Practice questions · Topic 3.1

Unit 3 · Practice for the Topic 3.1 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. Energy values on the profiles are in kJ/mol.

Question 1

Sucrase, an enzyme in the yeast cell, speeds up the reaction sucrose + water → glucose + fructose. In a flask of sucrose solution with sucrase added, the amount of sucrose falls over an hour while glucose and fructose build up.

Which substances are the reactants in this reaction, and how can you tell?

A. Sucrose and water; they are the starting substances that are changed B. Glucose and fructose; they are the substances that build up in the flask C. Sucrase; it is the substance added to start the reaction D. Sucrose and glucose; both are present in the flask during the hour

Question 2

Pectinase is an enzyme that breaks down pectin, the substance that holds the juice inside crushed fruit. A student adds pectinase to 50 g of apple pulp and collects 45 mL of juice in 15 minutes.

What is the rate of juice release?

A. 0.33 mL/min B. 3.0 mL/min C. 45 mL/min D. 675 mL/min

Question 3

Volume of juice collected from 50 g of apple pulp over 10 minutes: tube 1 (solid, four drops of pectinase) and tube 2 (dashed, one drop). Gridlines every 4 mL and every 2 minutes.

Two tubes each hold 50 g of the same apple pulp. Tube 1 receives four drops of pectinase and tube 2 receives one drop. The graph shows the volume of juice collected from each tube over 10 minutes.

What is the rate of juice release from tube 1 over the first 6 minutes, and how will the final volumes of juice from the two tubes compare once both have finished?

A. 4.0 mL/min; tube 1 ends with more juice than tube 2 B. 2.0 mL/min; tube 1 ends with more juice than tube 2 C. 2.0 mL/min; both tubes end with 24 mL, tube 2 later D. 4.0 mL/min; both tubes end with 24 mL, tube 2 later

Question 4

Energy profile for the breakdown of a plant pigment with no enzyme present. Energy in kJ/mol, gridlines every 10 kJ/mol.

The figure shows the energy profile for the breakdown of a plant pigment with no enzyme present, with energy in kJ/mol.

How much energy does the reaction release overall?

A. 25 kJ/mol B. 35 kJ/mol C. 50 kJ/mol D. 85 kJ/mol

Question 5

Two reactions run in separate flasks at the same room temperature, with no enzyme present. Reaction 1 has an activation energy of 30 kJ/mol; reaction 2 has an activation energy of 80 kJ/mol. Both release about the same energy overall.

Which reaction runs faster, and why?

A. Reaction 2; its molecules take in more energy at the start, so they move faster B. Reaction 2; a higher hump means the products sit lower, so the reaction is pulled forward C. Reaction 1; more collisions carry at least 30 kJ/mol than carry 80 kJ/mol D. Both run at the same rate, because they release the same energy overall

Question 6

A juice factory passes cloudy apple juice through a column packed with beads coated in pectinase. Over six weeks the column clears 40,000 liters of juice. At the end of the six weeks the beads carry the same mass of pectinase as at the start, and the juice leaving the column contains the same breakdown products as in week one.

What do these observations show about the pectinase?

A. The pectinase is used up in each reaction, and the beads are recoated as fast as it goes B. The pectinase becomes part of the breakdown products that leave in the juice C. The pectinase is a reactant that the juice's sugars remake after each reaction D. The pectinase is unchanged by each reaction, so each molecule acts again and again

Question 7

A fruit-eating bird's gut cells make an enzyme that splits a sugar found in ripe berries. In autumn, when the berries are ripe, the cells make large amounts of the enzyme and the sugar is digested. In spring the same cells make hardly any of it, and the same sugar passes through the gut unchanged.

What do these observations show about how the bird's cells control the digestion of this sugar?

A. The cells control the reaction by controlling how much of its enzyme they make B. The sugar reacts on its own in autumn because ripe berries hold more of it C. The enzyme made in autumn is used up by spring, so digestion stops D. In spring the cells make the enzyme, but the sugar in the berries changes shape

Question 8

An enzyme's active site is a pocket whose shape fits its substrate. The substrate carries a negative charge, and one R group lining the pocket carries a positive charge. Researchers make a version of the enzyme in which that R group is replaced by one carrying a negative charge. The pocket keeps its shape, yet the substrate stays unbound.

What stops the substrate binding to the changed enzyme?

