SCIENCE TEST

SCIENCE TEST
This document contains the science portion of the ACT practice test that is available for free
at http://www.act.org/aap/pdf/preparing.pdf. Explanations of answers (in red print) were
added by TJ Leone, and are not part of the official ACT prep material.
35 Minutes—40 Questions
DIRECTIONS: There are seven passages in this test. Each passage is followed by several
questions. After reading a passage, choose the best answer to each question and fill in the
corresponding oval on your answer document. You may refer to the passages as often as
necessary.
You are NOT permitted to use a calculator on this test.
Passage I
Earthquakes produce seismic waves that can travel long distances through Earth. Two types of seismic
waves are p-waves and s-waves. P-waves typically travel 6−13 km/sec and s-waves typically travel 3.5−7.5
km/sec. Figure 1 shows how p-waves and s-waves move and are
refracted (bent) as they travel through different layers of Earth’s interior. Figure 2 shows a seismograph (an
instru- ment that detects seismic waves) recording of p-waves and s-waves from an earthquake. Figure 3
shows, in general, how long it takes p-waves and s-waves to travel given dis- tances along the surface from
an earthquake focus (point of origin of seismic waves).

Earthquakes produce seismic waves that can travel
long distances through Earth. Two types of seismic waves
are p-waves and s-waves. P-waves typically travel
6−13 km/sec and s-waves typically travel 3.5−7.5 km/sec.
Figure 1 shows how p-waves and s-waves move and are
ACT-64E-PRACTICE
both p-waves and
s-waves received
at seismographs
103°
shadow zone:
neither p-waves nor
s-waves received
at seismographs
earthquake
focus
Note: The figure is not to scale.
0°
solid
inner
core
mantle
142° 142°
only p-waves received
at seismographs
Figure 1
42
Earth’s interior. Figure 2 shows a seismograph (an instrument
that detects seismic waves) recording of p-waves and
s-waves from an earthquake. Figure 3 shows, in general,
how long it takes p-waves and s-waves to travel given distances
along the surface from an earthquake focus (point of
origin of seismic waves).
liquid outer core
crust
Key
both p-waves and
s-waves received
at seismographs
103°
p-waves
s-waves
shadow zone:
neither p-waves nor
s-waves received
at seismographs
4 4
ch seismograph from
uake focus (min)
22
20
18
16
14
12
10
time
earthquake
starts at the
focus
s-waves
p-waves
first p-waves
arrive at
seismograph
Figure 2
first s-waves
arrive at
seismograph
GO ON TO THE NEXT PAGE.
1
minute
2. According to Figure 1, when p-waves encounter the
boundary between the mantle and the core, the
p-waves most likely:
F. stop and do not continue into the core.
G. enter the core and are refracted.
H. change to s-waves.
J. change to a third type of seismic wave.
3. Based on Figure 3, for a given seismograph, the time
elapsed between the arrival of the first p-waves and the
arrival of the first s-waves from an earthquake focus

Answer: G.
time to reach seismograph from
earthquake focus (min)
22
20
18
16
14
12
10
8
6
4
2
0
0
1,000
focus
2,000
3,000
distance along Earth’s surface from earthquake
focus to seismograph (km)
Figure 3
s-waves
p-waves
4,000
5,000
6,000
7,000
8,000
9,000
10,000
Figure 2
1. Figure 1 shows that a seismograph located at a point 125° around Earth from an earthquake’s focus
5. According to Figure 2, which o
would receive which type(s) of seismic waves, if either, from that earthquake?
ments best describes the relative
A. P-waves only
p-waves to arrive at the seism
B. S-waves only
s-waves to arrive at the seismogr
C. Both p-waves and s-waves
D. Neither p-waves 1. nor Figure s-waves 1 shows that a seismograph located at a point
125° around Earth from an earthquake’s focus would
receive which type(s) of seismic waves, if either, from
Answer: D.
that earthquake?
the first p-waves to arrive at the
A. smaller than the amplitude
arrive at the seismograph.
B. larger than the amplitude o
arrive at the seismograph.
A. P-waves only
C. nonzero, and the same as the
B. S-waves only
s-waves to arrive at the seism
In Figure 1, 0° is shown
C.
at
Both
the
p-waves
spot labeled
and s-waves
“earthquake focus”. Between 103°
D.
and
zero, as is the amplitude o
142° on either side ofD. theNeither earthquake p-waves focus nor s-waves are areas labeled “shadow zone: neither arrive at the seismograph.
p-­‐waves nor s-­‐waves received at seismographs”. Since 125° is between 103° and
142°, it is in the shadow zone, so it receives neither p-­‐waves nor s-­‐waves.
ACT-64E-PRACTICE
43
2. According to Figure 1, when p-waves encounter the boundary between the mantle and the core, the pwaves
most likely:
F. stop and do not continue into the core.
G. enter the core and are refracted.
H. change to s-waves.
GO ON TO T
J. change to a third type of seismic wave.
The boundary between the mantle and the core is the circle around the area labeled
“liquid outer core.” According to the key, the p-waves are solid lines with arrows to
indicate direction. The s-waves are dotted lines with arrows. The lines for s-waves stop
2. According to Figure 1, when p
boundary between the mantle
p-waves most likely:
F. stop and do not continue into
G. enter the core and are refract
H. change to s-waves.
J. change to a third type of seis
3. Based on Figure 3, for a given
elapsed between the arrival of th
arrival of the first s-waves from
10,500 km away would most like
A. less than 5 min.
B. between 5 min and 7 min.
C. between 8 min and 10 min.
D. more than 10 min.
4. Based on the information prov
quake starts at the focus” in Fi
which of the following points on
F. , 0 km, 0 min
G. 2,000 km, 5 min
H. 5,000 km, 12 min
J. 10,000 km, 20 min

when they reach the core boundary. The p-waves go through the boundary and bend. In
Passage I, we are told that “Figure 1 shows how p-waves and s-waves move and are
refracted (bent) as they travel through different layers of Earth’s interior.” This tells us
that “refracted” means “bent”. So the p-waves are refracted when they enter the core.
3. Based on Figure 3, for a given seismograph, the time elapsed between the arrival of the first p-waves and
the arrival of the first s-waves from an earthquake focus 10,500 km away would most likely be:
A. less than 5 min.
B. between 5 min and 7 min.
C. between 8 min and 10 min.
D. more than 10 min.
Answer: D.
In Figure 3, the horizontal axis of the graph shows “distance along Earth’s surface
from earthquake focus to seismograph (km)”. The vertical axis shows “time to reach
seismograph from earthquake focus (min)”. So the time elapsed between the arrival
of the first p-­‐waves and the first s-­‐waves at a given distance is the vertical difference
between the lines at that distance. For example, at 2,000 km, it takes 7 minutes for
the first s-­‐waves to reach the seismograph, and 4 minutes for the first p-­‐waves. The
time elapsed is 7-­‐4 = 3 minutes. At 3,000 km, the time elapsed is 10-­‐6 = 4 minutes.
Since 10,500 km is not on the graph, we have to guess. We see that the distance
between the s-­‐wave and p-­‐wave curves (time elapsed) is getting bigger. At 9,000
km, the distance between the curves is about 22-­‐11 = 11 minutes. At 10,000 km, we
can’t tell exactly since the s-­‐wave curve is already past 23 minutes, but it looks like
the distance between the curves continues to grow. We expect that at 10,500 km the
distance will be even greater, so time elapsed is probably greater than 11 minutes.
This makes D the best answer.
4. Based on the information provided, the “time earthquake starts at the focus” in Figure 2 corresponds to
which of the following points on Figure 3?
F. 0km,0min
G. 2,000 km, 5 min
H. 5,000 km, 12 min.
J. 10,000 km, 20 min
Answer: F.
We are looking for the point on Figure 3 that represents the “time earthquake starts
at focus”. The seismograph is at the earthquake focus when it is 0 km from the focus.
The time 0 min marks the start of the earthquake. So the “time earthquake starts at
focus” is represented by the point 0 km, 0 min (bottom left corner) on Figure 3.
5. According to Figure 2, which of the following statements best describes the relative amplitudes of the
first p-waves to arrive at the seismograph and the first s-waves to arrive at the seismograph? The amplitude
of the first p-waves to arrive at the seismograph is:
A. smaller than the amplitude of the first s-waves to arrive at the seismograph.
B. larger than the amplitude of the first s-waves to arrive at the seismograph.

