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Pyruvate dehydrogenase is an enzyme that converts pyruvate to acetyl- . Acetyl- 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).

Figure 1. Accumulation of acetyl- under different conditions

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

A The cytosol

B The lysosomes

C The nucleus

D The mitochondrial matrix

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Which of the following observations provides the best evidence that acetyl- negatively regulates
2.
pyruvate dehydrogenase activity?

The rate of the pyruvate dehydrogenase–catalyzed reaction is slower in the presence of a higher
A
concentration of acetyl- .

B The gene that encodes pyruvate dehydrogenase is transcribed when excess acetyl- is detected.

C The accumulation of acetyl- stops after seconds, regardless of the reaction mixture.

D Acetyl- is continuously broken down in the Krebs cycle.

3. The rate of transpiration, the flow of water through the stem, and leaf water potential are measured in a tree
during a 24-hour period under normal environmental conditions. The results from these measurements are
shown in the graphs below.

All of the following changes would be likely to decrease the rate of transpiration at 8 A.M. EXCEPT

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A causing the stomata to close

B increasing the humidity of the atmosphere

C increasing the water potential of the atmosphere

D increasing the water potential of the soil

E placing the plant in total darkness

4. A scientist designed an experiment to test an artificial membrane that mimics the phospholipid bilayer of a
cell.

The scientist built a tube that was divided by an artificial membrane and filled with distilled water. The
scientist put a known amount of a protein into the water on one side of the membrane. After some time, the
scientist measured the concentration of the protein on either side of the membrane but found that there had
been no change.

Which of the following experimental changes would allow the scientist to observe transport of a solute
across the artificial membrane?

A Increase the solute concentration in the solution

B Use a small, nonpolar solute instead of a protein

C Increase the temperature of the solution

D Add artificial aquaporins to the membrane

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5. Scientists compared the chemical structure of several molecules that various bacterial species use for
quorum sensing. Quorum sensing is an ability some bacteria have to detect the number of related cells
nearby. The chemical structure of some of these molecules found in certain species of bacteria are shown in
Figure 1.

Figure 1. The chemical structure of several molecules used for quorum sensing in three species of bacteria
Which of the following research questions would best guide an investigation of the link between the
structure of the signaling molecules and the evolution of quorum sensing?

A Do these molecules require the same receptors in each bacteria species to generate a response?

B Did these species evolve from a common ancestor that used a similar signaling molecule?

Do these species all perform the same action when the concentration of the signaling molecules is high
C
enough?

Did these species evolve from the same common ancestor that is still living today and uses the same
D
receptors?

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 of the cultures at -minute intervals as the bacteria metabolized the glucose into lactic acid. The data are
shown in Table 1.

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

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Average of Control Average of Treatment

Time (min)
( ) ( )

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

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

B The bacteria are creating a gradient to synthesize more .

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

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

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

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A Time

C Glucose concentration

D Lactic acid concentration

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

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9. What evolutionary advantage does compartmentalization of core metabolic processes offer eukaryotes?

A Evolution of the mitochondria allowed eukaryotes to perform respiration.

B With the evolution of mitochondria in eukaryotes, the Krebs cycle and electron transport chain also evolved.

Evolution of a nucleus in eukaryotes separates the processes of transcription and translation and they can be
C
regulated separately.

D A nucleus in bacteria provides separation of respiration from transcription.

10. Which of the following describes the most likely location of cholesterol in an animal cell?

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A Embedded in the plasma membrane

B Dissolved in the cytosol

C Suspended in the stroma of the chloroplast

D Bound to free ribosomes

Certain reef-building corals contain photosynthetic, symbiotic algae that have the ability to make
dimethylsulphoniopropionate ( ), a chemical involved in the marine sulfur cycle. is released into the
surrounding water, where it is converted to the gas dimethyl sulfide ( ) by microorganisms and enters the
atmosphere. Once in the atmosphere, it triggers the formation of sulfate aerosols, which induce cloud formation and
block sunlight from heating up the water.

The symbiotic algae produce when they are stressed by a high water temperature. If water temperature is
too high, corals will expel the symbiotic algae that produce . Researchers measured the amount of
produced by juvenile and adult coral and their symbionts under normal and thermally stressed conditions. The data
are shown in the graphs in Figure 1.

