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).
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).
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
0°
earthquake
focus
Key
p-waves
s-waves
both p-waves and
s-waves received
at seismographs
both p-waves and
s-waves received
at seismographs
core
outer
id
qu
li
solid
inner
core
103°
103°
shadow zone:
neither p-waves nor
s-waves received
at seismographs
shadow zone:
neither p-waves nor
s-waves received
at seismographs
mantle
crust
142°
142°
only p-waves received
at seismographs
4
Note: The figure is not to scale.
Figure 1
GO ON TO THE NEXT PAGE.
ACT-64E-PRACTICE
42
time
earthquake
starts at the
focus
first p-waves
arrive at
seismograph
first s-waves
arrive at
seismograph
Figure 2
22
ch seismograph from
uake focus (min)
20
18
s-waves
16
14
12
10
p-waves
4
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
focus
Figure 2
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
22
time to reach seismograph from
earthquake focus (min)
20
18
s-waves
16
14
12
10
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.
p-waves
8
6
4
2
0
0
00 00 00 00 00 00 00 00 00 00
1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0 10,0
distance along Earth’s surface from earthquake
focus to seismograph (km)
Figure 3
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
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
the first p-waves to arrive at the
C. Both p-waves and s-waves
1. nor
Figure
1 shows that a seismograph located at a point
D. Neither p-waves
s-waves
A. smaller than the amplitude
125° around Earth from an earthquake’s focus would
arrive at the seismograph.
receive which type(s) of seismic waves, if either, from
B. larger than the amplitude o
that earthquake?
Answer: D. 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 t the p-waves
spot labeled “earthquake focus”. Between 103° C. aBoth
and s-waves
D. and zero, as is the amplitude o
142° on either side of D.the Neither
earthquake are areas labeled “shadow zone: neither p-wavesfocus nor s-waves
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.
J. change to a third type of seismic wave.
Answer: G.
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
GO ON TO T
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 δ18Ο values
of groundwater taken from samples of the top 40 m of sediment at 3 sites along the same cross section.
δ18Ο is calculated from a ratio of 2 oxygen isotopes (18O and 16O) in the groundwater. Smaller δ18 O values
indicate cooler average temperatures.
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
δ18O values of groundwater taken from samples of the top
40 m of sediment at 3 sites along the same cross section.
sted between
11,700 and 9,500 years
δ18O is calculated from a ratio of 2 oxygen isotopes (18O
(see Figure 1).16The lake was formed
Manitoba
and O) in the groundwater. Smaller δ18O values
indicate
dammed several rivers. Groundwater
cooler average temperatures.
acial sediments provides information
maximum
he time the sediments were deposited.
extent of
ss section of the sediments (lake clay
Lake
edrock in the area. Figure 3 shows the
Agassiz
dwater taken from samples of the top
Site 1
•
3 sites along the same cross section.
Site 3
•
18
•
m a ratio of 2 oxygen isotopes ( O
•
•
North
dwater. Smaller δ18O values indicate
Grand
Dakota
atures.
Forks
4
Hudson
Bay
maximum
extent of
Lake
Agassiz
Site 1
Site 3
Hudson
North
Bay
Dakota
••
•
•
•
Winnipeg
Site 2
Great
Lakes
Grand
Forks
Winnipeg
Site 2
Great
Lakes
Figure 1
Figure 1
surface
surface
surface
250
200
Key
sediment/rock
lake clay
glacial till
bedrock
250
Key
sediment/rock
lake clay
glacial till
bedrock
150
Figure 2
ACT-64E-PRACTICE
44
0
S
200
S
150
Site 3
Site 2
surface
Grand Forks,
200 North Dakota
Site 2
Winnipeg,
Manitoba
Site 1
N
elevation (m above sea level)
Site 3
250
Grand Forks,
North Dakota
Site 1
Winnipeg,
Manitoba
N
0
0
Manitoba
GO ON TO THE NEXT PAGE.
150
Figure 2
44
GO ON TO THE NEXT PAGE.
Site 1
Site 2
Site 3
δ 18 O
δ 18 O
δ 18 O
smaller
larger
–26 –22 –18 –14
(surface) 0
10
20
lake
clay
30
40
10
30
glacial
till
40
glacial
till
– 1$
!"___________________________
O/ O of standard water sample #
18
Note: δ 18 O =
lake
clay
20
18
O/16O of groundwater sample
16
4
smaller
larger
–26 –22 –18 –14
(surface) 0
depth (m)
depth (m)
smaller
larger
–26 –22 –18 –14
(surface) 0
depth (m)
4
10
20
lake
clay
30
40
×
1,000
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.
6. According to Figure 2, the lake clay deposit is thinnest
9. According to Figure 2, which of the following graphs
elevation
(m above sea level)
elevation
(m above sea level)
elevation
(m above sea level)
elevation
(m above sea level)
6. According
Figure
2, the
lake
of the following
cities
sites?sea level, of
at which oftothe
following
cities
or clay
sites?deposit is thinnest at which
best represents
the elevations,
in morabove
the top of the glacial till layer at Sites 1, 2, and 3 ?
F.
Winnipeg
F. Winnipeg
G. Site
A.
225
C.
225
G.
Site1 1
H. Site 2
H.
