1Generalized eukaryotic cell test w solutions

BIOLOGY TOPICAL:
Generalized Eukaryotic Cell
Test 1
Time: 21 Minutes*
Number of Questions: 16

* The timing restrictions for the science topical tests are optional. If
you are using this test for the sole purpose of content
reinforcement, you may want to disregard the time limit.


MCAT

DIRECTIONS: Most of the questions in the following
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
H
1.0

2

2
He
4.0

3
Li
6.9

4
Be
9.0

5
B
10.8

6
C
12.0

7
N
14.0

8
O
16.0

9
F
19.0

10
Ne
20.2

11
Na
23.0

12
Mg
24.3

13
Al
27.0

14
Si
28.1

15
P
31.0

16
S

32.1

17
Cl
35.5

18
Ar
39.9

19
K
39.1

20
Ca
40.1

21
Sc
45.0

22
Ti
47.9

23
V
50.9

24
Cr
52.0

25
Mn
54.9

26
Fe
55.8

27
Co
58.9

28
Ni
58.7

29
Cu
63.5

30
Zn
65.4

31
Ga

69.7

32
Ge
72.6

33
As
74.9

34
Se
79.0

35
Br
79.9

36
Kr
83.8

37
Rb
85.5

38
Sr
87.6

39
Y
88.9

40
Zr
91.2

41
Nb
92.9

42
Mo
95.9

43
Tc
(98)

44
Ru
101.1

45
Rh
102.9

46
Pd

106.4

47
Ag
107.9

48
Cd
112.4

49
In
114.8

50
Sn
118.7

51
Sb
121.8

52
Te
127.6

53
I
126.9

54
Xe
131.3

55
Cs
132.9

56
Ba
137.3

57
La *
138.9

72
Hf
178.5

73
Ta
180.9

74
W
183.9

75
Re

186.2

76
Os
190.2

77
Ir
192.2

78
Pt
195.1

79
Au
197.0

80
Hg
200.6

81
Tl
204.4

82
Pb
207.2

83
Bi
209.0

84
Po
(209)

85
At
(210)

86
Rn
(222)

87
Fr
(223)

88
Ra
226.0

89
Ac †
227.0

104
Rf

(261)

105
Ha
(262)

106
Unh
(263)

107
Uns
(262)

108
Uno
(265)

109
Une
(267)

*

58
Ce
140.1

59
Pr

140.9

60
Nd
144.2

61
Pm
(145)

62
Sm
150.4

63
Eu
152.0

64
Gd
157.3

65
Tb
158.9

66
Dy
162.5

67
Ho
164.9

68
Er
167.3

69
Tm
168.9

70
Yb
173.0

71
Lu
175.0

†

90
Th
232.0

91
Pa
(231)

92
U
238.0

93
Np
(237)

94
Pu
(244)

95
Am
(243)

96
Cm
(247)

97
Bk
(247)

98
Cf
(251)

99
Es

(252)

100
Fm
(257)

101
Md
(258)

102
No
(259)

103
Lr
(260)

as developed by


Generalized Eukaryotic Cell Test 1
Passage I (Questions 1–6)
It has been proposed that the nervous tissue of the
electric eel, Electrophorus electricus, may be useful in the
repair of damaged nerves in humans. To isolate the
various subcellular structures within the nerve cells of
Electrophorus electricus, a researcher uses rate-zonal
centrifugation. This technique separates the various
subcellular structures by size and density. After

centrifugation, the large dense structures typically form a
pellet at the bottom of the test tube, while the lighter,
less dense structures remain suspended in the supernatant
(the fluid layer). By centrifuging at different speeds and
durations, different organelles can be isolated.
The eel nerve cells are suspended in a 0.25 M
sucrose solution, which is isotonic to the cells. Most of
the cells are broken open by stirring this suspension in a
high-speed blender to form a cell homogenate. The
homogenate is then filtered to remove any unbroken cells.
The filtered homogenate is centrifuged at increasing speeds
four times, with a pellet collected after each step, and the
supernatant transferred to each successive tube. This
procedure is outlined in Figure 1.
Centrifuge

Pour out:
15,000g x
5 min

600g x
10 min

Filtered
Pellet 1
homogenate
Supernatant
1

Pellet 2

Pour out:
100,000g x
60 min

Pellet 3

Supernatant
2

Supernatant
3

1 . Analysis of Pellet 2 showed that it possessed high
succinate cytochrome C reductase activity. This
enzyme oxidizes succinic acid to fumaric acid, both
intermediates of the Krebs cycle. Therefore, this pellet
most likely contains:
A.
B.
C.
D.

mitochondria.
plasma membrane.
endoplasmic reticulum.
cytoplasm.

