AS & A Level Biology (9700)
Practical skills (generic)
7 questions · about 1% of the marks we indexed · examined in 3 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Skill: conclusions & interpreting data · Core · Paper 3 style · 5 marksOne student in Islamabad skipped for different lengths of time and measured her pulse rate immediately afterwards, resting for 15 minutes between trials.0 min: 72 beats per minute1 min: 98 beats per minute2 min: 118 beats per minute3 min: 131 beats per minute4 min: 134 beats per minute
- (a) Describe the relationship between the duration of skipping and the pulse rate. [2]
- (b) Calculate the percentage increase in pulse rate after 2 minutes compared with resting. [2]
- (c) Suggest why these results cannot be used to draw a conclusion about all students. [1]
Show answer
- (a) Pulse rate increases as duration increases, from 72 beats per minute at rest to 134 after 4 minutes; the increase gets smaller each minute and the rate levels off between 3 and 4 minutes.
B1 positive relationship / increases; B1 levels off (increase of only 3 between 3 and 4 min) with data quote.
- (b) 64 %
M1 (118 − 72)/72 × 100; A1 63.9 ≈ 64 %.
- (c) Only one student was tested (no repeats, no range of ages, fitness or sex).
B1 single individual / small sample.
A student counted the stomata in ten fields of view on each surface of bean (Phaseolus) leaves.upper surface: mean 36 stomata per mm², standard deviation 9lower surface: mean 248 stomata per mm², standard deviation 22
- (a) Compare the stomatal density on the two surfaces. [2]
- (b) Suggest an advantage to the plant of having most stomata on the lower surface. [1]
Show answer
- (a) The lower surface has many more stomata (mean 248 per mm²) than the upper surface (mean 36 per mm²), about seven times as many; the lower surface also shows more variation (SD 22 compared with 9).
B1 lower surface has more stomata, with figures; B1 comparison of spread (SD) or ratio ≈ 7 times.
- (b) Less water loss by transpiration, because the lower surface is shaded and cooler / not in direct sunlight.
B1 reduces water loss (cooler, shaded, less air movement).
100 wheat grains were placed on filter paper soaked in sodium chloride solutions of different concentration. The percentage that germinated after 7 days was recorded.0 mmol dm⁻³: 96 %50 mmol dm⁻³: 90 %100 mmol dm⁻³: 71 %150 mmol dm⁻³: 38 %200 mmol dm⁻³: 9 %
- (a) State a conclusion that can be drawn from the data. [1]
- (b) Use your knowledge of water potential to explain the results. [3]
- (c) Farmers in parts of Punjab have soils that are becoming more saline. Suggest one further investigation that would help them choose which wheat variety to plant. [1]
Show answer
- (a) Increasing salt concentration decreases the percentage germination of wheat seeds (most sharply above 50 mmol dm⁻³).
B1 higher salt concentration → lower germination.
- (b) Salt lowers the water potential of the solution; the water potential gradient between the solution and the seed is reduced (or reversed), so less water enters the seed by osmosis; water is needed to activate enzymes / mobilise food stores (e.g. amylase hydrolysing starch) so fewer seeds germinate.
B1 solution has lower (more negative) water potential; B1 less water absorbed by osmosis / imbibition; B1 water needed for enzyme activity / hydrolysis of stores / cell expansion.
- (c) Compare germination (or growth) of several varieties at the same salt concentrations found in their soil.
B1 compare different varieties at a range of salt concentrations / test using their own soil samples.
A student measured the light intensity and the height of 12 plants of one species growing along a path from under trees into open ground. Spearman's rank correlation coefficient, rs, was calculated as 0.81.
- (a) State what the value of rs shows about the relationship between the two variables. [1]
- (b) The critical value of rs at p = 0.05 for 12 pairs of readings is 0.587. State what the student can conclude. [2]
- (c) The student then concluded that light intensity causes plants to grow taller. Explain why this conclusion is not fully justified by the data. [2]
Show answer
- (a) A strong positive correlation: plants receiving more light grew taller.
B1 strong positive correlation.