Site 2Forks
J. Grand
200
200
J. Grand Forks
175
175
Answer: F. 150
150
1 2 3
1 2 3
Site
Site
According
tothe Figure
3, at
Sites 1, 2,
and 3,
the smallest
In 7.Figure 2
, k
ey s
hows t
hat l
ake c
lay i
s i
ndicated w
ith l
ight g
ray. T
he v
ertical axis δ 18O value of the groundwater in the lake clay was
B.
225
D.
225
of the figure s labeled recorded
at a idepth
between:“elevation (m above sea level). This tells us, for example, that A. innipeg 0 m and 10
m. has lake clay between 230 m and 2200
the W
site 40 m above sea level. 200 In other B. 10 m and 20 m.
words, the lake clay is 240-­‐230 = 10 m “thick” at the W
innipeg s
ite. L
ooking at the C. 20
m and
30 m.
175
175
D. 30 m and 40 m.
figure, we can see that the lake clay (light gray area) is thinnest at the Winnipeg site. 150
150
1 2 3
1 2 3
Site
7. According to Figure 3, at Sites 1, 2, and 3, the smallest δ18Ο value of the groundwater
in the lake Site
clay
was recorded at a depth between:
10. Precipitation that falls at Sites 1, 2, and 3 soaks into
A. 0 m and 10 m.
the soil until it reaches the groundwater table about
8. B.
According
to Figure 2, as the thickness of the lake clay
3 m below the surface. Based on Figure 3, and assum10mand20m.
deposit increases from Grand Forks to Site 3, the
ing no alteration of the precipitation, the δ18O value of
C.
20mand30m.
thickness of the glacial till beneath it:
present-day precipitation in the study area is closest to:
D.
30mand40m.
F. increases.
F. −26.
remains the same.
G. −23.
G.
H. first increases and then decreases.
H. −20.
J. decreases.
J. −15.
Answer: C. ACT-64E-PRACTICE
TO v
THE
NEXT
PAGE.
In Figure 3, the smallest δ18O values are farthest to the left GO
and ON
have alues between 45
18
-­‐26 and -­‐14. At Site 1, the smallest δ O 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 δ18O 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:
8
lake
clay
20
30
40
O=
!"
F.
G.
H.
J.
depth (m)
depth (m)
10
10
20
lake
clay
30
glacial
till
40
increases.
remains the same.
first increases and then decreases.
decreases.
18
O/16O of groundwater
sample
___________________________
Answer: J. 18
O/16O of standard water sample
# – 1$
×
1,000
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 ry, and T. W. D. Edwards, “Isotopic Composition of Old Ground Water from Lake Agassiz:
thinner. o the thickness of the glacial till decreases. by the American Association
for theSAdvancement
of Science.
B.
C.
225
200
175
150
1
2 3
Site
D.
225
200
175
150
1
2 3
Site
elevation
(m above sea level)
elevation
(m above sea level)
A.
elevation
(m above sea level)
d 3, the smallest
lake clay was
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 ?
elevation
(m above sea level)
posit is thinnest
?
225
200
175
150
1
2 3
Site
1
2 3
Site
225
200
175
150
10. Precipitation that falls at Sites 1, 2, and 3 soaks into
Answer: the soilC. until it reaches the groundwater table about
of the lake clay
3 m below the surface. Based on Figure 3,18and assumto Site 3, the
ing no alteration of the precipitation, the δ O value of
Notice that the word elevation appears italics present-day
precipitation
in the study
area isin closest
to:in this question. In previous questions F. −26.about Figure 2, we’ve been look at the thickness of different layers. G. −23.wrote the question wants to make sure that we pay attention to elevation Whoever H. −20.
instead of thickness. J. −15.
In Figure 2, the elevation of a layer is shown with numbers along the vertical axis TOsea THE
NEXT
PAGE.
labeled “elevation GO
(m ON
above level)”. So, 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.
G. −23.
H. −20.
J. −15.
4
Passage III
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).
Experiment 2
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
Answer: J. We are told that present-­‐day precipitation m below the present-­‐day fixedsoaks points to 3incline
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 Table 2
the surface. Notice that the values on the vertical axes start at zero on the top left Roll tim
corner and get bigger as they go down. At all three sites, the δ18O value at around 3m before shaking a
can
below the surface Passage
is about III
-­‐15. Experiment 2
Trial
(sec)
angle of
The students added 1 L of the f
inclination
4
Some students tested their hypothesis that the pres4 They sealed
1.86 the can
an empty can.
Passage III
ence of bubbles in cans of various liquids would affect the
5
1.75 they fou
aside.
Fifteen
minutes
later
Some students tested their
of1bubbles
cans of slipping,
various liquids would affect
roll hypothesis
time (the that
timethe
it presence
took
a can
to roll, inwithout
Figure
can
before
and
immediately
after
the roll time (the time itdown
took an
a can
to roll,
without2slipping,
down see
an incline
2 fixed points; see
incline
between
fixed points;
Figurebetween
1).
Again they set the can aside. Two h
Figure 1).
the roll time of the can before and im
ing it (Trial 5). The results are shown
Identical 1.2 L aluminum cans were used in the first two
Experiment
3
fixed
points ofincline
experiments. The angle of
inclination
the incline was
The
students
added 1 L of the f
2.3° in all three experiments.
an empty 2 L clear plastic bottle a
Table
2 t
When they rolled the bottle
down
formed. They shook the bottle, cau
Roll min
tim
and set the bottle aside. Fifteen
bles were still visible, but after 2 ho
before shaking a
can
be seen.