Figure 1: concentration in juvenile and adult corals and their symbionts in normal and thermally-stressed
conditions. Error bars represent .

The researchers also measured the density of the symbiont as well as the photosynthetic yield in adult corals at the
two temperatures. Photosynthetic yield is an index measure of energy output compared to sunlight energy input in
which larger photosynthetic yield values represent photosynthetic organisms producing more energy.

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Figure 2: Variation in symbiont density and photosynthetic yield in adult corals grown in normal and thermally-
stressed conditions. Error bars represent .

11. Which of the following best describes the production of by coral and coral symbionts?

A A negative feedback mechanism that increases the environmental change

B A negative feedback mechanism that reverses the environmental change

C A positive feedback mechanism that increases the environmental change

D A positive feedback mechanism that reverses the environmental change

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12.

Figure 1. A model of an endocrine signaling pathway showing involved body parts and hormones. =
gonadotropin-releasing hormone, = luteinizing hormone, and = follicle-stimulating hormone.
Figure 1 shows a model of the endocrine signaling pathway that regulates ovulation. Which of the following
observations would provide evidence of a positive feedback mechanism in this system?

A Estrogen from the ovaries inhibits the release of from the hypothalamus.

B Progesterone from the ovaries stimulates the thickening of the uterine lining.

C Progesterone from the ovaries inhibits the release of and from the anterior pituitary.

Estrogen from the ovaries stimulates the hypothalamus and anterior pituitary to secrete more , ,
D
and .

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13. Researchers have identified a molecule produced by Ecteinascidia turbinata, a marine invertebrate, from
which the drug trabectedin is produced. Soft-tissue tumors treated with trabectedin rapidly decrease in size.
In a preliminary study, healthy cells and tumor cells sampled from skin cancer patients treated with
trabectedin were collected, and several characteristics of the cells were observed. The observed results of the
study are shown in the table.
Characteristic Healthy Cell Tumor Cell
(in picograms)

Nuclear envelope Thinning Present

Nucleolus Absent Present

Cyclin level High Low

General appearance Round, firm Shrunken

Which of the following best explains the most likely method by which this antitumor drug works?

A Trabectedin increases the production of cyclin proteins that signal the cancer cells to enter prophase.

B Trabectedin interferes with the plasma membrane, causing it to break down and expose the to damage.

Trabectedin interferes with the duplication of during interphase and thus prevents cancer cells from
C
passing the checkpoint.

Trabectedin interferes with the regulations of cyclin proteins, causing their levels to increase and creating
D
errors in .

Certain chemicals, including sodium fluoride ( ), are capable of inhibiting specific steps of glycolysis. Figure 1
shows the steps of the glycolysis pathway, indicating where various macromolecules enter the pathway as well as
the specific reaction inhibited by .

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Figure 1. Key steps in the metabolic pathway of glucose

Based on Figure 1, the net number of molecules produced during glycolysis from the metabolism of a
14.
single glucose molecule is closest to which of the following?

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15. Scientists were interested in testing the effects of rotenone, a broad-spectrum pesticide, on a cell culture.
Cell culture was used as a control, while culture was treated with rotenone. After a period of time, the
scientists measured the concentration of several metabolites in the mitochondria of cells in both cultures.
Their results are shown in the table below.
Concentration in Culture Concentration in Culture
Metabolites

Pyruvate

Based on the data in the table, which of the following best explains the effects of rotenone on cellular
respiration?

A Rotenone acts as an inhibitor of the enzymes in the Krebs cycle.

, produced during glycolysis, is not able to enter the mitochondria because transport proteins are
B
blocked from entering.

Treated cells are not able to break down because certain enzymes of the electron transport chain are
C
inhibited.

D Rotenone acts as an allosteric inhibitor of glycolytic enzymes, thus inhibiting cellular respiration.

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Scientists investigated the effect of oxygen levels on the net rate of carbon fixation in two types of plants. The plants were grown in either well-watered soil (control) or dry soil and
16.
then exposed to either or . The net rate of fixation for both types of plants was measured. Data are shown in Figure and Figure .

Figure 1. Net rate of fixation in two types Figure 2. Net rate of fixation in two types
of plants grown in wet (control) or dry soil of plants grown in wet (control) or dry soil at
at
Which of the following statements about the rate of fixation in the two types of plants is supported by the data shown in the figures?

