Task Verbs and Free-Response Technique: Exam Skills Guide (Biology)

How to use this guide

This guide is for anyone preparing for the free-response section of the biology exam. Many marks there are lost by answering a different question from the one asked: describing when the verb is explain, or predicting with no reason. Every question and data set here is original Marlbridge practice, not past exam material; organisms, compounds and numbers are fictional unless named. The marks shown are a suggested Marlbridge rubric to help you check your work; they are not official scoring.

The free-response section at a glance

Section II: 90 minutes, 6 questions, 50% of the exam. A four-function, scientific or graphing calculator is allowed. You view the questions in the Bluebook app but handwrite answers, and any graph, in a paper booklet.

Question	Type	Points	What the parts ask for
1	Interpreting and evaluating experimental results	9	(A) describe, 1; (B) methods or describe data, 3; (C) methods, data analysis or calculation, 3; (D) predict and justify, 2
2	Interpreting and evaluating experimental results, with graphing	9	(A) describe, 1; (B) graph the table’s data, 4; (C) analyse, calculate, state a null hypothesis or predict results, 2; (D) predict and justify, 2
3	Scientific investigation	4	describe; identify a procedure; state the null hypothesis or predict results; justify (1 each)
4	Conceptual analysis	4	describe; explain; predict causes or effects of a change; justify (1 each)
5	Analyse a model or visual representation	4	describe the visual; explain links within it; represent a relationship; connect to a wider principle (1 each)
6	Analyse data	4	describe data (two parts); use data to evaluate a hypothesis; link the results to wider biology (1 each)

The long questions carry 18 of the 34 points. Each short question draws on a different unit and big idea.

Task verbs: what each one asks for

Verb	What you must do	How much to write
Identify	Name the thing asked for. Reasons earn nothing extra.	A word or phrase
State (the null hypothesis)	Write the testable statement that the variable tested has no effect, or that the groups do not differ.	One sentence naming both variables
Describe	Give the relevant features of something or of the data. No reason needed.	One or two sentences, with numbers
Explain	Say how or why something happens, linking cause to effect. “How” wants the mechanism; “why” wants the reason.	Two to four linked sentences
Make a claim	Take a position that rests on data or on biology you know.	One sentence
Support a claim	Show, with reasoning, how given evidence backs a given claim.	Two or three sentences
Justify	Back a claim, choice or prediction with evidence and the reasoning that connects them.	Two or three sentences
Evaluate	Judge how much information matters, or how sound and accurate a claim is. Give a verdict and the deciding evidence.	Two or three sentences
Predict / Make a prediction	Say what a change to part of a system will cause, or what could have caused it, with a direction.	One sentence with a direction
Determine	Reach a value or decision by reasoning or calculation, and make it clear.	Short working, then the answer
Calculate	Show the steps: formula, values substituted, answer with units and sensible rounding.	Working lines and a final answer
Construct / Draw	Produce a graph, diagram or model showing a relationship. Label it whenever labels help.	The drawing, with a key
Represent	Show a relationship in a suitable form: graph, symbols, diagram, words or table.	Whatever the form needs

Describe vs explain: “the rate rises” describes; “the rate rises because more active sites are occupied” explains. Predict vs justify: the what versus the why; write them as separate sentences.

The verbs in four groups

Group 1: naming and describing (identify, state, describe)

A student counts oxygen bubbles released per minute by a water plant placed 10, 20, 30, 40 and 50 cm from a lamp.

• Identify the independent variable. “Distance from the lamp.” Nothing more.
• State the null hypothesis. “Distance from the lamp has no effect on the rate of bubble production.” “Closer plants make more bubbles” is a prediction instead.
• Describe the trend. “As the distance rises from 10 to 50 cm, the rate falls from 30 to 4 bubbles per minute.” Quote numbers with units; do not explain.

Group 2: arguing (make a claim, support a claim, justify, explain, evaluate)

After 10 minutes at 70 °C, an enzyme from a hot-spring bacterium keeps 90% of its activity; the same enzyme from a soil bacterium keeps 5%.

