Practice questions · Topic 3.2

Unit 3 · Practice for the Topic 3.2 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. Formulas: mean x̄ = Σxᵢ/n; standard deviation s = √(Σ(xᵢ − x̄)²/(n − 1)); standard error SE = s/√n; an error bar of ±2SE covers the range the true mean is likely to lie in. Rounding: carry three significant figures through your working and round only the final value you report.

Question 1

Honeybees keep the center of their hive near 35 °C. A purified digestive enzyme from a bee was tested with the same substrate concentration at 15 °C and at 35 °C. It made 3 μmol of product per minute at 15 °C and 11 μmol/min at 35 °C, and a fold test showed its fold intact at both temperatures.

Why is the rate higher at 35 °C?

A. The warmth lowers the activation energy of the reaction B. At 35 °C there are more enzyme molecules in the tube C. Enzyme and substrate molecules move faster, so they collide more often and with more energy D. The substrate breaks down on its own at 35 °C, with no help from the enzyme

Question 2

Rate of the same kind of reaction catalyzed by an enzyme from an Antarctic icefish (solid) and by an enzyme from a desert lizard (dashed), at temperatures from 0 to 60 °C. pH and substrate concentration were the same in every tube. Gridlines every 2 μmol/min and every 10 °C.

The graph shows the rate of the same kind of reaction catalyzed by two enzymes: one from an Antarctic icefish and one from a desert lizard.

What is the optimal temperature of each enzyme, and what is happening to the icefish enzyme at 35 °C?

A. Icefish 15 °C, lizard 40 °C; at 35 °C the icefish enzyme is denatured B. Icefish 15 °C, lizard 40 °C; at 35 °C the icefish enzyme is slowed by too few collisions C. Icefish 35 °C, lizard 40 °C; at 35 °C the icefish enzyme is at its optimum D. Both 40 °C; at 35 °C the icefish enzyme is still warming toward its optimum

Question 3

Equal samples of a lipase from milk were each held at one temperature for 20 minutes: 2, 20, 37 or 65 °C. Each sample's rate (μmol of fatty acid released per minute) was measured at its holding temperature and then again after every sample had been brought to 37 °C. Sample held at 2 °C: 3, then 20. Sample held at 20 °C: 12, then 20. Sample held at 37 °C: 20, then 20. Sample held at 65 °C: 1, then 1.

Which samples were only slowed, and what in the data shows it?

A. The 65 °C sample; its rate stayed low after the return to 37 °C B. The 2 °C sample only; it had the lowest rate while it was cold C. The 2, 20 and 65 °C samples; each had a lower rate than the 37 °C sample D. The 2 and 20 °C samples; their rates returned to 20 μmol/min at 37 °C

Question 4

A cola drink has a pH of about 3 and rainwater a pH of about 5.

How do their hydrogen ion (H⁺) concentrations compare?

A. The rainwater has ten times the H⁺ concentration of the cola B. The rainwater has a hundred times the H⁺ concentration of the cola C. The cola has ten times the H⁺ concentration of the rainwater D. The cola has a hundred times the H⁺ concentration of the rainwater

Question 5

An enzyme from inside a lysosome, where the pH is about 4.5, was purified and tested at 37 °C with the same substrate concentration at pH 4.5 and at pH 7.2, the pH of the cytosol. It made 30 μmol of product per minute at pH 4.5 and 2 μmol/min at pH 7.2. A fold test showed that the enzyme kept its overall fold at both pH values.

Why was the rate so much lower at pH 7.2?

A. Fewer H⁺ altered the charges on R groups at the active site, so the substrate bound less well B. The weak interactions holding the fold were disrupted, so the whole enzyme lost its shape C. Fewer H⁺ meant the enzyme and substrate collided less often each second D. At pH 7.2 the H⁺ broke the substrate down before it reached the enzyme

Question 6

A fixed amount of a purified enzyme was given its substrate at 1, 2, 4, 8 and 16 mmol/L, with temperature and pH the same in every tube. Its initial rates were 12, 22, 36, 44 and 45 μmol/min.

What rate is expected at 32 mmol/L of substrate, and why?

A. About 90 μmol/min, because doubling the substrate doubles the rate B. About 45 μmol/min, because the extra substrate has begun to denature the enzyme C. About 45 μmol/min, because nearly every active site is already occupied all the time D. About 22 μmol/min, because the extra substrate uses up the enzyme

Question 7

A weedkiller binds to a plant enzyme at a site away from the active site and changes the enzyme's shape. In leaf extracts with equal amounts of the enzyme, the weedkiller cuts the rate to about half at every substrate concentration tested, from 2 to 2,000 μM.

Why does raising the substrate concentration leave the weedkiller's effect unchanged?

A. The substrate and the weedkiller compete for the active site, and the weedkiller binds it more tightly B. The weedkiller and the substrate bind different sites, so more substrate cannot displace it C. More substrate denatures the enzyme, canceling any gain from the extra collisions D. The weedkiller has permanently denatured every enzyme molecule in the extract

Question 8

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 = √(Σ(xᵢ − 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 9

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 10

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

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: the five readings sum to 31.5 mg/min, and 31.5 divided by 5 is 6.30.

Write down the values in the question:

xᵢ = 6.2, 6.8, 5.9, 6.5, 6.1 mg/min
n = 5

Write down the equation:

      Σxᵢ
x̄ = ─────
       n

Substitute in the values, and calculate:

Σxᵢ = 6.2 + 6.8 + 5.9 + 6.5 + 6.1 = 31.5 mg/min
x̄ = 31.5 ÷ 5
x̄ = 6.30 mg/min

Check the box for each point your answer earns

x̄ = 31.5 ÷ 5 = 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.

