IGCSE Biology 0610/61 October/November 2025 Past Paper with Answers

Question 1

A student investigates the effect of temperature on the movement of blue dye (water) in celery stalks. In water-bath C the water was 4 C at the start and 15 C at the end. In water-bath W the water was 40 C at the start and 28 C at the end. What is the change in water temperature for water-bath C and for water-bath W?

1. water-bath C changed by 11 C and water-bath W changed by 12 C
2. water-bath C changed by 19 C and water-bath W changed by 68 C
3. water-bath C changed by 12 C and water-bath W changed by 11 C
4. water-bath C changed by 4 C and water-bath W changed by 28 C

Question 2

In the celery investigation the student cut 5 mm wide sections until the blue dye was no longer visible. The stalk from water-bath C gave 4 sections containing dye and the stalk from water-bath W gave 13 sections. The total distance moved by the dye is the number of sections multiplied by 5 mm. Which option describes a correct results table and the matching values?

1. a ruled table with a header line and column headings that include units, recording 4 sections and 20 mm for stalk C and 13 sections and 65 mm for stalk W
2. a ruled table with a header line and full headings, but recording 4 sections and 20 mm for stalk W and 13 sections and 65 mm for stalk C, with the two stalks swapped over
3. a ruled table with a header line and headings, recording 4 sections and 4 mm for stalk C and 13 sections and 13 mm for stalk W, with no multiplication by the 5 mm width
4. a list of the section counts with no ruled columns, no header line and no units, recording only that stalk C gave 4 and stalk W gave 13 with no distances worked out at all

Question 3

In water-bath W the blue dye moved a total distance of 65 mm during the 10 minutes that the celery stalk was in the dye. What is the rate of movement of the blue dye in the celery stalk in beaker W, including the correct unit?

1. 6.5 mm per minute
2. 65 mm per minute
3. 0.65 mm per minute
4. 6.5 minutes per mm

Question 4 · Answer A

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.

Question 5 · Answer A

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.

Question 6 · Answer A

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.

Question 7 · Answer A

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.

Question 8 · Answer A

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.

Question 9 · Answer A

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.

Question 10 · Answer A

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.

Question 11 · Answer A

Step 1: The leaf miner fly has a pair of wings while the aphid is wingless. Step 2: The fly shows only three legs from the side while the aphid shows six, and the fly is the larger insect. Step 3: So three correct differences are that the fly has wings, shows three legs and is larger than the aphid.

Question 12 · Answer A

Step 1: A biological drawing uses a single clear continuous outline with no shading and should fill more than half the available space. Step 2: It must show the real features in the right place: the V-shaped cleft on the upper right third of the blade. Step 3: It should also show three to five separate veins arising from the same level at the top of the leaf stalk.

Question 13 · Answer A

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.

Question 14 · Answer A

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.

Question 15 · Answer A

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.

Question 16 · Answer A

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.

Question 17 · Answer A

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.