Kaplan Full Length MCAT

Document Sample
Kaplan Full Length MCAT Powered By Docstoc
					01 MCAT FL Test1

06/26/2003

05:29 PM

Page 1

Physical Sciences
Time: 100 Minutes Questions 1–77

DO NOT BEGIN THIS SECTION UNTIL YOU ARE TOLD TO DO SO.

01 MCAT FL Test1

06/26/2003

05:29 PM

Page 2

PHYSICAL SCIENCES
DIRECTIONS: Most of the questions in the Physical Sciences test are organized into groups, with a descriptive passage preceding each group of questions. Study the passage, then select the single best answer to each question in the group. Some of the questions are not based on a descriptive passage; you must also select the best answer to these questions. If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain. Indicate your selection by blackening the corresponding circle on your answer sheet. A periodic table is provided below for your use with the questions.
PERIODIC TABLE OF THE ELEMENTS
1 H 1.0 3 Li 6.9 11 Na 23.0 19 K 39.1 37 Rb 85.5 55 Cs 132.9 87 Fr (223) 4 Be 9.0 12 Mg 24.3 20 Ca 40.1 38 Sr 87.6 56 Ba 137.3 88 Ra 226.0 21 Sc 45.0 39 Y 88.9 57 La * 138.9 89 Ac † 227.0 22 Ti 47.9 40 Zr 91.2 72 Hf 178.5 104 Rf (261) 58 Ce 140.1 90 Th 232.0 23 V 50.9 41 Nb 92.9 73 Ta 180.9 105 Ha (262) 59 Pr 140.9 91 Pa (231) 24 Cr 52.0 42 Mo 95.9 74 W 183.9 106 Unh (263) 60 Nd 144.2 92 U 238.0 25 Mn 54.9 43 Tc (98) 75 Re 186.2 107 Uns (262) 61 Pm (145) 93 Np (237) 26 Fe 55.8 44 Ru 101.1 76 Os 190.2 108 Uno (265) 62 Sm 150.4 94 Pu (244) 27 Co 58.9 45 Rh 102.9 77 Ir 192.2 109 Une (267) 63 Eu 152.0 95 Am (243) 64 G d 157.3 96 Cm (247) 65 T b 158.9 97 Bk (247) 66 Dy 162.5 98 Cf (251) 67 Ho 164.9 99 Es (252) 68 Er 167.3 100 Fm (257) 69 Tm 168.9 101 Md (258) 70 Y b 173.0 102 No (259) 71 Lu 175.0 103 Lr (260) 28 Ni 58.7 46 Pd 106.4 78 Pt 195.1 29 Cu 63.5 47 Ag 107.9 79 Au 197.0 30 Zn 65.4 48 Cd 112.4 80 Hg 200.6 5 B 10.8 13 Al 27.0 31 Ga 69.7 49 In 114.8 81 Tl 204.4 6 C 12.0 14 Si 28.1 32 Ge 72.6 50 Sn 118.7 82 Pb 207.2 7 N 14.0 15 P 31.0 33 As 74.9 51 Sb 121.8 83 Bi 209.0 8 O 16.0 16 S 32.1 34 Se 79.0 52 Te 127.6 84 Po (209) 9 F 19.0 17 Cl 35.5 35 Br 79.9 53 I 126.9 85 At (210) 2 He 4.0 10 Ne 20.2 18 Ar 39.9 36 K r 83.8 54 Xe 131.3 86 Rn (222)

*

†

GO ON TO THE NEXT PAGE. 2

01 MCAT FL Test1

06/26/2003

05:29 PM

Page 3

Passage I (Questions 1–6) The equation of state of an ideal gas is given by the ideal gas law: PV = nRT where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas. The gas particles in a container are constantly moving at various speeds. These speeds are characterized by the Maxwell shown in the figure below.

The average distance a particle travels between collisions is known as the mean free path l. Intuitively, the mean free path (mfp) could be expected to be larger for gases at low pressure, since there is a lot of space between particles. Similarly, the mfp should be larger when the gas particles are small. The following expression for the mfp shows this to be correct. l= kT 2πs2P

Equation 4 In this equation, s is the atomic diameter (typically on the order of 10–8), k is the Boltzmann constant, and P is the pressure. In addition to colliding with one another, gas particles also collide with the walls of their container. If the container wall has a pinhole that is small compared to the mfp of the gas, and a pressure differential exists across the wall, the particles will effuse (or escape) through this pinhole without disturbing the Maxwellian distribution of the particles. The rate of effusion can be described by:

fraction of molecules

speed

u

neff t

=

If two particles collide, their velocities change. However, if the gas is in thermal equilibrium, the velocity distribution of the gas as a whole will remain unchanged by the collision. The average kinetic energy (E) of a gas particle is given by: E = (1/2) mu2 Equation 1 where m is the mass of one particle and u is the root mean square speed (rms speed) of the gas particles: (i.e., u = [(v12 + v22 + ... + vn2)/N]1/2, where N is the number of gas particles; this is different from the average speed). For an ideal gas, the kinetic energy of all the particles is: Etotal = (3/2)nRT Equation 2 where n is the number of moles of gas. Combining these equations gives: u = (3RT/M)1/2 Equation 3 where M is the molar mass of the gas particles.

A(P – P1) 2 MRT

Equation 5 where neff is the number of moles of effusing particles, A is the area of the pinhole, P and P1 are the pressures on the inside and outside of the container wall respectively, and P > P 1.

1. Which of the following gives values for both standard temperature and pressure? A. B. C. D. 273 K and 760 Torr 273 K and 1 atm 0°C and 760 mm Hg All of the above

GO ON TO THE NEXT PAGE. 3

01 MCAT FL Test1

06/26/2003

05:29 PM

Page 4

2. If a pinhole were made in a container containing a mixture of equal amounts of H2, O2, N2 and CO2, which gas would have the fastest effusion rate? A. B. C. D. H2 O2 N2 CO2

5. The average kinetic energy of an ideal gas can be directly related to the: A. B. C. D. rms speed. temperature. Boltzmann constant. universal gas constant.

3. The mean free path of a gas will be longer if the : A. B. C. D. pressure of the gas is increased. number of gas particles per unit volume is increased. distance between collisions is decreased. pressure of the gas is decreased.

6. Which of the following will have the smallest root mean square speed at 298K? A. B. C. D. Cl2(g) O2(g) CO2(g) N2(g)

4. What is the relative rate of effusion for a mixture of two noble gases, GA and GB, which escape through the same pinhole? A. B. 1 PA MB PB MA MB MA PB MB PA MA

C.

D.

GO ON TO THE NEXT PAGE. 4

01 MCAT FL Test1

06/26/2003

05:29 PM

Page 5

Passage II (Questions 7–11) The periodic beating of the heart is controlled by electrical impulses that originate within the cardiac muscle itself. These pulses travel to the sinoatrial node and from there to the atria and the ventricles, causing the cardiac muscles to contract. If a current of a few hundred milliamperes passes through the heart, it will interfere with this natural system, and may cause the heart to beat erratically. This condition is known as ventricular fibrillation, and is life-threatening. If, however, a larger current of about 5 to 6 amps is passed through the heart, a sustained ventricular contraction will occur. The cardiac muscle cannot relax, and the heart stops beating. If at this point the muscle is allowed to relax, a regular heartbeat will usually resume. The large current required to stop the heart is supplied by a device known as a defibrillator. A schematic diagram of a defibrillator is shown below. This device is essentially a “heavy-duty” capacitor capable of storing large amounts of energy. To charge the capacitor quickly (in 1 to 3 seconds), a large DC voltage must be applied to the plates of the capacitor. This is achieved using a step-up transformer, which creates an output voltage that is much larger than the input voltage. The transformer used in this defibrillator has a step-up ratio of 1:50.

7. If the defibrillator has a capacitance of 10 µF, how much charge will build up on the two plates? A. B. C. D. 0.08 C 1.6 × 10–3 C 6.25 × 10–8 C 1.25 × 10–9 C

8. The resistance between the two electrodes when placed apart on the patient’s chest is 1,000 Ω when wetting gel is used. What is the initial current through the patient’s heart, assuming that all the current takes this path? A. B. C. D. 0.16 A 4A 6.25 A 8A

9. The plates of the capacitor are originally separated by a vacuum. If a dielectric κ > 1 is introduced between the plates of the capacitor, and the capacitor is allowed to charge up, which of the following statements is/are true? I. The capacitance of the capacitor will increase. II. The voltage across the capacitor plates will increase. III. The charge stored on the capacitor will increase. A. I only B. I and II only C. II and III only D. I and III only

The AC voltage that is obtained from the transformer must then be converted to DC voltage in order to charge the capacitor. This is accomplished using a diode, which allows current flow in one direction only. Once the capacitor is fully charged, the charge remains stored until the switch is moved to position B and the plates are placed on the patient’s chest. To cut down the resistance between the patient’s body and the defibrillator, the electrodes are covered with a wetting gel before use. Care must be taken to insure that the patient is not in electrical contact with the ground while the defibrillator is in use.

GO ON TO THE NEXT PAGE. 5

01 MCAT FL Test1

06/26/2003

05:29 PM

Page 6

10. Why is it important to insure that the patient is not in electrical contact with the ground while the defibrillator is in use? A. B. Contact with the ground will decrease the resistance across the patient’s body. The doctor administering the treatment will be in greater danger of receiving an electric shock if the patient is in electrical contact with the ground. Contact with the ground will cause a smaller current to pass through the patient’s heart. The patient receiving the treatment will be in greater danger of receiving burns due to the high current density if he is in electrical contact with the ground.

C. D.

11. If a dielectric was inserted between the plates of the capacitor in the defibrillator when the switch is in position A: A. B. C. D. the energy increase. the energy decrease. the electric increase. the electric decrease. stored in the capacitor would stored in the capacitor would field between the plates would field between the plates would

GO ON TO THE NEXT PAGE. 6

01 MCAT FL Test1

06/26/2003

05:29 PM

Page 7

Passage III (Questions 12–18) Many nutrients required by plants exist in soil as basic cations: Mg2+, Mn2+, and Ca2+. A soil’s cation-exchange capacity is a measure of its ability to adsorb these basic cations as well as exchangeable hydrogen and aluminum ions. The cation-exchange capacity of soil is derived from two sources: small clay particles called micelles consisting of alternating layers of alumina and silica crystals, and organic colloids. Replacement of A13+ and Si4+ by other cations of lower valence creates a net negative charge within the inner layers of the micelles. This is called the soil’s permanent charge. For example, replacement of an atom of aluminum by calcium within a section where the net charge was previously zero, as shown below, produces a net charge of –1, to which other cations can become adsorbed. O2–Al3+OH–

Due to the buffering effect of the soil’s cationexchange capacity, just measuring the soil solution’s pH will not indicate how much base is needed to change the soil pH. In another experiment, measured amounts of acid and base were added to 10-gram samples of well-mixed soil that had been collected from various locations in a field. The volumes of the samples were equalized by adding water. The results were recorded in Figure 2.

8 pH 6 4 0.8 0.4 0 0.4 0.8

→

meq acid O2–CA2+OH– Figure 2

meq base

Figure 1 A pH-dependent charge develops when hydrogen dissociates from hydroxyl moieties on the outer surfaces of the clay micelles. This leaves negatively-charged oxygen atoms to which basic cations may adsorb. Likewise, a large pH-dependent charge develops when hydrogen dissociates from carboxylic acids and phenols in organic matter. In most clays, permanent charges brought about by substitution account for anywhere from half to nearly all of the total cation-exchange capacity. Soils very high in organic matter contain primarily pH-dependent charges. In a research study, three samples of soil were leached with a 1 N solution of neutral KCl, and the displaced A13+ and basic cations measured. The sample was then leached again with a buffered solution of BaCl2 and triethanolamine at pH 8.2, and the displaced H+ measured. Table 1 gives results for three soils tested by this method. Table 1 (meq/100 g) pH Al3+ Basic Cations 1.9 16.3 9.8 H+ Total Cation Exchange Capacity 47.6 37.4 18.1 14. Which soil from Table 1 most likely has the highest percentage of organic matter? A. B. C. D. I II III Cannot be determined

12. Which column(s) in Table 1 represent(s) the permanent charge of the soil micelles? A. B. C. D. Al3+ H+ Al3+ and Basic Cations Al3+ and H+

13. What percentage of the cation exchange capacity of Sample I is base-saturated? A. B. C. D. 4% 6% 29% 40%

Sample I Sample II Sample III

4.5 5.3 6.0

11.7 1.6 0.5

34.0 19.5 7.8

GO ON TO THE NEXT PAGE. 7

01 MCAT FL Test1

06/26/2003

05:29 PM

Page 8

15. What would be the effect of leaching the three soil samples in Table 1 with a buffered BaCl2 solution at pH 9.5 instead of 8.3? A. B. C. D. The measured permanent charge would greater. The measured pH-dependent charge would greater. The measured permanent charge would smaller. The measured pH-dependent charge would smaller. be be be be

18. Anaerobic organisms are able to denitrify wet soils by the following metabolic pathway. HNO3 → HNO2 → H2N2O2 → N2O(g) → N2(g) If all the oxygen in the nitric acid is converted to water, how many additional equivalents of acid will be consumed during the production of 5 M of nitrogen? A. B. C. D. 20 30 40 50

16. The amount of soil on a particular one-acre field down to a depth of one furrow slice weighs 9 × 105 kilograms. Based on Figure 2, how many kilograms of CaCO3 would have to be added to this field to raise the pH from 5 to 6? A. B. C. D. 900 kg 1800 kg 9 × 105 kg 1.8 × 106 kg

Questions 19 through 24 are NOT based on a descriptive passage.
19. Solution X boils at 100.26°C and solution Y boils at 101.04°C. Both solutions are at atmospheric pressure and contain the same solute concentration. Which of the following conclusions can be drawn? A. The freezing point of solution X is lower than that of solution Y. The vapor pressure of solution X is higher than that of solution Y at 100.26°C. Solution X and solution Y are immiscible. The vapor pressure of solution X is lower than that of solution Y at 100.26°C.

17. Which of the following would probably NOT displace Al3+ in soil micelles? A. B. C. D. Na+ Mg2+ Si4+ Cr2+

B. C. D.

20. A converging lens has a focal length of 8 cm. If the object is 10 cm to the left of the lens, what are the position of the image formed and the magnification of the lens? A. B. C. D. 0.025 cm to the right of the lens and 0.0025× 4.4 cm to the right of the lens and 0.4× 40 cm to the right of the lens and 4× 40 cm to the left of the lens and 4×

GO ON TO THE NEXT PAGE. 8

01 MCAT FL Test1

06/26/2003

05:29 PM

Page 9

21. If 29 g of maleic acid (C4O4H4) is dissolved in 500 g of ammonia (NH3), what is the molality of the resulting solution? A. B. C. D. 0.05 m 0.10 m 0.25 m 0.50 m

24. A body is dropped from a height of 30 m on Earth and hits the ground with a velocity ve. The body is then taken to the Moon, which has a gravitational acceleration 1/6 that of Earth. It is again dropped from a height of 30 m, hitting the Moon with a velocity of vm. What is the ratio of vm/ve? A. B. C. D. 1/6 1/6 6 36

22. An electron travels in the plane of the page from left to right, perpendicular to a magnetic field that points into the page. The direction of the resulting magnetic force on the electron will be in the plane of the page and: A. B. C. D. upwards. downwards. to the left. to the right.

