ACT Past Papers with Answers

Paper 12

ACT April 2022

Form E25 · Science Test · 40 questions

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Paper 12 · ACT April 2022 · Form E25 · Science Test

Which of the polyols listed in Table 2 is least soluble in water at 40°C ?

AXylitolBMaltitolCErythritolDIsomalt

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Paper 12 · ACT April 2022 · Form E25 · Science Test

Consider the blood samples having an average total cholesterol of 420 mg/dL, as shown in Figure 2. Based on Figure 1, were those blood samples collected when the 13-lined ground squirrels were hibernating? were in an active bout during hibernation. were in a period of torpor during hibernation. were in an active bout, which does not occur during hibernation. were in a period of torpor, which does not occur during hibernation. 41 GO ON TO THE NEXT PAGE. ACT-E25 4 4 Passage III A study examined the production and consumption of methane (CH4) by soil bacteria in 4 areas of a particular forest: an area of healthy black spruce trees, an area of burned black spruce trees, an area of healthy jack pine trees, and an area of burned jack pine trees. Study On May 15, 4 identical soil gas collection cylinders were placed in the forest, one in each area. Each cylinder had an open end and a closed end; its open end was pushed into the soil, and then its aboveground surface was covered with shiny foil (see Figure 1). That same day, a gas sample was taken from each cylinder by inserting the needle of a 50 cm3 syringe into the closed end (through an airtight seal). The CH4 content of each gas sample was determined. Likewise, on the 15th day of June, July, August, and Sep- tember, a gas sample was taken from each cylinder and its CH4 content was determined. Figure 1 The difference in the CH4 contents of consecutive samples (for example, the samples taken on May 15 and June 15) from a cylinder indicated whether there had been a net production of CH4 (more CH4 had been produced than consumed) or a net consumption of CH4 (more CH4 had been consumed than produced) in the area over the month between the 2 sampling dates. From these differences, the average rate of CH4 production or consumption was calcu- lated, in milligrams per square meter per day (mg/m2 /day), for each area over each month (see Figure 2). Figure 2 Figure 2 adapted from Sami Ullah et al., “Greenhouse Gas Fluxes from Boreal Forest Soils during the Snow-Free Period in Quebec, Canada.” ©2009 by NRC Research Press.

AYes; the samples were collected while the squirrelsBYes; the samples were collected while the squirrelsCNo; the samples were collected while the squirrelsDNo; the samples were collected while the squirrels

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Paper 12 · ACT April 2022 · Form E25 · Science Test

The dielectric constant (ε) of a substance is a measure of the polarity of the substance; the greater the value of ε, the greater the polarity. If the ε of acetone is 21 and the ε of H2O is 80, was the solvent used in Experi- ment 1 less polar or more polar than the solvent used in Experiment 2 ? ment 1 contained a higher percent of acetone than did the solvent in Experiment 2. ment 1 contained a higher percent of H2O than did the solvent in Experiment 2. ment 1 contained a higher percent of acetone than did the solvent in Experiment 2. ment 1 contained a higher percent of H2O than did the solvent in Experiment 2. L _______ mol·sec L _______ mol·sec L _______ mol·sec L _______ mol·sec L _______ mol·sec L _______ mol·sec 45 GO ON TO THE NEXT PAGE. ACT-E25 4 4 Passage V When attacked by predators, tadpoles secrete Protein F into their surroundings. A scientist conducted 2 experiments with Protein F: one to examine how the activity of tadpoles changes immediately after a period of exposure to Protein F and the other to examine how depriv- ing tadpoles of food affects their activity during a subse- quent period of exposure to Protein F. Experiment 1 One hundred tadpoles were equally divided into 4 groups (Groups 1−4). The groups were separately placed into equal volumes of water having the Protein F concen- trations shown in Table 1. After 2 hr, the groups were separately placed into equal volumes of pure water. The tadpole activity of each group was then monitored at regular intervals over the next 90 min. (The tadpole activity of a group was the number of tadpoles that were swimming during a 5 sec observation period.) The results are shown in Figure 1. Figure 1 Figure 1 adapted from Michael E. Fraker, “The Dynamics of Preda- tion Risk Assessment: Responses of Anuran Larvae to Chemical Cues of Predators.” ©2008 by British Ecological Society. Experiment 2 Two hundred tadpoles were equally divided into 4 groups (Groups 5−8), and the groups were separately placed into equal volumes of pure water. Each group was deprived of food for a specific length of time: Group 5 for 24 hr, Group 6 for 1 hr, Group 7 for 12 hr, and Group 8 for 24 hr. At the end of their respective deprivation periods, the groups were separately placed into equal volumes of water having the Protein F concentrations shown in Table 2. The tadpole activity of each group was then monitored at regular intervals over the next 150 min (see Figure 2). Figure 2 Figure 2 adapted from Michael E. Fraker, “Risk Assessment and Anti-Predator Behavior of Wood Frog (Rana sylvatica) Tadpoles: A Comparison with Green Frog (Rana clamitans) Tadpoles.” ©2010 by Society for the Study of Amphibians and Reptiles. Table 1 Group Protein F concentration (pM*) 1 000 2 100 3 200 4 300 *picomoles per liter 12 10 8 6 4 2 0 time (min) number of tadpoles swimming Key Group 1 Group 2 Group 3 Group 4 0 15 30 45 60 75 90 Table 2 Group Protein F concentration (pM) 5 000 6 200 7 200 8 200 24 20 16 12 8 4 0 time (min) number of tadpoles swimming Key Group 5 Group 6 Group 7 Group 8 30 60 90 120 150 46 GO ON TO THE NEXT PAGE. ACT-E25 4 4

