1,296 ACT Practice Questions, 3rd Edition (2013)

Science Practice Section 1

SCIENCE REASONING TEST

35 Minutes—40 Questions

DIRECTIONS: There are seven passages in the following section. Each passage is followed by several questions. After reading a passage, choose the best answer to each question and blacken 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

A study was conducted regarding the fossil shells of a particular species of turtle that lives off the coast of the Opulasian Peninsula. Scientists discovered a continuous record of fossilized shells in the seabed off the coast dating back 120,000 years. In addition to examining the fossilized turtle shells, the scientists also examined the shells of living turtles.

From each layer of seabed, the scientists randomly selected five complete, unbroken fossilized shells. Each shell was carefully prepared, measured, and photographed. A bit of each shell was then clipped off and sent to a laboratory for radiocarbon dating to determine the precise age of each shell.

Study 1

All of the living turtles had a distinct band of hexagonal scutes (bony plates) running the length of their shells, from head to tail. The fossilized shells’ scutes were not visible to the naked eye; however upon application of a particular dye, a similar band of scutes from head to tail was observed in every shell.

Scutes extending greater than  of the length of the shell were labeled major (M), where scutes extending less than or equal to  of the length of the shell were labeled minor (m). The pattern of scutes was recorded for each fossil. For each time period, the percent of fossils exhibiting each pattern is given in Table 1.

Study 2

For each shell, the surface area of the shell, the height of the shell’s bridge (the part of the shell linking the upper and lower plates), and the total number of scutes were recorded (see Figure 1).

For the shells of each age, the average of each measurement was calculated. The results are presented in Figure 2.

  1. In a layer of seabed determined to be 250,000 years old, the scientists found fragments of twelve turtle shells, but no complete, intact shells. Which of the following is the most likely reason this layer of seabed was not included in the studies?

       A.  250,000 years is too old to obtain an accurate radiocarbon date.

       B.  Shells that were 250,000 years old would have been irrelevant to the studies.

       C.  Accurate measurements of the dimensions of the shells could have been impossible to obtain.

       D.  The scientists would not have been able to accurately determine the color of the shells.

  2. With regard to the descriptions given in Study 1, the shells with the M-M-m-m-M band of scutes probably most closely resembled which of the following?

       F.  

       G.  

       H.  

       J.   

  3. According to the results of Study 2, how do the average number of scutes and the average bridge height of living turtles of the Opulasian Peninsula compare to those of the turtles of the Opulasian Peninsula from 120,000 years ago? For the living turtles:

       A.  both the average number of scutes and the average bridge height are larger.

       B.  both the average number of scutes and the average bridge height are smaller.

       C.  the average number of scutes is larger, and the average bridge height is smaller.

       D.  the average number of scutes is smaller, and the average bridge height is larger.

  4. Suppose, in Study 1, the scientists had found another seabed layer with fossilized shells that were radiocarbon dated and found to be 86,000 years old. Based on the results of Study 1, the scute pattern percents for the group of shells would most likely have been closest to which of the following?

 

M-m-M-M-m

M-M-m-m-M

M-m-M-m-M

F.

100%

  0%

  0%

G.

  50%

25%

25%

H.

  36%

61%

  4%

J.

  26%

69%

  5%

  5. In Study 2, the average shell surface area of fossilized turtle shells that were 80,000 years old was closest to:

       A.  670 cm2

       B.  680 cm2

       C.  690 cm2

       D.  700 cm2

  6. Which of the following statements best describes how Study 1 differed from Study 2?

       F.  In Study 1, the scientists examined 3 characteristics regarding the shape and size of turtle shells; but in Study 2, the scientists examined the frequency of occurrence of different patterns of scutes on turtle shells.

       G.  In Study 1, the scientists examined the frequency of occurrence of different patterns of scutes on turtle shells; but in Study 2, the scientists examined the environment in which turtles live.

       H.  In Study 1, the scientists examined the frequency of occurrence of different patterns of scutes on turtle shells; but in Study 2, the scientists examined 3 characteristics regarding the shape and size of turtle shells.

       J.   In Study 1, the scientists examined 3 characteristics regarding the shape and size of turtle shells; but in Study 2, the scientists examined the environment in which turtles live.

Passage II

The 4 different blood types in sheep are A, B, AB, and O. The blood type of an offspring is determined by the blood types of its parents. Each parent contributes a single gene to its offspring, forming a pair of genes. The genotype of an offspring refers to the arrangement of the offspring’s new gene formed by the combination of the parents’ genes.

