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AP Biology – Data Analysis and FRQ Practice

Experimental design, statistics, long free-response questions.

10 questions  ·  39 marks  ·  answers included  ·  AP Biology (College Board)
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This worksheet covers:

  • Experimental design
  • Descriptive and inferential statistics
  • And long free-response question (FRQ) style problems
  • Matching the AP Biology Science Practices for data analysis

A calculator is required for the numerical questions; this set has 10 questions in the AP FRQ style.

Questions

Section A – Statistics and Experimental Design

Calculator required
1

A student measures the length, in mm, of seven leaves from a plant: $23, 25, 21, 27, 24, 26, 22$.(a) Calculate the mean leaf length.(b) Calculate the sample standard deviation, correct to 3 significant figures.(c) Calculate the standard error of the mean (SEM), correct to 3 significant figures.

Show answer
(a) Mean $= 24.0\,\text{mm}$.(b) $s = \sqrt{\dfrac{\sum(x-\bar{x})^2}{n-1}} = 2.16\,\text{mm}$.(c) $\text{SEM} = \dfrac{s}{\sqrt{n}} = \dfrac{2.16}{\sqrt{7}} = 0.816\,\text{mm}$.
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2

Explain why the standard error of the mean (SEM) is often a more useful statistic than the standard deviation when comparing the means of two different treatment groups in an experiment.

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The standard deviation describes the spread of the individual data points around the sample mean, whereas the SEM describes the precision of the sample mean itself as an estimate of the true population mean, and decreases as sample size increases. Since it is the means (not individual data points) that are being compared, the SEM (used to construct error bars/confidence intervals for the mean) is more directly relevant to judging whether two means are meaningfully different.
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3

In an experiment investigating the effect of light intensity on the rate of photosynthesis in an aquatic plant, explain the difference between a control group and a controlled variable, giving one example of each for this experiment.

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A control group is a group of samples that is set up like the experimental groups but is not exposed to the tested variable (or is exposed to a baseline/zero level of it), providing a baseline for comparison — for example, a plant kept in complete darkness (or under a standard reference light intensity). A controlled variable is any other factor besides the independent variable that is kept constant across all groups so it cannot affect the results — for example, water temperature, carbon dioxide concentration, or the species/size of plant used.
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4

A researcher compares the mean height of two plant populations. Population A has a mean of $24\,\text{cm}$ with $\text{SEM} = 0.9\,\text{cm}$; Population B has a mean of $27\,\text{cm}$ with $\text{SEM} = 1.1\,\text{cm}$. Using $\text{mean} \pm 2\,\text{SEM}$ as an approximate 95\% confidence interval for each population mean, determine whether these data provide clear evidence that the two population means are different, and explain your reasoning.

Show answer
Population A’s interval is $22.2$ to $25.8\,\text{cm}$, and Population B’s interval is $24.8$ to $29.2\,\text{cm}$. These two intervals overlap (between $24.8$ and $25.8\,\text{cm}$), so the data do not provide clear evidence that the two population means are significantly different; a formal statistical test (such as a $t$-test) would be needed to determine this more rigorously.
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5

A dihybrid genetic cross is predicted to produce offspring in a $9:3:3:1$ phenotypic ratio. Out of $556$ offspring observed, the counts for the four phenotype classes are $315$, $101$, $108$ and $32$.(a) Calculate the expected number of offspring for each phenotype class.(b) Calculate the chi-squared ($\chi^2$) statistic for these data.(c) The critical value of $\chi^2$ at $3$ degrees of freedom and $p = 0.05$ is $7.82$. State, with a reason, whether these data support the $9:3:3:1$ hypothesis.

Show answer
(a) Expected values $= 556 \times \dfrac{9}{16}, \dfrac{3}{16}, \dfrac{3}{16}, \dfrac{1}{16} = 312.75, 104.25, 104.25, 34.75$.(b) $\chi^2 = \sum\dfrac{(O-E)^2}{E} = 0.470$.(c) Since $0.470 < 7.82$, the calculated $\chi^2$ does not exceed the critical value, so we fail to reject the null hypothesis: the observed data are consistent with (support) the predicted $9:3:3:1$ ratio.
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6

State the null hypothesis being tested by the chi-squared test in the previous question, and explain what it means for the calculated $p$-value to be greater than $0.05$.