C. nonzero, and the same as the amplitude of the first s-waves to arrive at the seismograph.
D. zero, as is the amplitude of the first s-waves to arrive at the seismograph.
Answer: A.
The amplitude of a wave is half the distance between the bottom and the crest of the
wave. In the illustration below, the amplitude of the top wave is smallest, and the
amplitude of the bottom wave is largest:
In Figure 2, the amplitude of the first p-­‐waves to arrive at the seismograph is
smaller than the amplitude of the first s-­‐waves to arrive.
Passage II
Lake Agassiz existed between 11,700 and 9,500 years ago in North America (see Figure 1). The lake was
formed when a large glacier dammed several rivers. Groundwater trapped in lake and glacial sediments
provides information about the climate at the time the sediments were deposited. Figure 2 shows a cross
section of the sediments (lake clay and glacial till) and bedrock in the area. Figure 3 shows the δ 18 Ο values
of groundwater taken from samples of the top 40 m of sediment at 3 sites along the same cross section.
δ 18 Ο is calculated from a ratio of 2 oxygen isotopes ( 18 O and 16 O) in the groundwater. Smaller δ 18 O values
indicate cooler average temperatures.

ted between 11,700 and 9,500 years
(see Figure 1). The lake was formed
ammed several rivers. Groundwater
cial sediments provides information
e time the sediments were deposited.
s section of the sediments (lake clay
drock in the area. Figure 3 shows the
water taken from samples of the top
sites along the same cross section.
a ratio of 2 oxygen isotopes ( 18 O
water. Smaller δ 18 Lake Agassiz existed between 11,700 and 9,500 years
ago in North America (see Figure 1). The lake was formed
when a large glacier dammed several rivers. Groundwater
trapped in lake and glacial sediments provides information
about the climate at the time the sediments were deposited.
Figure 2 shows a cross section of the sediments (lake clay
and glacial till) and bedrock in the area. Figure 3 shows the
δ
Manitoba
maximum
extent of
Lake
Agassiz
Site 1
•
Site 3 •
•
O values indicate
North •
tures.
Dakota Grand
Forks
18 O values of groundwater taken from samples of the top
40 m of sediment at 3 sites along the same cross section.
δ 18 O is calculated from a ratio of 2 oxygen isotopes ( 18 O
and 16 O) in the groundwater. Smaller δ 18 O values indicate
cooler average temperatures.
surface
Site 1
Site 2
elevation (m above sea level)
Site 3
Winnipeg,
Grand Forks,
N Manitoba
200 North Dakota S
ACT-64E-PRACTICE
250
150
Figure 2
44
surface
250
200
150
Figure 1
Hudson
Bay
Winnipeg
Site 2
Great
Lakes
Key
sediment/rock
Figure 2
44
lake clay
glacial till
bedrock
Site 1
Site 3
North
Dakota
Winnipeg,
Grand Forks,
N Manitoba
North Dakota S
surface
Site 1
Site 2
Site 3
surface
Manitoba
4
maximum
extent of
Lake
Agassiz
250
200
150
• •
•
Grand
Forks
GO ON TO THE NEXT PAGE.
Figure 1
Hudson
Bay
Winnipeg
Site 2
Great
Lakes
Key
sediment/rock
lake clay
glacial till
bedrock
GO ON TO THE NEXT PAGE.

4 4
δ
–26
(surface) 0
–22 –18 –14
18 O
δ
–26
(surface) 0
–22 –18 –14
–26
(surface) 0
–22 –18 –14
18 O δ 18 O
smaller larger smaller larger smaller larger
depth (m)
6. According to Figure 2, the lake clay deposit is thinnest 9. According to Figure 2, which of the following graphs
6. According at which of to the Figure following 2, the cities lake or clay sites? deposit is thinnest at which best represents of the following the elevations, cities in m or above sites? sea level, of
F. Winnipeg
the top of the glacial till layer at Sites 1, 2, and 3 ?
G. Site 11
H. Site 2
H. J. Grand Site 2 Forks
J. Grand Forks
A. 225
200
C. 225
200
Answer: F.
In7. Figure According 2, to the Figure key 3, at shows Sites 1, 2, that and 3, lake the smallest clay is indicated with light gray. The vertical axis
δ
of the figure is labeled “elevation (m above sea level). This tells us, for example, that
the Winnipeg site has lake clay between 230 m and 240 m above sea level. In other
words, the lake clay is 240-­‐230 = 10 m “thick” at the Winnipeg site. Looking at the
figure, we can see that the lake clay (light gray area) is thinnest at the Winnipeg site.
18 O value of the groundwater in the lake clay was
B. 225
D. 225
recorded at a depth between:
A. 0 m and 10 m.
200
200
B. 10 m and 20 m.
C. 20 m and 30 m.
D. 30 m and 40 m.
175
175
7. According to Figure 3, at Sites 1, 2, and 3, the smallest δ 18 Ο value of the groundwater in the lake clay
was recorded at a depth between:
A. 0 m and 10 m.
10. Precipitation that falls at Sites 1, 2, and 3 soaks into
the soil until it reaches the groundwater table about
8. B. According 10mand20m. to Figure 2, as the thickness of the lake clay
deposit increases from Grand Forks to Site 3, the
C. thickness 20mand30m. of the glacial till beneath it:
3 m below the surface. Based on Figure 3, and assuming
no alteration of the precipitation, the δ
D. F. increases. 30mand40m.
G. remains the same.
H. first increases and then decreases.
J. decreases.
18 150
1 2 3
150
1 2 3
Site
Site
O value of
present-day precipitation in the study area is closest to:
F. −26.
G. −23.
H. −20.
J. −15.
Answer: C.
10
20
30
40
Site 1
lake
clay
glacial
till
depth (m)
10
20
30
40
Figure 3
Figures adapted from V. H. Remenda, J. A. Cherry, and T. W. D. Edwards, “Isotopic Composition of Old Ground Water from Lake Agassiz:
Implications for Late Pleistocene Climate.” ©1994 by the American Association for the Advancement of Science.
In Figure 3, the smallest δ18O values are farthest to the left and have values between
-­‐26 and -­‐14. At Site 1, the smallest δ18O value recorded is a little less than -­‐25 at a
depth of around 27 m. At Site 2, the smallest δ18O value recorded is around -­‐25 at a
depth of around 27 m or 28 m. At Site 3, the smallest δ18 ACT-64E-PRACTICE
GO ON TO THE NEXT PAGE.
45
O value recorded is a little
less than -­‐25 at a depth of around 28 m or 29 m. So the answer is C.
8. According to Figure 2, as the thickness of the lake clay deposit increases from Grand Forks to Site 3, the
thickness of the glacial till beneath it:
Site 2
lake
clay
glacial
till
elevation
(m above sea level)
elevation
(m above sea level)
175
150
1
depth (m)
Note: δ 18 O = – 1 × 1,000
18 O/ 16 O of groundwater sample
___________________________
18 O/ 16 O of standard water sample
2
Site
3
10
20
30
40
elevation
(m above sea level)
elevation
(m above sea level)
Site 3
175
150
lake
clay
1
2
Site
3

depth (m)
10
20
30
40
osit is thinnest
3, the smallest
lake clay was
f the lake clay
to Site 3, the
30
glacial
till
F. increases. 40
G. remains the same.
H. first increases and then decreases.
J. decreases.
Figure 3In
Figure 2, the key shows us that the glacial till is indicated with stripes. As we
move left from Grand Forks to Site 3, we see that the area of the glacial till gets
thinner. So the thickness of the glacial till decreases.
9. According to Figure 2, which of the following graphs
best represents the elevations, in m above sea level, of
the top of the glacial till layer at Sites 1, 2, and 3 ?
A.
B.
lake
clay
18 16
___________________________
O/ O of groundwater Answer: sample J.
18 16
O/ O of standard water sample
y, and T. W. D. Edwards, “Isotopic Composition of Old Ground Water from Lake Agassiz:
by the American Association for the Advancement of Science.
elevation
(m above sea level)
elevation
(m above sea level)
225
200
175
150
225
200
175
150
1
1
depth (m)
O = – 1 × 1,000
2
Site
2
Site
3
3
10
20
C.
D.
elevation
(m above sea level)
elevation
(m above sea level)
225
200
175
150
225
200
175
150
1
1
lake
clay
2
Site
2
Site
10. Precipitation that falls at Sites 1, 2, and 3 soaks into
Answer: the soil C. until it reaches the groundwater table about
3 m below the surface. Based on Figure 3, and assuming
no alteration of the precipitation, the δ
Notice that the word elevation appears in italics in this question. In previous
questions about Figure 2, we’ve been look at the thickness of different layers.
Whoever wrote the question wants to make sure that we pay attention to elevation
instead of thickness.
18 O value of
present-day precipitation in the study area is closest to:
F. −26.
G. −23.
H. −20.
J. −15.
In Figure 2, the elevation of a layer is shown with numbers along the vertical axis
labeled “elevation GO (m ON above TO sea THE level)”. NEXT So, PAGE. for example, the top of the glacial till is
45 230 m at Winnipeg and 200 m at Site 1. Looking at sites 1, 2, and 3, we see that the
elevation of the top of the glacial till is highest at Site 2 and lowest at Site 3, so C is
the only possible answer.
10. Precipitation that falls at Sites 1, 2, and 3 soaks into the soil until it reaches the groundwater table about
3 m below the surface. Based on Figure 3, and assuming no alteration of the precipitation, the δ18Ο value
of present-day precipitation in the study area is closest to:
F. −26.
3
3

