AP Biology

Practice Exam #2
and Notes
For the
Spring 2020
Exam
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2 AP Biology Practice Exam

Contents

I. Practice Exam
Exam Content and Format. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Administering the Practice Exam. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Answer Sheet for Multiple-Choice Section . . . . . . . . . . . . . . . . . . . . . . . . . . 6
AP® Biology Practice Exam. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7

II. Notes on the Practice Exam

Multiple-Choice Section. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
Answer Key and Question Alignment to Course Framework. . . . . . . . . 101
Free-Response Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

Contact Us. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

AP Biology Practice Exam 3

 Practice Exam

Exam Content and Format

The AP Biology Exam is approximately 3 hours in length. There are two sections.
• Section I is 1 hour, 30 minutes and consists of 60 multiple-choice questions,
accounting for 50 percent of the final score.

• Section II is 1 hour, 30 minutes and consists of 2 long free-response

questions and 4 short free-response questions, accounting for 50 percent
of the final score.

Administering the Practice Exam

This section contains instructions for administering the AP Biology Practice
Exam. You may wish to use these instructions to create an exam situation that
resembles an actual administration. If so, read the indented, boldface directions
to the students; all other instructions are for administering the exam and need
not be read aloud. Before beginning testing, have all exam materials ready for
distribution. These include test booklets and answer sheets. (Reminder: Final
instructions for every AP Exam are published in the AP Exam Instructions book.)

SECTION I: Multiple Choice

When you are ready to begin Section I, say:

Section I is the multiple-choice portion of the exam. Mark all of your

responses on your answer sheet, one response per question. If you need
to erase, do so carefully and completely. Your score on the multiple-
choice section will be based solely on the number of questions answered
correctly. You may use a four-function (with square root), scientific, or
graphing calculator. Are there any questions?

You have 1 hour and 30 minutes for this section. Open your Section I
booklet and begin.

Note Start Time here ________ . Note Stop Time here ________ . After 1 hour
and 20 minutes, say:

There are 10 minutes remaining.

After 10 minutes, say:

Stop working. I will now collect your Section I booklet and multiple-
choice answer sheet.

There is a 10-minute break between Sections I and II.

4 AP Biology Practice Exam

SECTION II: Free Response

After the break, say:

Section II is the free-response portion of the exam.

You have 1 hour and 30 minutes to answer the questions. You are
responsible for pacing yourself, and may proceed freely from one
question to the next. Write your answers on the lined pages provided for
each question. If you need more paper during the exam, raise your hand.
At the top of each extra piece of paper you use, be sure to write your
name and the number of the question you are working on. Are there any
questions? Open your Section II booklet and begin.

Note Start Time here ________ . Note Stop Time here ________ . After 1 hour
and 20 minutes, say:
There are 10 minutes remaining.

After 10 minutes, say:

Stop working and close your exam booklet. Put your exam booklet on
your desk, face up. Remain in your seat, without talking, while the exam
materials are collected.

If any students used extra paper for the free-response section, have those students staple
the extra sheet/s to the first page corresponding to that question in their exam booklets.
Collect a Section II booklet from each student and check that each student wrote answers
on the lined pages corresponding to each question. Then say:

The exam is over. You are now dismissed.

AP Biology Practice Exam 5

 Name:

AP® Biology
Answer Sheet
for Multiple-Choice Section

No. Answer No. Answer

1 31
2 32
3 33
4 34
5 35
6 36
7 37
8 38
9 39
10 40
11 41
12 42
13 43
14 44
15 45
16 46
17 47
18 48
19 49
20 50
21 51
22 52
23 53
24 54
25 55
26 56
27 57
28 58
29 59
30 60

6 AP Biology Practice Exam

AP Biology Exam
®

SECTION I: Multiple Choice

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.

Instructions
At a Glance
Section I of this exam contains 60 multiple-choice questions. Indicate all of your answers
Total Time to the Section I questions on the answer sheet. No credit will be given for anything written
1 hour and 30 minutes in this exam booklet, but you may use the booklet for notes or scratch work.
Number of Questions
60 Use your time effectively, working as quickly as you can without losing accuracy. Do not
Percent of Total Score spend too much time on any one question. Go on to other questions and come back to
50% the ones you have not answered if you have time. It is not expected that everyone will
Writing Instrument know the answers to all of the multiple-choice questions.
Pencil required
Electronic Device Your total score on Section I is based only on the number of questions answered correctly.
Calculator allowed Points are not deducted for incorrect answers or unanswered questions.

AP Biology Practice Exam 7

 AP® BIOLOGY EQUATIONS AND FORMULAS
Statistical Analysis and Probability
Mean Standard Deviation x = sample mean
n 2
1 Â (xi - x ) n = sample size
nÂ
x = xi s =
i=1 n -1
s = sample standard deviation (i.e., the sample-based
Standard Error of the Mean Chi-Square estimate of the standard deviation of the
population)
SE x =
s  o  e 2
n 2   e o = observed results

Chi-Square Table e = expected results

p Degrees of Freedom
value  = sum of all
1 2 3 4 5 6 7 8
0.05 3.84 5.99 7.81 9.49 11.07 12.59 14.07 15.51
Degrees of freedom are equal to the number of
0.01 6.63 9.21 11.34 13.28 15.09 16.81 18.48 20.09 distinct possible outcomes minus one.

Laws of Probability Metric Prefixes

If A and B are mutually exclusive, then:
Factor Prefix Symbol
P(A or B) = P(A) + P(B)
10 9 giga G
If A and B are independent, then: 10 6 mega M
P(A and B) = P(A)  P(B) 10 3 kilo k
10 – 1 deci d
Hardy-Weinberg Equations
10 – 2 centi c
p2 + 2pq + q2 = 1 p = frequency of allele 1 in a
10 – 3 milli m
population
p+q=1 10 – 6 micro μ
q = frequency of allele 2 in a 10 – 9 nano n
population 10 – 12 pico p

Mode = value that occurs most frequently in a data set

Median = middle value that separates the greater and lesser halves of a data set

Mean = sum of all data points divided by number of data points

Range = value obtained by subtracting the smallest observation (sample minimum) from the greatest (sample maximum)

8 AP Biology Practice Exam

 Rate and Growth Water Potential ( Y )
Rate dY = amount of change Y = Y P + YS
dY dt = change in time
dt YP = pressure potential
B = birth rate
Population Growth
D = death rate YS = solute potential
dN
= B-D N = population size
dt The water potential will be equal to the
Exponential Growth K = carrying capacity solute potential of a solution in an open
container because the pressure potential of
rmax = maximum per capita
dN the solution in an open container is zero.
= rmax N growth rate of population
dt
The Solute Potential of a Solution
Logistic Growth

( )
YS = -iCRT
dN K-N
= rmax N
dt K i = ionization constant (1.0 for sucrose
because sucrose does not ionize in
Simpson’s Diversity Index water)

( )
2
n C = molar concentration
Diversity Index = 1 - Â
N
R = pressure constant
݊ ൌ total number of organisms of a particular species
( R = 0.0831 liter bars/mole K)
ܰ ൌ total number of organisms of all species T = temperature in Kelvin (ºC + 273)

pH = – log[H+]
Surface Area and Volume

Surface Area of a Sphere Volume of a Sphere r = radius

SA  4 r 2 4
V   r3 l = length
3
Surface Area of a Rectangular Volume of a Rectangular Solid h = height
Solid V  lwh
w = width
SA  2lh  2lw  2wh
Volume of a Cylinder s = length of one
Surface Area of a Cylinder V   r 2h side of a
SA  2 rh  2 r 2 cube
Volume of a Cube SA = surface area
Surface Area of a Cube V  s3
SA  6s 2 V = volume

AP Biology Practice Exam 9

 BIOLOGY
SECTION I
Time—1 hour and 30 minutes
60 Questions
Directions: Each of the questions or incomplete statements below is followed by four suggested answers or
completions. Select the one that is best in each case and then enter the letter in the corresponding space on the
answer sheet.

1. Which of the following best describes the

hydrolysis of carbohydrates?

(A) The removal of a water molecule breaks a

covalent bond between sugar monomers.
(B) The removal of a water molecule forms a
covalent bond between sugar monomers.
(C) The addition of a water molecule breaks a
covalent bond between sugar monomers.
(D) The addition of a water molecule forms a
covalent bond between sugar monomers.

2. Which of the following best describes a

characteristic of DNA that makes it useful as
hereditary material?

(A) There are many different types of Figure 1. Pedigree of an inherited trait
nucleotide bases that can be incorporated
into DNA. 3. Based on the pedigree in Figure 1, which of the
following best explains the observed pattern of
(B) The nucleotide bases can also be used to
inheritance?
provide the energy needed for
reproduction.
(A) The trait is autosomal dominant, because
(C) Nucleotide bases can be randomly replaced the cross between individuals I-3 and I-4
with different nucleotide bases to increase produced an affected offspring.
variation.
(B) The trait is autosomal recessive, because
(D) Nucleotide bases in one strand can only be
the cross between individuals I-1 and I-2
paired with specific bases in the other
produced an affected offspring.
strand.
(C) The trait is sex-linked dominant, because
the cross between individuals II-5 and
II-6 produced an affected male.
(D) The trait is sex-linked recessive, because
the cross between individuals II-2 and
II-3 produced an affected female.

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10 AP Biology Practice Exam

Questions 4 - 8

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

Average pH of Control Average pH of Treatment

Time (min) (± 2 SE x ) (± 2 SE x )

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

4. Which of the following best describes the 5. Which of the following was the dependent
process by which the bacteria are breaking down variable in the researcher’s experiment?
the glucose to produce lactic acid?
(A) Time
(A) The bacteria are breaking down sugars in (B) pH
the absence of oxygen.
(C) Glucose concentration
(B) The bacteria are creating a H + gradient to (D) Lactic acid concentration
synthesize more ATP.
(C) The bacteria are using their mitochondria
to break down glucose in the presence of
oxygen.
(D) The bacteria are producing CO 2 in the
Krebs cycle that is then converted into
lactic acid.

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AP Biology Practice Exam 11

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

(A)

(B)

(C)

(D)

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12 AP Biology Practice Exam

7. Based on the data in Table 1, which of the 9. A mutation in the gene coding for a
following is the earliest time point at which single-polypeptide enzyme results in the
there is a statistical difference in average pH substitution of the amino acid serine, which has
between the control and treatment groups? a polar R group, by the amino acid
phenylalanine, which has a nonpolar R group.
(A) 5 minutes When researchers test the catalysis of the normal
enzyme and the mutated enzyme, they find that
(B) 15 minutes the mutated enzyme has much lower activity
(C) 20 minutes than the normal enzyme does.
(D) 35 minutes Which of the following most likely explains how
the amino acid substitution has resulted in
decreased catalytic activity by the mutated
8. According to the data, which of the following enzyme?
best explains the results of the experiment?
(A) The substitution decreased the mass of the
(A) The pH of the treatment culture was lower enzyme so that the mutated enzyme binds
than the pH of the control because the more weakly to the substrate than the
chemical increased the bacterial normal enzyme does.
metabolic rate. (B) The substitution altered the secondary and
(B) The pH of the treatment culture was tertiary structure of the enzyme so that
higher than the pH of the control because the mutated enzyme folds into a different
the chemical denatured bacterial enzymes shape than the normal enzyme does.
and decreased the metabolic rate. (C) The substitution caused many copies of the
(C) The chemical increased the metabolic rate mutated enzyme to cluster together and
of the bacteria because it lowered the pH . compete for substrate to bind.
(D) The chemical decreased the metabolic rate (D) The substitution caused the directionality
of the bacteria because it bound all of the enzyme to change such that the
available oxygen. amino terminus of the normal enzyme has
become the carboxy terminus of the
mutated enzyme.

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AP Biology Practice Exam 13

 10. Pitcher plants are carnivorous plants that grow in 11. A particular genetic disorder is associated with a
areas where the soil contains low levels of key single gene with two alleles. Individuals with
nutrients such as nitrogen. To obtain these
two recessive alleles are affected. The
nutrients, most pitcher plants capture prey using
traps containing a digestive fluid. The captured prevalence of the disorder is 1 in 6,600 .
prey are then broken down and digested, and the Assuming the population is in Hardy-Weinberg
pitcher plant absorbs the nutrients. equilibrium, which of the following is closest to
the frequency of carriers in the general
The traps of one species of pitcher plant, population?
Nepenthes hemsleyana, do not contain digestive
fluid. Instead they provide a suitable place for
(A) 0.00015
woolly bats (Kerivoula hardwickii) to sleep.
The feces from the bat are released into the trap (B) 0.01230
where nutrients in the feces are absorbed and (C) 0.02430
provide the plant with the nitrogen it needs.
(D) 0.98770
Which of the following best describes the
relationship between the pitcher plant and the
woolly bat?

(A) The relationship is an example of

parasitism because the bat is harmed
while the plant benefits.
(B) The relationship is an example of
mutualism because both the plant and the
bat benefit.
(C) The relationship is an example of
commensalism because the plant benefits
but the bat is unaffected.
(D) The relationship is an example of
commensalism because the plant is
unaffected while the bat benefits.

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14 AP Biology Practice Exam

 Figure 1. A signaling cascade triggers the relaxation of smooth muscle cells.
12. Blood vessels are surrounded by cells called smooth muscle cells. Nitric oxide triggers a signaling cascade in
smooth muscle cells that causes the cells to relax (Figure 1).
Which of the following is represented by the gradual increase in thickness of the arrows from the top to the
bottom of Figure 1 ?

(A) The rate at which nitric oxide triggers signaling gradually increases over time.
(B) The number of signaling molecules that are produced or activated increases with each step in the pathway.
(C) The size of the proteins in the pathway increases as the signaling cascade moves through the cell.
(D) The signaling pathway uses an increase in negative feedback to reduce intracellular Ca2+ levels and cell
sensitivity to Ca2+ .

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AP Biology Practice Exam 15

 Questions 13 - 18

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 Percent Change in Mass

Solution
(weight/volume)
Distilled water 0% 13.48%

NaCl 20% −23.39%

Tap water 0.8% 9.46%

Grape juice 2.1% 2.8%

Grape soda 13% −15.00%

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

Concentration of Small Concentration of

Location
Diffusible Solutes Nondiffusible Solutes
Inside grape 0.4 M 1.2 M

In solution 1.6 M 0.8 M

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16 AP Biology Practice Exam

13. Based on Table 1, which of the following best 16. Assuming a negligible pressure potential, which
explains the difference in water potential of the following best predicts the net movement
between certain solutions and the grapes? of the small diffusible solutes and water in the
second experiment (Table 2) ?
(A) NaCl and tap water have a lower water
potential because these two solutions (A) Small diffusible solutes will diffuse into
caused the grape to gain water. the grape cells, followed by water.
(B) Grape soda and NaCl have a lower water (B) Small diffusible solutes will diffuse out of
potential because these two solutions the grape cells and water will diffuse into
caused the grape to lose water. the cells.
(C) Tap water and grape juice have a lower (C) Small diffusible solutes will diffuse out of
water potential because these two the grape cells, followed by water.
solutions caused the grape to lose water. (D) Small diffusible solutes will diffuse into
(D) Grape soda and grape juice have a lower the grape cells and water will diffuse out
water potential because these two of the cells.
solutions caused the grape to gain water.