There are three possible alleles (forms) of this gene: the type-A blood allele (IA), the type-B blood allele (IB), and the type-O blood allele (IO). Both IA and IB are dominant to IO, and IO is recessive to IA and IB. This means that an individual with 1 IA and 1 IO will have type-A blood, and an individual with one IB and one IO will have type-B blood. When an individual has one IA and one IB allele, this individual will have type-AB blood, due to the codominance of the IA and IB alleles.

Table 1

Blood Type

Possible Genotypes

A

IAIA or IAIO

B

IBIB or IBIO

AB

IAIB

O

IOIO

To explore the inheritance patterns of blood types in sheep, researchers conducted 4 analyses. In each analysis, male and female sheep of differing blood types were mated and the resultant blood types of their offspring recorded.

Analysis 1

One thousand males with type-O blood were mated with 1,000 females with type-AB blood. The following blood types were observed in the offspring:

Type A: 50%

Type B: 50%

Analysis 2

Two hundred of the type-A offspring from Analysis 1 were mated with 200 type-O mates from no previous experiment. The following blood types were observed in the offspring:

Type A: 50%

Type O: 50%

Analysis 3

One hundred of the type-A offspring from Analysis 1 parented children with 100 type-B offspring from Analysis 1. The following blood types were observed in the offspring:

Type A: 25%

Type B: 25%

Type AB: 25%

Type O: 25%

Analysis 4

Twenty-five of the type-A offspring from Analysis 3 were mated with type-B mates with Genotype IBIB who were not from any previous analysis. The following blood types were observed in the offspring:

Type AB: 50%

Type B: 50%

  7. The ratio of blood types containing at least one IA allele to the blood types containing at least one IB allele produced in Analysis 3 was:

       A.  1:0.

       B.  1:1.

       C.  2:1.

       D.  3:1.

  8. An offspring whose blood type exhibits codominance has which of the following genotypes?

       F.  IBIB

       G.  IBIO

       H.  IAIB

       J.   IAIO

  9. To produce only offspring with AB blood, one would mate two sheep with which of the following sets of genotypes?

       A.  IAIB × IAIB

       B.  IAIB × IOIO

       C.  IAIA × IBIB

       D.  IBIB × IAIO

10. In Analysis 3, the offspring used from Analysis 1 most likely had which of the following genotypes?

       F.  IAIO and IBIB

       G.  IAIO and IBIO

       H.  IAIA and IBIB

       J.   IAIA and IBIO

11. Some or all of the offspring had 1 allele for type-O blood in Analyses:

       A.  1 and 2 only.

       B.  2 and 3 only.

       C.  1, 2, and 4 only.

       D.  1, 2, 3, and 4.

12. Suppose that 300 offspring were produced in Analysis 3. Based on the results, the number of offspring with type-B blood produced in Analysis 3 would most likely have been closest to:

       F.  25.

       G.  50.

       H.  75.

       J.   100.

Passage III

Vasoconstriction involves a narrowing of blood vessels that could lead to poor blood flow in the body if it persists over a long time. Ergotamine is a substance that can cause vasoconstriction. When ergotamine is injected into a normal blood vessel, vasoconstriction occurs quickly at the site of the injection (see Figure 1).

The diameter of the blood vessel at the site of vasoconstriction is less than the diameter of the normal blood vessel, so blood flow has a higher velocity through this narrow site. As a result, the blood pressure in the site of vasoconstriction is less than the blood pressure in the normal blood vessel. Moreover, the higher the velocity of the blood flow through the site of vasoconstriction, the lower the blood pressure at that site.

The percent change in blood pressure (%BP) can be defined as:

Blood vessel sections of similar diameters were isolated from laboratory rats and %BP was measured over three experiments. When the researchers needed to create a site of vasoconstriction for some of the experimental trials, they would inject ergotamine to induce vasoconstriction within the blood vessel section.

Experiment 1

An artificial heart, which mimics a human’s heartbeat, is used to move a constant volume of 500 mL of blood with each beat through four blood vessel sections. These four blood vessel sections were injected with the same amount of ergotamine, leading to sites of vasoconstriction of the same diameter. The rate at which the blood is pumped was varied for the four different blood vessel sections, and the %BP values that resulted were measured.

Table 1

Rate of artificial heart beat (beats per minute)

%BP

60

1.2

90

9.3

120

22.3

150

45.1

Experiment 2

The artificial heart used in Experiment 1 was then used to pump a constant volume of 500 mL of blood with each beat at a constant rate of 90 beats per minute through five other blood vessel sections. These blood vessel sections were injected with different amounts of ergotamine, resulting in sites of vasoconstriction with different diameters. The %BP values were then measured.

