Practice questions · Topic 3.2b

Unit 3 · Practice for the Topic 3.2b end-of-topic test

You’ve gone through everything in this topic. The summary video below recaps it all, so you’re ready for the questions.

Watch first: Topic 3.2b summary: five readings to an error bar

Video coming soon

Five readings give a mean, a standard deviation and a standard error; an error bar is the mean ± 2SE; ±2SE bars apart show a real difference, and bars that overlap show none.

These are practice questions in the shape of the topic test. Work through them before you take the test; every question tells you what it wanted.

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write short, simple sentences, each on its own line, and show any calculation; make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. The first free-response question walks you through the reasoning one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Use the mean, the standard deviation and the standard error of the mean exactly as the AP formula sheet writes them; the standard deviation divides by n − 1, and an error bar of ±2SE covers the range the true mean is likely to lie in. Rounding: keep the full value in your calculator and write at least three significant figures in your working; report a mean to one more decimal place than the readings, and a standard deviation or standard error to three significant figures. Where a graph's legend says its error bars represent ±2SE, the overlap rule applies.

Question 1

Five readings of an enzyme's rate were 8.0, 8.5, 7.5, 8.1 and 7.9 mg/min. Their mean is 8.00 mg/min. Use the formula on the AP sheet, $s = \sqrt{\frac{\sum {(x_{i} - 𝑥̄)}^{2}}{n - 1}}$ .

What is the standard deviation of these readings?

A. 0.13 mg/min B. 0.322 mg/min C. 0.361 mg/min D. 0.52 mg/min

Question 2

Sixteen trials of a reaction gave a mean rate of 9.6 μmol/min with a standard deviation of 0.80 μmol/min.

What is the standard error of the mean for these sixteen trials?

A. 0.050 μmol/min B. 0.20 μmol/min C. 0.80 μmol/min D. 3.2 μmol/min

Question 3

Mean rate of an enzyme from a salt-marsh plant at three salt concentrations, five trials each, same temperature and pH. Error bars represent ±2SE. Gridlines every 1 μmol/min.

The graph shows the mean rate of an enzyme from a salt-marsh plant at three salt concentrations, five trials each, with temperature and pH the same in every tube.

Which statement do the error bars support?

A. 20 g/L differs from 0 g/L, but 0 and 10 g/L are not shown to differ B. The 10 g/L rate is shown to be lower than the 0 g/L rate C. The 0 g/L rate and the 10 g/L rate are shown to be the same D. Every reading at 20 g/L lay below every reading at 0 g/L

Question 4

Six tubes of catalase at 20 °C break hydrogen peroxide down, and a student collects the oxygen each tube produces. The six tubes produce oxygen at 2.4, 3.1, 2.2, 2.8, 2.5 and 2.6 mL/min.

What mean rate should be reported for 20 °C?

A. 0.9 mL/min B. 2.60 mL/min C. 3.12 mL/min D. 15.6 mL/min

Question 5

A gardener grows two trays of six radish seedlings in two soils. In tray 1 the seedlings have a mean height of 48.0 mm with a standard deviation of 2.1 mm. In tray 2 they have a mean height of 52.0 mm with a standard deviation of 8.4 mm.

Which statement do these values support?

A. The larger standard deviation shows that tray 2's seedlings grew better B. Tray 1's six heights all lay within 2.1 mm of one another C. The wide spread of tray 2's heights makes its mean wrong D. Tray 2's seedling heights vary more from one seedling to the next

Question 6

Five tubes of catalase at 25 °C produced oxygen at a mean rate of 4.20 mL/min. The standard deviation of the five readings is 0.50 mL/min and the standard error of the mean is 0.224 mL/min.

Which value says how far this sample mean is likely to sit from the true mean?