17. Mercurial sulfhydryl is an inhibitor of

14. Based on Table 1, which of the following aquaporins. Which of the following is the most
percentages is closest to the solute concentration likely effect of adding mercurial sulfhydryl to
of the grape? the distilled water solution?

(A) 0.0% (A) The grape cells will burst because of

excess water entering by active transport.
(B) 1.3%
(B) The grape cells will gain more water
(C) 5.5% because of the activation of the transport
(D) 10.1% protein.
(C) The grape cells will shrink because active
15. A student hypothesizes that the solute transport has been inhibited.
concentration of grape juice is higher than the (D) The grape cells will gain water more
solute concentration of the actual grape because slowly because of a lack of facilitated
the grape juice has added sugar. diffusion.
Based on the data in Table 1, which of the
following best evaluates the student’s
hypothesis?

(A) The hypothesis is supported because the

mass of the grape decreased in the grape
juice.
(B) The hypothesis is supported because the
grape juice has a greater solute potential
than the grape has.
(C) The hypothesis is not supported because
the grape was isotonic to the grape juice.
(D) The hypothesis is not supported because
the mass of the grape increased in the
grape juice.

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AP Biology Practice Exam 17

 18. Which of the following best explains why larger
grapes have a different rate of water absorption
per gram of mass than smaller grapes do?

(A) The rate is slower because smaller grapes

have a larger surface-area-to-volume ratio
than the larger grapes do.
(B) The rate is slower because larger grapes
have a larger surface-area-to-volume
ratio than the smaller grapes do.
(C) The rate is slower because smaller grapes
can expand more than larger grapes to
hold excess water. Figure 1. Change in the relative concentration of
(D) The rate is slower because larger grapes product over time
have more volume to hold excess water
than smaller grapes do. 20. Figure 1 shows the amount of product produced
in an enzyme-catalyzed reaction over five
minutes. Which of the following best explains
how the rate of the reaction changes over time?

(A) The rate increases because more products

19. Which of the following best explains how some are made over time.
cells of an individual produce and secrete a (B) The rate increases because the ratio of
specific enzyme, but other cells of the same product to substrate increases.
individual do not?
(C) The rate decreases because the ratio of
(A) The cells contain different genes and product to substrate increases.
therefore do not make the same proteins. (D) The rate decreases because the enzyme is
(B) The cells have evolved under different used up as the reaction progresses.
selective pressures, resulting in some cells
making proteins that others cannot.
(C) The cells transcribe and translate different
combinations of genes, leading to the
production of different sets of proteins.
(D) The cells produce different types of
ribosomes that enable the translation of
different genes.

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18 AP Biology Practice Exam

21. Pyruvate kinase, a key enzyme in the glycolysis 22. Different species of bumblebee compete for
pathway, is inhibited by the amino acid alanine. flower nectar in their ecosystem. While the
The ability of alanine to inhibit the enzyme is flowers vary in the length of their petals,
not affected by increasing the concentration of bumblebees vary in the length of a mouth part
substrate. called a proboscis. Bumblebees demonstrate a
preference for flowers that correspond to the
Which of the following best explains the length of their proboscis. Species with
mechanism by which alanine inhibits pyruvate proboscises of similar length tend to occupy
kinase activity? different areas. Species that live in the same area
tend to have proboscises of different lengths.
(A) Alanine binds to an allosteric site of the
enzyme, changing the shape of the Which of the following best explains the
enzyme’s active site. relationship between the different bumblebee
(B) Alanine increases the enzyme-substrate species living in the same area?
binding until the enzyme becomes
saturated. (A) The bumblebees have undergone allopatric
speciation due to geographic isolation.
(C) Alanine is a competitive inhibitor that
reversibly binds to the active site of the (B) The bumblebees have undergone niche
enzyme. partitioning due to competition.
(D) Alanine binds to the substrate, preventing (C) The bumblebees have a mutualistic
the substrate from being able to bind to relationship with other bumblebee species
the active site of the enzyme. due to natural selection.
(D) The bumblebees have a commensalistic
relationship due to the variety of niches
that are available.

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AP Biology Practice Exam 19

 23. Researchers investigated whether time of day affects the rate at which certain mRNAs produced by plants are
broken down. At either 1 hour after the start of the light period (morning) or 8 hours after the start of the light
period (afternoon), the researchers treated identical plant seedlings with a compound that blocks transcription
(time = 0 min). The researchers measured the percent remaining of two mRNAs, mRNA G and mRNA H,
over the course of 120 minutes. The data are shown in Figure 1.

Figure 1. Degradation of mRNA G and mRNA H over time after exposure to light for 1 hour (morning) or 8
hours (afternoon)

Based on the data, which of the following best describes the relationship between light and the degradation of
mRNA G and mRNA H ?

(A) Exposure to light causes the degradation of both mRNA G and mRNA H.
(B) mRNA G and mRNA H degrade at the same rate during morning exposure to light.
(C) A longer exposure to light increases the rate of mRNA G degradation but not of mRNA H degradation.
(D) Exposure to more-intense afternoon light causes both mRNA G and mRNA H to degrade more rapidly
in the afternoon than in the morning.

24. Which of the following observations would

provide the strongest evidence that two plants
belong to different biological species?

(A) There are variations in their RNA

sequences.
(B) The leaves and flowers show
morphological differences.
(C) They produce viable but sterile offspring.
(D) They occupy unique habitats in an
ecosystem.

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20 AP Biology Practice Exam

Questions 25 - 29

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

Figure 2: Blood insulin levels in normal mice and Esp 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, OPG, that binds to neighboring osteoclasts. In response, the osteoclasts release protons (H + ) and

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AP Biology Practice Exam 21

 create an area of lower pH outside the cell. This low pH activates osteocalcin, a protein secreted in an inactive form by
osteoblasts.
The Esp 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 Esp mutation that prevented the
production of a functional Esp product (mutant). The researcher then exposed the mutant strain and a normal strain
that expresses Esp to glucose and compared the levels of insulin in the blood near the osteoblasts (Figure 2).

25. Based on the information provided, which of the 27. Which of the following claims is most consistent
following best justifies the claim that osteocalcin with the data shown in Figure 2 ?
is a hormone?
(A) Esp expression is necessary to prevent the
(A) The phosphorylation of the insulin receptor overproduction of insulin.
causes a response in osteoblast bone cells. (B) Esp protein does not regulate blood-sugar
(B) The osteoblasts in the bone secrete levels in normal mice.
osteocalcin, which causes cells in the (C) Normal mice require a higher blood
pancreas to change their activity. concentration of insulin than mutant mice
(C) The change in expression of Esp changes do.
the insulin receptor activity of the (D) Mutant mice have a cyclical pattern of
osteoblast. insulin secretion.
(D) The activation of the osteocalcin by a bone
cell is pH dependent.
28. Which of the following was a positive control in
the experiment?
26. Which of the following best describes the effect
of insulin binding to the receptor on the (A) Minutes after glucose injection
osteoblast cells? (B) Blood insulin
(C) Mutant strain
(A) Insulin binding ultimately increases
pancreatic secretion of additional insulin. (D) Normal strain
(B) Insulin binding blocks the release of
osteocalcin from the osteoblasts.
29. A researcher observes that mice from the mutant
(C) Insulin binding inhibits the expression of strain experience low blood sugar. Which of the
Esp . following best describes the feedback
(D) Insulin binding increases the pH of the mechanism in the pathway (Figure 1) causing
extracellular matrix. the low blood sugar in the mutant strain?

(A) The positive feedback of insulin production

(B) The negative feedback of inactive
osteocalcin production
(C) The positive feedback of the Esp protein
(D) The negative feedback of insulin-secreting
pancreatic cell proliferation

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22 AP Biology Practice Exam

30. A researcher hypothesizes that, in mice, two
autosomal dominant traits, trait Q and trait R,
are determined by separate genes found on the
same chromosome. The researcher crosses mice
that are heterozygous for both traits and counts
the number of offspring with each combination
of phenotypes. The total number of offspring
produced was 64 . The researcher plans to do a
chi-square analysis of the data and calculates the
expected number of mice with each combination
of phenotypes. Which of the following is the
expected number of offspring that will display
both trait Q and trait R?

(A) 4
(B) 12
(C) 36
(D) 48

31. Researchers studying new viruses analyzed the genetic material found in four different virus samples to
determine the percent nitrogen base composition of each virus. The data are shown in the table.

BASE DISTRIBUTION IN FOUR NUCLEIC ACID SAMPLES

Percent in Percent in Percent in Percent in

Nitrogen Base
Sample 1 Sample 2 Sample 3 Sample 4
Adenine 27 33 21 22
Cytosine 23 17 33 28
Guanine 23 17 27 28
Thymine 0 33 0 22
Uracil 27 0 19 0
Which of the following samples most likely contains a double-stranded RNA virus?

(A) Sample 1
(B) Sample 2
(C) Sample 3
(D) Sample 4

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AP Biology Practice Exam 23

 32. Goats and sheep belong to the same family but 33. Ethylene gas is an organic molecule that serves
different genera. While they often live together many cell signaling roles in flowering plants.
in the same pastures, the hybrid offspring that Which of the following best explains how a
are occasionally produced between the two positive feedback mechanism involving ethylene
species rarely survive. When such a hybrid does works?
survive, it is usually sterile.
(A) Cells of ripening fruit produce ethylene,
Which of the following best explains the which activates the ripening response in
mechanism that maintains reproductive isolation other fruit cells.
between goats and sheep?
(B) Low water stress causes cells to produce
ethylene, which binds to root cells and
(A) Gene flow is prevented because the two
initiates cell division.
species belong to different trophic levels
and therefore do not share a food source. (C) Cells damaged by leaf-eating insects
produce ethylene, which is released into
(B) Habitat isolation creates a prezygotic
the air, and repels insects.
barrier between the two species.
(D) Fertilized ovules produce ethylene, which
(C) The males of one species and the females
initiates apoptosis in flower petal cells.
of the other species are fertile at different
times.
(D) The two species have a different number of
chromosomes, resulting in a postzygotic
barrier.

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24 AP Biology Practice Exam

Questions 34 - 38

A student is investigating photosynthesis in plants. The student planted grass seeds in a tray with three sections and
grew the grass under artificial lights for 14 days (Figure 1). After 14 days, the student collected all of the grass
from section I and recorded its mass (Table 1). The student then placed a clear cover over section II and placed an
aluminum foil cover over section III (Figure 2). The student then placed the tray back under the artificial lights for
seven additional days. On day 21, the student collected and measured the mass of the grass from sections II and III
(Table 1).

TABLE 1. MASS OF GRASS GROWN IN EACH SECTION

Day Mass was Mass (g)

Section
Measured
I 14 5.1

II 21 9.6

III 21 4.2

34. Based on the data, which of the following is the

best approximation of the rate of growth per
week in section II from day 14 to day 21 ?

(A) 3.2 g/week

(B) 4.5 g/week
(C) 5.1 g/week
(D) 9.6 g/week

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AP Biology Practice Exam 25

 35. Which of the following best explains the 37. Based on the data, which of the following is the
observed difference in mass between the grass in most likely effect of a period of increased
section I after 14 days and the grass in section volcanic activity that significantly decreases the
amount of sunlight reaching a particular
III after 21 days?
ecosystem for several years?
(A) The grass in section III required more (A) There will be a decrease in the population
energy for metabolic processes than the sizes of most species and a decrease in
grass in section I . the number of trophic levels.
(B) The grass in section III performed more (B) The population sizes of plant species will
photosynthesis than the grass in section I . stay the same, but the average biomass of
(C) Only the grass in section III used energy individual plants will increase.
from the seed for growth and (C) There will be an increase in the population
development. sizes of omnivores because they can eat
other animals instead of the plants.
(D) Mass lost by cell respiration in section III
was not replaced by photosynthesis. (D) The biodiversity of the ecosystem will
increase as new niches arise and become
occupied by new species.
36. Which of the following best explains the
connection between energy, growth, and the
maintenance of an ordered system in the 38. Which of the following is an appropriate null
experiment? hypothesis for the student’s experiment?

(A) Energy input from light is required for the (A) The absence of light negatively affects the
grass to grow and maintain an ordered mass of the grass.
structure. (B) An increase in light promotes grass growth.
(B) The grass obtains the energy for growth (C) A change in light intensity changes the rate
and maintenance of order from nutrients of grass growth.
from the soil. (D) The presence of light has no effect on the
(C) The grass couples the release of energy mass of the grass.
from the light reactions with the
production of oxygen (O 2 ) used to
produce sugars for growth.
(D) Energy is required for the growth of the
grass but not the maintenance of order.

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26 AP Biology Practice Exam

 41. A researcher is investigating the effects of a
chemical that makes thylakoid membranes
permeable to hydrogen ions (H + ). Which of the
following is the most likely direct effect of
Figure 1. An amino acid adding the chemical to plant cells?
39. The amino acid in Figure 1 is found in a region (A) The plant cells will produce less NADPH .
of a polypeptide that folds away from water.
Which part of the amino acid most likely (B) The chloroplasts will generate less ATP.
contributes to the hydrophobic behavior of this (C) Chlorophyll will require less light energy
region of the polypeptide? to excite its electrons.
(D) The plant cells will split fewer water
(A) Amine (NH2) group molecules into hydrogen ions and
(B) Carboxyl (COOH) group oxygen.
(C) Methyl (CH3) group
(D) Hydrogen (H) atom

40. A researcher observes that when two

heterozygous plants with red flowers are
crossed, the resulting offspring include plants
with red, white, or pink flowers. The researcher
proposes the null hypothesis that flower color is
the result of independent assortment and
incomplete dominance. The researcher
calculates a chi-square value of 7.3. Assuming
two degrees of freedom, which of the following
is the correct interpretation of the chi-square
analysis, using a p -value of 0.05 ?

(A) The null hypothesis should be rejected

because the critical value is less than the
calculated value.
(B) The null hypothesis should not be rejected
because the critical value is less than the
calculated value.
(C) The null hypothesis should not be rejected
because the critical value is greater than
the calculated value.
(D) The null hypothesis should be rejected
because the critical value is greater than
the calculated value.

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AP Biology Practice Exam 27

 Figure 1. For paired groups of organisms, a comparison of the number of amino acid substitutions in cytochrome c with
the time since the groups diverged from a common ancestor
42. Researchers analyzed the amino sequence of the protein cytochrome c in various groups of organisms and
determined the number of amino substitutions that have occurred in the different groups of organisms. They
plotted the data with respect to the time since divergence of the members of paired groups from a common
ancestor (Figure 1). Based on the data, which of the following organisms are most distantly related?