Table 2

Diameter of site of vasoconstriction (cm)

%BP

0.4

40.3

0.6

18.6

0.8

9.3

1.0

4.6

1.2

2.5

Experiment 3

The artificial heart used in Experiment 1 was used to pump different volumes of blood at a constant rate of 90 beats per minute through five blood vessel sections with the same diameter at the site of vasoconstriction. The %BP values were then measured.

Table 3

Volume of blood pumped (mL)

%BP

400

8.4

450

8.8

500

9.3

550

9.7

600

10.2

13. Under the conditions described for Experiment 3, a %BP of 9.0 would most likely be obtained if the entering volume of blood equaled:

       A.  350 mL.

       B.  475 mL.

       C.  550 mL.

       D.  650 mL.

14. Based on the results of Experiment 1, if the rate of the artificial heart beat had been less than 60 beats per minute, then the %BP would most likely have been:

       F.  less than 1.2.

       G.  between 1.2 and 9.3.

       H.  between 9.3 and 22.3.

       J.   greater than 22.3.

15. Which of the following is the most likely explanation for the results of Experiment 1? As the rate of the artificial heart beat increases, %BP:

       A.  increases, because the velocity of blood through the site of vasoconstriction increases.

       B.  increases, because the velocity of blood through the site of vasoconstriction decreases.

       C.  decreases, because the velocity of blood flow through the site of vasoconstriction increases.

       D.  decreases, because the velocity of blood flow through the site of vasoconstriction decreases.

16. Consider blood flow through three regions of the same blood vessel, each of which has a different diameter. The velocity of blood flow is measured in milliliters per minute (mL/min) and the blood pressure is measured in millimeters of mercury (mmHg), and their values for each of the blood vessel regions are shown in the following table:

Location

Velocity of blood flow (mL/min)

Blood pressure (mmHg)

A

500

31

B

1,000

29

C

900

30

Based on the information in the passage about blood flow, which of the following diagrams best represents the relative diameters of the three blood vessel regions?

       F.  

       G.  

       H.  

       J.   

17. Based on the results of Experiments 1 and 2, what was the diameter of the site of vasoconstriction in the blood vessel section used in Experiment 3?

       A.  0.4 cm

       B.  0.6 cm

       C.  0.8 cm

       D.  1.0 cm

18. For the blood vessel sections used in Experiment 2 that had sites of vasoconstriction with diameters of 0.4, 0.8, and 1.2 cm, which of the following graphs best displays the comparison between blood pressure at each site of vasoconstriction and blood pressure in the normal region of the blood vessel leading to the site of vasoconstriction?

       F.  

       G.  

       H.  

       J.   

Passage IV

As the pressure on a gas is increased, the volume of that gas is expected to decrease by an inversely proportional amount. For example, if pressure is doubled the volume is halved. Under certain conditions, the volume of the gas will change by an amount that deviates from an inverse proportion. Various 10.00 L samples of gas were subjected to increases in pressure. Table 1 shows the resulting volume changes at 300°C, while Tables 2 and 3 show the volume changes at 25°C and −200°C, respectively. All pressures are measured in atmospheres (atm).

19. Which of the following gases shown in Tables 1−3 was compressed by the same amount each time the pressure was changed, regardless of its initial pressure?

       A.  Helium

       B.  Carbon Dioxide

       C.  Neon

       D.  Oxygen

20. Which of the following is the best explanation for the change in volume seen in any one of the samples of carbon dioxide in Table 1? As pressure on one sample of carbon dioxide was increased, the volume of that sample:

       F.  increased as the molecules of carbon dioxide were forced closer together.

       G.  increased as the molecules of carbon dioxide were forced farther apart.

       H.  decreased as the molecules of carbon dioxide were forced closer together.

       J.   decreased as the molecules of carbon dioxide were forced farther apart.

21. Based on Table 2, if the sample of nitrogen at a pressure of 4 atm were returned to its initial pressure of 2 atm, the volume would most likely:

       A.  decrease by 5.00 L.

       B.  decrease by 8.00 L.

       C.  increase by 5.00 L.

       D.  increase by 8.00 L.

22. Based on Table 3, if the pressure on a 10.00 L sample of neon gas is increased from 8 atm to 16 atm at a temperature of −200°C, the change in volume will most likely be closest to which of the following?

       F.  −5.12 L

       G.  −5.06 L

       H.  −5.03 L

       J.   −5.02 L

23. A scientist concludes that whenever the pressure on helium is increased, its volume will decrease. Based on Tables 2 and 3, is this a valid conclusion?

       A.  Yes; in every trial that the pressure of helium was increased, the change in volume was negative.

       B.  No; in every trial that the pressure of helium was increased, the change in volume was positive.

       C.  Yes; when the pressure on helium was increased from 1 to 2 atm, its change in volume was positive at 25°C and negative at −200°C.

