AP Biology 3.5 (Part 1) - Cellular Respiration

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AP Biology

Unit 3: Cellular Energetics

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1
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Which two cellular organelles in eukaryotes have both electron transport systems and chemiosmotic mechanisms?

Responses

A

Ribosomes and endoplasmic reticulum

B

Chloroplasts and endoplasmic reticulum

C

Chloroplasts and mitochondria

D

Mitochondria and nuclei

E

Nuclei and Golgi apparatus

C

Chloroplasts and mitochondria

2
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According to the chemiosmotic theory (chemiosmotic coupling), the energy required to move protons from the mitochondrial matrix to the intermembrane space against a concentration gradient comes most directly from

Responses

A

photons of red or blue light

B

the hydrolysis of ATP

C

the breakdown of high-energy fatty acids in
the mitochondrial matrix

D

electrons flowing along the electron transport
chain

E

substrate-level phosphorylation

D

electrons flowing along the electron transport
chain

3
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Pyruvate dehydrogenase is an enzyme that converts pyruvate to acetyl-CoA. Acetyl-CoA is further metabolized in the Krebs cycle. A researcher measured the accumulation of acetyl- in a reaction containing pyruvate and pyruvate dehydrogenase under several different conditions (Figure 1).

The figure presents a line graph. The horizontal axis is labeled Time, in seconds, and the numbers 0 through 90, in increments of 10, are indicated. The vertical axis is labeled “Acetyl-Co A, in micromoles”, and the numbers 0 through 70, in increments of 10, are indicated. Three lines of data are shown on the graph, labeled Condition 1, Condition 2, and Condition 3. The data represented in the graph are as follows. Note that all values are approximate. The line labeled Condition 1 begins at the origin and moves upwards and to the right at a constant rate until reaching the point 60 seconds comma 60 micromoles, after which it begins to level off to move horizontally to the right at the point 70 seconds comma 65 micromoles. It continues to move horizontally until it ends at the point 90 seconds comma 65 micromoles. The line labeled Condition 2 begins at the origin and curves gradually upwards and to the right, passing through the point 50 seconds comma 30 micromoles, after which it curves even more gradually upwards and to the right, ending at the point 90 seconds comma 38 micromoles. The line labeled Condition 3 begins at the origin and moves steeply upwards and to the right until reaching the point 20 seconds comma 55 micromoles, after which it begins to curve upwards and to the right, leveling off at 30 seconds comma 65 micromoles, and moving horizontally to the right until ending at the point 90 seconds comma 65 micromoles.

Figure 1. Accumulation of acetyl-CoA under different conditions

Which of the following best describes the cellular location where pyruvate dehydrogenase is most likely active

Responses

A

The cytosol

B

The lysosomes

C

The nucleus

D

The mitochondrial matrix

Answer D

Correct. Pyruvate dehydrogenase plays a role in the Krebs cycle, which occurs in the mitochondrial matrix.

4
New cards

Pyruvate dehydrogenase is an enzyme that converts pyruvate to acetyl-CoA. Acetyl-CoA is further metabolized in the Krebs cycle. A researcher measured the accumulation of acetyl-CoA in a reaction containing pyruvate and pyruvate dehydrogenase under several different conditions (Figure 1).

The figure presents a line graph. The horizontal axis is labeled Time, in seconds, and the numbers 0 through 90, in increments of 10, are indicated. The vertical axis is labeled “Acetyl-Co A, in micromoles”, and the numbers 0 through 70, in increments of 10, are indicated. Three lines of data are shown on the graph, labeled Condition 1, Condition 2, and Condition 3. The data represented in the graph are as follows. Note that all values are approximate. The line labeled Condition 1 begins at the origin and moves upwards and to the right at a constant rate until reaching the point 60 seconds comma 60 micromoles, after which it begins to level off to move horizontally to the right at the point 70 seconds comma 65 micromoles. It continues to move horizontally until it ends at the point 90 seconds comma 65 micromoles. The line labeled Condition 2 begins at the origin and curves gradually upwards and to the right, passing through the point 50 seconds comma 30 micromoles, after which it curves even more gradually upwards and to the right, ending at the point 90 seconds comma 38 micromoles. The line labeled Condition 3 begins at the origin and moves steeply upwards and to the right until reaching the point 20 seconds comma 55 micromoles, after which it begins to curve upwards and to the right, leveling off at 30 seconds comma 65 micromoles, and moving horizontally to the right until ending at the point 90 seconds comma 65 micromoles.

