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AP Biology Data Analysis FRQ Practice

127 Analyze Data questions written in the AP format. Pick one, read the full question, and write your response.

Your first scored response is included. A plan unlocks unlimited scoring.

New to the Data Analysis FRQ? Read the exam guide

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127 questions

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Enzyme thermal stability and protein folding mutations

Unit 1: Chemistry of Life

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Pectate lyase is an enzyme found in certain bacteria that degrades pectin in plant cell walls. The enzyme's function depends on its tertiary structure, which is maintained by interactions between amino acid R-groups.

Scientists identified a strain of bacteria with a mutation in the gene encoding pectate lyase. The mutation, labeled K128L, results in the substitution of lysine (which has a positively charged R-group) with leucine (which has a nonpolar R-group) at position 128 in the protein chain. To investigate the effect of this mutation, scientists purified both the wild-type (WT) and mutant (K128L) enzymes.

The scientists incubated both enzymes at a range of temperatures from 20°C to 60°C. They measured the relative enzymatic activity of each form (Figure 1A) and used spectroscopy to determine the percent of the protein that remained in its native, folded conformation at each temperature (Figure 1B).

Figure 1. Temperature dependence of wild-type and mutant pectate lyase

Figure 1
A.

Based on Figure 1A, identify the temperature at which the wild-type (WT) enzyme functions at its maximum relative activity.

B.

Based on Figure 1B, describe the difference in the percent of folded protein between the wild-type (WT) and mutant (K128L) enzymes at 40°C.

C.

Scientists hypothesize that the K128L mutation reduces the thermal stability of the enzyme's tertiary structure. Use the data in Figures 1A and 1B to support the scientists' hypothesis.

D.

Based on the chemical properties of the amino acid R-groups provided in the background, explain why the substitution of lysine with leucine results in the structural changes observed in Figure 1B.

What the Data Analysis FRQ asks

This is FRQ 6 on the AP Biology exam. You get data in a graph or table, describe it, use it to evaluate a hypothesis, and explain the biology behind the results.

10 min
suggested writing time
4 points
1 point for each part
50%
of your score, for all six FRQs
  1. Part A: Read the data

    Identify a value or pattern in the data

    1 pt

  2. Part B: Describe the data

    Describe a trend or comparison

    1 pt

  3. Part C: Evaluate

    Use the data to support or evaluate a hypothesis

    1 pt

  4. Part D: Biology principle

    Explain the biology behind the results

    1 pt

How Data Analysis FRQ practice works

Each question follows the exam’s format, from the full prompt to a score on every part.

  1. Read the full question

    The data and all four parts, A through D, laid out the way the exam shows them.

  2. Write on a 10-minute timer

    The time to plan for on exam day. Pause it or turn it off, and your response saves as you go.

  3. Submit for a score out of 4

    Your response is scored against the scoring guidelines written for that question. Your first score is included.

Feedback on every part

A summary at the top tells you what to work on next. Below it, each part shows whether you earned the point and what the scoring guidelines were looking for.

Detailed Feedback

Part A: 1/1 point

Correct. The student clearly identifies 'invaded soil' as the treatment with the lowest average seedling biomass, which matches the acceptable responses in the rubric. No numerical value was required, and the identification is unambiguous.

Part B: 1/1 point

Correct. The student describes that adding activated carbon to invaded soil increased mycorrhizal colonization to approximately 60%, compared to the much lower level in invaded soil alone. The direction of the change (increase) is clearly stated and an approximate value is provided, satisfying the rubric requirement.

Part C: 0/1 point

Insufficient. While the student notes that invaded soil had lower biomass and colonization than uninvaded soil, the response does not use the activated carbon treatment data to support the hypothesis. The key evidence needed is that adding activated carbon — which absorbs organic chemicals — restored both mycorrhizal colonization and seedling biomass toward uninvaded levels, implying a chemical from A. petiolata is responsible. Simply saying 'something in the invaded soil is harming the seedlings' without referencing the carbon treatment data does not earn this point.

Detailed feedback on a practice Data Analysis FRQ

Questions about the Data Analysis FRQ

How many responses can I get scored?

You can read every question without a plan. Your first scored response is included. A plan unlocks unlimited scoring.

How do I describe data so it earns the point?

Cite specific values and the direction of the change, like “increased from 2 to 5 mm per day.” Saying a value “went up” is usually too vague.

Do I need to calculate anything?

Sometimes, but rarely much. When you do, type your setup and the answer with its units.

How is this different from the Scientific Investigation FRQ?

This question centers on reading results and what they mean. The Scientific Investigation FRQ centers on how the experiment is designed.

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Scientific Investigation FRQ practice

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Conceptual Analysis FRQ practice

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Model Analysis FRQ practice

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AP Biology FRQs

Every FRQ type for AP Biology, with practice for each.

AP Biology study guides

Unit-by-unit review for the whole course.

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