Princeton Review MCAT Practice Test-
Graded A
Sickle-cell anemia (HbS) results from a substitution to valine from glutamic acid at
position 6 of the β chain of hemoglobin. Which of the following best explains why the
isoelectric point of HbS is higher than that of HbA?
A. The side chain of glutamic acid is less acidic than that of valine.
B. Glutamic acid is isoelectric at a lower pH than is valine.
C. Valine is isoelectric at a lower pH than is glutamic acid.
D. Glutamic acid has a net charge of 0 at its isoelectric point. - ANS-B. Glutamic acid is
isoelectric at a lower pH than is valine.
The isoelectric point is the pH at which an amphoteric molecule has a net electric
charge of zero. Glutamic acid is more acidic, not less acidic than valine (A is wrong).
Glutamate would require more acidic pH to neutralize its charge (C is wrong). D is a
true, but irrelevant statement. The best answer is B. The negative charge of glutamic
acid would require a more acidic pH to neutralize, making the isoelectric point of HbA
(with glutamic acid) lower than that of HbS (with valine).
A mixture of aspartate and phenylalanine is separated into its component molecules by
thin layer chromatography on a silica plate eluted with benzene.
Which of the following best explains why the separation occurs?
A. Aspartate will move farther with the mobile phase, because it has a polar side chain.
B. Aspartate will move farther with the mobile phase, because it has a nonpolar side
chain.
C. Phenylalanine will move farther with the mobile phase, because it has a polar side
chain.
D. Phenylalanine will move farther with the mobile phase, because it has a nonpolar
side chain. - ANS-D. Phenylalanine will move farther with the mobile phase, because it
has a nonpolar side chain.
The side chain on aspartate is -CH2COO-, which is very polar, while the side chain on
phenylalanine is -CH2Ph, which is nonpolar. Since "like dissolves like," and the mobile
phase is nonpolar, the molecule which is less polar (phenylalanine, in this case) will
move farther with the mobile phase.
SEE OTHERSIDE FIRST
D. Thickness of the membrane Correct Answer
,"Charge stored per unit voltage," Q / V, is the definition of capacitance, C. The equation
for the capacitance of a parallel-plate capacitor is C = κε0A / d, where κ is the dielectric
constant, ε0 is a universal constant (the permittivity of free space), A is the area of each
plate, and d is the distance between the plates. Of the choices given, only D,
decreasing the thickness of the membrane (that is, decreasing d), would increase the
capacitance, C. - ANS-Ion flow in neurons can be characterized as an electrical circuit
for both the resting neuron (Figure 1) and the active neuron (Figure 2). The membrane
is a capacitor, slow leakage channels are a 25 MΩ resistor, and the Na+/K+ pump is a
voltage generator. In a resting axon, there is no net transfer of charge across the axon
membrane. Figure 2 includes the additional Na+ influx (a 4 kΩ resistor) of an action
potential. Other ion fluxes are ignored. According to the figures, decreasing which of the
following would create the greatest increase in charge stored per unit voltage on an
axon membrane in its rest state?
A. Leakage channel resistance
B. Na+ channel resistance
C. Area of the membrane surfaces
D. Thickness of the membrane
Across the membrane of an axon in its rest state:
A. the potential is higher on the outside of the cell than the inside, and electric field lines
run from the outside to the inside.
B. the potential is higher on the outside of the cell than the inside, and electric field lines
run from the inside to the outside.
C. the potential is lower on the outside of the cell than the inside, and electric field lines
run from the outside to the inside.
D. the potential is lower on the outside of the cell than the inside, and electric field lines
run from the inside to the outside. - ANS-A. the potential is higher on the outside of the
cell than the inside, and electric field lines run from the outside to the inside.
The voltage across the membrane is -70 mV from the exterior to the interior of the cell;
thus, the potential is higher on the exterior of the cell (C and D are wrong). Since the
potential is higher on the exterior than in the interior, the electric field lines must point
from the outside to the inside (choice A). IN PHYSICS, FIELD LINES GO TOWARD
THE MORE NEGATIVE AREA.
Which of the following statements regarding RNA molecules is NOT true?