2 . To determine the activity of the enzyme lactate
dehydrogenase, the compound DPNH is added to each

sample of supernatant. For every molecule of
pyruvate that is reduced to lactate by lactate
dehydrogenase, a molecule of DPNH is oxidized:
Pyruvate + DPNH → Lactate + DPN
In which of the following samples would you expect
lactate dehydrogenase activity to be found?

Pour out:
300,000g x
2 hr

I.
II.
III.
IV.

Pellet 4
Supernatant
4

A.
B.
C.
D.

Supernatant 1
Supernatant 2
Supernatant 3
Supernatant 4

I only
III and IV only
II, III, and IV only
I, II, III, and IV

Figure 1

GO ON TO THE NEXT PAGE.

KAPLAN

3


MCAT
3 . Which of the following molecules would you expect
to find in the pellet containing the plama membrane?
O

I.
N
N

O
-O

P

NH

O

N

NH

4 . Which of the following structures would most likely
be found in a higher concentration in the plasma
membrane of eel nerve cells as compared to the
plasma membrane of eel epithelial cells?
A.
B.
C.
D.

Glycoproteins
Cholesterol
Acetylcholine receptors
Phospholipids

O

OH

H

H

OH

H
OH

O-

II.
O

P

O-

O
CH2

O

HC

O

C

H2C

O

C

(CH)16

(CH)16

CH3
CH3

O

NH2

III.
N
O

O
-O

P

O

N

O

O-

H

H

OH

H

H

A.
B.
C.
D.

H

I only
II only
III only
I and III only

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4

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Generalized Eukaryotic Cell Test 1
5 . The enzyme DNase is added to Pellet 1. After this
enzymatic digestion, which of the following would
most likely be found within the digested pellet?
A.

B.
C.
D.

mRNA
Histones
Rhodopsin
Elongation Factor 2

6 . The researcher believes that Pellet 4 contains
ribosomal subunits. Which of the following
observations would most support this hypothesis?
A . Pellet 4 was found to contain uncharged tRNA.
B . Pellet 4 was found to contain two molecules
with sedimentation coefficients of 40S and 60S.
C . Pellet 4 was found to contain the enzyme
peptidyl transferase.
D . Supernatant 4 was found to contain unprocessed
proteins.

Passage II (Questions 7–12)
The yeast Schizosaccharomyces pombe divides by
fission and is thus an ideal microorganism for the study of
cell division in higher eukaryotic cells. A scientist
believes that she has identified the region of the S. pombe
chromosome containing the DNA sequence that includes
the centromere. From this region she has isolated two
different fragments of DNA: Fragments I and II. The
centromere is the region of the chromosome that
maintains the attachment of sister chromatids during

meiosis I and includes the site of attachment to the
microtubules of the meiotic and mitotic spindles. During
division, the spindle must both assemble and disassemble
its tubulin subunits to properly function in the
segregation of chromosomes.
To test which of the two fragments contains the
centromere region, the scientist takes a mutant strain of S .
pombe cells that cannot grow unless adenine is provided
in the growth medium, and performs the following set of
experiments.
Experiment 1
Mutant cells are inoculated onto an agar plate
containing minimal medium.
Experiment 2
The normal gene for adenine biosynthesis (adenine+)
is inserted into a plasmid, which is then inserted into the
mutant cells. The cells are then inoculated onto agar plates
containing minimal medium. (A plasmid is a selfreplicating extrachromosomal DNA fragment.)
Experiments 3A and 3B
Fragment I and Fragment II are inserted into separate
plasmids containing the adenine+ gene. The plasmids are
then inserted into the mutant cells, which are inoculated
onto separate agar plates containing minimal medium. If
the plasmid contains the DNA fragment with the
functional centromere, it will be stably inherited by
daughter cells during cell division.
The experimental results are shown in Table 1.

GO ON TO THE NEXT PAGE.

KAPLAN

5


MCAT
Table 1
Expt.

Mutant S. pombe
on minimal
medium with:

1
2*
3A*

cells alone
cells + adenine+
cells + adenine+ +
Fragment I
3B*
cells + adenine+ +
Fragment II
*Cells with introduced plasmid

Increase in
cell # after
1st round
of division

No
Yes
Yes

Visible
colonies
after 20
rounds of
division
No
No
No

Yes

Yes

9 . Minimal medium contains only salts and sugar. The
primary reason that the cells used in Experiment 1
cannot grow on minimal medium is that these cells
are NOT able to synthesize:
A.
B.
C.
D.

protein.
essential fatty acids.
nucleic acids.
salts.

