Biology  by Bradford
support
AP Biology · Course and Exam Description · Unit 3

Unit 3: Cellular Energetics. Practice questions with scoring guidelines.

195 original AP-style questions on Unit 3, written to the CED learning objectives: 137 multiple-choice, 58 free-response. Below is a 20-mark practice set built from them, ready to assign as a unit check or homework, or to sit yourself and score against the guidelines. Print it, project it, or build a fresh one.

Topics in this unit

  1. 3.1 Enzymes 38 questions
  2. 3.2 Environmental Impacts on Enzyme Function 49 questions
  3. 3.3 Cellular Energy 34 questions
  4. 3.4 Photosynthesis 46 questions
  5. 3.5 Cellular Respiration 55 questions

In the bank for Unit 3

  • 137 multiple-choice
  • 58 free-response

Every question is original and tagged to a learning objective and science practice.

Make your own

The practice set

Take it online → Build a fresh paper Paper code BbB-IAAEAAAAABQAGLVY
AP Biology · practice quiz
MCQ + FRQ · unit practice · not an exam format
30 minutes20 marks

Paper code: this paper was generated, so the code is its recipe. Enter it at biologybybradford.com/exam-maker to rebuild this exact paper and its markscheme.

Covers Unit 3 (Cellular Energetics)
Name:
1.

An enzyme was extracted from the roots of bean seedlings, and its initial reaction rate was measured at pH 7 and 25 °C in buffer containing 0, 50, 100, or 200 mM sodium chloride (NaCl). Six replicate assays were run at each concentration. The figure shows the mean rates with error bars of ±2SE. The researchers' null hypothesis was that NaCl concentration has no effect on the rate. Comparing 0 mM with 50 mM, they conclude that 50 mM NaCl has no effect on the enzyme. Which statement best evaluates this conclusion?

[1]
Initial Reaction Rate of the Root Enzyme at Four NaCl Concentrations 0 5 10 15 20 25 30 0 50 100 200 NaCl concentration (mM) Mean initial rate (μmol product per min) error bars: ± 2SE, n = 6
  1. It goes too far: the overlap means only that the null hypothesis cannot be rejected, and more replicates might reveal a small effect.
  2. It is wrong, because the mean rate at 50 mM is higher than the mean at 0 mM, which shows that 50 mM NaCl activates the enzyme.
  3. It cannot be tested, because the ±2SE intervals at 50 mM and 100 mM overlap, so the effects of these two concentrations are alike.
  4. It is supported, because the overlap of the ±2SE intervals at 0 mM and 50 mM shows that the two true mean rates are equal.
2.

Which statement best compares the roles of photosynthesis and cellular respiration in the flow of energy through a plant cell?

[1]
  1. Both release energy from organic molecules.
  2. Photosynthesis releases energy from sugars.
  3. Photosynthesis stores it; respiration releases it.
  4. Both convert ATP directly into light energy.
3.

During the light-dependent reactions, protons (H⁺) accumulate on one side of the thylakoid membrane, and their movement powers ATP synthesis. Where do the protons accumulate, and in which direction do they then flow through ATP synthase?

[1]
  1. Outside the chloroplast; inward via aquaporins.
  2. In the matrix; they flow out to the cytosol.
  3. In the stroma; into the lumen via the chain.
  4. In the thylakoid lumen; back to the stroma.
4.

Muscle cells carrying out lactic acid fermentation and yeast cells carrying out alcoholic fermentation each break down one molecule of glucose without oxygen. Which comparison of the two pathways is correct?

[1]
  1. Muscle cells net about 2 ATP; yeast cells net about 30 ATP.
  2. Both net about 2 ATP; only the yeast cells release CO₂.
  3. Both net about 2 ATP; only the muscle cells release CO₂.
  4. Both release CO₂; only the yeast cells net any ATP at all.
5.Analyze Data

In the early 1960s, biologists knew that electron transport in mitochondria is linked to the synthesis of ATP, but two models of the link competed. In model 1, the energy released by electron transport is stored in a high-energy compound formed at a carrier of the chain, and this compound then reacts with ADP to form ATP. In model 2, the energy released by electron transport is used to move H⁺ across the inner mitochondrial membrane, and the flow of H⁺ back across the membrane through ATP synthase powers the synthesis of ATP.

