3Biology molecular genetics

MCAT Topical Tests
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Molecular Genetics

Topical Test 1
Time: 42 Minutes
Number of Questions: 30


MCAT

Molecular Genetics
DIRECTIONS: Most of the questions in the following
topical test are organized into groups, with a descriptive
passage preceding each group of questions. Study the
passage, then select the single best answer to each
question in the group. Some of the questions are not
based on a descriptive passage; you must also select the
best answer to these questions. If you are unsure of the
best answer, eliminate the choices that you know are
incorrect, then select an answer from the choices that
remain. Indicate your selection by blackening the
corresponding circle on your answer sheet. A periodic
table is provided below for your use with the questions.
PERIODIC TABLE OF THE ELEMENTS
1

2

H
1.0

He
4.0

3

4

5

6

7

8

9

10

Li
6.9

Be
9.0

B
10.8

C
12.0

N

14.0

O
16.0

F
19.0

Ne
20.2

11

12

13

14

15

16

17

18

Na
23.0

Mg
24.3

Al
27.0

Si
28.1

P
31.0

S
32.1

Cl
35.5

Ar
39.9

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

K
39.1

Ca
40.1

Sc
45.0

Ti
47.9

V
50.9

Cr
52.0

Mn
54.9

Fe
55.8

Co
58.9

Ni
58.7

Cu
63.5

Zn
65.4

Ga
69.7

Ge
72.6

As
74.9

Se
79.0

Br
79.9

Kr
83.8

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

Rb
85.5

Sr
87.6

Y
88.9

Zr
91.2

Nb
92.9

Mo
95.9

Tc
(98)

Ru
101.1

Rh
102.9

Pd
106.4

Ag
107.9

Cd
112.4

In
114.8

Sn
118.7

Sb
121.8

Te
127.6

I
126.9

Xe
131.3

55

56

57

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

Cs
132.9

Ba
137.3

La *
138.9

Hf
178.5

Ta
180.9

W
183.9

Re
186.2

Os
190.2

Ir
192.2

Pt
195.1

Au
197.0

Hg
200.6

Tl
204.4

Pb
207.2

Bi
209.0

Po
(209)

At
(210)

Rn
(222)

87

88

89

104

105

106

107

108

109

Fr
(223)

Ra
226.0

Ac †
227.0

Rf
(261)

Ha
(262)

Unh
(263)

Uns
(262)

Uno
(265)

Une
(267)

58

59

60

61

62

63

64

65

66

67

68

69

70

71

*

Ce
140.1

Pr
140.9

Nd
144.2

Pm
(145)

Sm
150.4

Eu
152.0

Gd
157.3

Tb
158.9

Dy
162.5

Ho
164.9

Er
167.3

Tm
168.9

Yb
173.0

Lu
175.0

90

91

92

93

94

95

96

97

98

99

100

101

102

103

†

Th
232.0

Pa
(231)

U
238.0

Np
(237)

Pu
(244)

Am
(243)

Cm
(247)

Bk
(247)

Cf
(251)

Es
(252)

Fm
(257)

Md
(258)

No
(259)

Lr
(260)

GO ON TO THE NEXT PAGE.

2

as developed by


Molecular Genetics Test 1
Passage I (Questions 1–10)
Acetylcholine (Ach) is one of the most commonly
found neurotransmitters in almost all animals, from fruit flies
to humans. The two major receptors for Ach in humans are
known as nicotinic and muscarinic. When nicotinic receptors
are stimulated at the neuromuscular junction muscle
contraction is produced. Muscarinic receptors can have
various functions.
Acetylcholinesterase (AchE) is the enzyme that breaks
down Ach into choline and an acetyl group, which can then
be reabsorbed by neurons where they are used to synthesize
new Ach molecules. AchE deactivates Ach in a synapse, thus
allowing a controlled length of stimulation and allowing the
cells to reset themselves before the next impulse. AchE
follows the general formula for enzyme kinetics:

k1

k3

ES

E+S

E+P

k2

competitive Ach inhibitors, like Galantamine, less Ach is
destroyed, and thus brain concentrations of the
neurotransmitter increase. Galantamine is a particularly
useful drug because it not only inhibits AchE, but it also is an
allosteric activator of nicotinic acetylcholine receptors,
increasing the amount and effect of Ach in the synapse.
Other types of AchE inhibitors like Tacrine are also used to
treat Alzheimer’s. Table 1 shows an experiment to determine
the effects of Tacrine.