Write down the values in the question:

s = 0.354 mg/min
n = 5

Write down the equation:

         s
SE = ─────
        √n

Substitute in the values, and calculate:

SE = 0.354 ÷ √5
SE = 0.354 ÷ 2.236
SE = 0.158 mg/min

Check the box for each point your answer earns

SE = 0.354 ÷ √5 = 0.354 ÷ 2.24 = 0.158 mg/min (accept 0.158 to 0.159).

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

Common slip: Dividing by n, 5, instead of by √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.

Write down the values in the question:

x̄ = 6.30 mg/min
SE = 0.158 mg/min

Write down the equation:

bar ends = x̄ ± 2SE

Substitute in the values, and calculate:

2SE = 2 × 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 · Conceptual Analysis · 4 points

An enzyme from a yogurt bacterium digests milk protein. Researchers measured its rate (mg of protein digested per minute) at 37 °C with the same protein concentration at five pH values: pH 3, 8 mg/min; pH 4, 30 mg/min; pH 5, 42 mg/min; pH 6, 20 mg/min; pH 7, 4 mg/min. A fold test showed the enzyme's overall fold intact at pH 5 and at pH 6. A sample was then moved to pH 8: its rate fell to 1 mg/min and the fold test found only 15% of its molecules folded normally. Returned to pH 5 for 30 minutes, the same sample ran at 38 mg/min and 92% of its molecules were folded normally.

(a) Identify the enzyme's optimal pH from the data, and describe how the rate changes on either side of it, using values. (1 point)

A full-credit answer: The optimal pH is 5, where the rate is greatest at 42 mg/min. On either side the rate falls: to 30 mg/min at pH 4 and 8 mg/min at pH 3, and to 20 mg/min at pH 6 and 4 mg/min at pH 7.

Check the box for each point your answer earns

the optimal pH is 5 (42 mg/min, the greatest rate), and the rate falls on both sides of it: to 30 and 8 mg/min at pH 4 and 3, and to 20 and 4 mg/min at pH 6 and 7.

Accept 'about pH 5' with at least two values quoted for the fall. Do not award the point for pH 7 as the optimum because it is neutral, or for an optimum with no description of the pattern.

Common slip: Assuming the optimum must be pH 7 because that is neutral. Different enzymes have different optimal pH values; this one peaks at pH 5, and at pH 7 it manages only 4 mg/min.

(b) Explain why the rate at pH 6 is lower than at pH 5, given that the fold test shows the enzyme's overall fold intact at both. (1 point)

A full-credit answer: At pH 6 the H⁺ concentration is a tenth of that at pH 5. The changed H⁺ concentration alters the charges on the R groups in and around the active site, so the milk protein's charges no longer match the pocket as well and it binds less well; the rate falls to 20 mg/min. The fold as a whole is intact, so the enzyme is slowed by mismatched charges, not denatured.

Check the box for each point your answer earns

at pH 6 there are fewer H⁺ than at pH 5, which alters the charges on the R groups in and around the active site, so the substrate's charges no longer match the pocket as well and it binds less well; the fold as a whole still holds, so this is not denaturation.

Accept 'the changed H⁺ concentration changes the charges in the active site so the protein binds less well'. Do not award the point for 'the enzyme is denatured at pH 6' (the fold test rules that out) or for 'fewer collisions'.

Common slip: Saying the enzyme is denatured at pH 6. The fold test shows the fold intact; near the optimum it is the charges in the active site that change, not the shape of the whole protein.

(c) Explain what the results at pH 8 and after the return to pH 5 show about this enzyme. (1 point)

A full-credit answer: At pH 8 the enzyme was denatured: only 15% of its molecules held their fold and the rate fell to 1 mg/min, because far from the optimum the hydrogen bonds and other weak interactions that hold the fold were disrupted and the active site lost its shape. Back at pH 5, 92% of the molecules were folded again and the rate returned to 38 mg/min, so the denaturation was reversible: the disrupting condition was removed before the unfolded chains tangled, and the same molecules refolded.

Check the box for each point your answer earns

at pH 8 the enzyme was denatured (only 15% folded, rate 1 mg/min), because far from the optimum the weak interactions holding the fold are disrupted and the active site loses its shape; the return of the fold (92%) and the rate (38 mg/min) at pH 5 shows that this denaturation was reversible: the disrupting condition was removed before the unfolded chains tangled, so the molecules refolded.

Accept 'denatured at pH 8, but reversibly'. Do not award the point for 'the enzyme was destroyed and replaced' (no cell is present to make new enzyme) or for 'the enzyme was only slowed at pH 8' (the fold test shows the fold lost).

Common slip: Stopping at 'the enzyme was denatured at pH 8'. The point needs the second half: the fold and the rate came back at pH 5, so this denaturation was reversible.

(d) At pH 5, the researchers double the amount of enzyme while keeping the same 20 mg of milk protein in the tube. Predict how the initial rate and the final amount of protein digested will compare with the original tube, and justify your prediction. (1 point)

A full-credit answer: The initial rate will be about twice as high, roughly 84 mg/min, because doubling the enzyme doubles the number of active sites the protein can meet. The final amount digested will be the same, 20 mg, because that is all the protein there is: more enzyme gets to the end sooner, not further.

Check the box for each point your answer earns

the initial rate roughly doubles (about 84 mg/min), because there are twice as many active sites, but the final amount of protein digested is the same, 20 mg, because it is set by the substrate available, so the richer tube finishes sooner.

Accept 'faster at first, same total in the end' with both reasons. Do not award the point for 'more protein is digested in the end' or for 'the rate is the same because the substrate is the same'.

Common slip: Predicting that more enzyme digests more protein in the end. The final amount is set by the substrate available; the enzyme changes only how quickly it is reached.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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