1 0 . Which of the following graphs best represents the
four growth curves obtained under the experimental
conditions summarized in Table 1?
Key:

7 . Based on the information in the passage, it can be
concluded that:

I. a drug that inhibits microtubule assembly is
added to the growth medium used in
Experiment 3B.
II. a drug that inhibits microtubule disassembly is
added to the growth medium used in
Experiment 3B.
III. a drug that inhibits cytokinesis is added to the
growth medium used in Experiment 3B.

A.
B.
C.
D.

log cell #
1

20

1

# of cell divisions

B.

# of cell divisions

20

D.
log cell #

8 . The results obtained in Experiment 3B can be changed
such that they are identical to the results obtained in
Experiment I if:

C.

A.

log cell #

Fragment I contains a centromere.
Fragment II contains a centromere.
Fragments I and II both contain centromeres.
the mutant S. pombe chromosomes do not
contain centromeres.

Expt 3A
Expt 3B

log cell #

A.
B.
C.
D.

Expt 1
Expt 2

1

# of cell divisions

20

1

# of cell divisions

20

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

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6

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Generalized Eukaryotic Cell Test 1
1 1 . The cells used in Experiment 2 are most likely able
to increase their cell number after one round of cell
division because:
A . the replicated adenine+ plasmid binds to
microtubules and segregates to daughter cells
during mitosis.
B . the replicated adenine+ plasmid binds to
microtubules and segregates to daughter cells
during meiosis.
C . the replicated adenine+ plasmid binds to
microfilaments and segregates to daughter cells
during cell division.
D . the replicated adenine+ plasmid randomly
segregates to daughter cells during cell division.

Questions 12 through 16 are
NOT based on a descriptive
passage.
1 2 . In vitro, the transcription factor SP1 binds nucleic
acids with a high affinity. Radio-labeled SP1 would
most likely be found in all of the following organelle
preparations EXCEPT:
A.
B.
C.

D.

nucleoli.
mitochondria.
ribosomes.
Golgi apparati.

1 3 . All of the following are involved in directing proteins
to their final destination within a cell EXCEPT:
A.
B.
C.
D.

amino acid sequence of the protein.
vesicles.
lysosomes.
endoplasmic reticulum.

1 4 . A researcher discovered that in order for the HAP2
protein to elder the nucleus and bind to the CYC1
gene, a molecule of ATP must be consumed. Which
of the following transport mechanisms must be
involved in the movement of HAP2 into the nucleus?
A.
B.
C.
D.

Active transport

Facilitated diffusion
Passive diffusion through the nuclear pores
Endocytosis

GO ON TO THE NEXT PAGE.

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7


MCAT
1 5 . Three funnels, each containing a different
concentration of a urea solution, are covered with a
semi-permeable membrane that is impermeable to
urea. The funnels are inverted and placed in a beaker
containing a different urea solution What is the
concentration of the solution?
A.
B.
C.
D.

2.0 M urea
1.5 M urea
1.0 M urea
0.0 M urea

dH20

1M
urea

2M
urea

dH20

1M
urea

Urea Solution

Urea Solution

Initial state

Final state

2M
urea

1 6 . Which of the following would most likely play a role
in cellular adhesion and recognition?
A . Carbohydrate molecules on the exterior of the
plasma membrane
B . The fluidity of the plasma membrane
C . Peripheral proteins on the interior of the plasma
membrane
D . The hydrophobic regions of the plasma

membrane

END OF TEST

8

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Generalized Eukaryotic Cell Test 1
ANSWER KEY:
1.
A
6.
2.
D
7.
3.
B
8.
4.
C
9.
5.
B
10.

KAPLAN

B

B
D
C
A

11.
12.
13.
14.
15.

D
D
C
A
C

16.

A

9


MCAT
GENERALIZED EUKARYOTIC CELL TEST 1 EXPLANATIONS
Passage I (Questions 1-6)
1.
The correct answer is choice A. Although the question stem may have scared you a little bit, this question is really
quite easy. The only thing you needed to get out of the question stem is that Pellet 2 was found to have one of the enzymes

