MCAT Practice Test Questions and Verified
Answers
Increased intermolecular attractions cause the ratio PV/(RT) to decline because
individual molecules:
A. eventually combine with other molecules, decreasing the number of particles in the
container.
B. lose kinetic energy to potential energy and strike the side of the container with less
force.
C. increase in speed due to electron repulsion and strike the side of the container with
increased force.
D. transfer electrons during collisions with other molecules in the container.
Correct Answer
B.
Intermolecular attractions increase the potential energy between molecules,
decreasing their kinetic energy. Note that the molecules do not undergo any
reactions, eliminating choices A and D. Choice C is incorrect because intermolecular
attractions do not result in electron repulsion.
Page 1 of 53
,Question 2
In a facultative anaerobe, which of the following processes occurs under both aerobic
and anaerobic conditions?
A. Fermentation
B. Krebs cycle
C. Glycolysis
D. Oxidative phosphorylation
Correct Answer
C.
A facultative anaerobe can survive through fermentation when oxygen is not
available but will use oxidative respiration when oxygen is available. Glycolysis will
occur under both aerobic conditions (in which case the pyruvate will go on to enter
the Krebs cycle as acetyl-CoA) and under anaerobic conditions (fermentation
reduces pyruvate to alcohol or lactate), so choice C is correct. Fermentation occurs
only in anaerobic conditions (choice A is wrong), while the Krebs cycle and oxidative
phosphorylation can occur only in aerobic conditions (choices B and D are wrong).
Question 3
1-Methylcyclohexanol reacts with HBr to form 1-bromo-1-methylcyclohexane. The
mechanism for this reaction is likely to be an:
A. SN1 reaction.
B. SN2 reaction.
C. Nucleophilic addition.
D. Addition-elimination.
Correct Answer
A.
Since no double bonds are formed or broken, this must be a substitution reaction
(eliminate choices C and D). Since the hydroxyl in the starting material is on a
tertiary carbon atom, the mechanism cannot occur by a bimolecular pathway
(eliminate choice B.) The protonated OH group (under acidic conditions) will leave
as water to yield a tertiary carbocation, which will be attacked by the bromide ion to
give the product.
Page 2 of 53
,Question 4
The standard potential for the reaction K+ + e- --> K(s) = -2.93 V, as referenced
against 2H+ + 2e- --> arrow H2(set to 0.0 V by definition). If solid potassium is placed
into an aqueous solution of HCl, then:
A. H2(g) and KCl(aq) are produced.
B. Cl2(g) and KCl(aq) are produced.
C. Cl2(g), H2(g), and KCl(aq) are produced.
D. no reaction occurs.
Correct Answer
A.
Since the reduction of K+ as given in the question has a potential more negative
than the reduction of H+, the oxidation of K(s) to K+ with transfer of electrons to
H+ will have a positive potential and be spontaneous. This eliminates choice D.
Hydrochloric acid (HCl, a strong acid) will dissociate nearly completely in water into
H+ and Cl-. In the presence of a reducing agent (K), H+ ions can accept electrons
and be reduced to H2, and K+ and Cl- will remain in solution. Since there is no
oxidant which can accept the extra electron from Cl-, Cl2 will not be formed,
eliminating choices B and C.
Question 5
An object is floating in a fluid of 1.5 specific gravity. If the volume of the fluid
displaced by the floating object is 5 × 10-3 m3, what is the object's mass?
A. 2.5 kg
B. 5.0 kg
C. 7.5 kg
D. Cannot be determined from the information given
Correct Answer
C.
Because the object is floating, the object's weight is balanced by the buoyant force;
that is, mg = ρfluidVsubg, or, after canceling the g's, m = ρfluidVsub. With ρfluid =
1.5ρH2O = 1500 kg/m3 and Vsub = 5 × 10-3 m3, we find that m = ρfluidVsub =
(1500 kg/m3)(5 × 10-3 m3) = 7.5 kg
Page 3 of 53
, Question 6
Can glucogenic amino acids be converted into glucose?
A. Yes: pyruvate and oxaloacetate can be converted directly into glyceraldehyde-3-P,
which is a major intermediate in both gluconeogenesis and glycolysis.
B. Yes: pyruvate and Krebs cycle intermediates can be converted into oxaloacetate,
then phosphoenolpyruvate, which can enter gluconeogenesis.
C. No: pyruvate and Krebs cycle intermediates are formed as part of glucose
breakdown and this process is important to generate ATP for the cell.
D. No: glucose is obtained from the diet and stored in the liver; it cannot be made as
a new molecule because cellular respiration has several steps with a -ΔG.
Correct Answer
B.
This is a typical two by two question. The passage says that glucogenic amino acids
are broken down into citric acid cycle intermediates or pyruvate. The Krebs cycle
regenerates oxaloacetate (OAA), and in the first step of gluconeogenesis, pyruvate
is also converted into OAA (by the enzyme pyruvate carboxylase). OAA is converted
into phosphoenolpyruvate (PEP) by the enzyme phosphoenolpyruvate
carboxykinase (or PEPCK). Gluconeogenesis can then continue to run, and will
generate glucose from these non-carbohydrate precursor molecules (choice B is
correct). Although glyceraldehyde-3-P is a major intermediate in both
gluconeogenesis and glycolysis, pyruvate and oxaloacetate cannot be converted
directly into this molecule (choice A is incorrect). Both choices C and D are incorrect
because they start with "No"; as discussed above, glucogenic amino acids can be
converted into glucose (choices C and D are wrong).
Question 7
ketogenic amino acids are broken down into
and can be used for what?
what are the 2 ketogenic AA?
