2021
AP
®
Biology
Scoring Guidelines
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AP® Biology 2021 Scoring Guidelines
Question 1: Interpreting and Evaluating Experimental Results 10 points
Polycystic kidney disease (PKD) is an inherited disease that causes water loss from the body and affects cell
division in the kidneys. Because water movement across cell membranes is related to ion movement, scientists
investigated the role of the
Na
+
/K
+
ATPase
(also known as the sodium/potassium pump) in this disease.
Ouabain, a steroid hormone, binds to the
+ +
Na /K ATPase
in plasma membranes. Individuals with PKD have a
genetic
mutation that
results in an increased binding of ouabain to the
+ +
Na /K ATPase
. The scientists treated
normal human kidney (NHK) cells and PKD cells with increasing concentrations of ouabain and measured the
number of cells (Figure 1)
and the activity of
the
+ +
Na /K ATPase
(Figure 2) after a period of time. The
scientists hypothesized that a signal transduction pathway that includes the protein kinases MEK and ERK
(Figure 3) may play a role in PKD symptoms.
Figure 1. Cell number compared with the number of
cells at
0 pM
ouabain. Normal human kidney
(NHK) cells and polycystic kidney disease (PKD)
cells were treated with increasing concentrations of
ouabain. Error bars represent
2SE
X
±
.
Figure 2. Percent
+ +
Na /K ATPase
activity of NHK
and PKD cells treated with increasing concentrations
of ouabain. Error bars represent
2SE
X
±
.
Figure 3. Signal transduction pathway hypothesized to play a role in the increased number of PKD cells
© 2021 College Board
AP® Biology 2021 Scoring Guidelines
(a)
Describe the characteristics of the plasma membrane that prevent simple diffusion of
Na
+
and
K
+
across the membrane.
1 point
Accept one of the following:
• The interior of the plasma membrane is hydrophobic/nonpolar.
• The phospholipid tails are hydrophobic/nonpolar.
• The exterior of the pla
sma membrane is hydrophilic/polar.
• The phospholipid heads are hydrophilic/polar.
Explain why ATP is required for the activity of the
+ +
Na /K ATPase
.
1 point
• The
+ +
Na /K ATPase
pumps ions against their concentration gradients. This requires
an input of (metabolic) energy.
Total for part (a) 2 points
(b) Identify a dependent variable in the experiment represented in Figure
1. 1 point
• The number of cells
Justify the use of normal human kidney NHK cells as a control in the experiments. 1 point
Accept one of the following:
• It allows the scientists to determine the effect of PKD on the cells’ responses to (various
concentrations of) ouabain.
• It allows the scientists to compare the responses of PKD cells and nor
mal cells (to
ouabain).
Justify the use of a range of ouabain concentrations in the experiment represented in
Figure 1.
1 point
Accept one of the following:
• The scientists need to determine whether different concentrations have different effects
on the
cell numbers.
• The scientists did not know at which concentrat
ion of ouabain there would be an effect.
Total for part (b) 3 points
(c)
Based on the data shown in Figure 2, describe the rel
ationship between the concentration of
ouabain and the
+ +
Na /K ATPase
activity both in normal human kidney (NHK) cells AND
in PKD cells.
1 point
Accept one of the following:
• Increasing concent
rations of ouabain result in decreasing ATPase activity (in both types
of cells).
• There is an inverse relationship/negative correlation between t
he concentration of
ouabain and the ATPase activity (in both types of cells).
The scientists de
termined that
+ +
Na /K ATPase
activity in PKD cells treated with
1 pM
ouabain is 150 units of ATP
hydrolyzed/sec
. Calculate the expected
+ +
Na /K ATPase
activity (units/sec) in PKD cells treated with
10
6
pM
ouabain.
1 point
• 45 (Accept between 40 and 50)
Total for part (c) 2 points
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AP® Biology 2021 Scoring Guidelines
(d)
In a third experiment, the scientists added an inhibitor of phosphorylated MEK (pMEK) to
th
e PKD cells exposed to
10
4
pM
ouabain. Based on Figure 3, predict the change in the
relative ratio of ERK to pERK in ouabain-treated PKD cells with the inhibitor compared
with ouabain-treated PKD cells without the inhibitor.
1 point
Accept one of
the following:
• Option 1: The ratio of ERK to pERK will increase in the cells with the inhibitor.
• Option 2: The ratio of ERK to pERK will stay the same in the cells with the inhibitor.