of the Krebs cycle. So as you can see, this question is simply asking you where the Krebs cycle occurs. Well from
introductory biology, you should know that the Krebs cycle occurs in the mitochondria, along with electron transport and
oxidative phosphorylation. Therefore choice A is the correct answer. Just to review briefly, in eukaryotic cells, glucose is
broken down to pyruvate in the cytoplasm in a series of reactions known as glycolysis. The pyruvate then enters the
mitochondria, where it is converted into acetyl CoA. The acetyl CoA then enters the Krebs cycle. All of the NADH and
FADH2 molecules generated during these processes are shuttled into the electron transport chain, which is coupled with
oxidative phosphorylation to produce ATP.
Anyway, back to the question. As we just learned, glycolysis, not the Krebs cycle, occurs in the cytoplasm, and so
choice D is incorrect. The other two choices have nothing to do with the Krebs cycle at all. The plamsa membrane surrounds
the cell and acts as a barrier between the cell and the outside environment and regulates the passage of material into and out of
the cell. So choice B is incorrect. Finally, the endoplasmic reticulum, choice C, is involved with protein processing and is
the site of ribosomal attachment during translation. Thus choice C is also wrong.
2.
Choice D is the correct answer. In order to answer this question you need to understand the basic principle of
centrifugation as explained in the passage, as well as the information provided in the question stem. From the question stem
you know that DPNH is used to monitor the conversion of pyruvate to lactate. Although you have probably never heard of
DPNH, you should recognize the reaction of pyruvate going to lactate as the fermentation step that occurs in most eukaryotic
cells under anaerobic conditions. Well, where in the cell does fermentation occur? Glycolysis occurs in the cytoplasm, while
the Krebs cycle and electron transport/oxidative phosphorylation occur in the mitochondria.
So now that we’ve deciphered the question stem, let’s see how centrifugation works. From the passage you know
that larger objects will form a pellet at the bottom of a tube, while lighter, less dense structures remain suspended in the
supernatant. You also are told in the passage and shown in Figure 1, that the supernatant from the first tube is transferred to
the second, and so on and so on. Therefore, cytoplasm will be present in ALL four supernatants and so choice D is the correct
answer.
3.
The correct answer is choice B. The question is essentially asking, “the plasma membrane is made up of what
molecules?” The plasma membrane is composed of phospholipids, a molecule that has a glyerol backbone attached to two
fatty acids and a phosphate group. The phospholipids are arranged in a bilayer, which is why the cell membrane is also
referred to as the “phospholipid bilayer.” Roman numeral II is a phospholipid, so must appear in the correct answer. Answer
choice C is the only one that includes Roman numeral II. Molecule I is an RNA molecule, and III is a DNA molecule.

Neither RNA or DNA make up the plasma membrane. Another possible molecule could be a protein, as the plasma
membrane has receptors and ion channels/pumps studded throughout its length.
4.
Choice C is the correct answer. You’re being asked to deduce what will be found in a higher concentration in the
plasma membrane of an eel nerve cell as compared to the membrane of an eel epithelial cell. Well, what are the normal
components of a generic plamsa membrane? A eukaryotic plasma membrane consists of a phospholipid bilayer. So
phospholipids are a common element of all plasma membranes and therefore would not be found in a higher concentration in
nerve cell membranes. Thus choice D is incorrect. Cholesterol molecules are embedded in the hydrophobic interior of the
bilayer. Therefore, cholesterol is also a common component of plasma membranes. So choice B is incorrect. Glycoproteins,
which are proteins that contain carbohydrate components, extend out of the plasma membrane and function in cell adhesion
and recognition. Therefore, glycoproteins are also common plasma membrane elements, and thus choice A is incorrect. So by
the process of elimination, choice C is the correct answer. if you think about it, this answer makes sense. Acetylcholine is
one of the principal neurotransmitters responsible for the transmission of a nerve impulse, and its message is transmitted
when it binds to acetylcholine receptors found on the plasma membrane of its target cells. Since acetylcholine is a
neurotransmitter, the plasma membrane of an epithelial cell would not be expected to have a high concentration of
acetylcholine receptors. But you would most likely find a high concentration of acetylcholine receptors on the plasma
membrane of eel nerve cells. Therefore, choice C is the correct answer.
5.
Choice B is the correct answer. Although you are not explicitly told what is in Pellet 1, you should have been able
to figure it out. First of all, you’re told that DNase is added to Pellet 1. Well, you know that the only place DNA is found in

10

as developed by


Generalized Eukaryotic Cell Test 1
an animal cell is within the nucleus and the mitochondria. And according to the passage, larger, denser organelles will pellet
out first. Since the nucleus is larger than any mitochondrion, it makes sense that the nucleus would probably be in Pellet 1,
while mitochondria would remain in supernatant 1. So now we know that Pellet 1 contains eel nerve cell nuclei. What effect