G. −23.
H. −20.
J. −15.
Answer: J.
We are told that present-­‐day precipitation soaks to 3m below the present-­‐day
surface. Since the data was recorded fairly recently (not thousands of years ago), we
assume that we can use the data recorded in Figure 3 for δ18O values at 3m below
the surface. Notice that the values on the vertical axes start at zero on the top left
corner and get bigger as they go down. At all three sites, the δ18 4Passage
III
Experiment 2
Some students tested their hypothesis that the presence
of bubbles in cans of various liquids would affect the
roll time (the time it took a can to roll, without slipping,
down an incline between 2 fixed points; see Figure 1).
The students added 1 L of the f
an empty can. They sealed the can
aside. Fifteen minutes later they fou
can before and immediately after
Again they set the can aside. Two h
the roll time of the can before and im
ing it (Trial 5). The results are shown
fixed points incline
4
O value at around 3m
Table 2
Roll tim
below the surfacePassage is about III -­‐15.
can
Experiment 2
before shaking a
angle of
Trial (sec)
The students added 1 L of the f
Some students
inclination
tested their hypothesis that the pres-
Passage III
an empty can. 4 They sealed 1.86 the can
ence of bubbles in cans of various liquids would affect the
Some students tested their roll hypothesis time (the that time the it presence took a can of bubbles to roll, in without cans of slipping, various liquids aside. would Fifteen affect 5 minutes later 1.75 they fou
Figure 1
the roll time (the time it down took an a can incline to roll, between without 2 slipping, fixed points; down see an incline Figure between 1). 2 fixed
can
points;
before
see
and immediately after
Again they set the can aside. Two h
Figure 1).
the roll time of the can before and im
Identical 1.2 L aluminum cans were used in the first two
experiments. The angle of fixed inclination points of incline the incline was
2.3° in all three experiments.
ing it (Trial 5). The results are shown
Experiment 3
The students added 1 L of the f
an empty 2 L clear plastic bottle a
When they rolled the bottle Table down 2 t
formed. They shook the bottle, cau
and set the bottle aside. Fifteen Roll min tim
angle of
can
bles were still visible, but after 2 ho
be seen. before shaking a
Trial (sec)
inclination
Figure 1
4 1.86
Adapted from David Kagan, “The Shaken
5 1.75
by The American Association of Physics T
Experiment 1
Identical 1.2 L aluminum cans were used in the first two experiments. The angle of inclination of the
Identical
incline was 2.3° in all three The experiments. students 1.2 L aluminum added 1 L cans of a were liquid—tap used in water the first contain- two Experiment 3
ing experiments. no bubbles—to The angle an empty of inclination can, sealed of the the can, incline and found was
2.3° in all three experiments.
The students added 1 L of the f
its roll time. Next, they added 1 L of the tap water to a
Experiment 1
an empty 2 L clear plastic bottle a
second empty can, sealed it, shook it, and immediately When they rolled the bottle down t
The students added 1 L found of a liquid—tap its roll time. water They contain- repeated ing no these bubbles—to procedures an empty usingcan,
sealed
formed.
the can,
They shook the bottle, cau
and found its roll time. soapy Next, they water added containing 1 L of the many tap water bubbles, to a second and a empty carbonated can, sealed it, and shook set the it, and bottle aside. Fifteen min
immediately found its roll beverage time. They that repeated contained these no procedures bubbles and using that soapy tasted water flat, containing bles many were still visible, but after 2 ho
bubbles, and a carbonated having beverage lost most that contained of its carbonation. no bubbles The and that results tasted are flat, shown having lost be most seen. of its
carbonation. The results
in
are
Table
shown
1.
11. In Experiment 3, what is the mos
in Table 1.
dents used the plastic bottle rat
Table 1
can? Compared to an aluminum
Adapted from David Kagan, “The Shaken
by The A. American rolled more Association rapidly of down Physics the T
B. made bubbles in the liquid e
C. contained a greater quantity
Experiment 1
Roll time
D. had thicker walls and was le
The students added 1 before L of a liquid—tap shaking after water shaking containing
Trial no bubbles—to Liquid an empty can, (sec) sealed the can, (sec) and found
its roll time. Next, they added 1 L of the tap water to a
second 1 empty tap water can, sealed it, shook 1.75 it, and immediately 1.75
found 2 its roll soapy time. water They repeated 1.97these procedures 2.15 using
soapy 3 water flat-tasting containing many 1.75 bubbles, and a carbonated 1.96
beverage that beverage contained no bubbles and that tasted flat,
having lost most of its carbonation. The results are shown
in Table 1.
12. Based on the results of Experim
of the following trials, before sh
age speeds of the cans the same?
F. Trials 1 and 2
G. Trials 2 and 3
H. Trials 2 and 4
J. Trials 3 and 5
11. In Experiment 3, what is the mos
dents used the plastic bottle rat
ACT-64E-PRACTICE
Table 1
46
can? Compared to an aluminum
GO ON TO T
A. rolled more rapidly down the
B. made bubbles in the liquid e
Roll time
C.
D.
contained a greater quantity
had thicker walls and was le

4
Experiment 2
The students added 1 L of the flat-tasting beverage to
sted their hypothesis that the pres- an empty can. They sealed the can, shook it, and set it
s of various liquids would affect the
Experiment 2 aside. Fifteen minutes later they found the roll time of the
ook a can to roll, without slipping,
The students added 1 L can of the before flat-tasting and immediately beverage to an after empty shaking can. They it sealed (Trial the 4).
en 2 fixed points; see Figure 1).
can, shook it, and set it
aside. Fifteen minutes later
Again
they
they
found
set
the
the
roll
can
time
aside.
of the
Two
can
hours
before
later
and immediately
they found
after shaking it
the roll time of the can before and immediately after shak-
(Trial 4). Again they set ing the it can (Trial aside. 5). Two The results hours later are they shown found in Table the roll 2. time of the can before and
immediately after shaking it (Trial 5). The results are shown in Table 2.
fixed points incline
gle of
clination
Figure 1
can
Experiment 3
The students added 1 L of the flat-tasting beverage to an empty 2 L clear plastic bottle and sealed the bottle.
um cans were used
When
in the
they
first
rolled
two
the bottle Experiment down the 3 incline, no bubbles formed. They shook the bottle, causing bubbles to
le of inclination of the incline was
ments.
form, and set the bottle aside. The Fifteen students minutes added later, 1 L some of the bubbles flat-tasting were still beverage visible, to but after 2 hours, no
bubbles could be seen. an empty 2 L clear plastic bottle and sealed the bottle.
When they rolled the bottle down the incline, no bubbles
Adapted from David Kagan, formed. “The Shaken-Soda They shook Syndrome.” the bottle, ©2001 causing by The American bubbles Association to form, of Physics Teachers.
and set the bottle aside. Fifteen minutes later, some bubbles
were still visible, but after 2 hours, no bubbles could
11. In Experiment 3, what be seen. is the most likely reason the students used the plastic bottle rather than an
aluminum can? Compared to an aluminum can, the plastic bottle:
A. rolled more rapidly down the incline.
B. made bubbles Adapted in the liquid from David easier Kagan, to see. “The Shaken-Soda Syndrome.” ©2001
by The American Association of Physics Teachers.
C. contained a greater quantity of liquid.
D. had thicker walls and was less likely to break.
Answer: B.
d 1 L of a liquid—tap water containempty
can, sealed the can, and found
ey added 1 L of the tap water to a
aled it, shook it, and immediately
ey repeated these procedures using
g many bubbles, and a carbonated
ed no bubbles and that tasted flat,
carbonation. The results are shown
12. Based on the results 11. of Experiments In Experiment 1 and 3, what 2, in is which the most of the likely following reason trials, the before stu- shaking, were the
average speeds of the cans the dents same? used the plastic bottle rather than an aluminum
F. Trials 1 and 2 can? Compared to an aluminum can, the plastic bottle:
G. Trials 2 and 3 A. rolled more rapidly down the incline.
Table 1
H. Trials 2 and 4 B. made bubbles in the liquid easier to see.
J. Trials 3 and 5 C. contained a greater quantity of liquid.
Roll time
D. had thicker walls and was less likely to break.
We are told that the plastic bottle is clear. We are also told that in Experiment 3, bubbles
are visible after 15 minutes, but not after 2 hours. There is no mention of seeing any
bubbles in the other experiments.
Answer: J.
Table 2
Roll time
before shaking after shaking
Trial (sec) (sec)
4 1.86 1.96
5 1.75 1.93
before shaking after shaking 12. Based on the results of Experiments 1 and 2, in which
(sec) (sec)
of the following trials, before shaking, were the aver-
1.75
1.97
1.75
age speeds of the cans the same?
Trials 1.75 3 and 5 both have speeds of 1.75 seconds before shaking.
2.15
F. Trials 1 and 2
13. In 1.96
G. Trials 2 and 3
Experiment 2, a result H. of Trials shaking 2 and the can 4 of flat-tasting beverage was that the:
A. number of bubbles J. in the Trials beverage 3 and immediately 5 decreased.
B. mass of the can of beverage increased.
C. roll time of the can of beverage decreased.
D. roll time of the can of beverage increased. GO ON TO THE NEXT PAGE.
46

