0610/61 Oct/Nov 2025 Worked Answers (IGCSE Biology 0610 Extended)

17 questions · 40 marks · 60 minutes

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1. Question 1(a)(i)

   1 marksCalculating temperature changes in a water-movement investigation
   Step 1: The change in temperature is the final value minus the starting value, ignoring the direction. Step 2: For water-bath C the change is 15 - 4 = 11 C, and for water-bath W the change is 40 - 28 = 12 C. Step 3: So C changed by 11 C and W changed by 12 C.
   Method:

   Subtract each starting temperature from its final temperature to find the change for C and W.
   Examiner tips

   • A change in temperature is always the final reading minus the starting reading.

2. Question 1(a)(ii)

   4 marksDesigning a results table for dye movement
   Step 1: A good results table has ruled columns, a header line and headings that include the units. Step 2: The total distance is the number of sections multiplied by 5 mm, so stalk C gives 4 x 5 = 20 mm and stalk W gives 13 x 5 = 65 mm. Step 3: So the table records 4 sections / 20 mm for C and 13 sections / 65 mm for W.
   Method:

   Recall the features of a good results table, then convert each section count to a distance in mm.
   Examiner tips

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

3. Question 1(a)(iii)

   2 marksCalculating the rate of dye movement
   Step 1: Rate of movement = total distance moved / time taken. Step 2: Here that is 65 mm / 10 minutes = 6.5 mm per minute. Step 3: The distance is on top and the time underneath, so the unit is mm per minute, giving 6.5 mm per minute.
   Method:

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

   • Rate = distance / time; the unit follows the calculation (mm per minute).

4. Question 1(a)(iv)

   1 marksConclusion: temperature and water movement
   Step 1: The warm water-bath W gave the greater distance (65 mm) and the cold water-bath C gave the smaller distance (20 mm). Step 2: So a higher temperature led to more water (dye) movement up the stalk. Step 3: The conclusion is that the higher the temperature, the greater the movement of water.
   Method:

   Relate the larger distance in the warm bath to the higher temperature to state the trend.
   Examiner tips

   • A conclusion links the measured result (distance) to the variable changed (temperature).

5. Question 1(a)(v)

   1 marksWhy repeat an investigation
   Step 1: Repeating a measurement gives several values for the same conditions. Step 2: This lets you spot anomalous results, the outliers that do not follow the trend, and leave them out. Step 3: It also lets you calculate a mean, making the result more reliable.
   Method:

   Recall that repeating improves reliability by exposing anomalies and allowing a mean.
   Examiner tips

   • Repeats let you identify anomalies and calculate a reliable mean.

6. Question 1(b)

   2 marksControl variables in the celery investigation
   Step 1: A controlled variable is kept the same in every test so the comparison is fair. Step 2: Both stalks were cut to 10 cm and both were left in the dye for 10 minutes, so these were kept constant. Step 3: The temperature was the variable being changed, and the number of sections and distance moved were results, so they were not controlled.
   Method:

   Separate the changed variable and the results from the things kept the same for both stalks.
   Examiner tips

   • Controlled variables are kept the same; do not confuse them with the changed variable or the results.

7. Question 1(c)(i)

   2 marksSource of error: uncontrolled temperature
   Step 1: The starting and final temperatures differ greatly (C rose from 4 to 15 C, W fell from 40 to 28 C), showing the water-bath temperature was not held steady. Step 2: This uncontrolled temperature is the source of error. Step 3: A thermostatically controlled water-bath keeps the temperature constant throughout, reducing the error.
   Method:

   Use the drifting temperatures in the table to name the uncontrolled variable and a way to control it.
   Examiner tips

   • Look at what changed when it should have stayed constant: here, the water-bath temperature.

8. Question 1(c)(ii)

   1 marksImproving precision of the dye position
   Step 1: With 5 mm sections the dye position is only known to the nearest 5 mm, which is not precise. Step 2: Cutting thinner sections, or cutting the stalk lengthwise, lets you locate the end of the dye much more exactly. Step 3: So thinner or longitudinal cuts reduce this source of error.
   Method:

   Reduce the size of each step along the stalk so the dye position is read more precisely.
   Examiner tips

   • Smaller cutting intervals (or a lengthwise cut) give a more precise position.

