End-of-topic test: Enzymes

Unit 3 · Topic 3.1 end-of-topic test

Suggested time: about 48 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.

Answer every question. For each multiple-choice question, pick one option. When you have answered every question, press Submit the test; the feedback then gives the reasoning for each. For the two 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. Then open the scoring guide and mark your own work against it. Energy values on the energy profiles are in kJ/mol.

Question 1

In red blood cells, carbonic anhydrase speeds up the reaction carbon dioxide + water → carbonic acid. As blood passes through a working muscle, the red blood cells take up carbon dioxide, and the amount of carbonic acid in each cell rises.

Which substance is a product of this reaction, and how can you tell?

A. Carbon dioxide; it is the gas the working muscle gives off B. Water; it is the liquid the reaction takes place in C. Carbonic acid; its amount rises as the reaction forms it D. Carbonic anhydrase; it is made by the red blood cell

Question 2

Yeast cells speed up the reaction glucose → alcohol + carbon dioxide. A flask of glucose solution with yeast in it releases 18 mL of carbon dioxide in 6.0 minutes.

What is the rate of reaction?

A. 0.33 mL/min B. 3.0 mL/min C. 12 mL/min D. 108 mL/min

Question 3

Two flasks each hold 100 mL of the same glucose solution. Flask 1 has twice as much yeast in it as flask 2. Flask 1 releases carbon dioxide at 4.0 mL/min and flask 2 at 2.0 mL/min. Both flasks are left until the bubbling stops.

How do the total volumes of carbon dioxide released by the two flasks compare, and which flask finishes first?

A. Flask 1 releases twice as much carbon dioxide in total, and finishes sooner B. Flask 2 releases more in total because it bubbles for longer C. Both flasks release the same total, and both finish at the same time D. Both flasks release the same total, and flask 1 finishes sooner

Question 4

Energy profile for the reaction fat + water → fatty acids + glycerol with no enzyme present.

The figure shows the energy profile for the reaction fat + water → fatty acids + glycerol with no enzyme present, with energy in kJ/mol.

What is the activation energy of the reaction?

A. 20 kJ/mol B. 50 kJ/mol C. 70 kJ/mol D. 80 kJ/mol

Question 5

Energy profiles for the reaction sucrose + water → glucose + fructose: curve 1 (solid) and curve 2 (dashed).

The figure shows two energy profiles for the reaction sucrose + water → glucose + fructose, with energy in kJ/mol. One curve is the reaction on its own; the other is the same reaction with sucrase present.

Which curve shows the reaction with sucrase, and why is the reaction faster with sucrase?

A. Curve 1; more collisions carry enough energy B. Curve 1; the reaction releases more energy overall C. Curve 2; more collisions carry enough energy D. Curve 2; the reaction releases more energy overall

Question 6

Paper is mostly cellulose. Cellulose and the oxygen in the air can react: cellulose + oxygen → carbon dioxide + water, releasing a great deal of energy. Yet a book kept on a shelf at room temperature lasts for centuries.

Why is the reaction so slow at room temperature?

A. The reaction releases too little energy overall to happen on its own B. Few collisions at room temperature carry at least the activation energy C. Cellulose molecules in paper are still, so they never collide with oxygen D. A catalyst must be present before any reaction can begin at all

Question 7

A dairy adds lipase to cream to split the fat: fat + water → fatty acids + glycerol. At the start the cream holds 300 g of fat per liter and 5.0 mg of active lipase per liter. A day later it holds 20 g of fat per liter and still 5.0 mg of active lipase per liter.

What do the measurements show about the lipase?

A. The lipase was used up in step with the fat B. The lipase became part of the fatty acids and glycerol that formed C. The lipase was turned into a product and then made again D. The lipase is still there, unchanged, after splitting most of the fat

Question 8

Papain, an enzyme from papaya, speeds up the reaction protein + water → amino acids. With papaya juice, the protein in a piece of meat is split within an hour. With no papaya juice, the same protein takes weeks to break down.