Figure 1. Accumulation of acetyl-CoA under different conditions

The maximum production rate of acetyl-CoA under condition is closest to which of the following?

Responses

A

1 micromolecule/ sec

B

24 micromolecules/ sec

C

35 micromolecules/ sec

D

65 micromolecules/ sec

Answer A

Correct. This value is closest to the steepest slope of the curve under condition .

5
New cards

Pyruvate dehydrogenase is an enzyme that converts pyruvate to acetyl-CoA. Acetyl-CoA is further metabolized in the Krebs cycle. A researcher measured the accumulation of acetyl-CoA in a reaction containing pyruvate and pyruvate dehydrogenase under several different conditions (Figure 1).

The figure presents a line graph. The horizontal axis is labeled Time, in seconds, and the numbers 0 through 90, in increments of 10, are indicated. The vertical axis is labeled “Acetyl-Co A, in micromoles”, and the numbers 0 through 70, in increments of 10, are indicated. Three lines of data are shown on the graph, labeled Condition 1, Condition 2, and Condition 3. The data represented in the graph are as follows. Note that all values are approximate. The line labeled Condition 1 begins at the origin and moves upwards and to the right at a constant rate until reaching the point 60 seconds comma 60 micromoles, after which it begins to level off to move horizontally to the right at the point 70 seconds comma 65 micromoles. It continues to move horizontally until it ends at the point 90 seconds comma 65 micromoles. The line labeled Condition 2 begins at the origin and curves gradually upwards and to the right, passing through the point 50 seconds comma 30 micromoles, after which it curves even more gradually upwards and to the right, ending at the point 90 seconds comma 38 micromoles. The line labeled Condition 3 begins at the origin and moves steeply upwards and to the right until reaching the point 20 seconds comma 55 micromoles, after which it begins to curve upwards and to the right, leveling off at 30 seconds comma 65 micromoles, and moving horizontally to the right until ending at the point 90 seconds comma 65 micromoles.

Figure 1. Accumulation of acetyl-CoA under different conditions

Pyruvate dehydrogenase deficiency is a genetic disease most commonly linked to a mutation in the -subunit of the mitochondrial enzyme that causes the enzyme to cease functioning. As a result of the mutation, affected individuals build up dangerous amounts of lactic acid. Which of the following best explains the buildup of lactic acid in individuals with the mutation?

Responses

A

Cells use lactic acid to shunt electrons from pyruvate to the electron transport chain in the mitochondria.

B

Cells undergo glycolysis because there is a buildup of pyruvate in affected individuals.

C

Cells cannot transport pyruvate to the mitochondria in the absence of pyruvate dehydrogenase activity, so the pyruvate is broken down to lactic acid and ethanol.

D

Cells undergo fermentation because pyruvate cannot be metabolized to proceed into the Krebs cycle.

Answer D

Correct. Pyruvate that cannot be broken down by the Krebs cycle will be further metabolized by fermentation.

6
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During respiration, most ATP is formed as a direct result of the net movement of

Responses

A

potassium against a concentration gradient

B

protons down a concentration gradient

C

electrons against a concentration gradient

D

electrons through a channel

E

sodium ions into the cell

B

protons down a concentration gradient

7
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The figures below illustrate the similarities between ATP synthesis in mitochondria and chloroplasts.

ATP synthesis in mitochondrion. A horizontal line across the figure is labeled inner mitochondrial membrane, with four oval structures sitting across the membrane in series, touching each other, and another large structure some distance away also sitting across the membrane. There are several hydrogen ions below the membrane. NADH above the membrane pushes an electron into the first oval structure, which passes it through the next three structures. Three of the ovals are pumping hydrogen ions from above to below the membrane. The electron returns above the membrane and produces an H two O molecule with the addition of a proton and O two. On the other structure ADP plus Pi form ATP across the structure as the structure pumps hydrogen ions from below the membrane to above the membrane. ATP synthesis in chloroplast. A horizontal line across the figure is labeled thylakoid membrane, with three oval structures sitting across the membrane in series, touching each other, and another large structure some distance away also sitting across the membrane. There are several hydrogen ions below the membrane. H two O above the membrane pushes an electron into the first oval structure and releases a proton and O two. The electron passes through the next two oval structures while one oval pumps hydrogen ions from above to below the membrane. The electron returns above the membrane into a reaction of NADP plus and a proton to NADPH. On the other structure ADP plus Pi form ATP across the structure as the structure pumps hydrogen ions from below the membrane to above the membrane.

