Practice questions · Topic 3.2b

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

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 tells you what a wrong choice assumed. For the free-response questions, write your answer in full sentences and show any calculation. 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 this mean?

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. The 20 g/L rate differs from the 0 g/L rate; 0 and 10 g/L are not shown to differ B. The 10 g/L rate is lower than the 0 g/L rate, because its mean is lower C. The 0 and 10 g/L rates are the same, because their error bars overlap D. Every reading at 20 g/L lay below every reading at 0 g/L, since the bars are apart

Question 4

Catalase is breaking hydrogen peroxide down in six tubes at 20 °C, and the oxygen produced is collected. The six tubes produced 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

Two trays of six radish seedlings were grown 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. Tray 2's seedlings grew better, because the standard deviation of their heights is larger B. Tray 1's six heights all lay within 2.1 mm of one another C. Tray 2's heights should be dropped, because so wide a spread makes their mean wrong D. Tray 2's seedling heights vary more from one seedling to the next

Question 6

Five tubes of an enzyme at 30 °C released product at 7.3, 8.1, 6.9, 7.5 and 7.2 μmol/min. Their mean is 7.40 μmol/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. 0.20 μmol/min B. 0.400 μmol/min C. 0.447 μmol/min D. 0.80 μmol/min

Question 7

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 8

Five tubes of an amylase at pH 7 released maltose at a mean rate of 6.10 mg/min. The standard deviation of the readings is 0.313 mg/min and the standard error of the mean is 0.140 mg/min.

Between which values does the ±2SE error bar on this mean run?

A. 5.96 to 6.24 mg/min B. 5.82 to 6.38 mg/min C. 6.10 to 6.38 mg/min D. 5.47 to 6.73 mg/min

Question 9

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.

Between which values does the 25 °C error bar run?

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

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 papain, an enzyme from papaya fruit, 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. In a later run of five trials at 70 °C the mean rate was 1.2 mg/min; that run is not on the chart.

Mean rate of protein digestion by papain 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 40 °C mean. (1 point)

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

A full-credit answer: The standard error of the 40 °C mean 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, 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 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 papain works faster at 60 °C than at 40 °C. Use the two error bars to evaluate the claim, and state 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 (5.98 to 6.62 mg/min) and the 60 °C bar (6.00 to 7.00 mg/min) overlap, so the gap between the means, 6.50 against 6.30, could be chance. These data do not show a difference between the two temperatures, so 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 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; the data do not show a difference, and the null hypothesis is not rejected.

Accept 'the claim is not supported by these data'. 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'. 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 run is 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 run already shows the rate collapsing, from 6.50 mg/min to 1.2 mg/min, even though warmer molecules collide more often, so papain is past its optimum: heat disrupts the hydrogen bonds and other weak interactions that hold its fold, the active site loses its shape, the milk protein no longer fits, and the denatured enzyme can no longer catalyze the reaction. At 80 °C that loss is more complete still, and it far outweighs the extra collisions.

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 and outweighs the extra collisions that 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 run.

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 it, and more heat makes the loss worse.

Free-response score: 0 of 5

Free response 2 · Analyze Data · 4 points

A protease from a bacterium that lives in a salt marsh is breaking protein down, and the amino acids released are measured each minute. Researchers ran 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 researchers plan to draw the two means with error bars representing ±2SE, and they propose a further run at 8% salt. A single trial at 6% salt, run the same way, gave 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 instead of two. The bar the researchers plan is ±2SE, the range the true mean is likely to lie in.

(b) One researcher claims that the protease is slower at 2% salt than at 0% salt. Use the two error bars to evaluate the claim, and state whether the null hypothesis of no difference between 0% and 2% salt is rejected. (1 point)

A full-credit answer: The 0% bar (5.30 to 5.90 μmol/min) and the 2% bar (4.74 to 5.46 μmol/min) overlap between 5.30 and 5.46, so the gap between the means, 5.60 against 5.10, could be chance. These data do not show a difference between 0% and 2% salt, so 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 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; the data do not show a difference, and the null hypothesis is not rejected.

Accept "the claim is not supported by these data". 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". Overlap means the data have not shown a difference; the true means may still differ.

(c) The researchers consider running twenty tubes at each salt concentration instead of five. Explain how that would change the standard error of each mean 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 is the standard deviation divided by the square root of n. With the same spread, going from five tubes to twenty makes the square root of n twice as large, from 2.24 to 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, and the shorter bars might no longer overlap.

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 falls to a quarter because there are four times as many tubes. The equation divides by the square root of n, and the square root of 20 is twice, not four times, the square root of 5.

(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%, 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. A test of how many molecules still hold their normal fold at 8% salt would show whether the shape itself had changed.

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 9 correct.

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