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IGCSE Biology 0610 Paper 6 Practical Skills Practice Paper (10 Questions, 100 Marks, 60 Minutes)
IGCSE Biology 0610 P6
IGCSE Biology 0610 Paper 6 practical skills: magnification, cell counts, drawing rules, variables, means and outliers, percentage change, graph rates, osmosis, food tests.
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Paper details Level: IGCSE | Subject: Biology 0610 | Paper: 6 (Practical Skills) | Time: 60 minutes | Marks: 100 | Questions: 10
Section A Microscopy and Biological Drawing (30 marks)
Questions 1-3
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Q1 A student uses a microscope with a x10 eyepiece and a x40 objective. A cell in a leaf cross-section has a diameter of 60 mm in the field of view. (a) Calculate the total magnification of the microscope. (2 marks) (b) Calculate the actual diameter of the cell in mm and in micrometres. (4 marks) (c) If a x10 eyepiece and a x100 oil-immersion objective were used instead, what would the diameter of the cell be in the field of view? Explain why oil immersion is used rather than relying only on the x40 objective. (2 marks) |
(a) Total magnification = 10 x 40 = x400. (2 marks) (b) Actual diameter = 60 / 400 = 0.15 mm; 0.15 mm x 1,000 = 150 micrometres. (4 marks: formula 1, 0.15 mm 1, 150 micrometres 1, correct units 1) (c) New magnification = 10 x 100 = x1,000, so the diameter becomes 150 micrometres x 1,000 / 1,000 = 150 mm. Immersion oil has a refractive index close to that of glass and replaces air between the objective and the slide, reducing refraction and scattering so more light enters the lens and resolution and brightness are kept at high power. (2 marks) Marking notes: magnifications are multiplied, not added; the stated value is the image diameter, so it must be divided by total magnification. |
| 2 | Q2 Cell counting: each large square of the counting chamber is 1 mm x 1 mm with a depth of 0.1 mm. A student counts 120 cells in 5 large squares; the sample was diluted 10 times with distilled water. (a) Calculate the volume of one large square in mm3 and in microlitres. (3 marks) (b) Calculate the mean number of cells per large square, then the number of cells per microlitre and per cm3 of the diluted sample. (4 marks) (c) Find the number of cells per cm3 in the original sample, and explain why 5 squares are counted rather than 1. (3 marks) |
(a) Volume = 1 x 1 x 0.1 = 0.1 mm3; since 1 mm3 = 1 microlitre, this is 0.1 microlitre. (3 marks) (b) Mean = 120 / 5 = 24 cells per square; density = 24 / 0.1 = 240 cells per microlitre; per cm3 = 240 x 1,000 = 2.4 x 10^5 cells. (4 marks) (c) Original sample = 2.4 x 10^5 x 10 = 2.4 x 10^6 cells per cm3. Counting 5 squares gives random sampling, reduces error from uneven distribution in one field of view, and the mean is more representative of the whole sample, improving reliability. (3 marks) Marking notes: apply the dilution factor only at the end; multiplying the diluted density and then again contradicts the working. |
| 3 | Q3 A student draws a root-tip cell 45 micrometres long; the drawing measures 90 mm. (a) Calculate the magnification of the drawing. (3 marks) (b) State three rules of biological drawing that students most often break. (6 marks) (c) At the same magnification, how long in mm would a 20 micrometre cell be drawn? (3 marks) |
(a) 90 mm = 90,000 micrometres; magnification = 90,000 / 45 = x2,000. (3 marks) (b) Any three: use single, continuous, sharp lines with no sketchy repeated lines or shading; keep everything in proportion so relative sizes are correct; label lines straight, not crossing, ending on the structure labelled; labels written horizontally outside the drawing; give a title and state the magnification; draw only what is observed, adding no imagined detail. (2 marks each, 6 total) (c) 20 micrometres x 2,000 = 40,000 micrometres = 40 mm. (3 marks) Marking notes: a vague answer such as “draw neatly” earns nothing; the rule must be checkable. |
Section B Experimental Design and Data Handling (36 marks)
Questions 4-6
| # | Question | Answer and marking notes |
|---|---|---|
| 4 | Q4 A student investigates the effect of temperature on the rate of fermentation by yeast: 5 g of yeast and 20 cm3 of glucose solution are placed in a conical flask kept at a set temperature in a water bath, and bubbles are counted for 10 minutes. (a) Identify the independent variable, the dependent variable and three variables that must be controlled. (6 marks) (b) Describe one treatment that would act as a control and state what it shows. (3 marks) (c) Explain why each temperature is repeated three times, and why the number of bubbles rather than bubble size is measured. (3 marks) |
