During conjugation, transfer of the F factor induced a spontaneous mutation in the F– cells that conferred streptomycin resistance.. During conjugation, the F+ cells transfer enzymes tha
Trang 1BIOLOGY TOPICAL:
Microbiology
Test 1
Time: 20 Minutes*
Number of Questions: 15
* 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
Trang 2DIRECTIONS: 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
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
Unq
(261)
105
Unp
(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)
Trang 3Passage I (Questions 1–5)
Conjugation occurs between bacterial cells of
different mating types “Maleness” in bacteria is
determined by the presence of a small extra piece of DNA
that can replicate independently of the larger chromosome
Male bacteria having this sex factor, known as the F
factor, are termed F+ if the sex factor exists
extrachromosomally F+ bacteria can conjugate only with
F– bacteria, the “female” counter-parts, which do not
possess the F factor Genes on the F factor determine the
formation of hair-like projections on the surface of the F+
bacterium, called sex pili The pili form cytoplasmic
bridges through which genetic material may be transferred
The pili also aid the F+ cell in adhering to the F– cell
during conjugation During conjugation of an F+ cell
with an F– cell, the DNA that is most likely to be
transferred to the female is the F factor itself Prior to
transfer, the F factor replicates The F– thus becomes an
F+ by receiving one copy of the F factor, while the
original F+ holds on to the other copy
If this were the only type of genetic exchange in
conjugation, all bacteria would become F+ and
conjugation would eventually cease However, in F+
bacterial cultures, a few bacteria can be isolated that have
the F factor incorporated into their chromosome These
bacteria, referred to as Hfr bacteria, may also conjugate
with F– cells They do not transfer their F factor during
conjugation, but they frequently transfer linear portions of
their chromosomes The transfer is interrupted by the
spontaneous breakage of the DNA molecule at random
sites, usually before the F factor crosses to the F– cell
This process is unidirectional; no genetic material from
the F– cell is transferred to the Hfr cell
1 Which of the following will most likely occur when
a suspension of Hfr cells are mixed with an excess of
F– cells?
A Most of the F– cells will become F+ cells
B The F– cells will produce sex pili that attach to
the Hfr cells
C Hfr cells will replicate the F factor independently
of their chromosomes
D Hfr chromosomal DNA will be transferred to F–
cells
2 A strain of the F+ bacteria is resistant to the antibiotic streptomycin If the F+ strain conjugates with a non-resistant F– strain, the F– strain becomes resistant to streptomycin Which of the following best accounts for this observation?
A During conjugation, transfer of the F factor
induced a spontaneous mutation in the F– cells that conferred streptomycin resistance
B During conjugation, the F+ cells transfer enzymes that activate the gene for streptomycin resistance in the F– cells
C The gene for streptomycin resistance in the F–
cells is carried on extrachromosomal F factor DNA
D The gene for streptomycin resistance in the F+
cells is carried on extrachromosomal F factor DNA
3 An Hfr strain of E coli was found to donate genes to
four F– cells in the following order:
F– cells Gene order
What is the map of these genes on the E coli Hfr
chromosome?
C
A
G T
L
P A
S
P A
L
S G
T
D
B
A S G
T P L T
P A
G S L
GO ON TO THE NEXT PAGE.
Trang 44 Prior to conjugation, the F factor in F+ bacteria is
replicated by the enzyme:
A integrase.
B DNA ligase.
C reverse transcriptase.
D DNA polymerase.
5 Which of the following conjugations will result in a
significant number of new F+ bacteria?
I F+ and F–
II F+ and Hfr
III Hfr and F–
IV F+, F–, and Hfr
A I only
B I and IV only
C I, II, and IV only
D II, III, and IV only
GO ON TO THE NEXT PAGE.
Trang 5Passage II (Questions 6–10)
A strain of penicillin-resistant pneumococci bacteria
exists, despite the absence of the bacterial beta-lactamase
gene that typically confers penicillin resistance
Additionally, half of the cells in this strain are unable to
metabolize the disaccharides sucrose and lactose A
microbiologist studying this strain discovered that all of
the cells in this strain were infected with two different
types of bacteriophage, Phage I and Phage II, both of
which insert their DNA into the bacterial chromosome
To determine if bacteriophage infection could give rise to
this new bacterial strain, the microbiologist infected
wild-type pneumococci with the two phage
Experiment 1
10µL of Phage I and 10µL of Phage II were added to
separate 5mL nutrient broth solutions containing actively
growing wild-type pneumococci In addition, a 5mL broth
solution containing only wild-type pneumococci was used
as a control After 20 minutes of room temperature
incubation, the microbiologist diluted 1µL of the broth
solutions in separate 1.999mL aliquots of sterile water
She plated these dilutions on three different agar plates
containing glucose, sucrose, and lactose, respectively The
plates were incubated for 24 hours at 37°C The results are
shown in Table 1
Table 1
I-infected cells
Phage II-infected cells
Wild-type cells
(+) denotes bacterial growth; (–) denotes no growth
Experiment 2
1µL of the broth solutions from Experiment 1 were
again diluted in separate 1.999mL aliquots of sterile water
These dilutions were plated on three different agar plates
containing tetracycline, penicillin, and no antibiotic,
respectively The plates were incubated for 24 hours at
37°C The results are shown in Table 2.