(A) Birds and mammals

(B) Mammals and reptiles
(C) Fish and land vertebrates
(D) Insects and vertebrates

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28 AP Biology Practice Exam

43. Which of the following best explains how
continuity of genetic information in cells is
ensured across generations?

(A) Replication uses a parental strand of DNA

as a template to create a new strand of
DNA.
(B) DNA molecules are shaped like a double
helix with a constant diameter
throughout.
(C) Transcription copies the information in
DNA into an RNA transcript.
(D) Cells contain different polymerases for
DNA replication and transcription. Figure 1. Key physiological points in the human
sleep-wake cycle
44. In the spring and summer, the fur of an arctic
fox contains a pigment called melanin that gives 45. The human sleep-wake cycle is regulated by
the fox’s fur a dark color. In the fall and winter, melatonin. The synthesis of melatonin is
the fur of the arctic fox is white. regulated by light exposure (Figure 1). The
human body typically develops a sleep-wake
Which of the following most likely explains how rhythm that does not respond quickly to change.
the changing seasons result in changing fur color
in an arctic fox? Long-term exposure to extended periods of
bright light after sunset is most likely to affect a
(A) Environmental factors cause changes in person in which of the following ways?
gene expression, resulting in seasonal
variations in pigment production. (A) Melatonin synthesis will be increased, and
(B) Environmental factors cause different the entire sleep-wake cycle will be shifted
mutations in DNA during different by several hours.
seasons, resulting in seasonal changes in (B) Melatonin synthesis will be increased, and
fur phenotype. the person will quickly fall asleep.
(C) Environmental factors cause proteins to be (C) Melatonin synthesis will be inhibited, and
translated using different genetic codes the person will have difficulty sleeping.
during different seasons, resulting in (D) Melatonin synthesis will be inhibited, and
variations in pigment production. the person will quickly fall asleep.
(D) Environmental factors cause enzymes to
react with different substrates, resulting in
the accumulation of different pigments.

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AP Biology Practice Exam 29

 46. A typical bag of fertilizer contains high levels of
nitrogen, phosphorus, and potassium but trace
amounts of magnesium and calcium. Which of
the following best matches the fertilizer
component with the molecule in which it will be
incorporated by organisms in the area?

(A) Nitrogen will be incorporated into nucleic

acids.
(B) Phosphorus will be incorporated into
amino acids.
(C) Potassium will be incorporated into lipids.
Figure 1. Change in energy over the course of four
(D) Magnesium will be incorporated into
chemical reactions
carbohydrates.
47. Based on the data in Figure 1, which of the
following most likely represents the change in
energy that occurs when ATP hydrolysis is
coupled with the phosphorylation of a substrate?

(A) Line 4 represents ATP hydrolysis, and

line 1 represents phosphorylation of a
substrate.
(B) Line 1 represents ATP hydrolysis, and
line 4 represents phosphorylation of a
substrate.
(C) Line 2 represents ATP hydrolysis, and
line 3 represents phosphorylation of a
substrate.
(D) Line 3 represents ATP hydrolysis, and
line 2 represents phosphorylation of a
substrate.

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30 AP Biology Practice Exam

Questions 48 - 52

The brown anole (Anolis sagrei) is a species of lizard native to Cuba that has been introduced into the southeastern
United States. The range of brown anoles in the United States has been expanding, and they are now competing with
native green anoles (Anolis carolinensis). Some of the characteristics of the green and brown anoles are shown in
Table 1.

TABLE 1. CHARACTERISTICS OF TWO ANOLE SPECIES

Limb Length Toe Pad Size

Anole Species Body Length Relative to Relative to Cold Tolerance
Body Length Body Length
A. sagrei 55 – 80 mm Long Intermediate 30 – 35°C
(brown anole)
A. carolinensis 45 – 85 mm 25 – 30°C
Short Very large
(green anole)

A cladogram showing the relationships between some anole species is shown in Figure 1.

Figure 1. Cladogram of some anole species

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AP Biology Practice Exam 31

 48. Based on the data in Table 1, which of the 50. Which of the following is the outgroup in the
following best explains the observed differences cladogram shown in Figure 1 ?
in cold tolerance between brown anoles
(A. sagrei) and green anoles (A. carolinensis) in (A) A. evermanni
the United States? (B) A. stratulus
(C) A. gundlachi
(A) Brown anoles developed a tolerance for
cold temperatures and passed on this (D) A. carolinensis
acquired tolerance.
(B) Brown and green anoles produced a hybrid
51. Based on Figure 1, which of the following
species that has greater tolerance to cold
would provide the strongest evidence to support
temperatures.
a claim that A. valencienni and A. lineatopus are
(C) Green anoles with greater tolerance for the two anole species most closely related to
cold had greater reproductive success in A. sagrei ?
areas with colder temperatures.
(D) New mutations occurred in brown anoles in (A) Comparing body structures, such as toe
response to exposure to colder pads and limb length
temperatures. (B) Analyzing homologous DNA sequences
among the species
(C) Determining which species occupy a
49. In anoles, small toe pads are better for gripping similar niche in Cuba
irregular surfaces such as the forest floor, and
large toe pads are better for gripping smooth (D) Observing which species have mating
surfaces such as leaves. Based on the data in behaviors most similar to those of A.
Table 1, if brown anoles are introduced into a sagrei
forest where green anoles currently live, which
of the following predictions about the two
species is most likely? 52. Based on the cladogram in Figure 1, which of
the following conclusions about the evolution of
(A) Brown anoles will occupy a different area the anole species is most reasonable?
of the forest from green anoles which will
minimize competition. (A) Each species evolved in a different
location, and all underwent convergent
(B) Brown anoles will mostly prey on flying
evolution.
insects, and green anoles will prey on
worms and grubs. (B) Each species evolved as a result of
different selective pressures.
(C) Brown anoles and green anoles will evolve
to have the same size toe pads because (C) Each species evolved as a result of a
they share the same niche. genetic bottleneck.
(D) Green anoles are better adapted to the (D) Each species evolved by inheriting random
forest habitat and will chase away the mutations from the common ancestor.
introduced brown anoles.

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32 AP Biology Practice Exam

53. Beta thalassemia is a disorder caused by 54. A common test for liver function involves
mutations in the HBB gene. Examination of the sprinkling sulfur powder onto a sample of urine
HBB protein in an individual with beta (mostly water with dissolved bodily waste).
Sulfur powder sprinkled on a sample from an
thalassemia shows that the protein is missing
individual with impaired liver function will sink
many amino acids at its carboxyl terminus. because the urine contains a high level of bile
salts, while the sulfur powder sprinkled on
Which of the following is the most likely
normal urine samples will float.
explanation for how a mutation in the DNA
could result in the loss of the carboxyl terminus Which of the following best explains why bile
of the HBB protein? salts cause the sulfur powder to sink?

(A) The mutation keeps the HBB gene (A) Bile salts decrease the surface tension of
wrapped tightly around histones, the urine sample.
preventing transcription of the gene. (B) Bile salts increase the water potential of
(B) The mutation changes the promoter the urine.
sequence of the gene such that different (C) Bile salts increase the density of the urine
transcription factors initiate transcription sample.
of the gene. (D) Bile salts decrease the strength of the
(C) The mutation changes a codon in the covalent bonds within a water molecule.
coding region of the HBB transcript to a
stop codon such that translation
terminates earlier than it should.
(D) The mutation results in hydrogen bonds
between adjacent amino acids instead of
covalent bonds, resulting in the
production of an unstable protein.

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AP Biology Practice Exam 33

 Figure 1. A karyotype obtained from a sample of human cells

55. A karyotype is a representation of all the metaphase chromosomes in a sample of cells from a particular
individual (Figure 1).

Which of the following most likely explains how the chromosomes circled in Figure 1 could cause a genetic
disorder in the person from whom the cells were obtained?

(A) The extra chromosome causes crowding in the nucleus of the cells and blocks RNA polymerase from
binding to and transcribing certain genes.
(B) The extra chromosome will affect the levels of RNA transcribed from certain genes and the amount of
protein produced from those genes in each cell.
(C) The cells will not divide and enable growth, because the extra chromosome will interfere with the pairing
of homologous chromosomes.
(D) The extra chromosome will cause other chromosomes in the cell to become triploid during future rounds
of cell division.

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34 AP Biology Practice Exam

56. Which of the following is the strongest evidence 57. Which of the following best explains how the
supporting the endosymbiont hypothesis? expression of a eukaryotic gene encoding a
protein will differ if the gene is expressed in a
(A) Mitochondria have their own DNA and prokaryotic cell instead of in a eukaryotic cell?
divide independently of the cell.
(B) Mitochondria can carry out hydrolytic (A) No transcript will be made, because
reactions on organic molecules. eukaryotic DNA cannot be transcribed by
(C) Mitochondria have a highly folded prokaryotic RNA polymerase.
membrane. (B) The protein will have a different sequence
(D) Mitochondria are found in both plants and of amino acids, because prokaryotes use a
animals. different genetic code.
(C) The protein will be made but will not
function, because prokaryotes cannot
remove introns.
(D) The protein will not be made, because
prokaryotes lack the ribosomes necessary
for translation.

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AP Biology Practice Exam 35

 58. Low concentrations of cadmium in the water negatively affect steelhead trout. In a river community, steelhead
trout are part of the trophic structure shown in Figure 1.

Figure 1. The trophic structure of a river community

Which of the following is the most likely immediate effect of an increase in runoff containing cadmium on the
trophic structure of the river community?

(A) The population of Cladophora will decrease, resulting in an increase in the trout population.
(B) There will be a large decrease in the trout population, resulting in an increase in damselfly nymphs.
(C) Increased stream volume will provide more area for the trout to reproduce, causing a large increase in the
population of algae.
(D) The population of trout will decrease because the population of damselfly nymphs will decline.

59. Poaching is the illegal hunting or capturing of

wild animals. Both male and female African
elephants have tusks that are a significant source
of ivory used for ornaments and jewelry.
Assuming that the size of the tusks is heritable,
which of the following is the most likely
long-term effect of poachers targeting African
elephants with large tusks?

(A) An increase in average tusk size because of

the need to provide more ivory for the
poaching hunters
(B) A decrease in average tusk size because
elephants with large tusks are less likely
to survive to reproduce
(C) An increase in average tusk size to provide
the elephants with a defensive mechanism
against poachers
(D) A decrease in average tusk size to reduce
the energy investment in the tusks

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36 AP Biology Practice Exam

60. A scientist adds a chemical to a culture of dividing cells in order to disrupt DNA replication. The replicated
DNA produced by the cells is double-stranded, but sections of it lack covalent bonds between adjacent
nucleotides (Figure 1).

Figure 1. Replicated DNA produced after a chemical is introduced

Which of the following claims is best supported by the data?

(A) The chemical prevents the formation of RNA primers.

(B) The chemical inhibits DNA ligase.
(C) The chemical blocks DNA polymerase.
(D) The chemical disrupts hydrogen bonding.

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AP Biology Practice Exam 37

 END OF SECTION I

IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY

CHECK YOUR WORK ON THIS SECTION.

DO NOT GO ON TO SECTION II UNTIL YOU ARE TOLD TO DO SO.

____________________________________________________________________________

MAKE SURE YOU HAVE DONE THE FOLLOWING:

• PLACED YOUR AP ID LABEL ON YOUR ANSWER SHEET

• WRITTEN AND GRIDDED YOUR AP ID CORRECTLY ON YOUR
ANSWER SHEET
• TAKEN THE AP EXAM LABEL FROM THE FRONT OF THIS BOOKLET AND
PLACED IT ON YOUR ANSWER SHEET

38 AP Biology Practice Exam

AP Biology Exam
®

SECTION II: Free Response

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.

At a Glance Instructions
Total Time The questions for Section II are printed in this booklet. You may use the unlined pages to
1 hour and 30 minutes organize your answers and for scratch work, but you must write your answers on the
Number of Questions labeled pages provided for each question.
6 Each answer should be written in paragraph form; an outline or bulleted list alone is not
Percent of Total Score acceptable. Do not spend time restating the questions or providing more than the number
50%
of examples called for. For instance, if a question calls for two examples, you can earn
Writing Instrument
Pen with black or dark credit only for the first two examples that you provide. Labeled diagrams may be used to
blue ink supplement discussion, but unless specifically called for by the question, a diagram alone
Electronic Device will not receive credit. Write clearly and legibly. Begin each answer on a new page. Do not
Calculator allowed skip lines. Cross out any errors you make; crossed-out work will not be scored.
Suggested Time
Approximately Manage your time carefully. You may proceed freely from one question to the next. You
25 minutes per long may review your responses if you finish before the end of the exam is announced.
question, and 10 minutes
per short question.
Weight
Approximate weights:
Questions 1 and 2:
26% each
Questions 3 6:
12% each

AP Biology Practice Exam 39

 AP® BIOLOGY EQUATIONS AND FORMULAS
Statistical Analysis and Probability
Mean Standard Deviation x = sample mean
n 2
1 Â (xi - x ) n = sample size
nÂ
x = xi s =
i=1 n -1
s = sample standard deviation (i.e., the sample-based
Standard Error of the Mean Chi-Square estimate of the standard deviation of the
population)
SE x =
s  o  e 2
n 2   e o = observed results

Chi-Square Table e = expected results

p Degrees of Freedom
value  = sum of all
1 2 3 4 5 6 7 8
0.05 3.84 5.99 7.81 9.49 11.07 12.59 14.07 15.51
Degrees of freedom are equal to the number of
0.01 6.63 9.21 11.34 13.28 15.09 16.81 18.48 20.09 distinct possible outcomes minus one.

Laws of Probability Metric Prefixes

If A and B are mutually exclusive, then:
Factor Prefix Symbol
P(A or B) = P(A) + P(B)
10 9 giga G
If A and B are independent, then: 10 6 mega M
P(A and B) = P(A)  P(B) 10 3 kilo k
10 – 1 deci d
Hardy-Weinberg Equations
10 – 2 centi c
p2 + 2pq + q2 = 1 p = frequency of allele 1 in a
10 – 3 milli m
population
p+q=1 10 – 6 micro μ
q = frequency of allele 2 in a 10 – 9 nano n
population 10 – 12 pico p

Mode = value that occurs most frequently in a data set

Median = middle value that separates the greater and lesser halves of a data set

Mean = sum of all data points divided by number of data points

Range = value obtained by subtracting the smallest observation (sample minimum) from the greatest (sample maximum)

40 AP Biology Practice Exam

 Rate and Growth Water Potential ( Y )
Rate dY = amount of change Y = Y P + YS
dY dt = change in time
dt YP = pressure potential
B = birth rate
Population Growth
D = death rate YS = solute potential
dN
= B-D N = population size
dt The water potential will be equal to the
Exponential Growth K = carrying capacity solute potential of a solution in an open
container because the pressure potential of
rmax = maximum per capita
dN the solution in an open container is zero.
= rmax N growth rate of population
dt
The Solute Potential of a Solution
Logistic Growth

( )
YS = -iCRT
dN K-N
= rmax N
dt K i = ionization constant (1.0 for sucrose
because sucrose does not ionize in
Simpson’s Diversity Index water)

( )
2
n C = molar concentration
Diversity Index = 1 - Â
N
R = pressure constant
݊ ൌ total number of organisms of a particular species
( R = 0.0831 liter bars/mole K)
ܰ ൌ total number of organisms of all species T = temperature in Kelvin (ºC + 273)

pH = – log[H+]
Surface Area and Volume

Surface Area of a Sphere Volume of a Sphere r = radius

SA  4 r 2 4
V   r3 l = length
3
Surface Area of a Rectangular Volume of a Rectangular Solid h = height
Solid V  lwh
w = width
SA  2lh  2lw  2wh
Volume of a Cylinder s = length of one
Surface Area of a Cylinder V   r 2h side of a
SA  2 rh  2 r 2 cube
Volume of a Cube SA = surface area
Surface Area of a Cube V  s3
SA  6s 2 V = volume

AP Biology Practice Exam 41

 BIOLOGY
SECTION II
Time—1 hour and 30 minutes
6 Questions

Directions: Questions 1 and 2 are long free-response questions that require about 25 minutes each to answer.
Questions 3 through 6 are short free-response questions that require about 10 minutes each to answer.
Read each question carefully and completely. Write your response in the space provided for each question. Only
material written in the space provided will be scored. Answers must be written out in paragraph form. Outlines,
bulleted lists, or diagrams alone are not acceptable.