[Ach]
(mM)
Tube 1
Tube 2
Tube 3
Tube 4
Tube 5

[E] (mM)
-8

2.0 x 10
9.5 x 10-5
1.9 x 10-4

1.9 x 10-4
2.2 x 10-2

-5

2.5 x 10
2.5 x 10-5
2.5 x 10-5
2.5 x 10-5
2.5 x 10-5

[Tacrine]
(mM)

Vo (mM/min)

0
0
1.0 x 10-4
1.0 x 10-2
1.0 x 10-4

X
1.75 x 10-1
1.92 x 10-1
1.10 x 10-3
1.92 x 10-1

Table 1

Formula 1

Equation 1
Where Km is the overall rate constant. The velocity of
the reaction is given by the Michaelis-Menten equation:

vo =

v max[ S ]
KM + [ S ]

and

kcat =

v max
[E]

AchE Activity

Km = k2 + k3
k1

32˚

37˚

42˚

Figure 1

Equation 2
For acetylcholinesterase acting on Ach, Km = 9.5 x 10-5
and Kcat = 1.4 x 104. Kcat is a measure of the catalytic ability
or the efficacy of an enzyme.
In addition to being an important tool in endogenous
negative feedback loops, AchE is the target of a variety of
drugs and synthetic compounds and its inhibition can have
many very different impacts.
Organophosphate pesticides bind tightly to AchE and
act as irreversible inhibitors, producing death in many
organisms. Nerve gases are also organophosphates and act in
the same way. These toxins can produce headaches, runny
nose, narrowing of the pupils, and a tightened chest. Lethal
doses produce uncontrollable muscle contractions and death
by asphyxiation.
GO ON TO THE NEXT PAGE.

On the other hand, AchE inhibitors are also used to
treat a number of diseases, most notably Alzheimer’s
Disease. In the later stages of Alzheimer’s, patients have low
levels of Ach in their brains. By employing reversible

KAPLAN

3


MCAT
1.

Which of the following changes in acetylcholinesterase
kinetics would occur after taking galantamine?
A.
B.
C.
D.

2.

5.

Increased Km
Decreased Vm
Decreased k2 + k3
Increased Vm

Some researchers hypothesize that manganese is a
cofactor for acetylcholinesterase. If this is true, which of
the following might be effects of a deficiency of
manganese in the diet?

A.
B.
C.
D.

6.

A. Partial muscle paralysis, pupillary dilation,

confusion
B. Difficulty breathing, muscle spasms, confusion
C. Vasodilatation, headaches, pupillary constriction
D. Bloody nose, difficulty breathing, headache

3.

7.1 x 10-2 mM/min
3.4 x 10-21 mM/min
3.7 x 103 mM/min
7.4 x 10-5 mM/min

Which of the following if true would strengthen the
theory that some of the effects Alzheimer’s disease are
associated with lower levels of acetylcholine in the
brain?
A. Alzheimer’s patients frequently experience
temporary partial paralysis.
B. Alzheimer's plaques are found in areas where Ach
concentration is high.
C. Higher levels of nicotinic Ach receptors are found in
Alzheimer’s affected brains.
D. Levels of endogenous acetylcholinesterase inhibitors
are higher in normal brains.

The acetylcholine in humans and insects are identical.
How would the graph in Figure 1 change for insects?
A.
B.
C.

D.

4.

Which of the following is the velocity of the reaction in
Tube 1?

It shifts to the left.
It shifts to the right.
It would be narrower.
It would be wider.

Based on the information from the passage and table
above, what type of effect does Tacrine demonstrate?
A.
B.
C.
D.

Irreversible inhibition
Reversible competitive inhibition
Reversible noncompetitive inhibition
Allosteric activation

7.

According to the passage, which of the following best
explains the effect of Galantamine on nicotinic
acetylcholine receptors?
A. Galantamine binds to the receptor protein and thus

changes the conformation of the acetylcholine
binding site.
B. Galantamine binds to the receptor and activates a
second messenger.
C. When Galantamine binds to the receptor, it shifts its
reaction velocity versus substrate curve to the right.
D. Galantamine binds to the receptor and stabilizes its
conformation before the receptor binds
acetylcholine.

GO ON TO THE NEXT PAGE.

4

as developed by


Molecular Genetics Test 1
8.

The
competitive
acetycholinesterase
inhibitor
Pyridostigmine bromide is used to effectively treat
myasthenia gravis, a disease that causes debilitating
muscle weakness. Which of the following is a possible
cause of the disease?

10. Which of the following graphs could represent an

allosteric enzyme? (x-axis = Increasing [S], y-axis =
Increasing reaction velocity.)
Inhibited
Uninhibited

A. Ineffective mutated acetylcholine secreted into the
synapse at the neuromuscular junction.
B. A loss of nicotinic receptors at the neuromuscular
junction.
C. Damaged dendrites of motor neurons innervating the
muscles.
D. Overproduction of the neurotransmitter and muscle
spasms.
A.

9.

Some
molecules
of
acetylcholinesterase
are
predominantly hydrophobic and some are hydrophilic. If
a mutation occurs which prevents the production of
hydrophilic acetylcholinesterase, which of the following
might occur?
A. After acetylcholine is released into the
neuromuscular junction, there will be a greater
concentration of it in the middle region of the space
than closer to either membrane.

B. After acetylcholine is released into the
neuromuscular junction, there will be a greater
concentration of it on the nerve and muscle sides of
the synapse than in the middle.
C. More acetylcholine will be released into the
neuromuscular junction, creating a stronger muscle
contraction.
D. More acetylcholine will be broken down, creating a
weaker muscle contraction.

B.

C.

D.

GO ON TO THE NEXT PAGE.