A. 4.20 mL/min, the mean itself B. 0.50 mL/min, the standard deviation C. 1.00 mL/min, twice the standard deviation D. 0.224 mL/min, the standard error

Question 7

Mean rate of fatty-acid release by a lipase at 15 °C and 25 °C, five tubes each, same pH and fat concentration. Error bars represent ±2SE. Gridlines every 0.1 μmol/min.

The graph shows the mean rate of fatty-acid release by a lipase at 15 °C and at 25 °C, five tubes each, with error bars.

What are the two ends of the 25 °C error bar?

A. 1.20 to 1.60 μmol/min B. 2.00 to 2.30 μmol/min C. 2.00 to 2.60 μmol/min D. 2.15 to 2.45 μmol/min

Question 8

A student measured the mass of six eggs from one hen: 58, 61, 55, 60, 57 and 59 g. The AP formula sheet writes the mean as $𝑥̄ = \frac{\sum x_{i}}{n}$ .

Which symbol stands for the mean mass of the six eggs?

𝑥̄

\sum

x_{i}

D. n

Question 9

A student plots the mean height of eight maize seedlings after ten days as a bar with an error bar through its top. The legend reads "error bars represent ±2SE".

Which of the following does the error bar show?

A. How far a single seedling's height typically sits from the mean B. The range the true mean is likely to lie in C. The range from the shortest seedling to the tallest D. The size of the mistake made in measuring each seedling

Question 10

A student plots the mean rate at which a protease from pineapple breaks protein down at pH 5, five tubes, as a bar with an error bar through its top. The error bar runs from 3.2 to 4.8 mg/min. The legend reads "error bars represent one standard deviation".

Which of the following does this error bar show?

A. The range the true mean is likely to lie in B. Two standard errors of the mean either side of the mean C. How widely the five readings spread either side of their mean D. The range from the slowest of the five tubes to the fastest

How to tackle the free-response questions. Read the verb first: describe asks what you see or know; explain asks why or how, so name the mechanism; predict asks what will happen and why; justify asks for the evidence that supports a claim. Each point is earned by one idea, stated in a sentence that names the thing and the mechanism. Extra words earn nothing; a wrong extra can lose the point. If there is a figure or table, use what it shows. When you finish, check the box for each point your answer earns and compare your sentences with the full-credit answer.

Free response 1 · Analyze Data · 5 points

Students measured the rate at which a protease purified from a fungus digests a milk protein, at three temperatures with five trials at each. The pH and the protein concentration were the same in every trial. Their bar chart, with error bars representing ±2SE, is shown for 20 °C and 60 °C. The five readings at 40 °C were 6.2, 6.8, 5.9, 6.5 and 6.1 mg of protein digested per minute; their standard deviation is s = 0.354 mg/min. The students have still to add the 40 °C bar to the chart. Five later trials at 70 °C gave a mean rate of 1.2 mg/min; those trials are not on the chart.

Mean rate of protein digestion by the fungal protease at 20 °C and 60 °C, five trials each; the 40 °C bar has not yet been drawn. Error bars represent ±2SE. Gridlines every 0.5 mg/min.

(a) Calculate the mean rate at 40 °C. (1 point)

Hint: Which formula on the AP sheet gives the mean, and how many readings does the sum have to be divided by?

A full-credit answer: The mean rate at 40 °C is 6.30 mg/min.

x_{i} = 6.2, 6.8, 5.9, 6.5, 6.1 mg/min

n = 5

𝑥̄ = \frac{\sum x_{i}}{n}

𝑥̄ = \frac{\sum x_{i}}{n}

\sum x_{i} = 6.2 + 6.8 + 5.9 + 6.5 + 6.1 = 31.5 mg/min

𝑥̄ = \frac{31.5}{5}

𝑥̄ = 6.30 mg/min

Check the box for each point your answer earns

$\bar{x} = 6.30$ mg/min.

Accept 6.3 mg/min. Do not award the point for 31.5 (the sum) or for 7.88 (the sum divided by 4).

Common slip: Dividing the sum by 4 instead of 5. The mean divides by n, the number of readings; it is the standard deviation that divides by n − 1.