- (b) 0.81 > 0.587 so the null hypothesis (no correlation) is rejected; the positive correlation is significant – the probability it is due to chance is less than 5 %.
B1 compares rs with critical value correctly; B1 reject null / significant correlation (not due to chance).
- (c) Correlation does not show cause; other factors that vary with light in the field (e.g. soil moisture, temperature, nutrient levels, trampling) were not controlled and may affect height.
B1 correlation ≠ causation; B1 named uncontrolled variable that could vary with position.
Potato cylinders were left in sucrose solutions for 24 hours. The mean percentage change in mass was:0 mol dm⁻³: +12.4 %0.2 mol dm⁻³: +5.8 %0.4 mol dm⁻³: −1.6 %0.6 mol dm⁻³: −7.9 %0.8 mol dm⁻³: −12.1 %
- (a) Estimate the sucrose concentration at which there would be no change in mass, assuming a straight line between the two readings either side of zero. Give your answer to 2 decimal places. [2]
- (b) Explain what this concentration tells you about the potato tissue. [2]
- (c) Explain why the results are given as percentage change in mass rather than change in mass. [1]
Show answer
- (a) 0.36 mol dm⁻³
M1 between 0.2 (+5.8 %) and 0.4 (−1.6 %): 0.2 + 5.8/7.4 × 0.2; A1 0.36 mol dm⁻³.
- (b) At this concentration the water potential of the sucrose solution equals the water potential of the potato cells, so there is no net movement of water by osmosis.
B1 water potential of tissue = water potential of solution; B1 no net osmosis / no net water movement.
- (c) The cylinders had different starting masses, so percentage change allows a fair comparison.
B1 different initial masses.
Students measured their reaction time using a ruler-drop test after drinking either an energy drink or water.energy drink: mean 0.250 s, standard error 0.007 swater: mean 0.270 s, standard error 0.0085 s
- (a) Calculate the 95 % confidence interval (mean ± 2 × SE) for each group. [2]
- (b) The company that makes the drink claims that it 'significantly improves reaction time'. Use the data to evaluate this claim. [3]
Show answer
- (a) Energy drink: 0.236 to 0.264 s; water: 0.253 to 0.287 s
B1 0.250 ± 0.014 = 0.236–0.264 s; B1 0.270 ± 0.017 = 0.253–0.287 s.
- (b) Mean reaction time is shorter after the drink (0.250 s compared with 0.270 s), but the confidence intervals overlap (0.253–0.264 s), so the difference is not shown to be significant; also the sample size, whether the trial was blind, and other variables (sleep, time of day, practice effect) are not stated.
B1 mean is lower with drink; B1 confidence intervals overlap so difference not significant; B1 further limitation (sample size / no placebo or blinding / practice effect / variables not controlled).
- (a) Using a 1.0 mol dm⁻³ sucrose solution and distilled water, describe how to make 10 cm³ of each of 1.0, 0.8, 0.6, 0.4 and 0.2 mol dm⁻³ sucrose solutions. Present the volumes in a table. [2]
- (b) Explain why a simple dilution was used here rather than a serial dilution. [1]
Show answer
- (a) 1.0 mol dm⁻³: 10.0 cm³ sucrose + 0.0 cm³ water; 0.8 mol dm⁻³: 8.0 cm³ sucrose + 2.0 cm³ water; 0.6 mol dm⁻³: 6.0 cm³ sucrose + 4.0 cm³ water; 0.4 mol dm⁻³: 4.0 cm³ sucrose + 6.0 cm³ water; 0.2 mol dm⁻³: 2.0 cm³ sucrose + 8.0 cm³ water
B1 correct volumes of stock (concentration ÷ 1.0 × 10 cm³); B1 correct water volumes making 10 cm³ each, table with headings and units.
- (b) The concentrations form an even (linear) series; a serial dilution gives concentrations that fall by the same factor each time (e.g. halving).
B1 equal intervals needed / serial dilution gives proportional (e.g. 1.0, 0.5, 0.25) concentrations.