will DNase have on the nucleus? DNase is an enzyme that degrades DNA by attacking the phosphodiester bonds that hold it
together. So adding DNase to the nucleus will destroy the DNA of eukaryotic chromosomes, leaving behind the DNA binding
proteins that are an integral part of eukaryotic chromosome structure. These DNA binding proteins are known as histones;
there are five major types of histones. Keep in mind that prokaryotes lack histones. Therefore choice B is correct. Choice A is
wrong because a DNase only degrades DNA, and therefore mRNA would not be an end product of its action. RNases degrade
RNA. Rhodopsin, choice C, is a protein found in the external segments of the rods of the retina. Like most proteins,
rhodopsin is not located within the nucleus of retinal cells, nor in that of nerve cells. Furthermore, DNase activity would not
yield rhodopsin. Therefore, choice C is incorrect. Choice D is also incorrect for the same reason. Elongation Factor 2 is a
protein that associates with ribosomes during the addition of each amino acid during translation.
6.
Choice B is the correct answer. Since eukaryotic cells were used, the researcher must believe that Pellet 4 contains
eukaryotic ribosomal subunits. But before we look at the answer choices, let’s briefly review the properties of eukaryotic
ribosomes. Eukaryotic ribosomes consist of a 60S subunit and a 40S subunit, which come together to form an 80S complex.
Both of the subunits are comprised of proteins and rRNA. Prokaryotic ribosomes, on the other hand, consist of a 30S subunit
combined with a 50S subunit, forming a 70S complex.
Now let’s look at the answer choices. Uncharged tRNA, choice A, is typically found in the cytoplasm. Uncharged
tRNA, which is tRNA without an amino acid, is only transiently associated with the whole ribosome after it has transferred
its amino acid to the growing peptide chain. Nor would uncharged tRNAs be associated with the individual ribosomal
subunits, except for the initiator methionine tRNA, which does bind to the 40S subunit. Therefore, choice A definitely does
NOT support the researcher’s hypothesis of ribosomal subunits in Pellet 4. Thus, choice A is incorrect. The presence of a
40S and a 60S molecule in the pellet would support the presence of the ribosomal subunits. Therefore this does support the
researcher’s hypothesis. And although it DOES support the hypothesis, it in no way PROVES that Pellet 4 contains
ribosomal subunits. But, so far, this seems like a pretty good answer. Let’s check out the remaining two choices. The
enzyme peptidyl transferase is responsible for the formation of peptide bonds during translation. And like the uncharged tRNA
molecules, peptidyl transferase is never bound to the ribosome itself, and would only be found in association with the whole
ribosome, not its individual subunits. Therefore choice C is also incorrect. The contents of the supernatant associated with
Pellet 4 does not give us any information about Pellet 4 itself, so it neither supports nor contradicts the notion that Pellet 4
contains ribosomal subunits. So choice D is also incorrect. Therefore, choice B must be the correct answer.
Passage II (Questions 7-11)
7.

The correct answer is choice B. To answer this question you must utilize information contained in the passage and
the data in Table 1. We see that in Experiment 1, the mutant cells cultured in minimal medium cannot grow. And the reason
why the mutant strain is unable to grow on minimal medium is that it lacks adenine. Remember, the passage specifically
states that the mutant strain CANNOT grow in the absence of adenine. Although the passage does not explicitly state that the
minimal medium used in the experiments lacks adenine, it is the most obvious conclusion that can be drawn given that the
mutant strain doesn’t grow in it. Minimal medium contains only the bare essentials for a wild-type organism to grow, and
since wild-type organisms are able to synthesize adenine on their own, there is no reason for minimal medium to be
supplemented with adenine.
In Experiments 2, 3A, and 3B, the mutant yeast cells that were inserted with plasmids containing the gene that codes
for adenine were able to complete at least one round of cell division on minimal medium, according to Table 1. Therefore we
can conclude that the mutant yeast cells CAN grow on minimal medium if they contain a plasmid with the adenine+ gene.
The passage also implies that if a plasmid does not have a centromere, the plasmid will not be stably maintained in the
mutant yeast cell colony. And if a cell doesn’t have the plasmid, it will die, since it can’t synthesize adenine on its own. Only
those cells with plasmids containing centromeric DNA and the adenine+ gene will be able to form visible colonies.
Now that we understand the logic behind the data presented in Table 1, let’s look at the answer choices. From the
table you see that in Experiment 3A, the yeast with Fragment I produced no visible colonies after 20 rounds of replication.
This means that Fragment I must NOT contain a centromere, and so choice A and choice C are incorrect. In Experiment 3B,
however, the yeast with Fragment II DO produce visible colonies after 20 rounds of cell division. Therefore Fragment II must
contain a centromere, and so choice B is correct. Finally, choice D is incorrect because if the mutant yeast’s chromosomes did
not contain centromeres, the mutant yeast strain could not live under any circumstances. Chromosomes without centromeres
would be lost during cell division, and all of the genetic information contained on the chromosomes would likewise be lost.
And without the genetic information, life is not possible.

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11


MCAT
8.