(b) Calculate the standard error of the mean for 40 °C. (1 point)

Hint: The AP sheet's standard error of the mean uses s and n. Which value in the question is s, and what is n?

A full-credit answer: The standard error of the mean for 40 °C is 0.158 mg/min: the standard deviation, 0.354 mg/min, divided by the square root of the five trials.

s = 0.354 mg/min

n = 5

S E_{𝑥̄} = \frac{s}{\sqrt{n}}

S E_{𝑥̄} = \frac{s}{\sqrt{n}}

S E_{𝑥̄} = \frac{0.354}{\sqrt{5}}

S E_{𝑥̄} = \frac{0.354}{2.236}

S E_{𝑥̄} = 0.158 mg/min

Check the box for each point your answer earns

SE = 0.354 divided by the square root of 5 (2.24) = 0.158 mg/min (accept 0.158 to 0.159).

Do not award the point for 0.354 divided by 5 = 0.0708 (dividing by n instead of the square root of n) or for 0.354 (the standard deviation given as the standard error).

Common slip: Dividing by n, 5, instead of by $\sqrt{n}$ . The equation divides the standard deviation by the square root of the number of readings.

(c) Calculate the two ends of the ±2SE error bar for the 40 °C mean. (1 point)

Hint: The caption says what each bar represents. How far above and below the mean does a bar of that kind reach?

A full-credit answer: The 40 °C error bar runs from 5.98 to 6.62 mg/min: two standard errors of the mean, 0.316 mg/min, below and above the mean of 6.30.

𝑥̄ = 6.30 mg/min

SE = 0.158 mg/min

bar ends = 𝑥̄ \pm 2 S E

bar ends = 𝑥̄ \pm 2 S E

2 S E = 2 \times 0.158 = 0.316 mg/min

lower end = 6.30 - 0.316 = 5.98 mg/min

upper end = 6.30 + 0.316 = 6.62 mg/min

Check the box for each point your answer earns

both ends: 2SE = 0.316 mg/min, so the bar runs from 6.30 − 0.316 = 5.98 to 6.30 + 0.316 = 6.62 mg/min (accept each end within ±0.01).

Accept the range written as 6.30 ± 0.32 mg/min. Do not award the point for a bar of ±1SE (6.14 to 6.46) or for a bar built from the standard deviation.

Common slip: Adding and subtracting one standard error of the mean instead of two. The bar the students drew is ±2SE, the range the true mean is likely to lie in.

(d) One student claims that the protease works faster at 60 °C than at 40 °C. Evaluate the claim using the two error bars, and determine whether the null hypothesis of no difference between 40 and 60 °C is rejected. (1 point)

Hint: Compare the top of the lower bar with the bottom of the higher bar. What does the caption say the bars represent, and what does the overlap rule say for bars of that kind?

A full-credit answer: The 40 °C bar runs from 5.98 to 6.62 mg/min and the 60 °C bar from 6.00 to 7.00 mg/min.
The two bars overlap.
So the gap between the two mean rates, 6.50 against 6.30 mg/min, could be chance.
Therefore the claim is not supported by these data.
Because the bars overlap, the null hypothesis of no difference is not rejected.
That is not the same as showing the two rates are equal.

Check the box for each point your answer earns

the judgement (the claim is not supported) with its ground (the two ±2SE bars overlap: 5.98 to 6.62 shares its range with 6.00 to 7.00), so the difference between the means could be chance, AND the decision that the null hypothesis of no difference between 40 and 60 °C is not rejected.

Evaluate needs the judgement and its ground; Determine needs the decision and what it rests on. Do not award the point for 'the rates are the same' (overlap shows no difference, not equality), or for a decision made from the two means alone.

Common slip: Reading overlapping bars as 'the rates are the same', or giving the judgement without the bars. Overlap means the data have not shown a difference; the true means may still differ.

