ACT Past Papers with Answers

Paper 13

ACT April 2022

Form Z08 · Science Test · 40 questions

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Paper 13 · ACT April 2022 · Form Z08 · Science Test

Table 1 shows the results of how many total tests?

A04B08C11D16

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Paper 13 · ACT April 2022 · Form Z08 · Science Test

When each of the 4 halide solutions is mixed with the Ca2+ test solution, only 1 precipitate forms. This pre- cipitate most likely has what chemical formula? 39 GO ON TO THE NEXT PAGE. ACT-Z08 4 4 Passage II A study performed at a site off the Mediterranean coast examined how an ocean storm during the winter affected wave height, current speed, and the rate of sedimentation. Study Before the winter storm season began, 3 types of instruments were anchored to the ocean floor in water 26 m deep (see Figure 1). A wave height buoy, which was attached to its anchor with an elastic cord, contained a device that measured its vertical movements in the waves. An acoustic profiler, fixed in position 2.5 m above the ocean floor, used sound waves to measure the horizontal speed of the ocean current at that depth. A pair of sediment traps, fixed at the same distance above the ocean floor as the acoustic profiler, collected sediment that settled out of the water. Figure 1 From November 4 to November 24, the average wave height, in meters (m), and the average current speed, in centimeters per second (cm/sec), were determined daily (see Figures 2 and 3, respectively). The mass of sediment, in grams (g), deposited per square meter (m2 ) of ocean floor per day (the sedimentation rate) was also determined daily over those 21 days (see Figure 4). A strong storm affected the site beginning the morning of November 12. Figure 2 Figure 3 sediment traps floats acoustic profiler elastic cord ocean floor water surface wave height buoy Note: Drawing is not to scale. anchors average wave height (m) 0 N o v . 4 N o v . 8 N o v . 1 2 N o v . 1 6 N o v . 2 0 N o v . 2 4 storm period date average current speed (cm/sec) 0 N o v . 4 N o v . 8 N o v . 1 2 N o v . 1 6 N o v . 2 0 N o v . 2 4 storm period date 40 GO ON TO THE NEXT PAGE. ACT-Z08 4 4 Figure 4 Figures adapted from Antoine Grémare et al., “The Effects of a Strong Winter Storm on Physical and Biological Variables at a Shelf Site in the Mediterranean.” ©2003 by Elsevier SAS.

ACaFBCa2FCCaF2DCaF3 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05

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Paper 13 · ACT April 2022 · Form Z08 · Science Test

The acoustic profiler was fixed in position at what depth ? sedimentation rate (g of sediment/m 2 /day) 400 350 300 250 200 150 100 50 0 N o v . 4 N o v . 8 N o v . 1 2 N o v . 1 6 N o v . 2 0 N o v . 2 4 storm period date 41 GO ON TO THE NEXT PAGE. ACT-Z08 4 4 Passage III Scientists studied how perceived predation (hearing the calls and sounds of predators), in the absence of direct predation, affects the production of offspring by female song sparrows (Melospiza melodia). M. melodia sparrows produce 2 clutches (groups) of eggs in a breeding season. Study At the beginning of a breeding season, 24 M. melodia nests without eggs were located in low-growing shrubs. Direct predation was eliminated by surrounding each shrub with an electric fence to protect the nest from ground predators and by covering each shrub with a net to protect the nest from aerial predators. The nests were evenly divided into 2 groups: a preda- tor treatment (PT) group and a nonpredator treatment (NPT) group. A speaker that broadcast the calls and sounds of predators of M. melodia was set up near each nest in the PT group. A speaker that broadcast the calls and sounds of nonpredators of M. melodia was set up near each nest in the NPT group. Throughout the breeding season, each speaker broad- cast the appropriate calls and sounds every few minutes on a 4-day-on, 4-day-off cycle. (The parents and offspring associated with each nest did not hear sounds emitted by speakers other than their own speaker.) The number of eggs and the number of hatchlings produced in each nest were recorded. The number of hatchlings that successfully left each nest (such hatchlings are called fledglings) was also recorded. The figure shows, for each group, the average number of offspring (as eggs, as hatchlings, and as fledg- lings) per nest in each clutch. Figure adapted from Liana Y. Zanette et al., “Perceived Predation Risk Reduces the Number of Offspring Songbirds Produce per Year.” ©2011 by American Association for the Advancement of Science. Key NPT group PT group first clutch second clutch July1 June1 M ay1 April1 average number of offspring per nest eggs eggs fledglings hatchlings hatchlings fledglings 42 GO ON TO THE NEXT PAGE. ACT-Z08 4 4

