Cellular Energy

Unit 3 · Topic 3.3 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 is measured in joules (J) and kilojoules (kJ; 1 kJ = 1,000 J).

Question 1

A drawing of an ATP molecule. Two regions are marked, 1 and 2.

The figure shows an ATP molecule with two regions marked, 1 and 2.

Which region of the figure is adenosine, and what does it consist of?

A. Region 1: the base joined to the sugar B. The left part of region 1: the base on its own C. Region 2: the chain of three phosphate groups D. Regions 1 and 2 together: the whole molecule

Question 2

A cell hydrolyzes a molecule of ATP.

How does the ADP that is left differ from the ATP?

A. It has one phosphate group instead of three B. It has lost all three phosphates, leaving adenosine on its own C. It has two phosphate groups instead of three D. It still has three phosphates, with a water molecule added onto the outer one

Question 3

A firefly's light organ hydrolyzes ATP to power each flash of light. With no ATP, the organ stays dark.

Where does the energy for the flash come from?

A. From light the organ absorbed during the day and stored as chemical energy B. From the whole reaction: ADP and Pi hold less energy than ATP and water did C. From the bond to the outer phosphate: the energy stored in it is released when it breaks D. From the water molecule, which is split apart and gives up the energy it held

Question 4

On an energy profile for ATP hydrolysis, drawn in relative units, the reactants (ATP + H₂O) sit at 100 and the products (ADP + Pi) at 70. There is a hump between them, as in any reaction.

What does the profile say about the reaction?

A. It takes in 30 units, since energy must be put in to break the phosphate bond B. It releases 100 units, the energy that was stored in the phosphate bond C. It releases 70 units, the energy left over in the products D. It releases 30 units, because the products hold less energy than the reactants

Question 5

A membrane protein moves a solute into a vesicle against its concentration gradient. Researchers tested it with ATP, with ADP plus Pi, and with a look-alike of ATP that binds the protein's ATP site but cannot be hydrolyzed. The solute accumulated inside only when real ATP was present.

What do the results show about how the protein uses ATP?

A. Binding ATP alone changes the protein; removing the phosphate is unnecessary B. ATP hydrolysis is coupled to the protein's change of shape; binding alone is not enough C. Moving the solute releases energy that turns ADP and Pi into ATP D. The protein uses ADP and Pi directly as its source of energy

Question 6

When the sodium–potassium pump hydrolyzes ATP, the outer phosphate group is transferred onto the pump protein itself: the pump is phosphorylated.

What does the added phosphate do to the pump?

A. Unfolds it, so that it is denatured B. Makes it split into two smaller proteins C. Changes its shape so that it pushes Na⁺ out of the cell D. Gives it a second active site for K⁺

Question 7

A muscle is given a poison that stops it from remaking ATP. It contracts normally for a few seconds and then stops, although its contracting proteins are undamaged.

Why does the muscle stop?

A. Its ATP runs out within seconds, so hydrolysis no longer drives the work B. The poison denatured the proteins that do the contracting C. The muscle's glucose supply is used up within a few seconds D. ADP builds up and blocks the proteins that do the contracting

Question 8

During a ten-second sprint a runner's leg muscles use ATP at about 100 units per second, yet the amount of ATP in the muscle barely falls over the ten seconds.

How is this possible?

A. The muscle stores enough ATP for many minutes of sprinting B. ATP is not used in a sprint; the muscle uses glucose directly C. Each ATP molecule can be hydrolyzed many times before it becomes ADP D. ATP is remade from ADP and Pi almost as fast as it is used, with energy from food

Question 9

A seedling germinated in total darkness grows for about a week and then dies. An identical seedling in the light keeps growing.

Why does the dark seedling die?

A. It created its own energy at first, then lost the ability B. Its only energy input was the food stored in the seed, which ran out C. Darkness lowered its temperature until its enzymes were denatured D. It took in energy from the soil, and the soil's energy was exhausted

Question 10

In one afternoon a leaf absorbs 900 kJ of light energy and stores about 40 kJ of it in the sugar it makes.

What happened to the other 860 kJ?