(a) Independent variable: temperature. Dependent variable: number of bubbles in 10 minutes, i.e. bubble rate per minute. Controlled variables (any three): mass of yeast, concentration and volume of glucose solution, equilibration time in the water bath, seal of the flask, counting method and observer. (6 marks: 2 each) (b) Boil the yeast first and let it cool before use, as a control: this shows the bubbles come from enzyme-catalysed reactions in living yeast, not from spontaneous breakdown of glucose or expansion of air. (3 marks) (c) Three repeats allow a mean to be calculated, reducing random error and improving reliability; bubble number is a countable discrete quantity that can be recorded objectively; bubble size depends on surface tension and the bore of the delivery tube, so it is hard to quantify and highly subjective. (3 marks) Marking notes: do not swap the independent and dependent variables; “keep the room at constant temperature” is not a controlled variable because temperature is what is being changed. |
| 5 | Q5 The table shows raw data for Q4 (bubbles per 10 minutes): 10 C: 6, 8, 7 20 C: 15, 14, 16 30 C: 27, 26, 25 40 C: 33, 34, 32 50 C: 9, 44, 8 (a) Calculate the mean for each temperature, identify the reading that should be questioned and give a reason. (6 marks) (b) If that reading were kept, what would the mean be? Why is it normally rejected? (3 marks) (c) Using the 50 C mean after rejecting the doubtful value, calculate the bubble rate per minute. (3 marks) |
(a) Means: 10 C = 7, 20 C = 15, 30 C = 26, 40 C = 33, 50 C = 8.5 after rejecting 44. The doubtful reading is 44 at 50 C: it is roughly five times the other two values in the same group, and 50 C is at or beyond the optimum for yeast enzymes, so denaturation should slow the reaction; 44 is most likely a counting or timing error, such as too short a timing period or bubbles counted from elsewhere. (6 marks: 1 for each of five means, 1 for identifying 44 with a valid reason) (b) Keeping 44 gives (9 + 44 + 8) / 3 = 20.3. It is normally rejected because a single outlier pulls the mean up and distorts the trend, making 50 C appear faster than 40 C; the correct procedure is to find the cause and repeat that group, not simply delete the value. (3 marks) (c) Rate = 8.5 / 10 = 0.85 bubbles per minute. (3 marks) Marking notes: rejecting a value needs a reason; “because it is too big” alone does not earn full credit. |
| 6 | Q6 Using the accepted means from Q5 (10 C = 7, 20 C = 15, 30 C = 26, 40 C = 33, 50 C = 8.5): (a) Calculate the percentage increase from 20 C to 40 C. (3 marks) (b) Calculate the percentage increase from 10 C to 30 C. (3 marks) (c) What is the percentage change from 40 C to 50 C? Explain the result in terms of enzyme behaviour and state the approximate optimum temperature. (6 marks) |
(a) (33 – 15) / 15 x 100 = 120%. (3 marks) (b) (26 – 7) / 7 x 100 = 271.4% to 2 decimal places. (3 marks) (c) (8.5 – 33) / 33 x 100 = -74.2%, a fall of about 74%. Above 40 C the enzyme progressively denatures: the three-dimensional shape of the active site is destroyed, substrate can no longer bind and the rate falls sharply. The optimum is roughly 30 to 40 C, since the rate is still rising from 30 to 40 C and has collapsed by 50 C, so the peak lies near 35 to 40 C. (6 marks: calculation 2, denaturation explanation 2, optimum range 2) Marking notes: percentage change needs a sign or an explicit rise or fall; “the enzyme died” without a structural explanation earns half credit only. |
Section C Graphs, Errors and Evaluation (34 marks)
Questions 7-10
| # | Question | Answer and marking notes |
|---|---|---|
| 7 | Q7 A leaf is left in the light and its mass is measured every 10 minutes (g): 0 min 0.0, 10 min 1.8, 20 min 3.4, 30 min 4.6, 40 min 5.4, 50 min 5.8 (cumulative change in mass). (a) Calculate the mean rate in g per minute for 0-10, 20-30 and 40-50 minutes. (6 marks) (b) Describe how the shape of the curve changes and explain why. (2 marks) (c) Why does the curve level off after 50 minutes? Give one specific way to extend the experiment to test your explanation. (2 marks) |