Trial
(sec)
angle of
inclination
4
1.86
Adapted from David Kagan, “The Shaken
5
1.75
by The American Association of Physics T
Figure 1
Experiment 1
Identical 1.2 L aluminum cans were used in the first two experiments. The angle of inclination of the
Identical
1.2 L aluminum
used inwater
the first
two
Experiment 3
The
students
added 1 Lcans
of a were
liquid—tap
containincline was 2.3° in all three
experiments.
experiments.
The angle
of inclination
was
ing
no bubbles—to
an empty
can, sealed of
thethe
can,incline
and found
The students added 1 L of the f
2.3°roll
in all
three
experiments.
its
time.
Next,
they added 1 L of the tap water to a
an empty 2 L clear plastic bottle a
Experiment 1
second empty can, sealed it, shook it, and immediately
When they rolled the bottle down t
The students added 1 Lfound
of a liquid—tap
water
containing nothese
bubbles—to
an empty
the can,
its roll time.
They
repeated
procedures
usingcan, sealed
formed.
They shook the bottle, cau
water
containing
many
bubbles,
and a empty
carbonated
and found its roll time.soapy
Next, they
added
1 L of the
tap water
to a second
can, sealed it,
shook
it, and
and set the
bottle aside. Fifteen min
beverage
that repeated
contained
noprocedures
bubbles and
that
tasted
flat,
immediately found its roll
time. They
these
using
soapy
water
containing
many
bles
were still visible, but after 2 ho
having
lost most
of its carbonation.
results
areflat,
shown
bubbles, and a carbonated
beverage
that contained
no bubbles The
and that
tasted
having lost
of its
bemost
seen.
11.
In Experiment
3, what is the mos
in are
Table
1. in Table 1.
carbonation. The results
shown
dents used the plastic bottle rat
can? Compared to an aluminum
Adapted from David Kagan, “The Shaken
A. American
rolled more
rapidly
the
by The
Association
of down
Physics
T
Table 1
B. made bubbles in the liquid e
C. contained a greater quantity
D. had thicker walls and was le
Roll time
Experiment 1
The students added 1 before
L of a liquid—tap
watershaking
containshaking after
ing
no
bubbles—to
an
empty
can,
sealed
the
can,
and
Trial
Liquid
(sec)
(sec) found
its roll time. Next, they added 1 L of the tap water to a
second
can, sealed it, shook
1 empty
tap water
1.75 it, and immediately
1.75
found
time.
They repeated
2 its roll
soapy
water
1.97these procedures
2.15 using
soapy
containing many 1.75
bubbles, and a carbonated
3 water
flat-tasting
1.96
beverage that
contained no bubbles and that tasted flat,
beverage
having lost most of its carbonation. The results are shown
in Table 1.
ACT-64E-PRACTICE
Table 1
Roll time
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
can? Compared to an aluminum
GO ON TO T
A. rolled more rapidly down the
46
B. made bubbles in the liquid e
C. contained a greater quantity
D. had thicker walls and was le
4
Experiment 2
The students added 1 L of the flat-tasting beverage to
ested their hypothesis that the presan empty can. They sealed the can, shook it, and set it
ns of various liquids would affect the
aside. Fifteen minutes later they found the roll time of the
Experiment
2
took a can to roll, without
slipping,
can
before
and immediately
it sealed
(Trialthe
4).can, shook it, and set it
students
of the
flat-tasting
beverage to anafter
emptyshaking
can. They
een 2 fixed points; seeThe
Figure
1). added 1 LAgain
they
set
the
can
aside.
Two
hours
later
they
found
aside. Fifteen minutes later they found the roll time of the can before and immediately after shaking it
the roll time of the can before and immediately after shak(Trial 4). Again they set
theit can
aside.
hours later
found
the roll
ing
(Trial
5). Two
The results
are they
shown
in Table
2. time of the can before and
immediately after shaking it (Trial 5). The results are shown in Table 2.
fixed points
incline
Table 2
Roll time
can
ngle of
nclination
Figure 1
Trial
before shaking
(sec)
after shaking
(sec)
4
5
1.86
1.75
1.96
1.93
Experiment 3
The students added 1 L of the flat-tasting beverage to an empty 2 L clear plastic bottle and sealed the bottle.
num cans were used When
in the they
firstrolled
two the bottle
Experiment
3 incline, no bubbles formed. They shook the bottle, causing bubbles to
down the
gle of inclination of the incline was
form, and set the bottle aside.
Fifteen
minutes
later,
bubbles
were still
visible, to
but after 2 hours, no
The
students
added
1 Lsome
of the
flat-tasting
beverage
ments.
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
formed.
They shook
the bottle,
bubblesAssociation
to form, of Physics Teachers.
Adapted from David Kagan,
“The Shaken-Soda
Syndrome.”
©2001causing
by The American
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
is the most likely reason the students used the plastic bottle rather than an
be seen.
aluminum can? Compared to an aluminum can, the plastic bottle:
A. rolled more rapidly down the incline.
from David
B. made bubblesAdapted
in the liquid
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.