A. RNAs can act as enzymes to catalyze reactions.
B. Some RNAs have more than four different types of bases.
C. Some RNAs are synthesized in the nucleolus.
D. RNAs are insusceptible to alkaline hydrolysis. - ANS-D. RNAs are insusceptible to
alkaline hydrolysis.
,RNA molecules have decreased stability compared to DNA in part because of their
susceptibility to alkaline hydrolysis due to the presence of hydroxyl group at 2'-C
position (choice D is not true of RNA and is the correct answer choice). Some RNAs
have enzymatic function (such as in telomerase) and they are termed ribozymes (A is
true). tRNA has unique and modified bases apart from the traditional four bases A,U,C,
and G (such as inosine, B is true). rRNA is synthesized in the nucleolus (C is true).
Two blocks are suspended from the ends of a massless meter stick; a 4kg weight from
one side and a 1kg weight from the opposite side. How far from the center of the stick
must the rope be attached in order to maintain rotational equilibrium?
A. 10 cm
B. 20 cm
C. 25 cm
D. 30 cm - ANS-D. 30 cm
A Q-switched laser can be used to treat skin blemishes and to remove tattoos. The Q
switch momentarily interrupts the inducing light creating a build-up of energy within the
crystal. This does not increase the overall energy of the laser, but concentrates it into
shorter time periods or pulses. A longer interruption with the Q-switch most likely would
increase the:
A. total amount of work done by the laser.
B. power of each laser pulse.
C. wavelength of the laser light.
D. frequency of the laser light. - ANS-B. power of each laser pulse.
Since the overall energy of the laser does not change, neither will the frequency,
wavelength, nor work done by the laser. This leaves choice B: Concentrating the energy
into a shorter time period increases the power of each pulse (since power equals
energy delivered per unit time, by definition).
A CO2 laser used as a laser scalpel produces a beam of laser light with a wavelength of
10.6 µm. Compared to the excited electrons in a laser with a wavelength of 1.06um, the
excited electrons in the CO2 laser most likely have:
A. greater mass.
B. less mass.
C. a greater energy difference between their normal state and their excited state.
D. a smaller energy difference between their normal state and their excited state. -
ANS-D. a smaller energy difference between their normal state and their excited state.
The wavelength of the CO2 laser is 10 times greater than the 1.06 wavelength laser.
Therefore, the frequency and the energy of the CO2 laser are 10 times lower. Since the
laser light is the energy released by transitions of electrons dropping to a lower energy
level, less emitted energy implies a smaller difference between the energy levels. The
, mass of an electron is independent of its atomic or molecular energy state (A and B are
wrong).
If the front mirror in a laser, which is made of glass of refractive index 3/2, were
repositioned so that the laser beam strikes at an angle of 30° to its normal, what would
be the angle of reflection?
A. sin-1(1/3)
B. 30°
C. sin-1(1/31/2)
D. 60° - ANS-B. 30°
If the laser beam strikes the mirror at an angle of 30° relative to the normal, then this is
the angle of incidence, and, by the Law of Reflection, it is also the angle of reflection.
The refractive index of the mirror is irrelevant.
SEE OTHERSIDE FIRST
The figure indicates that when [AgI] = 1.5 x 10-15 M, the concentration of NaI = 0.06 M.
In 500 mL (0.5 L), this solution should contain 0.030 mol NaI. Since the molar mass of
NaI is 150 g/mol, this amounts to 4.5 g of NaI. - ANS-How many grams of NaI should be
added to 500 mL of a saturated solution of AgI to make a solution that is 1.5 x 10-15 M
Ag+?
A. 9.0 g
B. 4.5 g
C. 0.06 g
D. 0.03 g
If a fully saturated solution of AgI, with precipitate present, were treated with NaCl
instead of NaI, which of the following observations is likely?
A. As NaCl is added, all precipitates are dissolved into the aqueous solution.
B. The decrease in [AgI] is even more drastic than with the addition of NaI in Figure 1.
C. There is no change in the amount of undissolved AgI.
D. The concentration of [I-] increases. - ANS-D. The concentration of [I-] increases.
Unlike NaI, NaCl does not have a common ion with AgI and will therefore NOT cause a
decrease in the solubility for AgI with increasing concentration (eliminate choice B). The
following will act as a competing reaction when [Cl-] concentrations become sufficiently
large:
Ag+ (aq) + Cl- (aq) → AgCl (s)
Graded A
Sickle-cell anemia (HbS) results from a substitution to valine from glutamic acid at
position 6 of the β chain of hemoglobin. Which of the following best explains why the
isoelectric point of HbS is higher than that of HbA?