23. How much solid NaOH is required to neutralize 700 mL of 2 N HNO3? A. B. C. D. 40 g 48 g 56 g 64 g

GO ON TO THE NEXT PAGE. 9

01 MCAT FL Test1

06/26/2003

05:29 PM

Page 10

Passage IV (Questions 25–30) When light in the ultraviolet region of the spectrum is shone on a type of material known as a phosphor, it fluoresces and emits light in the visible region of the spectrum. Lamps that utilize this property, known as fluorescent lamps, are very efficient light sources. The arrangement of a typical fluorescent lamp is shown below. The lamp is a glass tube whose inside walls are covered with a phosphor. The tube has an appreciable length-to-diameter ratio so as to reduce the power losses at each end, and it is filled with argon gas mixed with mercury vapor. Inside each end of the tube are tungsten electrodes covered with an emission material. Electrons are liberated at the cathode and accelerated by an applied electric field. These free electrons encounter the gas mixture, ionizing some mercury atoms and exciting others. Since it requires more energy to ionize the atoms than to excite the electrons, more excitation than ionization occurs. When the excited electrons revert to their ground state, they radiate ultraviolet photons with a wavelength of 253.7 nm. These photons impinge on the phosphor coating of the tube and excite electrons in the phosphor to higher energy states. The excited electrons in the phosphor return to their ground state in two or more steps, producing radiation in the visible region of the spectrum. Not every fluorescent lamp emits the same color of radiation; the color is dependent on the relative percentages of different heavy metal compounds in the phosphor. The fluorescent lamp shown operates at 100 volts and draws 400 milliamps of current during normal operation. Of the total power that the lamp consumes, only 25% is converted to light, while the remaining 75% is dissipated as heat. This energy keeps the lamp at its optimum working temperature of 40°C. In the lamp shown, the phosphor coating is calcium metasilicate, which emits orange to yellow light.

25. The photons emitted by the mercury vapor have energies: A. B. C. D. equal to the energies of the electric current. equal to the voltage across the tube. equal to the energy differences between electron orbitals in the mercury atom. less than or equal to the energy differences between the electron orbitals of the mercury atom.

26. If the fluorescent light is left on for 4 hours, how much useful energy is emitted as light? A. B. C. D. 144 kJ 432 kJ 576 kJ 900 kJ

27. As the excited electrons in the coating drop back to their ground states in more than one step, they will emit light of: A. B. C. D. higher frequency than the light absorbed. longer wavelength than the light absorbed. the same wavelength as the light absorbed. greater energy than the light absorbed.

GO ON TO THE NEXT PAGE. 10

01 MCAT FL Test1

06/26/2003

05:29 PM

Page 11

28. In the phosphor coating, an electron falls from an excited state to a lower energy state, emitting a photon with an energy of 2.07 eV. What is the wavelength of the light emitted by the fluorescent tube? (Note: Planck’s constant h = 4.14 10–15 eV•s, and c = 3 108 m/s.) A. B. C. D. 300 nm 600 nm 900 nm 1242 nm

30. The lamp also emits a small proportion of ultraviolet light in addition to the light emitted in the visible spectrum. This ultraviolet light is incident on a metal that has a work function, which is the minimum energy necessary to free an electron, of 2.00 eV. What will be the kinetic energy of an electron that is ejected from the metal if the frequency of the incident light is 1.2 1015 Hz? (Note: h = 4.14 10-15 eV•s.)? A. B. C. D. 9.936 eV 6.948 eV 4.968 eV 2.968 eV

29. Some fluorescent light bulbs are observed to glow for a short period after their power supply has been turned off. This glow is generated mainly by: A. B. C. D. the incandescence of the hot ionic gas within the bulb surface. emission of light stored as vibrational kinetic energy in the phosphor coating. the dissipation of electric charge built up on the bulb’s surface. electrons returning to the ground state from excited states after the power was shut off.

GO ON TO THE NEXT PAGE. 11

01 MCAT FL Test1

06/26/2003

05:29 PM

Page 12

Passage V (Questions 31–36) Every atomic orbital contains plus and minus regions, defined by the value of the quantum mechanical function for electron density. When orbitals from different atoms overlap to form bonds, an equal number of new molecular orbitals results. These are of two types: σ or π bonding orbitals, formed by overlap between orbital regions with the same sign, and antibonding σ* or π* orbitals, formed by overlap between regions with opposite signs. Bonding orbitals have lower energy than their component atomic orbitals, and antibonding orbitals have higher energy. The electron pairs reside in the lower-energy bonding orbitals; the higher-energy, less stable orbitals remain empty when the molecule is in its ground state. A benzene ring has six unhybridized pz orbitals (one from each carbon atom), which together from six molecular π orbitals, each one delocalized over the entire ring. Of the possible π orbital structures for benzene, the one with the lowest energy has the plus region of all six p orbital functions on one side of the ring. The six electrons occupying the orbitals fill the three most stable molecular orbitals, leaving the other three empty. Molecular orbitals are filled from the lowest to the highest energy level. The number of bonds between atoms is determined by the number of filled bonding orbitals minus the number of filled antibonding orbitals; each antibonding orbital cancels out a filled bonding orbital. For a diatomic molecule, orbitals in the n = 2 energy level are filled as follows: σ2s, σ*2s, σ2p , π2px and π2py (equal in z energy), π*2px and π*2py (equal in energy), σ*2pz. (The designation of the three p orbitals as px, py, and pz are interchangeable.) Absorption of a photon can raise an electron to a higherenergy molecular orbital. The excited electron does not immediately change its spin, which is opposite to that of the electron with which it was previously paired. This singlet state is relatively unstable: the molecule may interact with another molecule, or fluoresce and return to its ground state. Alternatively, there may be a change in spin direction somewhere in the system; the molecule then enters the so-called triplet state, which generally has lower energy. The molecule now cannot return quickly to its ground state, since the excited electron no longer has a partner of opposite spin with which to pair. It also cannot return to the singlet state, because the singlet has greater energy. Consequently, the triplet state, which has two unpaired electrons in separate orbitals, is long-lived by atomic standards, with a lifetime that may be ten seconds or more. During this period, the molecule is highly reactive.

31. Which of the following four depictions of molecular π orbitals represents the highest energy state for a 6carbon polyene molecule? (The signs given are the signs for the mathematical functions defining the p orbitals on one side of the molecule.) A. B. C. D. –––––– +++––– ++––++ +–+–+–

32. Among conjugated polyenes (molecules with alternating carbon-carbon double and single bonds) why are those that are longer able to absorb longer wavelengths of light? A. B. C. D. Larger molecular orbitals have a lower ground state. A longer wavelength is better able to interact with a longer molecular orbital. The larger number of molecular orbitals allows for smaller energy transitions. Larger molecular orbitals can absorb more energy.

33. Given the order in which orbitals are filled, which molecule is a triplet in its ground state? A. B. C. D. H2 O2 N2 F2

GO ON TO THE NEXT PAGE. 12

01 MCAT FL Test1

06/26/2003

05:29 PM

Page 13

34. Molecular orbitals in hydrocarbons are formed between the 1s atomic orbital of hydrogen and the sp, sp2, or sp3 hybrid atomic orbitals of carbon. Which choice correctly lists the energy level of the C-H bonds, from lowest to highest? A. B. C. D. C6H6, HC≡CH, CH4 H2C=CH2, CH4, C6H6 C6H6, CH4, H2C=CH2 HC≡CH, C6H6, CH4

36. The quantum number that distinguishes the px orbital from the py orbital is called the: A. B. C. D. azimuthal quantum number. magnetic quantum number. principal quantum number. spin quantum number.

35. Which of the following figures describes the shape of σ*2pz molecular orbital?

A.

B.

C.

D.

GO ON TO THE NEXT PAGE. 13

01 MCAT FL Test1

06/26/2003

05:29 PM

Page 14

Passage VI (Questions 37–42) A ski jump is an inclined track from which a ski jumper takes off through the air. After traveling down the track, the skier takes off from a ramp at the bottom of the track. The skier lands farther down on the slope. Figure 1 shows a ski jump, in which the ramp at the lower end of the track makes an angle of 30° to the horizontal. The track is inclined at an angle of to the horizontal and the slope is inclined at an angle of 45° to the horizontal. A ski jumper is stationary at the top of the track. Once the skier pushes off, she accelerates down the track, and then takes off from the ramp. The vertical height difference between the top of the track and its lowest point is 50 m, and the vertical height difference between the top of the ramp and its lowest point is 10 m.

37. How would the speed of a skier leaving the jump ramp change if the vertical height of the jump ramp were increased from its original height of 10 meters? A. B. C. D. increase decrease remain the same The answer depends on the incline angle of the jump ramp.

38. Another ski jumper sets off from a point farther down the jump track, and leaves the ramp at a speed of 16 m/s. If the time in flight is 4 s, what is the total horizontal distance traveled by the ski jumper after leaving the ramp? A. B. C. D. 4m 8 3m 32 3 m 48 m

39. Which of the following would increase the jump distance? I. Increasing the vertical height h of the jump track II. Increasing the angle of incline of the jump track III. Carrying extra weight to increase the total mass of the ski jumper A. I only B. I and II only C. II and III only D. I and III only

Figure 1 The distance traveled by the skier between leaving the ski jump ramp and making contact with the slope is called the jump distance. In some cases, in order to increase the jump distance a skier will jump slightly upon leaving the ramp, thereby increasing the vertical velocity. Unless otherwise stated, assume that friction between the skis and the slope is negligible, and ignore the effects of air resistance.

GO ON TO THE NEXT PAGE. 14

01 MCAT FL Test1

06/26/2003

05:29 PM

Page 15

40. How would the work done by gravity on the skier when she skis down the track compare with the work done by gravity on the skier if she fell the same vertical height? A. B. C. D. Less work would be done on the skier when she skis down the track. More work would be done on the skier when she skis down the track. Equal amounts of work would be done. The answer depends on the angle of the track.

42. If a skier uses skis of greater surface area, which of the following would occur? A. B. C. D. The normal force of the slope on the skier would increase. The normal force of the slope on the skier would decrease. The pressure exerted on the slope by the skis would increase. The pressure exerted on the slope by the skis would decrease.

41. What is the acceleration of an 80-kg skier going down the track if θ = 45°? A. B. C. D. 6.9 m/s2 9.8 m/s2 13.9 m/s2 80 m/s2

GO ON TO THE NEXT PAGE. 15

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 16

Questions 43 through 47 are NOT based on a descriptive passage.
43. Suppose an α-particle starting from rest is accelerated through a 5 megavolt potential difference. What is the final kinetic energy of the α-particle? (Note: Assume that e = 1.6 10–19 C.) A. B. C. D. 1.6 8.0 6.4 3.2 10–12 J 10–13 J 10–26 J 10–26 J

46. Which titration curve would be produced by titrating 25 mL of a 0.1 N weak base with a 0.1 N strong acid? A.

B.

44. Based on the table below, what is the cell voltage for the following reaction? Fe2O3 + 2 Al → 2 Fe + Al2O3 Half-Reaction Fe2+ + 2e– → Fe Fe3+ + 3e– → Fe 2H2O + 2e– → H2 + 2OH– Al3+ + 3e– → Al A. B. C. D. –1.33 V 1.99 V 1.33 V 1.62 V Standard Potential (V) –0.44 –0.037 –0.83 –1.66 D. C.

45. A particle of mass m moves in a circle of radius r at a uniform speed and makes 1 revolution per second. What is the energy of the particle? A. B. C. D. m2r2/4π2 2π2mr2 4π2mr2 mr2/2

GO ON TO THE NEXT PAGE. 16

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 17

47. Four charges of equal magnitude but different sign are arranged in the four corners of a square, as shown below. What is the direction of the electric field in the center of the square?

A. B. C. D.

A B C D

GO ON TO THE NEXT PAGE. 17

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 18

Passage VII (Questions 48–54) Several techniques have been developed to determine the order of a reaction. The rate of a reaction cannot be predicted on the basis of the overall equation, but can be predicted on the basis of the rate-determining step. For instance, the following reaction can be broken down into three steps. A+D→F+G Step 1 Step 2 Step 3 A→B+C (slow) B + D → E + F (fast) E+C→G (fast) Reaction 1 In this case, the first step in the reaction pathway is the rate-determining step. Therefore, the overall rate of the reaction must equal the rate of the first step, k1[A] where k1 is the rate constant for the first step. (Rate constants of the different steps are denoted by kx, where x is the step number.) In some cases, it is desirable to measure the rate of a reaction in relation to only one species. In a second-order reaction, for instance, a large excess of one species is included in the reaction vessel. Since a relatively small amount of this large concentration is reacted, we assume that the concentration essentially remains unchanged. Such a reaction is called a pseudo first-order reaction. A new rate constant, k', is established, equal to the product of the rate constant of the original reaction, k, and the concentration of the species in excess. This approach is often used to analyze enzyme activity. In some cases, the reaction rate may be dependent on the concentration of a short-lived intermediate. This can happen if the rate-determining step is not the first step. In this case, the concentration of the intermediate must be derived from the equilibrium constant of the preceding step. For redox reactions, the equilibrium can be correlated with the voltage produced by two half-cells by means of the Nernst equation. This equation states that at any given moment: E = E° – (RT/nF)ln([C]c[D]d/[A]a[B]b) Equation 1 when aA+bB→cC+dD Reaction 2 Note: R = 8.314 J/K•mol; F = 9.6485 × 104 C/mol.)

48. An enzyme, R, catalyzes the oxidation of A to B. Reacting various concentrations of A and B with a large excess of R produced the following results during the first few minutes of the reaction.

M

M

0.2

0.2

1M ; [B ]=

rate

2M ; [B ]=

[A ]=

Which of the following is the best tentative rate equation? A. B. C. D. Rate = k'[A]x Rate = k'[B]y Rate = k'[A]x[B]y Rate = k[A]x[[B]y[R]z

49. In a test of the rate of Step 3 of Reaction 1, a solution is prepared containing a 0.1 M concentration of E and a 50 M concentration of C. The rate is calculated after the reaction has gone 50% to completion. By what percent will the calculated rate differ from the true rate if we treat the reaction as pseudo first-order? A. B. C. D. 0.02% 0.05% 0.1% 0.2%

50. If Step 2 above were the rate-determining step of Reaction 1, which of the following equations would correctly define the rate? A. B. C. D. Rate = k1k2[D]/k–1[C] Rate = k1k2[D]/k–1k–2[C] Rate = k1k2[A][D]/k–1[C] Rate = k1k2[A][D]/k–1k–2[C]

18

[A ]=

time

GO ON TO THE NEXT PAGE.

[A ]=

2M ; [B ]=

0.1

M

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 19

51. Which of the following is true of a reaction at equilibrium? I. k1/k–1 = 1 II. E = E° III. ln([C]c[D]d/[A]a[B]b) = nFE°/RT A. I only B. III only C. I and II only D. I, II, and III

54. Catalysts are effective in increasing the rate of a reaction because they: A. B. C. D. increase the energy of the activated complex. increase the value of the equilibrium constant. decrease the number of collisions between reactant molecules. lower the activation energy.

52. What is the effect of increasing the concentration of reactants in a voltaic cell? A. B. C. D. The voltage increases, while the spontaneity of the reaction remains the same. The spontaneity of the reaction increases, but the voltage remains the same. Both the voltage and the spontaneity of the reaction increase. The reaction rate increases, but the voltage and spontaneity of the reaction are unchanged.