ALess polar, because the solvent used in Experi-BLess polar, because the solvent used in Experi-CMore polar, because the solvent used in Experi-DMore polar, because the solvent used in Experi-

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Paper 12 · ACT April 2022 · Form E25 · Science Test

Consider the statement “The activity of the tadpoles that had been exposed to 300 pM of Protein F increased faster than did the activity of the tadpoles that had been exposed to 100 pM of Protein F.” Do the results of Experiment 1 support this statement? was greater for Group 1 than it was for Group 3. was greater for Group 4 than it was for Group 2. was greater for Group 3 than it was for Group 1. was greater for Group 2 than it was for Group 4. 47 GO ON TO THE NEXT PAGE. ACT-E25 4 4 Passage VI A physics teacher asked 3 students to predict the changes, if any, to light’s energy, E; frequency, f (the number of wave peaks passing a given point each second); wavelength, λ (the distance between adjacent peaks of a light wave); and speed, vL, that occur when light travels from a vacuum into clear glass and then from the glass back into the vacuum. The teacher asked the students to base their predictions on one or both of the following equations: E = h × f, where h always has the same value vL = f × λ Student 1 When light enters the glass, f and E are unchanged. This occurs because light loses no energy when it collides with atoms of the glass. However, vL becomes less than c (the speed of light in a vacuum, 3 × 108 m/sec) due to these collisions, so λ must also decrease. As the light leaves the glass, both f and E are still unchanged. But upon reentering the vacuum, vL once again becomes c, so λ must increase. Student 2 When light enters the glass, both f and E decrease, because light loses energy when it collides with atoms of the glass. In addition, vL decreases due to these collisions, so the product f × λ must also decrease. However, λ can either decrease or increase, so long as any increase in λ is not so great as to cause f × λ to increase. When the light leaves the glass, neither f nor E changes, because there is nothing present in a vacuum that will cause f or E to increase. But vL increases to c, the speed of light in a vacuum, so λ must also increase. Student 3 When light enters the glass, both f and E decrease, because light loses energy when it collides with atoms of the glass. However, vL becomes greater than c due to the gravitational attraction between the glass atoms and the light, so the product f × λ must also increase. Thus, λ must increase, and its increase must be great enough to over- come the decrease in f. As the light leaves the glass, f and E will have the same values as they had inside the glass, because there is nothing present in a vacuum that will cause f and E to change. However, because of the gravitational attraction between the glass atoms and the light, vL decreases to c, so λ must also decrease.

AYes; the rate at which tadpole activity increasedBYes; the rate at which tadpole activity increasedCNo; the rate at which tadpole activity increasedDNo; the rate at which tadpole activity increased

Form E25 · Science Test · 40 questions

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