- (a) Describe a method to investigate the effect of pH on the activity of catalase in potato tissue. Catalase breaks down hydrogen peroxide into water and oxygen. [6]
Show answer
- (a) Use buffer solutions at pH 4, 5, 6, 7, 8, 9; cut potato into discs of equal size (cork borer, same diameter and thickness) or use same mass of blended potato; place discs with 5 cm³ of buffer in a boiling tube connected to a gas syringe / upturned measuring cylinder in water; add 5 cm³ of 3 % hydrogen peroxide (same volume and concentration); keep at 25 °C in a water bath; measure volume of oxygen collected in 2 minutes; repeat 3 times at each pH and calculate mean; calculate rate.
1 mark each for any six of: at least 5 pH values using buffers; same mass / surface area of potato; same volume and concentration of H₂O₂; constant temperature with water bath; collect O₂ in gas syringe or by displacement of water (or count bubbles / measure foam height); measure volume in fixed time / calculate rate; control (boiled potato or no potato); equilibrate tubes before mixing; 3 repeats and mean; safety: goggles as H₂O₂ is an irritant.
- (a) Describe a method to investigate the effect of temperature on the permeability of beetroot cell membranes, using a colorimeter to measure the amount of red pigment that leaks out. [6]
Show answer
- (a) Cut beetroot cylinders with a cork borer and cut to equal lengths (e.g. 1 cm); rinse in distilled water until no more pigment leaks; place one cylinder in 10 cm³ of distilled water in tubes in water baths at 20, 30, 40, 50, 60, 70 °C for 30 min; remove cylinders, shake tubes; zero colorimeter with distilled water using a blue-green filter; measure absorbance of each solution; 3 repeats at each temperature; calculate mean absorbance.
1 mark each for any six of: cylinders of same size / surface area (cork borer and ruler); rinsed to remove pigment from cut cells; same volume of water; at least 5 temperatures (range up to ~70 °C) in water baths; water equilibrated before adding cylinders; same time in water; colorimeter zeroed with distilled water / blank; appropriate filter (blue-green); absorbance (or transmission) recorded; repeats and mean.
- (a) Describe how the student could investigate the effect of light intensity on the rate of photosynthesis of the pondweed Hydrilla. Include how variables will be controlled and how the results will be processed. [7]
Show answer
- (a) Place a 5 cm piece of Hydrilla (cut end up) in a boiling tube of 0.2 % sodium hydrogencarbonate solution at 25 °C in a water bath; place a lamp at distances 10, 20, 30, 40, 50 cm (light intensity ∝ 1/d²); use a heat filter / glass of water between lamp and plant; darken the room; allow 5 min to adjust at each distance; count bubbles per minute or collect gas in a photosynthometer / measure length of gas bubble in capillary for 5 min; repeat 3 times and calculate mean rate; plot rate against 1/d².
1 mark each for any seven of: light intensity varied by lamp distance (≥ 5 distances) and relationship 1/d²; same length / mass of pondweed, same species; constant temperature (water bath / heat shield); excess CO₂ / same concentration of NaHCO₃; other light excluded (dark room); acclimatisation time at each intensity; measure O₂ volume (photosynthometer) or bubble count in fixed time; calculate rate; repeats ≥ 3 and mean; graph of rate against light intensity; safety (lamp hot, electricity near water).
- (a) Describe a method the ecologist could use to compare the abundance of a grass species in a grazed and an ungrazed area of a hillside in the Margalla Hills. [6]
Show answer
- (a) Mark out a 20 m × 20 m area in each site; use random number tables to generate coordinates; place a 0.5 m × 0.5 m quadrat at each coordinate; estimate percentage cover of the grass (or count plants / use a 100-point frame) in each quadrat; at least 20 quadrats per area; sites matched for slope, aspect, altitude and soil type; calculate mean percentage cover for each area; use a t-test to see whether the difference is significant.
1 mark each for any six of: two areas of equal size marked out; random sampling method (random numbers / coordinates on grid); quadrat size stated; abundance measure (percentage cover / frequency / density) described; at least 10 quadrats per area; areas matched for slope / aspect / soil / altitude; same time of year; calculate mean (and SD); t-test to compare means.