To test the models, researchers suspended isolated mitochondria in a weakly buffered solution that contained a respiratory substrate but no ADP and no O₂. The outer mitochondrial membrane lets H⁺ pass freely, so any H⁺ that crosses the inner membrane out of the matrix enters the solution, where an electrode measured it. At time 0, a small volume of liquid containing 5.0 nmol of O₂ was injected, enough for electron transport to run for only a few seconds. In treatment 1 the mitochondria were intact. In treatment 2 their inner membranes had first been broken open by a detergent that does not damage the proteins of the chain, and the injected O₂ was used up as fast as in treatment 1. The results are shown in Figure 1.

H⁺ added to the solution after an injection of O₂ 0 30 60 90 120 150 180 −20 0 20 40 60 80 100 120 time after O₂ injected / s H⁺ added to the solution / nmol 5.0 nmol O₂ injected at 0 s treatment 1: intact treatment 2: broken open
(A) Using Figure 1, calculate the number of H⁺ that entered the solution for each molecule of O₂ used by the intact mitochondria in treatment 1.
(B) Explain how the results of treatment 1 support model 2 rather than model 1.
(C) A scientist who favored model 1 suggested that the H⁺ came from reactions of the chain that release H⁺ into whatever solution surrounds them, rather than from H⁺ being moved across a membrane. Explain how the results of treatment 2 argue against this suggestion.
(D) In treatment 3, the researchers repeat treatment 1 with ADP and inorganic phosphate (Pᵢ) added to the solution. Predict whether the H⁺ in the solution will fall back toward 0 nmol faster or more slowly than in treatment 1, and justify your prediction using model 2.
6.Analyze Data

Enzyme H, from a soil bacterium, hydrolyzes compounds in which a ring-shaped group is joined by an ester bond to a hydrocarbon tail. When such a substrate binds, its tail lies in a narrow pocket of the active site and the ester bond sits at the mouth of the pocket, where it is broken. At the closed end of the pocket is amino acid 147, which has a large nonpolar side chain. Researchers made a mutant form of the enzyme, H-147G, in which amino acid 147 is replaced by glycine, whose side chain is a single hydrogen atom. They measured the initial rate of hydrolysis by each form with substrates whose tails were 2 to 12 carbon atoms long, each substrate supplied separately at the same concentration, with five replicate assays per substrate. The results are shown in Table 1.

Table 1. Initial Rate of Hydrolysis by Two Forms of Enzyme H with Substrates of Different Tail Lengths (μmol per min per mg of enzyme; mean ± 2SE, n = 5)
Tail length (carbon atoms)Wild type (mean ± 2SE)H-147G (mean ± 2SE)
26 ± 13 ± 1
438 ± 39 ± 2
6112 ± 627 ± 3
824 ± 394 ± 7
104 ± 186 ± 6
121 ± 115 ± 3
(A) Describe how the initial rate of the wild-type enzyme changes as the tail length of the substrate increases from 2 to 12 carbon atoms.
(B) Determine whether the rate of H-147G with the 8-carbon substrate differs significantly from its rate with the 10-carbon substrate. Justify your answer.
(C) Explain, in terms of the shape of the active site, why H-147G hydrolyzes substrates with 8-carbon and 10-carbon tails fastest, whereas the wild-type enzyme hydrolyzes the substrate with a 6-carbon tail fastest.
(D) Another mutant form of enzyme H has, at position 147, an amino acid whose side chain is larger than that of the wild type. Predict the tail length that this form would hydrolyze fastest, and justify your prediction.
7.Conceptual Analysis

Glycolysis breaks glucose down to pyruvate and makes a small amount of ATP. It is found in members of every major group of bacteria, archaea, and eukaryotes that has been studied, with largely the same intermediates formed in the same order. O₂-releasing photosynthesis, in contrast, is carried out only by cyanobacteria and by eukaryotes that have chloroplasts, such as plants and algae; it is absent from archaea, from most groups of bacteria, and from animals and fungi.

(A) Describe one feature of glycolysis that allows a cell to make ATP by glycolysis in an environment that has no O₂.
(B) Predict which arose earlier in the history of life, glycolysis or O₂-releasing photosynthesis.
(C) Justify your prediction in part B using the distribution of the two processes described.
(D) Explain how the evolution of O₂-releasing photosynthesis changed the way in which later cells could obtain energy from glucose.
8.Analyze Data

The body temperature of a desert lizard follows the temperature of its surroundings. At dawn its body is at about 20 °C; it then basks in the sun and warms up, and in the middle of the day its body temperature is about 40 °C. Researchers extracted an enzyme of glycolysis from the leg muscle of the lizard and measured the amount of product formed in the first 2 minutes of the reaction at a range of temperatures. Every assay contained the same amount of enzyme and an excess of substrate in buffer at pH 7.2. The means of four assays at each temperature are shown in Table 1.

