Practice questions · Topic 3.1

Unit 3 · Practice for the Topic 3.1 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 3.1 summary, part 1: enzymes

Video coming soon

Rate is the product formed divided by the time; an enzyme lowers the activation energy and comes out unchanged; the active site fits one substrate by shape and charge.

Watch next: Topic 3.1 summary, part 2: the enzyme experiment

Video coming soon

Six tubes of hydrogen peroxide with potato or liver, and one tube with no tissue as the control; the independent, dependent and control variables; the hypothesis, and the null hypothesis that predicts no difference.

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 one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. 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 energy profiles are in kJ/mol.

Question 1

Bacteria in a vinegar barrel speed up the reaction alcohol + oxygen → acetic acid + water. Over a month, the alcohol in the barrel is used up while the liquid turns sour as acetic acid builds up.

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

A. Alcohol and oxygen; they are the starting substances that are changed B. Acetic acid and water; they are the substances that build up in the barrel C. The bacteria; they are added to the barrel to start the reaction D. Alcohol and acetic acid; both are present in the barrel during the month

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 36 mL of juice in 15 minutes.

What is the rate of juice release?

A. 0.42 mL/min B. 2.4 mL/min C. 36 mL/min D. 540 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. The juice is already in the pulp; pectinase only releases it. 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 reaching it later D. 4.0 mL/min; both tubes end with 24 mL, tube 2 reaching it 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.

Which of the following is the energy the reaction releases overall?

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

Question 5

Cells in the pancreas make lipase. Lipase speeds up the reaction fat + water → fatty acids + glycerol in the small intestine, where the temperature is 37 °C.

Why does the body need lipase for this reaction?

A. Few collisions at 37 °C carry the activation energy, so without lipase the reaction is far too slow B. Without lipase, fat and water would react to form different products at 37 °C C. Lipase supplies the energy that the reactants must take in before they can react D. Without lipase, the reaction would release too little energy overall to be useful

Question 6

A juice factory passes cloudy apple juice through a column of beads coated in pectinase. After six weeks the beads carry the same mass of pectinase as at the start, and the juice leaving the column still 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 bird's gut cells make an enzyme that splits a sugar in ripe berries. In autumn 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 fits its substrate. The substrate carries a negative charge, and one R group lining the pocket carries a positive charge. A researcher replaces that R group with a negatively charged one. 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

Which of the following substances is an enzyme?

A. Nickel, a metal that speeds up the hardening of vegetable oil into margarine B. Lactase, a protein made by gut cells that speeds up the splitting of lactose in milk C. Keratin, a protein made by skin cells that hair and nails are made of D. Sucrose, a sugar made by plant cells that stores energy in a sugar beet

Question 10

A student compares how fast pectinase releases juice from apple pulp at 20, 30 and 40 °C. Each flask holds 50 g of the same pulp and receives four drops of pectinase. One extra flask of pulp at 30 °C receives no pectinase. She collects the juice from each flask for 10 minutes.

Which of the following is a control variable in this experiment?

A. The temperature of the pulp B. The flask with no pectinase C. The mass of pulp in each flask D. The volume of juice collected in 10 minutes

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, and it binds at the enzyme's active site.
Gelatin is the protein actinidin acts on.
The active site is 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, and accept 'gelatin and water'. Do not award the point for water alone, with no gelatin named, or for 'the enzyme's surface' with no active site named.

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

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

Hint: Which two levels on the energy profile does the activation energy lie between? Read both peaks off the gridlines.

A full-credit answer: Without actinidin the activation energy is 120 − 70 = 50 kJ/mol.
With actinidin the activation energy is 90 − 70 = 20 kJ/mol.
So the enzyme lowers the activation energy by 30 kJ/mol.

reactants = 70 kJ/mol

peak without actinidin = 120 kJ/mol

peak with actinidin = 90 kJ/mol

activation energy = peak - reactants' level

activation energy = peak - reactants' level

without actinidin: activation energy = 120 - 70 = 50 kJ/mol

with actinidin: activation energy = 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 the effect of actinidin on the reactants' level and the products' level of the energy profile, and calculate the energy the reaction releases overall. (1 point)

Hint: Compare the two curves from left to right. Read where each curve starts and where each curve ends. Between which two levels is the energy released overall measured?