How do the products and the overall energy released compare in the two cases?

A. Same products; less energy is released overall with papain B. Same products; more energy is released overall with papain C. Same products; the same energy is released overall D. Different products; the same energy is released overall

Question 9

Cells lining the small intestine make maltase, a protein folded into a specific shape. Maltase speeds up the reaction maltose + water → glucose. With no maltase present, maltose passes through the small intestine almost unchanged at body temperature.

What does maltase do that lets the reaction happen fast enough in the gut?

A. Maltase raises the temperature of the gut contents B. Maltase lowers the activation energy of the reaction C. Maltase supplies the energy the reactants must take in D. Maltase raises the energy released when maltose splits

Question 10

Muscle cells and skin cells from one person both contain a stored sugar. A researcher measured the enzyme that breaks this sugar down in both cell types: the muscle cells made large amounts of it; the skin cells made hardly any.

In which cells does the breakdown happen at a useful rate, and why?

A. Muscle cells only; the reaction is fast only where its enzyme is present B. Skin cells only; the muscle enzyme is used up as fast as it is made C. Both cell types; the sugar is present in both, so it reacts on its own D. Both cell types; every cell in one person carries the same working enzymes

Question 11

Maltase (the large shape) with a molecule of maltose (the small shape) near its surface. 1, 2 and 3 mark parts of the picture.

The figure shows a folded maltase molecule and a molecule of maltose near its surface, with three parts labeled 1, 2 and 3.

What are the parts labeled 1 and 2?

A. Active site at 1, substrate at 2 B. Product at 1, active site at 2 C. Substrate at 1, active site at 2 D. Active site at 1, product at 2

Question 12

An enzyme's active site (the pocket in the large shape), lined with negatively charged R groups, and the three molecules tested, X, Y and Z, each marked with the charge it carries.

An enzyme's active site is a pocket lined with R groups that carry negative charges. The figure shows the shape of the pocket and the three molecules tested, X, Y and Z, drawn to the same scale, with the charge each carries.

Judging by shape and by charge, which molecules are held in the active site?

A. X only B. Y only C. X and Y D. X and Z

Question 13

A three-panel model of an enzyme acting on its substrate. The large shape in each panel is the enzyme; the panels are numbered 1 to 3.

The figure is a three-panel model of an enzyme acting on its substrate. One substrate molecule enters the reaction, and two product molecules leave it.

Which panel shows the enzyme–substrate complex, and what is happening in it?

A. Panel 1; the substrate is drawn to the active site by its charges B. Panel 2; the enzyme is being used up as the products form C. Panel 3; the products stay bound so the enzyme can be reused D. Panel 2; the reaction happens there with a lowered activation energy

Question 14

Urease from soil bacteria has an active site that fits a urea molecule. Urease speeds up the reaction urea + water → ammonia + carbon dioxide. A student adds the same amount of urease to three tubes at 25 °C: one of urea, one of sucrose (table sugar), and one of protein.

In which tubes does urease speed up a reaction?

A. Urea only B. Urea and sucrose C. Urea and protein D. Urea, sucrose and protein

Question 15

A student asks whether the concentration of sucrose changes how fast yeast releases carbon dioxide. She puts 10 mL of 2%, 4%, 6% or 8% sucrose solution into four flasks at 30 °C, adds the same 1.0 g of yeast to each, and records the volume of carbon dioxide collected in 10 minutes.

What are the independent and dependent variables?

A. Independent: mass of yeast; dependent: carbon dioxide collected B. Independent: carbon dioxide collected; dependent: sucrose concentration C. Independent: sucrose concentration; dependent: time allowed D. Independent: sucrose concentration; dependent: carbon dioxide collected

Question 16

What the student adds to each tube of a 5.0 g gelatin cube in 10 mL of water at 25 °C.