53. What would be the cell emf of the following system at 298K? Zn(s)|Zn2+(0.2 M)||Cu2+(0.02 M|Cu(s) E°cell = +1.10 V A. B. C. D. 1.07 V 1.10 V 1.13 V 1.20 V

GO ON TO THE NEXT PAGE. 19

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 20

Passage VIII (Questions 55–61) X-rays are produced by a device which beams electrons with an energy between 103 and 106 eV at a metal plate. The electrons interact with the metal plate and are stopped by it. Much of the energy of the incoming electrons is released in the form of X-rays, which are highenergy photons of electromagnetic radiation. An example of such a device is shown below. Electrons are accelerated from the cathode towards the anode by an electric field.

55. An electron is accelerated through a distance of 0.1 m by a potential difference of 10,000 volts. What is the electron’s energy as it strikes the anode? A. B. C. D. 100 eV 1,000 eV 10,000 eV 1J

56. What is the direction of the electric field that accelerates the electrons? A. B. C. D. From the anode toward the cathode From the cathode toward the anode Into the page Out of the page

57. How does the wavelength of an X-ray produced from a K-alpha transition in molybdenum compare to that produced from a lower energy K-alpha transition in copper? There are two mechanisms by which the X-rays are produced within the metal. The first mechanism is called bremsstrahlung, which is German for “breaking radiation.” X-rays are emitted by the electrons as they are brought to rest by interactions with the positive nuclei of the anode. The second mechanism occurs when an incoming electron knocks an inner electron out of one of the metal atoms of the anode. This electron is replaced by an electron from a higher energy level of the atom, and a photon making up the energy difference is emitted. X-rays are absorbed by a material when they pass through it. The amount of X-rays absorbed increases with the density of the material. In addition, lower energy X-rays are more likely to be absorbed than higher energy X-rays. (Note: 1 eV = 1.6 10–19 J; Planck’s constant h = 4.1 10–15 eV•s; speed of light c = 3 108 m/s.) A. B. C. D. It is shorter. It is the same. It is longer. It depends on the energy of the incoming electron.

GO ON TO THE NEXT PAGE. 20

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 21

58. What is the minimum potential difference required to produce a 0.06 nm X-ray from an electron transition in a metal? A. B. C. D. 15,000 V 20,000 V 20,500 V 21,500 V

61. Which of the following graphs best represents the relationship between the amount of X-rays absorbed per unit length of material and the energy of the X-rays, for lead, bone, and air? A. C.

59. An X-ray source produces X-rays with a maximum frequency of 6 1018 Hz. If the cathode current is doubled so that twice as many electrons are emitted per unit time, what is the new maximum frequency of the X-rays produced? A. B. C. D. 3 1018 Hz 6 1018 Hz 12 1018 Hz 24 1018 Hz

B.

D.

60. In an X-ray tube, electrons of charge e are accelerated through a potential difference of V. The anode is cooled by water of mass m with specific heat c. If n electrons per second strike the anode, what is the maximum possible rise in the temperature of the water after 100 s? A. B. C. D. nVe/100mc 100Ve/mc 100n(Ve + mc) 100nVe/mc

GO ON TO THE NEXT PAGE. 21

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 22

Passage IX (Questions 62–67) Compound A researcher in a molecular biology lab planned to carry out an extraction procedure known as an alkaline plasmid prep, which is designed to purify plasmids, small pieces of the hereditary material DNA, from bacterial cells. The bacteria are first placed into a test tube containing liquid nutrient medium and allowed to grow until they reach a high population density. The culture, which consists of solid cells suspended in the medium, is then centrifuged; a solid pellet is formed. The supernatant is poured out, leaving the pellet behind, and the cells are resuspended in a mL of lysis buffer solution (50 mM glucose, 25 mM Tris buffer and 10 mM ethylenediaminetetraacetic acid (EDTA), with 5 mg of the enzyme lysozyme added). They are then incubated for 30 minutes at 0°C, during which time the bacterial cell walls break down and the cell contents are released into the solution. After incubation, 1 mL of 0.4 N sodium hydroxide and 1 mL of 2% sodium dodecyl sulfate (SDS) are added, and the solution is again incubated on ice for 10 minutes. 2 mL of 3 M sodium acetate are added and the mixture is incubated for 30 minutes at 0°C. The test tube is centrifuged once more and the supernatant is decanted into a clean tube, leaving behind the protein and most other cell components in the pellet. Finally, 10 mL of pure ethanol are added to the supernatant from the previous step to precipitate out the DNA, and the test tube is incubated at –20°C for 60 minutes, during which the mixture remains liquid. The mixture is centrifuged a final time and the supernatant removed. The translucent precipitate that results is washed with 70% ethanol (70% ethanol and 30% water by volume), allowed to dry, and resuspended in 1 mL of TE buffer (10 mM Tris, 1 mM EDTA). In preparation for this experiment, the researcher prepared stock solutions of the various chemicals that she will need in the experiment. Stock solutions are highly concentrated solutions of commonly used chemicals in water from which dilute solutions are prepared for daily use. Table 1 shows the chemicals, their molecular formulas and weights, and the composition of commonly used stock solutions. Tris EDTA

Table 1 Formula (CH2OH)3CNH2 (HOOCCH2)4(CNH2)2

MW

Stock

121 1M (pH 8) 292 0.5 M (pH 8) 40 5 N 288 10% 82 3 M (pH 5.2) 46 95%

Sodium hydroxide NaOH SDS Sodium acetate Ethanol C11H23CH2OSO3–Na+ CH3COO–Na+ CH3CH2OH

62. EDTA is available commercially in the form of a hydrated sodium salt, Na2EDTA • 2H2O. How much of this salt must be used to produce 1 L of a 0.5 M stock solution? A. B. C. D. 145 g 146 g 186 g 187 g

63. Tris (Tris(hydroxymethyl)aminomethane) is generally used as a buffer. If pH 8.0 is a good buffering region for Tris, then: I. the pKa of Tris must be near pH 8.0 II. if Tris is titrated with acid, the titration curve will possess a steep region near pH 8.0. III. a great deal of NaOH would have to be added to pH 8.0 Tris in order to significantly affect the pH. A. I only B. III only C. I and II only D. I and III only

64. What is the molality of a stock solution that is 10% SDS by mass? A. B. C. D. 0.028 m 0.100 m 0.347 m 0.385 m GO ON TO THE NEXT PAGE. 22

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 23

65. Pure ethanol (CH3CH2OH) is difficult to prepare and therefore expensive; 95% ethanol is much cheaper. Consequently, 95% ethanol is generally used in the preparation of dilute ethanol solutions. How much 95% ethanol would be needed to produce a 500 mL solution of 70% ethanol by volume in water? A. B. C. D. 333 mL 350 mL 368 mL 475 mL

67. What would be the pH of 100 mL of the sodium acetate stock solution after the addition of 3.6 g of HCl? (pKa of acetic acid = 4.74) A. B. C. D. 1.0 4.74 5.2 6.0

66. Which of the following conclusions can be reached based on the fact that DNA precipitates in the last step of the plasmid prep procedure? A. B. C. D. DNA dissolves better in water at lower temperatures. DNA is polar and therefore dissolves better in water than in a mixture of water and ethanol. DNA is nonpolar and therefore dissolves better in ethanol than in water. DNA dissolves well in ethanol and precipitates only because the solution is centrifuged.

GO ON TO THE NEXT PAGE. 23

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 24

Passage X (Questions 68–73) The simple harmonic motion of a mass suspended from vertical springs is investigated in two experiments. The springs used in both experiments have a spring constant k and a natural length L0. The material used to make the springs has a Young’s modulus of 2 1011 Pa. In the first experiment a mass m is suspended from a spring. The mass stretches the spring to a new length L, called the equilibrium length. In the second experiment the mass m is suspended from two identical springs as shown in Figure 2 below. When the mass m is in equilibrium, each spring is stretched from its natural length by the same amount xe. In both experiments the masses of the springs are negligible, and the elastic limits of the springs are never exceeded.

69. The mass in the first experiment is pulled down a distance A from its equilibrium position and then released from rest. The mass will then oscillate with simple harmonic motion. As the mass moves up and down, energy is dissipated due to factors such as air resistance and internal heating of the spring. The mass will no longer oscillate when the total energy dissipated equals: A. B. C. D. kL2/2 kA2/2 k(L + A)2/2 kL02/2

70. In the first experiment the mass is pulled down and set into motion. The position of greatest speed is: A. B. C. D. at the equilibrium position. at the position where the spring’s length is its natural length. at the lowest point in its motion. at the highest point in its motion.

71. In the first experiment, when a 5-kg mass is oscillating, the frequency of oscillation is 2 Hz. What is the value of the spring constant? A. B. C. D. 5/π2 N/m 20 N/m 40π2 N/m 80π2 N/m

68. In the first experiment, what is the mass of the object hanging from the spring? A. B. C. D. kL/g kL0/g k(L – L0)/g k/g

GO ON TO THE NEXT PAGE. 24

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 25

72. The two springs in Experiment 2 are replaced by a single spring having a spring constant k’ such that the equilibrium length xe does not change. What is the ratio of k’ to k? A. B. C. D. 1/2 1 2 2

73. If the spring in Experiment 1 was suspended from the ceiling of an elevator accelerating with acceleration a, how would the equilibrium length of the spring compare to the equilibrium length of the spring when the elevator is stationary? A. The equilibrium length of the spring would be greater when the elevator is accelerating upward. The equilibrium length of the spring would be greater when the elevator is stationary. The equilibrium length of the spring would be greater when the elevator is accelerating downward with acceleration smaller than the acceleration due to gravity. The equilibrium length of the spring doesn’t depend on the acceleration of the elevator.

B. C.

D.

GO ON TO THE NEXT PAGE. 25

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 26

Questions 74 through 77 are NOT based on a descriptive passage.
74. If the noise level is increased by 30 decibels, what is the ratio of the new intensity to the original intensity? A. B. C. D. 10 30 100 1000

77. Which of the following is NOT an intermolecular force? A. B. C. D. Dispersion forces Resonance Hydrogen bonding Dipole interactions

STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS SECTION ONLY.

75. A given volume of a diprotic acid is completely neutralized by twice that volume of a 0.3 N NaOH solution. What is the molarity of the acid? A. B. C. D. 0.15 M 0.30 M 0.60 M 1.20 M

76. A fireman of mass m slides down a vertical pole with an average acceleration a. If the acceleration due to gravity is g, what is the average frictional force exerted by the fireman? A. B. C. D. mg m(g + a) m(g – a) ma

26

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 27

Verbal Reasoning
Time: 85 Minutes Questions 78–137

DO NOT BEGIN THIS SECTION UNTIL YOU ARE TOLD TO DO SO.

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 28

VERBAL REASONING
DIRECTIONS: There are nine passages in this Verbal Reasoning Section. Each passage is followed by several questions. After reading a passage, select the one best answer to each question. If you are not certain of an answer, eliminate the alternatives that you know to be incorrect and then select an answer from the remaining alternatives. Indicate your selection by blackening the corresponding oval on your answer document.
Passage I (Questions 78–84)
40

5

10

15

In the early nineteenth century a large number of communal experiments, both secular and religious, sprang up in the northeastern United States. Perhaps the most famous secular commune was Brook Farm, founded by transcendentalists George Ripley and William H. Channing to promote the pursuit of leisure and culture through the proper application of time and labor. Its members (among the more notable were Nathaniel Hawthorne and Margaret Fuller) pursued field labor by day, art and philosophy by night. For a time the system worked so well that two afternoons a week were set aside for leisure and Brook Farm began outcompeting local farmers at the produce market. But by nature the Farm’s members were thinkers, not workers; despite their success they remained mainly interested in the theoretical and philosophical implications of the experiment. Thus, when a devastating fire brought the community considerable financial burdens in its fifth year, the members felt little compunction about closing shop and returning to their comfortable Boston homes. One of the most notable religious utopias was the Oneida community. Its founder, John Humphrey Noyes, believed that Christ’s second coming had already occurred and that everyone alive was favored by Divine grace, which Noyes saw as an imperative to live a better life. Perhaps surprisingly, the Oneidans embraced industry and commerce, achieving success in fruit packing, trap making, and silk thread winding. They owned everything communally, and this principle extended to each other. The Oneidans saw monogamy as a selfish act and asserted that the men and women of the community were united in one “complex” marriage; sex between any two consenting members was perfectly acceptable. The Oneidans maintained order solely through “criticism”—anyone acting out of line was made to stand before the other members and hear his or her faults recounted. Oneida remained viable for some thirty years, until the leadership devolved on Noyes’ son, an agnostic. The old religious fervor died out, and the dream degenerated into a joint stock company.

45

50

55

60 20

Doubtless the most successful communalists were the Shakers, so called for the early propensity to tremble ecstatically during religious worship. Their guiding light, Mother Ann, espoused four key principles: Virgin Purity, Christian Communism, Confession, and Separation from the World. Though the Shakers were less adamant on the last point—maintaining social relations and some commerce with their neighbors—they insisted on the other three, and renounced both personal property and sex. Men and women lived in a single large “Unitary Dwelling” and were considered complete equals, but they occupied separate wings and could speak together only if a third person were present. Despite their religious strictness, Shakers were known as simple, sincere, intelligent people, healthy and long-lived, producers of lovely books and hymns, and of furniture still prized for its quality and durability. In their heyday, six thousand Shakers lived in fifty-eight separate “families” throughout the Northeast. Later their celibacy, combined with their strict discipline, led to a decline in numbers, but even today a small number of elderly Shakers in two communities in Maine and New Hampshire continue to keep the faith.

25

30

78. The passage implies that the end of the Brook Farm experiment was probably brought on by: A. B. C. D. faltering commitment in the face of hardship. a failure to attract members of sufficient intellect or ability. the completion of the community’s aims. the incompetence of philosophers at field labor. GO ON TO THE NEXT PAGE. 28

35

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 29

79. According to the passage, the Oneidans believed that: A. B. C. D. men and women were equal in the eyes of God. monogamy was wrong in principle. rules and standards of behavior were unnecessary. they were destined to witness Christ’s second coming.

82. The Shakers resembled the Oneidans in their attitude toward: A. B. C. D. sexual practices. equality of men and women. personal property. contact with the outside world.

80. The passage implies that Brook Farm’s economic system: A. B. C. D. did not include the selling of produce outside the farm. was based on the hiring of farm hands. efficiently utilized time and labor. was primarily intended to maximize collective profit.

83. It can be inferred from the passage that the cohesion of a secular workers’ cooperative, based on the principles of collective ownership and the sharing of profits, would probably be weakened by: I. diminished contact with the outside world. II. increasing agnosticism. III. considerable economic losses. A. I only B. II only C. III only D. I and II only

81. According to the passage, all of the following were characteristic of the Oneida community EXCEPT: A. B. C. D. complex marriage. maintenance of order through social pressure. belief in present grace. shared living quarters.

84. If the passage were to continue, the next topic the author would discuss would probably be: A. B. C. D. a comparison between nineteenth and twentieth century communal living experiments. a theory explaining why communal living might become popular again. an analysis of why early communes attracted intellectuals and artists. an investigation into why the three communes discussed were successful to varying degrees.

