May/June 2025 Paper 62 Worked Answers (IGCSE Biology 0610 Extended)

20 questions · 40 marks · 60 minutes

Question papers and mark schemes are copyright Cambridge International. We do not reproduce them: the worked answers here are written by The Practice Book. Have the paper open alongside. Get the official paper from Cambridge

1. Question 1(a)(i)

   1 marksDiluting a stock solution for an osmosis investigation
   Step 1: Halving the concentration from 1.0 to 0.5 mol per dm3 means the stock must be diluted to twice its volume. Step 2: So half of the final 50 cm3 must be stock and half must be water. Step 3: That is 25 cm3 of salt solution S and 25 cm3 of distilled water W, which gives 50 cm3 in total.
   Method:

   Work out the dilution factor, then split the final volume between stock and water.
   Examiner tips

   • To halve a concentration, use equal volumes of stock and water.

2. Question 1(a)(ii)

   1 marksMeasuring the total length of potato discs
   Step 1: The discs are in pure water, which has a higher water potential than the potato cells. Step 2: Water moves into the cells by osmosis, so the discs swell slightly and the total length increases above the starting 100 mm. Step 3: Reading the ruler gives a final total length of about 102 mm.
   Method:

   Recall that pure water makes discs swell, then read the small increase above 100 mm.
   Examiner tips

   • Read the ruler carefully; in pure water the discs swell a little above 100 mm.

3. Question 1(a)(iii)

   3 marksDesigning a results table for an osmosis investigation
   Step 1: A results table needs at least three ruled columns with a header line: one for the concentration and one each for the start and final lengths. Step 2: Every heading must include its unit, here mol per dm3 for concentration and mm for length. Step 3: The start and final lengths must be recorded against the correct concentration so the change can be found; only the first option does all of this.
   Method:

   Check the table has columns, units, and lengths matched to the right concentration.
   Examiner tips

   • A good results table has ruled columns, headings with units, and every value in the right row.

4. Question 1(a)(iv)

   1 marksCalculating the change in total length
   Step 1: The change in length is the final length minus the start length. Step 2: For S1, 94 minus 102 is minus 8 mm, a decrease of 8 mm. Step 3: For S2, 98 minus 103 is minus 5 mm, a decrease of 5 mm, because the salt solutions draw water out of the discs.
   Method:

   Subtract each start length from its final length and note the size and direction of change.
   Examiner tips

   • Change in length = final length minus start length.

5. Question 1(a)(v)

   1 marksDrawing a conclusion about salt concentration and osmosis
   Step 1: A more concentrated salt solution has a lower water potential than the potato cells. Step 2: So water moves out of the discs by osmosis, and the more concentrated the solution, the more water is lost and the greater the decrease in length. Step 3: In pure water the discs gain water instead, which fits the conclusion that water movement depends on the salt concentration.
   Method:

   Relate the size and direction of the length change to the salt concentration.
   Examiner tips

   • A conclusion links the change in length to the salt concentration through osmosis.

6. Question 1(a)(vi)

   1 marksIdentifying the dependent variable
   Step 1: The dependent variable is the one measured in response to the change. Step 2: Here the student measures the total length of the potato discs after they have been in each solution. Step 3: So the dependent variable is the total length of the potato discs.
   Method:

   Identify the quantity measured in response to the changing concentration.
   Examiner tips

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

7. Question 1(a)(vii)

   1 marksJustifying the use of repeated units in a sample
   Step 1: The change in length of one thin disc is very small and hard to measure precisely. Step 2: Measuring the total length of 10 discs gives a much larger change that can be measured more accurately. Step 3: This reduces the percentage error and the effect of any one unusual disc, making the result more reliable.
   Method:

   Explain how a larger total length improves the accuracy of the measured change.
   Examiner tips

   • Measuring more units gives a larger change and a smaller percentage error.

8. Question 1(a)(viii)

   1 marksSafety when cutting potato tissue
   Step 1: The hazard here is the sharp blade used to cut the potato. Step 2: Cutting on a tile or flat surface and moving the blade away from the hand and body keeps the blade clear of the skin. Step 3: This reduces the risk of cutting yourself, which is the relevant safety precaution.
   Method:

   Pick the precaution that keeps the sharp blade away from the skin.
   Examiner tips

   • Cut on a flat surface with the blade moving away from your hand and body.

9. Question 1(b)

   1 marksTesting potato tissue for starch
   Step 1: A yellow-brown solution that turns blue-black on a food sample is iodine solution. Step 2: Iodine solution turns blue-black only when starch is present. Step 3: So the solution is iodine solution and the conclusion is that the potato disc contains starch.
   Method:

   Match the colour change to the food test and state the positive result.
   Examiner tips

   • Iodine solution: yellow-brown to blue-black means starch is present.