A student compares protease from pineapple and from papaya. Each tube holds a 5.0 g cube of gelatin (a protein) in 10 mL of water at 25 °C; the table shows what she adds to each tube. She records the mass of gelatin left in each tube after 30 minutes.

Which tube is the control, and what is one control variable?

A. Tube 1; the kind of fruit juice B. Tube 3; the kind of fruit juice C. Tube 3; the mass of gelatin D. Tube 2; the volume of liquid added

Question 17

In an experiment at 37 °C, a tube of hydrogen peroxide with a drop of blood in it released 14.0 mL of oxygen in 4 minutes. A tube of the same hydrogen peroxide with no blood released 0.5 mL in the same 4 minutes.

What does the no-blood tube show?

A. That the experiment worked, because oxygen was collected from both tubes B. That the peroxide broke down on its own, so the blood made no difference C. That temperature was kept the same for both tubes, which is what a control does D. How much oxygen the peroxide gives off on its own, with no blood added

Question 18

A student tests whether the concentration of a starch solution changes the mass of maltose that amylase forms in 5 minutes.

Which of the following statements is a hypothesis predicting that the concentration of starch affects the mass of maltose formed?

A. A higher concentration of starch gives a larger mass of maltose in 5 minutes B. Amylase splits starch into maltose C. More starch is better for the amylase D. The concentration of starch makes no difference to the mass of maltose formed in 5 minutes

Question 19

Which of the following substances is an enzyme?

A. Collagen, a protein made by skin cells that gives skin its strength and stretch B. Manganese dioxide, a black powder that speeds up the breakdown of hydrogen peroxide into water and oxygen C. Lysozyme, a protein made by tear-gland cells that speeds up the breakdown of bacterial cell walls D. Starch, a chain of glucose units that a potato cell stores for later use

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 · 4 points

Amylase, an enzyme in saliva, speeds up the reaction starch + water → maltose (a sugar made of two glucose units). A student stirs a drop of saliva into a tube of thick starch paste at 37 °C; within a few minutes the paste has thinned and tests show maltose. A second tube of the same starch paste with no saliva is still thick after a day. The figure shows the energy profile of the reaction with and without amylase, with energy in kJ/mol.

Energy profile for starch + water → maltose without amylase (solid) and with amylase (dashed).

(a) Describe the effect of amylase on the activation energy of the reaction and on the energy the reaction releases overall. (1 point)

A full-credit answer: Amylase lowers the activation energy.
The peak falls from 95 kJ/mol to 65 kJ/mol, so the activation energy drops from 45 kJ/mol to 15 kJ/mol.
Amylase leaves the reactants' level and the products' level unchanged: the reactants still sit at 50 kJ/mol and the products at 30 kJ/mol.
So the energy released overall, 20 kJ/mol, is unchanged.

reactants = 50 kJ/mol

peak without amylase = 95 kJ/mol

peak with amylase = 65 kJ/mol

products = 30 kJ/mol

activation energy = peak - reactants' level

energy released overall = reactants' level - products' level

activation energy = peak - reactants' level

without amylase: activation energy = 95 - 50 = 45 kJ/mol

with amylase: activation energy = 65 - 50 = 15 kJ/mol

energy released overall = reactants' level - products' level

energy released overall (both) = 50 - 30 = 20 kJ/mol

Check the box for each point your answer earns

amylase lowers the activation energy (the peak falls from 95 kJ/mol to 65 kJ/mol, so the activation energy drops from 45 kJ/mol to 15 kJ/mol) while the energy levels of the reactants (50 kJ/mol) and products (30 kJ/mol), and so the energy released overall (20 kJ/mol), are unchanged.

Accept a description in words with no numbers ("the peak is lower; the start and end are the same"). Do not award the point for an answer that says amylase lowers the products or changes the energy released.

Common slip: Saying the enzyme lowers the whole curve, or that less energy is released. Only the peak moves. The start level and the end level stay where they were.

