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IB Biology – Data-Based Questions (Paper 1B)

Unfamiliar experiments, statistics and evaluation practice.

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12 questions  ·  34 marks  ·  answers included  ·  IB Biology SL & HL (2016 curriculum, Paper 1B style)
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This worksheet covers:

  • Practice in the Paper 1B style
  • Analysing data from unfamiliar experiments
  • Applying statistics (mean, standard deviation, error bars, correlation)
  • Evaluating experimental design

A calculator is allowed; data are given as text rather than as figures. Allow about 40 minutes.

Questions

Section A – Reading and processing data

1

A student measured the lengths of six randomly sampled leaves from an oak tree, in cm: 4.2, 4.5, 3.9, 4.8, 4.1, 4.4.(a) Calculate the mean leaf length.(b) Calculate the range of the data.

Show answer
(a) Mean $=\dfrac{4.2+4.5+3.9+4.8+4.1+4.4}{6}=\dfrac{25.9}{6}=4.32\,\text{cm}$ (to 3 s.f.).(b) Range $=4.8-3.9=0.9\,\text{cm}$.
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2

In an enzyme investigation, the rate of reaction (arbitrary units) of an enzyme was measured at increasing pH: pH 6, rate $=20$; pH 7, rate $=35$; pH 9, rate $=8$.(a) Calculate the percentage increase in rate from pH 6 to pH 7.(b) Calculate the percentage decrease in rate from pH 7 to pH 9.

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(a) Percentage increase $=\dfrac{35-20}{20}\times100=75\%$.(b) Percentage decrease $=\dfrac{35-8}{35}\times100=77.1\%$ (to 3 s.f.).
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3

Explain why the pH data in the previous question suggest that this enzyme has an optimum pH close to 7, and suggest one further pH value the student should test to more precisely locate the optimum.

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Rate of reaction is highest at pH 7 (35 units) compared with the lower rates measured either side of it (20 at pH 6 and 8 at pH 9), suggesting the enzyme’s active site/tertiary structure is best suited to catalysis at a pH close to 7 (the optimum), with denaturation/reduced activity as pH moves away from this value. The student should also test intermediate values (e.g. pH 6.5 and pH 7.5, or pH 8) to narrow down the exact optimum, since the true peak could lie anywhere between the tested points.
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4

Define the term standard deviation, and state what a small standard deviation indicates about a set of repeated measurements compared with a large standard deviation.

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Standard deviation is a measure of the amount of variation or spread of a set of data values around the mean. A small standard deviation indicates that the data values are clustered closely around the mean (little variability/high precision, repeatable), whereas a large standard deviation indicates the data values are spread widely around the mean (more variability, less precision).
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Section B – Statistical analysis (exam-style)

5

Plant heights (cm) were recorded for two groups of 5 plants each, grown under different light intensities.Control group (A): 15, 18, 16, 17, 14.High-light group (B): 22, 25, 23, 24, 21.(a) Calculate the mean height for each group.(b) Calculate the standard deviation for group A (to 2 d.p.).

Show answer
(a) Mean A $=\dfrac{15+18+16+17+14}{5}=16.0\,\text{cm}$. Mean B $=\dfrac{22+25+23+24+21}{5}=23.0\,\text{cm}$.(b) Deviations from mean A: $-1,2,0,1,-2$; squared: $1,4,0,1,4$, sum $=10$. Sample variance $=\dfrac{10}{5-1}=2.5$. Standard deviation $=\sqrt{2.5}=1.58\,\text{cm}$ (to 2 d.p.).
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6

Using the standard deviations for groups A and B (both equal to 1.58 cm), the standard error of the mean is calculated as $SE=\dfrac{s}{\sqrt{n}}$.(a) Calculate the standard error for each group ($n=5$).(b) Calculate the range (mean $\pm$ 1 standard error) for each group.

Show answer
(a) $SE=\dfrac{1.58}{\sqrt{5}}=0.71\,\text{cm}$ for both groups.(b) Group A: $16.0\pm0.71 = 15.29$ to $16.71\,\text{cm}$. Group B: $23.0\pm0.71 = 22.29$ to $23.71\,\text{cm}$.
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7

Using your answer to the previous question, evaluate whether the difference in mean plant height between the control and high-light groups is likely to be statistically significant, explaining the rule you have applied.