10. Question 1(c)(ii)

    3 marksCalculating magnification from a measurement
    Step 1: Magnification = length of line PQ ÷ actual length. Step 2: Substitute the values: 63 ÷ 39 = 1.615... Step 3: To two significant figures this is ×1.6; magnification has no units.
    Method:

    Put the measured and actual lengths into the formula and round to two significant figures.
    Examiner tips

    • magnification = image length ÷ actual length.

11. Question 1(c)(iii)

    2 marksComparing the structure of two leaves
    Step 1: A valid difference compares a visible feature, here the potato leaf has more, and more branched, veins than the tomato leaf. Step 2: A valid similarity is a feature both leaves share, such as a midrib and a leaf stalk. Step 3: Only the first option pairs a true difference with a true similarity; the others contain statements that are not correct.
    Method:

    Pick a true visible difference and a true shared feature of the two leaves.
    Examiner tips

    • Compare a visible feature for the difference and a shared feature for the similarity.

12. Question 2(a)(i)

    1 marksMaintaining a constant temperature for an enzyme reaction
    Step 1: A constant temperature needs a source that holds the same temperature throughout. Step 2: A thermostatically controlled water-bath set to 40 C keeps the mixture at that exact temperature. Step 3: A Bunsen flame, the open bench or a refrigerator would all let the temperature change, so they do not keep it constant at 40 C.
    Method:

    Choose the method that holds the mixture at a steady, set temperature.
    Examiner tips

    • A thermostatically controlled water-bath holds a set temperature constant.

13. Question 2(a)(ii)

    2 marksIdentifying controlled variables in an enzyme investigation
    Step 1: A controlled variable is kept the same in every test so the comparison is fair. Step 2: The mass of crushed apple and the volume of pectinase solution must be the same each time, so only the pectinase concentration differs. Step 3: The concentration of pectinase is the variable being changed and the volume of juice is the result, so neither of those is controlled.
    Method:

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

    • Keep everything the same except the one variable you are testing.

14. Question 2(b)(i)

    4 marksPlotting a line graph of enzyme 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; only the first approach meets every requirement for a line graph.
    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.

15. Question 2(b)(ii)

    2 marksReading an estimate from a graph
    Step 1: A concentration of 0.5% lies just above the 0.4% point (12 cm3) and well below the 0.8% point (22 cm3). Step 2: Reading up from 0.5% to the line and across to the volume axis gives a value a little above 12 cm3. Step 3: Interpolating between the points gives an estimate of about 15 cm3.
    Method:

    Use the line to read the volume that corresponds to a concentration of 0.5%.
    Examiner tips

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

16. Question 2(b)(iii)

    1 marksDescribing the effect of enzyme concentration
    Step 1: From 0.1% to 0.8% the volume of juice rises as the concentration increases. Step 2: From 0.8% to 1.6% the volume stays at 22 cm3, so it has stopped increasing. Step 3: So increasing the concentration increases juice production up to a point, after which it levels off.
    Method:

    State the trend, then note where the volume stops changing.
    Examiner tips

    • Describe both the rise and the levelling off shown by the data.

17. Question 2(b)(iv)

    1 marksSuggesting further data to find an optimum
    Step 1: The optimum is the lowest concentration that gives the maximum juice, somewhere between 0.4% and 1.6%. Step 2: The gap between 0.4% and 0.8% is large, so the exact turning point is not known. Step 3: Testing more concentrations in small steps between these values, such as 0.7% and 0.9%, would show whether 0.8% is truly the optimum.
    Method:

    Identify the region of interest and add closely spaced concentrations there.
    Examiner tips

    • To find an optimum, test more values in small steps around the suspected best point.

18. Question 2(b)(v)

    3 marksChoosing apparatus to filter and measure a liquid
    Step 1: To separate the juice from the pulp the mixture is poured through a funnel lined with filter paper. Step 2: The filtered juice must run into a measuring cylinder so that its volume can be read from the scale. Step 3: So the correct apparatus is a funnel with filter paper over a measuring cylinder, with the parts labelled.
    Method:

    Pick apparatus that both separates the juice and measures its volume.
    Examiner tips

    • Filter paper in a funnel separates the juice; a measuring cylinder reads its volume.

19. Question 3

    6 marksPlanning an investigation into photosynthesis and respiration
    Step 1: Light intensity is the independent variable, so it must be changed over several values, for example by altering the lamp distance, while everything else is held constant. Step 2: The colour of the hydrogencarbonate indicator around the submerged plant is the dependent variable, and a heat screen keeps the temperature constant so only light intensity changes. Step 3: The rate of photosynthesis equals the rate of respiration at the intensity where the indicator stays red; only the first plan changes one variable, controls the rest and repeats the readings.
    Method:

    Check the plan changes only the light intensity, follows the indicator colour and controls the other variables.
    Examiner tips

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