IGCSE Biology 0610/62 May/June 2025 Past Paper with Answers

Question 1

A student needs to make 50 cm3 of a 0.5 mol per dm3 salt solution by diluting a 1.0 mol per dm3 stock salt solution S with distilled water W. Which volumes of S and W should the student mix together?

1. 25 cm3 of the 1.0 mol per dm3 salt solution S and 25 cm3 of distilled water W
2. 50 cm3 of the 1.0 mol per dm3 salt solution S and 50 cm3 of distilled water W
3. 5 cm3 of the 1.0 mol per dm3 salt solution S and 45 cm3 of distilled water W
4. 40 cm3 of the 1.0 mol per dm3 salt solution S and 10 cm3 of distilled water W

Question 2

After 20 minutes in a 0.0 mol per dm3 (distilled-water) bath, the 10 potato discs from beaker S3 are lined up against a ruler at their actual size. The discs had a starting total length of 100 mm. Reading the ruler, the final total length is just over 100 mm. Which value is the correct final total length of the 10 discs?

1. about 102 mm
2. about 94 mm
3. about 80 mm
4. about 150 mm

Question 3

A student records, for three salt concentrations, the start and final total lengths of 10 potato discs: at 1.0 mol per dm3 the lengths were 102 mm then 94 mm; at 0.5 mol per dm3 they were 103 mm then 98 mm; at 0.0 mol per dm3 they were 100 mm then the measured final length. Which option describes a correct results table for these data?

1. a table with at least three ruled columns and a header line, with headings and units such as final concentration / mol per dm3, start length / mm and final length / mm, recording all the start and final lengths against the correct concentration
2. a table with a single column of numbers and no headings or units, so that it is impossible to tell which figure is a concentration and which is a length
3. a table that records only the start lengths and leaves out every final length, so the change in each beaker cannot be worked out at all
4. a table with correct columns and units but in which the lengths are matched to the wrong concentrations, so the 1.0 mol per dm3 row holds the values for the 0.0 mol per dm3 beaker

Question 4 · Answer A

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.

Question 5 · Answer A

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.

Question 6 · Answer A

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.

Question 7 · Answer A

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.

Question 8 · Answer A

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.

Question 9 · Answer A

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.

Question 10 · Answer A

Step 1: A biological drawing uses single, clear, continuous lines with no shading or colouring. Step 2: It should be large, using at least half the available space. Step 3: It must show the observed detail, here the midrib as a double line and the small bumps across the base of the leaf; only the first option meets all of these.

Question 11 · Answer A

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.

Question 12 · Answer A

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.

Question 13 · Answer A

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.

Question 14 · Answer A

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.

Question 15 · Answer A

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.

Question 16 · Answer A

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.

Question 17 · Answer A

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.

Question 18 · Answer A

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.

Question 19 · Answer A

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.

Question 20 · Answer A

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.