Handling Data (M1): OCR A Level Biology Exam Questions by Topic

OCRMaths SkillsHandling DataExam Questions
Written by Tom Whitburn

OCR A M1 · 20 questions · mark schemes and examiners’ tips included

Handling data is the biggest group of maths skills in the OCR A Level Biology specification, and it turns up on every paper. It covers rounding to the right number of significant figures, means, histograms, probability, sampling, correlation, statistical tests, standard deviation and uncertainty. The biology changes from cloned goats to cheetahs to wood sorrel, but the maths underneath is the same each time.

The examiners' reports show that students lose marks here on the detail rather than the idea. They round 31.66 to 31.6, divide by the wrong value in a percentage change, name the t-test but don't say it compares means, ignore overlapping error bars, or forget that a length measured from two readings has two uncertainties. Every one of these is easy to fix once you've seen it.

Work through the twenty questions below with a calculator. They are grouped by skill, starting with significant figures and means and finishing with statistical tests, standard deviation and uncertainty. If you'd like to go through questions like these with me each week, my OCR A masterclasses run on Wednesdays at 5.15pm for Year 12 and on Tuesdays at 6.30pm or Thursdays at 5.15pm for Year 13.

What these questions cover (OCR maths skills M1: handling data)

The questions test these mathematical skills from the OCR A Level Biology A specification (H420):

  • M1.1 Use an appropriate number of significant figures
  • M1.2 Find arithmetic means
  • M1.3 Construct and interpret frequency tables and diagrams, bar charts and histograms
  • M1.4 Understand simple probability
  • M1.5 Understand the principles of sampling as applied to scientific data
  • M1.6 Understand the terms mean, median and mode
  • M1.7 Use a scatter diagram to identify a correlation between two variables
  • M1.8 Make order of magnitude calculations
  • M1.9 Select and use a statistical test
  • M1.10 Understand measures of dispersion, including standard deviation and range
  • M1.11 Identify uncertainties in measurements and use simple techniques to determine uncertainty when data are combined

How to use this page

Write your answer on paper before you open the mark scheme. Allow about 1.2 minutes per mark: the 20 questions are worth 46 marks, so about 55 minutes in total. Then mark your answer strictly against the mark scheme and read the examiners' tip underneath it. If you'd rather print the questions, download the PDF pack.

Significant figures and means (M1.1, M1.2)

Question 1

A level Paper 2 · June 2019 · Q19(c)(i) · 2 marks · Easy

Calculate the percentage of pregnancies that gave live births in cloned goats and mice.

Table 19 of the number of pregnancies and live births after cloning by SCNT in goats, monkeys, mice and sheep.
Show the mark scheme Mark scheme: goats 31% (or 30.8%) and mice 13% (or 12.8%); one mark only if the answers are not given to 2 or 3 significant figures.

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From the examiners' report: Most students got both marks. The marks that were lost were for rounding wrongly or giving too many significant figures, so match your answer to the precision of the data.

Question 2

A level Paper 1 · June 2023 · Q17(b) · 2 marks · Medium

Calculate the percentage change in mean root length compared with water, to 2 significant figures.

Table 17.1 giving a mean root length of 16.5 grid squares in the test solution and 13.9 grid squares in water.
Show the mark scheme Mark scheme: 16.5 minus 13.9 = 2.6, divided by 13.9 and multiplied by 100, gives 19%; only one mark if the answer is not to 2 significant figures.

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From the examiners' report: The common error was dividing the difference by the final length, which gives 16%. Always divide by the value you are comparing against (here water), then round to the 2 significant figures the question asks for. Show your working so a slip still earns a mark.

Question 3

AS Paper 2 · June 2023 · Q2(b)(ii) · 2 marks · Easy

Work out a mean distance and a rate of bubble movement, both to 1 decimal place.

Tables of the distance moved by a bubble in three trials for groups A, B and C after different numbers of sprays of water, with the mean and rate missing for group A.
Show the mark scheme Mark scheme: mean 31.7 mm and rate 10.6 mm per minute; one mark if both are given to more than 1 decimal place.

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From the examiners' report: Most students found the rate, but the usual mistake for the mean was 31.6, from rounding down instead of up. A recurring dot over 31.6 is not 1 decimal place either: write 31.7. Remember group A ran for 3 minutes, not 5.

Question 4

AS Paper 2 · June 2019 · Q4(b)(i) · 3 marks · Medium

Identify an anomalous reading and evaluate how much it has changed the mean.