9. Question 1(d)

   2 marksCalculating the volume of a xylem cylinder
   Step 1: The radius is half the diameter, so r = 0.06 / 2 = 0.03 mm. Step 2: v = pi r^2 h = 3.14 x 0.03^2 x 100 = 3.14 x 0.0009 x 100 = 0.2826 mm3. Step 3: Rounded to two decimal places this is 0.28 mm3.
   Method:

   Halve the diameter for the radius, square it, then multiply by pi and the length and round to 2 d.p.
   Examiner tips

   • Always convert diameter to radius (halve it) before using pi r^2 h.

10. Question 1(e)(i)

    3 marksMagnification calculation for the black aphid
    Step 1: Rearranging the formula, actual length = length of line AB / magnification. Step 2: That is 60 mm / 35 = 1.714 mm. Step 3: To one significant figure 1.714 mm rounds to 2 mm.
    Method:

    Divide the measured line length by the magnification and round to one significant figure.
    Examiner tips

    • Actual size = image size / magnification; an aphid is only a couple of mm long.

11. Question 2

    6 marksPlanning an investigation into pH and amylase activity
    Step 1: The independent variable is pH, set with buffers at several values, and only pH should change. Step 2: The dependent variable is the time for the starch to be digested, found by removing drops onto iodine on a white tile and timing until the iodine stays yellow-brown. Step 3: Temperature and the volumes and concentrations of amylase and starch are kept constant, and the test is repeated to find a reliable mean.
    Method:

    Identify pH as the only variable to change, choose the iodine timing as the measure, control the rest and repeat.
    Examiner tips

    • A good plan changes one variable (pH), measures a clear outcome (time), controls the rest and repeats.

12. Question 3(a)(i)

    4 marksPlotting a line graph of height against vital capacity
    Step 1: Both axes must be labelled with their units, here height of person / cm and vital capacity / dm3. Step 2: A suitable linear scale should make the plotted points fill at least half the grid in each direction, and all eight points must be plotted accurately. Step 3: A single suitable line is then drawn between the points to show the trend.
    Method:

    Apply the rules for plotting a line graph: labelled axes, sensible scale, accurate points and a single line.
    Examiner tips

    • Good graphs have labelled axes with units, a scale that fills the grid, accurate points and a clear line.

13. Question 3(a)(ii)

    1 marksDescribing the relationship between height and vital capacity
    Step 1: The values rise together: 1.9 dm3 at 130 cm up to 7.2 dm3 at 200 cm. Step 2: A taller person has a larger vital capacity throughout the data, with no fall. Step 3: So as the height increases, the vital capacity increases.
    Method:

    Compare the vital capacity at the smallest and largest heights to state the direction of the trend.
    Examiner tips

    • A relationship describes how one quantity changes as the other increases.

14. Question 3(a)(iii)

    2 marksEstimating a value from a line graph
    Step 1: A height of 175 cm lies halfway between 170 cm (4.1 dm3) and 180 cm (5.1 dm3). Step 2: Reading the line at 175 cm gives a value about halfway between 4.1 and 5.1 dm3. Step 3: That is about 4.6 dm3.
    Method:

    Locate 175 cm between the two nearest points and read the matching value off the line.
    Examiner tips

    • To estimate between points, read the line at the required height on the scale.

15. Question 3(b)

    1 marksReagent for the protein test
    Step 1: The chemical test for protein uses biuret reagent. Step 2: A positive result is a colour change from blue to purple (lilac). Step 3: Iodine tests for starch, Benedict’s for reducing sugars and DCPIP for vitamin C, so biuret is the reagent for protein.
    Method:

    Recall the standard food test reagents and match protein to biuret.
    Examiner tips

    • Biuret reagent tests for protein (blue to purple).