Table 2
I-infected cells
Phage II-infected cells
Wild-type cells
(+) denotes bacterial growth; (–) denotes no growth
6 Based on the experimental data, which of the
following statements is most likely true of Phage I?
A Phage I inserted its DNA into the bacterial
chromosome, making the bacterial cell wall impervious to the effects of penicillin
B Phage I reduced the penicillin on the agar plates,
therefore allowing the bacteria to grow
C Phage I contained the viral gene that coded for
beta-lactamase
D Phage I inhibited the growth of the bacteria.
7 Which of the following conclusions could be inferred
from the data in Table 1?
A Phage I inserted its DNA into the region of the
bacterial chromosome that codes for the enzymes
of glycolysis
B Phage I prevented larger molecules such as
lactose and sucrose from passing through the bacterial cell wall
C Phage II inserted its DNA into the region of the
bacterial chromosome that codes for enzymes that digest disaccharides
D Phage II utilized all of the sucrose and lactose and
starved out the bacteria
8 Which of the following best accounts for the results
of Experiment 2?
I Both Phage I DNA and Phage II DNA code for beta-lactamase
II Both Phage I DNA and Phage II DNA code for enzymes that inhibit tetracycline’s deleterious effects
III Both phage disrupted the wild-type bacteria’s ability to resist penicillin
IV The wild-type bacteria has no natural resistance to either penicillin or tetracycline
A I only
B I and IV only
C III and IV only
D I, II, and IV only
GO ON TO THE NEXT PAGE.
Trang 69 Which of the following best describes the appearance
of pneumococci, a streptococcal bacteria, when
stained and then viewed with a light microscope?
A Rod
B Helical
C Squamous
D Spherical
1 0 In which of the following cycles must Phage I be to
produce plaques, which are transparent areas within
the bacterial lawn caused by bacterial cell death?
A Lytic
B Lysogenic
C Mitotic
D Integration
GO ON TO THE NEXT PAGE.
Trang 7Questions 11 through 15 are
NOT based on a descriptive
passage
1 1 Which of the following is true of both a
bacteriophage and a retrovirus?
A Both can integrate their genetic material into the
host cell genome
B Both have genes that code for reverse
transcriptase
C Both can infect human cells.
D Both are immunosuppressive agents.
1 2 Addition of DNase to a bacterial cell results in
hydrolysis of the cell’s DNA, preventing protein
synthesis and causing cell death However, certain
viruses pre-treated with DNase continue to produce
new proteins following infection Which of the
following best accounts for this observation?
A The icosahedral protein coat of these viruses
denature DNase
B The genomes of these viruses contain multiple
reading frames
C The genomes of these viruses are comprised of
RNA
D The genomes of these viruses contain multiple
copies of their genes
1 3 In the treatment of an influenza viral infection, an
experimental drug that directly blocks the synthesis of
the viral protein coat is discovered This drug most
likely:
A affects synthesis of host cell proteins, and may
therefore cause side effects in the host
B has no affect on host cell protein synthesis, and
may therefore be an effective treatment
C blocks the cell surface receptors to which both
the viral protein coat and host hormones bind,
and may therefore cause side effects in the host
D increases the affinity of antibodies to the viral
protein coat antigens, and may therefore be an
effective treatment
1 4 All of the following are true of fungi EXCEPT:
A they can exist in either a haploid state or a
diploid state
B their cells contain a plasma membrane, but no
cell wall
C they contain masses of hyphae that form the
mycelium
D they are heterotrophs.
1 5 One form of hepatitis (inflammation of the liver) is
caused by a virus The serum of patients with hepatitis may contain the hepatitis B surface antigen, HBsAg, and/or the hepatitis B core antigen, HBcAg Which of the following is most likely true?
A HBcAg is probably composed primarily of lipids.
