Science reason english 6 doc

Distance traveled (cm) Log (Weight) Weight (Da) 0.5 4.81 65 000 1.0 4.55 35 000 1.8 4.23 17 000 2.1 4.16 14 000 2.5 4.02 11 000 2.7 3.91 8 000 3.3 3.79 6 000 76. Which statement characterizes the migration of SDS-associated proteins? f. Diffusion moves the proteins from a region of higher concentration to one of lower concentration. g. An electric field causes negatively charged objects to migrate toward the anode (positive end). h. The electrical resistance of negatively charged objects determines the speed of migration. j. Osmosis of water indirectly causes the migration of the proteins. 77. A protein of weight 45 000 Da would be expected to migrate to the region on the graph marked: a. W. b. X. c. Y. d. Z. 78. A protein essential for metabolism has just been discovered. SDS-PAGE reveals that this protein migrates a distance of 1.7 cm. Which statement best characterizes the new protein? f. The weight of the protein is somewhere between 6 000 Da and 11 000 Da. g. The weight of the protein is somewhere between 11 000 Da and 17 000 Da. h. The weight of the protein is somewhere between 14 000 Da and 17 000 Da. j. The weight of the protein is somewhere between 17 000 Da and 35 000 Da. 5 6 4 3 2 1 0 0 0.5 1 1.5 2 2.5 3 3.5 W X Y Z Distance traveled [cm] – ACT SCIENCE REASONING TEST PRACTICE– 309 79. Another essential protein in metabolism is made up of two units, each unit traveling a different dis- tance from the other. The combined weight of the two units is approximately 50 000 Da. Referring to the regions W, X, Y and Z on the graph, which combination will NOT give the possible weight of each unit? a. X + Z b. Y + Z c. X + Y d. W + Y 80. What would happen if an electric field were to be applied to SDS-PAGE for an indefinite length of time? f. Larger proteins will reach the anode before the smaller proteins. g. All proteins will eventually reach a limiting resistance in the matrix, at which point they cease to migrate further. h. Proteins associated with more SDS will reach the anode while proteins associated with less SDS will stop migrating due to resistance. j. All proteins will eventually reach the anode. – ACT SCIENCE REASONING TEST PRACTICE– 310  Practice Questions Answers and Explanations Passage I 1. d. One way to solve this problem is to draw a line through the graph along the 300 g of sugar per 100 g of water mark on the graph, as illustrated in the figure below. In the passage, a supersaturated solution was defined as one in which the amount of solute dissolved exceeds solubility at a given temperature. The line going through the 300 mark is above the solubility curve, at all temperatures listed in choices A, B, and C. At temperature D, however, 300 g sugar /100 g of water does not exceed solubility. Therefore, at 70 degrees Celsius, the solution is NOT supersaturated. 2. h. You could use the strategy described in problem 1. If you draw a line through the 250 mark, you will see that it crosses the solubility curve at about 45 degrees Celsius. Below that temperature (choices f and g), the sugar will not dissolve completely. At 65 degrees (choice j) the sugar will dissolve. Choice j is incorrect because 65 degrees is above the minimum temperature required to dissolve the sugar. 3. d. This question is asking you to extrapolate, make a prediction, based on the given data. The solubil- ity of sugar in water increases, as the temperature increases. You can assume that the trend will con- tinue. So you can rule out choices a and b. Draw a line through the 100 degree Celsius mark, and extend the solubility curve to that mark, following the trend, as illustrated in the figure below. This should help you rule out choice c, since it will require the shape of the curve to change. Solubility of Sugar in Water 100 150 200 250 300 350 400 450 500 550 600 50 0 0 102030405060708090100 110 Solubility [g of sugar/100 g of water] Temperature [degrees Celsius] Solubility of Sugar in Water 100 150 200 250 300 350 400 450 500 550 600 50 0 0 102030405060708090100 110 Solubility [g of sugar/100 g of water] Temperature [degrees Celsius] A B C D – ACT SCIENCE REASONING TEST PRACTICE– 311 4. f. The question could be answered by going back to the passage. Rock candy is made by first com- pletely dissolving the excess sugar, at a high temperature, then slowly cooling to room temperature. Choices g, h, and j don’t describe heating, followed by slow cooling. 5. d. You can solve this problem by drawing a line through the 45 degree Celsius mark. It intersects the solubility curve at about 250 g of solute per 100 g of solvent. In order for a solution to be supersatu- rated, the amount of sugar has to exceed solubility. Therefore, a total of more than 250 g is necessary. If a solution already contains 50 g of sugar, more than 200 grams are required. 