(e) The students propose repeating the experiment at 80 °C. Predict how the mean rate at 80 °C will compare with the mean at 60 °C, and justify your prediction in terms of the enzyme's structure. (1 point)

Hint: The 70 °C trials are the place to look. What must be happening to the enzyme molecules themselves when the rate falls even though the temperature has risen?

A full-credit answer: At 80 °C the mean rate will be far lower than at 60 °C, close to zero.
The 70 °C trials already show the rate collapsing, from 6.50 to 1.2 mg/min, so the protease is past its optimum.
Heat disrupts the hydrogen bonds and other weak interactions that hold its fold.
So the active site loses its shape and the milk protein no longer fits.
At 80 °C the loss is more complete still.

Check the box for each point your answer earns

the mean rate at 80 °C will be much lower than at 60 °C (near zero), because the fall from 6.50 mg/min at 60 °C to 1.2 mg/min at 70 °C shows that the enzyme is already past its optimum: heat disrupts the hydrogen bonds and other weak interactions that hold the protein's fold, so the active site loses its shape and the substrate no longer fits (denaturation), and at 80 °C this loss is more complete.

Accept with or without the point that this loss outweighs the extra collisions warming brings. Accept 'denatured' only with what it does to the active site or to substrate binding. Do not award the point for 'faster, because molecules move faster', for 'lower' with no structural reason, or for a prediction that ignores the 70 °C trials.

Common slip: Predicting a higher rate because hotter molecules collide more often. That holds only below the optimum. The fall from 6.50 mg/min at 60 °C to 1.2 mg/min at 70 °C shows that 70 °C is already past the optimum, and more heat makes the loss worse.

Free-response score: 0 of 5

Free response 2 · Analyze Data · 4 points

A protease from a soil bacterium is breaking protein down, and the amino acids released are measured each minute. Students set up five tubes at 0% salt and five at 2% salt, with temperature, pH and protein concentration the same in every tube. At 0% salt the mean rate was 5.60 μmol/min with a standard error of 0.150 μmol/min. At 2% salt the mean rate was 5.10 μmol/min with a standard error of 0.180 μmol/min. The students will report the two means with error bars representing ±2SE, and they plan five more tubes at 8% salt. A single tube at 6% salt, set up the same way, released amino acids at 2.10 μmol/min.

(a) Calculate the two ends of the ±2SE error bar for each mean. (1 point)

A full-credit answer: The 0% bar runs from 5.30 to 5.90 μmol/min, and the 2% bar from 4.74 to 5.46 μmol/min.

0% salt: 𝑥̄ = 5.60 μmol/min

0% salt: SE = 0.150 μmol/min

2% salt: 𝑥̄ = 5.10 μmol/min

2% salt: SE = 0.180 μmol/min

bar ends = 𝑥̄ \pm 2 S E

bar ends = 𝑥̄ \pm 2 S E

0%: 2 S E = 2 \times 0.150 = 0.300 μmol/min

0%: lower end = 5.60 - 0.300 = 5.30 μmol/min

0%: upper end = 5.60 + 0.300 = 5.90 μmol/min

2%: 2 S E = 2 \times 0.180 = 0.360 μmol/min

2%: lower end = 5.10 - 0.360 = 4.74 μmol/min

2%: upper end = 5.10 + 0.360 = 5.46 μmol/min

Check the box for each point your answer earns

both bars: 0% salt, 2SE = 0.300, so the bar runs from 5.30 to 5.90 μmol/min; 2% salt, 2SE = 0.360, so the bar runs from 4.74 to 5.46 μmol/min (accept each end within ±0.01).

Accept the bars written as 5.60 ± 0.30 and 5.10 ± 0.36 μmol/min. Do not award the point for bars of ±1SE.

Common slip: Adding and subtracting one standard error of the mean instead of two. The bar the students will report is ±2SE, the range the true mean is likely to lie in.