A. It was destroyed in the reactions that made the sugar B. It was transformed to heat and passed to the leaf's surroundings C. It became the matter, the atoms, of the new sugar D. It was stored in the leaf's ATP, to be used at night

Question 11

A student writes: 'Plants are producers because they make their own energy from sunlight.'

Which statement corrects the student?

A. Plants make energy from carbon dioxide and water rather than from sunlight B. Plants store sunlight itself inside their leaves for later C. Plants transform light energy into chemical energy in sugar; they create none D. Plants make energy from the minerals they take up from the soil

Question 12

As a cell breaks down glucose, about a third of the energy released ends up in ATP.

What happens to the rest of the energy?

A. It is destroyed, so the total amount of energy falls B. It spreads into the cell and its surroundings as heat and can no longer do work C. It is stored in the ADP and Pi that are left over D. It stays in the part of the glucose molecule that was not used

Question 13

A dead leaf on the ground crumbles and scatters over a few weeks. The living leaf on the tree keeps its shape, its molecules built and rebuilt every day.

Which idea from the second law of thermodynamics does the dead leaf show?

A. With no energy input its matter grows more disordered: its entropy rises B. Energy is being created as the leaf's molecules break apart C. The leaf's atoms are being destroyed as it crumbles to dust D. Its order increases as it dries out, shrinks and hardens

Question 14

A poison blocks a cell's ability to make ATP. Within hours the cell's ion gradients run down, its membranes leak and it dies.

Why does the cell die?

A. The poison directly dissolves the cell's membranes B. The poison denatures the cell's enzymes C. The cell still makes ATP but its proteins can no longer hydrolyze it D. With no ATP, the work that keeps its contents ordered stops, and order is lost

Question 15

An animal that stops eating first uses up its stored fat, then breaks down its own muscle proteins, and eventually dies.

Why does the animal break down its own tissues?

A. Heat still leaves it every moment, and its own molecules are the only energy source left B. Its own tissues create the new energy it needs for as long as no food is coming in C. Breaking down its own tissues stops energy from leaving its body as heat while it starves D. Its cells stop needing ATP once food stops arriving, so the tissue is spare to burn

Question 16

A metabolic pathway in a cell. Each arrow is one reaction, catalyzed by the enzyme written above it.

The figure shows a metabolic pathway in which W is converted, step by step, to Z.

Which molecule is both the product of enzyme E2's reaction and the reactant of enzyme E3's reaction?

A. W B. X C. Y D. Z

Question 17

A metabolic pathway in a cell. Each arrow is one reaction, catalyzed by the enzyme written above it.

In the pathway shown, a cell loses the ability to make enzyme E2. E1 and E3 are unchanged, and W keeps arriving.

What happens to the amounts of the pathway's molecules?

A. X builds up; Y and Z are no longer made B. W builds up; X, Y and Z are no longer made C. Y builds up; only Z is no longer made D. X and Y build up; only Z is no longer made

Question 18

A spoonful of sugar held in a flame burns in one flash of heat and light. A cell releases the same energy from the same amount of sugar in dozens of small enzyme-catalyzed steps.

What is the advantage of the many small steps?

A. The steps release more energy in total than burning the sugar does B. Each step releases a little energy that can be captured in ATP rather than lost as heat C. Many small steps break the sugar down faster than a flame does D. The steps release the energy with no loss at all as heat

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

A germinating seed kept in the dark lives on the starch stored in the seed. In one ten-minute interval, students measured the energy transferred from the seed's stores: 250 J was transferred from stored starch, 100 J of it ended up in newly made ATP, and 150 J left the seed as heat. Assume these values account for all the energy transferred in the interval. The seedling uses its ATP to build new cells and to move materials into them. If it stays in the dark, the seed's store will be used up in a few days.

(a) Describe what the first law of thermodynamics says about the 250 J, and calculate the percentage of it that was captured in ATP. (1 point)

A full-credit answer: The first law says energy is never created or destroyed, only transferred or transformed, so every joule of the 250 J is accounted for: 100 J in ATP and 150 J as heat, 250 J in all. The seed captured 40% of the transferred energy in ATP.