The figures can best assist in answering which of the following questions?

Responses

A

Do electron transport chains create a gradient so that ATP synthase can generate ATP molecules?

B

What are the sources of energy that drive mitochondrial and chloroplast electron transport systems?

C

What is the optimal temperature at which ATP synthase chemically converts ADP and a phosphate group into one molecule of ATP?

D

What is the evolutionary relationship between the ATP synthase in mitochondria and the ATP synthase in chloroplasts?

Answer A

This option is correct because both diagrams indicate that hydrogen ions are flowing back down their gradient through a channel in the transmembrane protein (ATP synthase) to phosphorylate ADP, forming ATP.

8
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A T P Synthase . The figure presents a mechanism of A T P synthase. There is an arrow running from the top of the mechanism through the center to the bottom of the mechanism. The top of the arrow is labeled “High, open bracket, H plus, close bracket.” The bottom of the arrow is labeled “Low, open bracket, H plus, close bracket.” On the bottom left of the mechanism, a curved arrow is present. The curve of the arrow is the only portion making contact with the mechanism. The beginning of the arrow is labeled “A D P plus P subscript i.” The end of the arrow is labeled “A T P.”

Which of the following questions will best direct an investigation of the mechanism of ATP synthase?

Responses

A

What is the source of the inorganic phosphate that is used to generate ATP from ADP?

B

Is the phosphorylation of ADP by ATP synthase dependent on the formation of a proton gradient?

C

Can ATP synthase use the energy released by phosphorylation of ADP to pump protons against a concentration gradient?

D

Can oxidative phosphorylation be uncoupled from the electron transport chain?

B

Is the phosphorylation of ADP by ATP synthase dependent on the formation of a proton gradient?

9
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To investigate bacterial metabolism, a researcher divided a population (culture) of Staphylococcus capitis bacteria into two sets of culture tubes containing glucose. The researcher added a chemical to one set of tubes and measured the pH of the cultures at 5-minute intervals as the bacteria metabolized the glucose into lactic acid. The data are shown in Table 1.

TABLE 1. AVERAGE CHANGE IN pH IN CONTROL AND TREATMENT GROUPS OVER A 40-MINUTE PERIOD

Time (min)

Average pH of Control(±2 SEx¯)

Average pH of Treatment(±2 SEx¯)

0

8.04±0.05

8.04⁢±⁢0.06

5

7.96±⁢0.03

7.91⁢±0.04

10

7.88⁢±⁢0.02

7.85±⁢0.04

15

7.82±⁢0.02

7.79⁢±⁢0.06

20

7.76⁢±⁢0.03

7.70±⁢0.04

25

7.71⁢±⁢0.04

7.67⁢±⁢0.02

30

7.63⁢±⁢0.03

7.63⁢±⁢0.02

35

7.65±⁢0.02

7.60±⁢0.02

40

7.65⁢±⁢0.01

7.59⁢±⁢0.02

Which of the following best describes the process by which the bacteria are breaking down the glucose to produce lactic acid?

Responses

A

The bacteria are breaking down sugars in the absence of oxygen.

B

The bacteria are creating a H+ gradient to synthesize more ATP.

C

The bacteria are using their mitochondria to break down glucose in the presence of oxygen.

D

The bacteria are producing CO2 in the Krebs cycle that is then converted into lactic acid.

Answer A

Correct. Lactic acid is one of the end products of anaerobic metabolism of sucrose.

10
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To investigate bacterial metabolism, a researcher divided a population (culture) of Staphylococcus capitis bacteria into two sets of culture tubes containing glucose. The researcher added a chemical to one set of tubes and measured the pH of the cultures at 5-minute intervals as the bacteria metabolized the glucose into lactic acid. The data are shown in Table 1.