KAPLAN

5


MCAT
Passage II (Questions 11–20)
For many years it was thought that all biological
catalysts were proteins. However that changed when Thomas
Cech discovered that the precursor of a ribosomal RNA in
Tetrahymena could undergo self-splicing to produce an
intron that was able to catalyze the transformation of other

RNA molecules. These RNA molecules with enzymatic
activity were dubbed ribozymes. This discovery provided a
mechanism for enzymatic activity before the development of
proteins early in the evolution of life.
Recently it has been hypothesized that ribozymes, in
addition to being catalysts, might also serve a metabolic
function by regulating the amount of enzymes produced.
Studies have suggested that the 3' end of mRNA of the gene
glmS, which encodes for amidotransferase, can be cleaved
into a ribozyme in a mechanism similar to what Cech
observed in Tetrahymena. Because Amidotransferase is an
enzyme that catalyzes the formation of glucosamine-6phosphate from glutamine and fructose-6-phosphate, its
splicing into a ribozyme would regulate the amount of
enzyme present.
In order to enhance the understanding of the mechanism
by which the glmS gene is converted into a ribozyme,
students in a biology class performed an experiment using a
strain of E. Coli that overexpressed glmS gene as their
control. Three mutants were made by transforming the
control with ampicillin resistant plasmids that overexpressed
fructose-6-phosphate, glutamine, and both fructose-6phosphate and glutamine to respectively produce mutant 1,
mutant 2, and mutant 3. These cells were then plated in
minimal media containing ampicillin and incubated for 24hrs
at 37C.

A

B

C

D

11. Which of the following would not have to be added to
the RT-PCR mixture?
A.
B.
C.
D.

Helicase
Reverse Transcriptase
Polymerase
dNTPs

12. Which of the following represents dish B?
A.
B.
C.
D.

Mutant 1
Mutant 2
Mutant 3
Wild-type

13. How many of the four trials will develop a band in the
southern blot?
A.
B.

C.
D.

0
1
2
3

14. If the formation of the ribozyme is triggered by the
binding of glucosamine-6-phosphate to a binding site on
glmS transcript:
A.
B.
C.
D.

Southern for Muatant 1 will have 2 bands
Southern for Muatant 2 will have 2 bands
Southern for Muatant 3 will have 2 bands
Southern for Wild-type will have 2 bands

Figure 1
Following the 24-hour incubation period, the RNA
from each dish was analyzed using reverse transcriptase
polymerase chain reaction (RT-PCR), Southern Blotting and
hybridization with labeled glmS gene.

15. According to the passage, the formation of the ribozyme
is an example of
A.

B.
C.
D.

Positive regulation
Negative regulation
Inducible system
Cannot be determined from the passage

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6

as developed by


Molecular Genetics Test 1
16. Suppose that the mRNA for glmS is polycistronic
(capable of coding for more than one peptide). How
would this change the results obtained from the southern
blot?
A. It would not change the results
B. Only half the number of ribozymes would be
formed.
C. It depends on the mechanism of ribozyme
formation.
D. It would change the results, the degree to which
depends on the mechanism.

19. Which of the following will not contribute to genetic

variance in E.Coli?
A.
B.
C.
D.

Translocation
Transformation
Transduction
Conjugation

20. How is a ribozyme different from a protein catalyst?

17. Why is RT-PCR a more suitable method than RNA
amplification?
A.
B.
C.
D.

RT-PCR is RNA amplification.
RNA cannot be amplified.
RNA cannot easily be separated and detected.
DNA is more stable than RNA.

A. A ribozyme is used up in the reaction whereas a
protein is not consumed.
B. A ribozyme can not form tertiary structures whereas
proteins can for tertiary and quaternary structures
C. Proteins can have a greater variety of interactions

with their substrates.
D. Proteins are less stable than ribozymes

18. It is possible for the RNA of the gene glmS to lose its
polarity when the 3’ end is cleaved off because
A. RNA is always single stranded and does not need to
be polarized.
B. RNA is normally single stranded and does not need
to be polarized.
C. Polarity in only necessary during replication.
D. The RNA does not loose its polarity.

GO ON TO THE NEXT PAGE.

KAPLAN

7


MCAT
Passage III (Questions 21–30)
Hormones have two main mechanisms of
communicating with cells. They may bind to a cell-surface
receptor or they may interact with a cytosolic receptor.
Those that bind to cell-surface receptors are unable to diffuse
into the cell; instead they activate receptors with cytosolic
domains, which in turn activate a second messenger cascade
that, depending on the hormone and receptor, will turn genes
on or off. Hormones that interact with cytosolic receptors
diffuse through the cell membrane and bind to their receptor.

This hormone-receptor complex then translates into the
nucleus to exert its effects directly.
One hormone that diffuses into the cell is estrogen.
Estrogen receptors are very similar to other intracellular
receptors. They have an N-terminal domain, which functions
as the transcription-activation domain. Near the center of the
protein is the zinc-finger, which binds the DNA. Estrogen
binds at the C-terminal.
The estrogen receptor is capable of diffusing in and out
of the nucleus. In the cytosol, the receptor is attached to
receptor-associated proteins.
Some of these receptorassociated proteins hide the DNA-binding domain of the
estrogen receptor. In the nucleus, in the absence of estrogen,
the estrogen receptor suppresses estrogen-regulated genes by
promoting histone deacetylation.
When estrogen binds to its receptor, the estrogen
receptor dissociates from its chaperone proteins. The
estrogen-receptor complex dimerizes and translocates into the
nucleus. The exact mechanism of translocation is unclear but
appears to occur through interaction of caveolin-1 (a
cytosolic protein) with the estrogen receptor. The dimer
binds to the estrogen response element in DNA, and
promotes hyper-acetylation of the histones. The complex
then recruits additional transcription factors, and the genes
under control of estrogen are transcribed.