The correct answer is choice D. This question uses the Roman numeral format and requires some information from
the passage. The passage states that both microtubule assembly and disassembly are necessary for proper chromosome and
plasmid segregation. This is exemplified by the chromosome movements that occur in metaphase of mitosis. During
metaphase, the chromosomes, which are attached to microtubules at the centromere, move back and forth across the midline
of the cell. The microtubules are capable of moving the chromosomes in a back and forth fashion because the microtubules
fluctuate between growing longer and then shorter. Because the passage tells you that chromosome segregation requires
microtubule assembly and disassembly, you can conclude that the experimental conditions described in both Roman numerals
I and II would disrupt proper chromosome segregation. If chromosome segregation is disrupted, the cells cannot successfully
divide and grow, and if these cells were grown on minimal medium, the result would be the same as for Experiment 1--no
increase in cell number after the first division, and no visible colonies after 20 divisions. The cells in Experiment 1 cannot
divide and grow on minimal medium because they have no source of adenine for DNA synthesis. Cells grown using the
experimental conditions described in Roman numerals I and II would not grow because the microtubules in these cells could
not properly perform their roles in chromosome segregation. So, because we know that both Roman numerals I and II are
correct, we can eliminate choices A and B. Let’s look at Roman numeral III. Given that cytokinesis is the final stage of cell
division, adding a cytokinesis inhibitor to the growth medium would also prevent cell growth on minimal medium, or any
kind of media for that matter, and would give the same results as seen for Experiment 1.
9.
The correct answer is choice C. This question requires that you combine information contained in the passage with
outside knowledge. The passage states that the mutant yeast cells cannot grow unless adenine is supplied in their growth
medium, which implies that this strain has a defect in the adenine biosynthesis pathway. The cells used in Experiment 1 did
not have adenine+ gene-containing plasmids inserted in them, and as we see in Experiment 1, these cells cannot grow on
minimal medium. So from the information contained in the passage we can conclude that the mutant cells cannot synthesize
adenine and that is why the mutant cells are unable to grow on minimal medium.
Now that we know this, all we have to do is figure out what role adenine plays in cell life. Adenine is a nitrogen
base contained in DNA and RNA. You should also remember that DNA and RNA are nucleic acids. Thus, the mutant cells
used in Experiment 1 cannot grow because they do not have a source of adenine and thus they cannot synthesize the DNA and
RNA that is needed for cell division to occur. Therefore choice C is the correct answer. Let’s look at the wrong answer
choices quickly. Choice A is incorrect because adenine is not a protein or protein precursor. Amino acids are the building
blocks of proteins, not nucleotides. Choice B is incorrect as well. Fatty acids are the building blocks of certain lipids such as
fats or triacylglycerides. Nothing in the passage suggests that the mutant yeast cells have an inability to synthesize fatty

acids. Finally, choice D is incorrect because, as is specified in the question stem, minimal medium contains salts.
Furthermore, living organisms do not synthesize salts.
10.
The correct answer is choice A. Let’s start analyzing this question by predicting what the growth curve should be for
the cells used in Experiment 3B. The cells in Experiment 3B will act like normal, wild-type, yeast cells cultured in growth
medium. You know this because as evidenced in Table 1, these cells continue to produce visible colonies at their 20th cell
division. Furthermore, you should recall that the growth curves of microorganisms are sigmoidal, or S-shaped, like the curves
shown for Experiment 3B in choices A and C. You should have immediately eliminated choices B and D because for
continuously exponential growth curves like those depicted in B and D to occur, the yeast would have to continuously grow
exponentially, never running out of nutrients, and never having to prepare metabolically for growth. Obviously, this is not in
the realm of the possible. Let’s briefly review why the growth curves for microorganisms are S-shaped. When cells are first
inoculated onto growth medium and the cell number is very low, the cells must activate the biochemical machinery required
for growth and cell division. While the cells are gearing up for growth and cell division, the culture is said to be in the lag
phase of growth. Growth during this period is occurring at a very low rate. Once the cells have completed the lag phase, the
cells begin to increase their numbers exponentially, which results in the log phase of growth (log is short for logarithmic).
This period of exponential growth is a consequence of the fact that each cell in the population divides into two cells. At a
certain point, the cells will enter the stationary phase of growth. During this phase, the growing microorganisms have
depleted the growth medium of an essential nutrient, or toxic byproducts of metabolism begin to accumulate, and the growth
rate becomes very low, as in the lag phase.
Anyway, back to the question. So, we know that the correct answer is either choice A or C. How do these two
graphs differ from each other? In choice A, the cells in Experiments 2 and 3A reach a higher cell number after 20 cell
divisions than the cells in Experiment 1. In choice C, the cells in Experiments 1, 2, and 3A all have a similar small cell
number after 20 cell divisions. Well, if you look at Table 1 you will realize that choice A is correct. The cells in Experiments
2 and 3A will reach a higher cell number than the cells in Experiment 1, because the cells in Experiments 2 and 3A have an
adenine+ plasmid. Even though the adenine+ plasmid has no centromere, it will most likely be maintained through a small