6. h. According to the passage, solubility is defined as the amount of solute that can be dissolved in a sol- vent at a given temperature. 7. b. The solubility of 200 grams of sugar/100 grams of water is 20 degrees Celsius. The solubility of 250 grams of sugar/100 grams of water is 40 degrees Celsius. Therefore the difference in temperature is 20 degrees Celsius. 8. h. According to the passage, the compound that is dissolved is the solute, while the liquid is the sol- vent. Therefore in sugar water, sugar is the solute and water is the solvent. Passage II 9. d. The graphs and the data tables both show that the temperature of the soil increases more quickly during the heating up period and decreases more quickly during the cooling off period. This indicates that the soil heats and cools faster. The correct choice is d. 10. f. The graphs and the data table show that the temperature of the soil increases more than the tem- perature of the water during the heating up period, and the soil reaches a higher maximum tempera- ture. 11. b. Changing the length of time for the heating up period would allow both the soil and the water to reach higher maximum temperature values. The soil will still heat faster than the water so it will still have a higher curve on the temperature versus time graph than the water. 12. g. Since soil heats faster, the air above land should then be heated faster by the heat radiated by the land. This narrows the selection to choices g and j. Since the soil also cools faster, the air above the land will cool faster as it comes to equilibrium with the cooler ground temperature by losing heat to the ground. This narrows the final choice to g. 13. a. Since the air above the land heats and cools faster it will get warmer faster during the day. This means during the day the air over the land will be warmer than the air over the ocean. At night, how- ever, the temperature of the land will cool faster than the temperature of the ocean. This means the air above the ocean will be warmer than the air above the land at night. 14. g. During the day, the air above the land is warmer than over the ocean since the land heats faster than the oceans (as seen by the soil heating faster than the water in this experiment). Since air will move from cooler regions to warmer regions, the cool air over the ocean will move over to the land. This cre- ates the sea breeze during the day. – ACT SCIENCE REASONING TEST PRACTICE– 312 15. c. It is not likely that unfiltered water or soil from your garden will heat differently than any other water or soil. Also, the size of the containers is not likely to affect the outcome of the experiment. However, if the heating lamp were faulty, it would cause your results to be inaccurate. 16. j. In the 13 th minute, the soil is 30.5 degrees Celsius and the water is 22.0. The difference in tempera- ture is 8.5 degrees Celsius. Passage III 17. c. The exercise stopped at 4 minutes, but the heart rate did not return to its resting rate until about 5 minutes 45 seconds. Remember that between each of the minute lines on the graph, are 60 seconds. So if a point falls halfway between 5 and 6 minutes, that is 5 minutes and 30 seconds. 18. j. The exercise started at the beginning of the 1 minute and by a quarter of a minute later (60 divided by 4 = 15 seconds) the heart rate started its steep incline. 19. b. The total time for the heart rate to reach its peak height was 1 minute 15 seconds, while the total time for the heart rate to go through recovery time was about 1 minute 30 seconds. 20. j. According to the text above Graph 1, the heart rate increases or decreases depending on the body’s need to transport waste and nutrients. Therefore during exercise the heart rate increases in order to transport more of these materials. 21. b. As seen in Graph 1, the heart rate decreased until it returned to the initial resting heart rate. Because the participant was continuing to rest, the heart rate would reflect that of a resting period. 22. f. Note that Graph 1 begins at minute 0, and ends at minute 6, therefore the only table that accurately reflects Graph 1 is choice f. 23. c. The graph would not look exactly alike because the male participant is likely to have different rest- ing and peak heart rates. There is no evidence to suggest that choices b or d are correct. Only c states what the graph is likely to look like. 24. g. The experiment is attempting to show the heartbeats per minute during rest and exercise, therefore it is reasonable that the title of the graph would be Heartbeats per Minute During Rest and Exercise. There is no mention of the gender of the participant on the graph, so choices f and h are incorrect, and the title Rest and Exercise is incomplete. Passage IV 25. b. Water in container 5, which has the largest radius, boils first. Water in other containers confirms this trend. You may have been tempted to choose d, because of the statement that Lorna was not able to collect quantitative data. However, there seemed to be a clear trend to support b and the statement that she obtained qualitative data means that she was confident that although the exact boiling times could be off, the trend she observed was real. 26. g. There is no mention of problems associated with f, h, and j in the passage. The last sentence in the passage should point you to the correct answer. 