Question 1 is on the following page.

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42 AP Biology Practice Exam

1. Electrons from the oxidation of glucose and other nutrients by glycolysis and the Krebs cycle are added to
the electron-carrying compound NAD+ to form NADH. To meet the continuous requirement for NAD+ by
cells, NADH is reduced back to NAD+ through the electron transport chain, in the presence (+) of oxygen,
or through fermentation, in the absence (−) of oxygen (Figure 1).

Figure 1. A simplified model of metabolism and the recycling of NAD+

MT -ND5 is a mitochondrial gene that encodes a subunit of NADH dehydrogenase, the enzyme that catalyzes the
initial oxidation of its substrate NADH to NAD+ and H + in the electron transport chain of mitochondria. A
mutation in MT -ND5 is associated with a rare genetic disorder that results in a buildup of lactic acid in the body. A
researcher hypothesizes that the mutated NADH dehydrogenase has decreased activity but is not completely
nonfunctioning and that, by increasing the pool of NADH in cells, the activity of NADH dehydrogenase will
increase.

To test this idea, the researcher treated a group of individuals with this disorder with a vitamin that is similar to
NADH and measured the concentration of NAD+ and lactic acid in the blood over the course of 20 weeks. The
results from a representative individual are shown in Figure 2.

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AP Biology Practice Exam 43

 Figure 2. The concentration of NAD+ (top) and lactic acid (bottom) in the blood of a representative treated individual

(a) Describe the pattern of inheritance that is most likely associated with a mutation in the MT -ND5 gene.
Explain why individuals are not typically heterozygous with respect to mitochondrial genes.
(b) Identify a dependent variable measured in the researcher’s experiment. Identify one control that the
researcher could use to improve the validity of the experiment. Justify the researcher analyzing blood samples
at many intermediate time points instead of at only the beginning and the end of the 20-week period.

(c) Describe the relationship between the concentration of NAD+ in the blood and the concentration of lactic
acid in the blood during the first 5 weeks of treatment with the vitamin. Based on Figure 2, calculate the
average rate of change in blood NAD+ concentrations from week 5 to week 17.

(d) The researcher performed a follow-up experiment to measure the rate of oxygen consumption by muscle
and brain cells. Predict the effect of the MT -ND5 mutation on the rate of oxygen consumption in muscle and
brain cells. Justify your prediction. The researcher had hypothesized that the addition of the vitamin that is
similar in structure to NADH would increase the activity of the mutated NADH dehydrogenase enzyme in
individuals with the disorder. Explain how the vitamin most likely increased the activity of the enzyme.

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44 AP Biology Practice Exam

THIS PAGE MAY BE USED FOR TAKING NOTES AND PLANNING YOUR ANSWERS.
NOTES WRITTEN ON THIS PAGE WILL NOT BE SCORED.
WRITE ALL YOUR RESPONSES ON THE LINED PAGES.

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AP Biology Practice Exam 45

 PAGE FOR ANSWERING QUESTION 1

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46 AP Biology Practice Exam

ADDITIONAL PAGE FOR ANSWERING QUESTION 1

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AP Biology Practice Exam 47

 ADDITIONAL PAGE FOR ANSWERING QUESTION 1

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48 AP Biology Practice Exam

ADDITIONAL PAGE FOR ANSWERING QUESTION 1

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AP Biology Practice Exam 49

 NO TEST MATERIAL ON THIS PAGE

50 AP Biology Practice Exam

22.2 Atherosclerosis is a disease that results when certain cells and proteins of an individual’s body adhere to and
damage blood vessels, especially those around the heart. Researchers continue to look for improved ways to treat
individuals with the disease. Data obtained from earlier experiments suggested that it might be possible to reduce
damage to the blood vessels by increasing the expression of FXR1, an RNA -binding protein in the muscle cells that
form the outer surface of the blood vessels. Additional data suggested that IL-19, a protein secreted by certain white
blood cells, might regulate expression of the FXR1 gene.

To investigate the regulation of the FXR1 gene by IL-19, a researcher added IL-19 to vessel-lining muscle cells
growing in the lab. The researcher measured the amount of FXR1 protein produced by the cells over a period of 48
hours in the presence of IL-19. The researcher then calculated the relative amount of FXR1 present at each time
point compared with the amount of FXR1 at Time 0, when IL-19 was first added to the cells. Data from three
replicate experiments are shown in Table 1.

TABLE 1. RELATIVE AMOUNT OF FXR1 PRODUCED BY CELLS

IN RESPONSE TO IL-19

Time in the Presence of Relative Amount of FXR1

IL-19 (h) Protein ±2SE x

0 0.98 ± 0.070

4 1.70 ± 0.220

8 1.60 ± 0.275

16 3.10 ± 0.800

24 2.15 ± 0.405

48 1.60 ± 0.520

(a) Describe how amino acids are categorized by their chemical properties. Explain how a change in the
amino acid sequence of the FXR1 protein could decrease the ability of the protein to bind to RNA .

(b) Using the template in the space provided for your response, construct an appropriately labeled graph that
represents the data shown in Table 1. Determine whether there is a statistical difference in the amount of
FXR1 protein produced by the cells after 16 and 24 hours in the presence of IL-19.

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AP Biology Practice Exam 51

 (c) Based on the data for the 48-hour period, describe the effect of IL-19 on FXR1 gene expression.
(d) The researcher hypothesizes that the FXR1 gene codes for a protein that binds to mRNAs that encode some
of the proteins that damage arteries. Individuals with a particular mutation of the FXR1 gene tend to have
high levels of these proteins. Based on this information, predict how the FXR1 protein most likely interacts
with the mRNAs.

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52 AP Biology Practice Exam

THIS PAGE MAY BE USED FOR TAKING NOTES AND PLANNING YOUR ANSWERS.
NOTES WRITTEN ON THIS PAGE WILL NOT BE SCORED.
WRITE ALL YOUR RESPONSES ON THE LINED PAGES.

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AP Biology Practice Exam 53

 PAGE FOR ANSWERING QUESTION 2
(data table reprinted for reference)

TABLE 1. RELATIVE AMOUNT OF FXR1 PRODUCED BY CELLS

IN RESPONSE TO IL-19

Time in the Presence of Relative Amount of FXR1

IL-19 (h) Protein ±2SE x

0 0.98 ± 0.070

4 1.70 ± 0.220

8 1.60 ± 0.275

16 3.10 ± 0.800

24 2.15 ± 0.405

48 1.60 ± 0.520

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54 AP Biology Practice Exam

ADDITIONAL PAGE FOR ANSWERING QUESTION 2

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AP Biology Practice Exam 55

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58 AP Biology Practice Exam

3. Researchers grew populations of identical Escherichia coli (E. coli ) bacteria in a growth medium that contained a
low concentration of glucose and a high concentration of citrate, a substance that is not typically consumed by
E. coli . For thousands of generations, the bacteria used only glucose as an energy source and grew relatively slowly
and to a low density because of the low concentration of glucose. After about 30,000 generations, one population
emerged that began to rapidly grow to a much higher density. The researchers hypothesized that the bacteria evolved
the ability to use citrate as an energy source and referred to them as Cit+. To test the hypothesis, the researchers grew
separate populations of the Cit+ bacteria and bacteria from the original population (Cit−) in a growth medium that
contained only citrate.

(a) Describe one outcome that would demonstrate that a given population has evolved.
(b) Identify the dependent variable measured in the experiments.

(c) Predict the results obtained by the researchers when they grew the Cit+ and Cit− bacteria in the medium
that contained only citrate.

(d) The researchers claim that the Cit+ mutation increases the fitness of the bacteria. Provide reasoning to
support the claim.

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AP Biology Practice Exam 59

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60 AP Biology Practice Exam

4. The Tasmanian devil (Sarcophilus harrisii) is a species of mammal whose only wild population is native to the
isolated island of Tasmania, off the coast of Australia. Tasmanian devils have a very low level of genetic diversity
across the entire species. Researchers hypothesize this lack of diversity is the result of several genetic bottlenecks in
the species’ history.

In 1996 a new disease called devil facial tumor disease (DFTD) appeared in the wild Tasmanian devil population.
DFTD can spread from one animal to another and causes the death of most affected animals. The impact of DFTD
on the Tasmanian devil population has resulted in a decrease in the total population of more than 85% over the last
two decades.

(a) Describe the process that maintained a stable Tasmanian devil population size before the appearance of
DFTD in 1996.
(b) Explain how the huge reduction of the Tasmanian devil population since 1996 affects the susceptibility of
the current population to new diseases in comparison with the susceptibility of the population before 1996.
(c) Tasmanian devils are top predators and are considered a keystone species in their community. Predict the
effect of the rapid reduction of the Tasmanian devil population on the rest of the community.
(d) Justify the prediction of part (c).

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AP Biology Practice Exam 61

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62 AP Biology Practice Exam

5. The p53 protein is a transcription factor that regulates a cell’s progression through the cell cycle after DNA is
damaged by ultraviolet (UV) radiation. The p53 pathway is shown in Figure 1.

Figure 1. The role of p53 in regulating cell cycle progression in response to DNA damage
(a) A skin cell completes one round of the cell cycle. Describe the products.

(b) Based on Figure 1, explain how p53 regulates the cell cycle in the presence of damaged DNA.

(c) Draw an X on the template in the space provided for your response to indicate the phase during which the
replication of damaged DNA would occur.
(d) Based on Figure 1, explain how a mutation to p53 may lead to an increased risk of cancer.

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AP Biology Practice Exam 63

 PAGE FOR ANSWERING QUESTION 5

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66 AP Biology Practice Exam

6. The human kidney filters the blood to retain useful substances and remove waste products and excess water as urine.
SLC3A1 and SLC7A9 are two different genes that each encode one part of a transporter protein complex typically
found in the plasma membrane of human kidney cells. This transporter complex moves certain amino acids across
the plasma membrane of the cells. Mutations in both the SLC3A1 and SLC7A9 genes are associated with increased
levels of the polar amino acid cysteine in the urine, a condition called cystinuria. A researcher identified seven
individuals with a family history of cystinuria. The researcher measured the concentration of cysteine in the urine of
the seven individuals and sequenced their SLC3A1 and SLC7A9 genes to determine whether they carry functional
(+) or mutant (−) alleles of genes. The results are shown in Table 1.

TABLE 1. CHARACTERISTICS OF SEVEN INDIVIDUALS WHO CARRY AT LEAST

ONE MUTANT ALLELE

Concentration of Cysteine
Individual in Urine Relative to SLC3A1 Alleles SLC7A9 Alleles
Normal Values
1 32.2 −/ − +/ −

2 29.8 +/ − −/ −

3 11.0 −/ − +/ +

4 8.8 +/ + −/ −

5 7.7 −/ − +/ +

6 2.3 +/ + +/ −

7 0.5 +/ − +/ +

+ indicates the presence of a functional allele and − indicates the

presence of a mutant allele.
(a) Identify the individual who most likely exhibits symptoms of cystinuria.
(b) Describe the relationship between the total number of mutant alleles in an individual and the concentration
of cysteine in the urine.

(c) Evaluate the hypothesis that mutations in SLC7A9 have a greater effect on the transport of cysteine across
the plasma membrane of kidney cells than do mutations in SLC3A1.
(d) Explain how the data support the claim that cysteine is a large polar molecule.

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AP Biology Practice Exam 67

 PAGE FOR ANSWERING QUESTION 6

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68 AP Biology Practice Exam

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AP Biology Practice Exam 69

 STOP

END OF EXAM

IF YOU FINISH BEFORE TIME IS CALLED,

YOU MAY CHECK YOUR WORK ON THIS SECTION.
___________________________________________

THE FOLLOWING INSTRUCTIONS APPLY TO THE COVERS OF THE SECTION II BOOKLET.

MAKE SURE YOU HAVE DONE THE FOLLOWING:

• COMPLETED THE IDENTIFICATION INFORMATION AS REQUESTED ON THE

FRONT AND BACK COVERS OF THE SECTION II BOOKLET
• CHECKED THAT YOUR AP ID LABEL IS IN THE BOX ON THE FRONT COVER

70 AP Biology Practice Exam

Notes on the AP Biology Practice Exam
Multiple-Choice Section
Course Framework Alignment and Rationales

Question 1
Skill Learning Objective Topic
1.A: Describe biological concepts SYI-1.B: Describe Introduction
and/or processes. the properties of the to Biological
monomers and the type Macromolecules
of bonds that connect the
monomers in biological
macromolecules.
(A) Incorrect. The removal of water is not involved in breaking the bond
between sugar monomers.
(B) Incorrect. Hydrolysis of carbohydrates does not result in the formation
of bonds between sugar molecules. Additionally, it does not involve
the removal of a water molecule.
(C) Correct. The hydrolysis of carbohydrates involves the addition of a
water molecule, which breaks the covalent bond between monomers.
(D) Incorrect. Although hydrolysis does involve the addition of a water
molecule, hydrolysis does not result in the formation of bonds
between sugar molecules.

Question 2
Skill Learning Objective Topic
1.A: Describe biological concepts IST-1.L: Describe the DNA and RNA
and/or processes. characteristics of DNA that Structure
allow it to be used as the
hereditary material.
(A) Incorrect. There are only four different types of nucleotide bases that
can be incorporated and the number of types doesn’t contribute to its
usefulness as hereditary material.
(B) Incorrect. Although nucleotide bases could be metabolized to provide
energy, this would not help maintain the sequence of bases making it
useful as hereditary material.
(C) Incorrect. The random replacement of nucleotide bases in a DNA
molecule would destroy the information contained in the sequence.
(D) Correct. The pairing of specific bases allows one strand to be used as
a template for replicating the other strand, ensuring that DNA can be
replicated with a relatively high accuracy.