• Make a claim. “The hot-spring enzyme is more heat-stable.”
• Support the claim. “It kept 90% of its activity at 70 °C against 5%, so far more of its folded shape and active site survived the heat.”
• Evaluate the claim. “The data support it at 70 °C, but only one temperature was tested and no repeats are shown, so the support is limited.”

Pattern: claim, evidence, reasoning. Evaluate adds how far the evidence goes.

Group 3: looking ahead (predict, determine)

On a fictional island, one bat species is the only pollinator of a cactus. A disease kills most of the bats.

• Predict the effect on new cactus seedlings. “The number of seedlings falls.” “The cactus is affected” earns nothing.
• Justify. “Fewer flowers are pollinated, so fewer seeds form and fewer seedlings grow.”
• Determine ends in a decision: for grass → grasshopper → shrew → hawk, “the hawk feeds at the fourth trophic level”.

Group 4: numbers and pictures (calculate, construct, represent)

• Calculate. A seedling’s mass rises from 2.40 g to 3.12 g in 6 days. Calculate the growth rate. Rate = (3.12 − 2.40) g ÷ 6 days = 0.72 g ÷ 6 days = 0.12 g day⁻¹.
• Construct a graph. Independent variable on the x-axis; axes labelled with variable and units; an even scale using most of the grid; accurate plotting; a line graph for continuous data, bars for categories; error bars if given. Suggested Marlbridge split of 4 graph points: type, axes and units, scale, plotting with error bars.
• Represent. In a food web, arrows point from the eaten to the eater, the way energy flows.

How to read a multi-part question

1. Read the stem and data before part (a). Note what was changed, what was measured, the units and any control.
2. Underline each verb and its object. “Explain the effect of temperature on the rate” names the variables your answer must use.
3. Count the points. Short-question parts are worth 1 point; long-question parts can be worth 2, 3 or 4, needing that many separate ideas.
4. Label every part in the booklet, and make each stand alone: re-name the variable instead of writing “as above”.
5. Use the scenario’s terms and numbers. A general answer rarely earns credit.
6. Stuck early? State an assumption and answer the later parts.

Planning your 90 minutes

There are 34 points in 90 minutes, about 2.6 minutes per point: roughly 24 minutes for a long question and 11 for a short one. A suggested Marlbridge plan:

Question	Points	Suggested minutes
1 (long)	9	22
2 (long, with graph)	9	24
3–6 (short)	4 each	10 each, 40 in total
Final check	—	4

The question 2 graph takes longer than you expect. If a short question passes 12 minutes, move on; come back in the final check.

Showing work on calculations

• Formula first, from the equations and formulas sheet.
• Substitute with units. For a 0.15 mol L⁻¹ sodium chloride solution at 25 °C (i = 2): Ψs = −(2)(0.15 mol L⁻¹)(0.0831 L bar mol⁻¹ K⁻¹)(298 K) = −7.43 bars.
• Answer with units, matching the data’s precision; round only at the end.
• Percentage change: divide the change by the starting value, not the final one.
• Chi-square: tabulate o, e and (o − e)² ÷ e. Give the degrees of freedom (categories − 1), the critical value at p = 0.05, and “reject” or “fail to reject” the null hypothesis. Never “prove”.
• Error bars: for mean ± 2 SE (roughly a 95% confidence interval), bars that do not overlap suggest a real difference; overlapping bars do not let you claim one.

What earns and loses credit

Earns credit	Loses credit
Doing what the verb asks	Describing when asked to explain
A prediction with a direction and a named variable	“It will be affected”
Reasoning linking evidence to claim	Restating data with no biology
Numbers from the data, with units	“It went up a lot”
Precise terms (protein, proton; gene, allele)	Look-alike terms, or an analogy instead of the biology
One clear answer	Two different answers that contradict each other

Worked example 1: Describe, explain, predict and justify (Unit 3)

Late in a long dive, a fictional diving mammal’s swimming muscles run out of oxygen, yet keep contracting at the same rate, and lactate builds up in them.