A. The changed pocket is now too small for the substrate to enter B. The substrate's own shape changed when the enzyme was altered C. The negative charges on the R group and on the substrate push apart D. The substrate now binds elsewhere on the enzyme's surface instead

Question 9

Lactase splits lactose, the sugar in milk, into two smaller sugars; its active site fits lactose. A student adds lactase to a solution of sucrose, table sugar, whose molecule has a different shape from lactose, and leaves the tube at 37 °C for an hour.

What happens to the sucrose?

A. The sucrose is split slowly, because it fits the active site less well than lactose B. The sucrose is unchanged, because it is never held in lactase's active site C. The sucrose is split, because lactase lowers the activation energy of any sugar's reaction D. The sucrose is split once the lactase has finished with any lactose present

Question 10

A student's hypothesis is that blending apple pulp, which breaks open more cells, gives a faster release of juice than cutting the pulp into cubes, when the same amount of pectinase is added to each.

Which statement is the null hypothesis for this experiment?

A. Blended pulp releases juice faster than pulp cut into cubes B. Pulp cut into cubes releases juice faster than blended pulp C. Adding pectinase makes no difference to how fast juice is released D. How finely the pulp is broken up makes no difference to how fast juice is released

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 · Conceptual Analysis · 5 points

Actinidin is an enzyme in kiwi fruit. It speeds up the reaction gelatin + water → smaller protein pieces (gelatin is a protein). A dessert made by stirring fresh kiwi juice into warm gelatin never sets: within an hour the gelatin has been broken into pieces too small to form a gel. The same gelatin with no kiwi juice sets firmly and is still firm a week later. The figure shows the energy profile of the reaction with and without actinidin, with energy in kJ/mol.

Energy profile for gelatin + water → smaller protein pieces, without actinidin (solid) and with actinidin (dashed). Energy in kJ/mol, gridlines every 10 kJ/mol.

(a) Identify actinidin's substrate, and identify the part of the enzyme where that substrate binds. (1 point)

Hint: Which molecule in the dessert does actinidin act on, and what is the name of the place on an enzyme where its substrate sits?

A full-credit answer: Actinidin's substrate is gelatin, the protein it acts on, and it binds at the enzyme's active site, the pocket on the enzyme's surface whose shape and charges fit the gelatin chain.

Check the box for each point your answer earns

gelatin (the protein) is the substrate, and it binds at the active site.

Accept 'the protein in the dessert' for the substrate. Do not award the point for water as the substrate, or for 'the enzyme's surface' with no active site named.

Common slip: Naming water as the substrate. Water is a reactant, but the substrate is the molecule the enzyme holds in its active site, and that is the gelatin.

(b) Calculate the activation energy of the reaction on its own and with actinidin. (1 point)

Hint: Between which two levels on the profile does the activation energy run? Read both peaks off the gridlines.

A full-credit answer: Without actinidin the activation energy is 50 kJ/mol, and with actinidin it is 20 kJ/mol: the enzyme lowers the barrier by 30 kJ/mol.

Write down the values from the figure:

reactants = 70 kJ/mol
peak without actinidin = 120 kJ/mol
peak with actinidin = 90 kJ/mol

Write down the equation:

activation energy = peak − reactants' level

Substitute in the values, and calculate:

without actinidin: 120 − 70 = 50 kJ/mol
with actinidin: 90 − 70 = 20 kJ/mol

Check the box for each point your answer earns

both values: 120 − 70 = 50 kJ/mol without actinidin and 90 − 70 = 20 kJ/mol with actinidin.

Accept values within ±2 kJ/mol read from the figure. Do not award the point for the peak heights (120 and 90 kJ/mol) given as the activation energies.

Common slip: Reading the height of each peak above zero, 120 and 90 kJ/mol. The activation energy is the climb from the reactants' level, 70 kJ/mol, up to the peak.

(c) Describe what actinidin leaves unchanged on the profile, and calculate the energy the reaction releases overall. (1 point)

Hint: Compare the two curves from left to right: which parts of the profile does actinidin move, and which does it leave where they were? Between which two levels is the energy released overall measured?

A full-credit answer: Actinidin leaves the start and the end of the profile where they were: the reactants stay at 70 kJ/mol and the products at 45 kJ/mol. The energy released overall is therefore 25 kJ/mol with or without the enzyme; only the hump moves.