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Generalized Eukaryotic Cell Test 1
number of cell divisions before being lost and will thus allow the cultures in Experiments 2 and 3A to reach an elevated cell
number with respect to the culture derived from Experiment 1. As we discussed previously, the mutant cells used in
Experiment 1 do not contain an adenine+ plasmid and thus they cannot grow at all on minimal medium. What do I mean by
“lost?” Well, according to the passage, plasmids are self-replicating; in other words, they replicate independently of the cell
cycle. If the plasmid does NOT have a centromere, during mitosis, it will not be able to bind to the mitotic spindle. This
means that when the cell undergoes cytokinesis, there is no guarantee that either of the daughter cells will inherit the plasmid
and its copy or copies. In fact, research has shown that such plasmids are inherited only 8-20% of the time. Anyway, choice
C is incorrect.
11.
The correct answer is choice D. To answer this question you must understand what is occurring in Experiment 2. In
Experiment 2, the mutant yeast cells contain an adenine+ plasmid that does not contain a centromere. Since the plasmid does
not contain a centromere, the replicated plasmid will not be able to bind to the microtubules of the mitotic spindle, and thus
the replicated plasmid will randomly segregate to daughter cells. By random segregation we mean that since the plasmid
CANNOT segregate to daughter cells by normal means, when the cells divide, the plasmid will get distributed to daughter
cells by chance, or not at all, and will not be faithfully distributed to ALL daughter cells. Due to this random segregation, the
plasmid will eventually be lost from almost all progeny cells. However, after one round of cell division, most of the yeast
cells will still have plasmids, and these cells will be able to complete a second round of cell division, thereby further
increasing the cell number of the culture. This explains the increase in cell number after one round of division observed in the
cells of Experiment 2. [For more detail on how a plasmid gets lost, refer to the explanation to Question 10.]
On the molecular level, random segregation of a plasmid means that the replicated plasmid segregates to daughter
cells without binding to the mitotic spindle. As stated in the passage, a plasmid must contain a centromere in order to bind to
microtubules. Thus choices A and B are incorrect. We can also immediately rule out choice B because the cell division
described in the passage is mitotic, not meiotic. Meiotic cell division would require mating and sporulation, and the passage
gives us no reason to conclude that such processes are occurring. Finally choice C is also incorrect. Microfilaments are
composed of actin molecules and are involved in formation of the contractile ring during cytokinesis, but are not involved
directly in the segregation of genetic material via binding to chromosomes or plasmids.
Discretes (Questions 12-16)
12.

The correct answer is choice D. From the question stem you know that the transcription factor will bind to nucleic
acids. This means that SP1 will bind to both DNA and RNA. So all you have to do to answer this question correctly is figure
out which organelle does not contain DNA or RNA. Well you should know that the nucleolus is the small dense region of
the nucleus where rRNA synthesis occurs. Thus, SP1 would bind to material within the nucleolus and so choice A must be
incorrect. The mitochondria is the organelle, bound by a double membrane, in which the reactions of the Krebs cycle, electron
transport, and oxidative phosphorylation take place. Mitochondria are the site of the production of the majority of the ATP
molecules produced during aerobic respiration. In addition to this function, mitochondria contain their own DNA, which
means that SP1 will bind to it. Therefore choice B is also incorrect. Choice C is also incorrect. Ribosomes are small
organelles composed of protein and rRNA and are the site of translation. Therefore SP1 would also bind to ribosomes. So
choice D is our correct answer. The Golgi apparatus is the organelle responsible for the processing, packaging, and
distribution of proteins. It is not composed of nucleic acids, nor does it process them. Since there are no nucleic acids within
the Golgi apparatus, SP1 would not bind to it. Therefore, choice D is the correct answer.
13.
The correct answer is choice C. In order to answer this question correctly you needed to know the functions of each
of the subcellular structures listed as choices. Let’s go through each one until we find one that is NOT responsible for
directing proteins to their final destinations within cells. The Golgi apparatus and endoplasmic reticulum sort proteins
according to their destination using signals inherent in the amino acid sequence of proteins. For example, the amino acid
sequence lys-asp-glu-leu at the N-terminus causes a protein to be transported from the Golgi apparatus to the endoplasmic
reticulum. So it is obvious that both the endoplasmic reticulum and the amino acid sequence of a protein play a role in
protein transport. Therefore choices A and D are incorrect. Vesicles, choice B, are formed from the fusing of membrane
surfaces. They are involved in transporting proteins from one location to another by virtue of their ability to move around the
cell and through membranes. They may fuse with a target membrane and release the protein at its destination. So choice B is
also incorrect. Therefore choice C must be the correct answer. Lysosomes are rather small, spherical membrane-enclosed
bodies that contain hydrolytic enzymes. If a protein were to come in contact with these enzymes it would be degraded. So
lysosomes are not involved in protein trafficking. Therefore, choice C is the correct answer.

