1.1 Estimate stomatal density per mm².
Reveal worked answer
Area = π(0.48/2)² ≈ 0.181 mm², so density ≈ 34/0.181 ≈ 188 mm⁻².
Original practice · not an official paper
Paper 1B skills practice
Eight independent scenarios, 24 short tasks. Work first, then reveal the answer and compare your reasoning.
These are original educational exercises, not reproduced or predicted IB examination questions.
0 of 24 tasks reviewed.
Microscopy and stomata
A leaf cast shows 34 stomata in a circular field of view 0.48 mm in diameter. A scale bar labelled 20 µm measures 15 mm on the displayed image; one guard-cell pair measures 27 mm.
1.2 Estimate the actual length represented by 27 mm on the image.
Reveal worked answer
Actual size = (27/15) × 20 µm = 36 µm.
1.3 How could the density estimate be made more reliable?
Reveal worked answer
Sample multiple independently selected fields or leaf regions using a consistent counting rule, then summarize the replicate densities.
Useful tools: Field-of-view density · Scale bar actual size · Mean, median, mode & range
Variation and error bars
Two treatments produce five replicate lengths each. Treatment A: 12.0, 12.4, 11.8, 12.1, 11.9 mm. Treatment B: 13.0, 13.2, 12.8, 13.3, 12.9 mm.
2.1 Which summary values would you report before a t-test?
Reveal worked answer
At minimum, sample size, mean and a measure of spread such as SD; SE may be used for uncertainty in the mean if clearly labelled.
2.2 What null hypothesis is suitable for a two-sample comparison?
Reveal worked answer
That the population means do not differ under the conditions represented by the two samples.
2.3 Why is “the error bars overlap” not a complete significance test?
Reveal worked answer
Overlap depends on what the bars represent and is not equivalent to the formal sampling distribution used by a t-test.
Useful tools: SD & standard error · Two-sample t-test · Error bar ranges
Species association
Quadrats are scored for presence/absence of two plant species. Observed counts are 18 both present, 7 A only, 8 B only and 17 neither.
3.1 What statistical test is appropriate for association in this 2×2 frequency table?
Reveal worked answer
A chi-squared test of association, provided the expected-frequency assumptions are reasonable.
3.2 What does the null hypothesis state?
Reveal worked answer
That presence of species A and species B is independent across the sampled quadrats.
3.3 If p < 0.05, what can you conclude?
Reveal worked answer
Reject the stated null at the 5% level; the data provide evidence of association, but the test alone does not establish the biological cause.
Useful tools: Chi-squared test · Quadrat population estimate · Simpson reciprocal index
Capture–mark–recapture
Researchers mark 80 beetles. Later they capture 70 beetles, of which 14 are marked.
4.1 Estimate the population size.
Reveal worked answer
N = 80 × 70 / 14 = 400 beetles.
4.2 Give one important assumption of the estimate.
Reveal worked answer
For example: marked individuals have mixed back into the population and have similar recapture probability to unmarked individuals.
4.3 Why would loss of marks tend to bias the estimate?
Reveal worked answer
Marked beetles could be counted as unmarked, reducing the recaptured marked count and inflating the calculated population estimate.
Useful tools: Lincoln index · Population growth
Correlation and calibration
A biological assay gives a nearly linear response to concentration over the measured standard range, with R² = 0.982. An unknown response falls just beyond the largest standard.
5.1 What does R² = 0.982 indicate?
Reveal worked answer
The linear model accounts for about 98.2% of the observed variation in response around the fitted trend.
5.2 Why is estimating the unknown outside the standards less secure?
Reveal worked answer
It is extrapolation: the relationship has not been directly established beyond the calibration range.
5.3 Does a high R² prove causation or absence of systematic bias?
Reveal worked answer
No. It describes fit to the chosen model, not causal mechanism or calibration validity.
Useful tools: Calibration curve · Regression, r & R² · Dilution
Genetics and frequencies
In a diploid population, genotype counts are AA = 36, Aa = 48 and aa = 16.
6.1 Calculate the frequency of allele A.
Reveal worked answer
p = (2×36 + 48)/(2×100) = 120/200 = 0.60.
6.2 Under Hardy–Weinberg equilibrium, what heterozygote frequency is predicted?
Reveal worked answer
2pq = 2(0.60)(0.40) = 0.48.
6.3 How could observed genotype counts be compared with equilibrium expectations?
Reveal worked answer
Convert predicted genotype frequencies to expected counts, then use an appropriate chi-squared goodness-of-fit test.
Useful tools: Allele frequency · Hardy–Weinberg · Chi-squared test
Rates and percentage change
A seedling grows from 24 mm to 36 mm over 6 days.
7.1 Calculate the average growth rate.
Reveal worked answer
(36 − 24)/6 = 2.0 mm day⁻¹.
7.2 Calculate the percentage change in length.
Reveal worked answer
(36 − 24)/24 × 100 = 50%.
7.3 Why is average rate not necessarily the instantaneous rate on every day?
Reveal worked answer
Growth can vary through time; the calculation is the secant gradient across the whole interval.
Useful tools: Rate of change · Percentage change & difference · Significant figures
Diversity and sampling
A sample contains five species with counts 12, 8, 5, 5 and 2.
8.1 What two aspects of community structure influence a diversity index?
Reveal worked answer
Species richness and evenness of abundance among species.
8.2 Why should index values from two habitats use comparable sampling effort?
Reveal worked answer
Different effort can alter detected richness and abundance patterns, confounding the comparison.
8.3 What does a larger Simpson reciprocal index indicate under this convention?
Reveal worked answer
Greater diversity in the sampled community.
Useful tools: Simpson reciprocal index · Mean, median, mode & range · Quadrat population estimate