TABLE 1. AVERAGE CHANGE IN pH IN CONTROL AND TREATMENT GROUPS OVER A 40-MINUTE PERIOD

Time (min)

Average pH of Control(±2 SEx¯)

Average pH of Treatment(±2 SEx¯)

0

8.04±0.05

8.04⁢±⁢0.06

5

7.96±⁢0.03

7.91⁢±0.04

10

7.88⁢±⁢0.02

7.85±⁢0.04

15

7.82±⁢0.02

7.79⁢±⁢0.06

20

7.76⁢±⁢0.03

7.70±⁢0.04

25

7.71⁢±⁢0.04

7.67⁢±⁢0.02

30

7.63⁢±⁢0.03

7.63⁢±⁢0.02

35

7.65±⁢0.02

7.60±⁢0.02

40

7.65⁢±⁢0.01

7.59⁢±⁢0.02

Which of the following was the dependent variable in the researcher’s experiment?

Responses

A

Time

B

pH

C

Glucose concentration

D

Lactic acid concentration

Answer B

Correct. The pH is being measured in this experiment and is therefore the dependent variable.

11
New cards

To investigate bacterial metabolism, a researcher divided a population (culture) of Staphylococcus capitis bacteria into two sets of culture tubes containing glucose. The researcher added a chemical to one set of tubes and measured the pH of the cultures at 5-minute intervals as the bacteria metabolized the glucose into lactic acid. The data are shown in Table 1.

TABLE 1. AVERAGE CHANGE IN pH IN CONTROL AND TREATMENT GROUPS OVER A 40-MINUTE PERIOD

Time (min)

Average pH of Control(±2 SEx¯)

Average pH of Treatment(±2 SEx¯)

0

8.04±0.05

8.04⁢±⁢0.06

5

7.96±⁢0.03

7.91⁢±0.04

10

7.88⁢±⁢0.02

7.85±⁢0.04

15

7.82±⁢0.02

7.79⁢±⁢0.06

20

7.76⁢±⁢0.03

7.70±⁢0.04

25

7.71⁢±⁢0.04

7.67⁢±⁢0.02

30

7.63⁢±⁢0.03

7.63⁢±⁢0.02

35

7.65±⁢0.02

7.60±⁢0.02

40

7.65⁢±⁢0.01

7.59⁢±⁢0.02

Which of the following graphs best represents the data in Table 1 ?

Responses

A

The figure presents a pie chart divided into four sections. Each section is approximately equal in size. Moving clockwise the sections are labeled as follows. 0 minute control, 0 minute treatment, 40 minute control, and 40 minute treatment.

B

The figure presents a graph of two lines in a plane. The horizontal axis is labeled time in minutes and the numbers 0 through 40, in increments of 5, are indicated. The vertical axis is labeled p H and the numbers 7.5 through 8.2, in increments of 0.1, are indicated. A key indicates that one line represents control p H and one line represents treatment p H. The line that represents Control p H has 9 data points as follows. 0 comma 8.04 plus or minus 0.05, 5 comma 7.96 plus or minus 0.03, 10 comma 7.88 plus or minus 0.02, 15 comma 7.82 plus or minus 0.02, 20 comma 7.76 plus or minus 0.03, 25 comma 7.71 plus or minus 0.04, 30 comma 7.63 plus or minus 0.03, 35 comma 7.65 plus or minus 0.02, and 40 comma 7.65 plus or minus 0.01. The line that represents Treatment p H has 9 data points as follows. 0 comma 8.04 plus or minus 0.06, 5 comma 7.91 plus or minus 0.04, 10 comma 7.85 plus or minus 0.04, 15 comma 7.79 plus or minus 0.06, 20 comma 7.70 plus or minus 0.04, 25 comma 7.67 plus or minus 0.02, 30 comma 7.63 plus or minus 0.02, 35 comma 7.60 plus or minus 0.02, and 40 comma 7.59 plus or minus 0.02.