GO ON TO THE NEXT PAGE. 29

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 30

Passage II (Questions 85–90)
50

The time has come to acknowledge the ascendancy of the humanistic psychology movement. The so-called “Third Stream” emerged at mid-century, asserting itself against the opposition of a pair of mighty, long-established 5 currents, psychoanalysis and behaviorism. The hostility between these two older schools, as well as divisiveness within each of them, probably helped enable humanistic psychology to survive its early years. But the movement flourished because of its wealth of insights into the nature 10 of this most inexact science. Of the three major movements in the course of 20thcentury psychology, psychoanalysis is the oldest and most introspective. Conceived by Sigmund Freud as a means of treating mental and emotional disorders, psychoanalysis is based on the theory that people experience unresolved emotional conflicts in infancy and early childhood. Years later, although these experiences have largely disappeared from conscious awareness, they may continue to impair a person’s ability to function in daily life. The patient experiences improvement when the psychoanalyst eventually unlocks these long-repressed memories of conflict and brings them to the patient’s conscious awareness. In the heyday of behaviorism, which occurred between the two world wars, the psychoanalytic movement was heavily criticized for being too concerned with inner subjective experience. Behavioral psychologists, dismissing ideas and feelings as unscientific, tried to deal only with observable and quantifiable facts. They perceived the human being merely as an organism which generated responses to stimuli produced by its body and the environment around it. Patients’ neuroses no longer needed analysis; they could instead by modified by behavioral conditioning. Not even babies were safe: B.F. Skinner devised a container in which infants could be raised under “ideal” conditions—if a sound-proof box can be considered the ideal environment for child-rearing. By mid-century, a number of psychologists had grown dissatisfied with both the deterministic Freudian perspective and the mechanistic approach of behaviorism. They questioned the idea that human personality becomes permanently fixed in the first few years of life. They wondered if the purpose of psychology was really to reduce people to laboratory specimens. Was it not instead possible that human beings are greater than the sum of their parts? That psychology should speak to their search for fulfillment and meaning in life? It is questions like these that members of the Third Stream have sought to address. While the movement can-

55

not be simplified down to a single theoretical position, it does spring from certain fundamental propositions. Humanistic psychologists believe that conscious experience, rather than outward behavior, is the proper subject of psychology. We recognize that each human being is unique, capable of change and personal growth. We see maturity as a process dependent on the establishment of a set of values and the development of self. And we believe that the more aspects of self which are satisfactorily developed, the more positive the individual’s self-image. Abraham Maslow, a pioneer of the Third Stream, articulated a hierarchy of basic human needs, starting with food, water and air, progressing upward through shelter and security, social acceptance and belonging, to love, esteem and self-expression. Progress toward the higher stages cannot occur until all of the more basic needs have been satisfied. Individuals atop the pyramid, having developed their potential to the highest possible extent, are said to be “self-actualized.” If this humanist theoretical perspective is aimed at empowering the individual, so too are the movement’s efforts in the practical realm of clinical psychology. Believing that traditional psychotherapists tend to lead patients toward predetermined resolutions of their problems, Carl Rogers pressed for objective evaluations of both the process and outcome of psychotherapeutic treatment. Not content to function simply as a reformer, Rogers also pioneered the development of “client-centered” or nondirective therapy, which emphasizes the autonomy of the client (i.e., patient). In client-centered therapy, clients choose the subjects for discussion, and are encouraged to create their own solutions to their problems.

60

15

65

20

70

25

75

30

80

35

40

85. If the author of this passage met a Freudian psychoanalyst who felt that it was important for patients to consider themselves capable of fundamental change, he would most likely conclude that the psychoanalyst was: A. B. C. D. opposed to the Third Stream. concerned only with conscious experience. influenced by humanist theory. rejecting Maslow’s hierarchy of human needs. GO ON TO THE NEXT PAGE. 30

45

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 31

86. The author states that “not even babies were safe” (line 35) most probably in order to: A. emphasize that the use of even very young subjects is considered valid among most psychologists. indicate the pervasive influence of behaviorists on the field of psychology. show that behaviorists were anxious to apply their theories to a wide range of subjects. warn of the dangers of psychoanalysis for children.

89. According to the passage, the ultimate goal of Carl Rogers’s client-centered therapy is: A. B. C. D. simplification of the Third Stream’s theoretical perspective. self-directed personal growth for the client. rejection of Maslow’s scheme of self-actualization. increased autonomy of psychotherapists.

B. C. D.

90. Psychoanalysts and humanistic psychologists would be most likely to disagree about: A. B. C. D. the effects of internal conflicts on childhood behavior. the necessity of proper training for psychologists. the relevance and utility of clinical psychology. the significance of conscious experience.

87. The author most probably believes that, in its early days, the humanistic psychology movement: I. benefited from dissension among psychologists. II. acknowledged Maslow and Rogers as its only leaders. III. was an offshoot of behaviorism. A. I only B. II only C. I and II only D. II and III only

88. B.F. Skinner is mentioned in the passage to support the point that: A. B. C. D. the ultimate goal of behaviorism is technological innovation. raising babies in isolation prevents childhood conflicts. stimulus-response conditioning was attempted on all sorts of individuals. behaviorists reject the scientific validity of subjective experience.

GO ON TO THE NEXT PAGE. 31

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 32

Passage III (Questions 91–97) Due to ever-increasing paranoia about the transmission of hepatitis and AIDS via blood transfusions and the frequent difficulty of procuring matching blood donors for patients, researchers have been working at a feverish pace 5 to produce disease-free and easy-to-use blood substitutes. The difficulty most synthetic blood researches have had is in formulating a substance that combines qualities of sterility, high capacity for carrying oxygen to body tissues, and versatility within the human body. Three major substi10 tute technologies have been developed to date; each has certain advantages and shortcomings. “Red blood,” the first of the blood substitute technologies, is derived from hemoglobin which has been recycled from old, dead, or worn-out red blood cells and modified 15 so that it can carry oxygen outside the red blood cell. Hemoglobin, a complex protein, is the blood’s natural oxygen carrier and is attractive to scientists for use in synthetic blood because of its oxygen-carrying capacity. However, hemoglobin can sometimes constitute a two-fold threat to 20 humans when it is extracted from the red blood cell and introduced to the body in its naked form. First, hemoglobin molecules are rarely sterile and often remain contaminated by viruses to which they were exposed in the cell. Second, naked hemoglobin is extremely dangerous to the kidneys, 25 causing blood flow at these organs to shut down and leading, ultimately, to renal failure. Additional problems arise from the fact that hemoglobin is adapted to operate optimally within the intricate environment of the red blood cell. Stripped of the protection of the cell, the hemoglobin 30 molecule tends to suffer breakdown within several hours. Although modification has produced more durable hemoglobin molecules which do not cause renal failure, undesired side effects continue to plague patients and hinder the development of hemoglobin-based blood sub35 stitutes. Another synthetic blood alternative, “white blood,” is dependent on laboratory-synthesized chemicals called perfluorocarbons (PFCs). Unlike blood, PFCs are clear oillike liquids, yet they are capable of absorbing quantities of 40 oxygen up to 50% of their volume, enough of an oxygencarrying potential for oxygen-dependent organisms to survive submerged in the liquid for hours by “breathing” it. Although PFCs imitate real blood by effectively absorbing oxygen, scientists are primarily interested in them as con45 stituents of blood substitutes because they are inherently safer to use than hemoglobin-based substitutes. PFCs do not interact with any chemicals in the body and can be manufactured in near-perfect sterility. The primary pitfall of PFCs is in their tendency to form globules in plasma 50 that can block circulation. Dissolving PFCs in solution can 32

mitigate globulation; however this procedure also seriously curtails the PFCs’ oxygen capacity. The final and perhaps most ambitious attempt to form a blood substitute involves the synthesis of a modified version of human hemoglobin by genetically-altered bacteria. Fortunately, this synthetic hemoglobin seems to closely mimic the qualities of sterility, and durability outside the cellular environment, and the oxygen-carrying efficiency of blood. Furthermore, researchers have found that if modified hemoglobin genes are added to bacterial DNA, the bacteria will produce the desired product in copious quantities. This procedure is extremely challenging, however, because it requires the isolation of the human gene for the production of hemoglobin, and the modification of the gene to express a molecule that works without support from a living cell. While all the above technologies have serious drawbacks and difficulties, work to perfect an ideal blood substitute continues. Scientists hope that in the near future safe synthetic blood transfusions may ease blood shortages and resolve the unavailability of various blood types.

55

60

65

70

91. The author mentions all of the following as weaknesses of synthetic bloods EXCEPT: A. B. C. D. naked hemoglobin can cause renal failure in humans. “red blood” can transmit viruses to a recipient. genetic engineering can be extremely difficult. “white blood” has a low oxygen-carrying potential.

92. According to the passage, PFCs are helpful in the synthesis of blood substitutes because they: I. mimic the oxygen-carrying capacity of blood. II. do not react with other body chemicals. III. break down in the blood within several hours. A. I only B. II only C. I and II only D. II and III only GO ON TO THE NEXT PAGE.

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 33

93. According to the passage, all of the following are reasons for research into the development of synthetic bloods EXCEPT: A. B. dangerous diseases can be transmitted by conventional blood transfusions. synthetic bloods have greater oxygen-carrying capacities than naturally-produced human blood. donor blood is sometimes in short supply. certain blood types are not readily available.

96. According to the passage, how much oxygen can be absorbed by a 300 cc sample of PFC? A. B. C. D. 50 cc 100 cc 150 cc 300 cc

C. D.

97. It can be inferred from the passage that the difficulty of producing an ideal blood substitute is compounded by all of the following EXCEPT: A. B. C. D. there is no known way to isolate the DNA responsible for hemoglobin. naked hemoglobin tends to break down in the bloodstream. non-globulating PFCs have significantly abbreviated oxygen-carrying capacities. the use of PFCs may lead to blood clotting.

94. We can infer that all of the synthetic blood technologies discussed in this passage: A. B. C. D. sustain submerged oxygen-dependent organisms. possess high oxygen-carrying capacities. maintain high standards of sterility. exhibit versatility in the human body.

95. Which of the following is mentioned in the passage as a problem specific to “red blood”? A. B. C. D. “Red blood” cannot be produced in large enough quantities. “Red blood” tends to form globules that block circulation. Hemoglobin does not carry oxygen effectively. “Red blood” exhibits poor durability in the bloodstream.

GO ON TO THE NEXT PAGE. 33

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 34

Passage IV (Questions 98–103) Muzak, the intentionally unobtrusive music that most people associate with elevators and dentists’ waiting rooms, represents the paradoxical success story of a product designed to be ignored. Although few people admit to 5 enjoying its blandly melodic sounds, Muzak reaches over 100 million listeners in 14 countries and has played in the White House, the Apollo lunar spacecraft, and countless supermarkets, offices, and factories. This odd combination of criticism and widespread acceptance is not surprising, 10 however, when one considers that Muzak is not created for the enjoyment of its listeners: rather, its purpose is to modify physiological and psychological aspects of an environment. In the workplace, Muzak is credited with increasing both productivity and profitability. Research into the rela15 tionship between music and productivity can be traced to the earliest days of the Muzak Corporation. Developed by a military officer in 1922 as a way of transmitting music through electrical wires, Muzak blossomed in the 1930’s following a study which reported that people work harder 20 when they listen to certain kinds of music. Impressed by these findings, the BBC began to broadcast music in English munitions factories during World War II in an effort to combat fatigue. When workers assembling weapons increased their output by 6 percent, the U.S. War 25 Production Board contracted the Muzak Corporation to provide uplifting music to American factories. Today, the corporation broadcasts its “Environmental Music” to countless businesses and institutions throughout the world. And while most people claim to dislike Muzak’s discreet 30 cadences, it seems to positively influence both productivity and job satisfaction. Researchers speculate that listening to Muzak and other soft music improves morale and reduces stress by modifying our physiology. Physiological changes such as 35 lowered heart rate and decreased blood pressure have been documented in hospital studies testing the effect of calming music on cardiac patients. In addition, certain kinds of music seem to effect one’s sense of emotional, as well as physical, well being. It is just this sort of satisfaction which 40 is thought to result in increased performance in the workplace. In a study of people performing repetitive clerical tasks, those who listened to music performed more accurately and quickly than those who worked in silence; those who listened to Muzak did better still. Moreover, while 45 Muzak was conceived as a tool for productivity, it also seems to influence a business’ profitability. In an experiment in which supermarket shoppers shopped to the mellow sounds of Muzak, sales were increased by as much as 12 percent.

50

55

60

65

70

What makes Muzak unique is a formula by which familiar tunes are modified and programmed. Careful instrumentation adds to an overall sound that is neither monotonous nor rousing. But it is the precisely timed programming that separates Muzak from other “easy listening” formats. At the core of the programming is the concept of the “Stimulus Progression.” Muzak programs are divided into quarter-hour groupings of songs, and are specifically planned for the time of day at which they will be heard. Each composition is assigned a mood rating between 1 and 6 called a stimulus value; a song with a rating of 2, for example, is slower and less invigorating than one with a value of 5. Approximately six compositions with ascending stimulus values play during any given quarter hour; each 15 minute segment ends in silence. Each segment of a 24-hour program is carefully planned. Segments that are considered more stimulating air at 11 a.m. and 3 p.m. (the times when workers typically tire), while more soothing segments play just after lunchtime and towards the end of the day, when workers are likely to be restless. From the point of view of management, then, Muzak is a useful tool in the effort to maximize both productivity and profits. However, some people object to its presence, labeling it as a type of unregulated air pollution. Still others see it as an Orwellian nightmare, a manipulation of the subconscious. But Muzak’s effectiveness seems to lie in the fact that most people never really listen to it. While it may be true that no one actually likes this carefully crafted aural atmosphere, many simply ignore it, allowing its forgettable sounds to soften the contours of the day.

75

80

98. According to the passage, a 15-minute segment of Muzak with an average stimulus value of 5 would most likely be broadcast at: A. B. C. D. 4:30 p.m. 8:15 a.m. 3:00 p.m. 1:15 p.m.

GO ON TO THE NEXT PAGE. 34

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 35

99. Of the following, the author is most interested in discussing: A. B. C. D. the origins of the Muzak Corporation. how Muzak modifies physical states and psychological atmospheres. how Muzak increases productivity in the workplace. the ways in which Muzak differs from other “easy listening” formats.

102. According to the passage, Muzak differs from other “easy listening” formats in that Muzak: I. produces measurable health benefits. II. improves workers’ job performances. III. is programmed in order to effect behavioral changes. A. I only B. II only C. III only D. II and III only

100. According to the passage, Muzak may provide all of the following benefits EXCEPT: A. B. C. D. increased work productivity. decreased blood pressure. increased business profitability. decreased job absenteeism.

103. It can be inferred from the statements in the passage that the author regards Muzak as: A. B. C. D. a paradoxical phenomenon. an unnecessary evil. a violation of privacy. a pleasurable diversion.

101. It can be inferred from the passage that some critics of Muzak believe that Muzak: A. B. C. D. is not significantly different from other “easy listening” programs. subtly manipulates the subconscious mind. is actually distracting to many workers. caters to the whims of supermarket consumers.

GO ON TO THE NEXT PAGE. 35

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 36

Passage V (Questions 104–110)
50

The Russian wheat aphid, Diuraphis noxia, is a small green insect discovered in southern Russia around the turn of the century. Agricultural researchers are not quite sure, but they believe the Russian aphid adapted itself to wheat 5 about ten thousand years ago, when the crop was first domesticated by man. What is not in doubt is the insect’s destructiveness. Spread by both wind and human transport, the Russian aphid has destroyed wheat fields throughout Asia, Africa, and Latin America. Until a few years ago, the United States had been free of this pest. But in the spring of 1986, a swarm of Russian aphids crossed the Mexican border and settled a few hundred miles north, in central Texas. From there, it quickly spread to other Western states, destroying wheat fields all 15 along its path. In fact, the level of destruction has been so great over the past five years that entomologists are calling the Russian aphid the greatest threat to American agriculture since the Hessian fly, Phytophaga destructor, was inadvertently brought to the colonies on ships by German 20 mercenary troops during the Revolutionary War.
10

may substantially curb the destructiveness of the Russian aphid in the future. For the time being, however, American farmers are left to their own devices when it comes to protecting their wheat crops.