A At , plant type has a lower rate of fixation than plant type does in both types of soil.

At , plant type has a higher rate of fixation than plant type does in the dry soil but not in the
B
control soil.

C Plant types and have a statistically different rate of fixation in both soil types at both oxygen levels.

D The rate of fixation is the same in both types of plants in the control soil at both oxygen levels.

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17.

Figure 1. A model of epinephrine signaling

Two types of cells, alpha and beta cells, produce signaling molecules that affect blood sugar levels in
opposite ways (Figure 1). Epinephrine is a chemical, often released during periods of exercise, that
ultimately causes an increase in blood sugar levels in the body.
Based on Figure 1, which of the following best explains how exercise causes blood glucose levels to rise?

Epinephrine inhibits alpha cells, causing the release of glucagon, and activates beta cells, blocking the
A
release of insulin.

Epinephrine activates alpha cells, blocking the release of glucagon, and inhibits beta cells, causing the
B
release of insulin.

Epinephrine activates alpha cells, causing the release of glucagon, and inhibits beta cells, blocking the
C
release of insulin.

Epinephrine inhibits alpha cells, blocking the release of glucagon, and activates beta cells, causing the
D
release of insulin.

18. Carbon dioxide most likely enters a cell through which of the following processes?

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A Simple diffusion through the membrane

B Facilitated diffusion through membrane proteins

C Active transport through membrane proteins

D Active transport through aquaporins

A student peeled the skins from grapes, exposing cells with membranes that are only permeable to water and small
diffusible solutes. The student measured the mass of the peeled grapes. The student then placed each peeled grape
into one of five solutions. After 24 hours, the student removed the peeled grapes from the solutions, measured their
final mass, and calculated the percent change in mass (Table 1).

TABLE 1. PERCENT CHANGE IN MASS OF PEELED GRAPES IN SOLUTIONS

Concentration of Solution
Solution Percent Change in Mass
(weight/volume)
Distilled water

NaCl

Tap water

Grape juice

Grape soda

In a second experiment (Table 2), the student placed a peeled grape into a solution containing both small diffusible
solutes and solutes to which the membrane is impermeable (nondiffusible solutes).

TABLE 2. CONCENTRATION OF SOLUTES IN SECOND EXPERIMENT

Location Concentration of Small Diffusible Solutes Concentration of Nondiffusible Solutes

Inside grape

In solution

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19. Based on Table 1, which of the following percentages is closest to the solute concentration of the grape?

20. Which of the following best explains how the phospholipid bilayer of a transport vesicle contributes to
cellular functions?

The phospholipid bilayer allows the vesicle to fuse with the Golgi apparatus and the plasma membrane,
A
allowing the exocytosis of proteins.

The phospholipid bilayer physically connects the nuclear envelope to the rough endoplasmic reticulum, thus
B
increasing the rate of transcription and translation.

The phospholipid bilayer of a transport vesicle contains chemicals that digest the proteins made in the rough
C
endoplasmic reticulum.

The phospholipid bilayer contains enzymes that catalyze the conversion of hydrogen peroxide to water and
D
oxygen.

21. Which of the following is the strongest evidence supporting the endosymbiont hypothesis?

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A Mitochondria have their own and divide independently of the cell.

B Mitochondria can carry out hydrolytic reactions on organic molecules.

C Mitochondria have a highly folded membrane.

D Mitochondria are found in both plants and animals.

Dichlorophenolindophenol ( ) is a chemical dye. When is chemically reduced, it changes color

from blue to clear. can be used as an electron acceptor in experiments that measure the rate of electron
transport through the electron transport chain. A student performed an experiment to study the effects of a
chemical, , on photosynthesis.

The student prepared four tubes with a liquid buffer and chloroplasts that had been extracted from spinach leaves.
The student then added to three of the tubes and added to one of them. Additionally, tube was
wrapped in tin foil. The contents of each tube are shown in the table. The student then incubated each tube for
minutes and measured the absorbance ( ) of each solution at five-minute intervals. The absorbance readings of
each solution are shown in Figure 1.

Tube : Buffer and water

Tube : , buffer, and water

Tube : , buffer, and water (wrapped in foil)

Tube : , buffer, and

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Figure 1. Absorbance readings of four prepared tubes with various solutions over a -minute period.

