Enzyme Structure

Unit 3 · Topic 3.1 end-of-topic test

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 two free-response questions, write your answer in full sentences (and show any calculation), then open the scoring guide and mark your own work against it. Energy values on the profiles are in kJ/mol.

Question 1

Catalase speeds up the reaction hydrogen peroxide → water + oxygen. In a sealed tube, the amount of hydrogen peroxide falls steadily over five minutes while bubbles of oxygen collect at the top.

Which substance is a reactant in this reaction, and how can you tell?

A. Hydrogen peroxide; its amount falls as it is changed into new substances B. Oxygen; it is the gas the reaction needs in order to run C. Water; it is the liquid the reaction takes place in D. Catalase; it is added to the tube to start the reaction

Question 2

A tube of hydrogen peroxide with a few drops of catalase releases 18 mL of oxygen in 6.0 minutes.

What is the rate of the reaction?

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

Question 3

Two tubes each hold 20 mL of the same hydrogen peroxide solution. Tube A releases oxygen at 3.0 mL/min and tube B at 1.0 mL/min. Both are left until the fizzing stops.

How do the total volumes of oxygen released by the two tubes compare?

A. Tube A releases three times as much oxygen in total B. Tube B releases more in total because it runs for longer C. Both release the same total, and both finish at the same time D. Both release the same total, and tube A finishes sooner

Question 4

Energy profile for the breakdown of hydrogen peroxide into water and oxygen with no enzyme present.

The figure shows the energy profile for the breakdown of hydrogen peroxide into water and oxygen 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 breakdown of hydrogen peroxide into water and oxygen: curve 1 (solid) and curve 2 (dashed).

The figure shows two energy profiles for the breakdown of hydrogen peroxide into water and oxygen, with energy in kJ/mol. One curve is the reaction on its own; the other is the same reaction with catalase present.

Which curve shows the reaction with catalase, and how does its rate compare with the reaction on its own?

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

Question 6

Sugar and the oxygen in the air can react to form carbon dioxide and water, releasing a great deal of energy. Yet a jar of sugar kept on a shelf at room temperature stays sugar for years.

Why does the reaction run so slowly at room temperature?

A. Few collisions at room temperature carry at least the activation energy B. The reaction releases too little energy overall to run on its own C. Sugar molecules in a jar 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 lactase to milk to split lactose, the sugar in milk. At the start the milk holds 12 g of lactose per liter and 1.0 mg of active lactase per liter. An hour later it holds 2 g of lactose per liter and still 1.0 mg of active lactase per liter.

What do the measurements show about the lactase?

A. It was used up in step with the lactose B. It became part of the smaller sugars that formed C. It was turned into a product and then made again D. It is still there, unchanged, after splitting most of the lactose

Question 8

With a drop of catalase, a tube of hydrogen peroxide gives off its oxygen in seconds. Left alone, the same peroxide takes months 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 catalase B. Same products; more energy is released overall with catalase 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 lactase, a protein folded into a specific shape. Lactase splits lactose into two smaller sugars. With no lactase present, lactose passes through the gut almost unchanged at body temperature.

What does lactase do that lets the splitting reaction run fast enough in the gut?

A. Lowers the activation energy of the reaction B. Raises the temperature of the gut contents C. Supplies the energy the reactants must take in D. Raises the energy released when lactose splits

Question 10

Muscle cells and skin cells from the same person both contain a stored sugar. Muscle cells make large amounts of the enzyme that breaks this sugar down; skin cells make hardly any of it.

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

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

Question 11

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

The figure shows a folded catalase molecule and a molecule of hydrogen peroxide 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 is a pocket lined with R groups that carry negative charges. Three molecules are tested. Molecule X has the pocket's shape and carries a positive charge. Molecule Y has the pocket's shape and carries a negative charge. Molecule Z has the wrong shape for the pocket and carries a positive 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

Three stages of catalase acting on hydrogen peroxide: the small shape is the hydrogen peroxide molecule; in panel 3 the two smaller shapes are the products.