Answer: D.
In Experiment 2, the roll time of the can of flat-tasting beverage increased from 1.75
seconds before shaking to 1.96 seconds after shaking.
14. In Trial 5, is it likely that bubbles were present in large numbers immediately before the can was
shaken?
F. Yes; based on the results of Experiment 1, the bubbles produced in Trial 4 probably lasted for less
than 15 min.
G. Yes; based on the results of Experiment 1, the bubbles produced in Trial 4 probably lasted for
more than 2 hr.
H. No; based on the results of Experiment 3, the bubbles produced in Trial 4 probably lasted for less
than 2 hr.
I. No; based on the results of Experiment 3, the bubbles produced in Trial 4 probably lasted for more
than 3 hr.
Answer: H.
In Experiment 2, we are told that Trial 5 occurred 2 hours after Trial 4. In Experiment 3,
we are told that 2 hours after the plastic bottle was set aside, no bubbles could be seen.
From this we can conclude that the presence of bubbles immediately before shaking in
Trial 5 was unlikely.
15. Suppose that in Experiment 2, two hours after the completion of Trial 5, the students had measured the
roll time of the can of liquid without first shaking the can. Based on the results of Trials 4 and 5, the
roll time would most likely have been:
A. less than 1.86 sec.
B. between 1.86 sec and 1.93 sec.
C. between 1.94 sec and 1.96 sec.
D. greater than 1.96 sec.
Answer: A.
After Trial 4, the roll time decreased from 1.96 seconds immediately after shaking 1.75
after setting the can aside for 2 hours. So 2 hours after Trial 5, one would expect the roll
time to decrease, probably back down to 1.75 seconds (the roll time for both Trial 3 and
Trial 5) and certainly below 1.86 seconds.
16. Based on the results of Trials 3−5 and Experiment 3, if the students had added 1 L of the flat-tasting
beverage to one of the empty aluminum cans, sealed the can, and shaken it, how long would it most
likely have taken for the number of bubbles in the beverage to become too few to affect the roll time?
F. Less than 5 min
G. Between 5 min and 14 min
H. Between 15 min and 2 hr
J. Over 2 hr
Answer: H.
From Trial 4 in Experiment 2, we saw that 15 minutes was not enough to bring the roll
time down to the baseline value of 1.75 seconds in Trial 3. We also saw in Experiment 3

that bubbles were still 4visible after 15 minutes. However, from Experiment 3 we also
learned that there were no bubbles visible after 2 hours, and in Trial 5 of Experiment 2,
we saw that after the can was set aside for 2 hours, roll time went back to the baseline
speed of 1.75 seconds. 4Passage IV
Figure 2 shows the average rat
various wavelengths as a percent o
The chemical reactions associated with photosynthesis photosynthesis at 670 nm.
Passage IV Passage can be summarized IV with the following chemical equation:
Figure 2 shows the average rat
The chemical reactions associated with photosynthesis can be summarized with the following various chemical 110
wavelengths as a percent o
equation:
6,CO The chemical reactions associated with photosynthesis photosynthesis at 670 nm.
2 + 12,H
can be summarized
2O + energy → C
with the following
6H12O6 + 6,O2 + 6,H
chemical equation:
2O
100
110
6,CO2 + 12,H2O + energy → C6H12O6 + 6,O2 + 6,H2O 90
100
80
Table 1 lists wavelength ranges Table for 1 visible lists wavelength light and the ranges color frequently for visible associated light and with each range. 90
the color frequently associated with each range.
70
80
Table 1 lists wavelength ranges for visible light and 60
the color frequently associated with each range.
70
Table 1
50
60
Table Wavelength
40
1
Color (nm)
50
30
Wavelength
40
Violet 380−430
20
Blue Color 430−500 (nm)
30
Green 500−565
10
Violet
Yellow
380−430
565−585
20
Blue
Orange
430−500
0
585−630
Green 500−565
10400
440 480 520 560
Red 630−750
Yellow 565−585
0
wavelength (
Orange 585−630
400 440 480 520 560
Table 1 adapted from Neil RedA. Campbell, 630−750 Jane B. Reece, and Lawrence
G. Mitchell, Biology, 5th ed. ©1999 by Benjamin/Cummings.
wavelength Figure 2 (
Table 1 adapted from Neil A. Campbell, Jane B. Reece, and Lawrence
Figure 1 shows the relative G. Mitchell, absorption Biology, of 5th light ed. by ©1999 chlorophyll by Benjamin/Cummings.
Figure 2
a and chlorophyll b versus the Figures wavelength 1 and of 2 adapted from Peter H.
light from 400 nm to 750 nm.
Susan E. Eichhorn, Biology of Plants, 4th
lishers, Inc.
Figure 1 shows the relative absorption of light by Figures 1 and 2 adapted from Peter H. R
chlorophyll a and chlorophyll b versus the wavelength of Susan E. Eichhorn, Biology of Plants, 4th
light lishers, Inc.
Figure
from 400
1 shows
nm to 750
the
nm.
relative absorption of light by
chlorophyll a and chlorophyll b versus the wavelength of
light from 400 nm to 750 nm.
Key
chlorophyll a
Key chlorophyll b
100
chlorophyll a
90
100
80
90
70
80
60
70
50
60
40
50
30
40
chlorophyll b
20
30
10
20
0
10
400
0
400
450
450
500 550 600 650
wavelength (nm)
500 550 600 650
wavelength Figure 1 (nm)
700
700
750
750
17. Based on Table 1 and Figure 1,
associated with the wavelength
17. Based the greatest on Table absorption 1 and Figure by chloro 1,
associated A. Blue with the wavelength
the B. greatest Green absorption by chloro
A. C. Blue Yellow
B. D. Green Red
C. Yellow
Figure 1
D. Red
relative relative absorption absorption
ACT-64E-PRACTICE
rate of rate photosynthesis of photosynthesis (as % of (as rate % of at rate 670 at nm) 670 nm)
GO ON TO T

Figure 1 shows the relative absorption of light by
chlorophyll a and chlorophyll b versus the wavelength of
light from 400 nm to 750 nm.
relative absorption
100
90
80
70
60
50
40
30
Key
chlorophyll a
chlorophyll b
Figures 1 and 2 adapted from Peter H.
Susan E. Eichhorn, Biology of Plants, 4th
lishers, Inc.
20
10
0
400 450 500 550 600 650
wavelength (nm)
Figure 1
700 750
4
17. Based on Table 1 and Figure 1,
associated with the wavelength
the greatest absorption by chloro
A. Blue
B. Green
C. Yellow
D. Red
Figure 2 shows the average Figure rate of photosynthesis 2 shows the average at various rate wavelengths of photosynthesis as a percent at
ACT-64E-PRACTICE
of the average rate of
various wavelengths as a percent of the average rate of
tions associated with
photosynthesis
photosynthesis
at 670 nm.
photosynthesis at 670 nm.
48
h the following chemical equation:
GO ON TO T
nergy → C6H12O6 + 6,O2 + 6,H2O 110
100
elength ranges for visible light and
sociated with each range.
Table 1
Wavelength
lor (nm)
let 380−430
e 430−500
en 500−565
low 565−585
nge 585−630
630−750
. Campbell, Jane B. Reece, and Lawrence
. ©1999 by Benjamin/Cummings.
the relative absorption of light by
rophyll b versus the wavelength of
50 nm.
Key
chlorophyll a
chlorophyll b
rate of photosynthesis (as % of rate at 670 nm)
90
80
70
60
50
40
30
20
10
0
400 440 480 520 560 600 640 680 720
wavelength (nm)
Figure 2
Figures 1 and 2 adapted from Peter H. Raven, Ray F. Evert, and
Susan E. Eichhorn, Biology of Plants, 4th ed. ©1986 by Worth Publishers,
Inc.