A02.5 mB10.5 mC23.5 mD25.5 m 1.4 1.2 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0

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Paper 13 · ACT April 2022 · Form Z08 · Science Test

The results of the study provide evidence of which of the following situations involving predators and prey? bodily contact contact bodily contact contact 43 GO ON TO THE NEXT PAGE. ACT-Z08 4 4 Passage IV As a banana ripens, the external peel changes color and the chemical composition of the internal pulp changes. Three experiments were performed to study the ripening of bananas. Experiment 1 On Day 0, 100 bananas were harvested from the same banana plant. All the banana peels were completely green, so the average peel color “completely green” was assigned to the group. Next, the bananas were stored under ripening conditions (16°C and 85% relative humidity) over the next 35 days. Every 7 days during that time period, the bananas were inspected and a new average peel color was assigned. The relationship between average peel color and time is shown in Table 1. Experiment 2 On each day that an average peel color was assigned, the following procedures were also performed: First, 3 bananas were randomly selected from the group and peeled. Then, the pulps of the 3 bananas were thoroughly mixed together, and a 100 g sample of the pulp mixture was collected. Next, an 80% solution of ethanol in H2O (by volume) was mixed with the sample to extract the starch and the sugars from the sample. Then, the mass, in g, of the starch in the sample and the mass, in g, of the sugars in the sample were determined (see Figure 1). Figure 1 Experiment 3 On each day that the mass of the sugars in a sample was determined, the sugars were further analyzed, and the mass, in g, of each of the 3 sugars—fructose, glucose, and sucrose—in the sample was determined (see Figure 2). Figure 2 Table and figures adapted from R. C. Adão and M. B. A. Glória, “Bioactive Amines and Carbohydrate Changes During Ripening of ‘Prata’ Banana (Musa acuminate x M. balbisiana).” ©2005 by Elsevier Ltd. Table 1 Day Average peel color 00 completely green 07 green with yellow streaks 14 yellow with green tips 21 completely yellow 28 yellow with brown spots 35 brown Day mass in sample (g) 16 14 12 10 8 6 4 2 0 0 7 14 21 28 35 Key starch sugars Day mass in sample (g) 7 6 5 4 3 2 1 0 0 7 14 21 28 35 Key fructose glucose sucrose 44 GO ON TO THE NEXT PAGE. ACT-Z08 4 4

APredators being affected by prey through non-BPredators being affected by prey through bodilyCPrey being affected by predators through non-DPrey being affected by predators through bodily

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Paper 13 · ACT April 2022 · Form Z08 · Science Test