104. Which of the following statements would be most in agreement with the statements in the passage? A. B. C. D. It is no longer economical to grow crops with low profit margins. Humans are powerless against the forces of nature. Regional ecosystems are often severely damaged when new organisms are introduced. It is more difficult to stop the spread of an insect that reproduces asexually than one that reproduces sexually.

A combination of several factors have made it particularly difficult to deal with the threat posed by this aphid. First, Russian aphids reproduce asexually at a phenomenal rate. This process, known as parthenogenesis, often results 25 in as many as twenty generations of insects in a single year. Although most generations remain in a limited geographic area because they have no wings, a few generations are born with wings, allowing the insect to spread to new areas. Second, because wheat is a crop with a very 30 low profit margin, most American farmers do not spray it with pesticides; it simply is not economical to do so. And since the Russian aphid has only recently entered the United States, it has no natural enemies among North American insects or animals. As a result, there have been 35 no man-made or natural obstacles to the spread of the Russian aphid in the United States. Agricultural researchers seeking to control the Russian aphid have looked to its place of origin for answers. In the Soviet Union, the Russian aphid has been kept in check by 40 predators which have evolved alongside it over many thousands of years. One species of wasp seems to be particularly efficient at destroying the aphid. The pregnant females of the species search the Russian aphid’s home, the interior of a wheat stalk, sting the aphid into paralysis, and then 45 inject an egg into its body. When the egg hatches the wasp larva feeds off of the aphid, killing it in the process. The introduction of predators like the wasp, coupled with the breeding of new strains of insect-resistant wheat, 36

105. According to the passage, which of the following statements is/are true of Russian wheat aphids? I. Most are capable of flight. II. They are resistant to pesticides. III. They are capable of spreading rapidly. A. II only B. III only C. I and II only D. II and III only

106. It can reasonably be inferred that the author of the passage is: A. B. C. D. a botanist with an interest in wheat production. an agriculturist with an interest in pest control. a pest exterminator with an interest in agriculture. an entomologist with an interest in asexual reproduction. GO ON TO THE NEXT PAGE.

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 37

107. The passage supplies information for answering all of the following questions EXCEPT: A. B. What measures were taken to combat the Hessian fly during the 18th century? Why does the Russian wheat aphid cause less damage in the Soviet Union than in other countries? Is it logical for American farmers to use pesticides in order to attempt to protect their wheat crops from the Russian aphid? What sorts of solutions have agricultural researchers investigated in their efforts to curb the destructiveness of the Russian wheat aphid?

109. According to the passage, the Russian wheat aphid and the Hessian fly are comparable with respect to: I. the amount of destruction they have caused. II. the means by which they reproduce. III. the ways in which they entered the United States. A. I only B. II only C. I and II only D. I and III only 110. The author most likely believes American farmers will:

C.

D.

108. The author suggests the best way to control the Russian aphid population in the United States is to: A. B. C. D. devote less acreage to the production of wheat. spray wheat fields with large quantities of pesticides. transplant its natural enemies from the Soviet Union. disrupt its reproductive process by sterilizing females.

A. B. C. D.

develop new types of aphid-resistant wheat. develop their own effective methods for dealing with the Russian aphid. stop producing wheat until the Russian aphid is brought under control. continue to lose a portion of their wheat crops for the foreseeable future.

GO ON TO THE NEXT PAGE. 37

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 38

Passage VI (Questions 111–117) Millenialism is, generally speaking, the religious belief that salvation and material benefits will be conferred upon a society in the near future as the result of some apocalyptic event. The term derives from the Latin word for 1,000; 5 in early Christian theology, believers held that Christ would return and establish his kingdom on earth for a period of a thousand years. Millenialist movements, Christian and non-Christian, have arisen at various points throughout history, usually in 10 times of great crisis or social upheaval. In “nativistic” millenialist movements, a people threatened with cultural disintegration attempts to earn its salvation by rejecting foreign customs and values and returning to the “old ways.” One such movement involving the Ghost Dance 15 cults, named after the ceremonial dance which cult members performed in hope of salvation, flourished in the late 19th century among Indians of the western United States. By the middle of the 19th century, western expansion and settlement by whites was seriously threatening Native 20 American cultures. Mining, agriculture and ranching encroached on and destroyed many Indian land and food sources. Indian resistance led to a series of wars and massacres, culminating in the U.S. Government’s policy of resettlement of Indians onto reservations which constituted 25 a fraction of their former territorial base. Under these dire circumstances, a series of millenialist movements began among western tribes. The first Ghost Dance cult arose in western Nevada around 1870. A Native American prophet named Wodzi30 wob, a member of a Northern Paiute tribe, received the revelation of an imminent apocalypse which would destroy the white man, restore all dead Indians to life, and return to the Indians their lands, food supplies (such as the vanishing buffalo), and old way of life. The apocalypse 35 was to be brought about with the help of a ceremonial dance and songs, and by strict adherence to a moral code which, oddly enough, strongly resembled Christian teaching. In the early 1870s, Wodziwob’s Ghost Dance cult spread to several tribes in California and Oregon, but soon 40 died out or was absorbed into other cults. A second Ghost Dance cult, founded in January 1889, evolved as the result of a similar revelation. This time Wovoka—another Northern Paiute Indian, whose father had been a disciple of Wodziwob—received a vision dur45 ing a solar eclipse in which he died, spoke to God, and was assigned the task of teaching the dance and the millennial message. With white civilization having pushed western tribes ever closer to the brink of cultural

50

disintegration during the previous twenty years, the Ghost Dance movement spread rapidly this time, catching on among tribes from the Canadian border to Texas, and from the Missouri River to the Sierra Nevadas—an area approximately one-third the size of the continental United States. Wovoka’s Ghost Dance doctrine forbade Indian violence against whites or other Indians; it also involved the wearing of “ghost shirts,” which supposedly rendered the wearers invulnerable to the white man’s bullets. In 1890, when the Ghost Dance spread to the Sioux Indians, both the ghost shirts and the movement itself were put to the test. Violent resistance to white domination had all but ended among the Sioux by the late 1880s, when government-ordered reductions in the size of their reservations infuriated the Sioux, and made them particularly responsive to the millenialist message of the Ghost Dance. As the Sioux organized themselves in the cult of the dance, an alarmed federal government resorted to armed intervention which ultimately led to the massacre of some 200 Sioux men, women and children at Wounded Knee, South Dakota in December of 1890. The ghost shirts had been worn to no avail, and Wounded Knee marked the end of the second Ghost Dance cult.

55

60

65

70

111. The passage implies that the second Ghost Dance cult gained widespread popularity quickly because: A. B. C. D. the U.S. government no longer attempted to suppress Native American religious practices. many Native Americans felt particularly threatened by white civilization. Wovoka was a more charismatic religious leader than Wodziwob had been. it was founded on the basis of a spiritual revelation.

112. The passage implies that a paradoxical element of the Ghost Dance cults was their: A. B. C. D. organized resistance to cultural change. mixture of anti-white sentiment and Christian morality. belief in the ability of “ghost shirts” to protect them in combat. combination of millenialist message and desire to revive the “old ways.” GO ON TO THE NEXT PAGE. 38

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 39

113. All of the following characteristics are described in the passage as common to all millenialist movements EXCEPT: A. B. C. D. the desire for salvation. the belief in imminent apocalypse. attempts to preserve cultural integrity. adherence to Christian doctrines.

116. Which of the following tribes would probably NOT have taken part in the Ghost Dance cults? A. B. C. D. The Potawatomi of Illinois The Eastern Shoshoni of Wyoming The Pawnee of Nebraska The Southern Arapaho of Oklahoma

114. According to the passage, white encroachment on Native American tribes involved all of the following EXCEPT: A. B. C. D. forced resettlement of Indians. Western migration by whites. justifications based on spiritual revelations. depletion of Indian food sources.

117. The author answers all of the following questions EXCEPT: A. B. C. D. What was the magical property attributed to the “ghost shirts”? Was there any connection between the prophets of the two Ghost Dance cults? What distinguishes “nativistic” millenialist movements from other millenialist movements? What caused the first Ghost Dance cult to die out?

115. Which of the following was NOT part of the spiritual revelation described in the fourth paragraph of the passage? A. B. C. D. Unity among all Indian tribes Restoration of traditional Indian ways Resurrection of the dead Return of the buffalo

GO ON TO THE NEXT PAGE. 39

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 40

Passage VII (Questions 118–124)
50

Our sense of smell is arguably the most powerful of our five senses, but it also the most elusive. It plays a vital yet mysterious role in our lives. Olfaction is rooted in the same part of the brain that regulates such essential func5 tions as body metabolism, reaction to stress, and appetite. But smell relates to more than physiological function: its sensations are intimately tied to memory, emotion, and sexual desire. Smell seems to lie somewhere beyond the realm of conscious thought, where, intertwined with emo10 tion and experience, it shapes both our conscious and unconscious lives. The peculiar intimacy of this sense may be related to certain anatomical features. Smell reaches the brain more directly than do sensations of touch, sight, or sound. When 15 we inhale a particular odor, air containing volatile odiferous molecules is warmed and humidified as it flows over specialized bones in the nose called turbinates. As odor molecules land on the olfactory nerves, these nerves fire a message to the brain. Thus olfactory neurons render a 20 direct path between the stimulus provided by the outside environment and the brain, allowing us to rapidly perceive odors ranging from alluring fragrances to noisome fumes. Certain scents, such as jasmine, are almost universally appealing, while others, like hydrogen sulfide (which emits 25 a stench reminiscent of rotten eggs), are usually considered repellent, but most odors evoke different reactions from person to person, sometimes triggering strong emotional states or resurrecting seemingly forgotten memories. Scientists surmise that the reason why we have highly personal asso30 ciations with smells is related to the proximity of the olfactory and emotional centers of our brain. Although the precise connection between emotion and olfaction remains a mystery, it is clear that emotion, memory, and smell are all rooted in a part of the brain called the limbic lobe. Even though we are not always conscious of the presence of odors, and are often unable to either articulate or remember their unique characteristics, our brains always register their existence. In fact, such a large amount of human brain tissue is devoted to smell that scientists sur40 mise the role of this sense must be profound. Moreover, neurobiological research suggests that smell must have an important function because olfactory neurons can regenerate themselves, unlike most other nerve cells. The importance of this sense is further supported by the fact that 45 animals experimentally denied the olfactory sense do not develop full and normal brain function.
35

55

60

influenced by pheromones, which are odors that induce psychological or behavioral changes and often provide a means of communicating within a species. These chemical messages, often a complex blend of compounds, are of vital importance to the insect world. Honeybees, for example, organize their societies through odor: the queen bee exudes an odor that both inhibits worker bees from laying eggs and draws drones to her when she is ready to mate. Mammals are also guided by their sense of smell. Through odors emitted by urine and scent glands, many animals maintain their territories, identify one another, signal alarm, and attract mates. Although our olfactory acuity can’t rival that of other animal species, human beings are also guided by smell. Before the advent of sophisticated laboratory techniques, physicians depended on their noses to help diagnose illness. A century ago, it was common medical knowledge that certain bacterial infections carry the musty odor of wine, that typhoid smells like baking bread, and that yellow fever smells like meat. While medical science has moved away from such subjective diagnostic methods, in everyday life we continue to rely on our sense of small, knowingly or not, to guide us.

65

70

118. According to the passage, the location of the olfactory and emotional centers of the brain helps explain all of the following EXCEPT: A. B. C. D. why smells can evoke distant memories. why odors elicit different reactions from person to person. why a substantial part of the brain is devoted to smell. which functions are rooted in the limbic lobe.

119. The author’s central concern in this passage is to: A. B. C. D. discuss both the physiological and emotional aspects of olfaction. explain why the sense of smell is more important than other senses. detail the biological mechanisms by which smells trigger long-forgotten memories. defend the view that human emotion is rooted in anatomical processes. GO ON TO THE NEXT PAGE. 40

The significance of olfaction is much clearer in animals than in human beings. Animal behavior is strongly

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 41

120. The passage implies that physicians no longer make diagnoses based on odors because: A. B. C. D. the human sense of smell has considerably diminished over time. the association of odors with disease proved largely fictitious. such subjective diagnostic methods were shown to be useless. the medical profession today favors more objective techniques.

123. It can be inferred from the passage that the emotional element of human olfaction would be better understood through investigation into: A. B. C. D. the components and functions of the limbic lobe. how pheromones regulate social behavior and organization. the composition of certain highly evocative odors. the pathway between outside environment and olfactory nerves.

121. The sense of smell in animals is different from olfaction in humans in that animals: A. B. C. D. are unable to make associations between smells and past experience. only use smell to communicate outside their own species. rely on olfaction only for mating purposes. more clearly exhibit behavioral changes in response to odors.

124. Which of the following evidence does NOT support the author’s statement that smell has an important physiological function? A. B. C. D. Olfaction and metabolic function are located in the same area of the brain. Animals with impaired olfaction frequently exhibit abnormal brain function. A considerable amount of human brain tissue is devoted to olfaction. Human beings with impaired olfaction are usually able to behave and function normally.

122. The author describes the sense of smell as elusive because: A. B. C. D. odiferous molecules are extremely volatile. the functions of smell are emotional rather than physiological. the function and effects of smell are not fully understood. olfactory sensations are more fleeting than those of other senses.

GO ON TO THE NEXT PAGE. 41

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 42

Passage VIII (Questions 125–131)
50

Bebop lives! cries the newest generation of jazz players. During the 1980s, musicians like Wynton Marsalis revived public interest in bebop, the speedy, angular music that first bubbled up out of Harlem in the early 1940s, 5 changing the face of jazz. That Marsalis and others thought of themselves as celebrating and preserving a noble tradition is, in one sense, inevitable. After the excesses of experimental or “free” jazz in the 1960s and the electronic jazz-rock “fusion” of the 70s, it is hardly surprising that 10 people should hearken back to a time when jazz was “purer,” perhaps even at the apex of its development. But the recent enthusiasm for bebop is also ironic in light of the music’s initial public reception. In its infancy, during the first two decades of the 20th century, jazz was played by small groups of musicians improvising variations on blues tunes and popular songs. Most of the musicians were unable to read music, and their improvisations were fairly rudimentary. Nevertheless, jazz attained international recognition in the 1920s. Two of the 20 people most responsible for its rise in popularity were Louis Armstrong, the first great jazz soloist, and Fletcher Henderson, leader of the first great jazz band. Armstrong, with his buoyant personality and virtuosic technical skills, greatly expanded the creative range and importance of the 25 soloist in jazz. Henderson, a pianist with extensive training in music theory, foresaw the orchestral possibilities of jazz played by a larger band. He wrote out arrangements of songs for his band members that preserved the spirit of jazz, while at the same time giving soloists a more struc30 tured musical background upon which to shape their solo improvisations. In the 1930s, jazz moved further into the mainstream with the advent of the Swing Era. Big bands in the Henderson mold, led by musicians like Benny Goodman, Count Basie and Duke Ellington, achieved unprece35 dented popularity with jazz-oriented “swing” music that was eminently danceable.
15

55

60

65

Played mainly by small combos rather than big bands, bebop was not danceable; it demanded intellectual concentration. Soon, jazz began to lose its hold on the popular audience, which found the new music disconcerting. Compounding public alienation was the fact that bebop seemed to have arrived on the scene in a completely mature state of development, without that early phase of experimentation that typifies so many movements in the course of Western music. This was as much the result of an accident of history as anything else. The early development of bebop occurred during a three-year ban on recording in this country made necessary by the petrol and vinyl shortages of World War II. By the time the ban was lifted, and the first bebop records were made, the new music seemed to have sprung fully-formed like Athena from the forehead of Zeus. And though a small core of enthusiasts would continue to worship bebop pioneers like Charlie Parker and Dizzy Gillespie, many bebop musicians were never able to gain acceptance with any audience and went on to lead lives of obscurity and deprivation.