17. Based on Table 1 and Figure 1, which color of light is associated with the wavelength of light that
results in the greatest absorption by chlorophyll b ?
A. Blue
B. Green
C. Yellow
D. Red
Answer: A.
Figure 1 shows that the greatest absorption by chlorophyll b occurs at a wavelength
between 475 nm and 500 nm. Table 1 shows that this falls in the wavelength range for
Blue (430 nm to 500 nm).
18. In eukaryotic organisms, the chemical reactions associated with the chemical equation shown in the
passage typically occur within which of the following structures?
F. Chloroplasts
G. Mitochondria
H. Lysosomes
J. Nuclei
Answer: F.
Since photosynthesis involves, chlorophyll, a logical guess is Chloroplasts.
19. In Figure 2, at which of the following wavelengths does the rate of photosynthesis exceed the rate of
photosynthesis at 670 nm?
A. 400 nm
B. 430 nm
C. 630 nm
D. 700 nm
Answer: B.
In Figure 2, the vertical axis shows the rate of photosynthesis as a percentage of the rate
the rate at 670 nm. The rate is exceeded at values greater than 100%. Of the possible
answers, only 430 nm has a rate of more than 100% of the rate for 670 nm.
20. In the chemical equation shown in the passage, the carbon in CO2 becomes part of which of the
following types of molecules?
F. Fat
G. Sugar
H. Protein
J. Nucleic acid
Answer: G.
For this question, it helps to know that plants produce glucose during photosynthesis.
Glucose is a sugar. You don’t need to know the chemical formula for glucose. Look at
the equation shown in the passage:
6 CO 2 + 12 H 2O + energy → C 6H 12O 6 + 6 O 2 + 6 H 2O

We can see that the carbon from CO 2 goes into C 6H 12O 6. Since O 2 is oxygen and H 2O is
water, C 6H 12O 6 must be a glucose, or sugar molecule.
21. Which of the following conclusions is best supported by Figures 1 and 2? The wavelength that results
in the highest rate of photosynthesis also results in the:
A. lowest relative 4absorption by chlorophyll a.
B. lowest relative absorption by chlorophyll b.
C. highest relative absorption by chlorophyll a.
D. highest relative absorption by chlorophyll b.
Passage V
Experiment 3
Answer: C.
A solid plastic bead was plac
Students performed the following experiments to
sample of each of Liquids 1−10 from
determine the density of common plastics.
If the bead stayed at the bottom,
From Figure 2, we see that the wavelengths that result in the highest rates of Table 3. If the bead rose, “R” was re
procedure was repeated for various p
photosynthesis are around 440 nm and 680 nm. These values correspond with the highest
relative absorption for chlorophyll a in Figure 1.
Passage V
Experiment 1
Table 3
Students performed the following A dry experiments 100 mL graduated to determine cylinder the density was placed of common on an plastics.
electronic balance and tared (the balance was reset to
0.000 g). H2O was added to the graduated cylinder until a
Experiment 1 certain mass was obtained. Ethanol was added to the gradu- Plastic 1 2 3 4
Li
5
A dry 100 mL graduated ated cylinder was until placed the on volume an electronic of liquid balance was and 50.0 tared mL. (the Thebalance
was reset to
0.000 g). H2O was added density to the of graduated the liquid cylinder was until then a calculated. certain mass The was procedure obtained. Ethanol Polybutylene was added to
the gradu- ated cylinder was until repeated the volume with of different liquid was amounts 50.0 mL. of The ethanol density and of the H2O VLDPE
liquid was then
calculated. The procedure
(see
was
Table
repeated
1).
LDPE
with different amounts of ethanol and H2O (see Table HDPE 1).
R
S
S
S
R
R
S
S
R
R
S
S
R
R
R
S
R
R
R
R
PA-11 S S S S S
PA-6 S S S S S
Table 1
Polycarbonate
PVC
S
S
S
S
S
S
S
S
S
S
Mass of Mass of Total
Liquid
H2O (g)
ethanol
(g)
mass
(g)
Density
(g/mL)
1 0 39.67 39.67 0.793
2 10.24 32.43 42.67 0.853
3 19.79 25.23 45.02 0.900
4 35.42 12.47 47.89 0.958
5 49.96 0 49.96 0.999
Experiment 2
A known mass of potassium iodide (KI) was dissolved in a known mass of H2O. A dry 100 mL graduated
cylinder was placed on the balance and tared. The solution was added to the graduated cylinder until the
volume was 50.0 mL. The Experiment density of 2 the liquid was then calculated. The procedure was repeated with
different amounts of KI and H2O (see Table 2).
A known mass of potassium iodide (KI) was dissolved
in a known mass of H2O. A dry 100 mL graduated cylinder
was placed on the balance and tared. The solution was
added to the graduated cylinder until the volume was
50.0 mL. The density of the liquid was then calculated. The
procedure was repeated with different amounts of KI and
H2O (see Table 2).
Table 2
Mass of
Mass of Mass of solution in
H 2O in KI in graduated
solution solution cylinder Density
Liquid (g) (g) (g) (g/mL)
6 97.66 7.36 52.51 1.05
7 95.41 15.52 55.70 1.11
22. In Experiment 1, the density of
be:
F. less than 0.793 g/mL.
G. 0.793 g/mL.
H. 0.999 g/mL.
J. greater than 0.999 g/mL.
23. Based on the results of Experime
PA-11 is most likely:
A. less than 0.793 g/mL.

Answer: G.
In Experiment 1, only Liquid 1 is pure ethanol. The density of the Liquid 1 was found to
be 0.793 g/mL, so this is the measure of pure ethanol that was taken.
23. Based on the results of Experiments 1−3, the density of PA-11 is most likely:
potassium iodide (KI) was dissolved
O. A dry 100 mL graduated
A.
cylinder
less than 0.793 g/mL.
lance and tared. The solution B. between was 0.853 g/mL and 0.958 g/mL.
ed cylinder until the volume C. between was 0.999 g/mL and 1.05 g/mL.
f the liquid was then calculated. D. greater Thethan
1.11 g/mL.
d with different amounts of KI and
Table 2
Answer: C.
Mass of
Mass of solution in
KI in graduated
solution cylinder Density
(g) (g) (g/mL)
in a known mass of H 2O. A dry 100 mL graduated cylinder
was placed on the balance and tared. The solution was
added to the graduated cylinder until the volume was
50.0 mL. The density of the liquid was then calculated. The
procedure was repeated with different amounts of KI and
H 2O (see Table 2).
Table 2
Mass of
Mass of Mass of solution in
H 2O in KI in graduated
solution solution cylinder Density
Liquid (g) (g) (g) (g/mL)
6 97.66
7 95.41
8 94.38
9 92.18
10 87.77
Experiment 3
7.36
15.52
20.68
29.08
41.31
52.51
55.70
57.53
60.63
64.64
1.05
4
1.11
1.15
1.21
1.29
A solid plastic bead was placed at the bottom of a
ed the following experiments Experiment 3 to sample of each of Liquids 1−10 from Experiments 1 and 2.
f common plastics.
A solid plastic bead was If ACT-64E-PRACTICE
placed the bead at the stayed bottom at of the a sample bottom, of each “S” of was Liquids recorded 1−10 from in Experiments 1 and
2. If the bead stayed at Table the bottom, 3. If the “S” bead was recorded rose, “R” in was Table recorded 3. If the in bead Table rose, 3. “R” The was 50recorded
in
Table 3. The procedure procedure was repeated was for repeated various for plastics. various plastics.
Table 3
raduated cylinder was placed on an
d tared (the balance was reset to
ed to the graduated cylinder until a
Liquid
ned. Ethanol was added to the gradu- Plastic 1 2 3 4 5 6 7 8 9 10
volume of liquid was 50.0 mL. The
was then calculated. The procedure
ferent amounts of ethanol and H2O Polybutylene
VLDPE
LDPE
R
S
S
R
R
S
R
R
S
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
HDPE S S S S R R R R R R
PA-11 S S S S S R R R R R
PA-6 S S S S S S S R R R
Table 1
Polycarbonate
PVC
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
R
S
R
S
Mass of Total
ethanol
(g)
39.67
32.43
25.23
12.47
mass Density
(g) 22. In (g/mL) Experiment 1, the density of ethanol was found to be:
F. less than 0.793 g/mL.
39.67 G. 0.793 0.793 g/mL.
42.67
H.
0.853
0.999 g/mL.
45.02 0.900
47.89 J. 0.958 greater than 0.999 g/mL.
0 49.96 0.999
22. In Experiment 1, the density of ethanol was found to
be:
F. less than 0.793 g/mL.
G. 0.793 g/mL.
H. 0.999 g/mL.
J. greater than 0.999 g/mL.
23. Based on the results of Experiments 1−3, the density of
22. In Experiment 1, the density of
be:
F. less than 0.793 g/mL.
G. 0.793 g/mL.
H. 0.999 g/mL.
J. greater than 0.999 g/mL.
23. Based on the results of Experime
PA-11 is most likely:
A. less than 0.793 g/mL.
B. between 0.853 g/mL and 0.9
C. between 0.999 g/mL and 1.0
D. greater than 1.11 g/mL.
GO ON TO T