       D.  No; when the pressure on helium was increased from 1 to 2 atm, its change in volume was negative at 25°C and positive at −200°C.

Passage V

There are four planets in our solar system called gas giants: Jupiter, Saturn, Uranus, and Neptune. They are so named because they are composed largely of gases rather than solids. Figure 1 shows how temperatures of the atmospheres of Jupiter, Neptune, and Saturn vary with altitude above the cloud tops. Table 1 gives the composition of the planets in both relative abundance of gases and the altitude at which those gases are most abundant. Table 2 gives what the temperature at the cloud tops would be without greenhouse warming.

Figure 1

Table 2

Planet

Temperature at cloud tops without greenhouse warming (K)

Jupiter
Neptune
Saturn

100
50
25

24. According to Figure 1, the temperature of Neptune remains the same as altitude above the highest cloud tops increases from:

       F.  −250 km to −200 km.

       G.  −150 km to −50 km.

       H.  0 km to 100 km.

       J.   150 km to 200 km.

25. According to Figure 1, the temperature of Jupiter changes the most between:

       A.  −150 km and −50 km.

       B.  −50 km and 50 km.

       C.  50 km and 100 km.

       D.  100 km and 200 km.

26. Considering only the gases listed in Table 1, which gas is more abundant in the atmosphere of Jupiter than in the atmosphere of either Neptune or Saturn?

       F.  H

       G.  CH3

       H.  NH3

       J.   He

27. Based on Table 2, the average temperature at Saturn’s cloud tops without greenhouse warming is how many degrees cooler than the temperature given in Figure 1?

       A.  5 K

       B.  25 K

       C.  75 K

       D.  150 K

28. Which of the following statements about H and He in the atmospheres of the 3 planets is supported by the data in Table 1?

       F.  Both Saturn and Neptune have a higher relative abundance of He than of H.

       G.  Both Saturn and Jupiter have a higher relative abundance of He than of H.

       H.  Both Jupiter and Neptune have an equivalent relative abundance of He and H.

       J.   Both Saturn and Neptune have a lower relative abundance of He than of H.

Passage VI

Nuclear fission occurs when the nucleus (central core) of an atom splits into multiple parts. This splitting is accompanied by the release of a large amount of energy, as in nuclear weapons and nuclear power plants.

A chemical element is said to be radioactive if it is prone to fission. Fission is often the result of the nucleus of a radioactive atom absorbing a free neutron (an uncharged nuclear particle). When a fission event occurs, the nucleus often splits into two new nuclei and produces free neutrons. This process generates the possibility of a chain reaction. If, on average, a fission event produces one neutron and that neutron causes another nucleus to fission, the reaction is said to be critical; that is, it will sustain itself, but not increase in magnitude. If one fission event releases more free neutrons than are required to initiate another fission event, the reaction is said to be supercritical; that is, it will sustain and increase in magnitude. If more neutrons are required to initiate a fission event than are released in fission, the reaction is said to be subcritical: the reaction will not sustain itself.

Many factors affect how many neutrons from each fission event will trigger another fission event. The most important factor is the mass (m) of the substance. The criticality of a substance also depends on the substance’s purity, shape, density, temperature, and whether or not it is surrounded by a material that reflects neutrons.

In a nuclear weapon, a radioactive substance is made highly supercritical. One of the primary challenges in building a nuclear weapon is keeping the radioactive material subcritical prior to detonation, then upon detonation, keeping it supercritical for a long enough period of time for all of the material to fission before it is blown apart by the energy of the blast. A fizzle occurs when a nuclear weapon achieves supercriticality but is blown apart before all of the radioactive material fissions.

The first nuclear weapons were made of enriched uranium, or U-235. The density (ρ) of U-235 under normal conditions is 19.1 g/cm3. For U-235 to attain a supercritical state, the product of its mass and density must exceed 106 g2/cm3. If it is assembled over too long a time (t), it will achieve slight supercriticality and then fizzle. Therefore, the speed of assembly (measured as t divided by ρ), must be less than 10−5 sec × cm3/g (Michelson’s Criterion).

Two schemes for the assembly of a supercritical amount of U-235 that avoid fizzle are discussed below.

Gun-Type Weapon

At one end of a tube, similar to a gun barrel, is a hollow, subcritical cylinder of U-235 with a mass of 48 kg; on the other end is a subcritical pellet of U-235 with a mass of 12 kg. The pellet is propelled by a small explosion down the tube and into the cylinder of U-235. The combined mass of the two pieces of U-235 is great enough to induce a supercritical state. Since the combined cylinder of U-235 is at or near normal density, the assembly process must be completed in less than 2 × 10−4sec to meet Michelson’s Criterion.

