2026 AAMC MCAT EXAM UPDATE | STUDY
GUIDE WITH PRACTICE QUESTIONS,
ANSWERS, AND RATIONALES
| GRADED A+ | GUARANTEED SUCCESS
Updated 2026 Questions and Answers
100% Verified Exam Prep and Comprehensive
Rationales Included
,C/P: The concentration of enzyme for each experiment A) 2.5 × 10-2 s-1
was 5.0 μM. What is kcat for the reaction at pH 4.5 with
NO chloride added when Compound 3 is the substrate? The answer to this question is A. The fact that the rate of product formation did not
vary over time for the first 5 minutes implies that the enzyme was saturated with
Rate of reaction = 125 nM/s substrate. Under these conditions, kcat = Vmax/[E] = (125 nM/s)/5.0 μM = 2.5 × 10-2
s-1.
A) 2.5 × 10-2 s-1
B) 1.3 × 102 s-1
C) 5.3 × 103 s-1
D) 7.0 × 105 s-1
kcat, Vmax, [E] kcat = Vmax/[E]
C/P: Absorption of ultraviolet light by organic molecules B) Excitation of bound electrons
always results in what process?
A) Bond breaking The answer to this question is B. The absorption of ultraviolet light by organic
B) Excitation of bound electrons molecules always results in electronic excitation. Bond breaking can subsequently
C) Vibration of atoms in polar bonds result, as can ionization or bond vibration, but none of these processes are
D) Ejection of bound electrons guaranteed to result from the absorption of ultraviolet light.
C/P: Four organic compounds: 2-butanone, n-pentane, A) n-Pentane → 2-butanone → n-butanol → propanoic acid
propanoic acid, and n-butanol, present as a mixture, are
separated by column chromatography using silica gel The answer to this question is A. The four compounds have comparable molecular
with benzene as the eluent. What is the expected order weights, so the order of elution will depend on the polarity of the molecule. Since
of elution of these four organic compounds from first to silica gel serves as the stationary phase for the experiment, increasing the polarity
last? of the eluting molecule will increase its affinity for the stationary phase and
increase the elution time (decreased Rf).
A) n-Pentane → 2-butanone → n-butanol → propanoic
acid
B) n-Pentane → n-butanol → 2-butanone → propanoic
acid
C) Propanoic acid → n-butanol → 2-butanone → n-
pentane
D) Propanoic acid → 2-butanone → n-butanol → n-
pentane
C/P: The half-life of a radioactive material is: D) the time it takes for half of all the radioactive nuclei to decay into their
daughter nuclei.
A) half the time it takes for all of the radioactive nuclei to
decay into radioactive nuclei. The answer to this question is D because the half-life of a radioactive material is
B) half the time it takes for all of the radioactive nuclei to defined as the time it takes for half of all the radioactive nuclei to decay into their
decay into their daughter nuclei. daughter nuclei, which may or may not also be radioactive.
C) the time it takes for half of all the radioactive nuclei to
decay into radioactive nuclei.
D) the time it takes for half of all the radioactive nuclei to
decay into their daughter nuclei.
,C/P: A person is sitting in a chair. Why must the person D) to keep the body in equilibrium while rising
either lean forward or slide their feet under the chair in
order to stand up? The answer to this question is D because as the person is attempting to stand, the
only support comes from the feet on the ground. The person is in equilibrium only
A) to increase the force required to stand up when the center of mass is directly above their feet. Otherwise, if the person did
B) to use the friction with the ground not lean forward or slide the feet under the chair, the person would fall backward
C) to reduce the energy required to stand up due to the large torque created by the combination of the weight of the body
D) to keep the body in equilibrium while rising (applied at the person's center of mass) and the distance along the horizontal
between the center of mass and the support point.
C/P: The side chain of tryptophan will give rise to the D) part of a fully folded protein
largest CD signal in the near UV region when:
A) present as a free amino acid The answer to this question is D because tryptophan has an aromatic side chain
B) part of an a-helix that will give rise to a significant CD signal in the near UV region if it is found in a
C) part of a B-sheet fully folded protein.
D) part of a fully folded protein
C/P: Which amino acid will contribute to the CD signal in C) Ala
the far UV region, but NOT the near UV region, when part
of a fully folded protein?
"Asymmetry resulting from tertiary structural features
causes the largest increase in CD signal intensity in the
near UV region of peptides. The side chains of amino acid
residues absorb in this region.
The peptide bond absorbs in the far UV region (190-250
nm). The CD signals of these bonds are dramatically
impacted by their proximity to secondary structural
elements."
