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MCAT AAMC Practice Exam 1 | 176 Questions with Correct Answers and Rationale | Graded A+ | New Update

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Voorbeeld 4 van de 72 pagina's

MCAT AAMC Practice Exam 1 | 176 Questions with Correct Answers and Rationale | Graded A+ | New Update

Voorbeeld van de inhoud

MCAT AAMC Practice Exam 1 | 176 Questions
with Correct Answers and Rationale | Graded A+ |
New 2026-2027 Update


Question 1
In the chromatography of the reaction mixture, water absorbed on cellulose functioned as the
stationary phase. What was the principal factor determining the migration of individual
components in the sample?
A. Hydrogen bonding
B. Solute concentration
C. Stationary phase concentration
D. Thickness of paper
Correct Answer: A. Hydrogen bonding
Rationale: The relative amount of hydrogen bonding to the stationary phase will determine the
relative rate of migration of the various components in the sample. In paper chromatography
with water as the stationary phase, compounds that form stronger hydrogen bonds with water
will move more slowly, while those with weaker hydrogen bonding will migrate faster.


Question 2
What assumption is being made if scientists conclude that aspartic acid was formed by the
prebiological synthesis in the passage?
A. Aspartic acid is unstable at temperatures below 150°C.
B. All of the malic acid underwent the dehydration reaction to form fumaric/maleic acid.
C. Compound A and cyanide were available on primitive Earth.
D. The reaction between ammonia and fumaric acid was catalyzed by the presence of water.
Correct Answer: C. Compound A and cyanide were available on primitive Earth.
Rationale: In order for the experimental reaction sequence to be relevant to the primordial
formation of aspartic acid, the starting materials used (Compound A and cyanide) are assumed
to have been available on primitive Earth. This is a fundamental assumption for prebiotic
synthesis experiments.

,Question 3
According to the developed chromatography plate shown below, what is the approximate Rf
value of aspartic acid?
aspartic acid = 2; solvent front = 10
A. 0.20
B. 0.50
C. 5
D. 10
Correct Answer: A. 0.20
Rationale: Rf is the ratio of the distance travelled by the analyte relative to the solvent front
during a chromatographic separation. Aspartic acid travelled two units, while the solvent front
travelled ten units, giving an Rf of 2/10 = 0.20 for aspartic acid.


Question 4
Which of the following statements does NOT correctly describe the dehydration of malic acid
to fumaric acid and maleic acid?
A. The reaction occurs most readily with tertiary alcohols.
B. The reaction involves the loss of a water molecule.
C. The reaction has a carbocation intermediate.
D. The reaction is stereospecific.
Correct Answer: D. The reaction is stereospecific.
Rationale: The fact that both fumaric and maleic acid are produced means that the dehydration
of malic acid is NOT stereospecific. A stereospecific reaction would produce only one
stereoisomer.


Question 5
What type of functional group is formed when aspartic acid reacts with another amino acid to
form a peptide bond?
A. An amine group
B. An aldehyde group

,C. An amide group
D. A carboxyl group
Correct Answer: C. An amide group
Rationale: The functional group that forms during peptide bond formation is known as an
amide group. A peptide bond forms between the carboxyl group of one amino acid and the
amino group of another, resulting in an amide linkage (-CO-NH-).


Question 6
If 2-pentanol replaces 1-pentanol in the reaction shown in Figure 3 (SN2), the rate of
substitution is less because:
A. the C-O bond in 2-pentanol is stronger than the C-O bond in 1-pentanol.
B. there is a competing elimination reaction that slows the rate of substitution.
C. there is more steric hindrance at the oxygen atom in 2-pentanol than in 1-pentanol, making
protonation less likely.
D. there is more steric hindrance at the 2-position of 2-pentanol than at the 1-position of 1-
pentanol.
Correct Answer: D. there is more steric hindrance at the 2-position of 2-pentanol than at the
1-position of 1-pentanol.
Rationale: The rate of substitution of protonated alcohols is subject to steric hindrance. The 2-
position of 2-pentanol is more sterically hindered than the 1-position of 1-pentanol, which
inhibits the ability of nucleophiles to collide with the reacting electrophilic center and slows the
rate of reaction.


Question 7
If a solution containing the compounds shown in Figure 4 is injected into a gas-liquid
chromatograph, the first peak observed in the GC trace is attributable to which compound?
A. Methyl-2-butanol
B. Methyl-2-butene
C. Chloro-2-methylbutane
D. Bromo-2-methylbutane
Correct Answer: B. Methyl-2-butene

, Rationale: 2-methyl-2-butene will exhibit the lowest molecular weight and also the weakest
intermolecular forces of attraction. This substance will therefore migrate the fastest and be the
first peak in the gas chromatograph (GC) trace.


Question 8
Acetic acid and ethanol react to form an ester product as shown below. In determining which
reactant loses the -OH group, which of the following isotopic substitutions would be most
useful?
A. Replace the acidic H of acetic acid with D
B. Replace the alcoholic H of ethanol with D
C. Replace the carbonyl oxygen of acetic acid with O-18
D. Replace the hydroxyl oxygen of ethanol with O-18
Correct Answer: D. Replace the hydroxyl oxygen of ethanol with O-18
Rationale: This experiment involves labeling a group which does not exchange with other
groups present prior to reaction and will therefore give information about the true identity of
the groups which are exchanged during the reaction. Labeling the hydroxyl oxygen of ethanol
with O-18 allows tracking of which molecule's oxygen ends up in the water product versus the
ester.


Question 9
A person whose eye has a lens-to-retina distance of 2.0 cm can only clearly see objects that
are closer than 1.0 m away. What is the strength S of the person's eye lens? (Note: Use the
thin lens formula 1/O + 1/I = S)
A. -50 D
B. -10 D
C. 51 D
D. 55 D
Correct Answer: C. 51 D
Rationale: The strength of the eye lens is equal to the inverse of the focal length of the eye lens.
Its numerical value is given by (1 m)⁻¹ + (0.02 m)⁻¹ = 1 D + 50 D = 51 D.


Question 10
Which statement correctly describes how enzymes affect chemical reactions? Stabilization of:

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