22. Which of the following best explains how affected the reaction?

A acts as a second buffer in the reaction.

B acts as an additional carrier of electrons from photosystem to .

C acts as an additional source of electrons to the light reaction of photosynthesis.

D acts as an inhibitor to the movement of electrons within the light reaction of photosynthesis.

23. Pyruvate kinase, a key enzyme in the glycolysis pathway, is inhibited by the amino acid alanine. The ability
of alanine to inhibit the enzyme is not affected by increasing the concentration of substrate.
Which of the following best explains the mechanism by which alanine inhibits pyruvate kinase activity?

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A Alanine binds to an allosteric site of the enzyme, changing the shape of the enzyme’s active site.

B Alanine increases the enzyme-substrate binding until the enzyme becomes saturated.

C Alanine is a competitive inhibitor that reversibly binds to the active site of the enzyme.

Alanine binds to the substrate, preventing the substrate from being able to bind to the active site of the
D
enzyme.

Figure 1. Pathway activated by insulin binding to the insulin receptor

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Figure 2: Blood insulin levels in normal mice and mutant mice after exposure to glucose

Hormones are chemical signals that are released by cells in one part of the body that travel through the bloodstream
to signal cells in another part of the body. Insulin is a hormone that is released by the pancreas that induces the
uptake of glucose molecules from the bloodstream into cells. In this way, insulin lowers the overall blood glucose
levels of the body. Osteoblasts and osteoclasts are two types of bone cells that play a role in regulating blood
glucose levels (Figure 1).

Binding of insulin to the insulin receptor on osteoblasts activates a signaling pathway that results in osteoblasts
releasing a molecule, , that binds to neighboring osteoclasts. In response, the osteoclasts release protons ( )
and create an area of lower outside the cell. This low activates osteocalcin, a protein secreted in an inactive
form by osteoblasts.

The gene encodes a protein that alters the structure of the insulin receptor on osteoblasts and interferes with
the binding of insulin to the receptor. A researcher created a group of osteoblasts with an mutation that
prevented the production of a functional product (mutant). The researcher then exposed the mutant strain and
a normal strain that expresses to glucose and compared the levels of insulin in the blood near the osteoblasts
(Figure 2).

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24. Which of the following claims is most consistent with the data shown in Figure 2 ?

A expression is necessary to prevent the overproduction of insulin.

B protein does not regulate blood-sugar levels in normal mice.

C Normal mice require a higher blood concentration of insulin than mutant mice do.

D Mutant mice have a cyclical pattern of insulin secretion.

In chloroplasts, is synthesized from plus inorganic phosphate in a reaction catalyzed by

25.
synthase molecules that are embedded in the thylakoid membrane.
Which of the following statements provides evidence to support the claim that no will be synthesized
in the absence of a proton gradient across the thylakoid membrane?

Blocking electron flow from one carrier to the next in the electron transport chains blocks formation of a
A
proton gradient in the thylakoid.

Increasing the proton concentration difference across the thylakoid membrane is not associated with a
B
parallel increase in the rate of synthesis.

No is synthesized when channel proteins that allow the free passage of protons are inserted into the
C
thylakoid membrane.

No is synthesized while the Calvin cycle is synthesizing carbohydrates and using and
D
at a high rate.

For following group of questions first study the description of the situation and diagrams and then choose the one
best answer to each question following it and fill in the corresponding oval on the answer sheet.

A student studied the effects of light intensity on oxygen production in green algae. The algae were suspended in
water inside a sealed glass jar, and the jar was placed into a constant-temperature, lightproof box containing a light
source. A probe was inserted into the jar to record the concentration of oxygen. The probe was connected to a
recording device. The setup is shown below.

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The student decreased the intensity of the light hourly and recorded the corresponding changes in oxygen
concentration. The graph below shows the results from the recording device.

26. Based on the data shown, changes in the light intensity resulted in changes in the rate of which of the
following processes?

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A Excretion

B Photosynthesis

C Respiration

D Translation

E Transcription

27. Oxygen consumption can be used as a measure of metabolic rate because oxygen is

A necessary for ATP synthesis by oxidative phosphorylation

B necessary to replenish glycogen levels

C necessary for fermentation to take place

D required by all living organisms

E required to break down the ethanol that is produced in muscles

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28.