A) Trp
B) Phe
C) Ala
D) Tyr
, C/P: Based on the relative energy of the absorbed B) An aromatic side chain; the absorbed photon energy is lower.
electromagnetic radiation, which absorber, a peptide
bond or an aromatic side chain, exhibits an electronic The answer to this question is B because aromatic side chains absorb in the near
excited state that is closer in energy to the ground state? UV region of the electromagnetic spectrum, which has longer wavelengths, and
hence lower energy, than peptide bonds. Because the energy of the photon
"Asymmetry resulting from tertiary structural features matches the energy gap between the ground and the excited state, this implies
causes the largest increase in CD signal intensity in the that the aromatic side chain has more closely spaced energy levels.
near UV region of peptides. The side chains of amino acid
residues absorb in this region.
The peptide bond absorbs in the far UV region (190-250
nm). The CD signals of these bonds are dramatically
impacted by their proximity to secondary structural
elements."
A) An aromatic side chain; the absorbed photon energy
is higher.
B) An aromatic side chain; the absorbed photon energy is
lower.
C) A peptide bond; the absorbed photon energy is
higher.
D) A peptide bond; the absorbed photon energy is lower.
C/P: What is the net charge of sT-loop at pH 7.2? C) 0
"A synthetic peptide with the amino acid sequence The answer to this question is C because at pH 7.2, the N-terminus will be
KTFCGPEYLA was generated as a mimic of the T-loop. positively charged and the C-terminus will be negatively charged. In addition, the
This synthetic T-loop (sT-loop) was incubated with 32P- lysine side chain will carry one positive charge and the glutamic acid side chain
labeled ATP in the presence of PDK1 for different time will carry one negative charge.
periods at 37 ° C and pH 7.2, and the amount of
radioactivity incorporated into sT-loop was measured by
detection of β- decay."
A) -2
B) -1
C) 0
D) +1
C/P: In designing the experiment, the researchers used D) δP-32 ATP
which type of P-32 labeled ATP?
A) aP32-ATP The answer to this question is C because the phosphoryl transfer from kinases
B) BP32-ATP comes from the γ-phosphate of ATP. Therefore, the experiment should require
C) γP32-ATP γ32P-ATP.
D) δP-32 ATP
GUIDE WITH PRACTICE QUESTIONS,
ANSWERS, AND RATIONALES
| GRADED A+ | GUARANTEED SUCCESS
Updated 2026 Questions and Answers
100% Verified Exam Prep and Comprehensive
Rationales Included
,C/P: The concentration of enzyme for each experiment A) 2.5 × 10-2 s-1
was 5.0 μM. What is kcat for the reaction at pH 4.5 with
NO chloride added when Compound 3 is the substrate? The answer to this question is A. The fact that the rate of product formation did not
vary over time for the first 5 minutes implies that the enzyme was saturated with
Rate of reaction = 125 nM/s substrate. Under these conditions, kcat = Vmax/[E] = (125 nM/s)/5.0 μM = 2.5 × 10-2
s-1.
A) 2.5 × 10-2 s-1
B) 1.3 × 102 s-1
C) 5.3 × 103 s-1
D) 7.0 × 105 s-1
kcat, Vmax, [E] kcat = Vmax/[E]
C/P: Absorption of ultraviolet light by organic molecules B) Excitation of bound electrons
always results in what process?
A) Bond breaking The answer to this question is B. The absorption of ultraviolet light by organic
B) Excitation of bound electrons molecules always results in electronic excitation. Bond breaking can subsequently
C) Vibration of atoms in polar bonds result, as can ionization or bond vibration, but none of these processes are
D) Ejection of bound electrons guaranteed to result from the absorption of ultraviolet light.
C/P: Four organic compounds: 2-butanone, n-pentane, A) n-Pentane → 2-butanone → n-butanol → propanoic acid
propanoic acid, and n-butanol, present as a mixture, are
separated by column chromatography using silica gel The answer to this question is A. The four compounds have comparable molecular
with benzene as the eluent. What is the expected order weights, so the order of elution will depend on the polarity of the molecule. Since
of elution of these four organic compounds from first to silica gel serves as the stationary phase for the experiment, increasing the polarity
last? of the eluting molecule will increase its affinity for the stationary phase and
increase the elution time (decreased Rf).
