step 1: A controlled variable is kept the same in every test so the comparison is fair. Step 2: The volume and concentration of the DCPIP must be the same each time, otherwise different amounts of vitamin C would be needed for reasons other than the vitamin C concentration. Step 3: So the volume and concentration of DCPIP are the controlled variables.
   Method:

   Pick the conditions that must stay the same so only the vitamin C concentration changes.
   Examiner tips

   • Controlled variables are kept the same in every test; here that is the DCPIP.

9. Question 2(b)(i)

   4 marksPlotting a line graph of results
   Step 1: Both axes are labelled with units and an even scale is chosen so the plotted points fill at least half the grid. Step 2: All five points are plotted accurately, to within half a small square. Step 3: A single suitable line is then drawn through the points; only the first approach meets all of these requirements.
   Method:

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

   • Line graphs need labelled axes with units, an even scale filling the grid, accurate points and a single line.

10. Question 2(b)(ii)

    2 marksReading an estimate from a graph
    Step 1: A volume of 1.2 cm3 lies between the 500 mg per dm3 point (1.1 cm3) and the 250 mg per dm3 point (1.4 cm3). Step 2: Reading along the line, a volume of 1.2 cm3 corresponds to a concentration a little below 500 mg per dm3. Step 3: Interpolating between the points gives an estimate of about 420 mg per dm3.
    Method:

    Use the line to read the concentration that corresponds to a volume of 1.2 cm3.
    Examiner tips

    • Read across from the given volume to the line, then down to the concentration axis.

11. Question 2(c)(i)

    1 marksDrawing a conclusion about storage and vitamin C
    Step 1: Each stored sample has a lower vitamin C concentration than when it was fresh. Step 2: This pattern holds across all the samples. Step 3: So the conclusion is that storing the apple juice decreases its vitamin C concentration.
    Method:

    Compare the fresh and stored values and state the overall trend.
    Examiner tips

    • A conclusion compares the measured values before and after the change.

12. Question 2(c)(ii)

    1 marksMeaning of an anomalous result
    Step 1: An anomalous result stands out from the others. Step 2: It does not follow the pattern or trend that the rest of the data show. Step 3: So an anomalous result is one that does not fit the pattern of the other results.
    Method:

    Define an anomaly as a result that breaks the trend of the others.
    Examiner tips

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

13. Question 2(c)(iii)

    1 marksIdentifying an anomalous result
    Step 1: Two of the stored results, 96 and 104 mg per dm3, are close together and form a clear pattern. Step 2: The value 170 mg per dm3 is much higher than these two. Step 3: Because it does not fit the pattern of the others, 170 mg per dm3 is the anomalous result.
    Method:

    Compare the three stored values and pick the one that does not fit the pattern.
    Examiner tips

    • The anomaly is the value that sits well away from the cluster of the others.

14. Question 2(c)(iv)

    3 marksCalculating a percentage change
    Step 1: Change in concentration = stored − fresh = 104 − 272 = −168 mg per dm3. Step 2: Percentage change = (change ÷ fresh value) × 100 = (−168 ÷ 272) × 100. Step 3: This gives −61.76… = −61.8 % to one decimal place; the minus sign shows it is a decrease.
    Method:

    Find the change, divide by the fresh value, multiply by 100 and keep the negative sign.
    Examiner tips

    • Percentage change = (final − initial) ÷ initial × 100; a fall gives a negative answer.

15. Question 2(d)(ii)

    3 marksCalculating actual size from magnification
    Step 1: Rearrange magnification = length of PQ ÷ actual diameter to give actual diameter = length of PQ ÷ magnification. Step 2: Actual diameter = 61 ÷ 1.8 = 33.88… mm. Step 3: To three significant figures this is 33.9 mm.
    Method:

    Rearrange the magnification formula, divide 61 by 1.8 and round to three significant figures.
    Examiner tips

    • actual size = image length ÷ magnification.

16. Question 2(e)

    2 marksTesting for reducing sugars
    Step 1: Reducing sugars are tested for with Benedict’s solution. Step 2: The mixture must be heated in a hot water-bath for the reaction to take place. Step 3: A positive result is a colour change from blue to brick-red; iodine tests for starch, biuret for protein and the ethanol emulsion test for fats.
    Method:

    Recall the reagent and the heating step needed to test for reducing sugars.
    Examiner tips

    • Benedict’s solution plus heat tests for reducing sugars (blue to brick-red).