Implosion-Type Weapon

A 15-kg sphere of U-235 is surrounded by explosives. When the explosives are simultaneously detonated, the U-235 is compressed in order to achieve supercriticality. The explosives are designed to compress the U-235 to a density of approximately 70 g/cm3 in less than 10−7 sec.

29. For both types of weapon, avoiding fizzle is difficult because:

       A.  the mass of U-235 must be large.

       B.  2 separate pieces of U-235 must be brought together.

       C.  U-235 is highly unstable.

       D.  of the speed with which the U-235 must be assembled.

30. Comparing the mass of uranium used in the two types of weapons reveals that:

       F.  the mass of U-235 used in the implosion-type weapon is less than the mass of U-235 used in the gun type weapon.

       G.  the mass of U-235 used in the implosion-type weapon is greater than the mass of U-235 used in the gun type weapon.

       H.  the mass of U-235 used in the implosion-type weapon is greater in some cases and less in some cases than the mass used in the gun-type weapon.

       J.   the mass of U-235 used in both weapons is approximately the same.

31. Both types of weapons use explosives in order to:

       A.  increase the heat of the U-235.

       B.  release the nuclear energy of the weapon from the confinement of the bomb’s casing.

       C.  achieve supercriticality of U-235.

       D.  generate neutrons to start the chain reaction.

32. For an implosion-type weapon, when U-235 has reached supercriticality, to which of the following is the value of ρ closest?

       F.  10−3 g/cm3

       G.  0.1 g/cm3

       H.  100 g/cm3

       J.   106 g/cm3

33. In the implosion-type weapon, the explosives are used to:

       A.  trigger the first fission events.

       B.  heat the U-235 so it will become supercritical.

       C.  increase the density of U-235.

       D.  produce additional damage.

34. In order to achieve a supercritical state just before detonation, both methods:

       F.  increase the product of the mass and density of the U-235.

       G.  decrease the product of the mass and density of the U-235.

       H.  increase the amount of U-235 in the weapon.

       J.   decrease the time necessary for all the U-235 to fission.

35. Scientists are trying to build a bomb using only 8 kg of U-235. Presently they can achieve a ρ of 150 g/cm3 with t = 10−2 sec. Which of the following changes would be the most likely to get the weapon to meet Michelson’s Criterion?

       A.  Decrease both t and ρ.

       B.  Decrease t and leave ρ the same.

       C.  Increase t and decrease ρ.

       D.  Increase t and leave ρ the same.

Passage VII

A scientist studying hemoglobin investigated the impact of temperature and carbon dioxide (CO2) concentrations on the binding capacity of oxygen (O2). The scientist observed the binding of oxygen to hemoglobin molecules as the pressure of oxygen was increased. The temperature and CO2 were varied to identify their direct impact on the binding capacity of O2.

Figure 1 displays the impact of changes in temperature on the binding (percent of hemoglobin saturated) of oxygen. Figure 2 displays the impact of varying carbon dioxide concentrations on oxygen binding. Under normal conditions, the core body temperature is 37°C and has carbon dioxide and oxygen concentrations of 40 mmHg and 100 mmHg respectively.

Figure 1

Figure 2

36. According to Figure 1, if the temperature is 42°C, which of the following changes in pressure of oxygen will cause the least increase in the percent of hemoglobin saturated with O2?

       F.  0−15 mmHg

       G.  15−30 mmHg

       H.  30−45 mmHg

       J.   45−60 mmHg

37. According to Figure 1, which of the following sets of temperature and pressure of oxygen results in the lowest hemoglobin saturation with oxygen?

 

Temperature (°C)

Pressure of Oxygen (mmHg)

A.

37

45

B.

37

60

C.

42

45

D.

42

60

38. According to Figure 1, if the pressure of oxygen is 100 mmHg and 65% of hemoglobin molecules are saturated with oxygen then the core body temperature is most likely within which of the following ranges?

       F.  Less than 30°C

       G.  30°C−37°C

       H.  37°C−42°C

       J.   Greater than 42°C

39. Based on Figure 2, if an individual has 70% of his hemoglobin molecules saturated at a pressure of 75 mmHg of oxygen, then the individual’s carbon dioxide pressure is most likely closest to which of the following?