125. According to the passage, which of the following is true about the bebop music of the 1940s? A. B. C. D. It followed the tradition of jazz from the 1920s. It differed markedly from the music of the Swing Era. It celebrated the songs of Tin-Pan Alley. It did not require great improvisational skill.

Against this musical backdrop, bebop arrived on the scene. Like other modernist movements in art and literature, bebop music represented a departure from tradition in 40 both form and content, and was met with initial hostility. Bebop tempos were unusually fast, with the soloist often playing at double time to the backing musicians. The rhythms were tricky and complex, the melodies intricate and frequently dissonant, involving chord changes and 45 notes not previously heard in jazz. Before bebop, jazz players had improvised on popular songs such as those produced by Tin-Pan Alley, but bebop tunes were often originals with which jazz audiences were unfamiliar.

126. According to the passage, which of the following is true about the jazz of the 1920s? A. B. C. D. It resembled the jazz played during the first two decades of the century. It placed greater demands on the improvisatory skills of its soloists. Its fast tempos foreshadowed those of bebop in the 1940s. It was primarily dance music.

GO ON TO THE NEXT PAGE. 42

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 43

127. Based on the information in the passage comparing bebop to other movements in the history of Western music, it is reasonable to conclude that: I. most movements in music history passed through a stage of experimentation before reaching mature expression. II. World War II prevented bebop from reaching a more appreciative audience. III. bebop did not go through a developmental stage before reaching mature expression A. I only B. III only C. I and II only D. II and III only

130. The author suggests that bebop seemed to represent a radical departure from earlier jazz in that it: A. B. C. D. grew to maturity before reaching a wide audience. attracted primarily a youthful audience. dispensed with written arrangements of songs. expressed the alienation of the musicians who played it.

131. The author mentions Wynton Marsalis and Charlie Parker as: A. B. C. D. pioneers of jazz-rock “fusion.” architects of the bebop movement. Swing Era musicians hostile to bebop. bebop musicians of different eras.

128. It can be inferred from the passage that the innovations of Fletcher Henderson (lines 27-34) were inspired primarily by: A. B. C. D. his admiration for Louis Armstrong. a hunger for international recognition. the realization that the public favored large bands over small combos. a desire to go beyond the structural limitations of early jazz music.

129. According to the passage, all of the following are characteristic of bebop music EXCEPT: A. B. C. D. eminently danceable tunes. dissonant melodies. complex rhythms. intellectual complexity.

GO ON TO THE NEXT PAGE. 43

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 44

Passage IX (Questions 132–137) Studies of photosynthesis began in the late eighteenth century. One scientist found that green plants produce a substance (later shown to be oxygen) that supports the flame of a candle in a closed container. Several years later it was discovered that a plant must be exposed to light in order to replenish this flame-sustaining “substance.” Soon another discovery showed that the oxygen is formed at the expense of another gas, carbon dioxide. In 1804, de Saussure conducted experiments revealing that equal volumes of carbon dioxide and oxygen are exchanged between a plant and the air surrounding it. De Saussure determined that the weight gained by a plant grown in a pot equals the sum of the weights of carbon derived from absorbed carbon dioxide and water absorbed through plant roots. Using this information, de Saussure was able to postulate that in photosynthesis carbon dioxide and water combine using energy in the form of light to produce carbohydrates, water, and free oxygen. Much later, in 1845, scientists’ increased understanding of concepts of chemical energy led them to perceive that, through photosynthesis, light energy is transformed and stored as chemical energy. In the twentieth century, studies comparing photosynthesis in green plants and in certain sulfur bacteria yielded important information about the photosynthetic process. Because water is both a reactant and a product in the central reaction, it had long been assumed that the oxygen released by photosynthesis comes from splitting the carbon dioxide molecule. In the 1930s, however, this popular view was decisively altered by the studies of C. B. Van Niel. Van Niel studied sulfur bacteria, which use hydrogen sulfide for photosynthesis in the same way that green plants use water, and produce sulfur instead of oxygen. Van Niel saw that the use of carbon dioxide to form carbohydrates was similar in the two types of organisms. He reasoned that the oxygen produced by green plants must derive from water—rather than carbon dioxide, as previously assumed—in the same way that the sulfur produced by the bacteria derives from hydrogen sulfide. Van Niel’s finding was important because the earlier belief had been that oxygen was split off from carbon dioxide, and that carbon then combined with water to form carbohydrates. The new postulate was that, with green plants, hydrogen is removed from water and then combines with carbon dioxide to form the carbohydrates needed by the organism. Later, Van Niel’s assertions were strongly backed by scientists who used water marked with a radioactive isotope of oxygen in order to follow photosynthetic reactions. When the photosynthetically-produced free oxygen was analyzed, the isotope was found to be present. 44

5

10

15

20

25

132. Which of the following can be inferred about the scientists discussed in the passage? A. B. C. D. They relied on abstract reasoning in the absence of physical data. They never came to understand the role of light in photosynthesis. Each contributed to our understanding of the production of oxygen by plants. They tended to undervalue previous scientific findings.

30

35

40

133. According to the passage, C. B. Van Niel’s experiments: A. B. C. D. provided the first model of photosynthesis. showed that the carbon dioxide molecule is split during photosynthesis. proved that some organisms combine hydrogen sulfide with carbon dioxide in photosynthesis. provided evidence that weakened the accepted model of photosynthesis. GO ON TO THE NEXT PAGE.

45

50

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 45

134. According to the passage, the study of organisms that require hydrogen sulfide for photosynthesis: A. B. C. D. proved that oxygen is not produced in photosynthesis. contradicted the notion that oxygen is needed to support a candle’s flame. disproved assumptions about the role of light energy in photosynthesis. clarified the role of water in photosynthesis among green plants.

137. It can be inferred from the passage that, in evaluating Van Niel’s hypothesis about the role of water in photosynthesis, scientists were: A. B. C. D. willing to overlook minor inconsistencies in Van Niel’s account. biased in favor of an older, more established explanation. brought to reluctant agreement after repeated tests. thoroughly convinced after conducting an independent experiment.

135. Which of the following statements about photosynthesis would most probably NOT have been made by de Saussure? A. B. C. D. It involves an exchange of equal quantities of gases. It results in the conversion of light energy to chemical energy. It produces oxygen. It requires light.

STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS SECTION ONLY.

136. The passage supplies information for answering all of the following questions EXCEPT: A. B. C. D. Why is oxygen necessary for a candle to burn? What was de Saussure’s explanation of the function of water in photosynthesis? What is the function of light in photosynthesis? Is water required for all photosynthetic reactions?

45

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 46

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 47

Writing Sample
Time: 60 Minutes 2 Items, Separately Timed: 30 Minutes Each

DO NOT BEGIN THIS SECTION UNTIL YOU ARE TOLD TO DO SO.

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 48

WRITING SAMPLE
DIRECTIONS: This section is a test of your writing skills. The section contains two parts. You will have 30 minutes to complete each part. Your responses to the prompts given in the Writing Sample will be written in the ANSWER DOCUMENT. Your response to Part 1 must be written only on the answer sheets marked “1,” and your response to Part 2 must be written only on the answer sheets marked “2.” You may work only on Part 1 during the first 30 minutes of the test and only on Part 2 during the second 30 minutes. If you finish writing on Part 1 before the time is up, you may review your work on that part, but do not begin writing on Part 2. If you finish writing on Part 2 before the time is up, you may review your work only on Part 2. Use your time efficiently. Before you begin writing a response, read the assignment carefully and make sure you understand exactly what you are being asked to do. You may use the space below each writing assignment to make notes in planning your responses. Because this is a test of your writing skills, your response to each part should be an essay composed of complete sentences and paragraphs, as well organized and clearly written as you can make it in the allotted time. You may make corrections or additions neatly between the lines of your responses, but do not write in the margins of the answer booklet. There are six pages in your answer booklet to write your responses, three pages for each part of the test. You are not required to use all of the pages, but to be sure that you have enough space for each essay, do not skip lines. Essays that are illegible cannot be scored. In addition, essays that are not written in English will not be scored.

GO ON TO THE NEXT PAGE. 48

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 49

Part 1
Consider the following statement: It is each citizen's duty to obey the laws of the nation. Write a unified essay in which you perform the following tasks. Explain what you think the above statement means. Describe a specific situation in which a citizen might not have a duty to obey a law. Discuss what you think determines when citizens have a duty to obey the laws of the nation and when they do not.

GO ON TO THE NEXT PAGE. 49

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 50

Part 2
Consider the following statement: Heroes are ordinary people made heroic by circumstance. Write a unified essay in which you perform the following tasks. Explain what you think the above statement means. Describe a specific situation in which someone might be heroic because of something other than circumstance. Discuss what you think determines when heroism is dependent on circumstance and when it is not.

GO ON TO THE NEXT PAGE. 50

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 51

Biological Sciences
Time: 100 Minutes Questions 138–214

DO NOT BEGIN THIS SECTION UNTIL YOU ARE TOLD TO DO SO.

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 52

BIOLOGICAL SCIENCES
DIRECTIONS: Most of the questions in the following test are organized into groups, with a descriptive passage preceding each group of questions. Study the passage, then select the single best answer to each question in the group. Some of the questions are not based on a descriptive passage; you must also select the best answer to these questions. If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain. Indicate your selection by blackening the corresponding circle on your answer sheet. A periodic table is provided below for your use with the questions.

PERIODIC TABLE OF THE ELEMENTS
1 H 1.0 3 Li 6.9 11 Na 23.0 19 K 39.1 37 Rb 85.5 55 Cs 132.9 87 Fr (223) 4 Be 9.0 12 Mg 24.3 20 Ca 40.1 38 Sr 87.6 56 Ba 137.3 88 Ra 226.0 21 Sc 45.0 39 Y 88.9 57 La * 138.9 89 Ac † 227.0 22 Ti 47.9 40 Zr 91.2 72 Hf 178.5 104 Rf (261) 58 Ce 140.1 90 Th 232.0 23 V 50.9 41 Nb 92.9 73 Ta 180.9 105 Ha (262) 59 Pr 140.9 91 Pa (231) 24 Cr 52.0 42 Mo 95.9 74 W 183.9 106 Unh (263) 60 Nd 144.2 92 U 238.0 25 Mn 54.9 43 Tc (98) 75 Re 186.2 107 Uns (262) 61 Pm (145) 93 Np (237) 26 Fe 55.8 44 Ru 101.1 76 Os 190.2 108 Uno (265) 62 Sm 150.4 94 Pu (244) 27 Co 58.9 45 Rh 102.9 77 Ir 192.2 109 Une (267) 63 Eu 152.0 95 Am (243) 64 G d 157.3 96 Cm (247) 65 T b 158.9 97 Bk (247) 66 Dy 162.5 98 Cf (251) 67 Ho 164.9 99 Es (252) 68 Er 167.3 100 Fm (257) 69 Tm 168.9 101 Md (258) 70 Y b 173.0 102 No (259) 71 Lu 175.0 103 Lr (260) 28 Ni 58.7 46 Pd 106.4 78 Pt 195.1 29 Cu 63.5 47 Ag 107.9 79 Au 197.0 30 Zn 65.4 48 Cd 112.4 80 Hg 200.6 5 B 10.8 13 Al 27.0 31 Ga 69.7 49 In 114.8 81 Tl 204.4 6 C 12.0 14 Si 28.1 32 Ge 72.6 50 Sn 118.7 82 Pb 207.2 7 N 14.0 15 P 31.0 33 As 74.9 51 Sb 121.8 83 Bi 209.0 8 O 16.0 16 S 32.1 34 Se 79.0 52 Te 127.6 84 Po (209) 9 F 19.0 17 Cl 35.5 35 Br 79.9 53 I 126.9 85 At (210) 2 He 4.0 10 Ne 20.2 18 Ar 39.9 36 K r 83.8 54 Xe 131.3 86 Rn (222)

*

†

GO ON TO THE NEXT PAGE. 52

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 53

Passage I (Questions 138–144) Hemoglobin (Hb) and myoglobin (Mb) are the O2- carrying proteins in vertebrates. Hb, which is contained within red blood cells, serves as the O2 carrier in blood and also plays a vital role in the transport of CO2 and H+. Vertebrate Hb consists of four polypeptides (subunits) each with a heme group. The four chains are held together by noncovalent attractions. The affinity of Hb for O2 varies between species and within species depending on such factors as blood pH, stage of development, and body size. For example, small mammals give up O2 more readily than large mammals because small mammals have a higher metabolic rate and require more O2 per gram of tissue. The binding of O2 to Hb is also dependent on the cooperativity of the Hb subunits. That is, binding at one heme facilitates the binding of O2 at the other hemes within the Hb molecule by altering the conformation of the entire molecule. This conformational change makes subsequent binding of O2 more energetically favorable. Conversely, the unloading of O2 at one heme facilitates the unloading of O2 at the others by a similar mechanism. Figure 1 depicts the O2-dissociation curves of Hb (Curves A, B, and C) and myoglobin (Curve D), where saturation, Y, is the fractional occupancy of the O2-binding sites. The fraction of O2 that is transferred from Hb as the blood passes through the tissue capillaries is called the utilization coefficient. A normal value is approximately 0.25.

138. The llama is a warm-blooded mammal that lives in regions of unusually high altitudes, and has evolved a type of Hb that adapts it to such an existence. If Curve B represents the O2-dissociation curve for horse Hb, which curve would most closely resemble the curve for llama Hb? A. B. C. D. Curve A Curve B Curve C Curve D

139. If Curve B represents the O2-dissociation curve for elephant Hb, which curve most closely resembles the curve for mouse Hb? A. B. C. D. Curve A Curve B Curve C Curve D

140. If Curve B represents the O2-dissociation curve for human adult Hb, which of the following best explains why Curve A most closely resembles the curve for fetal Hb? A. B. C. D. Fetal tissue has a higher metabolic rate than adult tissue. Fetal tissue has a lower metabolic rate than adult tissue. Fetal Hb has a higher affinity for O2 than adult Hb. Fetal Hb has a lower affinity for O2 than adult Hb.

Figure 1 Myoglobin facilitates O2 transport in muscle and serves as a reserve store of O2. Mb is a single polypeptide chain containing a heme group, with a molecular weight of 18 kd. As can be seen in Figure 1, Mb (Curve D) has a greater affinity for O2 than Hb.

GO ON TO THE NEXT PAGE. 53

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 54

141. The sigmoidal shape of the O2-dissociation curve of Hb is due to: A. B. C. D. the effects of oxidation and reduction on the heme groups within the Hb molecule. the concentration of carbon dioxide in the blood. the fact that Hb has a lower affinity for O2 than Mb. the cooperativity in binding among the subunits of the Hb molecule.