A full-credit answer: Without actinidin the reactants sit at 70 kJ/mol and the products at 45 kJ/mol; with actinidin the reactants sit at 70 kJ/mol and the products at 45 kJ/mol; so the energy released overall is 25 kJ/mol without actinidin and 25 kJ/mol with it.
Actinidin moves only the peak.
Actinidin leaves the start and the end of the energy profile where they were.

reactants = 70 kJ/mol (both curves)

products = 45 kJ/mol (both curves)

energy released overall = reactants' level - products' level

energy released overall = reactants' level - products' level

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 that with the enzyme more energy is released 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: Molecules react only when a collision brings at least the activation energy.
Without actinidin the activation energy is 50 kJ/mol; few collisions carry that much, so the plain gelatin stays firm for a week.
Gelatin bound in actinidin's active site forms an enzyme–substrate complex, and in the enzyme–substrate complex the activation energy is 20 kJ/mol.
So far more collisions carry enough energy, and far more succeed each second.
So 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 activation energy 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 activation energy, 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 structure of actinidin. (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.
A molecule is held in the active site only if its shape fits the pocket and its charges match the R groups lining it.
Starch has a different shape from a protein chain, so actinidin never binds the starch chain.
No enzyme–substrate complex forms, so the activation energy of the starch's 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 that molecule slowly. The enzyme does not act on that molecule 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 asks whether compost bacteria release ammonia from urea faster than garden-soil bacteria. 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; the results are in the table. In a separate check, 1.0 g of the same compost is stirred into water and filtered, so that no cells pass through. This cell-free extract added to 50 mL of urea solution gives off ammonia. The same extract added to 50 mL of glucose solution leaves the glucose unchanged.

Ammonia formed in each flask in 10.0 minutes at 25 °C.

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

A full-credit answer: The independent variable is the material added to the urea solution: compost, garden soil or sand.
The dependent variable is the mass of ammonia formed in 10.0 minutes.
The control is flask S.
Flask S was treated the same way as flasks C and G but received sand, which contains no living things.

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 control variable (the 25 °C temperature, the 50 mL of urea solution, the 10.0 minutes) is named as the control.

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

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

A full-credit answer: The hypothesis is that urea solution given compost forms a larger mass of ammonia in 10.0 minutes than urea solution given garden soil.
The null hypothesis is that 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

both: the hypothesis, that urea solution given compost forms a larger mass of ammonia in 10.0 minutes than urea solution given garden soil (or forms it faster); and the null hypothesis, that there is no difference in the mass of ammonia formed in 10.0 minutes (or in the rate of ammonia formation) between the two. Both statements must name the tested factor (the source of the bacteria) and the measured result (the mass of ammonia, or its rate).

Accept 'the source of the soil bacteria has no effect on the rate of ammonia formation' for the null hypothesis. Do not award the point for a hypothesis or a null hypothesis that names only one variable ('compost is better'), for a null hypothesis that predicts a difference in either direction, or for 'urease has no effect on urea'.

Common slip: Writing 'compost is better' as the hypothesis. A hypothesis names both variables: the source of the bacteria and the mass of ammonia formed in 10.0 minutes. The null hypothesis predicts no difference, and names both too.

(c) Calculate the rate of ammonia formation in each of the three flasks. Support the claim that most of the ammonia in flask C came from urease in the compost, using the rates. (1 point)

A full-credit answer: Flask C formed ammonia at 0.80 mg/min, flask G at 0.30 mg/min and flask S at 0.02 mg/min.
Flask S was treated the same way as the other two flasks but received sand with no living things in it.
So flask S shows that the urea solution on its own gives off only 0.02 mg/min at 25 °C.
Flask C gave 0.78 mg/min more than flask S.
Therefore that extra 0.78 mg/min is credited to the urease in the compost.

flask C: 8.0 mg of ammonia in 10.0 min

flask G: 3.0 mg of ammonia in 10.0 min

flask S: 0.2 mg of ammonia in 10.0 min

rate = \frac{Δ Y}{Δ t} = \frac{mass of ammonia formed}{time taken}

rate = \frac{Δ Y}{Δ t}

flask C: rate = \frac{8.0}{10.0} = 0.80 mg/min

flask G: rate = \frac{3.0}{10.0} = 0.30 mg/min

flask S: rate = \frac{0.2}{10.0} = 0.02 mg/min

flask C - flask S = 0.80 - 0.02 = 0.78 mg/min

Check the box for each point your answer earns

the three rates (0.80 mg/min for flask C, 0.30 for flask G and 0.02 for flask S) AND the evidence linked to the claim: flask S, treated the same way but with no living things, gives off only 0.02 mg/min, so the extra 0.78 mg/min in flask C is credited to the compost's urease.

Support a claim needs the evidence (the rates, with flask S as the comparison) AND the reasoning that links it to the claim. Accept a comparison of the masses (8.0 mg against 0.2 mg) given alongside the rates. Do not award the point for the rates alone, or for a comparison of flask C against flask G alone.

Common slip: Giving the rates without the link to the claim. Support a claim needs the evidence and the reasoning: flask S shows the urea's own rate, so the extra in flask C is the urease's.

(d) Explain why the cell-free extract of the compost releases ammonia from urea but leaves the glucose unchanged. (1 point)

A full-credit answer: The extract contains the compost's urease.
Urease has an active site whose shape and charges match urea.
So urea binds in the active site, and urease splits it into ammonia and carbon dioxide.
A glucose molecule has a different shape from urea.
So glucose 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

a glucose molecule has a different shape from urea, so it is never held in urease's active site and is left unchanged, while urea fits the active site by shape and charge and is split.

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 urease never binds it.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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