Write down the values from the figure:

reactants = 70 kJ/mol (both curves)
products = 45 kJ/mol (both curves)

Write down the equation:

energy released overall = reactants' level − products' level

Substitute in the values, and calculate:

energy released overall = 70 − 45 = 25 kJ/mol
the same with and without actinidin

Check the box for each point your answer earns

the reactants' level (70 kJ/mol) and the products' level (45 kJ/mol) are the same on both curves, so the energy released overall, 70 − 45 = 25 kJ/mol, is unchanged by the enzyme.

Accept 'the start and end of the curve stay where they were' with the 25 kJ/mol calculated. Do not award the point for an answer that has actinidin lowering the products or changing the energy released.

Common slip: Saying the enzyme makes the reaction release more energy because the curve is lower. Only the peak is lower; the products sit at 45 kJ/mol either way, so the energy released is the same.

(d) Explain how actinidin makes the gelatin break down within an hour when the gelatin with no kiwi juice is still firm a week later. (1 point)

Hint: The molecules in both desserts are colliding all the time. What decides whether one of those collisions ends in a reaction, and what did you find in (b) that changes it?

A full-credit answer: Reacting molecules must collide with at least the activation energy before their bonds can rearrange, and at room temperature only a tiny fraction of collisions carry 50 kJ/mol, so the plain gelatin stays firm for a week. When a stretch of gelatin binds in actinidin's active site, forming an enzyme–substrate complex, the barrier is only 20 kJ/mol, so a far larger fraction of the collisions already happening succeed, the pieces leave, and the unchanged actinidin binds the next stretch. The gelatin is broken down within the hour.

Check the box for each point your answer earns

reacting molecules must collide with at least the activation energy; with the barrier lowered from 50 to 20 kJ/mol, a far larger fraction of the collisions already happening carry enough energy, so many more succeed each second and the gelatin is broken down faster. Accept the fuller chain: gelatin binds at the active site, the enzyme–substrate complex lowers the barrier, the pieces leave and the unchanged actinidin binds the next stretch.

Do not award the point for 'actinidin lowers the activation energy' alone with no link to collisions succeeding, or for 'actinidin heats the dessert' or 'actinidin supplies energy'.

Common slip: Stopping at 'the enzyme lowers the activation energy'. The point needs the next step: with a lower barrier, more of the collisions carry enough energy, so more of them succeed each second.

(e) A cook stirs the same kiwi juice into a starch pudding (starch and water, no protein). Predict what happens to the starch over the next hour, and justify your prediction using the active site. (1 point)

Hint: What two things about a molecule decide whether it is held in an active site, and how does starch compare with gelatin on each?

A full-credit answer: The starch will be unchanged after an hour. A molecule is held in the active site only if its shape fits the pocket and its charges match the R groups lining it; a starch chain has a different shape from a protein chain, so it is never bound, no enzyme–substrate complex forms, and the activation energy of its reaction is not lowered. Actinidin does nothing to the pudding.

Check the box for each point your answer earns

the starch is unchanged (no reaction is sped up) because a molecule is held in the active site only if its shape and charges fit, and a starch chain has a different shape from a protein chain, so it is never bound and its activation energy is not lowered.

Accept 'nothing happens to the starch because it does not fit actinidin's active site'. Do not award the point for 'the starch breaks down more slowly' or for 'actinidin only works on protein' with no reference to fit at the active site.

Common slip: Predicting a slow breakdown instead of none. An enzyme that does not fit a molecule does not act on it slowly; it does not act on it at all.

Free-response score: 0 of 5

Free response 2 · Scientific Investigation · 4 points

Urease, an enzyme made by many soil bacteria, speeds up the reaction urea + water → ammonia + carbon dioxide. A student wants to know whether the bacteria in compost release ammonia from urea faster than the bacteria in garden soil. Three flasks each receive 50 mL of the same urea solution at 25 °C. Flask C gets 1.0 g of compost, flask G gets 1.0 g of garden soil, and flask S gets 1.0 g of clean sand, which contains no living things. She measures the mass of ammonia formed in each flask in 10.0 minutes: flask C, 8.0 mg; flask G, 3.0 mg; flask S, 0.2 mg. In a separate check, 1.0 g of the same compost added to 50 mL of glucose solution gives off no ammonia and leaves the glucose unchanged.