144. In sperm whales, the Mb content of muscle is about 0.004 moles/kg of muscle. If a sperm whale has 1000 kg of muscle, approximately how much O2 is bound to Mb, assuming that the Mb is saturated with O2? A. B. C. D. 4 moles 8 moles 12 moles 16 moles

142. A sample of human adult Hb is placed in an 8 M urea solution, resulting in the disruption of noncovalent interactions. After this procedure, the α chains of Hb are isolated. Which of the four curves most closely resembles the O2-dissociation curve for the isolated α chains? [Note: Assume that Curve B represents the O2-dissociation curve for human adult Hb in vivo.] A. B. C. D. Curve A Curve B Curve C Curve D

143. The utilization coefficient is continually being adjusted in response to physiological changes. Which of the following values most likely represents the utilization coefficient for human adult Hb during strenuous exercise? A. B. C. D. 0.0 0.125 0.25 0.75

GO ON TO THE NEXT PAGE. 54

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 55

Passage II (Questions 145–151) Just as the ingestion of nutrients is mandatory for human life, so is the excretion of metabolic waste products. One of these nutrients, protein, is used for building muscle, nucleic acids, and countless compounds integral to homeostasis. However, the catabolism of the amino acids generated from protein digestion produces ammonia, which, if not further degraded, can become toxic. Similarly, if the same salts that provide energy and chemical balance to cells are in excess, fluid retention will occur, damaging the circulatory, cardiac, and pulmonary systems. One of the most important homeostatic organs is the kidney, which closely regulates the excretion and reabsorption of many essential ions and molecules. One mechanism of renal function involves the secretion of antidiuretic hormone (ADH). Diabetes insipidus (DI), is the condition that occurs when ADH is ineffective. As a result, the kidneys are unable to concentrate urine, leading to excessive water loss. There are two types of DI—central and nephrogenic. Central DI occurs when there is a deficiency in the quantity or quality of ADH produced. Nephrogenic DI occurs when the kidney tubules are unresponsive to ADH. To differentiate between these two conditions, a patient’s urine osmolarity is measured both prior to therapy and after a 24-hour restriction on fluid intake. Exogenous ADH is then administered and urine osmolarity is measured again. The table below gives the results of testing on four patients. Assume that a urine osmolarity of 285 mOsm/L of H2O is normal. Table 1 Urine Osmolarity (mOsm/L of H2O) Patient A B C D Before Therapy After fluid restriction 285 765 180 765 180 180 180 180 After ADH 765 765 400 180

145. An elevated and potentially toxic level of ammonia in the blood (hyperammonemia) would most likely result from a defect in an enzyme involved in: A. B. C. D. glycolysis. fatty acid catabolism. the urea cycle. nucleic acid degradation.

146. According to the passage, the catabolism of amino acids produces ammonia. Therefore, after a proteinrich meal, would you expect a build-up of ammonia in the lumen of the small intestine? A. B. Yes, because the ammonia will not be able to diffuse into the intestinal epithelium. Yes, because the rate at which digestive enzymes degrade ammonia is slower than the rate at which ammonia is produced. No, because the ammonia will diffuse into the intestinal epithelium and will be excreted by the kidneys. No, because the ammonia is produced inside individual cells, not within the lumen of the small intestine.

C.

D.

147. Which of the following substances would NOT be found in appreciable quantity in the urine of a healthy individual? A. B. C. D. Albumin Sodium Urea Potassium

GO ON TO THE NEXT PAGE. 55

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 56

148. Which of the following would you most likely expect to find in a patient with diabetes insipidus? A. B. C. D. Decreased plasma osmolarity Increased urine osmolarity Increased urine glucose Increased urine output

151. What is the most likely cause of Patient B’s dilute urine before therapy? A. B. C. D. Excessive water intake Dehydration Nephrogenic DI Central DI

149. Based on the data in Table 1, which of the four patients most likely has central diabetes insipidus? A. B. C. D. Patient A Patient B Patient C Patient D

150. Based on the data in Table 1, which of the four patients most likely has nephrogenic diabetes insipidus? A. B. C. D. Patient A Patient B Patient C Patient D

GO ON TO THE NEXT PAGE. 56

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 57

Passage III (Questions 152–158) Electromagnetic radiation from space constantly bombards the earth. Most wavelengths are absorbed by the atmosphere; however, there are two “windows” of nonabsorption through which significant amounts of radiation reach the ground. The first transmits ultraviolet and visible light, as well as infrared light or heat; the second transmits radio waves. As a result, terrestrial organisms have evolved a number of pigments that interact with light in various ways: some capture light energy, some provide protection from light-induced damage, and some serve camouflage or signaling purposes. Among these compounds are many conjugated polyenes, which play important roles as photoreceptors. For every chemical compound, there are certain wavelengths of light whose quanta possess exactly the correct amount of energy to raise electrons from their ground state to higher-energy orbitals. For most organic compounds, these wavelengths are in the UV range. However, conjugated double bond systems stabilize the electrons, so that they can be excited by lower-frequency photons with wavelengths in the visible spectrum. Such a pigment, known as a chromophore, will then transmit the “subtraction color,” a color complementary to the one absorbed. For instance, carotene, a hydrocarbon compound with eleven conjugated double bonds, absorbs blue light and transmits orange. The wavelength that is absorbed generally increases with the number of conjugated bonds; rings and side-chains also affect wavelength. Wavelength 480 nm 580 nm 680 nm Color blue yellow red Subtraction Color orange violet green

152. The electrons that give color to a carotene molecule are found in: A. B. C. D. s orbitals. π orbitals. d orbitals. f orbitals.

153. Two pigments are identical except for the lengths of their conjugated polyene chains. The first transmits yellow light and the second red. What can be said about the sizes of the chromophores? A. B. C. D. The first is longer. The second is longer. One of the chromophores must be a dimer. The comparative lengths cannot be determined.

154. Why is benzene colorless? A. B. C. D. The absorption energy is of too high a frequency to be visible. The absorption energy is of too low a frequency to be visible. Benzene does not absorb light. Benzene is not conjugated.

Among the many biological molecules that are affected by light is DNA, the genetic material of living organisms. DNA absorbs ultraviolet light, and may be damaged by UVC (< 280 nm) and UVB (280-315 nm). UVA (315-400 nm) and visible light can actually repair light-induced damage to DNA by a process called photorepair. For this reason UVA, which also stimulates tanning, was once considered beneficial. However, there is now increasing evidence that UVA can damage skin.

GO ON TO THE NEXT PAGE. 57

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 58

155. Many crustaceans produce a blue or green caroteneprotein complex. What is the most likely cause of the color change from green to orange when a lobster is boiled? A. B. C. D. Heat causes the prosthetic group to become partially hydrated. The increase in temperature permits the prosthetic group to absorb shorter wavelengths. The protein is separated from the carotenoid pigment. Heat causes the prosthetic group to become oxidized.

156. The four compounds represented by the electronic spectra below were evaluated as potential sunscreens. What is the correct sequence of sunscreen strength, from strongest to weakest, among these four?

I.

II.

III.

IV. A. B. C. D. I, II, III, IV IV, III, II, I III, II, I, IV IV, I, II, III

GO ON TO THE NEXT PAGE. 58

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 59

157. Which of the following compounds would be most likely to produce color?

Questions 159 through 163 are NOT based on a descriptive passage.
159. An increase in heart rate, blood pressure, and blood glucose concentration are all associated with stimulation of the: A. B. C. D. parasympathetic nervous system. sympathetic nervous system. somatic nervous system. digestive system.

A.

B. NH2 O

160. Which of the following compounds share the same absolute configuration? I. H COOH OH CH3 Cl Cl II. H3C OH OH A. B. C. D. NH2 I and III II and IV I and II II, III, and IV IV. COOH HOOC H III. I COOH C2H5 CH2CH3

C.

O

H

OH CH3

D.

158. The color-producing quality of conjugated polyenes is attributable to: A. B. C. D. antibonding orbitals. resonance. polarity. optical activity.

161. Which of the following structures plays a role in both the male excretory and male reproductive systems, but in the female excretory system only? A. B. C. D. Epididymis Prostate Urethra Ureter

GO ON TO THE NEXT PAGE. 59

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 60

162. The reaction R—Br + Br*– → R—Br* + Br– is always accompanied by inversion. If this reaction is carried out on an optically pure sample of a chiral compound, which of the following statements will be true? [Note: Br* represents a radioactive isotope of bromine.] A. B. C. D. The rate of Br* incorporation is half the rate of racemization. The rate of Br* incorporation is equal to the rate of racemization. The rate of Br* incorporation is twice the rate of racemization. The relation between the rate of Br* incorporation and the rate of racemization cannot be determined.

163. Which of the following cell types does NOT contain the diploid number of chromosomes? A. B. C. D. Spermatogonium Spermatid Zygote Primary oocyte

GO ON TO THE NEXT PAGE. 60

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 61

Passage IV (Questions 164–168) Hemophilia is a genetically inherited disease that causes the synthesis of an abnormal clotting factor. As a result, hemophiliacs bleed excessively from the slightest injury. The figure below is a partial pedigree for the hemophilia trait in Queen Victoria’s descendants. The pedigree indicates no history of hemophilia for either parent prior to the F1 generation.

164. According to Figure 1, which of the following assumptions about the P1 generation must be true? A. B. C. D. Albert did not have the gene for hemophilia. Queen Victoria had two X chromosomes, each with the gene for hemophilia. Neither Albert nor Queen Victorian had the gene for hemophilia. Albert was a carrier of the hemophilia gene.

165. Which of the following best explains why Louis IV was NOT a hemophiliac? A. B. C. D. His son Frederick was a hemophiliac. He did not inherit the gene for hemophilia from his mother. His father-in-law, Albert, was not a hemophiliac. Only females can be carriers of the gene for hemophilia.

166. If Beatrice had married a hemophiliac and had a son, what is the probability that the son would have been a hemophiliac? A. B. C. D. 0% 25% 50% 100%

Figure 1

GO ON TO THE NEXT PAGE. 61

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 62

167. Theoretically, what percentage of Victoria Eugenia’s sons should have been hemophiliacs? A. B. C. D. 25% 33% 50% 75%

168. Based on the pedigree, what is the most reasonable explanation for Rupert’s hemophilia? A. B. C. D. A mutation occurred on the Y chromosome that he inherited from his father. His mother was a hemophiliac and transmitted the gene to him. His father was a carrier of the gene for hemophilia. His maternal grandfather was a hemophiliac.

GO ON TO THE NEXT PAGE. 62

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 63

Passage V (Questions 169–175) Aerobic respiration is the major process used by oxygen-requiring organisms to generate energy. During respiration, glucose is metabolized to generate chemical energy in the form of ATP: C6H12O6 + 6O2 → 6CO2 + 6H2O + 36ATP The biochemical machinery necessary for cellular respiration is found in the mitochondria, small organelles scattered throughout the cytoplasm of most eukaryotic cells. The number of mitochondria per cell varies by tissue type and cell function. Mitochondria are unusual in that they have their own genetic systems that are entirely separate from the cell’s genetic material. However, mitochondrial replication is still dependent upon the cell’s nuclear DNA to encode essential proteins required for replication. Despite this fact, mitochondria seem to replicate randomly, out of phase with both the cell cycle and other mitochondria. The nature of the mitochondrial genome and proteinsynthesizing machinery has led many researchers to postulate that mitochondria may have arisen as the result of the ingestion of a bacterium by a primitive cell millions of years ago. It is postulated that the two may have entered into a symbiotic relationship and eventually became dependent on each another; the cell sustained the bacterium, while the bacterium provided energy for the cell. Gradually, the two evolved into the present-day eukaryotic cell, with the mitochondrion retaining some of its own DNA. This is known as the endosymbiotic hypothesis. Because mitochondrial DNA is inherited in a nonMendelian fashion (mitochondria are inherited from the maternal parent, who supplies most of the cytoplasm to the fertilized egg), it has been used to look at evolutionary relationships among different organisms.

169. In which of the following phases of the cell cycle could mitochondrial DNA replicate? I. G1 II. S III. G2 IV. M A. IV only B. I and III only C. II and IV only D. I, II, III, and IV

170. Scientists have demonstrated that human mitochondrial DNA mutates at a fairly slow rate. Because mitochondria play such an important role in the cell, these mutations are most likely to be: A. B. C. D. point mutations. frameshift mutations. lethal mutations. nondisjunctions.

171. Which of the following mitochondrial genome characteristics differs most from the characteristics of the nuclear genome? A. B. C. D. Mitochondrial DNA is a double-helix. Some mitochondrial genes code for tRNA. Specific mutations to mitochondrial DNA can be lethal to the organism. Almost every base in mitochondrial DNA codes for a product.

GO ON TO THE NEXT PAGE. 63

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 64

172. What is the net number of ATP molecules synthesized by an obligate anaerobe per molecule of glucose? A. B. C. D. 2 ATP 6 ATP 8 ATP 36 ATP

175. Which of the following pieces of evidence would NOT support the hypothesis that mitochondria were once independent bacteria that eventually formed a symbiotic relationship with eukaryotic cells? A. B. Mitochondrial DNA is circular and not enclosed by a nuclear membrane. Mitochondrial ribosomes more closely resemble eukaryotic ribosomes than prokaryotic ribosomes. Many present-day bacteria live within eukaryotic cells, digesting nutrients that their hosts cannot and sharing the energy thus derived. Mitochondrial DNA codes for its own ribosomal RNA.

173. A mating type of a wild-type strain of the algae C. reinhardii is crossed with the opposite mating type of a mutant strain of the algae, which has lost all mitochondrial functions due to deletions in their mitochondrial genome. All of the offspring from this cross also lack mitochondrial functions. Based on information in the passage, this can best be explained by the: A. B. C. D. endosymbiotic hypothesis. non-Mendelian inheritance of mitochondrial DNA. recombination of mitochondrial DNA during organelle replication. presence of genetic material in the mitochondria that is distinct from nuclear DNA.

C.

D.

174. Four different human cell cultures—erythrocytes, epidermal cells, skeletal muscle cells, and intestinal cells—were grown in a medium containing radioactive adenine. After 10 days, the mitochondria were isolated via centrifugation, and their level of radioactivity was measured using a liquid scintillation counter. Which of the following cells would be expected to have the greatest number of counts per minute of radioactive decay? A. B. C. D. Erythrocytes Epidermal cells Skeletal muscle cells Intestinal cells

GO ON TO THE NEXT PAGE. 64

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 65

Passage VI (Questions 176–181) The mechanism for the acid-catalyzed esterification of a carboxylic acid, carried out with R'OH, is shown below. The tagged alcohol R'18OH is used to study the reaction mechanism. The resulting ester is separated from the reaction mixture; the water from the reaction mixture is then distilled off completely and collected as a separate fraction. O R OH H+ O R H OH

176. Assuming that only the forward reaction occurs, which of the following statements is correct? A. B. C. D. The ester will contain labeled oxygen, while the water fraction will not. The water fraction will contain labeled oxygen, while the ester will not. Both the water fraction and the ester will contain labeled oxygen. The location of the labeled oxygen cannot be determined.

177. The rate of the reaction is negligible without the acid catalyst. The catalyst: A. B. C. attacks the carbonyl oxygen, permitting the nucleophilic group to attack the carbonyl carbon. attacks the carbonyl carbon, permitting the nucleophilic group to attack the carbonyl oxygen. attacks the carbonyl oxygen, permitting the electrophilic group to attack the carbonyl carbon. attacks the carbonyl carbon, permitting the electrophilic group to attack the carbonyl oxygen.

R'OH D. OH R R' O OH H

178. Esterification may also occur between parts of the same molecule. Which of the following compounds would most easily undergo internal esterification to form a cyclic ester? A. B. C. D. COOHCH2CH2OH COOHCH2CH2CH2OH COOHCH2CH2CH2CH2OH COOHCH2CH2CH2CH2CH2OH

OH R R' O − H2O OH R OR' − H+ O R OR' OH2

GO ON TO THE NEXT PAGE. 65

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 66

179. Which of the following carboxylic acids will be most acidic? A. B. C. D. CH3CH2COOH ClCH2CH2COOH ClCH2CH2CH2COOH Cl2CHCH2COOH

180. Another method for forming esters is: RCOO– + RX → RCOOR’ + X– Why does this reaction occur? A. B. C. D. The halide is a poor leaving group. The halide acts as a good nucleophile. The halide is an electron-donating group. The carboxylate anion is highly nucleophilic.