21. The amino acid sequence of estrogen and testosterone
receptors will be least similar in the:
A.
B.

C.
D.

N-terminal domain
Center of the protein
C-terminal domain
In the tertiary domain

22. Acetylation of histones most likely:
A. Relaxes DNA to allow transcription factor binding.
B. Coils DNA to allow transcription factor binding.
C. Relaxes DNA to prevent transcription factor
binding.
D. Coils DNA to prevent transcription factor binding.

23. A scientist transforms a cell-line with 3 estrogenregulated genes. If the colony produced contains about
10,000 cells, and the scientist wishes to fully stimulate
the production of the estrogen-regulated proteins without
wasting estrogen, how many estrogen molecules must he
add to the media?
A.
B.
C.
D.

10,000
20,000
30,000
60,000

24. Loss of the zinc-finger domain of the estrogen receptor
will result in:
A. Constitutive transcription of estrogen-regulated
genes.
B. Inability to hyper-acetylate the histones of estrogenregulated genes.
C. Deacetylation of estrogen-regulated genes
D. Inability of the estrogen-receptor dimer complex to
translocate into the nucleus.

25. Hormones that bind to intracellular receptors are
typically:
A.
B.
C.
D.

Lipophilic
Anionic
Amino acids
Nucleic acids

GO ON TO THE NEXT PAGE.

8

as developed by


Molecular Genetics Test 1

26. Estrogen and testosterone are both produced from:
A.
B.
C.
D.

Fatty acids
Cholesterol
Amino acids
Glycogen

29. A mutation is made in which the estrogen receptor is
unable to dissociate from its chaperone protein. This
mutation would most likely lead to:
A. Over-production of estrogen regulated genes.
B. Inability of the estrogen-receptor complex to
dimerize.
C. Inability of the estrogen-receptor complex to bind to
DNA.
D. Inability of estrogen to bind to the receptor.

27. The onset of action of steroids is typically:
A. Short, because the steroids diffuse directly into the
nucleus
B. Short, because the steroids directly bind the DNA
and stimulate transcription
C. Long, because the steroids must diffuse into the
nucleus
D. Long, because the DNA must be transcribed and
translated before the effect occurs

28. Loss of the C-terminal domain of the estrogen receptor
leads to:
A. Constitutive activation of the estrogen response
elements.
B. Inability of the receptor to bind to DNA.
C. Inability of the receptor to translocate into the
nucleus.
D. Constitutive deacetylation of the histones in
estrogen response elements.

30. A mouse model missing the caveolin-1 gene is created.
Females of this model will exhibit:
I.Infertility.
II.An increase in FSH.
III.Enlarged ovaries secondary to increased estrogen
production.
A.
B.
C.
D.

I only
I and III only.
II and III only.
I, II, and III.

END OF TEST

KAPLAN

9


MCAT

THE ANSWER KEY IS ON THE NEXT PAGE

10

as developed by


Molecular Genetics Test 1
ANSWER KEY:
1.
A
2.
B
3.
D
4.
C
5.
D

11.
12.
13.
14.

15.

A
D
D
C
D

21.
22.
23.
24.
25.

C
A
D
B
A

6.
7.
8.
9.
10.

16.
17.
18.
19.

20.

A
D
D
A
C

26.
27.
28.
29.
30.

B
D
D
C
D

KAPLAN

C
A
B
A
D

11

MOLECULAR GENETICS TEST 1 TRANSCRIPT
Passage I (Questions 1–10)
1.
Choice A is the correct answer. The last paragraph tells us that Galantamine is a reversible competitive
inhibitor. A competitive inhibitor is one that can be overcome by adding high concentrations of substrate, so with
enough substrate the reaction will proceed just as fast as without the inhibitor. Therefore the Vmax does not change.
Eliminate (B) and (D). However, the slope of the inhibited curve will be less steep. In other words, to reach ½ Vmax
the concentration of substrate must be greater with the inhibitor than without. Km is the concentration at which ½
Vmax is reached. Therefore in competitive inhibition Km will be greater than without.
(B)
(C)
(D)

Distortion. In competitive inhibition Vm does not change. In non-competitive inhibition,
however, Vm would decrease.
Opposite. If k2 + k3 decreased then Km would decrease. Km increases here, so k2 + k3 must also
increase.
Distortion. Vm does not change here. Vm would never increase due to any type of inhibition.

2.
Choice B is the correct answer. If the cofactor for AchE were missing the enzyme would not function.
This effect would be similar to inhaling nerve gas, so look at paragraph 4. Only (B) and (D) make sense here.
Though not mentioned in the paragraph, since Ach is common signal in the CNS, confusion is a likely side effect.
Answer (D) is very tempting since it almost matches the list in the paragraph, however, these effects were for nerve
gas inhalation. However (D) says bloody nose, whereas the list specifies runny nose. Furthermore, the runny nose
is unlikely to be a direct effect of AchE inhibition since acetylcholine does not directly effect the nose—the gas
irritates the mucous membranes. (B) is the best answer.
(A)
(C)

(D)

Opposite. The muscles will contract uncontrollably, but paralysis will not occur. The pupils
would contract, not dilate.
Distortion. Though the other two occur, there’s no reason to believe that vasodilation would
occur. Vasoconstriction is more likely.
Distortion. Though a very tempting choice, runny nose is an unlikely effect of manganese
deficiency.

