0610/52 May/June 2025 Worked Answers (IGCSE Biology 0610 Core)

19 questions · 40 marks · 75 minutes

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1. Question 1(a)(i)

   1 marksPreparing salt solutions for an osmosis investigation
   Step 1: Diluting halves the concentration, so use concentration × volume before = concentration × volume after: 1.0 × V = 0.5 × 50. Step 2: V = 25 cm³ of the 1.0 mol per dm³ salt solution S. Step 3: The total volume must be 50 cm³, so add 50 − 25 = 25 cm³ of distilled water W.
   Method:

   Apply C1V1 = C2V2 to find the volume of stock, then top up to 50 cm³ with water.
   Examiner tips

   • Use concentration × volume is conserved on dilution: C1V1 = C2V2.

2. Question 1(a)(ii)

   4 marksRecording osmosis results in a table
   Step 1: A results table needs ruled columns with headings and correct units, here concentration in mol per dm³ and length in mm. Step 2: Both the initial and final length must be recorded for each of the three concentrations, giving six lengths. Step 3: The strongest salt solution (1.0 mol per dm³) draws out the most water by osmosis, so it shows the greatest change in length.
   Method:

   Recall the features of a good table, then use osmosis to predict which solution gives the biggest change.
   Examiner tips

   • Tables need ruled columns, headings and units, and must capture every measured value.

3. Question 1(a)(iii)

   1 marksCalculating the change in length
   Step 1: The change in length = final length − initial length. Step 2: S1 = 26 − 30 = −4 mm, S2 = 29 − 30 = −1 mm, S3 = 33 − 30 = +3 mm. Step 3: A negative value shows the discs shrank (lost water) and a positive value shows they grew (gained water).
   Method:

   Subtract the initial 30 mm from each final length and record the signed change.
   Examiner tips

   • Always do final − initial and keep the sign to show the direction of change.

4. Question 1(a)(iv)

   1 marksConclusion on salt concentration and osmosis
   Step 1: A decrease in length means the discs lost water by osmosis. Step 2: The discs in the most concentrated salt solution lost the most length, so they lost the most water. Step 3: In distilled water the discs gained water and grew, so higher salt concentration means more water leaves the discs.
   Method:

   Translate the length changes into water movement and relate it to the salt concentration.
   Examiner tips

   • A conclusion links the dependent variable (length change) to the independent variable (concentration).

5. Question 1(a)(v)

   1 marksIdentifying the dependent variable
   Step 1: The dependent variable is the quantity that is measured in response to the change. Step 2: Here the student measures the total length of the potato discs. Step 3: So the length of the discs is the dependent variable.
   Method:

   Identify which quantity is measured in response to the change in concentration.
   Examiner tips

   • Dependent = measured; independent = changed; controlled = kept the same.

6. Question 1(a)(vi)

   1 marksSources of error in an osmosis investigation
   Step 1: A source of error is a feature of the method that makes the measurement less accurate. Step 2: The discs are curved and hard to line up perfectly, so the total length read off the ruler is uncertain. Step 3: The other options are correct, careful parts of the method, not errors.
   Method:

   Test each option for whether it reduces the accuracy of the length measurement.
   Examiner tips

   • Errors reduce accuracy; controlled, careful steps are not errors.

7. Question 1(a)(vii)

   1 marksSafety when preparing potato discs
   Step 1: Cutting safely means controlling the blade so it cannot slip onto the body. Step 2: Cutting on a flat tile with the blade moving away from the hand keeps fingers clear of the blade. Step 3: The other options put the hand or body in the path of the blade or do not address the cutting hazard.
   Method:

   Choose the precaution that keeps the cutting blade away from the body.
   Examiner tips

   • Cut on a firm surface with the blade moving away from your hand.

8. Question 1(b)

   1 marksTesting for starch with iodine
   Step 1: Iodine solution is the test for starch. Step 2: When starch is present, orange-brown iodine turns blue-black. Step 3: Potato cells store starch, so the iodine turns blue-black, showing starch is present.
   Method:

   Recall the colour change of iodine with starch and apply it to the starchy potato.
   Examiner tips

   • Iodine turns blue-black when starch is present.