B HBsAg is probably composed primarily of
proteins
C Serum concentration of liver enzymes remains
the same during acute hepatitis infection
D Serum concentration of liver enzymes decreases
during acute hepatitis infection
END OF TEST
Trang 8ANSWER KEY:
Trang 9MICROBIOLOGY TEST 1 TRANSCRIPT
Passage I (Questions 1-5)
1 The correct answer is choice D This is a fairly straightforward question that can be answered with the information given in the passage From the second paragraph we know that Hfr bacteria have the F factor integrated into their chromosome, instead of located in an autonomous circular DNA element in the cytoplasm in other words a plasmid From the passage you should have been able to figure out that when a suspension of Hfr cells is mixed with an excess of F- cells, the Hfr cells attach to the F- cells via pili, conjugation begins, and Hfr chromosomal DNA will be transferred to the F- cells Therefore, choice D is right and choice B is wrong, since it has the things the other way around Since the F factor is part of the bacterium’s main chromosome, the F factor will get replicated along with the rest of the cell’s chromosome prior to conjugation; so choice C is wrong During conjugation, the transfer f Hfr DNA is interrupted by the spontaneous breakage of the DNA molecule at random points Typically, the chromosome is broken before the F factor is transferred to the F- cell Therefore, the conjugation of an Hfr cell with an F- cell does NOT usually result in an F+ cell The F factor usually remains
in the Hfr cell Thus, choice A is wrong Again, choice D is the correct answer
2 The correct answer is choice D How can this newfound resistance to streptomycin in a previously nonresistant F-strain be accounted for? Since resistance to an antibiotic is a genetic thing, the only way that this could have occurred was if the gene for streptomycin resistance was transferred from the F+ cells to the F- cells Well, from the passage you know that during conjugation between F+ cells and F- cells, the extrachromosomal DNA containing the F factor replicates, the F+ cells attach to the F- cells via pili, and the extrachromosomal DNA is transferred to the F- cells The only thing transferred between the two cells is this DNA Thus, the gene for streptomycin resistance must have been carried on the F factor Thus, choice D
is the correct answer I hope that you weren’t tricked into picking choice C, which if read too fast, looks kind of like choice
D Choice C is wrong because the F- cells were NOT resistant to the antibiotic before conjugation, which implies that they certainly could NOT have carried the gene for streptomycin resistance Choice B can also be ruled out since, as we just said, the F- cells of this strain could not have carried the gene for streptomycin resistance prior to conjugation, which would have had to been the case for choice B to be true And furthermore, we know that only DNA, not enzymes, are transferred from the F+ cell during conjugation Choice A is also incorrect because it is not possible to induce a spontaneous mutation A mutation may either be induced by something such as a sex factor or it may be spontaneous, meaning its cause is nothing but pure chance Also, spontaneous mutations are very rare and it would not be possible for an entire strain of F- cells to spontaneously mutate at the same time to confer resistance to the same drug Again, choice D is the correct answer
3 The correct answer is choice A This problem requires you to determine the map of the chromosomal marker genes based on the transfer of genes during conjugation given to you in the question Before you attack this problem, you should visualize what is going on in this question You know from the passage that Hfr cells have the F factor integrated in their chromosome When an F- cell is in close proximity to an Hfr cell, a sex pili reaches out from the Hfr cell to the F- cell, forming a cytoplasm-filled bridge between the two cells Since the F factor is part of the chromosomal DNA it gets replicated along with the chromosomal DNA prior to conjugation, although the replicated DNA is in a linear form It is a segment of this linear DNA that gets transferred to the F- cell The transfer is interrupted by the spontaneous breakage of the DNA molecule at random times The DNA that entered the F- cell will be composed of only those genes that crossed the pili before breakage stopped the transfer
Let’s now take a look at the problem There are different combinations of gene order in the four recipient F- cells because the conjugation was disrupted at different times Since the genetic information is transmitted linearly, the order itself
is NOT changed Now before we go any further, we can eliminate choices B and D Why? Because as you know from introductory biology, the chromosome in prokaryotic cells is circular Even though the DNA transferred during conjugation is linear, this is the replicated DNA; the original DNA in the donor Hfr cells is circular Therefore, choice B and D must be wrong because they both have the DNA maps of the E coli Hfr chromosome as being linear So now all you have to do is decide between the maps in choices A and C To find the map of the circular chromosome, you simply have to fit the fragments from the question stem together by examining the overlapping areas The first segment has the order TLPA The second segment, PASG, overlaps with the P and A from the first segment So the order is now TLPASG The third segment, GTLP, overlaps with the T, L, and P of the order So the chromosome order is now GTLPASG Finally the last segment, SAPL, is already found in our gene order, if you read it backwards, that is Remember that in contrast to a linear chromosome, circular chromosomes don’t have a unique start point and end point Therefore, choice A must be the correct answer because it illustrates the genes in the order TLPASG Or you can read it PASGTL, or SAPLTG; it all depends on