The figure is a three-panel model of catalase acting on a molecule of hydrogen peroxide.

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

Amylase from saliva has an active site that fits a short stretch of a starch chain, a chain of glucose units joined end to end. A student adds the same amount of amylase to three tubes at body temperature: one of starch, one of protein, and one of lactose, a sugar made of one glucose joined to one galactose.

In which tubes does amylase speed up a reaction?

A. Starch only B. Starch and lactose C. Starch and protein D. All three

Question 15

A student asks whether the concentration of hydrogen peroxide changes how fast catalase releases oxygen. She puts 10 mL of 1%, 2%, 3% or 4% hydrogen peroxide into four tubes at 25 °C, adds the same two drops of catalase solution to each, and records the volume of oxygen collected in 3 minutes.

What are the independent and dependent variables?

A. Independent: drops of catalase; dependent: oxygen collected B. Independent: oxygen collected; dependent: peroxide concentration C. Independent: peroxide concentration; dependent: time allowed D. Independent: peroxide concentration; dependent: oxygen collected

Question 16

A student compares catalase from potato and from liver. Tube 1: 10 mL of 3% hydrogen peroxide at 25 °C and 1.0 g of mashed potato. Tube 2: 10 mL of 3% hydrogen peroxide at 25 °C and 1.0 g of mashed liver. Tube 3: 10 mL of 3% hydrogen peroxide at 25 °C with no tissue added. She records the oxygen released from each tube in 4 minutes.

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

A. Tube 1; the source of the catalase B. Tube 3; the source of the catalase C. Tube 3; the volume of hydrogen peroxide D. Tube 2; the mass of tissue

Question 17

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

What does the tube with no liver show?

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

Question 18

A student's hypothesis is that a higher concentration of hydrogen peroxide gives a faster rate of oxygen release when the same amount of catalase is added.

Which statement is the null hypothesis for this experiment?

A. A higher concentration of hydrogen peroxide gives a slower rate of oxygen release B. The concentration of hydrogen peroxide makes no difference to the rate of oxygen release C. A higher concentration of hydrogen peroxide gives a faster rate of oxygen release D. Adding catalase makes no difference to the rate of oxygen release

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) Using the figure, describe what amylase changes about the energy profile and what it leaves unchanged. (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. 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.

Write down the values from the figure:

reactants = 50 kJ/mol
peak without amylase = 95 kJ/mol
peak with amylase = 65 kJ/mol
products = 30 kJ/mol

Write down the equation:

activation energy = peak − reactants' level

Substitute in the values, and calculate:

without amylase: 95 − 50 = 45 kJ/mol
with amylase: 65 − 50 = 15 kJ/mol
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 hump 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 hump moves; the start and end levels stay where they were.

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

A full-credit answer: Reacting molecules must collide with at least the activation energy before their bonds can rearrange, and at 37 °C only a small fraction of collisions carry 45 kJ/mol. When a stretch of starch binds in amylase's active site, forming an enzyme–substrate complex, the activation energy is only 15 kJ/mol, so a far larger fraction of the collisions already happening succeed and maltose forms faster. The products leave and the amylase comes out unchanged, ready to bind the next stretch of starch.

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 barrier, 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 "it lowers the activation energy". The point needs the next step: more of the collisions carry enough energy, so more of them succeed.

(c) The student adds the same amount of saliva to a third tube holding protein instead of starch, at 37 °C. Predict 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.

Check the box for each point your answer earns

the protein is unchanged; no reaction is sped up and the tube stays as it was, however long it is left.

Accept "nothing happens to the protein". 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) Justify your prediction using the active site. (1 point)

A full-credit answer: 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 protein chain has a different shape from a stretch of starch, so it is never bound, no enzyme–substrate complex forms, and the activation energy for breaking it down is not lowered. The amylase is also unchanged by the starch it split, so being used up is not the reason.

Check the box for each point your answer earns

a substrate is held only if its shape fits the active site and its charges match the R groups lining it; a protein chain has a different shape (and charge pattern) from a stretch of starch, so it is never bound, no enzyme–substrate complex forms, and the activation energy for its reaction is not lowered.