B.
ed 1 L of a liquid—tapAnswer:
water containempty can, sealed the can, and found
hey added 1 L of theWe
tap are
water
to that
a the plastic bottle is clear. We are also told that in Experiment 3, bubbles
told
ealed it, shook it, and immediately
are visibleusing
after 15 minutes, but not after 2 hours. There is no mention of seeing any
hey repeated these procedures
ng many bubbles, and
a
carbonated
bubbles in the other experiments.
ned no bubbles and that tasted flat,
s carbonation. The results are shown
12. Based on the results11.
of Experiments
1 and
2, in is
which
of thelikely
following
trials,
In Experiment
3, what
the most
reason
the before
stu- shaking, were the
dents
used the plastic bottle rather than an aluminum
average speeds of the cans the
same?
can? Compared to an aluminum can, the plastic bottle:
F. Trials 1 and 2
A. rolled more rapidly down the incline.
G. Trials 2 and 3
Table 1
B. made bubbles in the liquid easier to see.
H. Trials 2 and 4
C. contained a greater quantity of liquid.
J. Trials 3 and 5
D. had thicker walls and was less likely to break.
Roll time
r
before shaking
(sec)
1.75
1.97
1.75
Answer:
J.
after shaking
12. Based on the results of Experiments 1 and 2, in which
of the following trials, before shaking, were the average speeds of the cans the same?
Trials
3 and 5 both have speeds of 1.75 seconds before shaking.
1.75
F. Trials 1 and 2
2.15
G. Trials 2 and 3
13. In1.96
Experiment 2, a result
shaking
the can
H.ofTrials
2 and
4 of flat-tasting beverage was that the:
A. number of bubblesJ.in the
beverage
decreased.
Trials
3 andimmediately
5
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
(sec)
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
4
4
that bubbles were still visible 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
Passage IV
Figure 2 shows the average rat
speed of 1.75 seconds.
rate ofrate
photosynthesis
(as % of
at rate
670 at
nm)
of photosynthesis
(asrate
% of
670 nm)
various wavelengths as a percent o
The chemical reactions associated with photosynthesis
photosynthesis at 670 nm.
Passage IV
Passage
IV
Figure 2 shows the average rat
can be summarized
with the following chemical equation:
various
The chemical reactions associated with photosynthesis can be summarized with the following
chemical
110 wavelengths as a percent o
The chemical
reactions associated with photosynthesis
photosynthesis at 670 nm.
equation:
6,CO
2 + 12,H 2O + energy → C 6H 12O 6 + 6,O 2 + 6,H 2O
can be summarized with the following chemical equation:
100
110
90
6,CO2 + 12,H2O + energy → C6H12O6 + 6,O2 + 6,H2O 100
80
Table
lists wavelength
for visible
light and
90
Table 1 lists wavelength ranges
for1visible
light and the ranges
color frequently
associated
with each range.
70
the color frequently associated with each range.
80
60
Table 1 lists wavelength ranges for visible light and
70
the color frequently associated
with
each
range.
Table 1
50
60
40
Wavelength
Table
1
50
Color
(nm)
30
40
Wavelength
Violet
380−430
20
Color
(nm)
Blue
430−500
30
10
Green
500−565
Violet
380−430
20
Yellow
565−585
0
Blue
430−500
Orange
585−630
10400 440 480 520 560
Green
500−565
Red
630−750
Yellow
565−585
wavelength (
0
Orange
585−630
400 440 480 520 560
RedA. Campbell,
630−750
Table 1 adapted from Neil
Jane B. Reece, and Lawrence
Figure 2 (
wavelength
G. Mitchell, Biology, 5th ed. ©1999 by Benjamin/Cummings.
Table 1 adapted from Neil A. Campbell, Jane B. Reece, and Lawrence
G. Mitchell,
Biology
ed.by
©1999
by Benjamin/Cummings.
relative
absorption
of, 5th
light
chlorophyll
a and chlorophyll b versus
relative
absorption
relative
absorption
Figure 1 shows the
light from 400 nm to 750 nm.
Figure 1 shows the relative absorption of light by
chlorophyll a and chlorophyll b versus the wavelength of
light Figure
from 400
nm to 750
1 shows
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
chlorophyll a
100
chlorophyll b
90
100
80
90
70
80
60
70
50
60
40
50
30
40
20
30
10
20
0
10400
450 500 550 600 650 700 750
wavelength (nm)
0
400
450
500
550
600
650
700
750
wavelength
(nm)
Figure 1
Figure 1
ACT-64E-PRACTICE
Figure 2
1 andof2 adapted from Peter H.
theFigures
wavelength
Susan E. Eichhorn, Biology of Plants, 4th
lishers, Inc.
Figures 1 and 2 adapted from Peter H. R
Susan E. Eichhorn, Biology of Plants, 4th
lishers, Inc.
17. Based on Table 1 and Figure 1,
associated with the wavelength
the greatest
absorption
by chloro
17. Based
on Table
1 and Figure
1,
associated
A. Blue with the wavelength
the
B. greatest
Green absorption by chloro
C.
Yellow
A. Blue
D. Green
Red
B.
C. Yellow
D. Red
GO ON TO T
Figures 1 and 2 adapted from Peter H.
Susan E. Eichhorn, Biology of Plants, 4th
lishers, Inc.
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.
Key
chlorophyll a
chlorophyll b
100
90
relative absorption
80
70
60
50
40
30
20
10
0
400
450
500
550 600 650
wavelength (nm)
700
Figure 1
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
velength ranges for visible light and
ssociated with each range.