       A.  30 mmHg

       B.  50 mmHg

       C.  70 mmHg

       D.  90 mmHg

40. According to Figure 2, at a CO2 pressure of 90 mmHg, as the pressure of O2 is increased from 45 mmHg to 90 mmHg, the percent of hemoglobin saturated with oxygen:

       F.  remains constant, then increases.

       G.  remains constant, then decreases.

       H.  increases, then decreases.

       J.   decreases, then increases.

Science Practice
Section 1
Answers and Explanations

SCIENCE PRACTICE 1 ANSWERS

  1. C

  2. G

  3. A

  4. J

  5. A

  6. H

  7. B

  8. H

  9. C

10. G

11. D

12. H

13. B

14. F

15. A

16. J

17. C

18. G

19. D

20. H

21. C

22. F

23. A

24. H

25. A

26. H

27. C

28. J

29. D

30. F

31. C

32. H

33. C

34. F

35. B

36. F

37. C

38. J

39. B

40. F

SCIENCE PRACTICE 1 EXPLANATIONS

  1.  C  Nothing is mentioned in the passage about how far back radiocarbon dating is accurate, so choice (A) is incorrect. You are not told what motivation caused the scientists to perform the study, and there is nothing to suggest that they are only interested in a certain time period, so choice (B) can be eliminated. For choice (D), the scientists do not try to determine the color of the shells; rather, they use a dye to visualize the scutes so that they can determine their sizes and patterns. Without any complete shells, it would be very difficult for the scientists to determine any of the measurements used in Study 2—choice (C).

  2.  G  M hexagons are the longer ones; m hexagons are the smaller ones. Choice (G) shows a band with two large, then two small, and another large hexagon in a band down the middle of the shell.

  3.  A  120,000 years ago, the average number of scutes was about 25, and now (at 0—the furthest right point on the graph), it is about 32, so the average number of scutes is larger. 120,000 years ago, the average bridge height was about 2.1 cm, and now it is about 2.5 cm, so the average bridge height is also larger.

  4.  J  The hypothetical 86,000-year-old data point falls between the 85,000 and 87,000-year-old data points. Therefore, you should expect the percent of shells having each pattern to be between the percentages at 85,000 and 87,000 years old. For the first column (M-m-M-M-m), 25.5 is the average of 21 and 30, so choice (J) looks like a good fit. To be sure, check the other columns as well. The value in the second column (M-M-m-m-M) should be about 67−72, and the value in the third column should be around 3−7. Choice (J) works well.

  5.  A  The third graph in Figure 2 shows average shell surface area. Find the 80,000 years point on the x-axis and move vertically to the data point. Move horizontally to the y-axis to find the answer: 670 cm2.

  6.  H  The first study examines the relative frequency of occurrence of different patterns of scutes (M-m-M-M-m, etc.). The second study looks at shell diameter, shell bridge height and the number of scutes on shells—3 shape and size characteristics of turtle shells, making choice (H) the best answer.

  7.  B  Since all four blood types were produced in equal measure in Analysis 3, refer back to Table 1 to determine your ratio. There are two blood types that contain at least one IA allele: A and AB. Similarly, there are two blood types that contain at least one IB allele: B and AB. Since there are two possible blood types for each allele, the ratio of possibilities must be 1:1.

  8.  H  If you’re not sure what codominance is, look to the description of the experiment: When an individual has 1 IA and 1 IB allele, this individual will have type-AB blood, due to the codominance of the IA and IB alleles.Consequently, an offspring whose genotype exhibits codominance must have type-AB blood. Look to Table 1 to see that the genotype for type-AB blood is IAIB, choice (H).

  9.  C  Table 1 indicates that the only genotype that can produce type AB blood is IAIB. For this to happen in every offspring, and because each parent contributes only a single allele, one parent must only be capable of contributing IA while the other parent is only capable of contributing IB. The cross of IAIA × IBIB will always produce offspring with IAIB genotypes and type AB blood phenotypes. If you chose choice (A), be careful—think of all possible combinations of these two genotypes. If one parent contributes IB and the other parent contributes IB, the offspring will have type-B blood.

10.  G  Since both parents come from Analysis 1, refer back to the description there: One thousand males with type-O blood were mated with 1,000 females with type-AB blood. In Analysis 1, the males had only Genotype IOIO and the females had only Genotype IAIB. The only possible combinations of these genotypes are IAIO and IBIO. Since it is these offspring who are mated in Analysis 3, their blood types must be IAIO and IBIO.

11.  D  If you’re not sure how to find the answer, use process of elimination. The easiest place to look for blood-types containing IO alleles would be in analyses that produced offspring with type-O blood. This includes Analyses 2 and 3, so any answer choice that does not contain both can be eliminated—eliminate choices (A) and (C). Since one of the parents in Analysis 1 has type-O blood (Genotype IOIO), one of these alleles must have been passed on to each of the offspring—eliminate choice (B). This leaves only choice (D), so even if you’re not sure about Analysis 4, choice (D) is the only answer that contains all the others about which you are sure. Note: Analysis 4 works because the parents had genotypes IAIO and IBIB, so their offspring could have only genotypes IBIO and IAIB.