Table 1. Product Formed in the First 2 Minutes by an Enzyme from Lizard Leg Muscle (mean of 4 assays)
Temperature (°C)Product formed in 2 minutes (μmol)
151.0
201.5
252.4
303.6
355.2
406.8
457.4
502.2
(A) Calculate the initial reaction rate, in μmol per minute, at the lizard's body temperature at dawn (20 °C) and in the middle of the day (40 °C).
(B) Calculate the percent increase in the initial reaction rate when the lizard's body warms from 20 °C to 40 °C.
(C) Assume that the lizard can run fast enough to catch prey only once the initial rate of this enzyme, as measured in the assay, reaches 2.0 μmol per minute, and that the rate changes linearly between the temperatures tested. Calculate the lowest body temperature at which the lizard could catch prey.
(D) On hot afternoons the lizard moves into the shade before its body temperature passes about 45 °C. Using the data, explain why staying in the sun until its body reached 50 °C would harm the lizard.

Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · AP® is a trademark registered by the College Board, which was not involved in and does not endorse this site.
Rebuild or edit this exact paper (and its markscheme): biologybybradford.com/exam-maker?code=BbB-IAAEAAAAABQAGLVY

Show the markscheme

Markscheme BbB-IAAEAAAAABQAGLVY

Each point is worth 1 point and is credited independently. Accept any one of the listed alternatives per point; ( ) marks optional wording and / separates interchangeable wording.

1. 3.2.B [1]
  • A: the ±2SE intervals at 0 mM (22.0 to 26.0) and 50 mM (23.5 to 28.5) overlap, so the difference is not statistically significant and the null hypothesis cannot be rejected for 50 mM; that is not evidence that the effect is zero. With more replicates the standard errors would shrink, the error bars would narrow, and a small real effect could then be detected. The higher mean at 50 mM is within the error, the 50 mM and 100 mM intervals (17.0 to 21.0) do not overlap, and the falls at 100 mM and 200 mM (7.5 to 10.5) are significant, consistent with high ion concentrations disrupting the interactions that hold the enzyme in its active shape.
2. 3.3.A [1]
  • C: photosynthesis converts light energy into chemical energy stored in organic molecules, while respiration releases that stored energy to generate ATP.
3. 3.4.B [1]
  • D: the electron transport chain pumps H⁺ into the thylakoid lumen, and their flow back to the stroma through ATP synthase drives ATP synthesis (photophosphorylation).
4. 3.5.A [1]
  • B: in both pathways ATP comes only from glycolysis (net about 2 ATP per glucose); converting pyruvate to ethanol releases CO₂, whereas converting pyruvate to lactate releases none.
5. Analyze Data 3.5.B
(A)
5.A Using Figure 1, calculate the number of H⁺ that entered the solution for each molecule of O₂ used by the intact mitochondria in treatment 1.
Accept one of the following:
  • 100 nmol of H⁺ ÷ 5.0 nmol of O₂ = 20 H⁺ per O₂
  • About 20 H⁺ for each O₂ (accept 19 to 21)

Total for part (A): 1 point

(B)
6.B Explain how the results of treatment 1 support model 2 rather than model 1.
Accept one of the following:
  • Model 2 predicts that electron transport moves H⁺ out of the matrix across the inner membrane, and in treatment 1 a large amount of H⁺ (about 20 per O₂) entered the solution as soon as the O₂ let electrons flow along the chain; model 1 involves no movement of H⁺ across a membrane, so it does not predict this release.
  • As soon as electron transport could run, H⁺ was moved out of the mitochondria, which is the first step of model 2; in model 1 the energy passes to ADP through a chemical compound, which gives no reason for H⁺ to leave the matrix.

Total for part (B): 1 point

(C)
6.C A scientist who favored model 1 suggested that the H⁺ came from reactions of the chain that release H⁺ into whatever solution surrounds them, rather than from H⁺ being moved across a membrane. Explain how the results of treatment 2 argue against this suggestion.
Accept one of the following:
  • In treatment 2 the chain used O₂ as fast as in treatment 1 but no H⁺ entered the solution; if the reactions of the chain simply released H⁺, it would have appeared whether or not the membrane was intact, so the H⁺ release depends on an intact inner membrane, as expected if H⁺ is moved across it.
  • With the inner membranes broken open there is no closed compartment for H⁺ to be moved out of, and no H⁺ appeared even though electron transport still ran, so the H⁺ in treatment 1 came from transport across the intact membrane, not from the reactions themselves.