Table 1
olor
Wavelength
(nm)
olet
ue
een
low
ange
d
380−430
430−500
500−565
565−585
585−630
630−750
A. Campbell, Jane B. Reece, and Lawrence
d. ©1999 by Benjamin/Cummings.
the relative absorption of light by
orophyll b versus the wavelength of
750 nm.
Key
chlorophyll a
chlorophyll b
rate of photosynthesis (as % of rate at 670 nm)
Figure
2 shows the average
rate
of photosynthesis
at of the average rate of
Figure 2 shows the average
rate of photosynthesis
at various
wavelengths
as a percent
ACT-64E-PRACTICE
various wavelengths as a percent of the average rate of
48
ctions associated withphotosynthesis
photosynthesisat 670 nm.
photosynthesis at 670 nm.
th the following chemical
equation:
110
energy → C6H12O6 + 6,O2 + 6,H2O
100
90
80
70
60
50
40
30
20
10
0
400
440
480
520 560 600 640
wavelength (nm)
680
720
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.
GO ON TO T
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 CO2 + 12 H2O + energy → C6H12O6 + 6 O2 + 6 H2O
We can see that the carbon from CO2 goes into C6H12O6. Since O2 is oxygen and H2O is
water, C6H12O6 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 absorption by chlorophyll a.
B. lowest relative absorption by chlorophyll b.
C. highest relative absorption by chlorophyll a.
D. highest relative absorption by chlorophyll b.
Answer: C.
4
Passage V
Experiment 3
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.
the bead stayed at the bottom,
From Figure 2, we see that the wavelengths that result in the highest rates of If
Table 3. If the bead rose, “R” was re
photosynthesis are around 440 nm and 680 nm. These values correspond with
the highest
procedure
was repeated for various p
relative absorption for chlorophyll a in Figure 1.
Experiment 1
Passage V
A dryexperiments
100 mL graduated
cylinder
was placed
on an
Students performed the following
to determine
the density
of common
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 graduPlastic
A dry 100 mL graduated
cylinder
wasuntil
placed
an electronic
balance
ated
cylinder
theonvolume
of liquid
was and
50.0tared
mL.(the
Thebalance was reset to
density
the liquid
was until
then acalculated.
0.000 g). H2O was added
to the of
graduated
cylinder
certain massThe
wasprocedure
obtained. EthanolPolybutylene
was added to
VLDPE
with
ethanol
H 2liquid
O
the gradu- ated cylinderwas
untilrepeated
the volume
of different
liquid wasamounts
50.0 mL.of
The
densityand
of the
wasLDPE
then
(seewas
Table
1). with different amounts of ethanol and H O (see Table 1).
calculated. The procedure
repeated
2
HDPE
PA-11
PA-6
Polycarbonate
Table 1
PVC
Mass of
Mass of
Total
H 2O
ethanol
mass
Density
Liquid
(g)
(g)
(g)
(g/mL)
1
2
3
4
5
0
10.24
19.79
35.42
49.96
39.67
32.43
25.23
12.47
0
39.67
42.67
45.02
47.89
49.96
0.793
0.853
0.900
0.958
0.999
Table 3
Li
1
2
3
4
5
R
S
S
S
S
S
S
S
R
R
S
S
S
S
S
S
R
R
S
S
S
S
S
S
R
R
R
S
S
S
S
S
R
R
R
R
S
S
S
S
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
density of2 the liquid was then calculated. The procedure was repeated with
Experiment
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
Liquid
Mass of
H2O in
solution
(g)
Mass of
KI in
solution
(g)
Mass of
solution in
graduated
cylinder
(g)
Density
(g/mL)
6
7
97.66
95.41
7.36
15.52
52.51
55.70
1.05
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.
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).
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.
Table 2
Liquid
Mass of
H2O in
solution
(g)
Mass of
solution in
graduated
cylinder
(g)
Mass of
KI in
solution
(g)
Density
(g/mL)
4
23. Based on the results of Experime
6
97.66
7.36
52.51
1.05
PA-11 is most likely:
7
95.41
15.52
55.70
1.11
A. less than 0.793 g/mL.
8
94.38
20.68
57.53
1.15
B. between 0.853 g/mL and 0.9
9
92.18
29.08
60.63
1.21
C. between 0.999 g/mL and 1.0
10
87.77
41.31
64.64
1.29
D. greater than 1.11 g/mL.
Experiment 3
A solid plastic bead was placed at the bottom of a
med the following experiments
Experiment 3 to
sample of each of Liquids 1−10 from Experiments 1 and 2.
of common plastics. A solid plastic bead was
ACT-64E-PRACTICE
If
the bead
bottom,
“S”ofwas
recorded
in Experiments 1 and
GO ON TO T
placed
at thestayed
bottomatof the
a sample
of each
Liquids
1−10 from
3. If the
rose, “R”inwas
recorded
Table
3. “R”
The was
2. If the bead stayed at Table
the bottom,
“S”bead
was recorded
Table
3. If theinbead
rose,
50 recorded in
wasfor
repeated
various plastics.