12.  H  Analysis 3 had offspring with 4 possible phenotypes in equal frequency. If 300 offspring were identified, then 25% of 300 or 75 of them would be expected to have type B blood.

13.  B  Table 3 shows that a pumped blood volume of 450 mL yields a %BP of 8.8, while a pumped blood volume of 500 mL yields a %BP of 9.3. The %BP of 9.0 is between these other two %BP values, so it was likely produced by a pumped blood volume between 450 and 500 mL. Only choice (B) fits this requirement.

14.  F  Table 2 shows that when the artificial heartbeat pumps blood at 60 beats per minute, the %BP is 1.2. As the rate at which the blood is pumped increases, so does the %BP, so it follows that as the rate decreases, so would the %BP. Therefore, choice (F) is the best answer.

15.  A  Experiment 1 shows that as the artificial heart beat rate increases, so does the %BP, and only choices (A) or (B) show this. The passage states that a lower blood pressure follows from faster blood flow through the site of vasoconstriction. Since %BP is calculated by subtracting the pressure at the site of vasoconstriction from normal blood pressure, %BP increases with a lower pressure at the site of vasoconstriction. Thus, an increase in velocity causes an increase in %BP. Choice (A) summarizes this.

16.  J  According to the passage, a faster velocity of blood flow and a lower blood pressure are consistent with a narrow region of a blood vessel, while a slower velocity of blood flow and a higher blood pressure are consistent with a wide region of a blood vessel. Therefore, the measurement at location A would have been taken at the region with the widest diameter, while the measurement at location B would have been taken at the region with the narrowest diameter. Choice (J) best illustrates this. Note: Since the values in the chart are not listed in a consistent order, you can safely eliminate choices (F) and (G), which show the three diameters consistently increasing from A–C or decreasing from A–C, respectively.

17.  C  Experiments 2 and 3 use an artificial heart that pumps blood at a constant rate of 90 beats per minute. In Experiment 3, the %BP is measured as 9.3 for a blood volume of 500 mL, the amount that was used in Experiment 1 for a heart rate of 90 beats per minute. In Experiment 2, a %BP of 9.3 was measured when the diameter of the site of vasoconstriction was 0.8 cm. Therefore, it follows that the diameter of the site of vasoconstriction used throughout Experiment 3 was also 0.8 cm, so choice (C) is the best answer.

18.  G  As the diameter of the site of vasoconstriction increases, the %BP decreases. Refer back to the passage for the %BP formula. From this formula, you can deduce that a larger %BP will result from a larger difference between normal pressure and pressure at the site of vasoconstriction. Only the graph in choice (G) shows this pattern across the three given diameters. If you selected choice (F) be careful—you may have chosen them in the opposite order.

19.  D  For every trial involving oxygen in Table 1, the volume decreased by exactly 5.00 L. All the other choices had volume changes that varied.

20.  H  In Table 1, each trial involving carbon dioxide showed a negative volume change, or decrease in volume. Therefore, choices (F) and (G) are eliminated. The molecules will occupy less volume only if they are pushed closer together, eliminating choice (J).

21.  C  The passage states that increases in pressure lead to decreases in gas volume. Therefore, decreases in pressure would be expected to result in increases in gas volume, eliminating choices (A) and (B). In Table 2, the volume of carbon dioxide decreased by 5.00 L when the pressure was increased from 2 atm to 4 atm. Therefore, the volume would be expected to increase by 5.00 L upon returning to its initial pressure.

22.  F  In Table 3, the volume of neon decreases by an increasingly larger factor as the pressure changes increase. The question proposes an even larger pressure increase than those listed in Table 3, so the resulting volume change would also be expected to be larger, eliminating all choices except (F).

23.  A  No trials on any of the tables have positive changes in volume, so you can eliminate choice (B). Choices (C) and (D) can be eliminated because while the pressure changes at different temperatures produced differing results for volume, the volume was always decreasing.

24.  H  Choice (H) is the correct answer for this problem because it is the straightest part of the Neptune line. To solve this problem you should look at which answer choice represents no temperature change, so which answer choice represents a section of the line that is straight up and down. Choices (F), (G), and (J) all represent sections of the line that represent relatively large changes in Neptune’s temperature.

25.  A  Compare the changes in temperature for each range in the answer choices. The question does not specify an increase or decrease, so you’ll want to pick the answer choice containing the largest range of temperatures. Choice (A) ranges from about 150 K to 300 K; choice (B) ranges from about 150 K to 200 K; choice (C) does not have a range any wider than about 30 K; choice (D) has a range similarly small to that of choice (C). Choice (A) clearly contains the largest range of temperatures and so can be said to change the most.