Total for part (C): 1 point

(D)
6.E In treatment 3, the researchers repeat treatment 1 with ADP and inorganic phosphate (Pᵢ) added to the solution. Predict whether the H⁺ in the solution will fall back toward 0 nmol faster or more slowly than in treatment 1, and justify your prediction using model 2.
Accept one of the following:
  • Faster: with ADP and Pᵢ present, ATP synthase can make ATP, so H⁺ flows back down its gradient into the matrix through ATP synthase as well as by the slow leak seen in treatment 1, and the H⁺ outside falls more quickly.
  • More quickly, because in model 2 ATP synthesis is powered by H⁺ returning through ATP synthase; with no ADP that route is closed, but with ADP and Pᵢ the H⁺ moved out by the chain can return through it while ATP is made.

Total for part (D): 1 point

Total for question 5: 4 points

6. Analyze Data 3.1.A
(A)
4.B Describe how the initial rate of the wild-type enzyme changes as the tail length of the substrate increases from 2 to 12 carbon atoms.
Accept one of the following:
  • The rate increases from 6 to 112 μmol per min per mg as the tail lengthens from 2 to 6 carbons, is highest with the 6-carbon tail, and then falls sharply with longer tails, to 24 at 8 carbons and about 1 at 12 carbons.
  • It rises to a peak at a tail length of 6 carbon atoms and then drops steeply, so tails longer or shorter than 6 carbons are hydrolyzed much more slowly.

Total for part (A): 1 point

(B)
5.B Determine whether the rate of H-147G with the 8-carbon substrate differs significantly from its rate with the 10-carbon substrate. Justify your answer.
Accept one of the following:
  • No, the difference is not statistically significant: the ±2SE intervals, 87 to 101 (8 carbons) and 80 to 92 (10 carbons), overlap (from 87 to 92).
  • Not significant, because the error intervals overlap, so the higher mean with the 8-carbon tail could be due to chance.

Total for part (B): 1 point

(C)
6.C Explain, in terms of the shape of the active site, why H-147G hydrolyzes substrates with 8-carbon and 10-carbon tails fastest, whereas the wild-type enzyme hydrolyzes the substrate with a 6-carbon tail fastest.
Accept one of the following:
  • Replacing the large side chain at the closed end of the pocket with a hydrogen atom makes the pocket deeper, so tails of 8 to 10 carbons now fit fully and fill it with the ester bond at the mouth, while a 6-carbon tail fills the deeper pocket less well and binds less well; in the wild type the large side chain blocks the pocket after about 6 carbons, so longer tails cannot fit fully and their bond cannot sit at the mouth.
  • The active site is shaped to fit a tail of one particular length. The large side chain of amino acid 147 sets the depth of the wild-type pocket to fit about 6 carbons; with glycine there the pocket is longer, so its shape now matches tails of about 8 to 10 carbons, and shorter tails no longer fit it closely.

Total for part (C): 1 point

(D)
6.E Another mutant form of enzyme H has, at position 147, an amino acid whose side chain is larger than that of the wild type. Predict the tail length that this form would hydrolyze fastest, and justify your prediction.
Accept one of the following:
  • A tail shorter than 6 carbons, such as 4 carbons: a larger side chain at the closed end makes the pocket shallower, so a 6-carbon tail can no longer fit fully with its ester bond at the mouth, while a shorter tail fills the shallower pocket.
  • About 4 carbons (shorter than for the wild type), because the bigger side chain takes up more of the pocket, so only shorter tails match the shape of the active site.