(a) Describe the role of oxygen in cellular respiration. (b) Explain how the muscle cells keep making ATP after their oxygen runs out. (c) Predict how the muscles’ rate of glucose use changes once the oxygen runs out. (d) Justify your prediction.

Weak answer. (a) Oxygen is needed to make energy. (b) The cells switch to anaerobic respiration, which makes lactic acid. (c) Glucose use goes down. (d) The animal is not breathing, so less respiration happens.

What it misses. (a) never names oxygen’s job. (b) uses the wrong term (fermentation has no electron transport chain) and does not say how ATP keeps coming. (c) and (d) give the wrong direction, confusing breathing with cellular respiration.

Strong answer. (a) Oxygen is the final electron acceptor at the end of the electron transport chain. It accepts electrons and H⁺ to form water, which keeps the chain pumping H⁺ for ATP synthase. (b) Glycolysis still makes a net 2 ATP per glucose, but it needs NAD⁺. In lactic acid fermentation, NADH passes its electrons to pyruvate, forming lactate and regenerating NAD⁺, so glycolysis keeps running. (c) The rate of glucose use increases. (d) Fermentation gives only 2 ATP per glucose, against about 30 from aerobic respiration. To supply the same ATP, the cells must break down glucose much faster.

Part	Suggested Marlbridge rubric point
(a)	O₂ is the final electron acceptor of the electron transport chain, forming water
(b)	Fermentation regenerates NAD⁺, so glycolysis keeps making ATP
(c)	Glucose use increases
(d)	Fewer ATP per glucose, so more glucose for the same ATP

Worked example 2: State, calculate and determine (Unit 5)

In a fictional herb, purple petals (P) are dominant to white (p). Two heterozygous plants are crossed. The 240 offspring are 166 purple and 74 white.

(a) State the null hypothesis. (b) Calculate the chi-square value. (c) Determine whether the null hypothesis can be rejected at p = 0.05.

Weak answer. (a) Purple is dominant. (b) (166 − 180)² ÷ 166 + (74 − 60)² ÷ 74 = 1.18 + 2.65 = 3.83. (c) 3.83 < 3.84, so the hypothesis is proven.

What it misses. (a) is a fact, not a testable statement. (b) divides by observed counts, which pulls χ² below the critical value and flips the conclusion. (c) gives no degrees of freedom and says “proven”.

Strong answer. (a) There is no significant difference between the observed numbers and a 3 purple : 1 white ratio; any difference is due to chance. (b) Expected: 240 × ¾ = 180 purple and 240 × ¼ = 60 white.

Phenotype	o	e	(o − e)² ÷ e
Purple	166	180	(−14)² ÷ 180 = 1.09
White	74	60	14² ÷ 60 = 3.27

χ² = 1.09 + 3.27 = 4.36. (c) Degrees of freedom = 2 − 1 = 1, so the critical value at p = 0.05 is 3.84. Since 4.36 > 3.84, reject the null hypothesis: the difference from 3 : 1 is unlikely to be chance alone.

Part	Suggested Marlbridge rubric point
(a)	No difference from the expected 3 : 1 ratio
(b)	Correct expected values and χ² = 4.36
(c)	1 degree of freedom, 3.84 compared and “reject” stated

Worked example 3: Describe and evaluate with error bars (Unit 8)

Ecologists test the claim that a grazing mammal reduces tree-seedling density. At each of two sites they fence 10 plots and leave 10 open. The grazer lives only at Site 1. Values are mean seedlings per m² ± 2 SE.

Site	Fenced plots	Open plots
1 (grazer present)	24.0 ± 3.1	15.0 ± 2.8
2 (grazer absent)	22.0 ± 4.0	21.0 ± 4.2

(a) Describe the difference between fenced and open plots at Site 1. (b) Evaluate the claim using both sites. (c) Explain why Site 2 was included.