(b) Explain how amylase makes maltose form faster at 37 °C. (1 point)

A full-credit answer: Molecules react only when a collision carries at least the activation energy.
Without amylase the activation energy is 45 kJ/mol; few collisions at 37 °C carry that much, so few succeed.
Starch bound in amylase's active site forms an enzyme–substrate complex, in which the activation energy is only 15 kJ/mol.
Far more of the collisions already happening carry 15 kJ/mol, so far more succeed each second.
Therefore maltose forms faster.

Check the box for each point your answer earns

with a lower activation energy, a far larger fraction of the collisions already happening at 37 °C carry enough energy to react, so more collisions succeed each second and maltose forms faster. Accept the fuller chain: starch binds at the active site, the enzyme–substrate complex lowers the activation energy, the products leave and the unchanged amylase is reused.

Do not award the point for "amylase lowers the activation energy" alone, with no link to collisions or to how many succeed, or for "amylase heats the paste" or "amylase supplies energy".

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

(c) The student adds the same amount of saliva to a third tube holding protein instead of starch, at 37 °C. Make a claim about what happens in this tube over the next day. (1 point)

A full-credit answer: Nothing happens to the protein.
It is unchanged after a day, just as the starch paste with no saliva was unchanged.

Check the box for each point your answer earns

a correct, specific claim: nothing happens to the protein; it is unchanged after a day.

Make a claim earns the point for the assertion; the reasoning is scored in part (d). Do not award the point for "the protein breaks down more slowly" or "it breaks down once the amylase has finished with the starch".

Common slip: Predicting a slow reaction 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.

(d) Support your claim using the structure of amylase. (1 point)

A full-credit answer: A molecule is held in the active site only if its shape fits the pocket and its charges are opposite to those of the R groups lining it.
A protein chain has a different shape from a stretch of starch.
So amylase never binds the protein chain.
No enzyme–substrate complex forms, so the activation energy for breaking the protein down is not lowered.

Check the box for each point your answer earns

the evidence (a protein chain has a different shape from a stretch of starch) AND the reasoning that links it to the claim: a substrate is held only if its shape fits the active site and its charges are opposite to those of the R groups lining it, so the protein is never bound, no enzyme–substrate complex forms, and its activation energy is not lowered.

Support a claim needs the evidence and the link to the claim. Do not award the point for "amylase is specific to starch" with no reference to fit at the active site, or for "the amylase was used up on the starch".

Common slip: Writing "amylase only works on starch" as if that supported the claim. The support is the reason: the protein does not fit the active site, so it is never bound.

Free-response score: 0 of 4

Free response 2 · Scientific Investigation · 4 points

Catalase speeds up the reaction hydrogen peroxide → water + oxygen. A student compares catalase from two sources. She mashes 2.0 g of mushroom in 10 mL of water and, separately, 2.0 g of radish in 10 mL of water. Three tubes each receive 10 mL of 3% hydrogen peroxide at 25 °C. Tube M gets 1.0 mL of the mushroom mash, tube R gets 1.0 mL of the radish mash, and tube W gets 1.0 mL of water. She collects the oxygen released from each tube over 4.0 minutes; the results are in the table. In a separate check, 1.0 mL of the radish mash added to 10 mL of sugar solution gives off no gas and leaves the sugar unchanged.

Oxygen collected in 4.0 minutes from 10 mL of 3% hydrogen peroxide at 25 °C.

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

A full-credit answer: The independent variable is the source of the catalase: mushroom or radish.
The dependent variable is the volume of oxygen collected in 4.0 minutes.
The control is tube W.
Tube W was treated the same way as tubes M and R but received water instead of mash.

Check the box for each point your answer earns

all three: independent variable, the source of the catalase (mushroom or radish); dependent variable, the volume of oxygen collected in 4.0 minutes (or the rate of oxygen release); control, tube W (peroxide with water and no mash).

Accept "the tissue added" 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 10 mL of peroxide, the 4.0 minutes) is named as the control.