A. The side chain of glutamic acid is less acidic than that of valine.
B. Glutamic acid is isoelectric at a lower pH than is valine.
C. Valine is isoelectric at a lower pH than is glutamic acid.
D. Glutamic acid has a net charge of 0 at its isoelectric point. - ANS-B. Glutamic acid is
isoelectric at a lower pH than is valine.
The isoelectric point is the pH at which an amphoteric molecule has a net electric
charge of zero. Glutamic acid is more acidic, not less acidic than valine (A is wrong).
Glutamate would require more acidic pH to neutralize its charge (C is wrong). D is a
true, but irrelevant statement. The best answer is B. The negative charge of glutamic
acid would require a more acidic pH to neutralize, making the isoelectric point of HbA
(with glutamic acid) lower than that of HbS (with valine).
A mixture of aspartate and phenylalanine is separated into its component molecules by
thin layer chromatography on a silica plate eluted with benzene.
Which of the following best explains why the separation occurs?
A. Aspartate will move farther with the mobile phase, because it has a polar side chain.
B. Aspartate will move farther with the mobile phase, because it has a nonpolar side
chain.
C. Phenylalanine will move farther with the mobile phase, because it has a polar side
chain.
D. Phenylalanine will move farther with the mobile phase, because it has a nonpolar
side chain. - ANS-D. Phenylalanine will move farther with the mobile phase, because it
has a nonpolar side chain.
The side chain on aspartate is -CH2COO-, which is very polar, while the side chain on
phenylalanine is -CH2Ph, which is nonpolar. Since "like dissolves like," and the mobile
phase is nonpolar, the molecule which is less polar (phenylalanine, in this case) will
move farther with the mobile phase.
SEE OTHERSIDE FIRST
D. Thickness of the membrane Correct Answer
,"Charge stored per unit voltage," Q / V, is the definition of capacitance, C. The equation
for the capacitance of a parallel-plate capacitor is C = κε0A / d, where κ is the dielectric
constant, ε0 is a universal constant (the permittivity of free space), A is the area of each
plate, and d is the distance between the plates. Of the choices given, only D,
decreasing the thickness of the membrane (that is, decreasing d), would increase the
capacitance, C. - ANS-Ion flow in neurons can be characterized as an electrical circuit
for both the resting neuron (Figure 1) and the active neuron (Figure 2). The membrane
is a capacitor, slow leakage channels are a 25 MΩ resistor, and the Na+/K+ pump is a
voltage generator. In a resting axon, there is no net transfer of charge across the axon
membrane. Figure 2 includes the additional Na+ influx (a 4 kΩ resistor) of an action
potential. Other ion fluxes are ignored. According to the figures, decreasing which of the
following would create the greatest increase in charge stored per unit voltage on an
axon membrane in its rest state?
A. Leakage channel resistance
B. Na+ channel resistance
C. Area of the membrane surfaces
D. Thickness of the membrane
Across the membrane of an axon in its rest state:
A. the potential is higher on the outside of the cell than the inside, and electric field lines
run from the outside to the inside.
B. the potential is higher on the outside of the cell than the inside, and electric field lines
run from the inside to the outside.
C. the potential is lower on the outside of the cell than the inside, and electric field lines
run from the outside to the inside.
D. the potential is lower on the outside of the cell than the inside, and electric field lines
run from the inside to the outside. - ANS-A. the potential is higher on the outside of the
cell than the inside, and electric field lines run from the outside to the inside.
The voltage across the membrane is -70 mV from the exterior to the interior of the cell;
thus, the potential is higher on the exterior of the cell (C and D are wrong). Since the
potential is higher on the exterior than in the interior, the electric field lines must point
from the outside to the inside (choice A). IN PHYSICS, FIELD LINES GO TOWARD
THE MORE NEGATIVE AREA.
Which of the following statements regarding RNA molecules is NOT true?
A. RNAs can act as enzymes to catalyze reactions.
B. Some RNAs have more than four different types of bases.
C. Some RNAs are synthesized in the nucleolus.
D. RNAs are insusceptible to alkaline hydrolysis. - ANS-D. RNAs are insusceptible to
alkaline hydrolysis.