AP Biology Practice Exam 71

 Question 3
Skill Learning Objective Topic
2.B: Explain relationships SYI-3.C: Explain how Chromosomal
between different characteristics chromosomal inheritance Inheritance
of biological concepts or generates genetic variation
processes represented visually- in sexual reproduction.
a. In theoretical contexts. b. In
applied contexts.
(A) Incorrect. Although autosomal dominance pattern of inheritance can
explain individuals I-3 and I-4 producing an affected offspring, it cannot
explain individuals I-1 and I-2 producing an affected offspring.
(B) Correct. Only an autosomal recessive pattern of inheritance can explain
individuals I-1 and I-2, who are unaffected, producing an affected
offspring. The autosomal recessive pattern is also consistent with the
complete pedigree.
(C) Incorrect. If the trait was a sex-linked dominant trait, then either
individual II-5 or II-6 would have to be affected.
(D) Incorrect. If the trait was a sex-linked recessive trait, then individual
II-3 would have to be affected.

Question 4
Skill Learning Objective Topic
1.A: Describe biological concepts ENE-1.L: Cellular Respiration
and/or processes. Explain how cells
obtain energy
from biological
macromolecules
in order to power
cellular functions.
(A) Correct. Lactic acid is one of the end products of anaerobic metabolism
of sucrose.
(B) Incorrect. The creation of a H+ gradient to synthesize ATP is not
associated with glycolysis or fermentation.
(C) Incorrect. While mitochondria do produce ATP in the presence of
oxygen, bacteria do not have mitochondria.
(D) Incorrect. While CO2 is produced in the Krebs cycle during aerobic
respiration, CO2 is not converted to lactic acid.

72 AP Biology Practice Exam

Question 5
Skill Learning Objective Topic
3.C: Identify experimental ENE-1.L: Explain how Cellular
procedures that are aligned cells obtain energy from Respiration
to the question, including- a. biological macromolecules
Identifying dependent and in order to power cellular
independent variables. b. functions.
Identifying appropriate controls.
c. Justifying appropriate controls.
(A) Incorrect. Time is an independent variable in this experiment because
time is controlled.
(B) Correct. The pH is being measured in this experiment and is therefore
the dependent variable.
(C) Incorrect. While pH decreases as glucose is metabolized, glucose
concentration was not measured in this experiment and cannot be a
dependent variable.
(D) Incorrect. While pH decreases as lactic acid is produced, lactic acid
concentration was not measured in this experiment and cannot be a
dependent variable.

Question 6
Skill Learning Objective Topic
4.B: Describe data from a table or ENE-1.L: Explain how Cellular
graph, including- a. Identifying cells obtain energy from Respiration
specific data points. b. Describing biological macromolecules
trends and/or patterns in the in order to power cellular
data. c. Describing relationships functions.
between variables.
(A) Incorrect. While the pie chart does indicate two time periods used in the
experiment, it provides no indication of pH.
(B) Correct. This graph shows the decreasing pH values and error bars over
time for both the control and treatment cultures.
(C) Incorrect. While the stacked bar chart does indicate pH values
consistent with the data from the treatment group, the control data are
hard to interpret because only the difference between the treatment and
control values are shown. Also, time is not shown.
(D) Incorrect. While the range of pH values for the control and treatment
groups may be represented by this graph, it does not show the change in
pH over time.

AP Biology Practice Exam 73

 Question 7
Skill Learning Objective Topic
4.B: Describe data from a table or ENE-1.L: Explain how Cellular
graph, including- a. Identifying cells obtain energy from Respiration
specific data points. b. Describing biological macromolecules
trends and/or patterns in the in order to power cellular
data. c. Describing relationships functions.
between variables.
(A) Incorrect. The pH values ±2SE x of the control and treatment at
5 minutes overlap, so the values are statistically indistinguishable.
(B) Incorrect. The pH values ±2SE x of the control and treatment at
15 minutes overlap, so the values are statistically indistinguishable.
(C) Incorrect. The pH values ±2SE x of the control and treatment at
20 minutes overlap, so the values are statistically indistinguishable.
(D) Correct. The first time the pH values ±2SE x of the control and
treatment do not overlap is at 35 minutes.

Question 8
Skill Learning Objective Topic
2.B: Explain relationships ENE-1.L: Explain how Cellular
between different characteristics cells obtain energy from Respiration
of biological concepts or biological macromolecules
processes represented visually- in order to power cellular
a. In theoretical contexts. b. In functions.
applied contexts.
(A) Correct. An increased bacterial metabolic rate results in the production
of more lactic acid, which lowers the pH of the cultures.
(B) Incorrect. The pH of the treatment culture was never higher than that of
the control culture.
(C) Incorrect. Because the initial pH of both cultures was the same,
there is no indication that the compound directly affected the pH .
(D) Incorrect. The bacteria in both cultures were using anaerobic
metabolism to break down the glucose. Oxygen is not required for this
process, so binding all available oxygen to another molecule should not
impact the rate of metabolism.

74 AP Biology Practice Exam

Question 9
Skill Learning Objective Topic
1.C: Explain biological concepts, SYI-1.B: Describe Properties
processes, and/or models in the properties of the of Biological
applied contexts. monomers and the type Macromolecules
of bonds that connect the
monomers in biological
macromolecules.
(A) Incorrect. The mass of the enzyme should not impact the affinity of the
enzyme for its substrate.
(B) Correct. The replacement of a polar amino acid with a nonpolar amino
acid will affect the folding of the enzyme because there will be different
interactions between the R groups.
(C) Incorrect. The replacement of the amino acid should not result in
competitive inhibition between enzymes.
(D) Incorrect. The replacement of the amino acid will not change the
directionality of the chain of amino acids.

Question 10
Skill Learning Objective Topic
1.C: Explain biological concepts, ENE-4.B: Explain how Community
processes, and/or models in interactions within and Ecology
applied contexts. among populations
influence community
structure.
(A) Incorrect. There is nothing to suggest that the bat is harmed by this
relationship.
(B) Correct. The pitcher plant benefits from the nitrogen from the bat, and
the bat benefits from the safe place to sleep.
(C) Incorrect. The bat benefits from the safe place to sleep provided by the
plant.
(D) Incorrect. The plant is affected by this relationship in that it benefits
from the nitrogen received from the bat feces.

AP Biology Practice Exam 75

 Question 11
Skill Learning Objective Topic
5.A: Perform mathematical EVO-1.K: Describe the Hardy-
calculations, including- a. conditions under which Weinberg
Mathematical equations in the allele and genotype Equilibrium
curriculum. b. Means. c. Rates. frequencies will change in
d. Ratios. e. Percentages. populations.
(A) Incorrect. This is the value for q2, which is the frequency of individuals
with the disorder.
(B) Incorrect. This is the value for q, which is the frequency of the recessive
allele.
(C) Correct. This is the value for 2pq, which is the frequency of individuals
who are heterozygous for the disorder.
(D) Incorrect. This is the value for p, which is the frequency of the dominant
allele.

Question 12
Skill Learning Objective Topic
2.A: Describe characteristics of IST-3.D: Describe the role Introduction
a biological concept, process, or of components of a signal to Signal
model represented visually. transduction pathway Transduction
in producing a cellular
response.
(A) Incorrect. Nitric oxide is only involved in the first step of the signaling
cascade and there is no indication that the nitric oxide triggered
signaling increases over time.
(B) Correct. One of the benefits of a signaling cascade is that a relatively few
initial signaling molecules can result in a large change, because at each
step, a signaling molecule activates many other signaling molecules.
(C) Incorrect. The thickness of the arrows represents the increase in the
number of signaling molecules. Just increasing the size of the proteins in
the pathway would not create a signaling cascade.
(D) Incorrect. There is no indication of any negative feedback in this
pathway.

76 AP Biology Practice Exam

Question 13
Skill Learning Objective Topic
6.D: Explain the relationship ENE-2.H: Explain how Tonicity and
between experimental results concentration gradients Osmoregulation
and larger biological concepts, affect the movement
processes, or theories. of molecules across
membranes.
(A) Incorrect. The mass of the grape decreased in NaCl solution, indicating
that the grape lost water.
(B) Correct. The mass of the grape decreased in both the grape soda and
the NaCl solutions, indicating that water left. Since water moves from
higher water potential to lower water potential, the grape has a higher
water potential than either solution has.
(C) Incorrect. The mass of the grape increased in both the tap water and the
grape juice, indicating the grape gained water.
(D) Incorrect. The mass of the grape decreased in the grape soda, indicating
that the grape lost water.

Question 14
Skill Learning Objective Topic
4.B: Describe data from a ENE-2.H: Explain how Tonicity and
table or graph, including- a. concentration gradients Osmoregulation
Identifying specific data affect the movement
points. b. Describing trends of molecules across
and/or patterns in the data. membranes.
c. Describing relationships
between variables.
(A) Incorrect. The distilled water had a concentration of 0.0%, and the mass
of the grape increased by 13.48% indicating the concentration of the
grape and solution were not equal.
(B) Incorrect. The grape gained mass in the grape juice which has a
concentration of 2.1%, suggesting that the concentration of the grape is
higher than 2.1% not lower.
(C) Correct. The grape had a small positive change in mass in the 2.1%
solution and a large negative change in mass in the 13% solution,
suggesting the concentration of the grape is somewhere between those
two percentages but closer to 2.1%.
(D) Incorrect. Although 10.1% is between the concentration of the solution
with the smallest decrease in mass (13%) and the concentration of the
solution with the smallest increase in mass (2.1%), 10.1% is closer to
13% than 2.1%, indicating the change in mass would be closer to –15%
than 0.0%.

AP Biology Practice Exam 77

 Question 15
Skill Learning Objective Topic
5.D: Use data to evaluate a ENE-2.H: Explain how Tonicity and
hypothesis (or prediction), concentration gradients Osmoregulation
including- a. Rejecting or failing affect the movement
to reject the null hypothesis. of molecules across
b. Supporting or refuting the membranes.
alternative hypothesis.
(A) Incorrect. According to the data provided, the mass of the grape
increased in the grape juice.
(B) Incorrect. The mass of the grape increased in the grape juice, suggesting
the solute potential of the grape was higher than that of the grape juice.
(C) Incorrect. The mass of the grape increased while in grape juice.
The mass would have stayed the same if the grape and the grape juice
were isotonic.
(D) Correct. The mass of the grape increased in the grape juice, suggesting
the grape juice is hypotonic compared with the grape and thus allowed
water to move into the grape. Extra sugar would have likely made the
grape juice hypertonic.

Question 16
Skill Learning Objective Topic
3.B: State the null and alternative ENE-2.E: Describe the Membrane
hypotheses or predict the results mechanisms that organisms Transport
of an experiment. use to maintain solute and
water balance.
(A) Correct. Small diffusible solutes would diffuse into the cell because
there is a lower concentration inside the grape. Combined with the
concentration of nondiffusible solutes, the interior of the grape will now
be hypertonic to the exterior and water would move in.
(B) Incorrect. There is a higher concentration of small diffusible solutes
outside the grape than inside the grape. Solutes move from high
concentration to low concentration.
(C) Incorrect. There is a higher concentration of small diffusible solutes
outside the grape than inside the grape. Solutes move from high
concentration to low concentration.
(D) Incorrect. While small diffusible solutes will move into the grape
increasing the osmolarity of the grape, water will not move out of the
grape.

78 AP Biology Practice Exam

Question 17
Skill Learning Objective Topic
6.E: Predict the causes or effects ENE-2.G: Explain how the Facilitated
of a change in, or disruption to, structure of a molecule Diffusion
one or more components in a affects its ability to pass
biological system based on- a. through the plasma
Biological concepts or processes. membrane.
b. A visual representation of a
biological concept, process, or
model. c. Data.
(A) Incorrect. Active transport is not used to move water through the
membrane.
(B) Incorrect. Mercurial sulfhydryl inhibits the transport protein. It does not
activate the transport protein.
(C) Incorrect. Although mercurial sulfhydryl inhibits aquaporins, water is
not transported actively.
(D) Correct. While small amounts of water can pass through the membrane
by simple diffusion, aquaporins are channel proteins that allow larger
amounts of water to move over a shorter period of time.

Question 18
Skill Learning Objective Topic
1.B: Explain biological concepts ENE-1.B: Explain the effect Cell Size
and/or processes. of surface area-to-volume
ratios on the exchange of
materials between cells
or organisms and the
environment.
(A) Correct. A grape with a smaller surface-area-to-volume ratio will have a
slower rate of water absorption per gram.
(B) Incorrect. A grape with a larger surface-area-to-volume ratio would
have a higher rate of water absorption per gram, not a lower one.
(C) Incorrect. The ability to expand is not related to the rate of water
absorption per gram. Additionally, there is nothing to suggest that
smaller grapes would be able to expand more than larger grapes can.
(D) Incorrect. Larger grapes may have a larger volume, but the rate of water
absorption per gram is more dependent on the surface area.

AP Biology Practice Exam 79

 Question 19
Skill Learning Objective Topic
1.C: Explain biological concepts, IST-2.D: Explain the Gene
processes, and/or models in connection between Expression
applied contexts. the regulation of gene and Cell
expression and phenotypic Specialization
differences in cells and
organisms.
(A) Incorrect. The majority of the cells in an organism have the same genes.
(B) Incorrect. Cells do not evolve. They may develop and specialize for
different functions, but this is not an evolutionary process.
(C) Correct. The process by which cells specialize for certain functions
involves activating different sets of genes.
(D) Incorrect. Eukaryotic organisms have only two types of ribosomes,
a larger ribosome in the cytosol and a small ribosome in the
mitochondria. These two types of ribosomes do not account for the
diverse sets of genes that are activated in different types of cells.

Question 20
Skill Learning Objective Topic
2.B: Explain relationships ENE-1.G: Explain how the Environmental
between different characteristics cellular environment affects Impacts on
of biological concepts or enzyme activity. Enzyme
processes represented visually- Function
a. In theoretical contexts. b. In
applied contexts.
(A) Incorrect. Although the total amount of product increases over time, the
decreasing slope of the line indicates that the reaction rate is decreasing
over time not increasing.
(B) Incorrect. The decreasing slope of the line indicates that the reaction
rate is decreasing. Additionally, an increase in the ratio of products to
substrate should result in a decrease in the rate of the reaction not an
increase.
(C) Correct. The decreasing slope of the line indicates that the reaction
rate is decreasing over time. As the reaction proceeds and the ratio of
product to substrate increases, the reaction rate decreases.
(D) Incorrect. Although the decreasing slope of the line indicates that the
reaction rate is decreasing over time, enzymes are not used up by the
reaction.

80 AP Biology Practice Exam

Question 21
Skill Learning Objective Topic
1.C: Explain biological concepts, ENE-1.G: Explain how the Environmental
processes, and/or models in cellular environment affects Impacts on
applied contexts. enzyme activity. Enzyme
Function
(A) Correct. Changing the shape of the enzyme’s active site will prevent
it from catalyzing the reaction regardless of the concentration of the
substrate.
(B) Incorrect. Increasing substrate-enzyme binding would increase the
activity of the enzyme, not inhibit the enzyme.
(C) Incorrect. If alanine were competing with the substrate for the active
site, then increasing the concentration of the substrate would increase
the chances of the substrate binding and decrease the effect of the
inhibitor.
(D) Incorrect. Alanine binds to and affects the enzyme, not the substrate.