3.
Choice D is the correct answer.
Don’t be fooled by the question stem; it’s asking about
acetylcholinesterase, not acetylcholine. All you need to know here is how cold blooded animals compare to warm
blooded ones. AchE in insects needs to work over a larger range of temperatures because the body temperatures of
the insects change with the ambient temperature. We cannot be sure which way the graph shifts—it would depend
on the average temperature of the insect’s environment—but we know that it must be wider. (D) is the correct
answer.
(A)
(B)
(C)

Distortion. It may or may not shift, depending on environment; “cold blooded” does not mean
their bodies are cold.
Distortion. It may or may not shift, depending on environment.
Opposite. The enzyme must function in various temperatures, so this graph would be wider than
in humans.

4.
The correct answer to this question is choice C. Look carefully at the table. In Tube 2, [S] = Km, so V = ½
Vmax. Therefore when [Ach] is doubled in Tube 3, V would reach Vmax if not for the inhibitor. (This is also another

way to find Vmax = 0.35 mM/min). Now, if this were a competitive inhibitor, if sufficient substrate were added, Vmax
would be reached. However, if we look at Tube 5 compared with Tube 3, [Ach] is increased by 2 orders of
magnitude, but the velocity does not change—Vmax for the enzyme with Tacrine has been reached and it’s smaller
than the uninhibited enzyme’s Vmax. This means that Tacrine must be noncompetitive. (C) is the correct answer.
(A)

Distortion. The passage tells us that reversible inhibitors, not irreversible ones, are used to treat
Alzheimer’s. Tacrine is used to treat Alzheimer’s and thus it must not be irreversible.

1


Kaplan MCAT Biological Sciences Test 2 Transcript

(B)
(D)

Opposite.
In competitive inhibition, the inhibitor can be overwhelmed by adding excess
substrate. The inhibitor reaches a lower Vmax than the uninhibited enzyme.
FUD. The passage says Tacrine is an inhibitor, so by definition it can’t be an activator.
Noncompetitive inhibitors are usually allosteric, however.

5.
The correct answer is choice D. Go back to the equations given in the passage. To find Vo, we must
combine the two equations. Substitute kcat [E] for Vmax.

vo =

=

kcat[ E ][ S ]
Km+[ S ]

(1.4 × 104 )(2.5 × 10−5 )(2.0 × 10−8 )
7.0 × 10−9
=
9.5 × 10−5 + 2.0 × 10−8
9.502 × 10−5

Now we need to approximate,

=

7
7
≅
= 7 × 10−5
95, 020 100, 000

We increased the denominator, so the answer should be a bit bigger than 7 x 10-5. Answer (D), 7.4 x 10-5 mM/min,
is the correct answer.
(A)
(B)
(C)

Miscalculation. Order of magnitude error from miscalculating exponents.
Miscalculation. Results from switching Km and kcat.
Miscalculation. Results if the denominator is multiplied instead of added.

6.
The correct answer is choice C. If the levels of acetylcholine in the brain dropped, the body would attempt
to mitigate the impact of such a change. One way it might do that is to produce more receptors for the Ach to bind
to, thus increasing the efficacy of the remaining neurotransmitter. (C) is the correct answer. You can also answer
this question by process of elimination-(A)

(B)

(D)

Out of scope. While Alzheimer’s does affect the brain severely, memory is its major victim. You
might arrive at this answer if you are misled by the fact that Ach is the primary neurotransmitter at
the neuromuscular junction. Partial paralysis could occur if the innervating neurons could not
release a signal to the muscle, but we are specifically talking about the brain here.
Opposite. If plaques are closely involved with either the cause or effect of the disease, then they
should be associated with other causes/effects. If the Ach levels are high in areas that the disease
has effect, this would suggest that Alzheimer’s leads to increased levels of Ach.
Opposite. This is a tricky option. The third paragraph tells us that AchE is subject to negative
feedback loops. Less inhibition means more enzyme, which means less acetylcholine. So this
could strengthen the hypothesis. However, if the body was producing less acetylcholine, it is
likely that endogenous AchE inhibitors would increase in order to retain as much of the Ach as
possible. Therefore, lower levels of inhibitor could suggest that Ach levels in the brain are too
high. This discovery would not conclusively strengthen or weaken the theory.

7.
Choice A is the right answer. The last paragraph states that Galantamine is an allosteric activator of the
nicotinic acetylcholine receptor. This means that it binds to the protein at a secondary site and changes the
receptor’s shape to give it a greater affinity for acetylcholine. (A) is true and is the correct answer.
(B)

Out of scope. This is actually true. When Galantamine binds to the nicotinic Ach receptor, it does
activate a second messenger which causes the cell to express more receptors. However, this is not
discussed in the passage, and you are not expected to bring outside knowledge of Galantamine to
the passage.