KAPLAN

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MCAT
14.
Choice A is the correct answer. The key information in the question stem is that this protein, HAP2, enters the
nucleus and a molecule to ATP is consumed in the process. All of the other information is just a distraction and I hope that it
did not throw you off. Furthermore, the specific name of the protein--HAP2--is not important in answering the question, and
you are not expected to know anything about this molecule. So all you have to do to answer this question is figure out which
type of transport uses ATP. Well, passive diffusion is the net movement of dissolved particles down their concentration
gradient from a region of higher concentration to a region of lower concentration. This process does not require ATP, so
choice C is incorrect. Facilitated diffusion is the net movement of dissolved particles down their concentration gradient with
the help of carrier molecules. This process, like passive diffusion, does not require ATP. Therefore, choice B is also incorrect.
Active transport is the net movement of particles AGAINST their concentration gradient with the help of carrier molecules.
Unlike facilitated diffusion or passive diffusion, active transport requires energy. And as you know, ATP stores energy so that
cells can perform various functions. Thus, choice A is the correct answer. Endocytosis, choice D, is a process in which the
cell membrane invaginates, forming a vesicle that contains extracellular medium. So this process, though ATP-dependent,
would not account for a cellular protein moving from the cytoplasm into the nucleus. Thus choice D is incorrect.
15.
The correct answer is choice C. To answer this question you need to determine the concentration of the solution in
the beaker. How can you do this? Well if you compare the solution levels in the three funnels in the initial state to the
solution levels in the final state, you can determine the concentration of urea in the beaker. Since you know from the question
stem that the funnels are covered with a semi-permeable membrane that is impermeable to urea, the urea itself will not be
able to flow from the solutions in the funnels to the solution in the beaker or vice versa. This means that water is the only
substance that will move between the funnels and the beaker; in other words, osmosis will occur. Osmosis is the movement
of water across a semi-permeable membrane from a region of lower solute concentration to a region of higher solute
concentration, until the solute concentrations on both sides of the membrane are equal.
So now that we know that we’re talking about osmosis, let’s look at the diagram. The distilled water funnel had a
decrease in water. For the water to move from the funnel to the solution, the solution must have a higher solute concentration
than distilled water. Therefore, choice D is incorrect. The level of solution in the 1 M urea solution remains unchanged; there
was no net movement of water. This means that the urea concentration in the beaker must be equal to the urea concentration
in the funnel. In other words, the solution in the beaker is isotonic to the solution in the funnel. Therefore, choice C must be

the correct answer. The level of solution in the 2 M funnel increases, indicating that water has moved from the beaker into the
funnel. Thus, the concentration of urea must be greater in the funnel, and so the concentration of urea in the beaker must be
less than 2 M. Thus, choices A and B are incorrect.
16.
Choice A is the correct answer. This question is about cellular adhesion and recognition. This refers to the adhesion
of cells to one another and to the recognition of molecules that interact with the cell, such as hormones, antibodies, and
viruses. So from this you know that we’re dealing with phenomenon of the cell surface, not the cell interior. With this in
mind you can eliminate choices C and D, since they do not refer to elements of the plasma membrane related to its exterior.
Choice C refers to the side of the plasma membrane that is in contact with neither the exterior cell surface nor the cytoplasm.
The fluidity of the plasma membrane, choice B, refers to the fact that both lipids and proteins are able to diffuse throughout
the entire plasma membrane. This fluidity enables the lipid bilayer to act as a solvent for membrane proteins. Although some
membrane proteins do play a role in adhesion and recognition, MOST do not, so they would only play a minor role in the
actual adhesion and recognition process. Thus, choice B is incorrect. So by the process of elimination, choice A must be the
correct answer. Let’s see why. In order for a cell to adhere to another cell, or for many molecules to interact with a cell, they
must first bind to receptors or ligands on the cell surface. Therefore, carbohydrate molecules on the exterior of the plasma
membrane are in a perfect position to interact with other cell and molecules, and in fact they do. Thus, choice A is the correct
answer.

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