PA-11 sank in liquids 1 through 5 and rose in liquids 6-10. The density of the liquids
increased from 1 to 10. The density of PA-11 has to be greater than the density of liquid
5, since it sank in liquid 5, and less than liquid 6, since it rose in liquid 6. So the density
of PA-11 is between 0.999 g/mL (the density of liquid 5) and 1.05 g/mL (the density of
liquid 6)
24. Suppose that a sixth KI/H 2O solution had been measured in Experiment 2 and the mass of the solution
in the graduated cylinder was 67.54 g. The density of this solution would most likely have been closest to
which of the following?
F. 1.25 g/mL
G. 1.30 g/mL
H. 1.35 g/mL
J. 1.40 g/mL
4 4
Answer: H.
Looking
24. Suppose
at Table
that a
2,
sixth
we see
KI/H
that the mass of solution for liquid 10 is 64.64 g, so a mass of
2O solution had been mea- 26. In Experiments 1 and 2, the students tared the grad
67.54 sured would in Experiment be an increase 2 and of the 3.10 mass g. of Between the solution liquids in 9 and 10, ated density cylinder goes in each up about trial so they could more easi
0.02 the g/mL graduated per gram cylinder of mass was 67.54 of solution g. The in density graduated of thiscylinder
determine: since mass goes from 60.63
solution would most likely have been closest to which
g to 64.64 g (about 4 g) when density goes from 1.21 g/mL to F. 1.29 the g/mL. mass The of the substances added to the graduate
of the following?
cylinder.
relationship F. 1.25 between g/mL mass of solution in cylinder and density G. looks the linear, density so of we the can graduated cylinder.
guess G. that 1.30 it will g/mLhold
as we go from 64.64 g to 67.54 g. So a rise H. when of 3.10 the g total in mass volume of the added substances w
should H. cause 1.35 g/mL a rise of about 0.06 g/mL in density, and 1.29+0.06=1.35. equal to 50.0 mL.
J. 1.40 g/mL
J. when all of the KI was dissolved in the H2O. 25. A plastic bead was tested as in Experiment 3 using
Liquids 1−4. Which of the following is NOT a plausible
set of results for the plastic?
Answer: B.
Since liquids get more dense as we go from 1 to 4, it is not plausible that a bead would
rise in the (lower density) liquids 1 and 2 and the sink in the (higher density) liquids 3
and 4.
26. In Experiments 1 and 2, the students tared the graduated cylinder in each trial so they could more easily
determine:
F. the mass of the substances added to the graduated cylinder.
G. the density of the graduated cylinder.
H. when the total volume of the added substances was equal to 50.0 mL.
J. when all of the KI was dissolved in the H 2O.
Answer: F.
Liquid
1 2 3 4
A. R R R R
B. R R S S
C. S S R R
D. S S S S
27. A student claimed that polycarbonate is more dens
than PA-6. Do the results of Experiments 1−3 suppo
his claim?
A. No, because in Liquid 8, polycarbonate stayed
the bottom and PA-6 rose.
B. Yes, because in Liquid 8, polycarbonate stayed
the bottom and PA-6 rose.
C. No, because in Liquid 8, polycarbonate rose an
PA-6 stayed at the bottom.
D. Yes, because in Liquid 8, polycarbonate rose an
PA-6 stayed at the bottom.

The word tared is italicized and defined in the description of Experiment 1. This is a clue
that it will be used in one of the questions (we saw the same thing with the word
refracted in Passage I). The definition given for tared is “the balance was reset to 0.000
g)”. Why would the students reset the balance to 0 after the graduated cylinder was
placed on it? Because they are interested in the weight of the liquid added to the cylinder,
not the weight of the liquid plus the weight of the cylinder.
27. A student claimed that polycarbonate is more dense than PA-6. Do the results of Experiments 1−3
support his claim?
A. No, because in Liquid 8, polycarbonate stayed at the bottom and PA-6 rose.
B. Yes, because in Liquid 8, polycarbonate stayed at the bottom and PA-6 rose.
C. No, because in Liquid 8, polycarbonate rose and PA-6 stayed at the bottom.
D. Yes, because in Liquid 8, polycarbonate rose and PA-6 stayed at the bottom.
4
Answer: B.
From Table 3, we see Passage that polycarbonate VI and PA-6 are both plastics used to make Experiment beads 2
for Experiment 3. If polycarbonate sinks in a particular liquid and PA-6 rises in that Synergism occurs when 2
Bacteria break down sugars by fermentation. To study
liquid, then polycarbonate is more dense that PA-6. This is exactly what happens together for to ferment a sugar by using
2 fermentation pathways, researchers performed 2 experi-
species can use alone. To investiga
liquid 8.
ments using broth that contained either the sugar sucrose or ment 1 was repeated, except that diff
the sugar lactose. One of the fermentation pathways pro- species were added to each large test
duces CO2 gas and increases the acidity (lowers the pH) of
Passage VI the solution. The other pathway produces acid but not CO2. Bacteria break down sugars by fermentation. To study 2 fermentation pathways, researchers performed 2
experiments using broth that contained either the sugar sucrose or the sugar lactose. One of the
Table 2
fermentation pathways produces CO2 gas and increases the acidity (lowers the pH) of the solution. The
other pathway produces acid but not CO2. Experiment 1
Sucrose broth
Species
Experiment 1
Sucrose broth was added to 5 large test tubes. Next,
added acid CO2 Sucrose broth was added phenol to 5 large red (a test pH tubes. indicator Next, phenol that is red yellow (a pH if indicator pH < 7, that red is ifyellow
if pH < 7, red
pH ≥ 7) was added to each large test tube. A Durham tube
A and B − −
if pH ≥ 7) was added to
(a
each
small
large
test
test
tube)
tube.
was
A Durham
placed,
tube
inverted,
(a small
in
test
each
tube)
large
was
test
placed, inverted, in
A and C + +
each large test tube to collect tube to CO collect 2 (see CO Figure 1). 2 (see Figure 1).
B and D + +
C and D + +
Durham
tube
broth
(red)
Figure 1
The large test tubes were capped, The large heated test until tubes the solutions were capped, were sterile, heated then until cooled. theOne
of 4 bacterial
species (Species A−D) solutions was added were to each sterile, of 4 of then the large cooled. test tubes. One The of 4 procedure bacterial was repeated using
lactose broth instead of species sucrose (Species broth. The A−D) 10 large was test added tubes to (all each containing of 4 of the solutions large at a pH of 7) were
then incubated at 37°C
test
for 48
tubes.
hr.
The procedure was repeated using lactose broth
instead of sucrose broth. The 10 large test tubes (all containing
solutions at a pH of 7) were then incubated at 37°C
for 48 hr.
28. In Experiment 1, which of the
The large test tubes and Durham tubes were exam- mented lactose?
ined. If acid was produced, the solution was yellow. If no F. Species B only
acid was produced, the solution remained red. If CO2 was G. Species C only
produced, a gas bubble was observed at the top of the H. Species B and Species D on
Durham tube (see Table 1).
J. Species C and Species D on
29. Suppose that in Experiment 2