Table 4.1 of the distance moved by an air bubble in 5 minutes over six trials in still air, with a fan and with the lower leaf surface covered in petroleum jelly, with means.
Show the mark scheme Mark scheme: 57 (trial 4 with petroleum jelly) is anomalous; it makes the mean higher; without it the mean is 29.4 mm, so it raises the mean by 4.6 mm (15.6%).

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From the examiners' report: Most students spotted 57 and said it made the mean higher. Full marks came from backing this up with a number: recalculate the mean without it (29.4 mm) or give the difference it makes.

Frequency diagrams and averages (M1.3, M1.6)

Question 5

A level Paper 2 · June 2022 · Q19(a)(i) · 5 marks · Medium

Plot the body lengths of Atlantic cod as a histogram, including one wider class.

Table of the frequency of Atlantic cod in body length classes from 100 to 800 mm, with the last class twice as wide as the others, above a blank grid.
Show the mark scheme Mark scheme: labelled axes with units, linear scales filling half the grid, ruled touching bars, the first five bars at 10, 48, 121, 130 and 119, and the last bar twice as wide with a height of 23.

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From the examiners' report: Most students labelled the axes, but some used a category scale on the x axis or drew a bar chart with gaps. Continuous data needs a histogram with touching bars. Very few spotted that the 600–800 mm class is twice as wide, so its bar must be twice as wide and half the height.

Question 6

A level Paper 3 · June 2019 · Q3(c)(iii) · 2 marks · Hard

Evaluate a claim that the mean is a better measure than the median or mode for the trypsin results.

Information from earlier in this question:

Earlier part of the same exam question, giving the information this part uses. Table 3 of the temperature giving the fastest reaction in five replicates for human and icefish trypsin, with the mean, mode and median for each.
Show the mark scheme Mark scheme: agree, because icefish trypsin has two modes; disagree, because the mean for human trypsin includes the outlier of 10 °C, and the median and mode are not affected by outliers.

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From the examiners' report: Definitions of mean, median and mode earned nothing. Use the table: the 10 °C outlier drags the human mean down, the median and mode ignore it, and icefish trypsin has two modes. Give a point on each side when you evaluate.

Probability and sampling (M1.4, M1.5)

Question 7

A level Paper 3 · June 2018 · Q4(b)(i) · 1 mark · Hard

Work out how many of six children of two carriers you would expect to have Vici syndrome or be carriers.

Part (b)(i): Vici syndrome is recessive; two carriers have six children, with answer lines for the number expected to have the syndrome and to be carriers.
Show the mark scheme Mark scheme: 1 or 2 with Vici syndrome and 3 carriers; 1.5 is not allowed.

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From the examiners' report: Most students got this right. The error was writing 1.5 children: a quarter of six is 1.5, but you can only have whole children, so round to 1 or 2.

Question 8

A level Paper 3 · June 2018 · Q4(b)(ii) · 1 mark · Medium

Calculate the probability that a daughter of the carriers and a male carrier have a child with Vici syndrome.

Part (b)(ii): a daughter of the two carriers has a child with a male carrier; calculate the probability that the child has Vici syndrome.
Show the mark scheme Mark scheme: 0.25 (25% or 1 in 4); 25 with no % sign and 1:3 are ignored.

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From the examiners' report: This was a stretch question, but many students reached 25%. Write a probability as 0.25, 25% or 1 in 4, not as a ratio like 1:3 or a bare 25.

Question 9

A level Paper 3 · November 2020 · Q5(a) · 3 marks · Hard

Describe how to choose the number and position of samples across three habitats of different areas.

Diagram of a study area made of farmland (800 m squared), peat bog (800 m squared) and grassland (400 m squared).
Show the mark scheme Mark scheme: stratified sampling, with the number of samples in each habitat in proportion to its area (40%, 40% and 20%, for example 8, 8 and 4 samples), placed randomly within each habitat.

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From the examiners' report: Most students knew samples should be in proportion to the area. Marks were lost for not naming stratified sampling or leaving out the calculation the question asked for, such as 8 samples in farmland, 8 in peat bog and 4 in grassland.

Correlation and orders of magnitude (M1.7, M1.8)

Question 10

A level Paper 1 · June 2024 · Q21(c)(ii) · 3 marks · Medium

Evaluate a conclusion that hydrogen peroxide changes sperm cell membranes and so reduces motility.

Fig. 21.2: sperm motility and the percentage of cells with changed membrane phospholipids against hydrogen peroxide concentration from 0 to 200 micromoles per dm cubed, with error bars.
Show the mark scheme Mark scheme: supports, as motility falls and membrane changes rise with concentration; does not support, as correlation does not prove causation, only 10 men, no statistical test, fertility clinic samples may not be representative, and some error bars overlap.