Total for part (D): 1 point

Total for question 6: 4 points

7. Conceptual Analysis 3.3.B, 3.4.A, 3.5.A
(A)
1.A Describe one feature of glycolysis that allows a cell to make ATP by glycolysis in an environment that has no O₂.
Accept one of the following:
  • None of the reactions of glycolysis uses O₂: ATP is made directly by enzymes in the cytosol, and the NAD⁺ that glycolysis needs can be regenerated by fermentation
  • Glycolysis does not require O₂ (or a membrane-bound organelle such as a mitochondrion), so it can run without O₂ as long as NAD⁺ is regenerated, for example by fermentation

Total for part (A): 1 point

(B)
6.E Predict which arose earlier in the history of life, glycolysis or O₂-releasing photosynthesis.
Accept one of the following:
  • Glycolysis arose earlier
  • Glycolysis arose before O₂-releasing photosynthesis

Total for part (B): 1 point

(C)
6.C Justify your prediction in part B using the distribution of the two processes described.
Accept one of the following:
  • Glycolysis is found, with the same steps, in every major group of bacteria, archaea, and eukaryotes, so it was most likely present in their common ancestor and inherited by all of them; O₂-releasing photosynthesis is found in only a few groups and is absent from whole groups such as the archaea, so it most likely arose later, within one lineage
  • A process shared by all groups is best explained by inheritance from the ancestor of all of them, whereas a process confined to a few groups most likely evolved after those groups had separated; if O₂-releasing photosynthesis had been present in the common ancestor, it would have to have been lost separately in many lineages

Total for part (C): 1 point

(D)
1.C Explain how the evolution of O₂-releasing photosynthesis changed the way in which later cells could obtain energy from glucose.
Accept one of the following:
  • O₂ released by photosynthesis built up in the environment, so O₂ became available as the final electron acceptor of an electron transport chain; cells could then oxidize the pyruvate from glycolysis completely (Krebs cycle and oxidative phosphorylation) and make far more ATP from each glucose than glycolysis alone
  • Once photosynthesis had added O₂ to the air and water, cells could carry out aerobic respiration, passing electrons from NADH to O₂ and using the resulting H⁺ gradient to make much more ATP per glucose

Total for part (D): 1 point

Total for question 7: 4 points

8. Analyze Data 3.2.B, 3.2.A
(A)
5.A Calculate the initial reaction rate, in μmol per minute, at the lizard's body temperature at dawn (20 °C) and in the middle of the day (40 °C).
Accept one of the following:
  • 20 °C: 1.5 ÷ 2 = 0.75 μmol per minute; 40 °C: 6.8 ÷ 2 = 3.4 μmol per minute.
  • 0.75 μmol/min at 20 °C and 3.4 μmol/min at 40 °C.

Total for part (A): 1 point

(B)
5.A Calculate the percent increase in the initial reaction rate when the lizard's body warms from 20 °C to 40 °C.
Accept one of the following:
  • (3.4 − 0.75) ÷ 0.75 × 100 = 353 % increase (accept 350 % to 355 %).
  • (6.8 − 1.5) ÷ 1.5 × 100 ≈ 353 %.

Total for part (B): 1 point

(C)
5.A Assume that the lizard can run fast enough to catch prey only once the initial rate of this enzyme, as measured in the assay, reaches 2.0 μmol per minute, and that the rate changes linearly between the temperatures tested. Calculate the lowest body temperature at which the lizard could catch prey.
Accept one of the following:
  • The rate is 3.6 ÷ 2 = 1.8 μmol per minute at 30 °C and 5.2 ÷ 2 = 2.6 at 35 °C; 2.0 is 0.2 ÷ 0.8 = one quarter of the way from 1.8 to 2.6, so 30 + 0.25 × 5 = about 31 °C (accept 31 °C to 31.5 °C).
  • About 31 °C: 2.0 μmol per minute corresponds to 4.0 μmol in 2 minutes, which lies one quarter of the way between 3.6 (30 °C) and 5.2 (35 °C).

Total for part (C): 1 point

(D)
6.D On hot afternoons the lizard moves into the shade before its body temperature passes about 45 °C. Using the data, explain why staying in the sun until its body reached 50 °C would harm the lizard.
Accept one of the following:
  • Between 45 °C and 50 °C the rate falls from 3.7 to 1.1 μmol per minute (by about 70 %), because heat disrupts the interactions that hold the enzyme in its folded shape, so its active site no longer binds substrate (it is denatured); glycolysis in the muscle would slow, so less ATP would be made for running.
  • Above about 45 °C the enzyme is denatured: its active site loses its shape, so the rate at 50 °C (1.1 μmol per minute) is lower even than at 25 °C (1.2), and the muscle could not break down glucose fast enough to supply energy for movement.

Total for part (D): 1 point

Total for question 8: 4 points

Other AP Biology units

All eight units →  ·  IB Biology subtopics →  ·  Open the exam maker →