(a) Identify the independent variable, the dependent variable, and the flask that is the control. (1 point)

A full-credit answer: Independent variable: the material added to the urea solution, compost, garden soil or sand. Dependent variable: the mass of ammonia formed in 10.0 minutes. Control: flask S, which was treated the same way as C and G but received sand with no living things in it.

Check the box for each point your answer earns

all three: independent variable, the source of the material added (compost, garden soil or sand); dependent variable, the mass of ammonia formed in 10.0 minutes (or the rate of ammonia formation); control, flask S (urea solution with sand, no living things).

Accept 'what is added to the urea' for the independent variable. Do not award the point if the independent and dependent variables are reversed, or if a controlled condition (the 25 °C temperature, the 50 mL of urea solution, the 10.0 minutes) is named as the control.

Common slip: Naming a controlled variable, such as the 25 °C temperature or the 50 mL of urea solution, as 'the control'. Controlled variables are kept the same in every flask; the control is the flask that lacks the factor under test.

(b) State the null hypothesis for the comparison between compost and garden soil. (1 point)

A full-credit answer: There is no difference in the mass of ammonia formed in 10.0 minutes between urea solution given compost and urea solution given garden soil.

Check the box for each point your answer earns

there is no difference in the mass of ammonia formed in 10.0 minutes (or in the rate of ammonia formation) between urea solution given compost and urea solution given garden soil. The statement must name the tested factor and the measured result.

Accept 'the source of the soil bacteria has no effect on the rate of ammonia formation'. Do not award the point for a prediction of a difference in either direction, or for 'urease has no effect on urea'.

Common slip: Writing the student's own prediction ('compost is faster') or its opposite as the null. The null hypothesis predicts no difference, and it names both the factor changed and the quantity measured.

(c) Calculate the rate of ammonia formation in flask C and in flask S, and use them to justify the claim that most of the ammonia in flask C came from urease in the compost. (1 point)

A full-credit answer: Flask C formed ammonia at 0.80 mg/min and flask S at 0.02 mg/min. Flask S is the control: it shows that the urea solution on its own gives off only 0.02 mg/min at 25 °C. The flasks were treated the same way except for the compost, so the extra 0.78 mg/min in flask C is credited to the urease in the compost.

Write down the values in the question:

flask C: 8.0 mg of ammonia in 10.0 min
flask S: 0.2 mg of ammonia in 10.0 min

Write down the equation:

        mass of ammonia formed
rate = ────────────────────────
              time taken

Substitute in the values, and calculate:

flask C: rate = 8.0 ÷ 10.0 = 0.80 mg/min
flask S: rate = 0.2 ÷ 10.0 = 0.02 mg/min
difference: 0.80 − 0.02 = 0.78 mg/min

Check the box for each point your answer earns

rates of 0.80 mg/min for flask C (8.0 ÷ 10.0) and 0.02 mg/min for flask S (0.2 ÷ 10.0), AND the justification that flask S shows how much ammonia the urea solution gives off on its own, so, because the flasks were treated the same way except for the compost, the extra 0.78 mg/min (7.8 mg) is credited to the compost's urease.

Accept a comparison of the masses (8.0 mg against 0.2 mg) given alongside the rates. Do not award the point for the two rates with no comparison against flask S, or for a comparison against flask G in place of flask S.

Common slip: Comparing flask C with flask G and stopping there. Flask G tells you which soil is faster; only flask S, with no living things, shows how much ammonia the urea gives off by itself.

(d) Explain why the compost releases ammonia from urea but leaves the glucose unchanged. (1 point)

A full-credit answer: The compost's urease has an active site whose shape and charges match urea, so urea binds there and is split into ammonia and carbon dioxide. A glucose molecule has a different shape, so it is never held in the active site, no enzyme–substrate complex forms, and the glucose is left unchanged.

Check the box for each point your answer earns

urease's active site fits urea by shape and charge, so urea binds there and is split into ammonia and carbon dioxide, while a glucose molecule has a different shape, is never held in the active site, and so is left unchanged.

Accept 'glucose does not fit urease's active site, so urease does nothing to it'. Do not award the point for 'urease is specific' or 'urease only works on urea' with no reference to fit at the active site.

Common slip: Writing 'urease is specific to urea' and stopping. The point is earned by the reason: glucose does not fit the active site, so it is never bound.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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