A) n-Pentane → 2-butanone → n-butanol → propanoic
acid
B) n-Pentane → n-butanol → 2-butanone → propanoic
acid
C) Propanoic acid → n-butanol → 2-butanone → n-
pentane
D) Propanoic acid → 2-butanone → n-butanol → n-
pentane
C/P: The half-life of a radioactive material is: D) the time it takes for half of all the radioactive nuclei to decay into their
daughter nuclei.
A) half the time it takes for all of the radioactive nuclei to
decay into radioactive nuclei. The answer to this question is D because the half-life of a radioactive material is
B) half the time it takes for all of the radioactive nuclei to defined as the time it takes for half of all the radioactive nuclei to decay into their
decay into their daughter nuclei. daughter nuclei, which may or may not also be radioactive.
C) the time it takes for half of all the radioactive nuclei to
decay into radioactive nuclei.
D) the time it takes for half of all the radioactive nuclei to
decay into their daughter nuclei.
,C/P: A person is sitting in a chair. Why must the person D) to keep the body in equilibrium while rising
either lean forward or slide their feet under the chair in
order to stand up? The answer to this question is D because as the person is attempting to stand, the
only support comes from the feet on the ground. The person is in equilibrium only
A) to increase the force required to stand up when the center of mass is directly above their feet. Otherwise, if the person did
B) to use the friction with the ground not lean forward or slide the feet under the chair, the person would fall backward
C) to reduce the energy required to stand up due to the large torque created by the combination of the weight of the body
D) to keep the body in equilibrium while rising (applied at the person's center of mass) and the distance along the horizontal
between the center of mass and the support point.
C/P: The side chain of tryptophan will give rise to the D) part of a fully folded protein
largest CD signal in the near UV region when:
A) present as a free amino acid The answer to this question is D because tryptophan has an aromatic side chain
B) part of an a-helix that will give rise to a significant CD signal in the near UV region if it is found in a
C) part of a B-sheet fully folded protein.
D) part of a fully folded protein
C/P: Which amino acid will contribute to the CD signal in C) Ala
the far UV region, but NOT the near UV region, when part
of a fully folded protein?
"Asymmetry resulting from tertiary structural features
causes the largest increase in CD signal intensity in the
near UV region of peptides. The side chains of amino acid
residues absorb in this region.
The peptide bond absorbs in the far UV region (190-250
nm). The CD signals of these bonds are dramatically
impacted by their proximity to secondary structural
elements."
A) Trp
B) Phe
C) Ala
D) Tyr
, C/P: Based on the relative energy of the absorbed B) An aromatic side chain; the absorbed photon energy is lower.
electromagnetic radiation, which absorber, a peptide
bond or an aromatic side chain, exhibits an electronic The answer to this question is B because aromatic side chains absorb in the near
excited state that is closer in energy to the ground state? UV region of the electromagnetic spectrum, which has longer wavelengths, and
hence lower energy, than peptide bonds. Because the energy of the photon
"Asymmetry resulting from tertiary structural features matches the energy gap between the ground and the excited state, this implies
causes the largest increase in CD signal intensity in the that the aromatic side chain has more closely spaced energy levels.
near UV region of peptides. The side chains of amino acid
residues absorb in this region.
The peptide bond absorbs in the far UV region (190-250
nm). The CD signals of these bonds are dramatically
impacted by their proximity to secondary structural
elements."
A) An aromatic side chain; the absorbed photon energy
is higher.
B) An aromatic side chain; the absorbed photon energy is
lower.
C) A peptide bond; the absorbed photon energy is
higher.
D) A peptide bond; the absorbed photon energy is lower.
C/P: What is the net charge of sT-loop at pH 7.2? C) 0
"A synthetic peptide with the amino acid sequence The answer to this question is C because at pH 7.2, the N-terminus will be
KTFCGPEYLA was generated as a mimic of the T-loop. positively charged and the C-terminus will be negatively charged. In addition, the
This synthetic T-loop (sT-loop) was incubated with 32P- lysine side chain will carry one positive charge and the glutamic acid side chain
labeled ATP in the presence of PDK1 for different time will carry one negative charge.
periods at 37 ° C and pH 7.2, and the amount of
radioactivity incorporated into sT-loop was measured by
detection of β- decay."
A) -2
B) -1
C) 0
D) +1
C/P: In designing the experiment, the researchers used D) δP-32 ATP
which type of P-32 labeled ATP?
A) aP32-ATP The answer to this question is C because the phosphoryl transfer from kinases
B) BP32-ATP comes from the γ-phosphate of ATP. Therefore, the experiment should require
C) γP32-ATP γ32P-ATP.
D) δP-32 ATP