9. Question 1(c)(ii)

   3 marksCalculating magnification
   Step 1: Magnification = length in the picture ÷ actual length = 63 ÷ 39. Step 2: 63 ÷ 39 = 1.615…. Step 3: To two significant figures this is 1.6.
   Method:

   Divide the measured 63 mm by the actual 39 mm and round to two significant figures.
   Examiner tips

   • magnification = image length ÷ actual length.

10. Question 1(c)(iii)

    2 marksComparing leaf features
    Step 1: A valid difference must be a visible feature, other than size, that is not the same in both leaves: the potato edge is smooth while the tomato edge is lobed. Step 2: A valid similarity is a feature shared by both, such as a central midrib. Step 3: Only the first option pairs a true difference with a true similarity.
    Method:

    Find one genuine visible difference (other than size) and one shared feature.
    Examiner tips

    • Differences must not be size; similarities must be true of both leaves.

11. Question 2(a)(i)

    1 marksMaintaining a constant temperature
    Step 1: A constant temperature needs a controlled heat source that holds one set value. Step 2: A thermostatically controlled water-bath set to 40 °C keeps the mixture steadily at that temperature. Step 3: The bench, a Bunsen burner and a fridge all give the wrong or a changing temperature.
    Method:

    Choose the method that holds one steady temperature over the whole 30 minutes.
    Examiner tips

    • Use a thermostatic water-bath to hold a constant temperature.

12. Question 2(a)(ii)

    2 marksControlling variables in an enzyme experiment
    Step 1: A fair test keeps everything constant except the variable being investigated, which here is the pectinase concentration. Step 2: Each run must use the same mass of crushed apple and the same volume of pectinase solution. Step 3: So the mass of apple and the volume of pectinase are the variables kept constant.
    Method:

    Identify the quantities that must stay the same so only pectinase concentration is tested.
    Examiner tips

    • Hold the apple mass and the volume of enzyme constant so only concentration varies.

13. 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 point 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.

14. Question 2(b)(ii)

    2 marksReading a value from a graph
    Step 1: A concentration of 0.5% lies between the 0.4% result (12 cm³) and the 0.8% result (22 cm³). Step 2: It is just above 0.4%, so the volume is a little above 12 cm³. Step 3: Reading between the points gives an estimate of about 15 cm³.
    Method:

    Locate 0.5% between the nearest known concentrations and read off the matching volume.
    Examiner tips

    • Interpolate between the two nearest points using the trend.

15. Question 2(b)(iii)

    1 marksEffect of enzyme concentration
    Step 1: From 0.1% to 0.8% the volume rises, so more enzyme produces more juice. Step 2: From 0.8% to 1.6% the volume stays at 22 cm³, so it has levelled off. Step 3: So the volume increases with concentration and then levels off.
    Method:

    Describe how the volume changes across the whole range, noting where it stops increasing.
    Examiner tips

    • Describe the rise and then the plateau when an enzyme effect levels off.

16. Question 2(b)(iv)

    1 marksImproving the resolution of data
    Step 1: The optimum is the concentration giving the most juice, and the present data jump from 0.4% to 0.8% to 1.6%. Step 2: To find the exact optimum, more concentrations are needed close to 0.8%, with smaller intervals. Step 3: So extra readings between 0.4% and 1.6% around 0.8% are required.
    Method:

    Identify that more closely spaced concentrations around 0.8% are needed.
    Examiner tips

    • Narrow the intervals around the suspected optimum to locate it precisely.

17. Question 2(b)(v)

    3 marksApparatus for filtering and measuring volume
    Step 1: Filtering separates the juice from the pulp using a funnel lined with filter paper. Step 2: Collecting the filtrate in a measuring cylinder lets the volume be read off a scale. Step 3: So a funnel with filter paper over a labelled measuring cylinder both filters and measures the juice.
    Method:

    Choose apparatus that both removes the pulp and lets the volume be measured.
    Examiner tips

    • Filter with funnel and filter paper; measure volume with a measuring cylinder.

18. Question 3

    6 marksPlanning a photosynthesis investigation
    Step 1: Light intensity is the independent variable, so it must be changed over several values, for example by altering the lamp distance. Step 2: The indicator colour is the dependent variable; the indicator stays red when photosynthesis equals respiration, so record the light intensity at which this happens. Step 3: Keep temperature, plant type and indicator volume constant and repeat the readings; only the first plan does all of this.
    Method:

    Check each plan changes only light intensity, uses the indicator colour and is repeated.
    Examiner tips

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