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From the examiners' report: Strong answers gave supporting points and then points against. A correlation between two variables doesn't show that one causes the other, so say so, and read the axis labels and key carefully before you describe the trends.

Question 11

AS Paper 2 · June 2018 · Q5(a)(ii) · 2 marks · Hard

Calculate how many orders of magnitude smaller the elephant population will be in 2213 than in 1913.

Part (a)(ii): the world elephant population fell from 10,000,000 in 1913 to 500,000 in 2013; assume the same rate of decline every 100 years.
Show the mark scheme Mark scheme: a 95% fall (5% remain) each century gives 25,000 in 2113 and 1,250 in 2213; 1.25 × 10 to the 3 compared with 1 × 10 to the 7 is 4 orders of magnitude smaller.

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From the examiners' report: Only the strongest students reached 4. Write the numbers in standard form (1 × 10⁷ and 5 × 10⁵), apply the same 5% that remains each century, then compare the powers of ten.

Choosing and using a statistical test (M1.9)

Question 12

A level Paper 3 · June 2019 · Q3(c)(iv) · 2 marks · Easy

Choose a statistical test to compare the reaction rates of human and icefish trypsin and explain why.

Part (c)(iv): the students compare the mean of five replicates for human trypsin with the mean of five for icefish trypsin at 30 °C.
Show the mark scheme Mark scheme: an (unpaired) Student's t-test, because they are comparing two means; a paired t-test is not allowed.

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From the examiners' report: Most students named the t-test but lost the second mark by writing 'two data sets'. Say it compares two means. Chi squared, standard deviation and Spearman's rank were the usual wrong answers.

Question 13

A level Paper 2 · June 2024 · Q20(a)(i) · 3 marks · Medium

Calculate chi squared for the number of wood sorrel plants near ash, oak and sycamore trees.

Table of the number of wood sorrel plants within 3 m of an ash tree (44), an oak (56) and a sycamore (20), with the null hypothesis that tree species has no effect.
Show the mark scheme Mark scheme: each expected value is 40, giving (O − E)²/E values of 0.4, 6.4 and 10, so chi squared = 16.8.

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From the examiners' report: About half of students got all three marks. The most common error was the expected values: under the null hypothesis the 120 plants are shared equally, so E = 40 for each tree.

Question 14

A level Paper 2 · June 2024 · Q20(a)(ii) · 3 marks · Medium

Use the critical value of 5.991 to discuss what the chi squared result shows.

Part (a)(ii) giving the critical value for chi squared at p = 0.05 as 5.991.
Show the mark scheme Mark scheme: the difference between observed and expected is significant, with less than a 5% probability that it is due to chance, so reject the null hypothesis: tree species is associated with the wood sorrel population.

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From the examiners' report: This went better than in earlier years. It is the difference between observed and expected that is significant, not the 'results' or the 'data'. If your value in (i) was wrong you could still score full marks here.

Question 15

A level Paper 3 · November 2021 · Q5(c)(i) · 3 marks · Hard

Calculate Spearman's rank correlation coefficient for soil water content and photosynthesis in maize.

Table 5.2 of soil water content and mean rate of photosynthesis at six sites for Calotropis procera and maize, and Table 5.3 of critical values of rs.
Show the mark scheme Mark scheme: the sum of d squared is 2 and n(n² − 1) is 210, so rs = 1 − (6 × 2) ÷ 210 = 0.943.

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From the examiners' report: Few students got all three marks. Rank each variable separately, find d for each site and square it, and use n = 6. Then compare your value with the table: 0.943 shows a significant positive correlation.

Standard deviation and range (M1.10)

Question 16

A level Paper 2 · June 2023 · Q17(b)(i) · 2 marks · Easy

Calculate a standard deviation from three repeat readings, to match the table.

Tables of the volume of oxygen produced over 180 seconds in three repeats at 2, 8 and 20 arbitrary units of hydrogen peroxide, with means and standard deviations, one missing.
Show the mark scheme Mark scheme: the mean is 76.7 and the sum of squared differences is 10.67, giving a standard deviation of 2.3; one mark for 2.3 to more than 1 decimal place.

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From the examiners' report: Most students got both marks. One mark was lost for 2.31: the other standard deviations in the table are to 1 decimal place, so yours should be too.

Question 17

A level Paper 2 · June 2023 · Q17(b)(ii) · 2 marks · Medium

Use the standard deviations to discuss how repeatable the results are.

Part (b)(ii) asking students to use the standard deviations to discuss repeatability.
Show the mark scheme Mark scheme: standard deviations are lower at lower concentrations and earlier times, so results there are more repeatable.