test tubes. The procedure was repeated using lactose broth
instead of sucrose broth. The 10 large test tubes (all containing
solutions at a pH of 7) were then incubated at 37°C
for 48 hr.
28. In Experiment 1, which of the
The large test tubes and Durham tubes were exam- mented lactose?
ined. If acid was produced, the solution was yellow. If no F. Species B only
acid was produced, the solution remained red. If CO2 was
The large test tubes and Durham tubes were examined. If acid was produced, the solution was yellow. G. Species If no C only
produced, a gas bubble was observed at the top of the H. Species B and Species D onl
acid was produced, the Durham solution tube remained (see red. Table If 1). CO2 was produced, a gas bubble was observed at
J.
the
Species
top
C and Species D onl
of the Durham tube (see Table 1).
29. Suppose that in Experiment 2
Species C had been added to a l
Table 1
ing sucrose broth and to a larg
lactose broth. Which of the f
likely depict the results?
In these experiments, bacteria breaks down one of two sugars—sucrose or lactose—in
two pathways. Both fermentation pathways increase acidity, but only one pathway
produces CO 2. Increasing acidity means lowering pH.
The pH indicator phenol red is yellow if pH < 7 (acidity is high) or red if pH ≥ 7 (acidity
is low). Solutions turn yellow if acid is produced. Otherwise, they remain red.
Synergism occurs when two bacterial species work together to ferment a sugar in a way
that neither can do alone.
bes were capped, heated 28. In Experiment until the 1, which of the bacterial species fermented lactose?
, then cooled. One of
F.
4 bacterial
Species B only
was added to each of 4 of the large
re was repeated using lactose G. Species broth C only
th. The 10 large test tubes H. (all Species con- B and Species D only
H of 7) were then incubated J. at Species 37°C C and Species D only
Answer: H.
Species
Sucrose broth Lactose broth
added acid CO2 acid CO2 Sucrose broth Lactose b
A
B
C
D
Experiment
None
2
−
−
+
+
−
−
−
+
−
−
−
+
−
+
−
−
+
−
−
− 4
A.
B.
C.
D.
acid
–
+
+
–
CO2 –
+
+
–
acid
+
–
+
–
C
Synergism occurs when 2 bacterial species act
wn sugars by fermentation. To study
Experiment 2 together to ferment a sugar by using a pathway that neither
ys, researchers performed 2 experi-
Synergism occurs when
species
2 bacterial
can
species
use alone.
act together
To investigate
to ferment
synergism, ACT-64E-PRACTICE
a sugar by using
Expericontained
either the sugar sucrose or
a pathway that neither
ment 1 was repeated, except that different pairs of bacterial
of the fermentation species pathways can use proalone.
To
species
investigate
were added
synergism,
to each
Experiment
large test
1
tube
was repeated,
(see Table
except
2).
that 52different
pairs of
reases the acidity (lowers bacterial the species pH) ofwere
added to each large test tube (see Table 2).
pathway produces acid but not CO2. GO ON TO T
s added to 5 large test tubes. Next,
ator that is yellow if pH < 7, red if
each large test tube. A Durham tube
placed, inverted, in each large test
e Figure 1).
urham
tube
broth
(red)
Figure 1
bes and Durham tubes were examuced,
the solution was yellow. If no
solution remained red. If CO 2 was
le was observed at the top of the
e 1).
Table 1
crose broth Lactose broth
Table 2
Sucrose broth Lactose broth
Species
added acid CO2 acid CO2 A and B − − + +
A and C + + − −
B and D + + + +
C and D + + + +
28. In Experiment 1, which of the bacterial species fermented
lactose?
F. Species B only
G. Species C only
H. Species B and Species D only
J. Species C and Species D only
29. Suppose that in Experiment 2 both Species B and
Species C had been added to a large test tube containing
sucrose broth and to a large test tube containing
lactose broth. Which of the following would most
likely depict the results?

of 4 of the large
ng lactose broth
t tubes (all concubated
at 37°C
es were examas
yellow. If no
red. If CO 2 was
the top of the
e broth
CO 2
−
+
−
− −
28. In Experiment 1, which of the bacterial species fermented
lactose?
For a F. sugar Species to be B fermented, only it must produce acid alone, or acid and CO2. Under the
G. Species C only
heading H. Species “Lactose B and broth” Species in Table D only1,
we see that acid and CO2 were produced by species
B, and J. acid Species alone C and was Species produced D only by species D.
29. 29. Suppose Suppose that that in Experiment in Experiment 2 both 2 Species both B Species and Species B and C had been added to a large test tube
Species C had been added to a large test tube contain-
containing
ing sucrose
sucrose
broth
broth and
and
to
to
a large
a large
test
test
tube
tube
containing
containing
lactose broth. Which of the following would
most likely lactose depict broth. the results? Which of the following would most
likely depict the results?
Answer: C.
GO ON TO THE NEXT PAGE.
52
Comparing Table 1 and Table 2, we see that when two species are combined, they
perform the fermentation they can do on their own plus any synergistic fermentation. For
example, from Table 1 we see that species C ferments sucrose by producing both acid
and CO2, and species D ferments lactose by producing both acid and CO2. In Table 2, we
see that together they ferment both sucrose and lactose by producing both acid and CO2. There is a similar relationship between B and C, so we expect a similar result.
30. Suppose a scientist isolates a bacterial species that is 1 of the 4 species used in Experiment 1. She adds
the species to sucrose broth and observes that neither acid nor CO2 is produced. She then adds the species
to lactose broth and observes that both acid and CO2 are produced. Based on the results of Experiment 1,
the species is most likely:
F. Species A.
G. Species B.
H. Species C.
J. Species D.
Answer: G.
Sucrose broth Lactose broth
acid CO 2 acid CO 2
A. – – + +
B. + + – –
C. + + + +
D. – – – –
From Table 1, we see that there are three species that produce neither acid nor CO 2 in a
sucrose broth when they are alone in the broth: species A, species B and species D. Of
these, only species B produces both acid and CO 2 in a lactose broth.
31. What is the evidence from Experiments 1 and 2 that Species C and Species D acted synergistically in
Experiment 2?
A. No acid was produced when each species was alone in the sucrose broth, but acid was produced
when the 2 species were together in the sucrose broth.
B. No acid was produced when each species was alone in the lactose broth, but acid was produced
when the 2 species were together in the sucrose broth.
C. No CO2 was produced when each species was alone in the sucrose broth, but CO2 was produced
when the 2 species were together in the sucrose broth.
D. No CO2 was produced when each species was alone in the lactose broth, but CO2 was produced
when the 2 species were together in the lactose broth.

Answer: D.
Once again, we see an italicized word from a passage (synergism) being used in a
question. From the description in Experiment 2, “synergism occurs when 2 bacterial
species act together to ferment a sugar by using a pathway that neither species can use
alone.” So, for evidence of synergism, we would need to see evidence of a pathway
(production of acid or production of acid and CO 2) that occurs when C and D are together
(Table 2) that we don’t see when they are separate (Table 1). What we see in Table 1 is
that species C does not ferment lactose and species D only produces acid in the lactose
broth. However, when they are added together as shown in Table 2, they produce both
acid and CO 2 in the lactose broth.
4
erial species that is 32. Which 32. Which of the following of the following figures best figures illustrates best the illustrates results of Experiment the 1 for Species D in the sucrose
ent 1. She adds the broth? results of Experiment 1 for Species D in the sucrose
ves that neither acid broth?
s the species to laccid
and CO2 are pro- F.
H.
Experiment 1, the
iments 1 and 2 that
synergistically in
each species was
t acid was produced
ther in the sucrose
each species was
acid was produced
ther in the sucrose
each species was
t CO 2 was produced
ther in the sucrose
each species was
CO 2 was produced
ether in the lactose
G.
33. Is the hypothesis that Species A and Species C acted
synergistically supported by the results of Experiment
2 ?
Answer: G.
Durham
tube
broth
(red)
Durham
tube
broth
(yellow)
J.
Durham
tube
broth
(yellow)
Durham
tube
broth
(red)
gas
bubble
gas
bubble
In the sucrose A. Yes, broth, because species both D acid produces and COacid but no CO2. Acid turns the broth yellow. If
2 were produced
no CO2 is produced, from sucrose. then no gas bubble is formed.
B. Yes, because both acid and CO 2 were produced
from lactose.
C. No, because only acid, not CO 2, was produced
from both sucrose and lactose.
D. No, because neither acid nor CO 2 was produced
from lactose.