Show answer
Null hypothesis: there is no significant difference between the observed and expected offspring ratios (the observed data fit the predicted $9:3:3:1$ ratio). A $p$-value greater than $0.05$ means that, if the null hypothesis were true, a deviation between observed and expected values at least this large would be expected to occur simply by chance more than $5\%$ of the time, so there is not enough evidence to reject the null hypothesis.
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7

State two conditions that should ideally be met for an independent-samples t-test to be an appropriate statistical test for comparing two sample means.

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The two samples should be independent of one another (not paired); each sample should be a random sample from its population; the underlying populations should be approximately normally distributed (especially important for small samples); and the two populations should have similar variances (any two).
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Section B – Extended Free-Response Practice

8

A biologist hypothesises that increasing the concentration of a nitrogen-based fertiliser increases the growth rate of tomato plants.Design an experiment to test this hypothesis. In your answer, include:(a) the independent and dependent variables,(b) two variables that should be controlled, and why,(c) how a control group should be included,(d) how you would use the collected data (including an appropriate statistical test) to determine whether fertiliser concentration has a significant effect on growth rate.

Show answer
(a) Independent variable: fertiliser concentration. Dependent variable: plant growth rate (e.g. increase in height or biomass per week).(b) Controlled variables should include, for example, light intensity and water volume/frequency (so that any difference in growth can be attributed to fertiliser concentration rather than to differences in these other factors); the same plant species/starting size should also be used in each group.(c) A control group of plants should receive no fertiliser (or a $0\%$ concentration solution) but otherwise be treated identically to the experimental groups, providing a baseline growth rate for comparison.(d) Multiple replicate plants should be used at each fertiliser concentration, and the mean growth rate (with standard error) calculated for each group; a statistical test such as ANOVA (or a $t$-test between pairs of groups) should be used to determine whether differences in mean growth rate between concentration groups are statistically significant (i.e. unlikely to be due to chance alone, typically $p<0.05$).
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9

A group of ecology students wants to compare species diversity between two woodland habitats. Explain how they should use random sampling and an appropriate sample size to obtain a reliable estimate of species diversity in each habitat, and describe one statistical test they could use to compare the two habitats’ diversity.

Show answer
Sampling points (e.g. quadrat locations) within each habitat should be chosen randomly (for example using randomly generated coordinates), rather than being chosen by the students, to avoid bias in which areas are sampled. A sufficiently large number of quadrats/samples should be taken in each habitat (and a species-area curve could be used to check that enough samples have been taken to capture most of the species present) so that the estimate of diversity is representative of the whole habitat rather than being skewed by a few unusual samples. To compare the two habitats statistically, a diversity index (such as Simpson’s Index) could be calculated for each habitat from the sample data, and a $t$-test (or chi-squared test comparing species composition) could be used to test whether the difference in diversity index between the two habitats is statistically significant.
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10

The activity of an enzyme was measured at a range of temperatures from $10\,^\circ\text{C}$ to $60\,^\circ\text{C}$. The rate of product formation increased steadily from $10\,^\circ\text{C}$ up to a maximum at $40\,^\circ\text{C}$, and then decreased sharply between $40\,^\circ\text{C}$ and $60\,^\circ\text{C}$.(a) Describe the relationship shown by these data.(b) Explain, in terms of enzyme structure, why the activity decreases sharply above the optimum temperature.

Show answer
(a) Enzyme activity increases with temperature up to an optimum of $40\,^\circ\text{C}$ (where the rate is highest), and then decreases sharply as temperature increases further beyond $40\,^\circ\text{C}$.(b) Above the optimum temperature, the increased kinetic energy of the enzyme molecule disrupts the hydrogen bonds and other interactions maintaining its tertiary structure, causing the enzyme (including its active site) to change shape (denature). Since the substrate can no longer bind effectively to the altered active site, the rate of reaction decreases sharply as more enzyme molecules become denatured.
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