Experimental Design and Data Analysis: Exam Skills Guide (Biology)

How to use this guide

This guide is for any Biology student preparing for the free-response section. The same experiment and data skills turn up in every unit, so it covers the whole course. Read the explanations, study the worked examples, then try Practise it before opening the model answers.

Every example here is original Marlbridge practice, not past exam material, and all organisms and data are fictional. The marks shown are a suggested Marlbridge rubric, not official scoring.

The free-response section at a glance

The free-response section has 6 questions in 90 minutes and is worth 50% of the exam. You read the questions on screen and write your answers, including graphs, by hand in a paper booklet. A four-function, scientific or graphing calculator is allowed, and a formula sheet is provided.

Question	Type	Points	What the parts ask for
1	Interpreting and Evaluating Experimental Results	9	Describe a concept; identify methods or describe data; analyse data or calculate; make and justify predictions
2	Interpreting and Evaluating Experimental Results with Graphing	9	Describe a concept; construct a graph (4 points); analyse, calculate or state a null hypothesis; make and justify predictions
3	Scientific Investigation	4	Describe a concept; identify procedures; state a null hypothesis or predict; justify
4	Conceptual Analysis	4	Describe and explain a concept; predict the effect of a disruption; justify
5	Analyze Model or Visual Representation	4	Describe and explain a model; represent a relationship; link to a wider principle
6	Analyze Data	4	Describe data; use data to evaluate a hypothesis; link results to biology

A sensible Marlbridge time plan: about 22 minutes for each long question and 11 for each short one.

Variables, controls and hypotheses

Label the set-up before you answer.

Term	Meaning	Quick test
Independent variable (IV)	The factor deliberately changed	“What differs between groups?”
Dependent variable (DV)	What is measured	“What number goes in the table?”
Controlled variables	Factors kept the same	Anything else that could change the DV
Control group	A baseline for comparison	Usually the untreated condition
Negative control	No effect expected	Shows the effect needs the factor tested (e.g. boiled enzyme)
Positive control	An effect is known to happen	Shows the method can detect an effect

Be specific. Name the quantity and how it is measured: “volume of oxygen collected in 2 minutes (mL)”, not “how well the enzyme works”.

Null and alternative hypotheses. H₀ says the independent variable has no effect on the dependent variable. The alternative says there is an effect, often with a direction. Name both variables:

• H₀: The concentration of compound K has no effect on the percentage of root-tip cells in mitosis.
• Alternative: Increasing the concentration of compound K decreases the percentage of root-tip cells in mitosis.

Predicting and justifying. A prediction gives a direction (“will be lower than the control”). A justification adds the mechanism linking cause to effect. One sentence of each is usually enough.

Graphing data

Question 2 always asks for a graph. Points come from the elements below; a title earns nothing but still helps the reader.

Element	What a strong graph does
Type	Line or scatter graph when both variables are continuous (time, concentration). Bar graph for categories (treatments, species). Histogram for the spread of one variable
Axes	IV on x, DV on y, both with units. A legend when there is more than one series
Scale	Even intervals using at least half the grid; include negative values if the data have them
Plotting	Accurate points; error bars whenever the table gives SE or ± 2 SE
Trend line	Join points or draw a best-fit line. Do not extend it past the data or to the origin; use a dashed line for any prediction

Reading error bars. Bars of mean ± 2 SE are roughly a 95% interval. If two bars do not overlap, reject H₀; if they overlap, fail to reject H₀. This is an approximation, weakest with few repeats.

The calculations you must be able to do

The formula sheet gives most equations. Always show the formula, the substituted numbers and the answer with units.

Calculation	How	Common trap
Rate	change in quantity ÷ change in time	Wrong time interval; no units
Percent change	(new − original) ÷ original × 100	Dividing by the new value; losing the minus sign
Mean	sum of values ÷ number of values	Including an outlier without comment
Standard deviation (s)	spread around the mean, using n − 1	Using n instead of n − 1
Standard error (SE)	s ÷ √n	Forgetting the square root; confusing SE with s
Chi-square (χ²)	Σ (o − e)² ÷ e with counts, df = categories − 1	Using percentages instead of counts; df = sample size − 1
Hardy–Weinberg	p + q = 1; p² + 2pq + q² = 1; q² = recessive phenotype frequency	Forgetting the square root for q
Water potential	Ψ = Ψp + Ψs and Ψs = −iCRT (i = 1 for sucrose, T in kelvin)	Using °C; forgetting the minus sign

For chi-square, compare χ² with the critical value at p = 0.05 for your degrees of freedom. If χ² is greater, reject H₀: the difference is unlikely to be chance alone. If it is smaller, fail to reject H₀.

Claim, evidence and reasoning

For “justify”, “support a claim” or “evaluate”, build three pieces:

1. Claim: a direct answer (“Above 35 °C the enzyme loses activity because it denatures”).
2. Evidence: numbers from the data, with units (“Oxygen fell from 11.4 mL at 35 °C to 0.6 mL at 55 °C”).
3. Reasoning: the principle linking evidence to claim (“Heat breaks hydrogen bonds holding the active site in shape, so substrate cannot bind”).