Table 3. The procedureprocedure
was repeated
variousfor
plastics.
raduated cylinder was placed on an
nd tared (the balance was reset to
ded to the graduated cylinder until a
ined. Ethanol was added to the gradue volume of liquid was 50.0 mL. The
was then calculated. The procedure
fferent amounts of ethanol and H 2O
Table 1
Mass of
ethanol
(g)
39.67
32.43
25.23
12.47
0
Table 3
Liquid
Plastic
1
2
3
4
5
6
7
8
9
10
Polybutylene
VLDPE
LDPE
HDPE
PA-11
PA-6
Polycarbonate
PVC
R
S
S
S
S
S
S
S
R
R
S
S
S
S
S
S
R
R
S
S
S
S
S
S
R
R
R
S
S
S
S
S
R
R
R
R
S
S
S
S
R
R
R
R
R
S
S
S
R
R
R
R
R
S
S
S
R
R
R
R
R
R
S
S
R
R
R
R
R
R
R
S
R
R
R
R
R
R
R
S
Total
mass Density
Experiment 1, the density of ethanol was found to be:
(g) 22. In (g/mL)
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
greater than 0.999 g/mL.
47.89 J. 0.958
49.96
0.999
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:
f potassium iodide (KI) was dissolved
A. less than 0.793 g/mL.
2O. A dry 100 mL graduated cylinder
B. between
alance and tared. The solution
was 0.853 g/mL and 0.958 g/mL.
C. between
ted cylinder until the volume
was 0.999 g/mL and 1.05 g/mL.
of the liquid was then calculated.
Thethan 1.11 g/mL.
D. greater
ed with different amounts
of KI and
Answer: C.
Table 2
Mass of
KI in
solution
(g)
Mass of
solution in
graduated
cylinder
(g)
Density
(g/mL)
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
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/H2O 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
Answer: H.
4
Looking
at Table
we see
the mass of solution for liquid
is 64.64 g,1soand
a mass
24. Suppose
that a2,sixth
KI/Hthat
26. In10Experiments
2, theof
students tared the grad
2 O solution had been mea67.54sured
would
be
an
increase
of
3.10
g.
Between
liquids
9
and
10,
density
goes
up
about
in Experiment 2 and the mass of the solution in
ated cylinder in each trial so they could more easi
graduated
cylinder
was 67.54
g. The in
density
of this cylinder determine:
0.02 the
g/mL
per gram
of mass
of solution
graduated
since mass goes from 60.63
solution would most likely have been closest to which
F.
mass The
of the substances added to the graduate
g to 64.64
g
(about
4
g)
when
density
goes
from
1.21
g/mL
to
1.29theg/mL.
of the following?
cylinder.
relationship
mass of solution in cylinder and density G.
looks
F. 1.25 between
g/mL
thelinear,
densityso
of we
the can
graduated cylinder.
1.30
g/mLhold as we go from 64.64 g to 67.54 g. So a rise
H. when
thegtotal
volume of the added substances w
guessG.that
it will
of 3.10
in mass
H. cause
1.35 g/mL
equal to 50.0 mL.
should
a rise of about 0.06 g/mL in density, and 1.29+0.06=1.35.
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?
Liquid
A.
B.
C.
D.
1
2
3
4
R
R
S
S
R
R
S
S
R
S
R
S
R
S
R
S
Answer: B.
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.
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 H2O.
Answer: F.
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.
Answer: B.
4
Passage
VI
Experiment
From Table 3, we see
that polycarbonate
and PA-6 are both plastics used to make
beads2
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
together
liquid, then polycarbonate
is more
dense that
PA-6. This
is exactly
what happens
forto ferment a sugar by using
2 fermentation
pathways,
researchers
performed
2 experispecies can use alone. To investiga
ments using broth that contained either the sugar sucrose or
liquid 8.
ment 1 was repeated, except that diff
the sugar lactose. One of the fermentation pathways prospecies 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
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
Species
Sucrose broth was added to 5 large test tubes. Next,
Experiment 1
added
phenol
red (a
indicator
that is red
yellow
pH < 7, that
red isif yellow if pH < 7, red
Sucrose broth was added
to 5 large
testpH
tubes.
Next, phenol
(a pHif indicator
pH ≥ 7) was added to each large test tube. A Durham tube
A and B
if pH ≥ 7) was added to(aeach
large
testtube)
tube.was
A Durham
(a small
tube)
wastest
placed, inverted, in
small
test
placed,tube
inverted,
intest
each
large
A and C
each large test tube to collect
Figure
1).
tube toCO
collect
(see
Figure
1).
B and D
2 (see CO
2
C and D
Table 2
Sucrose broth
acid
CO2
−
+
+
+
−
+
+
+
Durham
tube
broth
(red)
Figure 1
The large
were capped,
heated
theOne of 4 bacterial
The large test tubes were capped,
heatedtest
untiltubes
the solutions
were sterile,
thenuntil
cooled.
cooled.
One The
of 4procedure
bacterial
species (Species A−D)solutions
was added were
to eachsterile,
of 4 of then
the large
test tubes.
was repeated using
A−D)
wastest
added
each
of 4 of the
large at a pH of 7) were
lactose broth instead ofspecies
sucrose(Species
broth. The
10 large
tubesto(all
containing
solutions
tubes.
then incubated at 37°Ctest
for 48
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
mented lactose?
The large test tubes and Durham tubes were examined. 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
mented lactose?