Weak answer. (a) Fenced plots have more seedlings. (b) The claim is right: fenced plots have more seedlings at both sites. (c) For more data.

What it misses. (a) has no numbers. (b) ignores the error bars and treats a 1-seedling difference at Site 2 as real. (c) misses Site 2’s purpose.

Strong answer. (a) Open plots average 15.0 seedlings per m², 9.0 (37.5%) fewer than fenced plots (24.0). (b) At Site 1 the intervals (20.9–27.1 and 12.2–17.8) do not overlap, so the difference is probably not chance. At Site 2 they (18.0–26.0 and 16.8–25.2) overlap widely: no clear difference. Fencing matters only where the grazer lives, which supports the claim, though the sites may also differ in soil or light. (c) Site 2 controls for the fence itself: it shows fencing does not raise seedling numbers by another route, such as shelter, when the grazer is absent.

Part	Suggested Marlbridge rubric point
(a)	Direction with values from the table
(b)	Site 1 bars separate, Site 2 bars overlap, verdict linked to the grazer
(c)	Site 2 rules out an effect of the fence itself

Common mistakes

• Describing when the verb is explain.
• Stating a null hypothesis as a prediction (“more light gives more bubbles”).
• Dividing by the observed value in chi-square, or leaving out degrees of freedom.
• Saying a test “proves” a hypothesis.
• Claiming a difference when the ±2 SE error bars overlap.
• Working out percentage change from the final value.
• Drawing food-web or pathway arrows the wrong way.

Practise it

Task 1 (Unit 1). A fictional desert beetle is coated in wax made of long, nonpolar hydrocarbon chains. Explain how this layer reduces water loss from the beetle’s body.

Model answer

The hydrocarbon chains are nonpolar, so they cannot form hydrogen bonds with water. Water does not dissolve in or pass easily through the wax, which separates the moist body from the dry air, so less water evaporates.

Task 2 (Unit 2). Calculate the water potential of 0.25 mol L⁻¹ glucose in an open beaker at 27 °C. Predict whether plant cells with Ψ = −7.0 bars gain or lose water in it, and justify.

Model answer

T = 27 + 273 = 300 K. Ψs = −iCRT = −(1)(0.25)(0.0831)(300) = −6.23 bars. In an open beaker Ψp = 0, so Ψ = −6.23 bars. The cells gain water, because water moves from higher water potential (−6.23 bars) to lower (−7.0 bars).

Task 3 (Unit 4). A gland releases hormone H when blood calcium falls; H raises blood calcium, and high calcium stops H release. A mutation stops the gland detecting high calcium. Predict the effect on blood calcium and justify.

Model answer

Blood calcium rises above normal and stays high. The gland never gets the signal that switches H off, so H release continues: the negative feedback loop is broken.

Task 4 (Unit 6). A DNA change turns the mRNA codon GAA into GAG. Identify the type of mutation and explain why the phenotype may not change.

Model answer

A silent substitution. GAA and GAG both code for glutamic acid, so the amino acid sequence, and the protein’s shape and function, are unchanged.

Task 5 (Unit 7). In a beetle population in Hardy–Weinberg equilibrium, 49 of 400 beetles show the recessive green colour. Determine the number of heterozygous beetles.

Model answer

q² = 49 ÷ 400 = 0.1225, so q = 0.35 and p = 1 − 0.35 = 0.65. 2pq = 2 × 0.65 × 0.35 = 0.455. Heterozygotes = 0.455 × 400 = 182 beetles.

Where to practise next

• Cellular respiration study guide: the electron transport chain and fermentation.
• Mendelian genetics practice: chi-square.
• Membrane permeability practice: standard error and error bars.
• Community ecology study guide: species interactions.
• Tonicity and osmoregulation practice: water potential.
• Hardy–Weinberg equilibrium practice: allele and genotype frequencies.