33. Is the hypothesis that Species A and Species C acted synergistically supported by the results of
Experiment 2 ?
A. Yes, because both acid and CO2 were produced from sucrose.
B. Yes, because both acid and CO2 were produced from lactose.
C. No, because only acid, not CO2, was produced from both sucrose and lactose.
D. No, because neither acid nor CO2 was produced from lactose.
Answer: D.
Species C produces 4both acid and CO2 in sucrose, so we would expect both to be
produced in a mix of A and C, even without synergy. For evidence of synergy, we would
need to see the production of acid or both acid and CO2 when both A and C are present,
since neither species alone produces acid or CO2 in lactose.
Passage VII
34. Which of the following stateme
with the DNA Hypothesis? The
Passage VII
In the 1940s, scientists thought all genetic material generally increase from cell ty
In the 1940s, scientists was thought contained all genetic in structures material was called contained chromosomes in structures and called thatchromosomes
number and of:
that chromosomes had chromosomes been found only had in the been nucleus found of only a cell in (not the in nucleus the cytoplasm): of a cell
(not in the cytoplasm):
F. amino acids in the nucleus in
to cell type.
cytoplasm
G. amino acids in the cytoplas
type to cell type.
H. chromosomes in the nucleu
type to cell type.
J. chromosomes in the cytopla
nucleus
type to cell type.
chromosomes
35. By referring to the observatio
exclusively in the nucleus whi
throughout the cell, the scientis
Hypothesis implies that genes a
because which of the following
the nucleus?
A. Amino acids
Chromosomes are composed
Chromosomes
of 2 types
are
of molecules,
composed
proteins
of 2 types
and
of
deoxyribonucleic
molecules, pro-
acid (DNA). B. Proteins
teins and deoxyribonucleic acid (DNA). Proteins are com-
are com- posed of subunits called amino acids. DNA consists of chains of subunits called nucleotides. C. Gametes
posed of subunits called amino acids. DNA consists of
The
parts of chromosomes that are responsible for the transmission of genetic information are called D. genes. Chromosomes
chains of subunits called nucleotides. The parts of chromo-
Two scientists in the 1940s somes debate that whether are responsible genes are for made the of transmission proteins or DNA. of genetic
information are called genes.
36. According to the passage, a sim
Protein Hypothesis
and proteins is that both types of
Genes are made only of proteins. Two scientists Proteins make in the up 1940s 50% or debate more of whether a cell’s genes dry weight. are Cells contain F. are 20 found only in gametes.
made of proteins or DNA.
different amino acids that can be arranged in a virtually infinite number of ways to make different G. proteins. are abundant in the cytoplasm
The number and arrangement of different amino acids within a protein form the codes that contain H. contain 20 different amino a
hereditary information.
J. are composed of smaller sub
Protein Hypothesis
In contrast, only 4 different
Genes
nucleotides
are made
make up
only
the
of
DNA
proteins.
found in
Proteins
cells, and
make
they are
up
believed 37. to According form to the Protein Hypoth
50% or more of a cell’s dry weight. Cells contain 20 differ-
chains only in certain ratios. As a result, the number of different combinations that DNA can carry lowing is much observations provides t
ent amino acids that can be arranged in a virtually infinite
smaller than the number that proteins can carry.
that genes are NOT composed of
number of ways to make different proteins. The number
and arrangement of different amino acids within a protein A. DNA is composed of only 4
DNA Hypothesis form the codes that contain hereditary information.
B. DNA is composed of smal
proteins.
Genes are made only of DNA. DNA is found exclusively in the cell’s nucleus, whereas proteins are found
throughout the nucleus and cytoplasm. In contrast, Additionally, only 4 different the amount nucleotides of protein make in a up cell the C. DNA is abundant in both
varies from cell type to
DNA found in cells, and they are believed to form chains
cytoplasm.
cell type, even within the
only
same
in certain
animal.
ratios. As a result, the number of different D. The concentration of DNA
combinations that DNA can carry is much smaller than the
from cell to cell.
Though DNA is less abundant number than that proteins, proteins the can amount carry. is consistent from cell type to cell type within the
same animal, except for the gametes (the reproductive cells). Gametes have half the amount 38. of DNA Mitochondria as are organelles loc
other cells in the body. Gametes also have half the typical number of chromosomes. Thus, the amount that are of responsible for energy tr
DNA Hypothesis
Genes are made only of DNA. DNA is found exclusively
in the cell’s nucleus, whereas proteins are found
After the 1940s, it was observ
contain their own genes. This o
evidence stated in which hypothe
throughout the nucleus and cytoplasm. Additionally, the
amount of protein in a cell varies from cell type to cell
type, even within the same animal.
F. The DNA Hypothesis, beca
of DNA, the observation wo
present outside the nucleus.

DNA in a cell is correlated with the number of chromosomes in the cell. No such correlation is found for
proteins.
Chromosomes
Proteins DNA
amino
acids
nucleotides
Genes are some part of chromosomes that transmit genetic information. Two scientists debate whether
genes are made of proteins or DNA.
34. Which of the following statements is most consistent with the DNA Hypothesis? The amount of DNA
will generally increase from cell type to cell type as the number of:
F. amino acids in the nucleus increases from cell type to cell type.
G. amino acids in the cytoplasm increases from cell type to cell type.
H. chromosomes in the nucleus increases from cell type to cell type.
J. chromosomes in the cytoplasm increases from cell type to cell type.
Answer: H.
DNA is made up of nucleotides, not amino acids, so we can eliminate choices F and G.
At the time of this debate, chromosomes had only been found in the nucleus, not the
cytoplasm. Since genes are part of chromosomes, we expect the number of genes to go up
with the number of chromosomes. If the DNA Hypothesis were true, then the amount of
DNA would rise with the number of genes, which rises with the number of chromosomes.
35. By referring to the observation that DNA is found exclusively in the nucleus while proteins are found
throughout the cell, the scientist supporting the DNA Hypothesis implies that genes are made only of
DNA because which of the following are also found only in the nucleus?
A. Amino acids
B. Proteins
C. Gametes
D. Chromosomes
Answer: D.
In the first sentence of the passage we are told that chromosomes had only been found in
the nucleus at the time of the debate. This is relevant to the debate since the DNA
argument is that DNA composes genes which are found in chromosomes.
36. According to the passage, a similarity between DNA
and proteins is that both types of molecules:
F. are found only in gametes.
G. are abundant in the cytoplasm.
H. contain 20 different amino acids.
J. are composed of smaller subunits.
Answer: J.
DNA is composed of nucleotides and proteins are composed of amino acids.

37. According to the Protein Hypothesis, which of the following observations provides the strongest
evidence that genes are NOT composed of DNA ?
A. DNA is composed of only 4 types of nucleotides.
B. DNA is composed of smaller subunits than are proteins.
C. DNA is abundant in both the nucleus and the cytoplasm.
D. The concentration of DNA is generally consistent from cell to cell.
Answer: A.
In the second paragraph of the Protein Hypothesis, we see “only 4 different nucleotides
make up the DNA found in cells, and they are believed to form chains only in certain
ratios. As a result, the number of different combinations that DNA can carry is much
smaller that the number that proteins can carry.”
The size of the subunits that compose DNA doesn’t support or weaken either hypothesis.
Since proteins are also abundant in both the nucleus and the cytoplasm, C doesn’t help
the Protein Hypothesis.
Statement D supports the DNA Hypothesis.
38. Mitochondria are organelles located in the cytoplasm that are responsible for energy transformation in
a cell. After the 1940s, it was observed that mitochondria contain their own genes. This observation
contradicts evidence stated in which hypothesis?
F. The DNA Hypothesis, because if genes are made of DNA, the observation would show that DNA
is present outside the nucleus.
G. The DNA Hypothesis, because if genes are made of DNA, the observation would show that DNA
is present inside the nucleus.
H. The Protein Hypothesis, because if genes are made of proteins, the observation would show that
proteins are present outside the nucleus.
J. The Protein Hypothesis, because if genes are made of proteins, the observation would show that
proteins are present inside the nucleus.
Answer: F.
In the second sentence of the DNA Hypothesis we see the argument that “DNA is found
exclusively in the cell’s nucleus.” Since mitochondria are in the cytoplasm, the presence
of genes in the mitochondria means that genes are present outside the nucleus.
Answer G makes no sense since the observation shows that DNA is present outside the
nucleus, not inside it. Also, the observation of DNA inside the nucleus would not
contradict the DNA Hypothesis.
Answers H and J make no sense since the observation says nothing about the location of
proteins.
39. The scientist who describes the DNA Hypothesis implies that the Protein Hypothesis is weakened by
which of the following observations?
A. For a given organism, the amount of protein in the gametes is half that found in other types of
cells.

B. For a given organism, the amount of protein in different types of cells is not the same.
C. Protein molecules are composed of many subunits.
D. Proteins are found only in the nucleus.
Answer: B.
At the end of the first paragraph in the DNA Hypothesis, the pro-DNA scientist says, “the
amount of protein in a cell varies from cell type to cell type, even within the same
animal.” In the second paragraph, he goes on to say that “the amount [of DNA] is
consistent from cell type to cell type with the same animal.” This implies that the lack of
consistency in amount of protein from cell to cell is important.
Answer A gives an argument that was put forward about the amount of DNA in gametes
(in the second paragraph of the DNA Hypothesis), not the amount of protein.
For answer C, the fact that protein molecules are composed of many subunits was given
as an observation that strengthens the Protein Hypothesis.
For answer D, both scientists accepted the observation that proteins 4are not found only in
the nucleus.
40. Which of the following illustrations of a portion of a DNA molecule is consistent with the description
describes the DNA Hypothesis 40. Which of the following illustrations of a portion of a
otein Hypothesis is in weakened the passage? by DNA molecule is consistent with the description in the
ing observations?
passage?
anism, the amount of protein in the
that found in other types of cells.
anism, the amount of protein in difcells
is not the same.
les are composed of many subunits.
Key
AA - amino acid
N - nucleotide
nd only in the nucleus.
F.
AA N AA N
G.
H.
J.
N
AA
N
AA AA
AA
N
Answer: J.
END OF TEST 4
We are told in STOP! the passage DO that NOT DNA RETURN is composed TO ANY of nucleotides. OTHER TEST. Amino acids compose
proteins.
[See Note on page 56.]
AA
N
N
AA
N
END OF TEST 4
STOP! DO NOT RETURN TO ANY OTHER TEST.