(a) 0-10 min: 1.8 / 10 = 0.18 g per minute; 20-30 min: (4.6 – 3.4) / 10 = 0.12 g per minute; 40-50 min: (5.8 – 5.4) / 10 = 0.04 g per minute. (6 marks: 2 each) (b) The curve becomes steadily less steep, the rate falling from 0.18 to 0.04 g per minute, because the reaction slows with time: substrate is used up, or changes in water content and stomatal state limit further change. (2 marks) (c) Levelling off means equilibrium has been reached or substrate is exhausted, so mass no longer rises appreciably. To test this, add more substrate at 60 minutes, for example supply extra carbon dioxide or re-immerse in fresh solution: if mass rises again the exhaustion explanation is supported, and if it stays flat equilibrium is indicated. (2 marks) Marking notes: rate must be interval change divided by interval time; dividing the cumulative value by total time underestimates the later rate. |
| 8 | Q8 Potato strips, each 50 mm long initially, are placed in 0.0, 0.2, 0.4, 0.6 and 0.8 mol per dm3 sucrose solutions for 30 minutes. The changes in length are +4, +1, -1, -5 and -9 mm respectively. (a) Calculate the percentage change in length for each concentration. (5 marks) (b) Estimate the concentration of the potato cell sap (the isotonic point) and explain how you obtained it from the data. (4 marks) (c) Explain the change seen in the cells at 0.8 mol per dm3 and suggest two ways to make the results more reliable. (3 marks) |
(a) +8%, +2%, -2%, -10%, -18%, each change divided by 50 mm and multiplied by 100. (5 marks: 1 each) (b) The isotonic point lies between 0.2 and 0.4 mol per dm3, where the change goes from +2% to -2%; linear interpolation gives 0.3 mol per dm3 at zero change. At this point the external solution and the cell sap have the same water potential, so there is no net movement of water. (4 marks: between 0.2 and 0.4 for 2, interpolation to 0.3 for 1, equal water potential for 1) (c) 0.8 mol per dm3 is hypertonic: the cells have a higher water potential, so water leaves by osmosis, the vacuole shrinks and the membrane pulls away from the cell wall (plasmolysis), so length falls markedly by 18%. Improvements: use five strips per concentration and take a mean; use strips of the same variety, region and thickness and measure mass on an electronic balance, which is more sensitive than length; control soaking time and temperature; blot surface water before measuring. (3 marks) Marking notes: -18% means a shortening of 18%, not a length of -18 mm; the isotonic point must be interpolated, not guessed as a middle concentration. |
| 9 | Q9 A student must identify the nutrients present in four unknown food samples. (a) Match each reagent to the nutrient it tests for: Benedict’s solution, iodine solution, Biuret reagent, ethanol emulsion test. (4 marks) (b) Describe the positive result and the necessary condition for the Benedict’s test. (2 marks) |
(a) Benedict’s solution for reducing sugar such as glucose; iodine solution for starch; Biuret reagent for protein; ethanol emulsion test for fat. (4 marks: 1 each) (b) A positive result is a blue solution turning green, then yellow, orange and finally a brick-red precipitate, the exact colour depending on the reducing sugar concentration. The necessary condition is heating in a boiling water bath for about 5 minutes; without heating there is no colour change even when reducing sugar is present. (2 marks) Marking notes: heating is the key point for Benedict’s; iodine tests for starch, not sugar, and gives a blue-black colour. |
| 10 | Q10 Several students use the following procedures in the same experiment. (a) Classify each as reducing accuracy, precision or reliability: (i) an uncalibrated thermometer reads 0.5 C too high every time; (ii) the experiment is done only once; (iii) the meniscus is judged by eye and three readings differ by plus or minus 0.5 cm3; (iv) three repeats are done in the same water bath in the same position. (4 marks) (b) Suggest one specific remedy for (i). (2 marks) |
(a) (i) reduces accuracy: a systematic error shifting every reading the same way; (ii) reduces reliability, since without repeats the size of random error cannot be judged; (iii) reduces precision, as readings scatter and repeat poorly; (iv) does not reduce reliability through lack of repeats, but keeping one fixed position leaves a position effect, a systematic error that threatens accuracy. (4 marks: 1 each for (i), (ii), (iii), 1 for (iv)) (b) Calibrate the thermometer in ice water at 0 C and in boiling water at 100 C corrected for the day’s pressure, record the offset and apply the correction to all readings, or use a calibrated digital thermometer. (2 marks) Marking notes: accuracy is closeness to the true value, precision is closeness of repeat readings to each other, reliability is consistency across repeats; the three are not interchangeable. |
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