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From the examiners' report: Only about a third got both marks. Use the word 'repeatable' and make it comparative: more repeatable at low concentrations and early times, where the standard deviations are smaller. Just defining the terms scored nothing.

Question 18

AS Paper 1 · November 2020 · Q24(b) · 2 marks · Medium

Draw a conclusion from a bar chart of water loss from tomato and watermelon leaves with standard deviation error bars.

Fig. 24.2: bar chart of the volume of water collected after 6 hours per area of leaf for tomato (about 0.031 cm cubed per cm squared) and watermelon (about 0.023), with large overlapping error bars.
Show the mark scheme Mark scheme: there is probably no significant difference between the transpiration rates of tomato and watermelon, because the difference between the means is very small and the standard deviations are large (the error bars overlap).

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From the examiners' report: Many students ignored how small the difference between the means was and how much the error bars overlapped. When the standard deviations are large compared with the difference, conclude that there is probably no significant difference.

Uncertainty in measurements (M1.11)

Question 19

AS Paper 1 · June 2019 · Q21(c) · 1 mark · Medium

State how a student could reduce the uncertainty of their measurements.

Part (c) asking how the student could reduce the uncertainty of their data.
Show the mark scheme Mark scheme: use more precise apparatus, such as callipers or a micrometer.

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From the examiners' report: Many students didn't know what uncertainty means. Repeating and taking a mean earned no credit: uncertainty depends on the measuring instrument, so use one with a higher resolution.

Question 20

A level Paper 2 · June 2019 · Q16(b)(i) · 2 marks · Hard

Calculate the percentage uncertainty in a distance measured with a ruler read to 0.5 mm.

Fig. 16: a visking tubing bag of solution X on a capillary tube in solution Y; the liquid level rose from 10.5 mm to 26.5 mm, read on a ruler to the nearest 0.5 mm.
Show the mark scheme Mark scheme: (2 × 0.5) ÷ (26.5 − 10.5) × 100 = 6.25% or 6.3%.

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From the examiners' report: Only a few students got both marks. The distance comes from two readings, so the uncertainty is 2 × 0.5 mm, divided by the 16 mm change. Many used only one reading's uncertainty, or gave too many significant figures.

Where students lose marks (from OCR examiners' reports and mark schemes)

Each point below is followed by wording that gained marks.

  • Round up from 0.66: A mean of 31.66 was often written as 31.6, or as 31.6 with a recurring dot, which is not 1 decimal place. Answers with too many figures, such as a standard deviation of 2.31 in a table of 1 decimal place values, also lost a mark.

    ✔ Gains credit: "mean = 31.7" and "correct answer to 1 decimal place": 2.3.

  • Divide by the original value: In a percentage change, dividing the difference by the final value instead of the starting value gave 16% instead of 19%.

    ✔ Gains credit: "percentage change = [2.6 ÷ 13.9] × 100 ... 19 (%) to 2 sig. figs."

  • Whole children only: Using a 1 in 4 probability for six children, many students wrote 1.5 children with the syndrome.

    ✔ Gains credit: "syndrome 1 or 2 and carriers 3" ("DO NOT ALLOW 1.5").

  • Histogram bars touch: For continuous data, students drew bar charts with gaps or used a category scale on the x axis, and very few made the wider 600–800 mm class twice as wide.

    ✔ Gains credit: "histogram plotted with ruled lines and touching bars" and "6th bar twice the width and height 23".

  • A t-test compares means: Most students named the t-test but justified it with 'two data sets', which was too vague, and some chose a paired t-test.

    ✔ Gains credit: "(Student's)(unpaired) t-test" because "(they are) comparing means (of two data sets)".

  • It's the difference that is significant: Answers said the 'results' or 'data' were significant, or got the expected values wrong before they started.

    ✔ Gains credit: "differences (between observed and expected) are significant", "(less than) 5% chance that (difference is) due to chance", "reject null hypothesis".

  • Overlapping error bars mean caution: Students claimed one plant lost more water than the other, ignoring the small difference between the means and the large overlap of the standard deviation error bars.

    ✔ Gains credit: "there is (probably) no (significant) difference" because the difference is "very small" and the "standard deviations (very) large / data very spread out".

  • Two readings, double the uncertainty: For a distance worked out from two ruler readings, most students used the uncertainty of one reading only. Others said repeats and a mean would reduce uncertainty, which gained nothing.

    ✔ Gains credit: "(2 × 0.5) / (26.5 – 10.5) × 100" = 6.25%, and "use more, accurate / precise apparatus".

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Download the PDF pack (20 questions, mark schemes at the end)

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Questions, mark schemes and examiners' report extracts © OCR.

Tom Whitburn
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