C

The figure presents a bar graph with 9 stacked bars, numbered 1 through 9. The vertical axis is labeled p H and the numbers 7.3 through 8.2, in increments of 0.1, are indicated. A key indicates that the lower section of each bar represents treatment p H and the top section that extends beyond the lower section represents Control p H. Bar 1 represents both samples, extends to p H 8.04, and has two sets of error bars, one that is plus or minus 0.05 and one that is plus or minus 0.06. The lower portion of bar 2 extends to 7.91 with error bars of plus or minus 0.04 and the upper portion extends to 7.96 with error bars of plus or minus 0.03. The lower portion of bar 3 extends to 7.85 with error bars of plus or minus 0.04 and the upper portion extends to 7.88 with error bars of plus or minus 0.02. The lower portion of bar 4 extends to 7.79 with error bars of plus or minus 0.06 and the upper portion extends to 7.82 with error bars of plus or minus 0.02. The lower portion of bar 5 extends to 7.70 with error bars of plus or minus 0.04 and the upper portion extends to 7.76 with error bars of plus or minus 0.03. The lower portion of bar 6 extends to 7.67 with error bars of plus or minus 0.06 and the upper portion extends to 7.71 with error bars of plus or minus 0.04. The lower portion of bar 7 extends to 7.63 with error bars of plus or minus 0.02 and the upper portion extends to 7.63 with error bars of plus or minus 0.03. The lower portion of bar 8 extends to 7.60 with error bars of plus or minus 0.02 and the upper portion extends to 7.65 with error bars of plus or minus 0.02. The lower portion of bar 9 extends to 7.59 with error bars of plus or minus 0.02 and the upper portion extends to 7.65 with error bars of plus or minus 0.01.

D

The figure presents a box and whisker plot. The vertical axis is labeled p H and the numbers 7.5 through 8.1, in increments of 0.1, are indicated. A key indicates that one box represents Control p H, and the other box represents Treatment p H. There are two whiskers extending from each box. The box representing control p H begins at 7.65 p H and extends to 7.92. The whisker from the bottom of the box extends to 7.64. The whisker from the top of the box extends to 8.05. The box representing treatment p H begins at 7.62 and extends to 7.88. The whisker from the bottom of the box extends to 7.59. The whisker from the top of the box extends to 8.05.

Answer B

Correct. This graph shows the decreasing pH values and error bars over time for both the control and treatment cultures.

12
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To investigate bacterial metabolism, a researcher divided a population (culture) of Staphylococcus capitis bacteria into two sets of culture tubes containing glucose. The researcher added a chemical to one set of tubes and measured the pH of the cultures at 5-minute intervals as the bacteria metabolized the glucose into lactic acid. The data are shown in Table 1.

TABLE 1. AVERAGE CHANGE IN pH IN CONTROL AND TREATMENT GROUPS OVER A 40-MINUTE PERIOD

Time (min)

Average pH of Control(±2 SEx¯)

Average pH of Treatment(±2 SEx¯)

0

8.04±0.05

8.04⁢±⁢0.06

5

7.96±⁢0.03

7.91⁢±0.04

10

7.88⁢±⁢0.02

7.85±⁢0.04

15

7.82±⁢0.02

7.79⁢±⁢0.06

20

7.76⁢±⁢0.03

7.70±⁢0.04

25

7.71⁢±⁢0.04

7.67⁢±⁢0.02

30

7.63⁢±⁢0.03

7.63⁢±⁢0.02

35

7.65±⁢0.02

7.60±⁢0.02

40

7.65⁢±⁢0.01

7.59⁢±⁢0.02

Based on the data in Table 1, which of the following is the earliest time point at which there is a statistical difference in average pH between the control and treatment groups?

Responses

A

5 minutes

B

15 minutes

C

20 minutes

D

35 minutes

Answer D

Correct. The first time the pH values ±2 SEx¯ of the control and treatment do not overlap is at 35 minutes

13
New cards

To investigate bacterial metabolism, a researcher divided a population (culture) of Staphylococcus capitis bacteria into two sets of culture tubes containing glucose. The researcher added a chemical to one set of tubes and measured the pH of the cultures at 5-minute intervals as the bacteria metabolized the glucose into lactic acid. The data are shown in Table 1.