Provide reasoning to justify your predicti
on. 1 point
• The justification must indicate that the pMEK inhibitor blocks further phosphorylation
of
ERK AND one of the following:
Option 1:
• The amount of pERK will not increase as it does in cells without the inhibitor.
• The amount of ERK will not decrease as it does in cells without the inhibitor.
• The cell continues to synthesize ERK.
• Phosphorylated ERK is being dephosphorylated to ERK.
Option 2:
• No additional ERK is synthesized/pERK is not being dephosphorylated.
Using the data in Figure 1 AND the signal transduction pathway represented in Figure 3,
explain why the concentration of cyclin proteins may increase in PKD cells treated with
10
4
pM
ouabain.
1 point
• The cell number increases to a maximum at
10
4
pM
ouabain. The signaling pathway
stimulates transcription of genes involved in cell division. The target genes likely
include those for cyclins because cyclins regulate the cell cycle.
Total for part (d) 3 points
Total for question 1 10 points
© 2021 College Board
AP® Biology 2021 Scoring Guidelines
Question 2: Interpreting and Evaluating Experimental Results with Graphing
8 points
Geneticists investigated the mode of inheritance of a rare disorder that alters glucose metabolism and first shows
symptoms in adulthood. The geneticists studied a family in which some individuals of generations
II
and
III
are known to have the disorder. Based on the pedigree (Figure 1), the geneticists concluded that the disorder
arose in individual
II-2
and was
caused by a mutation in mitochondrial
DNA.
Figure 1. Pedigree of a family showing individuals with the glucose metabolism disorder. A question mark
indicates that the phenotype is unknown.
TABLE 1. AVERAGE BLOOD GLUCOSE LEVELS OF INDIVIDUALS IN GENERATION
IV
Individual
Average Blood Gl
ucose Level
(
mg/dL ± 2SE
X
)
IV − 1 170 ± 15
IV − 2 190 ± 10
IV − 3 145 ± 5
IV − 4 165 ± 15
IV − 5 110 ± 15
IV − 6 125 ± 5
IV − 7 105 ± 15
IV − 8 120 ± 10
TABLE 2. PHENOTYPIC CLASSIFICATIONS BASED ON BLOOD GLUCOSE LEVELS
Phenotype
Blood Glucose Level (
mg/dL
)
Normal < 140 mg/dL
At risk 140 − 199 m
g/dL
Affected ≥ 200 mg/dL
© 2021 College Board
AP® Biology 2021 Scoring Guidelines
(a)
The disorder alters glucose metabolism. Describe the atoms AND types of bonds in a
gl
ucose molecule.
1 point
• The atoms are carbon, hydrogen, and oxygen (C, H, and O) and are held together by
cova
lent bonds.
(b)
Use the template provided to construct an appropriately labeled graph based on the data in
Table 1.
3 points
• Point distribution: Axis labels; plotting in a bar graph or modified bar graph; error bars
Determine one individual who is both at risk of developi
ng the disorder and has a
significantly different blood glucose level from that of individual
IV-1
.
1 point
•
IV-3
Total for part (b) 4 points
(c) B
ased on the pedigree, identify all individuals in generation
IV
who can pass on the
mutation to their children.
1 point
•
IV-1, IV-2, IV-4
(d)
Based on the fact that individual
II-2
is affected, a student claims that the disorder is
inherited in an X-linked recessive pattern. Based on the student’s claim, predict which
individuals of generation
III
will be affected by the disorder.
1 point
•
III-4
and
III-8
Based on the pedigree, justify why the data do NOT support the student’s claim. 1 point
Accept one of the following:
• The data do not support the claim because females
III-2
and
III-6
have the disorder
and, if inheritance is X-linked recessive, they could only do so if their father
II-1
had
the disorder, which he does not.
• The data instead support mitochondrial inheritance, because all of the offspring of
indi
vidual
II-2
, not only the sons, have the disorder.
Total for part (
d) 2 points
Total for question 2 8 points
© 2021 College Board
AP® Biology 2021 Scoring Guidelines
Question 3: Scientific Investigation 4 points
Researchers hypothesize that the plant compound resveratrol improves mitochondrial function. To test this
hypothesis, researchers dissolve resveratrol in dimethyl sulfoxide (DMSO). The solution readily passes through
cell membranes. They add the resveratrol solution to mammalian muscle cells growing in a nutrient-rich
solution (culture medium) that contains glucose. They measure ATP production at several time points after the
addition of the resveratrol solution and find an increase in ATP production by the muscle cells.
(a)
Describe the primary advantage for a mammalian muscle cell in using aerobic respiration
over fermentation.
1 point
• More ATP (per glucose molecule) is produced by aerobic respiration.