Evidence alone is a description; reasoning alone is a textbook sentence. You need both.

Worked example 1: Variables, controls and rates (Unit 3)

Question. An enzyme from a fictional soil fungus breaks hydrogen peroxide into water and oxygen. A student adds 1 mL of extract to 10 mL of hydrogen peroxide in pH 7 buffer and collects the oxygen released in 2 minutes at five temperatures. Mean results (three trials): 15 °C, 4.2 mL; 25 °C, 7.8 mL; 35 °C, 11.4 mL; 45 °C, 6.0 mL; 55 °C, 0.6 mL. (a) Identify the independent and dependent variables. (b) Describe a suitable negative control. (c) Calculate the rate at 35 °C. (d) Predict the result if the 55 °C enzyme is cooled to 35 °C and retested, and justify your prediction.

Weak answer.

(a) IV is temperature, DV is the enzyme. (b) Use no temperature. (c) 11.4. (d) It will work again because it is back at the best temperature.

What it misses. The DV must be measured, not “the enzyme”. “No temperature” is impossible. The rate is not divided by time and has no units. The prediction ignores denaturation.

Strong answer.

(a) IV: temperature of the reaction mixture (°C). DV: volume of oxygen collected in 2 minutes (mL). (b) Repeat each temperature with enzyme extract that has been boiled and cooled, keeping all volumes the same. No oxygen should be released, showing that oxygen comes from enzyme activity. (c) Rate = 11.4 mL ÷ 2 min = 5.7 mL min⁻¹. (d) Oxygen release will stay low, near 0.6 mL in 2 minutes, not return to 11.4 mL. At 55 °C the hydrogen bonds holding the enzyme’s tertiary structure break, so the active site changes shape. This denaturation is usually not reversed by cooling, so the substrate still cannot bind.

Part	Suggested Marlbridge rubric (1 point each)
(a)	Both variables named with units or method of measurement
(b)	Boiled (or no) enzyme with everything else the same, and its purpose
(c)	5.7 with units of mL min⁻¹
(d)	Prediction of low activity and justification using denaturation of the active site

Worked example 2: Graphing and water potential (Unit 2)

Question. Cylinders cut from a fictional root vegetable were weighed, left for 24 hours in open beakers of sucrose solution at 22 °C, and reweighed. Mean percent change in mass: 0.0 M, +18.0%; 0.1 M, +9.6%; 0.2 M, +2.4%; 0.3 M, −4.8%; 0.4 M, −11.2%; 0.5 M, −16.0%. (a) Describe the graph you would draw. (b) Estimate the sucrose concentration at which the tissue would show no change in mass. (c) Calculate the water potential of the root tissue.

Weak answer.

(a) A bar graph of the results. (b) About 0.2 M because that is the smallest change. (c) Ψ = iCRT = 0.2 × 0.0831 × 22 = 0.37.

What it misses. Concentration is continuous, so a bar graph is wrong. The answer needs the point where the line crosses zero. The calculation uses °C, drops the minus sign and gives no units.

Strong answer.

(a) A line graph: sucrose concentration (mol L⁻¹) on the x-axis, from 0 to 0.5 in even steps; mean percent change in mass (%) on the y-axis, running from −20 to +20 so that negative values show. Plot the six points and join them or draw a line of best fit. (b) The line crosses 0% between 0.2 M (+2.4%) and 0.3 M (−4.8%), at about 0.23 M. At this concentration there is no net water movement. (c) The beakers are open, so Ψp of the solution is 0 and Ψ = Ψs. Ψs = −iCRT = −(1)(0.233 mol L⁻¹)(0.0831 L bar mol⁻¹ K⁻¹)(295 K) = −5.7 bar. With no net water movement, the tissue has the same water potential, about −5.7 bar.

Part	Suggested Marlbridge rubric
(a)	1 point: line graph with concentration on x and percent change on y, both with units; 1 point: even scales that include negative values
(b)	1 point: 0.22–0.25 M read from where the line crosses zero
(c)	1 point: correct substitution with T = 295 K and i = 1; 1 point: −5.7 bar (−5.4 to −6.0 accepted if consistent with (b)) with units

Worked example 3: Chi-square and the null hypothesis (Unit 5)

Question. In a fictional plant, flower colour shows incomplete dominance. Red × white gives pink offspring. Two pink plants are crossed and produce 200 offspring: 41 red, 114 pink and 45 white. (a) State the null hypothesis. (b) Calculate χ². (c) Decide whether to reject the null hypothesis.

Weak answer.

(a) The flowers are 1 : 2 : 1. (b) χ² = (20.5 − 25)² ÷ 25 + (57 − 50)² ÷ 50 + (22.5 − 25)² ÷ 25 = 2.04. (c) 2.04 is small so the hypothesis is proven correct.

What it misses. H₀ must say differences are due to chance. Percentages replace counts, which changes the value. “Proven” is never acceptable, and there is no df or critical value.

Strong answer.