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Kaplan MCAT Biological Sciences Test 2 Transcript

(C)
(D)

Opposite. In allosteric activation, this graph would actually shift to the left, that is, the activator
makes the protein do more work with less substrate because it has a greater affinity for substrate.
Distortion. Galantamine stabilizes the receptor-transmitter complex. It makes the empty receptor
less stable and thus increases its affinity for the transmitter—it needs the transmitter to become
stable again.

8.
The correct answer is choice B. Myasthenia gravis is an autoimmune disease in which antibodies attack
the nicotinic receptors on the postsynaptic membrane in the neuromuscular junction. The antibodies prevent
acetylcholine from binding to the receptors (and sometimes destroying them) and thus block the transmission and
prevent muscle contraction. Pyridostigmine bromide inhibits acetylcholinesterase and thus increases the
concentration of the acetylcholine produced at the nerve terminal. This helps to ease effect of lost (or blocked)
receptors by making the most use of the available ones. None of the other answers could both cause the symptoms
and be eased by the treatment. (B) is the correct answer.
(A)
(C)
(D)

Distortion. If the acetylcholine was not usable, preventing its breakdown and increasing its
concentration would not ease the symptoms of the disease.
Distortion. Again, this might cause the symptoms, but increasing Ach in the synapse will not
make up for damage in transmission in the dendrites.
Opposite. Too much Ach could result in spasms, which could result in weakness, but preventing
breakdown of Ach would only increase its concentration and thus could not solve this problem.

9.
Choice A is the correct answer. This is a complicated question. First we need to consider the differences
between the two types of enzyme. A protein that is largely hydrophobic is not going to be found in the cytoplasm or
in the synapse—it will be imbedded in the cell membrane. The hydrophilic enzyme will be free in fluid—likely in
the synapse as paragraph 2 discusses reabsorbing the metabolites from Ach. So if the hydrophilic enzyme is
knocked out, degradation of Ach would occur in the synapse close to the presynaptic or postsynaptic membranes.
More Ach would be present away from the enzymes that degrade it than near them. Thus there would be a
concentration gradient after neurotransmitter release, with more Ach in the middle of the gap than on either side.
(A) is the correct answer.
(B)
(C)
(D)

Opposite. The gradient would more likely occur in the opposite direction.
Distortion. Less degradation will not mean that more neurotransmitter is released.
Opposite. Less Ach will be broken down, most likely, because less enzyme will be present.

10.
The correct answer to this question is choice D. Allosteric enzymes have multiple active sites and their
reaction velocity depends on more than the concentration of one compound, hence the s-shaped curve. (Think of
Hb, an allosteric protein with cooperative binding.) Now to decide between (C) and (D). The choice is clear—an
inhibited enzyme will reach Vmax at higher concentrations of substrate.

(A)
(B)
(C)

Distortion. This enzyme shows no coopertivity. It is more likely to have only one active site.
Distortion. This enzyme shows no coopertivity. It is more likely to have only one active site.
Opposite. Inhibition would result is less product produced per unit of concentration of substrate.
The inhibition curve must shift right.

Passage II (Questions 11–20)
11.
The correct answer is choice A. Since the RNA has to be reverse transcribed into DNA using RT-PCR;
reverse transcriptase, polymerase, dNTPs and primers would be needed. Helicase on the other hand is an ATP
driven protein that unwinds the helix as the DNA is transcribed and is not necessary to carry out PCR.
(B)
(C)
(D)

180°. RT-PCR requires reverse transcriptase to convert the RNA into DNA.
180°. Polymeraes is required to elongate the DNA chains.
180°. dNTPs are needed to form the DNA.

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Kaplan MCAT Biological Sciences Test 2 Transcript

12.
The correct answer is choice D. According to the passage, the mutants were made by transforming the
wild-type bacteria with three different plasmids, each containing ampicillin resistant genes. Since the wild-type

bacteria was not transformed with such a plasmid, it would be unable to grow in a plate containing ampicillin.
(A)
(B)
(C)

180°. Mutant 1 was transformed with an ampicillin resistant plasmid.
180°. Mutant 2 was transformed with an ampicillin resistant plasmid.
180°. Mutant 3 was transformed with an ampicillin resistant plasmid.

13.
The correct answer is choice D. In the southern blot, the RNA from the plates are reverse transcribed to
DNA and then hybridized with the glmS gene. Thus plates A, C and D which represent mutants 1,2, and 3 (not
necessarily in that order), all of which overexpress glmS DNA will develop bands in the southern blot. In the event
where mRNA were to be cleaved in order to produce a ribozyme, both the cleaved portions of the mRNA would
hybridize with the labeled gene. Wild-type however will not develop a band because it does not grow in the
ampicillin-medium (dish B).
14.
The correct answer is C. Binding of glucosamine-6-phosphate to a binding site on glmS transcript will
cause the glmS transcript to be spliced into two pieces, both of which will hybridize with the labeled glmS in the
southern blot producing two bands. Mutant 3 is the only mutant that overexpresses both glutamine and fructose-6phosphate, both of which are necessary to produce glucosamine-6-phosphate and thus is the only mutant that will
have 2 bands in the southern.
(A)
(B)
(D)

180°. Mutant 1 only expresses fructose-6-phosphate, and so is not able to produce glucosamine6-phosphate. Without glucosamine-6-phosphate, glmS is not split into 2.
180°. Mutant 2 only expresses glutamine, and so is not able to produce glucosamine-6-phosphate.
Without glucosamine-6-phosphate, glmS is not split into 2.
180°. Wild-type does not express fructose-6-phosphate nor glutamine, and so is not able to
produce glucosamine-6-phosphate. Without glucosamine-6-phosphate, glmS is not split into 2.