181. Which of the following alkyl halides would be most likely to react with sodium butanoate (CH3CH2CH2COO–Na+) to form an ester? A. B. C. D. CH3CH2CH2Cl CH3Cl (CH3)2CHCl CH3CH2Cl

GO ON TO THE NEXT PAGE. 66

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 67

Passage VII (Questions 182–189) Four major blood types exist in the human ABO blood system: types A, B, AB, and O; and there are three alleles that code for them. The A and B alleles are codominant, and the O allele is recessive. Blood types are derived from the presence of specific polysaccharide antigens that lie on the outer surface of the red blood cell membrane. The A allele codes for the production of the A antigen; the B allele codes for the production of the B antigen; the O allele does not code for any antigen. While there are many other antigens found on red blood cell membranes, the second most important antigen is the Rh antigen. Rh is an autosomally dominant trait coded for by 2 alleles. If this antigen is present, an individual is Rh+; if it is absent, an individual is Rh–. For example, a person with type AB blood with the Rh antigen is said to be AB+. These antigens become most important when an individual comes into contact with foreign blood. Because of the presence of naturally occurring substances that closely mimic the A and B antigens, individuals who do not have these antigens on their red blood cells will form antibodies against them. This is inconsequential until situations such as blood transfusion, organ transplant, or pregnancy occur. Erythroblastosis fetalis is a condition in which the red blood cells of an Rh+ fetus are attached by antibodies produced by its Rh– mother. Unlike ABO incompatibility, in which there are naturally occurring antibodies to foreign antigens, the Rh system requires prior sensitization to the Rh antigen before antibodies are produced. This sensitization usually occurs during the delivery of an Rh+ baby. So while the first baby will not be harmed, any further Rh+ fetuses are at risk. The Coombs tests provide a method for determining whether a mother has mounted an immune response again her baby’s blood. The tests are based on whether or not agglutination occurs when Coombs reagent is added to a sample. Coombs reagent contains antibodies against the anti-Rh antibodies produced by the mother. The indirect Coombs test takes the mother’s serum, which contains her antibodies but no red blood cells, and mixes it with Rh+ red blood cells. Coombs reagent is then added. If agglutination occurs, the test is positive, and the mother must be producing anti-Rh antibodies. The direct Coombs test mixes the baby’s red blood cells with Coombs reagent. If agglutination occurs, the test is positive, and the baby’s red blood cells must have been attacked by its mother’s antiRh antibodies.

182. In a paternity case, the mother has type A+ blood and her son has type O– blood. If the husband has type B+ blood, which of the following is true? A. B. C. D. The husband could be the father. The husband could not be the father. The husband could not be the father of an O– son, but could be the father of an O– daughter. The husband is definitely the father.

183. A couple decide to have a child. If the father’s genotype is AO and the mother has type B blood of unknown genotype, which of the following are possible blood types for their child? I. A II. B III. AB IV. O A. I and II only B. I, II, and III only C. I, II, and IV only D. I, II, III, and IV

184. A new virus has been discovered that evades detection by the immune system of only those individuals with type A or type AB blood. Which of the following best accounts for this observation? A. B. C. D. The viral antigens resemble the A antigen. The viral antigens resemble the B antigen. The viral antigens are Rh+. The viral antigens are too small to elicit an immune response.

GO ON TO THE NEXT PAGE. 67

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 68

185. If a man with type AB blood needed a transfusion of red blood cells, which of the following individuals could safely donate blood? A. B. C. D. A man with type A blood A man with the genotype BO A woman with the genotype AB All four blood types are equally safe

188. A woman who has never been pregnant has type B– blood. Which of the following antibodies would you expect to find in her serum? A. B. C. D. Anti-B antibody Anti-A antibody Anti-Rh antibody Both anti-A and anti-Rh antibodies

186. How might one most practically assess the risk of erythroblastosis fetalis in a pregnant woman? A. B. C. D. Test all women for the presence of anti-Rh antibodies. Test all fetuses for the presence of the Rh antigen within the first trimester of pregnancy. Test only Rh– mothers for the presence of antiRh antibodies. Test all mothers of Rh+ children for the presence of anti-Rh antibodies.

189. A medical student suggested giving Rh– mothers of Rh+ fetuses a specific exogenous substance prior to delivery to prevent an immune response. Which of the following substances would likely be the safest and most effective? A. B. C. D. Rh antigen An immunosuppressive drug Anti-Rh antibody Iron pills

187. Based on information in the passage, what does the reaction below represent?

A. B. C. D.

Negative direct Coombs test Positive direct Coombs test Positive indirect Coombs test Negative indirect Coombs test

GO ON TO THE NEXT PAGE. 68

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 69

Questions 190 through 194 are NOT based on a descriptive passage.
190. A certain chemical is found to inhibit the synthesis of all steroids. The synthesis of which of the following hormones would NOT be affected when a dose of this chemical is administered to a laboratory rat? A. B. C. D. Cortisol Aldosterone Epinephrine Testosterone

193. Which of the following products may be formed in the reaction below? O H2O O H3O+

A. B. C. D.

CH3CH=CHCHO HOOCCH2CH(CH3)2 CH3CH(CH3)2 HOCH2CH(CH3)2

191. A biochemist grows two cultures of yeast—one aerobically and the other anaerobically—and measures the amount of ATP produced by each culture. He finds that the aerobically-grown yeast produce about 18 times as much ATP as the anaerobically-grown yeast. These observations are consistent with the fact that in the aerobically grown yeast: A. B. C. D. oxygen is converted into ATP. oxygen is necessary to convert glucose into pyruvate. oxygen is the final electron acceptor of the respiratory chain. oxygen is necessary for the reduction of pyruvate into lactate.

194. Which of the following statements is supported by the table below? Solubility per 100g H2O para-nitrophenol meta-nitrophenol ortho-nitrophenol 1.7 1.4 0.2 Melting point (°C) 114 97 44

A. B. C.

192. Growth hormone decreases the sensitivity of cellular receptors to insulin. Therefore, a patient with acromegaly, which is caused by the oversecretion of growth hormone, would be expected to have: A. B. C. D. a low blood glucose concentration. a high blood glucose concentration. a decreased urine volume. a decreased cardiac output.

D.

Ortho-nitrophenol has the greatest intramolecular hydrogen bonding. Para-nitrophenol has the weakest intermolecular hydrogen bonding. Meta- and para-nitrophenol form intramolecular hydrogen bonds. Ortho-nitrophenol does not form intermolecular hydrogen bonds.

GO ON TO THE NEXT PAGE. 69

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 70

Passage VIII (Questions 195–198) A student was given a sample of an unknown liquid and asked to determine as much as possible about its structure. He was told that the compound contained only carbon, hydrogen, and oxygen, and had only one type of functional group. The student found its boiling point to be 206°C. Using mass spectroscopy, he determined its molecular weight to be 138 g/mol. Finally, he took the infrared spectrum of the compound, which is shown below.

197. Assuming that all of the student’s deductions were correct, which of the following could be the structure of the unknown compound? COOH CH2OH

A.

CH2CH3

C.

CH2OH COOH

OCH3

B.

OCH3

D.

198. The student decides to carry out some simple tests on the compound in order to confirm his identification. Which of the following statements is NOT true? From this spectrum, the student quickly reached a conclusion about the functional group. He then turned his attention to the fingerprint region of the compound, which generally has a complicated pattern of peaks that are determined by the structure of the hydrocarbon portion of a molecule. The student decided that the large peak at 750 cm–1 must indicate that this was a disubstituted aromatic compound. A. He could distinguish between a phenol and a benzoic acid by seeing if the unknown can be extracted with a weak base. He could distinguish between a benzyl aldehyde and a benzyl ketone by seeing if the unknown will react with cold KMnO4. He could distinguish between a benzyl alcohol and a phenol by attempting to dissolve the unknown in an aqueous solution of HCl. He could distinguish between a benzyl alcohol and a benzoic ester by attempting to dissolve the unknown in an aqueous solution of NaOH.

B.

C.

D. 195. The correct formula for this compound could be: A. B. C. D. C7H10O3. C8H10O2. C9H13O. C7H21O2.

196. The overlapping set of peaks near 3000 cm–1 includes one peak at 2850 cm–1. What type of functional group could this indicate? A. B. C. D. Methyl Phenol Carboxyl Aldehyde carbonyl GO ON TO THE NEXT PAGE. 70

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 71

Passage IX (Questions 199–203) Although individual organisms have only two alleles for any given trait, it is possible for a trait to have more than two alleles coding for it. This phenomenon is known as multiple alleles. Multiple alleles are created when a single gene undergoes several distinct mutations. These alleles may have different dominance relationships with one another; for example, there are three alleles coding for the human blood groups, the IA, IB, and i alleles. Both the IA and IB alleles are dominant to the i allele, but IA and IB are codominant to each other. A multiple-allele system has recently been discovered in the determination of hair coloring in a species of wild rat. The rats are found to have one of three colors: brown, red, or white. Let B = the gene for brown hair; b = the gene for red hair; and w = the gene for white hair. The results from nine experimental crosses are shown below. The males and females in Crosses 1, 2, and 3 are all homozygous for hair color. Cross 1 2 3 4 5 6 7 8 9 Male brown brown red brown brown red red brown brown Female red white white brown brown red red red red Offspring all brown all brown all red 3 brown : 1 red all brown all red 3 red : 1 white 2 brown : 1 red : 1 white 1 brown : 1 red

201. Based on the experimental results, what is the genotype of the female in Cross 5? A. B. C. D. Bb BB or Bb BB or Bw BB, Bb, or Bw

202. A white male is crossed with the heterozygous red female from Cross 9. What is the expected ratio of red to white offspring? A. B. C. D. 3:1 1:3 1:1 2:1

203. If it were discovered that the alleles for red and white hair were actually incompletely dominant and produced a pink hair color in rats with one copy of each allele, what would be the expected phenotypic ratio in a cross between a Bb male and a pink female? A. B. C. D. 2 brown : 1 red : 1 white 2 brown : 1 red : 1 pink 1 brown : 2 white : 1 pink 1 brown : 1 white

199. Based on the experimental results, what is the genotype of the male in Cross 6? A. B. C. D. bw bb bw or bb Bb or bw

200. If a large number of brown offspring from Cross 8 are mated with each other, what is the expected percentage of white offspring? A. B. C. D. 6.25% 8.33% 12.5% 25%

GO ON TO THE NEXT PAGE. 71

01 MCAT FL Test1

06/26/2003

05:30 PM

Page 72

Passage X (Questions 204–209) Compounds containing a hydroxyl group attached to a benzene ring are called phenols. Derivatives of phenols, such as naphthols and phenanthrols, have chemical properties similar to those of phenols, as do most of the many naturally-occurring substituted phenols. Like other alcohols, phenols have higher boiling points than hydrocarbons of similar molecular weight. Like carboxylic acids, phenols are more acidic than their alcohol counterparts. Phenols undergo a number of different reactions; both their hydroxyl groups and their benzene rings are highly reactive. A number of chemical tests can be used to distinguish phenols from alcohols and carboxylic acids.
CH3 OH

204. Reaction A is an example of: A. B. C. D. a free radical substitution. an electrophilic aromatic substitution. an electrophilic addition. a nucleophilic aromatic substitution.

205. Comparing the Ka values for cyclohexanol (Ka =10–18) and phenol (Ka = 1.3 x 10–10) reveals that phenol is more acidic than cyclohexanol. Which of the following explain(s) the acidity of phenol? I. The exceptionally strong hydrogen bonding possible with phenol facilitates the loss of a proton, making it more acidic than cyclohexanol. II. Phenol’s conjugate base, phenoxide, is stabilized by resonance to a greater extent than phenol itself. III. The negative charge of the oxygen atom on the phenoxide ion is delocalized over the benzene ring. A. I only B. II only C. II and III only D. I, II, and III

OH OH CH(CH3)2 thymol

naphthol

phenanthrol

Thymol, a naturally occurring phenol, is an effective disinfectant that is obtained from thyme oil. Thymol can also be synthesized from m-cresol, as shown in Reaction A below. Thymol can then be converted to menthol, another naturally-occurring organic compound; this conversion is shown in Reaction B.
CH3 + OH H3PO4 OH CH(CH3)2 m-cresol thymol

206. Which of the following shows the order of decreasing acidity among the four compounds below?
OH OH NO2 OH O 2N OH NO2

NO2 I

NO2 II

CH3 III

NO2 IV

Reaction A

thymol

H2/Ni high pressure OH

A. B. C. D.

I, III, IV, II IV, I, II, III IV, III, II, I IV, II, I, III

menthol

Reaction B GO ON TO THE NEXT PAGE. 72

01 MCAT FL Test1

06/26/2003

05:31 PM

Page 73

207. The reaction of phenol with dilute nitric acid produces which of the following compounds? OH OH NO2 +

208. What simple chemical test could be used to distinguish between the following two compounds? I OH II CH2OH

A.

NO2 OH

CH3 A. B. C. D. Compound II’s solubility in NaHCO3 Compound I’s solubility in NaOH Compound I’s ability to decolorize a bromine solution Compound I’s solubility in NaHCO3

B.

NO2

OH

C. OH

NO2

OH O2N +

D.

NO2

GO ON TO THE NEXT PAGE. 73

01 MCAT FL Test1

06/26/2003

05:31 PM

Page 74

209. Compound X (C10H14O) dissolves in aqueous sodium hydroxide but is insoluble in aqueous sodium bicarbonate. The proton NMR spectrum of compound X is as follows: δ 1.3 (9H) singlet δ 4.8 (1H) singlet δ 7.1 (4H) multiplet Which of the following is the structure of Compound X?

COOH A.

OH B. Br

OH

C.

Br

O D.

GO ON TO THE NEXT PAGE. 74

01 MCAT FL Test1

06/26/2003

05:31 PM

Page 75

212. Exocrine secretions of the pancreas:

Questions 210 through 214 are NOT based on a descriptive passage.
210. Which of the following compounds readily undergoes E1, SN1, and E2 reactions, but not SN2 reactions? A. B. C. D. CH3CH2CH2Cl (CH3)3COH CH3CH2CH3 (CH3CH2)3CBr

A. B. C. D.

raise blood glucose levels. lower blood glucose levels. regulate metabolic rate. aid in protein and fat digestion.

213. Destroying the cerebellum of a cat would cause significant impairment of normal: A. B. C. D. urine formation. sense of smell. coordinated movement. thermoregulation.

211. A certain drug inhibits ribosomal RNA synthesis. Which of the following eukaryotic organelles would be most affected by the administration of this drug?

214. A cell with a high intracellular K+ concentration, whose plasma membrane is impermeable to K+, is placed in an ATP-rich medium with a low K+ concentration. After several minutes, it is determined that the extracellular concentrations of both K+ and ATP have decreased, while the intracellular K+ concentration has increased. What is the most likely explanation for this phenomenon? A. B. C. A. B. C. D. 1 2 3 4 The K+ passively diffused from the medium into the cell. The K+ entered the cell by way of facilitated transport. The ATP formed a temporary lipid-soluble complex with the K+, thus enabling the potassium to enter the cell. The K+ entered the cell by way of active transport.

D.

STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS SECTION ONLY.

75

01 MCAT FL Test1

06/26/2003

05:31 PM

Page 76

01 MCAT FL Test1

06/26/2003

05:31 PM

Page 77


				
DOCUMENT INFO
Description: Medical School aides -- getting in.