TABLE 1. AVERAGE CHANGE IN pH IN CONTROL AND TREATMENT GROUPS OVER A 40-MINUTE PERIOD

Time (min)

Average pH of Control(±2 SEx¯)

Average pH of Treatment(±2 SEx¯)

0

8.04±0.05

8.04⁢±⁢0.06

5

7.96±⁢0.03

7.91⁢±0.04

10

7.88⁢±⁢0.02

7.85±⁢0.04

15

7.82±⁢0.02

7.79⁢±⁢0.06

20

7.76⁢±⁢0.03

7.70±⁢0.04

25

7.71⁢±⁢0.04

7.67⁢±⁢0.02

30

7.63⁢±⁢0.03

7.63⁢±⁢0.02

35

7.65±⁢0.02

7.60±⁢0.02

40

7.65⁢±⁢0.01

7.59⁢±⁢0.02

According to the data, which of the following best explains the results of the experiment?

Responses

A

The pH of the treatment culture was lower than the pH of the control because the chemical increased the bacterial metabolic rate.

B

The pH of the treatment culture was higher than the pH of the control because the chemical denatured bacterial enzymes and decreased the metabolic rate.

C

The chemical increased the metabolic rate of the bacteria because it lowered the pH.

D

The chemical decreased the metabolic rate of the bacteria because it bound all available oxygen.

Answer A

Correct. An increased bacterial metabolic rate results in the production of more lactic acid, which lowers the pH of the cultures.

14
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Which of the following statements about mitochondrial chemiosmosis is NOT true?

Responses

A

A proton gradient is established across the inner membrane of the mitochondrion.

B

The potential energy released from the mitochondrial proton gradient is used to produce ATP.

C

The mitochondrial proton gradient provides energy for muscle contraction.

D

Proteins embedded in the inner mitochondrial membrane play an important role in ATP synthesis.

E

Heat energy is required to establish the electron transport chain.

E

Heat energy is required to establish the electron transport chain.

15
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Which of the following best describes the function
of the coenzymes NAD+ and FAD in eukaryotic
cellular respiration?

Responses

A

They participate in hydrolysis reactions by
accepting protons from water molecules.

B

They participate directly in the
phosphorylation of ADP to ATP.

C

They serve as final electron acceptors in the
electron transport chain.

D

They aid vitamins such as niacin in the breakdown
of glucose.

E

They accept electrons during oxidation-reduction
reactions.

E

They accept electrons during oxidation-reduction
reactions.

16
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The following questions refer to the following diagram. For each phrase or sentence, select the labeled part to which it is most closely related. Each option may be used once, more than once, or not at all for each group.

 

The figure shows a three dimensional diagram of a plant cell. A compartment labeled A is a small oval-shaped compartment near the cell membrane. It has an inner membrane that folds over on itself creating a layered structure inside the compartment. Compartment B is a medium-sized oval-shaped compartment filled with small, discreet ovals that are packed tightly and stacked, filling the compartment. There is a dark, round compartment in the cell that is partially surrounded by a maze-like structure with black dots all over the surface. C is a maze-like structure above this structure with no black dots. D is the black line that is the innermost circle surrounding the entire cell. E is a very large circular compartment that takes up about one-third of the cell diagram.

Site of conversion of chemical energy of glucose to ATP

Responses

A

B

C

D

E

A

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The figure presents a diagram of the electron transport chain and A T P synthase in the membrane of mitochondria. The mitochondrial membrane is shown with 6 embedded or associated molecules. From left to right, the molecules are as follows: a transmembrane molecule labeled 1, a molecule that is embedded in the lower half of the membrane and is labeled 2, a transmembrane molecule labeled 3, a small molecule sitting on the outer surface of the top half of the membrane, a transmembrane molecule labeled 4, and separated a bit from the first five molecules, a final transmembrane molecule that extends well below the lower half of the membrane. Many protons are shown above the membrane while somewhat fewer are shown below the membrane. Starting from the left side of the figure, N A D H is shown being converted below the membrane to N A D with a positive charge and two protons by molecule 1. Electrons from N A D H are being transferred to molecule 1. A proton is also shown moving through molecule 1 from below the membrane to above the membrane. Moving to the right, F A D H 2 is shown being converted below the membrane to F A D and two protons by molecule 2. Electrons from F A D H 2 are being transferred to molecule 2. Moving to the right, protons are shown moving through molecules 3 and 4 from below the membrane to above the membrane. A long arrow runs from molecule 1 through molecule 2, then molecule 3, then the unnumbered small molecule on the outer surface of the top of the membrane, then molecule 4, and finally to the space below the membrane. The end of the arrow points to another arrow below the membrane showing that O 2 plus 4 protons are converted to 2 H 2 O. On the far right of the membrane, a proton is shown moving through the final transmembrane molecule from above the membrane to below the membrane. A second arrow below the membrane touches the base of this transmembrane molecule and shows that A D P plus P I are converted to A T P.