(a) H₀: The offspring occur in a 1 red : 2 pink : 1 white ratio, and any difference between observed and expected numbers is due to chance. (b) Expected: 200 × ¼ = 50 red, 200 × ½ = 100 pink, 50 white. χ² = (41 − 50)² ÷ 50 + (114 − 100)² ÷ 100 + (45 − 50)² ÷ 50 = 1.62 + 1.96 + 0.50 = 4.08. (c) There are 3 categories, so df = 2 and the critical value at p = 0.05 is 5.99. As 4.08 < 5.99, fail to reject H₀. The data are consistent with incomplete dominance of a single gene with two alleles.

Part	Suggested Marlbridge rubric (1 point each)
(a)	Null hypothesis naming the 1 : 2 : 1 ratio and chance
(b)	Expected counts and χ² = 4.08
(c)	df = 2, critical value 5.99, and “fail to reject” with the comparison stated

Common mistakes

• Naming the dependent variable as an object (“the plant”) instead of a measured quantity with units.
• Describing a control as “keeping everything the same”; that defines controlled variables, not a control group.
• Writing a null hypothesis that already predicts an effect.
• Drawing a bar graph for continuous data, leaving out axis units, or using uneven scales.
• Ignoring error bars, or saying overlapping ± 2 SE bars show a significant difference.
• Using percentages, not counts, in chi-square, or setting df to the sample size minus one.
• Taking q directly as the frequency of the recessive phenotype instead of its square root.
• Writing “the data prove”, or quoting a trend with no number from the data.

Practise it

Task 1 (Unit 1). A student measures how high water rises in glass tubes of internal diameter 0.5, 1.0, 1.5 and 2.0 mm, five tubes per size, and records mean height ± 2 SE. Describe the graph to draw, and give one controlled variable.

Model answer

A scatter (or line) graph: tube diameter (mm) on x, mean water height (mm) on y, even scales. Plot each mean with vertical ± 2 SE error bars and a trend line that stops at the data, not at the origin. Controlled variable: water temperature (or type of glass). Height should fall as diameter rises: adhesion to the glass and cohesion through hydrogen bonds lift a narrow column higher.

Task 2 (Unit 4). Root tips are grown with or without a fictional compound K. The percentage of cells in mitosis in five root tips is: control 12, 14, 11, 13, 15; compound K 4, 9, 6, 8, 8. State the null hypothesis, calculate the mean and SE for compound K (control: mean 13, SE 0.71), and decide whether to reject the null hypothesis.

Model answer

H₀: Compound K has no effect on the percentage of root-tip cells in mitosis. Compound K: mean = 35 ÷ 5 = 7%. Deviations are −3, 2, −1, 1, 1; squares sum to 16; s = √(16 ÷ 4) = 2.0; SE = 2.0 ÷ √5 = 0.89. Mean ± 2 SE: control 11.6–14.4%; compound K 5.2–8.8%. The intervals do not overlap, so reject H₀. Compound K appears to reduce cell division, although five root tips is a small sample.

Task 3 (Unit 6). Researchers think a fictional transcription factor, TF-K, switches on a heat-response gene, hspB, in yeast at 37 °C. Identify the independent variable, the dependent variable and a suitable control, and predict the result if the idea is correct.

Model answer

IV: presence of a working TF-K gene (normal cells against cells with TF-K deleted). DV: amount of hspB mRNA after a set time at 37 °C. Control: normal yeast with working TF-K, same medium, temperature and time. Prediction: cells lacking TF-K will have much less hspB mRNA. Justification: a transcription factor binds DNA near the promoter and helps RNA polymerase start transcription, so without it, hspB transcription falls.

Task 4 (Unit 7). In a population of 500 fictional beetles, 80 show the recessive green colour. Assuming Hardy–Weinberg equilibrium, calculate p, q and the expected number of heterozygous beetles. State one assumption you made.

Model answer

q² = 80 ÷ 500 = 0.16, so q = √0.16 = 0.4 and p = 1 − 0.4 = 0.6. Heterozygote frequency 2pq = 2 × 0.6 × 0.4 = 0.48, so the expected number is 0.48 × 500 = 240 beetles. Assumption: mating is random (or no migration, no selection, no mutation, a large population).

Task 5 (Unit 8). An island population of a fictional lizard has 400 individuals. In one year there are 120 births and 60 deaths. Calculate dN/dt and the percent change in population size. Predict how the growth rate will change as the population nears the island’s carrying capacity, and justify your prediction.

Model answer

dN/dt = B − D = 120 − 60 = 60 lizards per year. Percent change = 60 ÷ 400 × 100 = 15%, giving 460 lizards. As the population nears carrying capacity, the growth rate will slow towards zero. Justification: more lizards share the same food and space, so competition raises deaths and lowers births; in the logistic model (K − N) ÷ K approaches 0.

Where to practise next

• Membrane permeability practice: standard error and ± 2 SE decisions.
• Tonicity and osmoregulation study guide: water potential calculations in full.
• Environmental impacts on enzyme function practice: variables, controls and predictions with enzymes.
• Mendelian genetics practice: more chi-square tests.
• Hardy–Weinberg equilibrium study guide: allele and genotype frequencies.
• Population ecology study guide: growth rates and carrying capacity.
• Unit 4 review: mixed questions on cell communication and the cell cycle.