Question 22
Skill Learning Objective Topic
1.C: Explain biological concepts, ENE-4.B: Explain how Community
processes, and/or models in interactions within and Ecology
applied contexts. among populations
influence community
structure.
(A) Incorrect. The bumblebee species are not geographically isolated from
one another.
(B) Correct. The bumblebee species rely on different flowers depending on
the length of their proboscis. This allows multiple species to occupy the
same area while decreasing competition.
(C) Incorrect. There may be a mutualistic relationship between bumblebees
and the plants they pollinate, but not between different bumblebee
species.
(D) Incorrect. No bumblebee species appears to benefit from the presence of
the other species.

AP Biology Practice Exam 81

 Question 23
Skill Learning Objective Topic
4.B: Describe data from a SYI-3.B: Explain how the Environmental
table or graph, including- a. same genotype can result in Effects on
Identifying specific data multiple phenotypes under Phenotype
points. b. Describing trends different environmental
and/or patterns in the data. conditions.
c. Describing relationships
between variables.
(A) Incorrect. The rate of mRNA H degradation is independent of the
length of light exposure, which suggests that light does not cause its
degradation.
(B) Incorrect. mRNA H degrades more rapidly than mRNA G during
morning exposure to light.
(C) Correct. The rate of degradation of mRNA G after the plant has been
exposed to light for 8 hours is greater than the rate of degradation of
mRNA G after the plant has been exposed to light for only 1 hour. The
rate of mRNA H degradation is unaffected by the length of time the
plant has been exposed to light.
(D) Incorrect. The rate of degradation of mRNA H is not affected by the
time of day.

Question 24
Skill Learning Objective Topic
1.A: Describe biological EVO-3.D: Describe the Speciation
concepts and/or processes. conditions under which
new species may arise.
(A) Incorrect. Individuals of the same species can have differences between
their RNA sequences.
(B) Incorrect. Individuals of the same species can be morphologically
distinct.
(C) Correct. Based on the biological species concept, organisms of different
species cannot produce hybrid offspring that are capable of reproducing.
(D) Incorrect. A single species can occupy multiple habitats.

82 AP Biology Practice Exam

Question 25
Skill Learning Objective Topic
6.C: Provide reasoning to justify a IST-3.B: Explain how cells Cell
claim by connecting evidence to communicate with one Communication
biological theories. another over short and
long distances.
(A) Incorrect. Phosphorylation of the insulin receptor does not provide
evidence that osteocalcin is a hormone.
(B) Correct. Hormones are signaling molecules that can be secreted from
one part of the body and act on another part of the body.
(C) Incorrect. Activation of intracellular pathways is not necessarily
dependent on hormone signaling.
(D) Incorrect. The pH dependency of osteocalcin activation does not
provide evidence that osteocalcin is a hormone.

Question 26
Skill Learning Objective Topic
2.A: Describe characteristics of IST-3.C: Describe the Introduction
a biological concept, process, or components of a signal to Signal
model represented visually. transduction pathway. Transduction
(A) Correct. Insulin binding increases the secretion of additional insulin via
osteocalcin signaling.
(B) Incorrect. Insulin binding stimulates the production of osteocalcin.
(C) Incorrect. Insulin binding is inhibited by Esp protein.
(D) Incorrect. Insulin binding causes osteoclasts to release protons and
create an area of lower pH outside the cell.

AP Biology Practice Exam 83

 Question 27
Skill Learning Objective Topic
6.A: Make a scientific claim IST-3.F: Describe the Signal
different types of cellular Transduction
responses elicited by
a signal transduction
pathway.
(A) Correct. Mutant mice that do not express Esp have a higher blood
insulin level twenty minutes after injection and beyond.
(B) Incorrect. Normal mice that produce Esp protein maintain a regulated
blood insulin level after injection.
(C) Incorrect. Insulin levels in normal mice are never statistically greater
than insulin levels in mutant mice.
(D) Incorrect. The data do not show a cyclical pattern of insulin secretion for
the mutant mice.

Question 28
Skill Learning Objective Topic
3.C: Identify experimental IST-3.E: Describe the role Signal
procedures that are aligned of the environment in Transduction
to the question, including- a. eliciting a cellular response.
Identifying dependent and
independent variables. b.
Identifying appropriate controls.
c. Justifying appropriate controls.
(A) Incorrect. Minutes after glucose injection is an independent variable
that is controlled by the researcher.
(B) Incorrect. Blood insulin is the dependent variable in the experiment that
is measured.
(C) Incorrect. The response of the mutant strain of mice to glucose was
unknown.
(D) Correct. Treatment of the normal strain of mouse has a known,
expected result to ensure the experiment is functioning and is therefore
the positive control.

84 AP Biology Practice Exam

Question 29
Skill Learning Objective Topic
2.A: Describe characteristics of ENE-3.C: Explain how Feedback
a biological concept, process, or positive feedback affects
model represented visually. homeostasis.
(A) Correct. The release of insulin from the pancreas ultimately leads to
more insulin being released. The excess insulin production results in too
much sugar being removed from the bloodstream and the low blood
sugar seen in the mutant mice.
(B) Incorrect. Inactive osteocalcin does not affect insulin production and
removal of sugar from the bloodstream.
(C) Incorrect. The Esp protein is not part of a positive feedback loop.
(D) Incorrect. Insulin-secreting pancreatic cell proliferation is not
dependent on a negative feedback pathway.

Question 30
Skill Learning Objective Topic
5.C: Perform chi-square IST-1.I: Explain the Mendelian
hypothesis testing. inheritance of genes and Genetics
traits as described by
Mendel’s laws.
(A) Incorrect. Four is the number of offspring expected to have the recessive
trait for both genes.
(B) Incorrect. Twelve is the number of offspring expected to have the
dominant trait for one gene and the recessive trait for the other gene.
(C) Correct. In a dihybrid cross of mice that are heterozygous for both
traits, nine out of sixteen offspring are expected to have the dominant
9
phenotype for both traits. × 64 = 36
16
(D) Incorrect. 48 is the number of offspring expected to have the dominant
phenotype in a monohybrid cross of heterozygotes.

AP Biology Practice Exam 85

 Question 31
Skill Learning Objective Topic
4.B: Describe data from a IST-1.L: Describe the DNA and RNA
table or graph, including- a. characteristics of DNA that Structure
Identifying specific data allow it to be used as the
points. b. Describing trends hereditary material.
and/or patterns in the data.
c. Describing relationships
between variables.
(A) Correct. A double-stranded nucleic acid will contain equal amounts of
complementary bases. Sample one contains 23 percent of both cytosine
and guanine and 27 percent of both adenine and uracil, consistent with
Chargaff’s rule. The presence of uracil instead of thymine indicates that
the nucleic acid is an RNA molecule, not a DNA molecule.
(B) Incorrect. While Sample 2 shows base percentages consistent with
a double-stranded nucleic acid, with a 17 percent frequency of both
cytosine and guanine and 33 percent frequency of both adenine
and thymine, the presence of thymine instead of uracil indicates the
molecule is DNA, not RNA.
(C) Incorrect. While the presence of uracil and absence of thymine indicates
that Sample 3 is RNA not DNA, the uneven distribution of the four
bases indicates that this is a single-stranded nucleic acid.
(D) Incorrect. While the relative frequencies of the four bases in Sample 4
suggest that this is a double-stranded molecule, the presence of thymine
instead of uracil indicates that this is DNA, not RNA.

Question 32
Skill Learning Objective Topic
1.C: Explain biological concepts, EVO-3.F: Explain the Speciation
processes, and/or models in processes and mechanisms
applied contexts. that drive speciation.
(A) Incorrect. Using different food sources does not necessarily prevent gene
flow between two species.
(B) Incorrect. The information indicates that goats and sheep often share a
habitat, so they are not isolated. Additionally, while hybrids are sterile
or do not survive, some are produced, indicating that the barrier is
postzygotic.
(C) Incorrect. Different times of fertility would be a prezygotic barrier, and
the information indicates that zygotes, and sometimes viable offspring,
are formed.
(D) Correct. Closely related species might be able to mate and even produce
a viable offspring, but if the chromosome numbers of the two species
are not the same, the offspring will not have homologous pairs of
chromosomes and will not be able to produce functional gametes.

86 AP Biology Practice Exam

Question 33
Skill Learning Objective Topic
1.C: Explain biological concepts, ENE-3.C: Explain how Feedback
processes, and/or models in positive feedback affects
applied contexts. homeostasis.
(A) Correct. This is an explanation of positive feedback. The production of
ethylene by some cells results in other cells producing ethylene, causing
a rapid increase in the ripening process and increasing the overall level
of ethylene present.
(B) Incorrect. This is an explanation of negative feedback. The production
of ethylene in response to low water stress results in root growth,
increasing the opportunity to find water and decrease low water stress.
(C) Incorrect. This is an explanation of negative feedback. The production
of ethylene in response to insect damage repels insects, decreasing the
insect damage. A decrease in the insect population would lead to a
decrease in ethylene production.
(D) Incorrect. This is an explanation of negative feedback. The production of
ethylene by fertilized ovules in a flower results in controlled death of the
flower petals.

Question 34
Skill Learning Objective Topic
5.A: Perform mathematical ENE-1.O: Explain how the Energy Flow
calculations, including- a. activities of autotrophs and through
Mathematical equations in the heterotrophs enable the Ecosystems
curriculum. b. Means. c. Rates. flow of energy within an
d. Ratios. e. Percentages. ecosystem.
(A) Incorrect. The average weekly growth rate for the whole three-week
period is 3.2 g/week (total growth rate divided by three weeks).
(B) Correct. The approximate growth rate of section II is 4.5 g/week
for days 14 to 21. Assuming the mass at day 14 of section II is the same
as that of section I , then the rate for days 14 through 21 is
9.6 – 5.1 g/week.
(C) Incorrect. The mass of section I on day 14 is 5.1g.
(D) Incorrect. The total mass of grass in section II on day 21 is 9.6g.

AP Biology Practice Exam 87

 Question 35
Skill Learning Objective Topic
6.D: Explain the relationship ENE-1.M: Describe the Energy Flow
between experimental results strategies organisms use to through
and larger biological concepts, acquire and use energy. Ecosystems
processes, or theories.
(A) Incorrect. The grass in section III did not require more energy than the
grass in section I, but the grass in section III did not store any energy
from photosynthesis during days 14 through 21.
(B) Incorrect. The grass in section III performed no photosynthesis after day
14 since the grass did not receive any light. This is also supported by the
decrease in biomass relative to section I at two weeks.
(C) Incorrect. The grass in all three sections used up the energy stored in
the seeds during germination before the grass was able to perform
photosynthesis.
(D) Correct. Grass in all sections used stored mass for cellular respiration,
but during days 14 through 21 the grass in section III was not able to
produce biomass through photosynthesis.

Question 36
Skill Learning Objective Topic
1.C: Explain biological concepts, ENE-1.H: Describe the Cellular Energy
processes, and/or models in role of energy in living
applied contexts. organisms.
(A) Correct. The plant is considered an ordered system, and without the
input of new energy by photosynthesis, the plant draws on its stored
energy reserves for maintenance at the cost of growth.
(B) Incorrect. Nutrients from the soil are incorporated into plant matter
synthesized for growth, but the energy source is light and the material
for growth is primarily water and carbon dioxide.
(C) Incorrect. The oxygen released during the light-dependent reactions is a
byproduct of photosynthesis and is not incorporated into the synthesis
of sugars but is instead released into the atmosphere.
(D) Incorrect. The loss of mass in section II reflects the plant’s use of stored
energy for maintenance.

88 AP Biology Practice Exam

Question 37
Skill Learning Objective Topic
6.E: Predict the causes or effects ENE-1.N: Explain Energy Flow
of a change in, or disruption to, how changes in through
one or more components in a energy availability Ecosystems
biological system based on- a. affect populations and
Biological concepts or processes. ecosystems.
b. A visual representation of a
biological concept, process, or
model. c. Data.
(A) Correct. The data establish that a lack of adequate light leads to a loss of
biomass. This loss would narrow the base of the energy pyramid, with
less energy available throughout the ecosystem, leading to a decrease in
population sizes of most species and possible extinction of some species
and the loss of one or more of the top trophic levels.
(B) Incorrect. The data establish that without light, biomass decreases rather
than increases.
(C) Incorrect. About 90 percent of the energy obtained by an omnivore is
lost to heat and maintenance. The consumer trophic levels would rapidly
use up energy stores without an influx of new energy from producers.
(D) Incorrect. Niche availability due to loss of energy-starved consumers
does not help other consumers, who would also suffer from the lack of
energy.

Question 38
Skill Learning Objective Topic
3.B: State the null and alternative ENE-1.M: Describe the Energy Flow
hypotheses or predict the results strategies organisms use to through
of an experiment. acquire and use energy. Ecosystems
(A) Incorrect. Predicting a negative effect is still an alternate hypothesis
as it is stating that changing the independent variable will affect the
dependent variable.
(B) Incorrect. This would be an alternate hypothesis that can be tested by
modifying the protocol of the experiment. It predicts a change in the
dependent variable based on a change in the independent variable.
(C) Incorrect. This would be an alternate hypothesis that can be tested by
modifying the protocol of the experiment. It predicts a change in the
dependent variable based on a change in the independent variable.
(D) Correct. A null hypothesis states that a change in the designated
independent variable would have no effect on the dependent variable. In
this experiment, light is the independent variable and the mass of grass
is the dependent variable.

AP Biology Practice Exam 89

 Question 39
Skill Learning Objective Topic
2.A: Describe characteristics of SYI-1.B: Describe Properties
a biological concept, process, or the properties of the of Biological
model represented visually. monomers and the type Macromolecules
of bonds that connect the
monomers in biological
macromolecules.
(A) Incorrect. All amino acids have an amine group in this position.
This amine group forms a covalent peptide bond with the carboxyl
group of another amino acid and does not contribute significantly to
hydrophobic behavior.
(B) Incorrect. All amino acids have a carboxyl group in this position.
This carboxyl group forms a covalent peptide bond with an amine
group of another amino acid and does not contribute significantly to
hydrophobic behavior.
(C) Correct. The methyl group is one of the functional groups that
differentiate the twenty amino acids found in proteins. Since the methyl
group is nonpolar, it is likely to make this region of the polypeptide
more hydrophobic.
(D) Incorrect. Hydrogen atoms, especially those bonded to carbon
atoms, do not on their own add either hydrophobic or hydrophilic
characteristics to their molecule.

Question 40
Skill Learning Objective Topic
5.C: Perform chi-square IST-1.J: Explain deviations Non-Mendelian
hypothesis testing. from Mendel’s model of the Genetics
inheritance of traits.
(A) Correct. The critical value for two degrees of freedom and a p-value of
0.05 is 5.99, which is less than the calculated value of 7.3.
(B) Incorrect. The null hypothesis is rejected if the critical value is less than
the calculated value.
(C) Incorrect. The critical value for two degrees of freedom and a p-value of
0.05 is 5.99, which is less than the calculated value of 7.3.
(D) Incorrect. The critical value for two degrees of freedom and a p-value of
0.05 is 5.99, which is less than the calculated value of 7.3.