Figure 1. Diagram of the electron transport chain and ATP synthase in the membrane of mitochondria

On average, more ATP can be produced from an NADH molecule than can be produced from a molecule of FADH2.

Based on Figure 1, which of the following best explains the difference in ATP production between these two molecules?

Responses

A

NADH contributes more electrons to the electron transport chain than FADH2 does and therefore provides more energy to pump protons.

B

The electrons of FADH2 are transferred through three complexes of the electron transport chain whereas those of NADH are transferred through all four complexes.

C

FADH2 contributes more protons to the mitochondrial matrix, which decreases the proton gradient.

D

The protons contributed by FADH2 are combined with O2 to make water and are not pumped across the membrane.

Answer B

Correct. The electrons from FADH2 are contributed to complex II rather than complex I and therefore are transferred through fewer complexes and have a lesser effect on the proton gradient than the electrons from NADH do.

18
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The figure shows two models of a mitochondrion. The first diagram is labeled “A Mitochondrion of Certain Fat Cells Under Normal Conditions,” and it shows a mitochondrion with ten folds of the inner membrane. The second diagram is labeled “A Mitochondrion of Certain Fat Cells Under Cold-Induced Stress,” and it shows a mitochondrion with approximately twice as many folds of the inner membrane.

Figure 1. Model of a mitochondrion in cells under normal conditions and in cells experiencing cold-induced stress

In a certain type of fat cells, cold temperatures trigger a series of events that include changes to the structure of the mitochondria. A model of a mitochondrion from these fat cells under normal environmental conditions and when experiencing cold-induced stress is shown in Figure 1.

Which of the following best explains how the changes in mitochondrial structure observed when the cell experiences cold-induced stress affect cellular respiration?

Responses

A

The volume of the mitochondrial intermembrane space decreases, preventing the reactions of the Krebs cycle from occurring.

B

The surface area of the inner membrane decreases, preventing the transport of oxygen molecules into the mitochondria.

C

The volume within the mitochondria increases, providing more space for the reactions of glycolysis to take place.

D

The surface area of the inner membrane increases, providing more space for the components of the electron transport chain.


Answer D

Correct. The increased folding of the inner membrane in cold-induced mitochondria results in an increase in surface area. This allows for more components of the electron transport chain (ETC) that is localized to the inner mitochondrial membrane. This in turn enables an increased rate of cellular respiration that can help maintain the body temperature of an endotherm exposed to cold temperatures.

19
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Newborn babies and hibernating animals contain a large amount of brown adipose (fat) tissue (BAT). Certain proteins in the BAT cells increase the permeability of the inner mitochondrial membrane to protons, disrupting the proton gradient.

Which of the following best predicts the effect of disrupting the proton gradient in BAT?

Responses

A

The ATP of the matrix will increase, allowing the production of more ATP per gram of substrate.

B

The pH of the intermembrane space will decrease, allowing a steeper proton gradient to form.

C

Electron transport and oxidative phosphorylation will be decoupled, generating more heat but less ATP.

D

The number of protons available to pass through ATP synthase will increase, resulting in more ATP.

Answer C

Correct. The protons pumped into the matrix because of electron transport will diffuse back into the intermembrane space and will not be available for oxidative phosphorylation. As a result, more energy will be converted to heat and less will be available for production.

20
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Melanocytes are skin cells that can become cancerous and develop into a cancer known as melanoma. Some cancerous melanocytes have developed resistance to the drugs currently used to treat melanoma. As a result, researchers are investigating the effects of a new compound (drug X) on four different melanoma cell lines. Researchers analyzed cell survival in two cell lines (Figure 1) and oxygen consumption in the presence of drug X in all four cell lines (Figure 2). Figure 3 shows the proposed mechanism by which drug X affects cells.