Paramecia are unicellular protists that have contractile vacuoles to remove excess intracellular water. In an
experimental investigation, paramecia were placed in salt solutions of increasing osmolarity. The rate at
which the contractile vacuole contracted to pump out excess water was determined and plotted against
osmolarity of the solutions, as shown in the graph. Which of the following is the correct explanation for the
data?

A At higher osmolarity, lower rates of contraction are required because more salt diffuses into the paramecia.

The contraction rate increases as the osmolarity decreases because the amount of water entering the
B
paramecia by osmosis increases.

The contractile vacuole is less efficient in solutions of high osmolarity because of the reduced amount of
C
ATP produced from cellular respiration.

In an isosmotic salt solution, there is no diffusion of water into or out of the paramecia, so the contraction
D
rate is zero.

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29.

Figure 1. Activity levels of two digestive enzymes over a range of

Trypsin and pepsin are enzymes that function in different areas of the digestive tract. One functions in the
stomach, where the is between and , while the other functions in the small intestines, where the
is between and .
Based on Figure 1, which of the following best describes where each enzyme functions?

A Pepsin works in the intestines because the optimal for pepsin is basic.

B Trypsin works in the stomach because the optimal for trypsin is basic.

C Pepsin works in the stomach because the optimal for pepsin is acidic.

D Trypsin works in the stomach because the optimal for trypsin is acidic.

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30.

Figure 1. Diagram of the cell cycle with key checkpoints

Which of the following describes a mutation that would lead to an increase in the frequency of
nondisjunction?

A A mutation affecting checkpoint proteins that forces cells to enter

B A mutation affecting checkpoint proteins that allows cells to divide with damage

C A mutation affecting checkpoint proteins that prevents attachment of spindle fibers

D A mutation affecting checkpoint proteins that prevents duplication of the chromosomes

31. Read each question carefully. Write your response in the space provided for each part of each question.
Answers must be written out in paragraph form. Outlines, bulleted lists, or diagrams alone are not
acceptable and will not be scored.

Students in a class are studying patterns of inheritance using genes involved in determining the body
color and wing shape of Drosophila flies. Each of the genes has only two alleles, one of which is
completely dominant to the other.

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Each student in the class performed a parental cross between a fly that is true-breeding for ebony body
and vestigial wings and a fly that is true-breeding for gray body and long wings. Each student then
crossed several pairs of the flies and determined the phenotypes of 500 of the resulting flies with
respect to body color and wing shape. The students in the class averaged their data for the frequencies
of the four possible phenotypes (Table 1).

Table 1. Averaged phenotypic data of flies

Number of Flies
Fly Phenotype

Ebony body and long wings

Ebony body and vestigial wings

Gray body and long wings

Gray body and vestigial wings

The students performed a second cross. The parental cross was between flies that are true-breeding for
gray bodies and long wings and flies that are true-breeding for ebony bodies and curly wings. They
crossed pairs of flies and determined the phenotypes of the resulting flies. The students found an
approximate ratio of flies with the dominant phenotype (gray bodies and long wings) to flies with
the recessive phenotype (ebony bodies and curly wings). Only a few of the flies expressed the dominant
phenotype of one trait and the recessive phenotype of the other trait.

(a) In the first analysis, all of the flies from the students’ crosses have the identical phenotype with
respect to body color and wing shape, but the flies have four different phenotypes. Describe how
fertilization contributes to this genetic variability.

Please respond on separate paper, following directions from your teacher.

(b) Using the template, construct an appropriately labeled graph, including error bars, to represent the
data in Table 1. Based on the data in Table 1, determine whether there is a significant difference
between the number of flies in each of the four phenotypes.

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Please respond on separate paper, following directions from your teacher.

(c) Based on the data in Table 1, describe why the dominant alleles for body color and wing shape are
the alleles that produce a gray body and long wings, respectively. Based on the data, describe why the
two genes are most likely on different chromosomes or why they are most likely on the same
chromosome. Calculate the probability of producing flies that have gray bodies and vestigial wings if a
cross is performed between one of the flies from the first analysis and a fly that is homozygous for a
gray body and vestigial wings.

Please respond on separate paper, following directions from your teacher.

(d) Predict the most likely cause of the ratio obtained by the students in the second analysis
between parental flies that are true-breeding for gray bodies and long wings and flies that are true-
breeding for ebony bodies and curly wings. Provide reasoning to justify your prediction.