Trang 104 The correct answer is choice D This question is very straightforward, requiring you to recall the primary enzyme responsible for replicating strands of DNA DNA polymerase DNA polymerase, choice D, is an enzyme that can synthesize a new DNA strain using a template DNA strand Thus, DNA polymerase must be responsible for the replication of the F factor
in DNA in F+ cells before conjugation occurs Don’t fall into the trap of thinking, “if it’s too obvious, then it must be incorrect.” There are often questions on the MCAT that are designed to be easy, questions that most people get right So let’s look at the other answer choices Integrase, choice A, is an enzyme that you are not required to be familiar with Integrase is a retroviral enzyme that integrates provirus DNA in host genomes So choice A is incorrect This choice was just thrown in as
a distraction Since this is not a common enzyme and it is not discussed in the passage, chances are it’s a wrong choice DNA ligase, choice B, catalyzes the formation of a phosphodiester bond to link two adjacent bases separated by a nick in one strand
of double-helical DNA Since DNA ligase does not replicate DNA, choice B is incorrect Reverse transcriptase, choice C, is another retroviral enzyme Reverse transcriptase synthesizes DNA from an RNA template Because DNA replication involves the synthesis of DNA from a DNA template, choice C is also incorrect Again, choice D is the correct answer
5 The correct answer is choice B The key to getting this one right is noting from the passage what is required for a cell to become an F+, or male, cell Recall that maleness is determined by the presence of a small extra piece of self-replicating DNA that contains the F factor With this in mind let’s review the listed conjugations In Roman numeral I, F+ bacteria conjugate with F- bacteria This will certainly result in new F+ cells As you know from the passage, the extrachromosomal F factor will be transferred to the F- cell, turning it into a F+ cell, where the F factor DNA will exist indefinitely as an extrachromosomal entity with the ability to replicate autonomously Since Roman numeral I is a correct choice, choice D can be eliminated In Roman numeral II, an F+ cell conjugates with an Hfr cell Since both cell types already contain the F factor, no conjugation will occur Thus no new F+ bacteria will result Thus Roman numeral II is incorrect, and choice C can be eliminated In Roman numeral III, an Hfr cell conjugates with a F- cell As you know from the passage, Hfr cells rarely transfer the F factor to F- cells during conjugation Therefore there will NOT be a significant number
of new F+ bacteria and Roman numeral III is incorrect Now if you really had your thinking cap on, you wouldn’t even have examined Roman numeral III, since you previously eliminated choice D, the only choice that contained Roman numeral III
In Roman numeral IV, new F+ cells will be formed because the two necessary ingredients F+ cells and F- cells are present Even though Hfr cells cannot be converted themselves or convert F- cells to F+ cells, the F- cells that happen to conjugate with F+ cells will be converted to F+ Therefore Roman numeral IV is also correct So Roman numerals I and IV are correct and choice B is the correct answer
Passage II (Questions 6-10)
6 The correct answer is choice C If you look at the answer choices you will see that they all refer to penicillin resistance and Phage I So this means that you need to look at Table 2 for your answer From Table 2 you know that the bacteria infected with Phage I were able to grow on agar plates containing penicillin You also know from the passage and Table 2 that the uninfected wild-type bacterial cells do not contain the gene that codes for beta-lactamase, which confers penicillin resistance, since no bacterial growth was observed when the control dilution was incubated in the presence of penicillin So, combining these two pieces of information, you can conclude that Phage I must be responsible for the observed penicillin resistance in Experiment 2 With this in mind, let’s look at the answer choices
Right away you should have eliminated choice D, since bacterial growth did occur in both Experiments 1 and 2 when the bacteria were infected with Phage I Choices A, B, and C, however, are all possible ways in which the bacterial cells could have become resistant to penicillin So, how do you decide between these three? Well, you’re asked to determine your answer based on the experimental data Thus, choice C is the only one that could be concluded on this basis the penicillin resistance observed in the bacteria infected with Phage I in Experiment 2 must have been conferred by the insertion of Phage I DNA that contained the gene that codes for beta-lactamase Thus choice C is the correct answer The passage does not state the exact mechanism by which this gene functions and therefore you are not expected to know this Perhaps beta-lactamase does function via the mechanisms proposed in either choice A or B by effecting the bacterium cell wall, or by reducing the penicillin on the agar plate However, neither of these can be concluded from the experimental data The only thing you DO know about penicillin resistance is that the gene for beta-lactamase must be present Why? Because the only difference between these bacterial cells and the wild-type cells of the control group is the Phage I infection Therefore, choices A and B are incorrect and choice C is the correct answer
7 The correct answer is choice C From the question stem you know that you will need to interpret the data in Table 1 According to Table 1, both the control bacteria and the bacteria infected with Phage I were able to grow in the presence of all