Figure 1 presents a graph with three curves showing the percent survival of normal melanocytes and cancerous melanocyte (melanoma) lines 1 and 2 after treatment with different concentrations of drug X. The horizontal axis is labeled Concentration of Drug X, micromolar, and the numbers 0 through 200, in increments of 50, are indicated. The vertical axis is labeled Percent Cell Survival and the numbers 0 through 100, in increments of 10, are indicated. All data points are approximate. A curve representing the survival of normal melanocytes starts at the point 0 comma 100 and moves horizontally and only very slightly downward to end at the point 180 comma 98. A second curve representing cell line 1 starts at 0 comma 100 and moves steeply down and to the right, passing through data points that include 25 comma 62, 40 comma 19 and 52 comma 10. The curve starts to level off as it approaches the horizontal axis with points at 120 comma 2 and finally 170 comma 4. A third curve representing cell line 2 starts at the point 0 comma 92 and moves approximately horizontally to the point 20 comma 92, after which it moves steeply down and to the right, passing through data points that include 30 comma 71 and 50 comma 42. The curve then moves more slowly down as it continues to the right, passing through data points that include 80 comma 30, 105 comma 28, 140 comma 27 and finally 180 comma 26.

Figure 1. Percent survival of normal melanocytes and cancerous melanocyte (melanoma) lines 1 and 2 after treatment with different concentrations of drug

Figure 2 presents a bar graph of oxygen consumption per cell in four melanoma lines after treatment with either solvent alone or solvent containing drug x. The horizontal axis is labeled Cell Line and the numbers 1 through 4 are shown. The vertical axis is labeled “Picomoles O 2 Consumed per Cell after Treatment”, and the numbers 0.0 through 0.6, in increments of 0.2 are indicated. All data points are approximate. For cell line 1, the bar representing solvent alone reaches 0.2 with an error bar from 0.18 to 2.2 and the bar representing solvent with drug reaches 0.05 with an error bar from 0.025 to 0.075. For cell line 2, the bar representing solvent alone reaches 0.32 with an error bar from 0.26 to 0.38 and the bar representing solvent with drug reaches 1.0 with an error bar from 0.065 to 1.035. For cell line 3, the bar representing solvent alone reaches 0.26 with an error bar from 0.2 to 0.32 and the bar representing solvent with drug reaches 0.09 with an error bar from 0.08 to 1.0. For cell line 4, the bar representing solvent alone reaches 0.23 with an error bar from 0.18 to 0.28 and the bar representing solvent with drug reaches 0.05 with an error bar from 0.01 to 0.06.

Figure 2. Oxygen consumption per cell in four melanoma lines after treatment with either solvent alone or solvent containing drug X. Error bars represent ±2 SEx¯.

Figure 3 shows a model of a pathway leading to cell survival, growth, and proliferation and the likely effect of drug X. The components of the pathway are Molecule A, Complex B, and Molecule C. A description of the pathway is as follows: Molecule A is represented on the far left as an oval. An arrow points from Molecule A to Complex B that is represented as a large oval plus two smaller ovals. A horizontal line with a perpendicular bar at its right end is positioned between Complex B and Molecule C to its immediate right. Molecule C is also represented as an oval. An arrow points from Molecule C to the words Cell survival, growth and proliferation. Below the pathway, two arrows extend from a box labeled “Drug X” to Complex B. A note next to Drug X states “Drug X leads to activation of complex B.”

Figure 3. Pathway leading to cell survival, growth, and proliferation and the likely effect of drug X

Based on the information presented, which of the following best explains why the researchers measured oxygen consumption as an indicator of the effectiveness of drug X?

Responses

A

Oxygen provides the source of electrons for cellular respiration and is necessary for energy production.

B

Oxygen consumption increases the mutation rate and causes cells to become cancerous.

C

Oxygen activates apoptosis, which results in the death of melanoma cells.

D

Oxygen accepts electrons in oxidative phosphorylation, a process necessary for melanoma cell survival.

Answer D

Correct. Melanoma cells require oxygen for normal metabolism, so oxygen consumption is a good indicator of metabolism.