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Please respond on separate paper, following directions from your teacher.

Part A

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

0 1

The response indicates that fertilization joins gametes with different allele combinations of the genes for body color
and wing shape.

Part B (i)

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

0 1 2 3

The sketched bars meet all of the criteria below.

Correct axis labeling

Correct scale and unit
Correctly plotted bar graph

Part B (ii)

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

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0 1

The response indicates that based on error bars, the number of flies with the phenotype ebony body and long wings
is the same as the number of flies with the phenotype gray body and vestigial wings. Based on the error bars, the
numbers of flies with the two other phenotypes are significantly different from each other and from those of the
first two phenotypes.

Part C (i)

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

0 1

The response indicates that flies in the largest fraction of the generation have these two traits, suggesting that
the alleles for these traits are dominant.

Part C (ii)

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

0 1

The response indicates that the ratio of phenotypes is , which is characteristic of a dihybrid cross with
two genes that are on separate chromosomes.

Part C (iii)

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

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 AP Biology Scoring Guide

Summative 2

0 1

The response indicates that the probability is 0.5. The probability of flies having gray bodies is 1.0, because the
gray color is dominant and one of the flies in the cross is homozygous for a gray body. The probability of flies
having vestigial wings is 0.5, because vestigial wings are recessive and one of the flies is homozygous for vestigial
wings, while the fly is heterozygous for this trait. The probability of flies having the two traits is calculated by
multiplying the two individual probabilities together: .

Part D (i)

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

0 1

The response indicates that in the second analysis the genes for gray or ebony body color and long or curly wings
are linked on the same chromosome.

Part D (ii)

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

0 1

The response indicates that if the genes are close together/linked, the combination of parental alleles will remain
unchanged except for a small percent of new combinations that result from limited crossing over in the flies as
they produce gametes. The flies are heterozygous, with one chromosome that has both dominant alleles and one
chromosome that has both recessive alleles. A cross between them is like a monohybrid cross. Approximately
of the flies will be homozygous dominant for both genes, will be heterozygous for the two genes, and
will be homozygous recessive for the two genes. This gives a phenotypic ratio of dominant to recessive for
both alleles.

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 AP Biology Scoring Guide

Summative 2

32. Read each question carefully. Write your response in the space provided for each part of each question.
Answers must be written out in paragraph form. Outlines, bulleted lists, or diagrams alone are not
acceptable and will not be scored.

Paramecia are single‑celled aquatic eukaryotes that can survive in a wide range of salinities. Paramecia
use contractile vacuoles to help maintain homeostasis under these varying salt conditions. The
contractile vacuoles fill with fluid and then contract to remove the fluid from the paramecia.

Scientists wished to determine the effect of salinity on contractile vacuole activity. They maintained
different groups of paramecia in water of different salinities for one month. They then measured the
average amount of fluid ejected from the contractile vacuoles in each group of paramecia each time the
vacuoles contracted (Table 1).

Table 1. Contractile Fluid Output by Paramecia Adapted to Different Salinities

Salinity of Maintenance Water ( ) Average Fluid Output of Contractile Vacuole ( )

10 (freshwater) 1.3

250 0.25

750 0.1

1000 0.05

(a) Describe how the phospholipids of a plasma membrane regulate the movement of large or polar
molecules across the membrane. Explain how osmosis will affect animal cells when the cells are placed
into an environment with a low water potential (high solute concentration) compared to the intracellular
water potential.

Please respond on separate paper, following directions from your teacher.

(b) Use the template to construct an appropriately labeled graph that represents the data in Table 1.

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 AP Biology Scoring Guide

Summative 2

Please respond on separate paper, following directions from your teacher.

(c) Describe how increasing salinity affects the amount of fluid ejected each time a contractile vacuole
contracts. Calculate the water potential ( ) of an animal cell without contractile vacuoles if water
enters the cell and creates a solute potential ( ) of . Assume that the pressure potential ( ) in the
cell is 0.

Please respond on separate paper, following directions from your teacher.

Part A (i)

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

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 AP Biology Scoring Guide

Summative 2

0 1

The response indicates that the hydrophobic tails of the phospholipid fatty acids block the movement of large and
polar molecules, which require protein channels to cross the membrane.