Correct Answer
acetyl-CoA
which can be used in fatty acid biosynthesis or ketogenesis
lysine and leucine
Page 4 of 53
Answers
Increased intermolecular attractions cause the ratio PV/(RT) to decline because
individual molecules:
A. eventually combine with other molecules, decreasing the number of particles in the
container.
B. lose kinetic energy to potential energy and strike the side of the container with less
force.
C. increase in speed due to electron repulsion and strike the side of the container with
increased force.
D. transfer electrons during collisions with other molecules in the container.
Correct Answer
B.
Intermolecular attractions increase the potential energy between molecules,
decreasing their kinetic energy. Note that the molecules do not undergo any
reactions, eliminating choices A and D. Choice C is incorrect because intermolecular
attractions do not result in electron repulsion.
Page 1 of 53
,Question 2
In a facultative anaerobe, which of the following processes occurs under both aerobic
and anaerobic conditions?
A. Fermentation
B. Krebs cycle
C. Glycolysis
D. Oxidative phosphorylation
Correct Answer
C.
A facultative anaerobe can survive through fermentation when oxygen is not
available but will use oxidative respiration when oxygen is available. Glycolysis will
occur under both aerobic conditions (in which case the pyruvate will go on to enter
the Krebs cycle as acetyl-CoA) and under anaerobic conditions (fermentation
reduces pyruvate to alcohol or lactate), so choice C is correct. Fermentation occurs
only in anaerobic conditions (choice A is wrong), while the Krebs cycle and oxidative
phosphorylation can occur only in aerobic conditions (choices B and D are wrong).
Question 3
1-Methylcyclohexanol reacts with HBr to form 1-bromo-1-methylcyclohexane. The
mechanism for this reaction is likely to be an:
A. SN1 reaction.
B. SN2 reaction.
C. Nucleophilic addition.
D. Addition-elimination.
Correct Answer
A.
Since no double bonds are formed or broken, this must be a substitution reaction
(eliminate choices C and D). Since the hydroxyl in the starting material is on a
tertiary carbon atom, the mechanism cannot occur by a bimolecular pathway
(eliminate choice B.) The protonated OH group (under acidic conditions) will leave
as water to yield a tertiary carbocation, which will be attacked by the bromide ion to
give the product.
Page 2 of 53
,Question 4
The standard potential for the reaction K+ + e- --> K(s) = -2.93 V, as referenced
against 2H+ + 2e- --> arrow H2(set to 0.0 V by definition). If solid potassium is placed
into an aqueous solution of HCl, then:
A. H2(g) and KCl(aq) are produced.
B. Cl2(g) and KCl(aq) are produced.
C. Cl2(g), H2(g), and KCl(aq) are produced.
D. no reaction occurs.
Correct Answer
A.
Since the reduction of K+ as given in the question has a potential more negative
than the reduction of H+, the oxidation of K(s) to K+ with transfer of electrons to
H+ will have a positive potential and be spontaneous. This eliminates choice D.
Hydrochloric acid (HCl, a strong acid) will dissociate nearly completely in water into
H+ and Cl-. In the presence of a reducing agent (K), H+ ions can accept electrons
and be reduced to H2, and K+ and Cl- will remain in solution. Since there is no
oxidant which can accept the extra electron from Cl-, Cl2 will not be formed,
eliminating choices B and C.
Question 5
An object is floating in a fluid of 1.5 specific gravity. If the volume of the fluid
displaced by the floating object is 5 × 10-3 m3, what is the object's mass?
A. 2.5 kg
B. 5.0 kg
C. 7.5 kg
D. Cannot be determined from the information given
Correct Answer
C.
Because the object is floating, the object's weight is balanced by the buoyant force;
that is, mg = ρfluidVsubg, or, after canceling the g's, m = ρfluidVsub. With ρfluid =
1.5ρH2O = 1500 kg/m3 and Vsub = 5 × 10-3 m3, we find that m = ρfluidVsub =
(1500 kg/m3)(5 × 10-3 m3) = 7.5 kg
Page 3 of 53
, Question 6
Can glucogenic amino acids be converted into glucose?
A. Yes: pyruvate and oxaloacetate can be converted directly into glyceraldehyde-3-P,
which is a major intermediate in both gluconeogenesis and glycolysis.
B. Yes: pyruvate and Krebs cycle intermediates can be converted into oxaloacetate,
then phosphoenolpyruvate, which can enter gluconeogenesis.
C. No: pyruvate and Krebs cycle intermediates are formed as part of glucose
breakdown and this process is important to generate ATP for the cell.
D. No: glucose is obtained from the diet and stored in the liver; it cannot be made as
a new molecule because cellular respiration has several steps with a -ΔG.
Correct Answer
B.
This is a typical two by two question. The passage says that glucogenic amino acids
are broken down into citric acid cycle intermediates or pyruvate. The Krebs cycle
regenerates oxaloacetate (OAA), and in the first step of gluconeogenesis, pyruvate
is also converted into OAA (by the enzyme pyruvate carboxylase). OAA is converted
into phosphoenolpyruvate (PEP) by the enzyme phosphoenolpyruvate
carboxykinase (or PEPCK). Gluconeogenesis can then continue to run, and will
generate glucose from these non-carbohydrate precursor molecules (choice B is
correct). Although glyceraldehyde-3-P is a major intermediate in both
gluconeogenesis and glycolysis, pyruvate and oxaloacetate cannot be converted
directly into this molecule (choice A is incorrect). Both choices C and D are incorrect
because they start with "No"; as discussed above, glucogenic amino acids can be
converted into glucose (choices C and D are wrong).
Question 7
ketogenic amino acids are broken down into
and can be used for what?
what are the 2 ketogenic AA?
Correct Answer
acetyl-CoA
which can be used in fatty acid biosynthesis or ketogenesis
lysine and leucine
Page 4 of 53