(b) Evaluate the claim that the protease is slower at 2% salt than at 0% salt, using the two error bars, and determine whether the null hypothesis of no difference between 0% and 2% salt is rejected. (1 point)

A full-credit answer: The 0% bar runs from 5.30 to 5.90 μmol/min and the 2% bar from 4.74 to 5.46 μmol/min.
The two bars overlap between 5.30 and 5.46.
So the gap between the two mean rates, 5.60 against 5.10 μmol/min, could be chance.
Therefore the claim is not supported by these data.
Because the bars overlap, the null hypothesis of no difference is not rejected.
That is not the same as showing the two rates are equal.

Check the box for each point your answer earns

the judgement (the claim is not supported) with its ground (the two ±2SE bars overlap: 5.30 to 5.90 shares the range 5.30 to 5.46 with 4.74 to 5.46), so the difference between the means could be chance, AND the decision that the null hypothesis of no difference between 0% and 2% salt is not rejected.

Common slip: Reading overlapping bars as 'the rates are the same', or giving the judgement without the bars. Overlap means the data have not shown a difference; the true means may still differ.

(c) Suppose twenty tubes were set up at each salt concentration instead of five. Explain how that would change the standard error of the mean for each salt concentration and the length of each error bar, if the spread of the readings stayed the same. (1 point)

A full-credit answer: The standard error of the mean is the standard deviation divided by the square root of n.
The spread of the readings stays the same, so s stays the same.
Going from five tubes to twenty makes the square root of n twice as large, from $\sqrt{5} = 2.24$ to $\sqrt{20} = 4.47$ .
So each standard error halves, and each ±2SE error bar is half as long.
Each mean is then a surer estimate of the true mean.

S E_{𝑥̄} = \frac{s}{\sqrt{n}}

\sqrt{5} = 2.236

\sqrt{20} = 4.472

\frac{\sqrt{20}}{\sqrt{5}} = 2

The divisor doubles, so each standard error halves and each ±2SE bar is half as long.

Check the box for each point your answer earns

the standard error is the standard deviation divided by the square root of n, and the square root of 20 (4.47) is twice the square root of 5 (2.24), so each standard error would halve and each ±2SE bar would be half as long; the means would be surer estimates of the true means (and the shorter bars might then separate).

Accept "SE halves, bars half as long" with the square root of n as the reason. Do not award the point for "the standard error falls to a quarter" (that divides by n) or for "the standard deviation falls" (the spread of readings is unchanged).

Common slip: Saying the standard error of the mean falls to a quarter because there are four times as many tubes. The equation divides by the square root of n. The square root of 20 is twice the square root of 5, not four times.

(d) Predict how the mean rate at 8% salt will compare with the mean rate at 2% salt, and justify your prediction in terms of the enzyme's structure. (1 point)

A full-credit answer: The mean rate at 8% salt will be much lower than at 2% salt, and may be close to zero.
Temperature and pH are unchanged, so any fall comes from the salt itself.
A high salt concentration can disrupt the interactions that hold the protease's fold.
So the active site loses its shape, the protein no longer fits, and the enzyme can no longer catalyze the reaction.

Check the box for each point your answer earns

the mean rate at 8% salt will be much lower than at 2%, because a high salt concentration can disrupt the interactions that hold the protease's fold, so the active site loses its shape and the protein substrate binds less well or no longer fits, and the rate falls even with temperature and pH unchanged.

Accept "the fold is disrupted, so the active site is lost" for the mechanism; accept a note that a test of how many molecules are still folded would confirm the change in shape. Do not award the point for "about the same as at 2%" (the 6% trial already shows the fall), for "the salt uses up the substrate", for "the salt cools the tubes", or for "lower" with no structural reason.

Common slip: Predicting a faster rate because salt adds ions that collide with the enzyme, or saying the salt uses up the substrate. Dissolved salt acts on the enzyme's fold. It is the shape of the active site that changes.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

Take the end-of-topic test