Part A (ii)

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

0 1

The response indicates that osmosis will cause an overall movement of water out of the cells and that the cells will
shrink.

Part B

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

0 1 2 3

The sketched bars/curve meet(s) all of the following criteria.

Correct axis labeling

Correct scale and unit
Correctly plotted bar or line graph

Part C (i)

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 AP Biology Scoring Guide

Summative 2

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

0 1

The response indicates that the amount of fluid ejected decreases with increasing solute concentration. There is a
large change in ejection volume when comparing paramecia in freshwater to paramecia in 250 milliosmoles of
. With paramecia in higher salinities, the amount of fluid ejected continues to decrease as salinity
increases.

Part C (ii)

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

0 1

The response indicates that , because pressure potential solute potential ( )equals water
potential ( ) .

33. Read each question carefully. Write your response in the space provided for each part of each question.
Answers must be written out in paragraph form. Outlines, bulleted lists, or diagrams alone are not
acceptable and will not be scored.

Many types of cancer are treated with a combination of therapies. In lung cancer, some tumors respond
well to the drug paclitaxel followed by radiation treatment. Paclitaxel is a chemical that disrupts
mitosis. Instead of spindle fibers originating from the two sides (poles) of the cell, paclitaxel-treated
cells develop three poles and then divide into three cells (tripolar division). Radiation therapy is more
effective on tumor cells that have undergone tripolar division than on cells that have undergone normal
mitosis.

Researchers treated cancer cells in the lab with different concentrations of paclitaxel for hours. The
researchers then determined the average percent of mitotic cells that were tripolar. The results are
shown in Table 1.

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 AP Biology Scoring Guide

Summative 2

TABLE 1. EFFECT OF PACLITAXEL CONCENTRATION ON PERCENT OF MITOTIC CELLS THAT WERE TRIPOLAR

Average Percent of Mitotic Cells that were Tripolar

Concentration of Paclitaxel ( )
( )

The gene encodes an enzyme that helps assemble the spindle fibers, which signals the cells to
continue through mitosis. When researchers analyzed the levels of protein in different types
of cancer cells, they found that cancer cells expressing high levels of protein had more
tripolar divisions when treated with paclitaxel, than did cancer cells expressing low levels of
protein.

(a) Describe the situations in which a normal human cell would enter the cell cycle and undergo mitotic
cell division. Explain how spindle fibers help ensure the products of mitosis are two identical cells with
a full set of chromosomes.

Please respond on separate paper, following directions from your teacher.

(b) Using the template in the space provided for your response, construct an appropriately labeled
graph that represents the data shown in Table 1. Based on the data, determine the concentration(s) of
paclitaxel that is (are) most effective in causing tripolar cell division.

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 AP Biology Scoring Guide

Summative 2

Please respond on separate paper, following directions from your teacher.

(c) Based on the data, identify the lowest level of paclitaxel that will allow for at least of the cells
to be tripolar. From the start codon through the stop codon, the length of the fully processed
is nucleotides. Calculate the number of amino acids in the polypeptide chain coded for
by the .

Please respond on separate paper, following directions from your teacher.

(d) Predict the effect of a mutation that prevents the expression of on a normal
(noncancerous) cell.

Please respond on separate paper, following directions from your teacher.

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 AP Biology Scoring Guide

Summative 2

Part A

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

0 1 2

The response includes both of the following.

· A description that cells divide by mitosis when the organism is growing or repairing tissues

· An explanation that spindle fibers attach to the center of each duplicated chromosome and assist in pulling one
chromatid to each pole of the cells so that, when the cell divides, each daughter cells contains a copy of each
chromosome

Part B

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

0 1 2 3 4

The response includes all of the following.

· Correct axes and labeling

· Correctly plotted points and scaling

· Correctly plotted error bars

· The determination that the concentration of paclitaxel that is most effective in causing tripolar cell division is
between

Part C

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

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 AP Biology Scoring Guide

Summative 2

0 1 2

The response includes both of the following.

· An identification of as the lowest level of paclitaxel that will allow for at least of the cells to be
tripolar

· A calculation of as the number of amino acids in the polypeptide chain coded for by the

Part D

Select a point value to view scoring criteria, solutions, and/or examples and to score the response.

0 1

The response includes the prediction that the cell will be unable to undergo mitosis.

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