15.
The correct answer is choice D. According to the passage, the 3' end of mRNA of the gene glmS is spliced
in a similar fashion to ribosomal RNA in Tetrahymena . The passage does not discuss what induces this splicing and
so we do not know whether this is positive regulation or negative regulation. Also, the passage does not discuss
repressors or inducers, so answer choice C can be eliminated.
(A)
(B)
(C)

Out of Scope. Regulation of ribozyme production is not discussed in the passage.
Out of Scope. Regulation of ribozyme production is not discussed in the passage.
Out of Scope. Regulation of ribozyme production is not discussed in the passage.

16.
The correct answer is choice A. mRNA that is polycistronic is RNA that can code for more than one
polypeptide. Since the sequence of nucleotides is the same it will be spliced in the same manner to produce a
ribozyme. Therefore, if mRNA for glmS is polycistronic, it would still produce the same strand of DNA when it
undergoes RT-PCR to yield the same results. The way in which mRNA is translated does not affect the results of
the PCR.
(B)
(C)
(D)

180°. The way in which mRNA is translated does not affect the results of the PCR.
180°. The way in which mRNA is translated does not affect the results of the PCR.
180°. The way in which mRNA is translated does not affect the results of the PCR.

17.
The correct answer is choice D. Although there are methods to amplify RNA, RT-PCR is used to analyze

that transcription because DNA is much more stable and easier to work with than RNA. RT-PCR however does not
technically amplify RNA, it transcribes the RNA to DNA then amplifies that DNA. Thus answer choice D is the
correct answer.
(A)

Distortion. RT-PCR however does not technically amplify RNA, it transcribes the RNA to DNA
then amplifies that DNA.

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Kaplan MCAT Biological Sciences Test 2 Transcript

(B)
(C)

180°. RNA can be amplified.
180°. RNA can be detected in a manner similar to Southern blot, called Northern blot.

18.
The correct answer is choice D. All nucleic acids are polarized. It is impossible for any nucleic strand to
not have both a 3’ end and a 5’ end. Thus answer choice D is the correct answer.
(A)
(B)
(C)

180°. Polarity refers to the presence of a 3' and 5' end. When the 3' end is cleaved, a new 3' end
is created.
180°. Polarity refers to the presence of a 3' and 5' end. When the 3' end is cleaved, a new 3' end is
created.

180°. Polarity refers to the presence of a 3' and 5' end. When the 3' end is cleaved, a new 3' end
is created.

19.
The correct answer is choice A. Translocation is the transfer of a chromosomal segment to a
nonhomologous chromosome. Since prokaryotes have only one circular chromosome, they cannot undergo
translocation.
(B)
(C)
(D)

180°. Transformation is the incorporation of a plasmid by a bacterial cell. It leads to genetic
variation by the incorporation of new genes into the chromosome.
180°. Transduction is the transfer of genetic material from one bacterial cell to another through a
bacteriophage. It leads to genetic variation by the spreading of new genes to new cells.
180°. Conjugation is the transfer of genetic material between two bacterial cells that are
temporarily joined. It leads to genetic variation by the spread of new genes to new cells.

20.
The correct answer is choice C. A ribozyme is an RNA catalyst that catalyzes reactions without itself
being consumed; therefore, it is a true enzyme. Although a ribozyme can form tertiary structures it has only 4
different types of nucleic acids whereas proteins have over 20 types of amino acids giving proteins a greater variety
of possible interactions. Finally, ribozymes comprised of RNA are much less stable than proteins.
(A)
(B)
(D)

180°. Enzymes are not used up in the reaction. Ribozymes are defined as enzymes, so they are
not used in up in reactions.
180°. Ribozymes can form tertiary structures.

180°. Proteins are more stable than RNA.

Passage III (Questions 21–30)
21.
The correct answer is choice C. The passage states that the intracellular receptors for hormones are similar.
The N-terminal domains contain the transcription-activation domain, and the middle of the protein contains the
DNA-binding domain. The C-terminal domain is specific for the hormone-this is the site where the hormone binds.
Testosterone and estrogen receptors will be least similar in this domain.
(A)
(B)
(D)

Opposite. The N-terminal domains of the estrogen and testosterone receptors will be similar
because they contain the transcription-activation domain.
Opposite. The center of the estrogen and testosterone receptors will be similar because they
contain the DNA-binding domain.
Out of Scope. The tertiary structure of the proteins will probably be similar; however, the
question asks about the amino acid sequence, which is the primary structure.

22.
The correct choice is A. When DNA is relaxed it is called euchromatin, or true chromatin. It is in this
form that it can be transcribed. The passage states that when the estrogen-receptor complex binds, the histones are
acetylated and transcription is activated. Therefore, acetylation promotes transcription factor binding. (A) is the
correct answer.
(B)

Opposite. Coiled DNA (heterochromatin) cannot bind transcription factors.