Write down the values in the question:

energy transferred from starch = 250 J
energy captured in ATP = 100 J
energy released as heat = 150 J

Write down the equation:

                       energy in ATP
percentage in ATP = ──────────────────── × 100
                     energy transferred

Substitute in the values, and calculate:

percentage in ATP = (100 ÷ 250) × 100
percentage in ATP = 40%
check: 100 J + 150 J = 250 J, all accounted for

Check the box for each point your answer earns

energy is neither created nor destroyed, only transferred or transformed, so the 250 J is all accounted for as 100 J in ATP plus 150 J of heat (100 + 150 = 250); AND 100 ÷ 250 × 100 = 40% captured in ATP.

Accept "the energy in equals the energy out" for the first law. Do not award the point for 40% alone with no statement of the law, or for a statement that energy was used up or lost.

Common slip: Describing the 150 J of heat as energy that was "lost" or "used up". It left the seed, but it still exists; the first law counts it.

(b) Explain why only part of the energy transferred from the starch ended up in ATP. (1 point)

A full-credit answer: Every energy transfer spreads some energy out as heat that can no longer do work, so no transfer is fully efficient; that is the second law of thermodynamics. As the seed breaks its starch down in many small steps, each step captures part of the energy released in ATP and the rest, here 150 J of the 250 J, warms the seed and its surroundings.

Check the box for each point your answer earns

by the second law of thermodynamics, in every energy transfer or transformation some energy spreads out as heat that can no longer do work, so no transfer is fully efficient; the pathway that breaks the starch down releases the energy in many small steps and captures a portion of each in ATP, while the rest leaves as heat.

Accept an answer that names the second law and heat without mentioning the steps. Do not award the point for "the seed wasted energy" with no mention of heat, or for "some energy was destroyed".

Common slip: Saying the seed "could not use" the 150 J and stopping. The point needs the second law's idea: some energy spreads out as heat at every transfer, so no transfer is fully efficient.

(c) Predict what will happen to the seedling if it stays in the dark after the seed's store of starch is used up. (1 point)

A full-credit answer: Once the starch is gone the seedling has no energy input at all: no food, and no light in the dark. It can no longer remake ATP, so the work of building new cells and pumping materials stops, its cells lose their order, and it dies.

Check the box for each point your answer earns

with no starch left and no light, the seedling has no energy input, so it can no longer make ATP; the building and pumping that keep its cells ordered stop, its order breaks down and it dies.

Accept "it stops growing and dies" with the reason that it can no longer make ATP. Do not award the point for "it stops growing" alone, or for a prediction that it keeps living on light it is not receiving.

Common slip: Predicting only that growth slows or stops. The prediction has to follow the chain to its end: no input, no ATP, order lost, death.

(d) Justify your prediction using what living systems must do with energy to stay alive. (1 point)

A full-credit answer: A living system stays ordered only by taking in more energy than it loses. Order costs energy every moment, paid for by ATP hydrolysis coupled to the work of building and pumping, and heat leaves at every transfer. In the dark, with the starch gone, the seedling's input is zero while its losses go on, so the order cannot be paid for; a significant loss of order, or an interruption of the energy flow, results in death.

Check the box for each point your answer earns

keeping a living system ordered costs energy continuously, paid for by ATP hydrolysis coupled to the work, while some energy leaves as heat at every step, so a living system must take in more energy than it loses; the seedling's losses continue in the dark but its input has fallen to zero, so its order cannot be maintained and a significant loss of order results in death.

Accept a justification built on "input must exceed loss" or on the two laws applied to the seedling. Do not award the point for restating the prediction, or for a justification that has the seedling creating energy from its own tissues.

Common slip: Justifying with "it has no energy" and stopping. The point needs the reason order depends on energy: work must be paid for continuously and heat is always leaving, so input must exceed loss.

Free-response score: 0 of 4

Free response 2 · Analyze Model or Visual Representation · 4 points

The model shows how a cell pays for its sodium–potassium pump. Arrow 1 is ATP hydrolysis; arrow 2 remakes ATP from ADP and inorganic phosphate. In a living cell the pump keeps working for as long as ATP is available. In an experiment, a cell is given a look-alike of ATP that binds the pump's ATP site but cannot be hydrolyzed; the pump stops.

A model of how a cell pays for its sodium–potassium pump. Arrow 1 is ATP hydrolysis; arrow 2 remakes ATP from ADP and inorganic phosphate (Pi).