- (a) For each investigation, name the most suitable statistical test and give a reason.(i) comparing the mean body length of two populations of snails(ii) testing whether the number of hours of sunshine is correlated with the number of butterflies seen(iii) testing whether the observed numbers of red-flowered and white-flowered plants fit an expected 3 : 1 ratio [3]
- (b) State the null hypothesis for investigation (i). [1]
Show answer
- (a) (i) t-test – compares two means of continuous, normally distributed data; (ii) Spearman's rank correlation – tests for a correlation between two variables; (iii) chi-squared test – compares observed and expected frequencies of categories.
B1 each test with reason.
- (b) There is no significant difference between the mean body lengths of the two snail populations.
B1 no significant difference between the means of the two named populations.
A student counted the number of seeds in six pea pods from one plant: 23, 27, 25, 31, 22, 28.
- (a) Calculate the mean number of seeds per pod. [1]
- (b) Calculate the standard deviation using s = √[Σ(x − x̄)² ÷ (n − 1)]. Give your answer to 2 decimal places. [3]
- (c) State what the standard deviation shows about the data. [1]
Show answer
- (a) 26
B1 156/6 = 26.
- (b) 3.35
M1 deviations: -3, 1, -1, 5, -4, 2; M1 Σ(x − x̄)² = 56; A1 √(56/5) = 3.35.
- (c) It shows the spread of the values around the mean (about 68 % of values lie within one SD of the mean).
B1 spread / variation of data about the mean.
A student measured the mean length of leaf hairs (µm) on plants of one species growing at different heights above the valley floor (m) along a hillside in the Swat valley.height 120 m: hair length 14.0 µmheight 250 m: hair length 13.1 µmheight 380 m: hair length 16.2 µmheight 510 m: hair length 18.5 µmheight 640 m: hair length 17.9 µmheight 770 m: hair length 21.4 µmheight 900 m: hair length 24.0 µmheight 1030 m: hair length 22.8 µm
- (a) Calculate Spearman's rank correlation coefficient, rs, using rs = 1 − 6Σd² ÷ (n³ − n). Show your working. Give your answer to 2 decimal places. [3]
- (b) The critical value at p = 0.05 for 8 pairs of readings is 0.738. State and explain the conclusion. [2]
- (c) Explain why Spearman's rank was a suitable test for these data. [1]
Show answer
- (a) rs = 0.93
B1 ranks for height 1–8; ranks for leaf length 2, 1, 3, 5, 4, 6, 8, 7; M1 Σd² = 6; A1 rs = 1 − 6 × 6/(512 − 8) = 0.93.
- (b) rs = 0.93 > 0.738, so reject the null hypothesis: there is a significant positive correlation between height above the valley floor and leaf hair length.
B1 compares with critical value; B1 significant positive correlation / reject null hypothesis.
- (c) The investigation looks for a correlation between two variables measured on the same plants / data may not be normally distributed; at least 5 (here 8) pairs.
B1 testing for correlation between two variables (non-normally distributed data / paired readings).
Wing lengths (in arbitrary units) of fruit flies reared at two temperatures.28 °C: n = 7, mean 68.4, standard deviation 4.118 °C: n = 9, mean 74.9, standard deviation 5.3
- (a) Calculate the value of t using t = |x̄₁ − x̄₂| ÷ √(s₁²/n₁ + s₂²/n₂). Give your answer to 2 decimal places. [2]
- (b) Calculate the number of degrees of freedom. [1]
- (c) The critical value of t at p = 0.05 for this number of degrees of freedom is 2.145. State and explain what the result shows. [2]
Show answer
- (a) t = 2.77
M1 √(4.1²/7 + 5.3²/9) = 2.350; A1 6.5/2.350 = 2.77.
- (b) 14
B1 7 + 9 − 2 = 14.
- (c) t = 2.77 > 2.145, so the null hypothesis is rejected: the mean wing length of flies reared at 18 °C is significantly greater than at 28 °C (less than 5 % probability that the difference is due to chance).
B1 t greater than critical value; B1 significant difference / reject null hypothesis, with direction.
A student investigates the effect of hydrogen peroxide concentration on the rate of oxygen production by catalase in pieces of liver.