Common slip: Naming a control variable, such as the 25 °C temperature, as "the control". Control variables are the conditions kept the same in every tube. The control is the tube that lacks the factor under test.

(b) State the null hypothesis for the comparison between mushroom and radish. (1 point)

A full-credit answer: There is no difference in the volume of oxygen released in 4.0 minutes between hydrogen peroxide given mushroom catalase and hydrogen peroxide given radish catalase.

Check the box for each point your answer earns

there is no difference in the volume of oxygen released in 4.0 minutes (or in the rate of oxygen release) between catalase from mushroom and catalase from radish. The statement must name the tested factor (the source of the catalase) and the measured result (oxygen released or its rate).

Accept "the source of the catalase has no effect on the rate of oxygen release". Do not award the point for a prediction of a difference in either direction ("radish is faster", "mushroom is slower"), for a null hypothesis about the water tube alone ("the mash has no effect compared with tube W"), or for "catalase has no effect on peroxide".

Common slip: Writing the opposite prediction ("mushroom releases oxygen faster") as the null hypothesis. The null hypothesis predicts no difference. It names both the factor changed and the quantity measured.

(c) Explain why the radish mash releases oxygen from hydrogen peroxide but leaves the sugar unchanged. (1 point)

A full-credit answer: Catalase's active site is a pocket whose shape and charges match hydrogen peroxide.
So hydrogen peroxide binds in the active site, and catalase splits it into water and oxygen.
A sugar molecule has a different shape from hydrogen peroxide.
So the sugar molecule is never held in the active site, and the sugar is left unchanged.

Check the box for each point your answer earns

catalase's active site fits hydrogen peroxide by shape and charge, so peroxide binds there and is split into water and oxygen, while a sugar molecule has a different shape, is never held in the active site, and so is left unchanged.

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

Common slip: Writing "catalase is specific to peroxide" and stopping. The point is earned by the reason: the sugar does not fit the active site, so it is never bound.

(d) Support the claim that most of the oxygen in tube R came from the catalase in the radish. Use the rate of oxygen release in each of the three tubes as your evidence, showing how you calculated each rate. (1 point)

A full-credit answer: Tube M released oxygen at 1.5 mL/min, tube R at 3.5 mL/min and tube W at 0.1 mL/min.
Tube W was treated the same way as the other two tubes but received no tissue.
So tube W shows that the peroxide on its own gives off only 0.1 mL/min at 25 °C.
Tube R gave 3.4 mL/min more than tube W.
Therefore that extra 3.4 mL/min is credited to the catalase in the radish.

tube M: 6.0 mL of oxygen in 4.0 min

tube R: 14.0 mL of oxygen in 4.0 min

tube W: 0.4 mL of oxygen in 4.0 min

rate = \frac{Δ Y}{Δ t} = \frac{volume of oxygen}{time}

rate = \frac{Δ Y}{Δ t}

tube M: rate = \frac{6.0}{4.0} = 1.5 mL/min

tube R: rate = \frac{14.0}{4.0} = 3.5 mL/min

tube W: rate = \frac{0.4}{4.0} = 0.1 mL/min

tube R - tube W = 3.5 - 0.1 = 3.4 mL/min

Check the box for each point your answer earns

the three rates (1.5 mL/min for tube M, 3.5 for tube R and 0.1 for tube W) AND the evidence linked to the claim: tube W, treated the same way but with no tissue, releases only 0.1 mL/min, so the extra 3.4 mL/min in tube R is credited to the radish's catalase.

Support a claim needs the evidence (the rates, with tube W as the comparison) AND the reasoning that links it to the claim. Accept a comparison of the volumes (14.0 mL against 0.4 mL) given alongside the rates. Do not award the point for the rates alone, or for a comparison of tube R against tube M alone.

Common slip: Giving the rates without the link to the claim. Tube W shows how much oxygen the peroxide gives off by itself, so the extra in tube R is the catalase's.

Free-response score: 0 of 4

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Multiple choice checked: 0 of 19 correct.

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