Accept "protein does not fit amylase's active site, so it is not bound". 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 were the reason. The point is earned by saying why: the protein does not fit the active site.

Free-response score: 0 of 4

Free response 2 · Scientific Investigation · 4 points

A student compares catalase from two sources. She mashes 2.0 g of potato in 10 mL of water and, separately, 2.0 g of liver in 10 mL of water. Three tubes each receive 10 mL of 3% hydrogen peroxide at 25 °C. Tube P gets 1.0 mL of the potato mash, tube L gets 1.0 mL of the liver 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 liver 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) Explain why the liver 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 peroxide binds there and is split into water and oxygen. A sugar molecule has a different shape, so it is never held in the active site and 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.

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

A full-credit answer: Independent variable: the source of the catalase, potato or liver. Dependent variable: the volume of oxygen collected in 4.0 minutes. Control: tube W, which was treated the same way as P and L but received water instead of mash.

Check the box for each point your answer earns

all three: independent variable, the source of the catalase (potato or liver); 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 controlled condition (the 25 °C temperature, the 10 mL of peroxide, the 4.0 minutes) is named as the control.

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

(c) State the null hypothesis for this experiment. (1 point)

A full-credit answer: There is no difference in the volume of oxygen released in 4.0 minutes between hydrogen peroxide given potato catalase and hydrogen peroxide given liver catalase. (A null for the water comparison also earns the point: adding tissue mash makes no difference to the oxygen released compared with tube W.)

Check the box for each point your answer earns

a no-difference statement about either of the experiment's two comparisons: there is no difference in the volume of oxygen released in 4.0 minutes (or in the rate of oxygen release) between catalase from potato and catalase from liver; OR adding tissue mash (potato or liver) makes no difference to the volume of oxygen released compared with water (tube W). The statement must name the tested factor and the measured result (oxygen released or its rate).

Accept "the source of the catalase has no effect on the rate of oxygen release" and "the mash has no effect on the oxygen released compared with tube W". Do not award the point for a prediction of a difference in either direction ("liver is faster", "potato is slower") or for "catalase has no effect on peroxide".

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

(d) Calculate the rate of oxygen release in tube L and in tube W, and use them to justify the claim that most of the oxygen in tube L came from the catalase in the liver. (1 point)

A full-credit answer: Tube L released oxygen at 3.5 mL/min and tube W at 0.1 mL/min. Tube W is the control: it shows that the peroxide on its own gives off only 0.1 mL/min at 25 °C. The two tubes were treated the same way except for the liver mash, so the extra 3.4 mL/min in tube L is credited to the catalase in the liver.

Write down the values in the question:

tube L: 14.0 mL of oxygen in 4.0 min
tube W: 0.4 mL of oxygen in 4.0 min

Write down the equation:

        volume of oxygen
rate = ──────────────────
              time

Substitute in the values, and calculate:

tube L: rate = 14.0 ÷ 4.0 = 3.5 mL/min
tube W: rate = 0.4 ÷ 4.0 = 0.1 mL/min
difference: 3.5 − 0.1 = 3.4 mL/min

Check the box for each point your answer earns

rates of 3.5 mL/min for tube L (14.0 ÷ 4.0) and 0.1 mL/min for tube W (0.4 ÷ 4.0), AND the justification that tube W shows how much oxygen the peroxide releases on its own, so, because the two tubes were treated the same way except for the liver mash, the extra 3.4 mL/min (13.6 mL) is credited to the liver's catalase.

Accept a comparison of the volumes (14.0 mL against 0.4 mL) given alongside the rates. Do not award the point for the two rates with no comparison against tube W, or for a comparison against tube P in place of tube W.

Common slip: Comparing tube L with tube P and stopping there. Tube P tells you which source is faster; only tube W, with no catalase, shows how much oxygen the peroxide gives off by itself.

Free-response score: 0 of 4

Multiple choice checked: 0 of 18 correct.

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