,RNA molecules have decreased stability compared to DNA in part because of their
susceptibility to alkaline hydrolysis due to the presence of hydroxyl group at 2'-C
position (choice D is not true of RNA and is the correct answer choice). Some RNAs
have enzymatic function (such as in telomerase) and they are termed ribozymes (A is
true). tRNA has unique and modified bases apart from the traditional four bases A,U,C,
and G (such as inosine, B is true). rRNA is synthesized in the nucleolus (C is true).
Two blocks are suspended from the ends of a massless meter stick; a 4kg weight from
one side and a 1kg weight from the opposite side. How far from the center of the stick
must the rope be attached in order to maintain rotational equilibrium?
A. 10 cm
B. 20 cm
C. 25 cm
D. 30 cm - ANS-D. 30 cm
A Q-switched laser can be used to treat skin blemishes and to remove tattoos. The Q
switch momentarily interrupts the inducing light creating a build-up of energy within the
crystal. This does not increase the overall energy of the laser, but concentrates it into
shorter time periods or pulses. A longer interruption with the Q-switch most likely would
increase the:
A. total amount of work done by the laser.
B. power of each laser pulse.
C. wavelength of the laser light.
D. frequency of the laser light. - ANS-B. power of each laser pulse.
Since the overall energy of the laser does not change, neither will the frequency,
wavelength, nor work done by the laser. This leaves choice B: Concentrating the energy
into a shorter time period increases the power of each pulse (since power equals
energy delivered per unit time, by definition).
A CO2 laser used as a laser scalpel produces a beam of laser light with a wavelength of
10.6 µm. Compared to the excited electrons in a laser with a wavelength of 1.06um, the
excited electrons in the CO2 laser most likely have:
A. greater mass.
B. less mass.
C. a greater energy difference between their normal state and their excited state.
D. a smaller energy difference between their normal state and their excited state. -
ANS-D. a smaller energy difference between their normal state and their excited state.
The wavelength of the CO2 laser is 10 times greater than the 1.06 wavelength laser.
Therefore, the frequency and the energy of the CO2 laser are 10 times lower. Since the
laser light is the energy released by transitions of electrons dropping to a lower energy
level, less emitted energy implies a smaller difference between the energy levels. The
, mass of an electron is independent of its atomic or molecular energy state (A and B are
wrong).
If the front mirror in a laser, which is made of glass of refractive index 3/2, were
repositioned so that the laser beam strikes at an angle of 30° to its normal, what would
be the angle of reflection?
A. sin-1(1/3)
B. 30°
C. sin-1(1/31/2)
D. 60° - ANS-B. 30°
If the laser beam strikes the mirror at an angle of 30° relative to the normal, then this is
the angle of incidence, and, by the Law of Reflection, it is also the angle of reflection.
The refractive index of the mirror is irrelevant.
SEE OTHERSIDE FIRST
The figure indicates that when [AgI] = 1.5 x 10-15 M, the concentration of NaI = 0.06 M.
In 500 mL (0.5 L), this solution should contain 0.030 mol NaI. Since the molar mass of
NaI is 150 g/mol, this amounts to 4.5 g of NaI. - ANS-How many grams of NaI should be
added to 500 mL of a saturated solution of AgI to make a solution that is 1.5 x 10-15 M
Ag+?
A. 9.0 g
B. 4.5 g
C. 0.06 g
D. 0.03 g
If a fully saturated solution of AgI, with precipitate present, were treated with NaCl
instead of NaI, which of the following observations is likely?
A. As NaCl is added, all precipitates are dissolved into the aqueous solution.
B. The decrease in [AgI] is even more drastic than with the addition of NaI in Figure 1.
C. There is no change in the amount of undissolved AgI.
D. The concentration of [I-] increases. - ANS-D. The concentration of [I-] increases.
Unlike NaI, NaCl does not have a common ion with AgI and will therefore NOT cause a
decrease in the solubility for AgI with increasing concentration (eliminate choice B). The
following will act as a competing reaction when [Cl-] concentrations become sufficiently
large:
Ag+ (aq) + Cl- (aq) → AgCl (s)