Show answer
The error bar ranges (mean $\pm$ 1 SE) for group A (15.29–16.71 cm) and group B (22.29–23.71 cm) do not overlap. When error bars do not overlap, this suggests the difference between the means is likely to be statistically significant (i.e. unlikely to be due to chance alone), so high light intensity is likely associated with a genuine increase in plant height. (To confirm this rigorously, a statistical test such as a $t$-test should be carried out and compared against a critical value at the 5% significance level.)
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8

A student carried out a $t$-test comparing the two groups in the previous questions and obtained a calculated $t$-value of 8.6, with a critical value of 2.31 at $p=0.05$ for the relevant degrees of freedom. State the conclusion the student should draw, and explain the meaning of the $p=0.05$ significance level in this context.

Show answer
Since the calculated $t$-value (8.6) is greater than the critical value (2.31), the null hypothesis (that there is no difference between the means of the two groups) is rejected; there is a statistically significant difference between the mean heights of the control and high-light groups. A significance level of $p=0.05$ means there is less than a 5% probability that the observed difference between the groups arose by chance alone (i.e. we can be more than 95% confident the difference is real).
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Section C – Evaluating experimental design (challenge)

9

In the plant height investigation, only 5 plants were used per group, all grown in the same room but at different distances from a single lamp.(a) Identify one limitation of this experimental design.(b) Suggest one improvement, and explain how it would increase confidence in the conclusion.

Show answer
(a) The sample size is small (only 5 plants per group), so the results may not be representative and could be strongly affected by natural variation between individual plants (or other variables, such as differences in distance from the lamp, may not be controlled between plants within the same group).(b) Increase the sample size (e.g. to 20-30 plants per group) and/or ensure all other variables (e.g. temperature, water, $\text{CO}_2$, soil type) are controlled/kept constant between groups. A larger sample size reduces the impact of individual variation on the mean, giving a more reliable estimate of the true population mean and a smaller standard error, increasing confidence that any observed difference reflects the effect of light intensity rather than chance or an uncontrolled variable.
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10

A separate study measured leaf width (mm) against light intensity (arbitrary units) for 10 plants and calculated a Pearson correlation coefficient of $r=0.92$.(a) Describe the strength and direction of this correlation.(b) Explain why this result does not, by itself, prove that light intensity causes an increase in leaf width.

Show answer
(a) $r=0.92$ indicates a strong, positive correlation between light intensity and leaf width (as light intensity increases, leaf width tends to increase).(b) Correlation shows only that two variables change together; it does not establish causation, since a third, confounding variable (for example, temperature or $\text{CO}_2$ concentration, which may also vary with distance from the light source) could be responsible for the observed relationship in both variables, or the relationship could even be coincidental. A controlled experiment, varying only light intensity while keeping other factors constant, would be needed to establish a causal relationship.
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11

A student concludes from the leaf width data that “increasing light intensity beyond the range tested will continue to increase leaf width indefinitely.” Evaluate this conclusion.

Show answer
This conclusion is not justified, because it extrapolates beyond the range of light intensities actually tested; the relationship observed within the tested range cannot be assumed to continue indefinitely. In reality, biological responses like leaf growth typically have a limiting factor and optimum, beyond which further increases in light intensity would have no additional effect, or could even become damaging (e.g. through photoinhibition or overheating), so leaf width would be expected to plateau or decrease rather than increase indefinitely.
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12

Outline one strength and one weakness of using a large sample size ($n=100$) rather than a small sample size ($n=5$) in a biological investigation such as this one.

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Strength: a larger sample size gives a more reliable/representative estimate of the true population mean, reduces the effect of anomalous results or individual variation, and gives a smaller standard error, increasing confidence in any conclusions drawn (e.g. about a significant difference between groups). Weakness: a larger sample size requires substantially more time, resources, space and/or organisms, and may not always be practical or ethical (for example, if the investigation involves live animals or destructive sampling).
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