The large test tubes and Durham tubes were examined. 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
The large test tubes andproduced,
Durham tubes
were
examined.
acid was at
produced,
was yellow.
If no C only
a gas
bubble
was If
observed
the topthe
ofsolution
the
H. Species B and Species D onl
acid was produced, theDurham
solution tube
remained
If 1).
CO2 was produced, a gas bubble was observed at
top
(see red.
Table
J. theSpecies
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
ing sucrose broth and to a larg
Table 1
lactose broth. Which of the f
likely depict the results?
Sucrose broth Lactose broth
Species
Sucrose broth Lactose b
added
acid
CO2
acid
CO2
4
acid
A
−
−
−
−
B
−
−
+
+
A.
–
C
+
+
−
−
B.
+
D
+
−
+
−
C.
+
−
−
−
−
ExperimentNone
2
D.
–
Synergism
occurs
when
2
bacterial
species
act
own sugars by fermentation. To study
together to ferment a sugar by using a pathway that neither
Experiment
2
ays, researchers performed
2 experispecies
can species
use alone.
To investigate
ExperiACT-64E-PRACTICE
Synergism
occurs
2 bacterial
act together
to fermentsynergism,
a sugar by using
a pathway that neither
contained either the sugar
sucrose
or whenment
1 was repeated, except that different pairs of bacterial
can use
investigate
synergism,
Experiment
was repeated,
different pairs of
e of the fermentationspecies
pathways
pro-alone. To
species
were added
to each
large test1 tube
(see Tableexcept
2). that52
creases the acidity (lowers
the species
pH) of were added to each large test tube (see Table 2).
bacterial
r pathway produces acid but not CO2.
Table 2
as added to 5 large test tubes. Next,
cator that is yellow if pH < 7, red if
each large test tube. A Durham tube
s placed, inverted, in each large test
ee Figure 1).
Durham
tube
broth
(red)
Figure 1
Species
added
A and B
A and C
B and D
C and D
Sucrose broth
Lactose broth
acid
CO2
acid
CO2
−
+
+
+
−
+
+
+
+
−
+
+
+
−
+
+
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 CO2. 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.
ubes were capped, heated
until the 1, which of the bacterial species fermented lactose?
28. In Experiment
e, then cooled. One of F.
4 bacterial
Species B only
) was added to each of 4 of the large
G.
Species
ure was repeated using lactose
broth C only
H.
Species
oth. The 10 large test tubes (all con- B and Species D only
J. at
Species
pH of 7) were then incubated
37°C C and Species D only
Answer: H.
ubes and Durham tubes were examduced, the solution was yellow. If no
e solution remained red. If CO2 was
ble was observed at the top of the
le 1).
Table 1
ucrose broth
Lactose broth
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?
CO2
acid
–
+
+
–
+
–
+
–
C
GO ON TO T
of 4 of the large
ing lactose broth
st tubes (all conncubated at 37°C
28. In Experiment 1, which of the bacterial species fermented lactose?
Species
only
For aF.sugar
to beB fermented,
it must produce acid alone, or acid and CO2. Under the
G. Species C only
heading
“Lactose
broth”
in Table
H. Species
B and
Species
D only1, we see that acid and CO2 were produced by species
J. acid
Species
C and
D only
B, and
alone
wasSpecies
produced
by species D.
bes were examwas yellow. If no
red. If CO2 was
t the top of the
29.Suppose
Suppose
in Experiment
both B
Species
B and
29.
thatthat
in Experiment
2 both 2Species
and Species
C had been added to a large test tube
Species C had been added to a large test tube containcontaining
sucrose
broth
and
to
a
large
test
tube
containing
lactose
ing sucrose broth and to a large test tube containing broth. Which of the following would
most likely
depict
the results?
lactose
broth.
Which of the following would most
likely depict the results?
se broth
CO2
−
+
−
−
−
Sucrose broth
Lactose broth
acid
CO2
acid
CO2
–
+
+
–
–
+
+
–
+
–
+
–
+
–
+
–
A.
B.
C.
D.
Answer: C.
52
GO ON TO THE NEXT PAGE.
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.
From Table 1, we see that there are three species that produce neither acid nor CO2 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 CO2 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 CO2) 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 CO2 in the lactose broth.
4
terial species that is 32. Which
32. Which
of the following
best
the
of the following
figures bestfigures
illustrates
theillustrates
results of Experiment
1 for Species D in the sucrose
ment 1. She adds the broth? results of Experiment 1 for Species D in the sucrose
rves that neither acid
broth?
ds the species to lac- acid and CO2 are proF.
H.
Experiment 1, the
Durham
tube
Durham
tube
broth
(red)
broth
(yellow)
gas
bubble
J.
G.
riments 1 and 2 that
d synergistically in
Durham
tube
Durham
tube
n each species was
ut acid was produced
ether in the sucrose
broth
(yellow)
broth
(red)
gas
bubble
n each species was
33. IsG.the hypothesis that Species A and Species C acted
t acid was produced Answer:
synergistically supported by the results of Experiether in the sucrose
ment 2 ?
broth,
species
produces
but no CO2.
n each species was In the sucrose
A. Yes,
because
bothDacid
and COacid
2 were produced
ut CO2 was produced no CO is produced,
from sucrose.
then
no
gas
bubble
is
formed.