27. c. This question required you to remember that it’s important to keep the experimental conditions unchanged throughout the experiment. Different hotplates, just like different ovens, may differ in their – ACT SCIENCE REASONING TEST PRACTICE– 313 heating efficiency and could affect the boiling times she was trying to measure. As long as all water used in the experiment came from the same source, it shouldn’t matter whether it was distilled or not. Stirring is not necessary since there is nothing to mix. There is nothing wrong with setting up a data sheet before the experiment. 28. f. The statement that the volume change was greatest in the container with the largest radius, and barely detectable in the container with the smallest radius should provide you with the right answer. 29. d. The scientist used the graduate cylinder to check whether and by how much the volume in the con- tainer had changed. 30. j. Looking at the unfilled table provided in the text, a container with a 7.0 cm radius has a radius that is smaller than that of container 5, but larger than that of container 4. That tells you that the order in which the water in the 7.0 cm radius container boiled would be between container 5 and container 4. In the text you were told that container 5 boiled first, so the container with a 7.0 cm radius would boil after the water in container 5. Passage V 31. b. In the impetus theory, impetus is a property of the object imparting the motion. In the theory of inertia, the property of inertia is a property of the moving object itself. 32. j. The theory of inertia correctly predicts the parabolic path of a projectile. This is because the projec- tile continues to move with constant velocity in the horizontal direction since there is no net force in the horizontal direction. The net force in the vertical direction is the gravitational force of the earth on the object. This causes the object to fall toward the earth as it travels horizontally creating the para- bolic path. In the impetus theory, however, the impetus of the projectile would run out abruptly which would then predict that the projectile should keep going in a straight line until it uses up the impetus and then it would be predicted to fall straight down. 33. c. According to the inertia theory, the net force that acts to slow down the arrow is the force of gravity on the arrow. The force of the bow on the arrow is what causes the arrow to begin moving. This means that selection I is false, the force of the bow on the arrow is not the net force acting to slow down the arrow, but selection III is true since the force of gravity is the net force acting to slow down the arrow according to the inertia theory. Selection II indicates that the arrow receives an infinite amount of impetus, but according to the explanation of the motion of the arrow using the impetus theory the arrow only receives a certain amount of impetus from the bow and when it uses up this impetus it will fall to the ground. This means selection II is false, but selection IV which says that the impetus imparted to the arrow by the bow is used up and that is why the arrow falls to the ground is true. 34. f. For an object to continue moving forever in a straight line with constant velocity, the impetus the- ory requires that the object be given an infinite amount of impetus. 35. d. The inertia theory states that an object will continue moving in a straight line with constant veloc- ity as long as no net force acts on it. 36. g. As defined by the impetus theory, impetus is the property of motion that is imparted to the object by whatever is acting on it. From the example in the reading, the impetus is a property motion of the bow that is transferred to the arrow. – ACT SCIENCE REASONING TEST PRACTICE– 314 . traveled (cm) Log (Weight) Weight (Da) 0.5 4.81 65 000 1.0 4.55 35 000 1.8 4.23 17 000 2.1 4. 16 14 000 2.5 4.02 11 000 2.7 3.91 8 000 3.3 3.79 6 000 76. Which statement characterizes the migration. protein is somewhere between 17 000 Da and 35 000 Da. 5 6 4 3 2 1 0 0 0.5 1 1.5 2 2.5 3 3.5 W X Y Z Distance traveled [cm] – ACT SCIENCE REASONING TEST PRACTICE– 309 79. Another essential protein. Water 100 150 200 250 300 350 400 450 500 550 60 0 50 0 0 102030405 060 708090100 110 Solubility [g of sugar/100 g of water] Temperature [degrees Celsius] Solubility of Sugar in Water 100 150 200 250 300 350 400 450 500 550 60 0 50 0 0