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Kaplan MCAT Biological Sciences Test 2 Transcript

(C)
(D)

Opposite. Relaxed DNA allows transcription factor binding.
Opposite. Acetylation allows relaxation of the DNA into euchromatin to allow transcription
factor binding.

23.
The correct choice is D. Each estrogen-regulated gene requires 2 molecules of estrogen to promote
transcription because the estrogen-receptor complex must dimerize before it translocates into the nucleus. If a cell
contains 3 estrogen-regulated genes, 6 molecules of estrogen are required per cell. For 10,000 cells, 60,000
molecules of estrogen are required.
(A)
(B)
(C)

Miscalculation. Only allows for 1 molecule of estrogen per cell.
Miscalculation. Allows for the required estrogen dimer, but does not account for the 3 genes in
each cell.
Miscalculation. Allows for the 3 genes in each cell, but does not account for the required
estrogen dimer.

24.
Choice B is the correct answer. The zinc-finger domain of the estrogen receptor is the DNA binding
domain. If the receptor does not have a zinc-finger domain then the receptor cannot bind to the DNA to acetylate
(when bound to estrogen) or deacetylate (when not bound to estrogen) the histones. So the answer is (B).
(A)

(C)
(D)

Opposite. Constitutive transcription of estrogen-regulated genes would result if the estrogen
receptor were always in the active conformation. Loss of the zinc-finger domain would result in
constitutive deactivation of the estrogen-regulated genes.
Opposite. Estrogen receptor not bound to estrogen deacetylates the estrogen regulated genes.
However, it still requires the zinc finger domain to do so. Therefore, loss of the zinc-finger
domain would result in inability to deacetylate the estrogen-regulated genes.
Out of Scope. The passage does not discuss how the estrogen-receptor dimer translocates into the
nucleus. In fact, scientists do not know how this translocation occurs.

25.
Choice A is the correct answer. Hormones that bind to intracellular receptors must be lipophilic in order to
cross the lipophilic cell membrane. They cannot be charged, nor can they be very polar molecules.
(B)
(C)
(D)

Opposite. Anions (negatively charged ions) cannot cross the cell membrane.
Opposite. Amino acids are polar molecules that cannot cross the cell membrane.
Opposite. Nucleic acids are polar molecules that cannot cross the cell membrane.

26.
Choice B is the correct answer Steroid hormones, such as estrogen and testosterone, are produced from
cholesterol.
(A)
(C)
(D)

Distortion. Fatty acids are used in energy production and storage.
Distortion. Amino acids are used in the synthesis of proteins.
Distortion. Glycogen is used for energy production and storage.

27.
Choice D is the correct answer. Steroid hormones typically exert long-lasting, delayed effects because the
proteins must be produced before the steroid takes effect. It could therefore by several days before the release of the
steroid into the bloodstream results in the production of proteins regulated by estrogen.
(A)
(B)
(C)

Distortion. Although the steroids diffuse directly into the nucleus, the onset of action is the time
until the protein is produced, which is long.
Distortion. The onset of action is the time it takes to produce the protein, which is long starting
from transcription.
Distortion. The length of time for the steroid to diffuse into the nucleus is not the ratedetermining step.

28.
Choice D is the right answer. The C-terminal domain contains the estrogen-binding domain. If the
estrogen-binding domain is absent, the estrogen receptor will never be able to bind estrogen. It will therefore
always behave as it does when no estrogen is present and continuously deacetylate the histones in estrogen response
elements.

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Kaplan MCAT Biological Sciences Test 2 Transcript

(A)

(B)
(C)

Opposite. Constitutive activation of the estrogen response elements would result from a mutation
that allowed estrogen to remain bound or a mutation that altered the conformation of the receptor
to the estrogen bound state.
Distortion. The zinc-finger domain is the DNA-binding domain.
Distortion. The receptor is capable of diffusing into the nucleus in the absence of estrogen; loss
of the C-terminal domain is unlikely to effect this.

29.
Choice C is the right answer. The passage only gives one function for the estrogen receptor chaperone
protein: hiding the DNA binding site. Therefore, if the receptor is unable to dissociate from the chaperone, the most
likely consequence will be inability to bind to DNA. Although (B) and (D) may occur, the passage gives no
evidence to that effect.
(A)
(B)
(D)

Opposite. A mutation in the chaperone would lead to underproduction of estrogen regulated
genes.
Out of Scope. Although this could occur, the passage gives no evidence or support of this
mechanism.
Out of Scope. Although this could occur, the passage gives no evidence or support of this
mechanism.

30.
Choice D is the right answer. If caveolin-1 is missing, the estrogen-receptor complex cannot reach the

nucleus and exert estrogen's effects on protein production. The body will perceive the defect as a lack of estrogen,
and try to compensate by over-producing estrogen. The pituitary will increase its production of FSH (follicle
stimulating hormone) to augment estrogen production, and the ovaries may enlarge due to cell production in
attempts to increase estrogen production. However, oocytes and the endometrium will be unable to respond to the
estrogen, resulting in infertility.

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