2
ether in the sucrose
B. Yes, because both acid and CO 2 were produced
from lactose.
n each species was
C. No, because only acid, not CO 2 , was produced
t CO2 was produced
from both sucrose and lactose.
gether in the lactose
D. No, because neither acid nor CO 2 was produced
from lactose.
Acid turns the broth yellow. If
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.
4
Species C produces both 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, scientistswas
thought
all genetic
material was
contained
in structures
called
and of:
contained
in structures
called
chromosomes
and
that chromosomes
number
had
been
foundofonly
in (not
the in
nucleus
of a cell
that chromosomes had chromosomes
been found only
in the
nucleus
a cell
the cytoplasm):
F. amino acids in the nucleus in
(not in the cytoplasm):
to cell type.
G. amino acids in the cytoplas
type to cell type.
H. chromosomes in the nucleu
cytoplasm
type to cell type.
J. chromosomes in the cytopla
type to cell type.
nucleus
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
composed proteins
of 2 types
molecules, pro-acid (DNA).
B.Proteins
Proteins
Chromosomes are composed
of 2 typesare
of molecules,
andofdeoxyribonucleic
teins and deoxyribonucleic acid (DNA). Proteins are comC. Gametes
are com- posed of subunits
called
amino
acids.
DNA
consists
of
chains
of
subunits
called
nucleotides.
The
posed of subunits called amino acids. DNA consists of
D. Chromosomes
parts of chromosomes chains
that areof
responsible
for the nucleotides.
transmission The
of genetic
subunits called
parts information
of chromo- are called genes.
Two scientists in the 1940s
debate
genes arefor
made
proteins or DNA.
somes
that whether
are responsible
the oftransmission
of genetic
information are called genes.
36. According to the passage, a sim
and proteins is that both types of
Protein Hypothesis
Two scientists
in the
whether
are Cells contain
Genes are made only of proteins.
Proteins make
up 1940s
50% ordebate
more of
a cell’s genes
dry weight.
20 found only in gametes.
F. are
made of proteins or DNA.
G.proteins.
are abundant in the cytoplasm
different amino acids that can be arranged in a virtually infinite number of ways to make different
H. contain 20 different amino a
The number and arrangement of different amino acids within a protein form the codes that contain
J. are composed of smaller sub
hereditary information.Protein Hypothesis
chromosomes
are made
only
proteins.
up believed
37. to
According
to the Protein Hypoth
In contrast, only 4 different Genes
nucleotides
make up
the of
DNA
found inProteins
cells, andmake
they are
form
50% or more of a cell’s dry weight. Cells contain 20 differlowing
observations
provides t
chains only in certain ratios.
As a result,
the number
differentincombinations
that DNA can carry is much
ent amino
acids that
can be of
arranged
a virtually infinite
that
genes
are
NOT
composed
of
smaller than the number
that proteins
can to
carry.
number
of ways
make different proteins. The number
A. DNA is composed of only 4
and arrangement of different amino acids within a protein
B. DNA is composed of smal
form the codes that contain hereditary information.
DNA Hypothesis
proteins.
Genes are made only of DNA. DNA is found exclusively in the cell’s nucleus, whereas proteins C.
are found
DNA is abundant in both
In
contrast,
only
4
different
nucleotides
make
up
the
throughout the nucleus and cytoplasm. Additionally, the amount of protein in a cell varies from cell type
to
cytoplasm.
DNA found in cells, and they are believed to form chains
cell type, even within the
same
animal.
D.
The
concentration of DNA
only in certain ratios. As a result, the number of different
from
cell to cell.
combinations that DNA can carry is much smaller than the
Though DNA is less abundant
number than
that proteins,
proteins the
canamount
carry. is consistent from cell type to cell type within the
same animal, except for the gametes (the reproductive cells). Gametes have half the amount38.
of DNA
as
Mitochondria
are organelles loc
other cells in the body. Gametes also have half the typical number of chromosomes. Thus, the amount
ofresponsible for energy tr
that are
DNA Hypothesis
After the 1940s, it was observ
contain their own genes. This o
Genes are made only of DNA. DNA is found excluevidence stated in which hypothe
sively in the cell’s nucleus, whereas proteins are found
throughout the nucleus and cytoplasm. Additionally, the
F. The DNA Hypothesis, beca
amount of protein in a cell varies from cell type to cell
of DNA, the observation wo
type, even within the same animal.
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.
4
For answer D, both scientists accepted the observation that proteins are not found only in
the nucleus.
40. Which of the following illustrations of a portion of a DNA molecule is consistent with the description
o describes the DNA Hypothesis
40. Which of the following illustrations of a portion of a
the passage?
rotein Hypothesis isinweakened
by
DNA molecule is consistent with the description in the
wing observations? passage?
ganism, the amount of protein in the
Key
f that found in other types of cells.
AA - amino acid
ganism, the amount of protein in difN - nucleotide
cells is not the same.
ules are composed of many subunits.
und only in the nucleus.
F.
AA
AA
N
N
G.
H.
J.
Answer: J.
N
AA
AA
N
AA
AA
AA
AA
N
N
N
N
END OF TEST 4
We are told in STOP!
the passage
DNA
is composed
of nucleotides.
Amino acids compose
DO that
NOT
RETURN
TO ANY
OTHER TEST.
proteins.
[See Note on page 56.]
END OF TEST 4
STOP! DO NOT RETURN TO ANY OTHER TEST.