- (a) Identify the independent variable and the dependent variable. [2]
- (b) State three variables that must be standardised. [2]
- (c) Explain why temperature must be kept constant. [1]
Show answer
- (a) Independent: concentration of hydrogen peroxide. Dependent: volume of oxygen produced in 1 minute (rate of reaction).
B1 IV H₂O₂ concentration; B1 DV volume of O₂ per unit time.
- (b) Any three of: temperature; pH; mass / surface area of liver or potato (concentration of catalase); volume of hydrogen peroxide; time of measurement; source of enzyme.
Three correct = 2 marks; two correct = 1 mark.
- (c) Temperature affects the kinetic energy of molecules and hence the rate of enzyme activity, so a change would affect the volume of oxygen produced.
B1 temperature changes rate of collisions / enzyme activity, altering the DV.
A student investigates the effect of temperature on the percentage germination of okra seeds using Petri dishes placed in incubators at 15, 20, 25, 30 and 35 °C.
- (a) For each of the following variables, describe how the student could keep it constant.(i) the volume of water available to the seeds(ii) light(iii) the type of seed [3]
- (b) Explain why at least 20 seeds should be used in each dish. [1]
Show answer
- (a) (i) add the same volume (e.g. 10 cm³) of water to each dish using a syringe / measuring cylinder, and cover dishes to prevent evaporation; (ii) keep all dishes in the dark (or same light) inside incubators; (iii) use seeds of the same variety from the same packet, of similar size.
B1 each variable with a practical method.
- (b) Germination varies between individual seeds; a larger sample reduces the effect of a few non-viable seeds, making the percentage more reliable.
B1 reduces effect of individual variation / non-viable seeds, results more representative.
A researcher compares the resting heart rate of adults in Lahore who walk for at least 30 minutes a day with those who do not exercise.
- (a) Identify two variables, other than exercise, that could affect resting heart rate in this study, and explain why each should be taken into account. [4]
- (b) Suggest why it is difficult to standardise all variables in investigations using humans. [1]
Show answer
- (a) e.g. age – heart rate changes with age; sex – mean resting rates differ; caffeine / smoking – stimulates heart rate; body mass / fitness level; time of day; stress; medication. Each must be matched or recorded because it would change heart rate independently of exercise.
For each of two: B1 variable; B1 explanation of its effect on heart rate / why it must be matched.
- (b) People differ genetically and in lifestyle, and it is unethical or impractical to control their diet and activity completely.
B1 natural variation / ethical or practical limits on controlling lifestyle.
A student tested the hydrolysis of starch by amylase.Tube A: starch + amylaseTube B: starch + amylase that had been boiled and cooledTube C: starch + distilled waterSamples were tested with iodine solution every minute.
- (a) Explain the purpose of tube B and of tube C. [2]
- (b) The student found that starch disappeared slowly in tube B as well as tube A. Suggest an explanation and how the method could be improved. [2]
Show answer
- (a) Tube B (boiled amylase) shows that the disappearance of starch is due to active enzyme and not to the heat treatment / other substances in the solution; tube C (no enzyme, water) shows that starch does not break down on its own (and that the iodine test works).
B1 B: confirms the effect is due to active (non-denatured) enzyme; B1 C: shows starch is stable without enzyme / acts as a colour standard.
- (b) The amylase in tube B was not fully denatured (not boiled long enough, or allowed to cool and some enzyme remained active); boil for at least 10 minutes in a water bath / check the boiled solution reached 100 °C.
B1 enzyme not completely denatured; B1 improvement: boil for longer / at 100 °C for a set time.
A student cut one potato cylinder for each sucrose concentration, weighed it on a balance reading to 0.1 g, left it in an open beaker of solution for 2 hours, then dried it with a paper towel and weighed it again.
- (a) Identify two sources of error in this procedure and suggest how each could be reduced. [4]
Show answer
- (a) e.g. cylinders blotted differently, so different amounts of surface water → blot each cylinder in the same way (roll once on paper towel); cylinders cut to slightly different lengths / some with skin → use a ruler and scalpel, remove skin; balance reading only to 0.1 g → use a balance reading to 0.01 g; solutions evaporated → cover tubes; only one cylinder per concentration → use three cylinders per concentration.