The variation in the mass of starch and in the mass of sugars in the banana pulps over the 35 days, as shown in Figure 1, depended on the process of osmosis. Which of the following statements about the transfer of water or soluble substances in the bananas is consistent with this information? mass of starch in the pulps decreased and the mass of sugars in the pulps increased. mass of starch in the pulps increased and the mass of sugars in the pulps decreased. to pulp, the mass of starch in the pulps decreased and the mass of sugars in the pulps increased. to pulp, the mass of starch in the pulps increased and the mass of sugars in the pulps decreased. 80 mL ethanol ____________ _ 20 mL H2O 20 mL H2O ____________ _ 80 mL ethanol 20 mL H2O ________________________ 80 mL ethanol + 20 mL H2O 80 mL ethanol ________________________ 80 mL ethanol + 20 mL H2O 45 GO ON TO THE NEXT PAGE. ACT-Z08 4 4 Passage V Spheres W, X, Y, and Z, initially at rest at the starting line on an inclined plane, will be released one at a time and allowed to roll, without sliding, all the way to the finish line at the bottom of the incline. The masses and radii of Spheres W and X are compared in the following table, as are the masses and radii of Spheres Y and Z. A teacher asks 2 students to predict whether Sphere W or Sphere X, if either, will reach the finish line in less time, and whether Sphere Y or Sphere Z, if either, will reach the finish line in less time. Student 1 Sphere W will reach the finish line in less time than Sphere X. Sphere W weighs more than Sphere X, so the force of gravity exerted on Sphere W is greater than that exerted on Sphere X. Because force = mass × acceleration, Sphere W will have the greater acceleration down the incline and, therefore, the greater average speed. Sphere Y will reach the finish line in less time than Sphere Z. For 2 spheres having the same mass, as sphere radius increases, sphere density decreases, so the energy required for a sphere to complete a rotation decreases. The mass of Sphere Y is spread over a greater volume than is the mass of Sphere Z, so Sphere Y is less dense than Sphere Z and will have the greater acceleration and the greater average speed. Student 2 Spheres W and X will reach the finish line in the same amount of time. The acceleration of a sphere due to gravity is independent of the sphere’s mass, so the 2 spheres will have the same acceleration and, therefore, the same aver- age speed. Sphere Z will reach the finish line in less time than Sphere Y. For spheres having the same mass, as sphere radius increases, the energy required for a sphere to com- plete a rotation increases. This is because as sphere radius increases, the distance traveled by the sphere to complete a rotation increases. Therefore, because Sphere Y has the greater radius, it will have the lesser acceleration and the lesser average speed.

AAs water was transferred from peel to pulp, theBAs water was transferred from peel to pulp, theCAs soluble substances were transferred from peelDAs soluble substances were transferred from peel

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Paper 13 · ACT April 2022 · Form Z08 · Science Test

Suppose the spheres were tested on the Moon instead of on Earth. Based on Student 2’s prediction, which sphere, Sphere W or Sphere X, if either, would reach the finish line in less time? Sphere W would be greater than the mass of Sphere X. would be less on the Moon than on Earth. would be the same on the Moon as it is on Earth. sphere due to the Moon’s gravity would be inde- pendent of the sphere’s mass. 47 GO ON TO THE NEXT PAGE. ACT-Z08 4 4 Passage VI Microtubules (MTs), which are structural components of cells, continuously elongate and shorten. An MT in catastrophe stops elongating and starts to shorten. An MT in rescue stops shortening and starts to elongate. When a cell enters mitosis from interphase, these transitions con- tribute to the rearrangement of the MT structure to form the mitotic spindle. Table 1 shows the values of 3 parameters that were set in 6 computer-simulation models (CSMs A−F) of MT pop- ulations. It also shows the known values of the parameters for interphase cells and mitotic cells. Figure 1 compares, for 3 MT characteristics, the aver- age values predicted by the CSMs with the known average values for the 2 cell types. Figure 1 Table and figure adapted from N. R. Gliksman et al., “How the Transition Frequencies of Microtubule Dynamic Instability (Nucle- ation, Catastrophe, and Rescue) Regulate Microtubule Dynamics in Interphase and Mitosis: Analysis Using a Monte Carlo Computer Simulation.” ©1993 by The American Society for Cell Biology. Table 1 Cell type or Model Number of MTs Catastrophe frequency* (number/sec) Rescue frequency† (number/sec) Interphase 0,500 0.019 0.044 Mitotic 2,000 unknown unknown CSM A 0,500 0.019 0.044 CSM B 2,000 0.019 0.044 CSM C 0,500 0.056 0.044 CSM D 2,000 0.056 0.044 CSM E 0,500 0.019 0.000 CSM F 2,000 0.019 0.000 *number of MTs in catastrophe per second †number of MTs in rescue per second *μm = 10−6 m 100 80 60 40 20 0 average MT length (μm*) interphasecells 20 15 10 5 0 average MT elongation velocity (μm*/min)

ASphere W, because on the Moon, the mass ofBSphere X, because the weight of each sphereCNeither sphere, because the mass of each sphereDNeither sphere, because the acceleration of each

Form Z08 · Science Test · 40 questions

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