90 AP Biology Practice Exam

Question 41
Skill Learning Objective Topic
3.B: State the null and alternative ENE-1.J: Explain how cells Photosynthesis
hypotheses or predict the results capture energy from light
of an experiment. and transfer it to biological
molecules for storage and
use.
(A) Incorrect. The production of NADPH is not directly tied to the proton
gradient across the thylakoid membrane.
(B) Correct. The phosphorylation of ADP to form ATP is dependent on the
flow of hydrogen ions through ATP synthase. Making the membrane
permeable to hydrogen ions would disrupt the proton gradient and
fewer protons would be available to flow through ATP synthase.
(C) Incorrect. The amount of light energy required to excite the electrons in
chlorophyll is not related to a proton gradient.
(D) Incorrect. While the splitting of a water molecule creates protons, this
reaction does not depend on a proton gradient to occur.

Question 42
Skill Learning Objective Topic
4.B: Describe data from a table or EVO-1.N: Explain Evidence for
graph, including- a. Identifying how morphological, Evolution
specific data points. b. Describing biochemical, and geologic
trends and/or patterns in the data provide evidence that
data. c. Describing relationships organisms have changed
between variables. over time.
(A) Incorrect. The data indicate that birds and mammals diverged
approximately 270 million years ago.
(B) Incorrect. The data indicate that reptiles and mammals diverged
approximately 325 million years ago.
(C) Incorrect: The data indicate that fish and land vertebrates diverged
approximately 400 million years ago.
(D) Correct. The data indicate that insects and vertebrates diverged
approximately 600 million years ago.

AP Biology Practice Exam 91

 Question 43
Skill Learning Objective Topic
1.B: Explain biological concepts IST-1.M: Describe the Replication
and/or processes. mechanisms by which
genetic information is
copied for transmission
between generations.
(A) Correct. The use of a parental DNA template strand ensures the new
DNA double helix produced will have the same sequence of bases as the
parental strand.
(B) Incorrect. Although DNA molecules are shaped like a double helix, this
information does not best explain the continuity of genetic information
across generations.
(C) Incorrect. Although the formation of an RNA transcript is important to
cellular functions such as protein synthesis, this does not help explain
the continuity of genetic information across generations.
(D) Incorrect. Although cells do contain different polymerases, this does not
help explain the continuity of genetic information across generations.

Question 44
Skill Learning Objective Topic
1.C: Explain biological concepts, SYI-3.B: Explain how the Environmental
processes, and/or models in same genotype can result in Effects on
applied contexts. multiple phenotypes under Phenotype
different environmental
conditions.
(A) Correct. Environmental factors can influence gene expression, in this
case resulting in a reduction of melanin during the fall and winter.
(B) Incorrect. While environmental factors can cause mutations, these
mutations are random and do not result in specific changes from season
to season.
(C) Incorrect. The is only one universal genetic code used for the translation
of all proteins.
(D) Incorrect. Enzymes are specific to their substrate because of their
structure. Environmental factors associated with a change in the seasons
will not affect the substrates that can bind to an enzyme.

92 AP Biology Practice Exam

Question 45
Skill Learning Objective Topic
6.E: Predict the causes or effects ENE-3.D: Explain how Responses
of a change in, or disruption to, the behavioral and/or to the
one or more components in a physiological response Environment
biological system based on- a. of an organism is related
Biological concepts or processes. to changes in internal or
b. A visual representation of a external environment.
biological concept, process, or
model. c. Data.
(A) Incorrect. Bright light inhibits melatonin production rather than
stimulating it.
(B) Incorrect. Bright light inhibits melatonin production rather than
stimulating it.
(C) Correct. Melatonin promotes sleep, and its production is inhibited
by light. Exposure to light after sunset will likely prevent melatonin
secretion and negatively impact sleep.
(D) Incorrect. While light does inhibit melatonin synthesis, the lack of
melatonin would likely prevent the person from falling asleep.

Question 46
Skill Learning Objective Topic
1.A: Describe biological ENE-1.A: Describe Elements of Life
concepts and/or processes. the composition of
macromolecules required
by living organisms.
(A) Correct. All the bases in the nucleotides that compose nucleic acids
contain nitrogen.
(B) Incorrect. All amino acids contain carbon, oxygen, nitrogen, and
hydrogen. A few amino acids contain sulfur. No unmodified amino
acids contain phosphorus.
(C) Incorrect. All lipids contain carbon, hydrogen, and oxygen. Some lipids
contain nitrogen and phosphorus. Lipids generally do not contain
potassium.
(D) Incorrect. All carbohydrates have carbon, hydrogen, and oxygen.
Carbohydrates generally do not contain magnesium.

AP Biology Practice Exam 93

 Question 47
Skill Learning Objective Topic
2.A: Describe characteristics of ENE-1.H: Describe the Cellular Energy
a biological concept, process, or role of energy in living
model represented visually. organisms.
(A) Incorrect. The hydrolysis of ATP is exergonic (releases energy)
and line 4 represents an endergonic (requires energy) reaction.
Additionally, the phosphorylation of a substrate is endergonic and
Line 1 represents an exergonic reaction.
(B) Correct. Line 4, an endergonic (requires energy) reaction that could
represent the phosphorylation reaction, is coupled to Line 1, an
exergonic reaction (requires energy) that could represent the hydrolysis
reaction. Additionally, the total energy of the reactants represented by
lines 1 and 4 is less than that of the products, because some energy is
converted to heat.
(C) Incorrect. Although line 2 could represent the hydrolysis of ATP,
an exergonic reaction (releases energy), it cannot be coupled to the
reaction represented by line 3. Since no reactions are 100% efficient,
some energy is converted to heat. The total energy of the products must
be less than the total energy of the reactants.
(D) Incorrect. The hydrolysis of ATP is exergonic (releases energy)
and line 3 represents an endergonic reaction (requires energy).
Additionally, the phosphorylation of a substrate is endergonic and
Line 2 represents an exergonic reaction.

Question 48
Skill Learning Objective Topic
6.D: Explain the relationship EVO-1.E: Describe the Natural
between experimental results importance of phenotypic Selection
and larger biological concepts, variation in a population.
processes, or theories.
(A) Incorrect. Although an individual organism may develop a tolerance
for cold temperatures, this adaptation does not involve a change in the
nucleotide base sequence in DNA. Acquired characteristics cannot be
passed on to new generations based on different sequences of base pairs.
(B) Incorrect. There is no information in Table 1 to suggest that a hybrid
species has formed.
(C) Correct. In a population of anoles that showed variation for
cold tolerance, individuals with an increased tolerance for colder
temperatures would have had an advantage over others in a colder
environment. This advantage would have led to increased reproductive
success and an increase in the frequency of those alleles permitting
increased cold tolerance in future generations.
(D) Incorrect. Mutations are random and do not occur in response to colder
temperatures. Existing mutations in individual organisms could affect
the phenotype of the organisms and result in these organisms having a
greater reproductive success.

94 AP Biology Practice Exam

Question 49
Skill Learning Objective Topic
6.E: Predict the causes or effects ENE-4.B: Explain how Community
of a change in, or disruption to, interactions within and Ecology
one or more components in a among populations
biological system based on- a. influence community
Biological concepts or processes. structure.
b. A visual representation of a
biological concept, process, or
model. c. Data.
(A) Correct. Since green anoles have large toe pads, they are likely to occupy
a niche on plants. The brown anoles have smaller toe pads than green
anoles do, suggesting that they would be less successful on plants.
(B) Incorrect. There is no direct information about the feeding habits of the
two lizards in the table. If brown anoles feed mainly at the ground level
and green anoles feed mainly while on plants, then it is more likely that
brown anoles will prey on worms and grubs and green anoles will prey
on flying insects.
(C) Incorrect. Based on toe pad size, they do not share the same niche and
competition is likely to help maintain this niche separation.
(D) Incorrect. There is no information in the table to indicate that one of the
two lizards will be better adapted to the forest habitat.

Question 50
Skill Learning Objective Topic
2.A: Describe characteristics of EVO-3.B: Describe the Phylogeny
a biological concept, process, or types of evidence that
model represented visually. can be used to infer an
evolutionary relationship.
(A) Incorrect. A.evermanni is more closely related to certain species than
to others. It is not the outgroup.
(B) Incorrect. A.stratulus is more closely related to certain species than to
others. It is not the outgroup.
(C) Incorrect. A.gundlachi is more closely related to certain species than to
others. It is not the outgroup.
(D) Correct. Based on the cladogram, A.carolinensis is more distantly
related to all of the other species in the cladogram than the other species
are related to each other.

AP Biology Practice Exam 95

 Question 51
Skill Learning Objective Topic
6.B: Support a claim with EVO-3.B: Describe the Phylogeny
evidence from biological types of evidence that
principles, concepts, processes, can be used to infer an
and/or data. evolutionary relationship.
(A) Incorrect. Dimensions of body parts, which may be under strong
selective pressures, are not reliable indicators of relatedness.
Additionally, small changes in DNA can result in large morphological
changes.
(B) Correct. The rate that mutations appear in the genomes of species is
relatively constant, and therefore the number of differences between two
species can serve as a molecular clock to measure how long ago species
diverged.
(C) Incorrect. Many unrelated species can occupy similar niches.
(D) Incorrect. Similar mating behaviors may have evolved by convergent
evolution and do not indicate a close evolutionary relationship.

Question 52
Skill Learning Objective Topic
6.A: Make a scientific claim EVO-3.B: Describe the Phylogeny
types of evidence that
can be used to infer an
evolutionary relationship.
(A) Incorrect. Convergent evolution increases the similarities between
species rather than the differences suggested by the cladogram.
(B) Correct. Different populations of anoles likely experienced different
selective pressures, which selected for different traits.
(C) Incorrect. There is nothing in the information provided to suggest that
the anole species experienced a genetic bottleneck.
(D) Incorrect. While species do inherit DNA from their common ancestor,
it is the selective pressures that each species experiences that cause
divergence.

96 AP Biology Practice Exam

Question 53
Skill Learning Objective Topic
1.C: Explain biological concepts, IST-1.O: Describe how the Translation
processes, and/or models in phenotype of an organism
applied contexts. is determined by its
genotype.
(A) Incorrect. If the mutation kept the gene wrapped around the histones,
no transcription would occur and, instead of a shortened protein, no
protein would be produced.
(B) Incorrect. Transcription by different transcription factors might affect
when and in which cells the protein would be produced, but it would
not affect the length of the protein.
(C) Correct. Changing a codon in the coding region to a stop codon would
cause the ribosome to stop translating before the polypeptide was fully
formed.
(D) Incorrect. Mutations could affect which amino acids are introduced
into the protein, but they would not affect the type of bond that forms
between adjacent amino acids.

Question 54
Skill Learning Objective Topic
1.C: Explain biological concepts, SYI-1.A: Explain how the Structure of
processes, and/or models in properties of water that Water and
applied contexts. result from its polarity and Hydrogen
hydrogen bonding affect its Bonding
biological function.
(A) Correct. If bile salts interfere with the hydrogen bonds between surface
water molecules, then the surface tension will be reduced and the sulfur
power will sink.
(B) Incorrect. Although water molecules will move from an area of higher
water potential to an area of lower water potential, this does not explain
the floating or sinking of the sulfur powder.
(C) Incorrect. If bile salts increased the density of the urine sample, then
materials such as sulfur powder would have a greater tendency to float
and not sink.
(D) Incorrect. Although bile salts may affect the hydrogen bonds between
water molecules, bile salts are not likely to affect the covalent bonds of
the water molecules.

AP Biology Practice Exam 97

 Question 55
Skill Learning Objective Topic
2.B: Explain relationships IST-4.A: Explain how Mutations
between different characteristics changes in genotype
of biological concepts or may result in changes in
processes represented visually- phenotype.
a. In theoretical contexts. b. In
applied contexts.
(A) Incorrect. While there are many chromosomes, an extra chromosome
would not cause the chromosomes to be so crowded that RNA
polymerase could not access the genes.
(B) Correct. Having an additional copy of all genes on chromosome 21
would result in an increase in the amount of RNA transcribed from
those genes. The protein levels for these genes would be higher than is
typical.
(C) Incorrect. Homologous chromosomes only pair during meiosis, not
mitosis which is necessary for cell division and growth.
(D) Incorrect. The presence of an extra chromosome would not affect the
number of copies of other chromosomes.

Question 56
Skill Learning Objective Topic
6.B: Support a claim with EVO-1.B: Describe Origins of Cell
evidence from biological the relationship Compartmentalization
principles, concepts, processes, between the
and/or data. functions of
endosymbiotic
organelles and their
free-living ancestral
counterparts.
(A) Correct. Existing bacteria all have a DNA genome and the ability to
divide. In order for mitochondria to have been free-living bacteria, as
suggested by the endosymbiont hypothesis, they would need these traits.
(B) Incorrect. The ability to carry out hydrolytic reactions on organic
molecules does not suggest that mitochondria were once free-living
bacteria, as suggested by the endosymbiont hypothesis.
(C) Incorrect. A highly folded membrane does not suggest that
mitochondria were once free-living bacteria, as suggested by the
endosymbiont hypothesis.
(D) Incorrect. Plants and animals are both eukaryotes. The presence of
mitochondria in these organisms does not support that mitochondria
were once free-living bacteria, as suggested by the endosymbiont
hypothesis.

98 AP Biology Practice Exam

Question 57
Skill Learning Objective Topic
1.C: Explain biological concepts, IST-1.N: Describe the Transcription
processes, and/or models in mechanisms by which and RNA
applied contexts. genetic information flows Processing
from DNA to RNA to
protein.
(A) Incorrect. There are no structural differences between eukaryotic DNA
and prokaryotic DNA that would prevent prokaryotic RNA polymerase
from transcribing a eukaryotic gene.
(B) Incorrect. Although DNA sequences can be unique to a species,
the genetic code used to translate an mRNA transcript into a protein is
universal to all species.
(C) Correct. The prokaryotic cell will be able to produce a protein from the
transcript, but it will have a different amino acid sequence and therefore
a different function, because only eukaryotes remove introns from
mRNA transcripts.
(D) Incorrect. While there are structural differences between the ribosomes
of prokaryotes and eukaryotes, prokaryotes contain ribosomes for
translation.

Question 58
Skill Learning Objective Topic
6.E: Predict the causes or effects ENE-4.B: Explain how Community
of a change in, or disruption to, interactions within and Ecology
one or more components in a among populations
biological system based on- a. influence community
Biological concepts or processes. structure.
b. A visual representation of a
biological concept, process, or
model. c. Data.
(A) Incorrect. Cladophora are the producers of this trophic structure.
If the Cladophora population decreases, there would be less food
available to the entire trophic structure resulting in a decrease in the
populations at all levels.
(B) Correct. Increased cadmium in the river will negatively impact the
steelhead trout. With fewer trout feeding on the damselfly nymphs, the
nymph population will increase.
(C) Incorrect. An increase in the steelhead trout population would
result in fewer damselfly nymphs to consume the midge larvae.
Then there would be more midge larvae to consume the Cladophora.
The Cladophora population would decrease, not increase.
(D) Incorrect. The information provided does not suggest the cadmium
has any direct impact on the damselflies. While the steelhead trout
population is expected to decrease, it is not because of a decrease in
damselflies.