For each of two: B1 error; B1 improvement.
A student mixed starch and amylase and every 30 seconds placed a drop of the mixture into iodine solution on a white tile. She recorded the time when the drop no longer turned blue-black.
- (a) Explain why the timing of the end point is a source of error. [2]
- (b) Suggest two improvements that would make the measurement of the end point more accurate. [2]
Show answer
- (a) Deciding when the blue-black colour has disappeared is subjective, as the colour changes gradually; samples are only taken every 30 s, so the true end point could be up to 30 s earlier.
B1 judging colour change is subjective / gradual; B1 sampling interval means end point time is imprecise (±30 s).
- (b) Use a colorimeter to measure absorbance of samples; take samples more frequently (e.g. every 10 s); compare with a colour standard (iodine in water); have the same person judge the end point against a white tile.
1 mark each for any two.
The time taken for a piece of milk agar to clear after protease was added was measured five times at 35 °C: 42 s, 45 s, 44 s, 61 s, 43 s.
- (a) Identify the anomalous result and calculate the mean of the remaining results. [2]
- (b) Calculate the percentage error in the mean if it had been calculated including the anomalous result. [2]
- (c) Suggest one possible cause of the anomalous result in this experiment. [1]
Show answer
- (a) 61 s is anomalous; mean = 43.5 s
B1 61 s identified (much higher than others); B1 (42 + 45 + 44 + 43)/4 = 43.5 s.
- (b) 8.0 %
M1 mean with 61 = 47; (47 − 43.5)/43.5 × 100; A1 8.0 %.
- (c) e.g. timer started late / end point misjudged; temperature of the water bath dropped; less enzyme added in that repeat; tube not mixed.
B1 any plausible cause linked to the procedure.
A student heated test tubes of glucose solution with Benedict's solution and recorded the time taken for the mixture to turn brick-red.0.5 % glucose: 150, 142, 149 seconds1.0 % glucose: 97, 94, 98 seconds2.0 % glucose: 60, 63, 59 seconds4.0 % glucose: 38, 36, 40 seconds
- (a) Prepare a table to record these results, including the mean time for each concentration (to the nearest second). [4]
Show answer
- (a) Columns: concentration of glucose / % | time to turn brick-red / s: trial 1, trial 2, trial 3 | mean time / s. Means: 0.5 % 147 s; 1.0 % 96 s; 2.0 % 61 s; 4.0 % 38 s.
B1 independent variable in first column with units in heading only; B1 dependent variable heading with unit (s), three trial columns; B1 all raw data entered correctly; B1 means correct to the nearest second, consistent decimal places.
A student recorded the mean length of bean shoots grown at different light intensities in this table (columns separated by | ):Results | Light4 cm | low5.5 cm | medium6.25 cm | high
- (a) Identify three faults in the way the table has been drawn and state how each should be corrected. [3]
Show answer
- (a) 1 Units written in the body of the table (cm) – units should be in the column heading only; 2 inconsistent number of decimal places (4, 5.5, 6.25) – record all to the same precision (e.g. 1 d.p.); 3 the independent variable is not in the first column – put light intensity first; 4 the heading 'Results' is not a variable name – use 'mean length of shoot / cm'.
1 mark each for any three faults with correction.
Percentage change in mass of onion tissue in salt solutions:0 mol dm⁻³: +8.2 %0.1 mol dm⁻³: +4.9 %0.2 mol dm⁻³: +2.1 %0.3 mol dm⁻³: −1.0 %0.4 mol dm⁻³: −3.8 %0.5 mol dm⁻³: −6.9 %
- (a) Describe how you would draw a graph of these results. [3]
- (b) Describe how you would use your graph to find the concentration of salt solution at which there is no change in mass, and explain why the line of best fit should be used rather than the two plotted points nearest to zero. [2]
Show answer
- (a) Concentration (mol dm⁻³) on the x-axis, percentage change in mass on the y-axis, with the x-axis crossing at 0 % and the y-axis extending to negative values; axes labelled with units; scales even and using more than half the grid; points plotted as small crosses; ruled line of best fit (or smooth curve) through the points.