26.  H  Compare the relative abundances of each gas in the answer choices. NH3, Choice (H), is the only gas that has a higher relative abundance in Jupiter than in either Neptune or Saturn. Choice (F) cannot be correct because H has a higher relative abundance in Saturn than in Jupiter. Similarly, choice (G) cannot be correct because Neptune has a higher relative abundance of CH3 than Jupiter, as is the case with He in choice (J).

27.  C  To solve this problem, you must look at Figure 1 and see what temperature Saturn is when the altitude above cloud tops equals 0 km, then compare to Table 2. Saturn is about 100 K at a 0 km altitude above the cloud tops, so the answer must be choice (C), because 100 K − 25 K = 75 K.

28.  J  Read the answer choices carefully. Since each answer choice is comparing the relative abundance of He and H on the three planets, see if you can identify any general trends in Table 1. You’ll notice that for each planet, the H values are much higher than the He values, so in no case will the relative abundance of He be greater than or equivalent to that of H. Only choice (J) has He and H in the proper relationship to one another.

29.  D  In the middle of the fifth paragraph is this sentence: If it [the U-235] is assembled over too long a time (t), it will achieve slight supercriticality and then fizzle.

30.  F  The implosion-type weapon uses 15 kg of U-235. The gun-type weapon uses a 48-kg cylinder and a 12-kg pellet for a total of 60 kg of U-235.

31.  C  In the fifth paragraph of the passage, notice that to achieve supercriticality, the product (multiplication) of the mass and density of U-235 must be a large value. Therefore, supercriticality could be achieved by increasing either the mass or the density of the U-235. In the gun-type weapon, the explosives are used to propel one piece of U-235 into another (increasing the total mass). In the implosion-type weapon, the explosives are used to increase the density of the U-235. In both cases the goal is to increase the product of mass and density in order to achieve supercriticality.

32.  H  ρ stands for density. Read the implosion-type weapon paragraph and notice that the density after compression is 70 g/cm3, which is closest to 100 g/cm3.

33.  C  In the section on implosion-type weapons, the passage says that explosives are used to compress the U-235. To compress is to decrease the amount of space a substance occupies while preserving its mass, which is to increase its density.

34.  F  Again, the passage says that to achieve a supercritical state the product of mass and density must be greater than some value. This is why the gun-type weapon combines two pieces of U-235 to increase the mass, and it is why the implosion-type weapon compresses the U-235.

35.  B  Michelson’s Criterion says that t divided by ρ must be less than some number. Therefore to meet this criterion, t could be decreased or ρ could be increased. Choice (B) is the only answer choice that fits with this.

36.  F  Examine Figure 1; the least increase in the percent of hemoglobin saturated with O2 occurs at the lowest temperatures where lines just begin to increase. Choice (F) results in a change of approximately 5%. Choices (G) through (J) will all result in changes greater than 10%, and since you are looking for the range of pressures that gives the least increase, choice (F) is the best answer. If you picked choice (J), be careful—this one has the greatest increase.

37.  C  First, examine Figure 1 closely. At higher temperatures, oxygen binding decreases, so answer choices (A) and (B) may be eliminated since their binding will be greater at 37oC than at 42oC for the same oxygen pressures. Furthermore, at lower pressures of oxygen, less hemoglobin binding occurs, so you can eliminate choice (D). Choice (C) has the best match of a high temperature and a low pressure.

38.  J  Look at Figure 1. At 65% saturation of the hemoglobin, any pressure of oxygen greater than 75 mmHg requires a temperature greater than 42oC. Subsequently, at a pressure of 100 mmHg, the temperature must be considerably greater than 42oC, correlating to answer choice (J).

39.  B  Examine Figure 2. The closest data point you have to those given in the problem is 75 mmHg pressure of O2 at approximately 80% hemoglobin saturation. Since the problem asks for a 70% saturation, note the relationship between CO2 pressures and percent of hemoglobin saturation—it is clear from the graph that as oxygen is kept at constant pressure, hemoglobin saturation increases with decreasing CO2 pressures. You will need something slightly larger than 40 mmHg, so you can eliminate choice (A), and you can eliminate choices (C) and (D) because they go too far in the other direction. Only (B) is appropriately close to the given data point.

40.  F  Look at the curve for a carbon dioxide pressure of 90 mmHg in Figure 2; the hemoglobin fails to saturate at oxygen pressures below 70 mmHg. This means that the percent of hemoglobin saturated remains constant at any pressure below 70 mmHg. This question asks for how the percentage changes as O2 pressure changes from 45 mmHg to 90 mmHg, and hemoglobin binding remains constant at first and then increases.