AP Biology Practice Exam 99

 Question 59
Skill Learning Objective Topic
6.E: Predict the causes or effects EVO-1.F: Explain how Artificial
of a change in, or disruption to, humans can affect diversity Selection
one or more components in a within a population.
biological system based on- a.
Biological concepts or processes.
b. A visual representation of a
biological concept, process, or
model. c. Data.
(A) Incorrect. Tusk size would not increase due to the need of poachers. If
poachers targeted the elephants with the largest tusks, fewer of these
elephants would survive to reproduce and future generations would
have fewer elephants with large tusks.
(B) Correct. If poachers targeted the elephants with the largest tusks,
then more elephants with small tusks would survive to reproduce and
future generations would see an increase in the number of small-tusked
elephants.
(C) Incorrect. Natural selection is the result of selective pressures and is
not goal directed. Elephants cannot change the size of their tusks just
because it would be beneficial.
(D) Incorrect. Although a decrease in average tusk size is expected, this
would be due to the alleles of the elephants that survive to reproduce,
not the need to reduce energy investment in the tusks.

Question 60
Skill Learning Objective Topic
6.B: Support a claim with IST-1.M: Describe the Replication
evidence from biological mechanisms by which
principles, concepts, processes, genetic information is
and/or data. copied for transmission
between generations.
(A) Incorrect. Because DNA replication has occurred and DNA polymerase
requires an RNA primer to initiate DNA replication, RNA primers must
have formed.
(B) Correct. The missing bonds in the backbone of the lagging strand of
DNA indicate that DNA ligase was not able to join the fragments.
(C) Incorrect. The double-stranded DNA indicates that DNA polymerase
was able to complete the replication of the DNA.
(D) Incorrect. Because complementary bases are held in place by hydrogen
bonding, disruption of hydrogen bonding would have resulted in the
dissociation of the two strands.

100 AP Biology Practice Exam

Answer Key and Question Alignment to Course Framework
Multiple-Choice Answer Skill Learning Topic
Questions Objective
1 C 1.A SYI-1.B Introduction to Biological
Macromolecules
2 D 1.A IST-1.L DNA and RNA Structure
3 B 2.B SYI-3.C Chromosomal Inheritance
4 A 1.A ENE-1.L Cellular Respiration
5 B 3.C ENE-1.L Cellular Respiration
6 B 4.B ENE-1.L Cellular Respiration
7 D 4.B ENE-1.L Cellular Respiration
8 A 2.B ENE-1.L Cellular Respiration
9 B 1.C SYI-1.B Properties of Biological Macromolecules
10 B 1.C ENE-4.B Community Ecology
11 C 5.A EVO-1.K Hardy-Weinberg Equilibrium
12 B 2.A IST-3.D Introduction to Signal Transduction
13 B 6.D ENE-2.H Tonicity and Osmoregulation
14 C 4.B ENE-2.H Tonicity and Osmoregulation
15 D 5.D ENE-2.H Tonicity and Osmoregulation
16 A 3.B ENE-2.E Membrane Transport
17 D 6.E ENE-2.G Facilitated Diffusion
18 A 1.B ENE-1.B Cell Size
19 C 1.C IST-2.D Gene Expression and Cell Specialization
20 C 2.B ENE-1.G Environmental Impacts on Enzyme
Function
21 A 1.C ENE-1.G Environmental Impacts on Enzyme
Function
22 B 1.C ENE-4.B Community Ecology
23 C 4.B SYI-3.B Environmental Effects on Phenotype
24 C 1.A EVO-3.D Speciation
25 B 6.C IST-3.B Cell Communication
26 A 2.A IST-3.C Introduction to Signal Transduction
27 A 6.A IST-3.F Signal Transduction
28 D 3.C IST-3.E Signal Transduction
29 A 2.A ENE-3.C Feedback
30 C 5.C IST-1.I Mendelian Genetics

AP Biology Practice Exam 101

 Multiple-Choice Answer Skill Learning Topic
Questions Objective
31 A 4.B IST-1.L DNA and RNA Structure
32 D 1.C EVO-3.F Speciation
33 A 1.C ENE-3.C Feedback
34 B 5.A ENE-1.O Energy Flow through Ecosystems
35 D 6.D ENE-1.M Energy Flow through Ecosystems
36 A 1.C ENE-1.H Cellular Energy
37 A 6.E ENE-1.N Energy Flow through Ecosystems
38 D 3.B ENE-1.M Energy Flow through Ecosystems
39 C 2.A SYI-1.B Properties of Biological Macromolecules
40 A 5.C IST-1.J Non-Mendelian Genetics
41 B 3.B ENE-1.J Photosynthesis
42 D 4.B EVO-1.N Evidence for Evolution
43 A 1.B IST-1.M Replication
44 A 1.C SYI-3.B Environmental Effects on Phenotype
45 C 6.E ENE-3.D Responses to the Environment
46 A 1.A ENE-1.A Elements of Life
47 B 2.A ENE-1.H Cellular Energy
48 C 6.D EVO-1.E Natural Selection
49 A 6.E ENE-4.B Community Ecology
50 D 2.A EVO-3.B Phylogeny
51 B 6.B EVO-3.B Phylogeny
52 B 6.A EVO-3.B Phylogeny
53 C 1.C IST-1.O Translation
54 A 1.C SYI-1.A Structure of Water and Hydrogen
Bonding
55 B 2.B IST-4.A Mutations
56 A 6.B EVO-1.B Origins of Cell Compartmentalization
57 C 1.C IST-1.N Transcription and RNA Processing
58 B 6.E ENE-4.B Community Ecology
59 B 6.E EVO-1.F Artificial Selection
60 B 6.B IST-1.M Replication

102 AP Biology Practice Exam

Free-Response Section
Scoring Guidelines

Question 1: Interpreting and Evaluating Experimental Results 10 points

Learning Objectives: IST-1.J ENE-1.K ENE-1.L

(A) Describe the pattern of inheritance that is most likely associated with a 1 point
mutation in the MT-ND5 gene. 1.A

• The mutant allele is only inherited from the maternal parent.

Explain why individuals are not typically heterozygous with respect to 1 point
mitochondrial genes. 1.C

• MT-ND5 is a mitochondrial gene and mitochondria only carry a single,

unpaired chromosome.

Total for part (A) 2 points

(B) Identify a dependent variable measured in the researcher’s experiment. 1 point

3.C
Accept one of the following:
• Concentration of lactic acid in the blood
• Concentration of NAD+ in the blood

Identify one control that the researcher could use to improve the validity of 1 point
the experiment. 3.C

• Measure NAD+ and lactic acid levels in a group of other affected individuals
treated with placebo for comparison.
Justify the researcher analyzing blood samples at many intermediate time 1 point
points instead of at only the beginning and the end of the 20-week period. 3.C

• By collecting more data, the researcher will see a more accurate trend.
Total for part (B) 3 points

AP Biology Practice Exam 103

 (C) Describe the relationship between the concentration of NAD+ in the blood 1 point
and the concentration of lactic acid in the blood during the first 5 weeks of 4.B

treatment with the vitamin.

• The concentration of NAD+ increased and the concentration of lactic acid
decreased.

Based on Figure 2, calculate the average rate of change in blood NAD+ 1 point
concentrations from week 5 to week 17. 5.A

Accept one of the following:

• 20 μmol/1/week based on the calculation [( 60 − 300 ) / (17 − 5 )]
• ‒20 μmol/1/week based on the calculation [( 60 − 300 ) / (17 − 5 )]

Total for part (C) 2 points

(D) The researcher performed a follow-up experiment to measure the rate of 1 point
oxygen consumption by muscle and brain cells. Predict the effect of the MT-ND5 3.B

mutation on the rate of oxygen consumption in muscle and brain cells.

• The mutation will decrease the rate of oxygen consumption by the electron
transport chain.
Justify your prediction. 1 point
• The mutation likely inhibits the rate of oxygen consumption because the 6.B

electron transport chain will be less efficient, so less oxygen will be required
as a terminal electron acceptor.

The researcher had hypothesized that the addition of the vitamin that is 1 point
similar in structure to NADH would increase the activity of the mutated NADH 6.D
dehydrogenase enzyme in individuals with the disorder. Explain how the
vitamin most likely increased the activity of the enzyme.
• The enzyme had decreased activity but still functioned. Increasing the amount
of substrate should increase the amount of product produced. Because the
vitamin was similar to NADH, it could bind to the active site of the enzyme
and effectively increase the substrate concentration.
Total for part (D) 3 points
Total for question 1 10 points

104 AP Biology Practice Exam

Question 2: Interpreting
and Evaluating Experimental Results with Graphing 8 points

Learning Objectives: SYI-1.B SYI-1.C IST-2.A IST-2.D IST-2.A

(A) Describe how amino acids are categorized by their chemical properties. 1 point
1.A
• Based on their R groups, amino acids are categorized as hydrophobic,
hydrophilic, or ionic (acidic or basic).

Explain how a change in the amino acid sequence of the FXR1 protein could 1 point
decrease the ability of the protein to bind to RNA. 1.C

• A change in amino acid sequence changes the shape of a protein, so that

RNA-binding sites formed by the interactions of certain amino acids may
change or be lost entirely.
Total for part (A) 2 point

(B) Using the template in the space provided for your response, construct an 3 points
4.A
appropriately labeled graph that represents the data shown in Table 1.
One point for each of the following:
• Correct axes and labeling
• Correctly plotted means
• Correctly plotted error bars
Determine whether there is a statistical difference in the amount of FXR1 1 point
protein produced by the cells after 16 and 24 hours in the presence of IL-19. 5.B

• No statistical difference

Total for part (B) 4 points

(C) Based on the data for the 48-hour period, describe the effect of IL-19 on FXR1 1 point
gene expression. 4.B

• IL-19 causes a small increase in expression of the FXR1 gene.

Total for part (C) 1 point

AP Biology Practice Exam 105

 (D) The researcher hypothesizes that the FXR1 gene codes for a protein that binds to 1 point
mRNAs that encode some of the proteins that damage arteries. Individuals with 6.E

a loss-of-function mutation of the FXR1 gene tend to have high levels of these
proteins. Based on this information, predict how the FXR1 protein most likely
interacts with the mRNAs.
Accept one of the following:
• The FXR1 protein must inhibit translation of the mRNAs.
• The FXR1 protein must degrade the mRNAs.
Total for part (D) 1 points
Total for question 2 8 points

106 AP Biology Practice Exam

Question 3: Scientific Investigation 4 points
Learning Objectives: EVO-1.L EVO-1.E EVO-1.D

(A) Describe one outcome that would demonstrate that a given population has 1 point
evolved. 1.A

• There is a change in allele frequency.

(B) Identify the dependent variable measured in the experiments. 1 point

3.C
Accept one of the following:
• Growth rate of the bacteria
• Density of the bacteria

(C) Predict the results obtained by the researchers when they grew the Cit+ and Cit- 1 point
bacteria in the medium that contained only citrate. 3.B

• Only the Cit+ bacteria grew.

(D) The researcher claims that the Cit+ mutation increases the fitness of the 1 point
bacteria. Provide reasoning to support the claim. 6.C

• The Cit+ bacteria have an energy source unavailable to other strains and could
grow and reproduce at a more rapid rate due to reduced competition for
resources.
Total for question 3 4 points

AP Biology Practice Exam 107

 Question 4: Conceptual Analysis 4 points
Learning Objectives: SYI-1.H SYI-3.F SYI-3.G

(A) Describe the process that maintained a stable Tasmanian devil population size 1 point
before the appearance of DFTD in 1996. 1.A

• The Tasmanian devil population would have been maintained at a stable

population size by density-dependent factors (e.g., birth rate equaling death
rate, the populations reaching carrying capacity).

(B) Explain how the huge reduction of the Tasmanian devil population since 1 point
1.C
1996 affects the susceptibility of the current population to new diseases in
comparison with the susceptibility of the population before 1996.
• The population underwent a genetic bottleneck and thus has reduced
genetic diversity. The current population is more likely to be susceptible to a
new disease (less likely to harbor natural resistance) than it would have been
in the past.

(C) Tasmanian devils are top predators and are considered a keystone species in 1 point
their community. Predict the effect of the rapid reduction of the Tasmanian devil 6.E

population on the rest of the community.

Accept one of the following:
• The community will collapse or be destabilized.
• The population sizes of some species will increase and others will decrease.
• Species diversity of the community will decrease.

(D) Justify the prediction of part (c). 1 point

6.C
• The prey species of the Tasmanian devil would have increased in number as
the Tasmanian devil population decreased. This in turn would have likely led to
population decreases of the food sources for these prey species.
Total for question 4 4 points

108 AP Biology Practice Exam

Question 5: Analyze Model
or Visual Representation of a Biological Concept or Process 4 points
Learning Objectives: IST-1.C IST-1.D IST-1.B IST-1.E

(A) A skin cell completes one round of the cell cycle. Describe the products. 1 point
1.A
• Two diploid cells with identical sets of DNA

(B) Based on Figure 1, explain how p53 regulates the cell cycle in the presence of 1 point
2.B
damaged DNA.
• p53 stimulates expression of the CDK inhibitor, preventing progression of the
cell cycle. (The CDK inhibitor blocks the interaction between the CDK and
cyclins.)

(C) Draw an X on the template in the space provided for your response to indicate 1 point
the phase during which the replication of damaged DNA would occur. 2.D

• Must place an X on any portion of S

(D) Based on Figure 1, explain how a mutation to p53 may lead to an increased risk 1 point
of cancer. 2.C

• A mutation may prevent activation of p53, which will then not stimulate the
CDK inhibitor, allowing the cell cycle to progress through the cell cycle with
damaged or mutated DNA.
Total for question 5 4 points

AP Biology Practice Exam 109

 Question 6: Analyze Data 4 points
Learning Objectives: ENE-2.G ENE-2.C

(A) Identify the individual who most likely exhibits symptoms of cystinuria. 1 point
4.B
• Individual 1

(B) Describe the relationship between the total number of mutant alleles in an 1 point
4.B
individual and the concentration of cysteine in the urine.
• The more mutant alleles an individual has, the higher the concentration of
cysteine in the urine.

(C) Evaluate the hypothesis that mutations in SLC7 A9 have a greater effect on the 1 point
transport of cysteine than do mutations in SLC3 A1. 5.D

• The hypothesis is not supported by the data. While the individual with a single
mutant allele for SLC7 A9 has a higher concentration than the individual with
a single mutant allele for SLC3 A1, the individual with two mutated SLC7 A9
alleles does not have a higher concentration than one of the individuals with
two mutant SLC3 A1 alleles.

(D) Explain how the data support the claim that cysteine is a large polar molecule. 1 point
6.D
• Cysteine builds up in the urine when the transporter proteins are mutated,
which suggests that cysteine requires specific channel proteins through which
to move, as do most large polar molecules.
Total for question 6 4 points

110 AP Biology Practice Exam

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AP Biology Practice Exam 111

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