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CIE IGCSE Biology 0610 Paper 6 Experimental Skills Practice Paper (6 Questions, 50 Marks, 60 Minutes)
IGCSE Biology 0610 Paper 6 Skills
CIE IGCSE Biology 0610 Paper 6 experimental skills practice paper: 6 original questions worth 50 marks in 60 minutes, covering microscope calibration and magnification.
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Paper details Level: IGCSE | Subject: Biology 0610 | Time: 60 minutes | Marks: 50 | Questions: 6
Section A Microscopy, enzymes and osmosis (25 marks)
Questions 1-3
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | (a) A student observes onion epidermal cells with a microscope. The eyepiece graticule has 100 divisions and the stage micrometer is 1 mm divided into 100 divisions. With the x40 objective, 40 eyepiece divisions line up with 10 stage-micrometer divisions. Calculate the real length of one eyepiece division in micrometres, and the real length of a cell that measures 48 eyepiece divisions. (5 marks) (b) The drawing of this cell is 36 mm long. Calculate the magnification of the drawing and explain why iodine solution is used. (4 marks) |
Answer: (a) Each stage-micrometer division = 1 mm / 100 = 0.01 mm = 10 um; 40 eyepiece divisions = 10 stage divisions = 100 um, so one eyepiece division = 100 / 40 = 2.5 um; cell length = 48 x 2.5 = 120 um (0.12 mm). (b) Magnification = drawing length / real length = 36 mm / 0.12 mm = 300. Iodine stains the nucleus and cytoplasm yellow-brown and starch grains blue-black, which increases contrast so the cell wall and nucleus are easier to see. Marking notes: correct conversion, unit in um, magnification has no unit. |
| 2 | (a) A student investigates catalase activity using potato discs and hydrogen peroxide, measuring the volume of oxygen collected in 30 s at different temperatures (cm3): 10 C 1.6, 20 C 3.4, 30 C 5.7, 40 C 7.1, 50 C 3.0, 60 C 0.4. Calculate the rate of reaction at each temperature in cm3 per minute and state the optimum temperature. (5 marks) (b) Explain why the rate falls sharply above 40 C and state two variables that must be kept constant. (3 marks) |
Answer: (a) Rate = volume / 0.5 min, so volume x 2: 10 C 3.2, 20 C 6.8, 30 C 11.4, 40 C 14.2, 50 C 6.0, 60 C 0.8 cm3/min. The highest rate is at 40 C, so the optimum is about 40 C. (b) Above the optimum the enzyme protein is denatured: the active site changes shape, substrate no longer fits and the rate falls from 14.2 to 0.8 cm3/min (a 94% drop). Keep constant: surface area or mass of potato, concentration and volume of hydrogen peroxide, pH. |
| 3 | (a) Six potato cylinders of equal size are placed in 0.0, 0.2, 0.4, 0.6, 0.8 and 1.0 mol/dm3 sucrose solution. After 30 minutes the percentage change in mass is +6.2, +3.1, -0.4, -4.8, -9.5 and -14.1 per cent. Estimate the concentration inside the potato cells by interpolation and explain why mass increases in 0.0 mol/dm3. (5 marks) (b) Explain why the cylinders are taken from the same variety, cut to the same length, and blotted dry before weighing. (3 marks) |
Answer: (a) No change lies between 0.2 and 0.4 mol/dm3: interpolating from +3.1% to -0.4%, concentration = 0.2 + 0.2 x 3.1 / (3.1 + 0.4) = 0.38 mol/dm3. In 0.0 mol/dm3 the water potential outside is higher than in the cell sap, so water enters by osmosis, the cells become turgid and the mass rises by 6.2%. (b) The same variety and length give a similar surface area and tissue, reducing individual variation; blotting removes surface solution so the mass change shows only water moved by osmosis. |
Section B Photosynthesis, respiration and food tests (25 marks)
Questions 4-6
| # | Question | Answer and marking notes |
|---|---|---|
| 4 | (a) A student uses pondweed to study the effect of light intensity on photosynthesis, placing a lamp at 10, 20, 30, 40 and 50 cm and counting bubbles released per minute: 32, 15, 9, 6 and 4. Light intensity is inversely proportional to the square of the distance; taking 10 cm as 1.00, calculate the relative intensity at each distance and describe the relationship. (5 marks) (b) Explain why the curve flattens beyond 40 cm and describe how temperature is controlled. (4 marks) |
Answer: (a) Relative intensity = (10 / distance)^2: 1.00 at 10 cm, 0.25 at 20 cm, 0.11 at 30 cm, 0.063 at 40 cm and 0.04 at 50 cm. Bubbles fall from 32 to 4, so more light gives faster photosynthesis, but the fall is not proportional (intensity drops to 0.25 while bubbles only fall to 15) because another factor also limits the rate. (b) Beyond 40 cm the intensity is only 0.063, light is the limiting factor and the counts approach the minimum, so differences fall within counting error. Temperature is controlled by keeping the tube in a water bath with a thermometer and recording it every 2 minutes at 25 C. |
| 5 | (a) A respirometer measures oxygen uptake by germinating seeds. Manometer readings in mm are 0 at 0 min, 1.8 at 2 min, 3.4 at 4 min, 5.0 at 6 min, 6.6 at 8 min and 8.2 at 10 min. Calculate the average rate over 0 to 10 minutes and the rate between 4 and 8 minutes in mm per minute. (5 marks) (b) Explain why potassium hydroxide is placed in the tube, why a control is set up, and why the apparatus stands for 5 minutes before timing starts. (3 marks) |
Answer: (a) Average rate = (8.2 – 0) / 10 = 0.82 mm/min; rate from 4 to 8 min = (6.6 – 3.4) / 4 = 0.80 mm/min. The two are close, so uptake is roughly steady. (b) Potassium hydroxide absorbs the carbon dioxide released, so the pressure fall measures oxygen uptake only. The control uses boiled dead seeds or glass beads to show the change is caused by living seeds and not by temperature or pressure fluctuations. Standing for 5 minutes lets the apparatus reach the water-bath temperature and the gases equilibrate, avoiding false readings from expansion. |
| 6 | (a) A student tests four foods with Benedict’s solution, iodine solution, Biuret reagent and the emulsion test. Results: food A gives a brick-red colour with Benedict’s and no change with iodine; food B turns iodine blue-black and does not react with Benedict’s; food C gives a purple colour with Biuret and a white emulsion; food D shows no reaction in any test. Deduce the contents of each food and state which has the highest energy density, with a reason. (5 marks) (b) Explain why the Benedict’s test is heated in an 80 C water bath for 5 minutes and why equal volumes of sample are used in each test. (3 marks) |
Answer: (a) A contains a reducing sugar such as glucose; B contains starch; C contains protein and lipid; D contains no reducing sugar, starch, protein or lipid, so it is mainly water and fibre. C has the highest energy density because lipid releases about 39 kJ per gram on oxidation, more than the 17 kJ per gram from carbohydrate or protein. (b) Benedict’s solution reacts with reducing sugar only when heated in a water bath at 80 C for 5 minutes, changing from blue through green and yellow to a brick-red precipitate; equal sample volumes and equal reagent volumes keep the reactant proportions the same so colour intensity can be compared semi-quantitatively. |
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