1 mark each for any three of: correct axes with labels and units; negative values shown / x-axis at y = 0; suitable linear scales; accurate plotting; line of best fit (not extended beyond data).
- (b) Find where the line of best fit crosses the line of 0 % change in mass (the x-axis) and read the concentration vertically down from that point; the line of best fit uses all six readings, so it reduces the effect of an error in any single reading.
B1 read the x-value where the line crosses 0 % change (draw a vertical guide line to the x-axis); B1 line of best fit is based on all the data / averages out random errors in individual points (answer about 0.27 mol dm⁻³).
A student in Faisalabad grew tomato plants with three different fertilisers, A, B and C, and calculated the mean yield per plant and the standard error for each.
- (a) State the type of chart that should be drawn and describe how the error bars should be added. [3]
- (b) The error bars for fertilisers A and B overlap but those for B and C do not. State what this suggests. [2]
Show answer
- (a) A bar chart (fertiliser type is categoric); bars of equal width, not touching; each bar height = mean yield; error bars drawn as a vertical line from mean − 2SE to mean + 2SE (95 % CI) with short horizontal caps.
B1 bar chart because categoric independent variable; B1 bars separated, labelled axes with units; B1 error bars extend 2 × SE (or 1 SD as stated) above and below the mean.
- (b) The difference between A and B may not be significant (could be due to chance); the difference between B and C is likely to be significant.
B1 A and B not significantly different; B1 B and C probably significantly different.
A piece of carrot tissue had a mass of 2.45 g. After 4 hours in distilled water its mass was 2.71 g.
- (a) Calculate the percentage change in mass. Give your answer to 1 decimal place. [2]
- (b) Calculate the mean rate of mass change in g per hour over the 4 hours. [1]
Show answer
- (a) +10.6 %
M1 (2.71 − 2.45)/2.45 × 100; A1 10.6 % (increase).
- (b) 0.065 g per hour
B1 0.26/4 = 0.065 g h⁻¹.
Numbers of five species of invertebrate found in leaf litter samples from a garden in Multan: 34, 21, 12, 8, 5.
- (a) Calculate Simpson's index of diversity, D = 1 − Σ(n/N)². Give your answer to 2 decimal places. [3]
- (b) A second plot near a busy road had D = 0.42. Compare the diversity of the two plots. [1]
Show answer
- (a) D = 0.71
B1 N = 80; M1 Σ(n/N)² = 0.2859; A1 D = 0.71.
- (b) The first plot (0.71) is more diverse than the plot near the road (0.42), because D closer to 1 means higher diversity.
B1 first plot more diverse; higher D = greater diversity.
Students cut samples of potato, apple and onion with scalpels, crushed them with water, and tested the extracts with iodine solution and with Benedict's solution in a boiling water bath.
- (a) Identify three hazards in this practical and state a precaution for each. [3]
Show answer
- (a) e.g. scalpel – sharp, cut on a tile away from the body; hot water bath / boiling water – scalds, use tongs and a test-tube holder; Benedict's solution (copper sulfate) – irritant, wear goggles and wash skin; iodine solution – stains / irritant, wear gloves and goggles; glassware breaking – handle with care, sweep up with brush.
1 mark each for any three hazards with matching precautions.
A photomicrograph of a leaf epidermis has a scale bar labelled 100 µm that is 25 mm long. One epidermal cell measures 38 mm long on the photomicrograph.
- (a) Calculate the actual length of the cell, in µm. [2]
- (b) Calculate the magnification of the photomicrograph. [1]
- (c) Suggest why measuring several cells and finding the mean gives a better estimate of cell length. [1]
Show answer
- (a) 152 µm
M1 38/25 × 100; A1 152 µm.
- (b) ×250
B1 25 mm = 25000 µm; 25000/100 = ×250.
- (c) Cells vary in size and may be cut at different angles; a mean reduces the effect of this variation.
B1 variation between cells / sectioning angle.
