MCAT Practice Test Princeton Review Exam
Questions with Complete Solutions (Latest
2025/026).
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. - CORRECT ANSWERS 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. -
CORRECT ANSWERS 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. - CORRECT ANSWERS 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. - CORRECT ANSWERS 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. - CORRECT ANSWERS 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 - CORRECT ANSWERS 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. - CORRECT ANSWERS 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. - CORRECT ANSWERS
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° - CORRECT ANSWERS 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. - CORRECT ANSWERS 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. - CORRECT ANSWERS 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)
With this in mind, there will be no situations wherein the solution is free of precipitate (eliminate choice
A). As the dissolved [NaCl] concentration increases, AgCl will be precipitated from solution, which will
enable additional AgI to dissolve (eliminate choice C). The increased dissolution of AgI will cause the
increase in [I-], even as [Ag+] levels remain low.
SEE OTHERSIDE FIRST
This is a two by two question. In the reaction shown in figure 1, raclopride-H+ donates a proton in the
first step (acting as a Brønsted-Lowry acid). Lewis bases are electron pair donors, and as H+ has no
electrons choices B and D can be eliminated. Raclopride-H+ has a hydrophobic halogenated aromatic
portion, and a hydrophilic side chain (protonated amine), making it an amphiphilic or amphipathic
molecule. The maximum number of stereoisomers a compound may have is given by the formula 2n,
where n is the number of chiral centers present. Raclopride-H+ has one chiral center, so it has two
stereoisomers (C is wrong and A is correct). - CORRECT ANSWERS Raclopride-H+ can best be
described as:
A. an amphiphilic molecule that has two stereoisomers.
B. a Lewis base that has two stereoisomers.
C. an amphiphilic molecule that has four stereoisomers.
D. a Lewis base that has four stereoisomers.
Questions with Complete Solutions (Latest
2025/026).
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. - CORRECT ANSWERS 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. -
CORRECT ANSWERS 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. - CORRECT ANSWERS 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. - CORRECT ANSWERS 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. - CORRECT ANSWERS 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 - CORRECT ANSWERS 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. - CORRECT ANSWERS 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. - CORRECT ANSWERS
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° - CORRECT ANSWERS 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. - CORRECT ANSWERS 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. - CORRECT ANSWERS 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)
With this in mind, there will be no situations wherein the solution is free of precipitate (eliminate choice
A). As the dissolved [NaCl] concentration increases, AgCl will be precipitated from solution, which will
enable additional AgI to dissolve (eliminate choice C). The increased dissolution of AgI will cause the
increase in [I-], even as [Ag+] levels remain low.
SEE OTHERSIDE FIRST
This is a two by two question. In the reaction shown in figure 1, raclopride-H+ donates a proton in the
first step (acting as a Brønsted-Lowry acid). Lewis bases are electron pair donors, and as H+ has no
electrons choices B and D can be eliminated. Raclopride-H+ has a hydrophobic halogenated aromatic
portion, and a hydrophilic side chain (protonated amine), making it an amphiphilic or amphipathic
molecule. The maximum number of stereoisomers a compound may have is given by the formula 2n,
where n is the number of chiral centers present. Raclopride-H+ has one chiral center, so it has two
stereoisomers (C is wrong and A is correct). - CORRECT ANSWERS Raclopride-H+ can best be
described as:
A. an amphiphilic molecule that has two stereoisomers.
B. a Lewis base that has two stereoisomers.
C. an amphiphilic molecule that has four stereoisomers.
D. a Lewis base that has four stereoisomers.