(b)
Identify an appropriate negative control for this experiment that would allow the
rese
archers to conclude that ATP is produced in response to the resveratrol treatment.
1 point
Accept one of the following:
• The researchers must run the experiment without adding resveratrol.
• The researchers must treat the cells with DMSO alone.
(c)
Predict the effect on short
-term ATP production when resveratrol-treated mammalian
muscle cells are grown in a culture medium that lacks glucose or other sugars.
1 point
Accept one of the following:
• No ATP production
• Reduced ATP production
(
d)
The researchers find
that resveratrol stimulates the production of components of the
electron transport chain. The researchers claim that treatment with resveratrol will also
increase oxygen consumption by the cells if glucose is not limiting. Justify the claim.
1 point
• More electrons can be transferred so that more oxygen is required as the final electron
accept
or.
Total for question 3 4 points
© 2021 College Board
AP® Biology 2021 Scoring Guidelines
Question 4: Conceptual Analysis 4 points
In 1981 a single immature male Geospiza conirostris finch flew more than 100 kilometers from the Galápagos
island of Española to the Galápagos island of Daphne Major, where no G. conirostris finches were living. The
immigrant finch bred with a female G. fortis, a species of finch common on Daphne Major. The
F
1
finches and
later generations interbred only within their lineage. By 2012 scientists counted 23 individuals, including eight
breeding pairs, within this hybrid lineage on Daphne Major. The hybrid lineage became known as Big Bird.
Birds with different beak shapes and sizes eat different types of food. The dimensions of the Big Bird beaks
rel
ative to the beaks of the major competitor finch species on Daphne Major are shown in Figure 1.
Figure 1. The dimensions of the beaks of the Big Bird lineage and of its major competitor species in 2012 on
Daphne Major. Each symbol represents the beak dimensions of a single bird.
(a)
The Big Bird line
age became reproductively isolated from G. fortis. Describe one
prezygotic mechanism that likely contributed to the reproductive isolation of the Big Bird
lineage from G. fortis.
1 point
Accept one of the following:
• Beak shape/size or song or behavior or mechanical/chemical differences or time of
mat
ing or location on the island or primary food source differs between the Big Bird
lineage and G. fortis.
• Description of another mechanism that prevents males and females from different
populations from encountering each other/recognizing each other as potential mates.
(b)
Based on the data in Figure 1, explain why the Big Bird population has been able to survive
and reproduce on Daphne Major.
1 point
• The birds have a beak size/shape that differs from the beaks of the competitor finches
on the
island. Thus, they probably do not compete with the other finch species for food
but instead, eat food that the other finches do not consume.
© 2021 College Board
AP® Biology 2021 Scoring Guidelines
(c)
A virus infects and kills all G. magnirostris on Daphne Major but does not affect the other
fi
nch species. Assuming food type and availability stay the same, predict the most likely
change in the beak phenotype of the Big Bird population after six more generations.
1 point
Accept one of the following predictions:
• Option 1: The (mean) beak size will increase (in the population).
• Option 2: The (average) b
eak (in the population) will be longer and deeper.
• Option 3: The frequency of large beaks will increase (in the population).
• Option 4: The (mean) beak size will stay the same (in the population).
(d) Provide reasoning to ju
stify your prediction in part (c). 1 point
Accept one of the following:
• Justification for options 1, 2, and 3: There will be directional selection for larger beaks
becau
se larger seeds are more accessible.
• Justification for option 4:
There is little genetic diversity because all birds are
descended from a single pair, and the birds are only six generations from the founder.
Total for question 4 4
points
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AP® Biology 2021 Scoring Guidelines
Question 5: Analyze a Model or Visual Representation 4 points
Annual plants complete their life cycle, including germination, seed production, and death, within one year.
Ambrosia trifida (giant ragweed) is an annual plant that readily colonizes any land that has had a disturbance
such as plowing. The plant is considered an invasive species in regions outside of its native range. In a particular
region, the seeds of A. trifida germinate from early March through the end of the summer, while the seeds of
other annual plants require warmer soil temperatures and thus germinate from late April through the end of the
summer.
Researchers studied the influence of A. trifida on the biodiversity of other annual plant species that grow in the
sam
e field. In early spring, the researchers marked off identical plots of land in a field that had been plowed the
previous fall and not replanted with new crops. All plants that grew on one half of the plots were left untouched
(Figure 1A), while all germinating A. trifida seedlings were removed from the other half of the plots throughout
the spring and summer (Figure 1B). In late summer, the researchers counted and identified all plants that grew in
the plots. The distribution of plants is represented by the symbols in Figures 1A and 1B.