(a) Describe what happens in arrow 1, naming the reactants and the products. (1 point)

A full-credit answer: In arrow 1, ATP hydrolysis, ATP reacts with water: ATP + H₂O → ADP + Pi. The outermost phosphate group is removed, leaving ADP, adenosine with two phosphates, and one free inorganic phosphate, and the reaction releases energy.

Check the box for each point your answer earns

ATP reacts with water (ATP + H₂O → ADP + Pi): the outermost of the three phosphate groups is removed, leaving ADP (adenosine diphosphate) and a free inorganic phosphate, Pi, and energy is released.

Accept "ATP is split by water into ADP and a phosphate". Do not award the point for products named as "ADP and energy" with no phosphate, or for ATP described as losing all three phosphates.

Common slip: Leaving water out of the reactants, or the phosphate out of the products. Hydrolysis uses water, and the phosphate that leaves is a product, Pi.

(b) Explain why the reaction in arrow 1 releases energy that the pump can use. (1 point)

A full-credit answer: The reaction releases energy because its products, ADP and inorganic phosphate, hold less energy than ATP and water did. The pump can use that energy only because the hydrolysis is coupled to its work: the outer phosphate is transferred onto the pump, phosphorylating it, and the pump changes shape and pushes Na⁺ out.

Check the box for each point your answer earns

the products, ADP and inorganic phosphate, hold less energy than ATP and water did, so the reaction as a whole releases energy; the pump can use it because the hydrolysis is coupled to the pump's work, the outer phosphate being transferred onto the pump (phosphorylation), which changes the pump's shape.

Accept "the products sit lower in energy than the reactants" for the source. Do not award the point for "energy stored in the high-energy phosphate bond is released when the bond breaks".

Common slip: Saying energy was stored in the bond to the outer phosphate and came out when the bond broke. Breaking a bond takes energy in; the release comes from the whole reaction, whose products hold less energy than its reactants.

(c) Represent arrow 2 as an equation, in words or symbols, and state what supplies the energy it needs. (1 point)

A full-credit answer: Arrow 2 is ADP + Pi + energy → ATP + H₂O, the reverse of hydrolysis. Because ATP holds more energy than ADP and Pi, remaking it needs an input of energy, and the cell supplies it from the energy released as it breaks down food molecules such as glucose.

Check the box for each point your answer earns

ADP + Pi + energy → ATP (+ H₂O), the reverse of hydrolysis, which needs an input of energy; the energy comes from the breakdown of food molecules such as glucose.

Accept the equation without water. Do not award the point for an equation with no energy input, or for the energy coming from oxygen, from Pi, or from the pump.

Common slip: Writing the equation with no energy term. Arrow 2 runs uphill; without the energy from food it does not happen.

(d) Explain how the model, together with the look-alike experiment, relates to the principle that a living system needs a continuous input of energy. (1 point)

A full-credit answer: The look-alike binds the pump but is never hydrolyzed, so no energy is released and nothing is coupled to the pump's shape change; the pump stops even though ATP-shaped molecules are present. What pays for the pump is the hydrolysis itself, and a cell holds only seconds' worth of ATP, so arrow 2 must run all the time, charging ADP back up with energy from food. The pump, and with it the cell's ion gradients and order, are kept only while energy keeps flowing in from outside.

Check the box for each point your answer earns

the look-alike binds but cannot be hydrolyzed, so no energy is released, nothing is coupled to the pump's shape change and the pump stops, showing that it is the hydrolysis, not the presence of ATP, that pays for the work; because a cell holds only seconds' worth of ATP, it must remake ATP continuously (arrow 2) from the energy in food, so the ion gradients and the rest of the cell's order can be kept only while energy keeps flowing in.

Accept an answer that links the pump stopping to "no hydrolysis, no energy" and the cycle to "ATP must be remade from food all the time". Do not award the point for restating that the pump stopped, or for treating ATP as a long-term store of energy.

Common slip: Stopping at "the pump stops because the look-alike is not ATP". The point needs the chain: no hydrolysis, no energy released, no coupled shape change; and then why the cell must keep remaking ATP from food.

Free-response score: 0 of 4

Multiple choice checked: 0 of 18 correct.

Next: Where sugar's energy goes