IGCSE Biology 0610 Paper 6 Alternative to Practical Practice Paper

Level: Year 10–11 (IGCSE) | Subject: Biology 0610 | Time: 60 minutes | Total: 40 marks | Parts: 25

Paper at a glance

Paper	Time	Marks	Weighting
Paper 6 (Alternative to Practical)	1 h	40	20%

Questions

Question 1 (8 marks) — Food tests and controls

(a) State the reagent and the treatment used to show that sample X contains a reducing sugar. (1 marks)

(b) After heating with Benedict’s solution, sample X gives a brick-red precipitate. State what this shows. (1 marks)

(c) Describe the test for starch (reagent and positive result) and the test for protein (reagent and positive result). (2 marks)

(d) The student concludes “sample X contains glucose”. Give two weaknesses of this conclusion. (2 marks)

(e) To compare the reducing sugar content of samples X and Y, state two conditions that must be kept the same. (2 marks)

Question 2 (8 marks) — Enzyme activity and temperature

(a) State the independent variable in this experiment. (1 marks)

(b) Explain why the time taken for the starch to disappear falls as the temperature rises from 10 °C to 40 °C. (2 marks)

(c) At 40 °C the starch disappears in 30 s. Calculate the rate of reaction at that temperature in per minute, and explain why this temperature was chosen. (2 marks)

(d) Why is the rate lower at 60 °C? (1 marks)

(e) At 70 °C no colour change is seen even after 300 s. Give the reason, and state one variable other than temperature that must be controlled. (2 marks)

Question 3 (8 marks) — Counting bubbles in photosynthesis

(a) Name the gas collected and give one test for it. (1 marks)

(b) At 20 cm, 60 bubbles are counted in 2 minutes. Calculate the rate per minute and describe the relationship between distance and rate. (2 marks)

(c) When the distance doubles from 10 cm to 20 cm, what fraction of the original light intensity remains? Hence comment on whether the rate ratio of 42 to 30, i.e. 1.40, is expected, and explain. (2 marks)

(d) Suggest two ways of making the results of this experiment more reliable. (2 marks)

(e) Why count bubbles rather than collect and measure the volume of gas? (1 marks)

Question 4 (8 marks) — Osmosis in potato cylinders

(a) In one solution the mass of the potato cylinder does not change at all. What is this solution called? (1 marks)

(b) In 0.8 mol/dm3 sucrose the mass falls from 2.00 g to 1.72 g. Calculate the percentage change in mass to one decimal place. (2 marks)

(c) Explain why the mass of the cylinder increases in 0.0 mol/dm3 (distilled water). (2 marks)

(d) Estimate the sucrose concentration of the potato tissue from the data, and describe how it could be found more precisely. (2 marks)

(e) Give one precaution in the procedure and explain why it is needed. (1 marks)

Question 5 (8 marks) — Magnification and biological drawing

(a) A cell measures 60 mm on a micrograph; its actual length is 0.15 mm. Calculate the magnification. (1 marks)

(b) Convert the actual length to micrometres and recalculate the magnification in the same units to show that the answer matches part (a). (2 marks)

(c) In the same micrograph a chloroplast has an actual length of 5 um. Calculate its length on the image and give it in mm. (2 marks)

(d) State two conventions for a biological drawing. (2 marks)

(e) Why is a scale bar more reliable than a stated magnification alone? (1 marks)

Answers and marking notes

Question 1 — Food tests and controls

(a) Benedict’s solution, heated in a water bath for 3 to 5 minutes (1). [1 marks]

(b) The sample contains a reducing sugar such as glucose, maltose or fructose (1). [1 marks]

(c) Iodine solution turns blue-black with starch (1); Biuret reagent turns purple with protein (1). [2 marks]

(d) Benedict’s solution reacts with all reducing sugars, so it cannot identify which one (1); the test is qualitative and no standards of known concentration were compared, so no amount can be stated (1). [2 marks]

(e) The same volume of each sample, e.g. 2 cm3 (1); the same volume of Benedict’s solution, the same heating time and the same water-bath temperature (1). [2 marks]

Question 2 — Enzyme activity and temperature

(a) The temperature, from 10 to 60 °C (1). [1 marks]

(b) The enzyme and starch molecules gain kinetic energy (1), so collisions between enzyme and substrate are more frequent and more enzyme-substrate complexes form, giving a faster reaction (1). [2 marks]

(c) Rate = 60 / 30 = 2.00 per minute (1); the time is the shortest and the rate the highest, so 40 °C is closest to the optimum temperature (1). [2 marks]

(d) The enzyme is denatured: the tertiary structure of the active site is destroyed so the substrate can no longer bind (1). [1 marks]

(e) The enzyme is completely denatured before the reaction starts, so no reaction occurs (1); the volumes and concentrations of starch and amylase, or the pH (buffer), must be kept the same (1). [2 marks]

Question 3 — Counting bubbles in photosynthesis

(a) Oxygen; a glowing splint placed in the gas relights (1). [1 marks]

(b) 60 / 2 = 30 per minute (1); as the distance increases the light intensity falls and the rate decreases (1). [2 marks]

(c) Intensity is inversely proportional to the square of the distance, so one quarter (0.25) remains (1); if light were the only limiting factor the rate would fall to a quarter, but it only falls from 42 to 30 (ratio 1.40), so carbon dioxide concentration or temperature is limiting at high light intensity (1). [2 marks]

(d) Repeat each distance three times and calculate a mean (1); let the pondweed equilibrate for 2 minutes at each distance and use the same length or mass of pondweed (1). [2 marks]

(e) The bubbles are small and not all the same size, and the total volume collected is too small to measure accurately (1). [1 marks]

Question 4 — Osmosis in potato cylinders

(a) An isotonic solution, with the same water potential as the cell sap (1). [1 marks]

(b) (1.72 – 2.00) / 2.00 x 100 (1) = -14.0% (1). [2 marks]

(c) The water potential of distilled water is higher than that of the cell sap (1), so water enters the cells through the partially permeable membrane by osmosis and the cells become turgid (1). [2 marks]

(d) The concentration giving no change in mass is 0.4 mol/dm3 (1); repeat with smaller intervals such as 0.30, 0.35 and 0.45, plot a graph and read the concentration where the line crosses zero (1). [2 marks]

(e) Blot the cylinders dry with paper before weighing, otherwise surface water makes the mass too high (1). [1 marks]

Question 5 — Magnification and biological drawing

(a) 60 / 0.15 = x400 (1). [1 marks]

(b) 0.15 mm = 150 um and 60 mm = 60 000 um (1); 60 000 / 150 = x400, the same as part (a) (1). [2 marks]

(c) 5 x 400 = 2 000 um (1) = 2.0 mm (1). [2 marks]

(d) Use a sharp pencil with single, continuous lines and no shading (1); label lines are straight, do not cross, and the magnification or scale bar is given (1). [2 marks]

(e) A scale bar stays valid if the image is enlarged or reduced, whereas a stated magnification no longer applies (1). [1 marks]

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