Figure 1. Representations of plant identity and distribution in experimental plots in late summer. Each box
represents one typical experimental plot, and each symbol represents 10 individual plants.
(a) Describe a cause of logistic growth of the ragweed population. 1 point
Accept one of the following:
• A factor that becomes limiting would cause the population size to stabilize.
• Space/sunlight/herbivory/phosphorus/nitr
ogen/other density-dependent factor becomes
limiting, and the population stabilizes.
(b)
Based on the r
epresentation in Figure 1, explain why the scientists claim that plot B would
be more resilient than plot A in response to a sudden environmental change.
1 point
• (Plot B is more resilient) b
ecause it has much greater (species) diversity than plot A
does.
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AP® Biology 2021 Scoring Guidelines
(c)
In a third group of plots, the researchers removed all seedlings of all plants that germinated
befo
re June 1. All plants that germinated after June 1 were left untouched. Using the
template in the space provided for your response and the symbols shown in Figure 1,
represent the expected plant species that would be found in this third group of plots three
months later. Draw no more than 12 symbols. Assume all other environmental conditions
are the same as for the initial study described.
1 point
• All four species, including A. trifida, must be added to the template.
(d)
Explain how an invasive species such as ragweed affects ecosystem biodiversity, as
illus
trated in Figure 1.
1 point
• The explanation requires a process or relationship - and must state that biodiversity
decreas
es.
• Examples of appropriate responses include:
o There are no predators of the invasive species, so its population grows faster and
reduces biodiversity.
o The invasive species germinates earlier, uses up resources, and reduces
biodi
versity.
o The invasive species outcompetes other species and reduces biodiversity.
Total for question 5 4 points
© 2021 College Board
AP® Biology 2021 Scoring Guidelines
Question 6: Analyze Data 4 points
The small invertebrate krill species Thysanoessa inermis is adapted to cold (
4C°
)
seawater. Over the past ten
years, there has been a gradual increase in the water temperature of the krill’s habitat. A sustained increase in
water temperature may ultimately affect the ability of the krill to survive.
One effect of higher temperatures is protein misfolding within cells. Krill have several hsp genes that code for
heat-sho
ck proteins (HSPs). These proteins help prevent protein misfolding or help to refold proteins to their
normal shapes.
Scientists conducted experiments on T. inermis to detect chan
ges in the expression of hsp genes when the krill
were exposed to temperatures above 4°C. An experimental group of krill was maintained in tanks with
4C°
seawater and then placed into tanks with
10 ° C
seawater for approximately three hours. The krill were then
given a six-hour recovery period in the
4C°
seawater tanks. A control group of krill was moved from a tank of
4C°
seawater to another tank of
4C°
seawater for approximately three hours and then returned to the original
tank. The scientists analyzed hsp gene expression by measuring the concentrations of three mRNAs (
I, II, III
)
transcribed from certain hsp genes in both the heat-shocked krill (Figure 1) and the control krill. For the control
krill, no transcription of the hsp genes was detected throughout the test period (data not shown).
Figure 1. Average concentration of three mRNAs (
I, II, III
) transcribed from hsp genes in krill heat shocked at
10 C°
. Error bars represent
±2SE
X
.
(a) Identify the hsp mRNA that has t
he slowest rate of concentration increase in response to
heat-shock treatment.
1 point
• (mRNA)
III
(b)
Describe the trend in the average concentration of mRNA
I
throughout the experiment.
1 point
• (No change in concentration from 1 to 3 hours) increased concentration (slightly)
betw
een 3 and 6 hours/during the heat shock, increased concentration at a greater rate
from 6 to 10 hours/for 4 hours after the heat shock, and then decreased concentration
after hour 10.
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AP® Biology 2021 Scoring Guidelines
(c)
The scientists hypothesized that the heat-shock protein (HSP) translated from mRNA
I
plays a greater role in refolding proteins than does the HSP translated from mRNA
II
.
1 point
Use the data to support the hypothesis.
• mRNA
I
is still expressed at a high level after the heat-shock period, while mRNA
II
levels decrease after the heat shock, when proteins would need to be refolded.
(d)
mRNAs
I
and
II
are transcribed from the same gene. Explain how a cell can produce
two different mRNAs from the same gene.
1 point
Accept one of the following:
• The cell expresses different exons/performs alternative splicing.
• The cell uses diff
erent transcription termination sites (poly(A) sites).
• The cell uses different promoters.
Total for question 6 4
points
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