Sample Examination Questions for Exam 3 Material - Solved
Biology 3300 / Dr. Jerald Hendrix
Warning! These questions are posted solely to provide examples of past test questions.
There is no guarantee that any of these questions will be on any examination in the future.
Students are responsible for all of the material covered in lectures, assigned readings,
textbook problems, laboratories, and any other assigned work. Since these samples have
been taken from several past exams, some questions may be very similar or identical. On
short answer, essay questions, and genetics problems, the point values from previous
exams have been included to give an indication of approximately how much “weight”
was given to a question in the past; however, there is no guarantee that any particular
question, format, or point distribution will be used on any examination.
5.
Which of the following enzymes catalyze the attachment of an amino acid to
tRNA in the formation of aminoacyl tRNA?
(a)
(b)
(c)
(d)
(e)
15.
In the formation of an aminoacyl tRNA
(a)
(b)
(c)
(d)
(e)
18.
RNA polymerase
aminoacyl tRNA synthetase
Type I restriction endonuclease
Type II restriction endonuclease
reverse transcriptase
the amino terminus of the amino acid is directly attached to the 5' end of
the tRNA.
the carboxyl terminus of the amino acid is directly attached to the 5' end of
the tRNA.
the amino terminus of the amino acid is directly attached to the 3' end of
the tRNA.
the carboxyl terminus of the amino acid is directly attached to the 3' end of
the tRNA.
the side chain group of the amino acid is directly attached to the anticodon
loop of the tRNA.
Which of the following statements is true?
(a)
(b)
(c)
(d)
(e)
Ribosomes read mRNA from the 5' to the 3' end and synthesize the
nascent protein chain from the carboxyl to the amino terminus.
Ribosomes read mRNA from the 3' to the 5' end and synthesize the
nascent protein chain from the amino to the carboxyl terminus.
Ribosomes read mRNA from the 5' to the 3' end and synthesize the
nascent protein chain from the amino to the carboxyl terminus.
Ribosomes read mRNA from the 3' to the 5' end and synthesize the
nascent protein chain from the carboxyl to the amino terminus.
Ribosomes read DNA from the 5' to the 3' end and synthesize the nascent
RNA chain from the 3' to the 5' end.
Page 1
22.
In prokaryotic genes, a promoter consensus sequence located about 10 bases
upstream from the start of transcription is the
(a)
(b)
(c)
(d)
(e)
24.
CAAT box.
Hogness box.
Shine-Dalgarno box.
Meselson-Stahl box.
Pribnow box.
A promoter
(a)
(b)
(c)
(d)
(e)
is a protein that binds to RNA polymerase to enhance the process of
transcription.
is a DNA sequence to which RNA polymerase binds.
is a DNA sequence to which Type I or Type II restriction endonucleases
bind.
is the location where DNA replication begins.
is an initiation factor in the process of translation.
26.
Energy for the process of elongation during translation comes from the hydrolysis
of
(a)
ATP.
(b)
CTP.
(c)
GTP.
(d)
TTP.
(e)
UTP.
30.
The enzyme responsible for forming the final phosphodiester bond between two
DNA fragments during DNA replication is
(a)
(b)
(c)
(d)
(e)
33.
DNA ligase
DNA polymerase I
DNA polymerase III
DNA-directed RNA polymerase
reverse transcriptase
Which of the following statements best describes the relationship between the
nucleotide sequences of the template strand and the nontemplate strand in the
DNA of a gene?
(a)
(b)
(c)
(d)
(e)
They are identical to each other.
They are parallel to each other.
They are covalently attached to each other.
They are complementary to each other.
U in the coding strand is found in place of the T in the noncoding stand.
Page 2
38.
Which of the following best describes the function of the stop codons UAA,
UAG, and UGA?
(a)
(b)
(c)
(d)
(e)
39
Okasaki fragments
(a)
(b)
(c)
(d)
(e)
40.
addition of a 5' methylated guanosine cap
addition of a 3' poly-A tail
splicing of RNA segments
all of the above
none of the above
Amino acids are assembled into proteins
(a)
(b)
(c)
(d)
(e)
1.
are fragments of the discontinuous strand of nascent DNA during
replication.
are fragments of the continuous strand of nascent DNA during replication.
are fragments of the template DNA during replication.
are fragments of E. coli DNA polymerase I that have 5' exonuclease
activity.
are fragments of E. coli DNA polymerase I that have 3' exonuclease
activity.
Which of the following processes occurs to eukaryotic mRNA during posttranscriptional processing?
(a)
(b)
(c)
(d)
(e)
48.
Termination of DNA replication
Origin of DNA replication
Termination of transcription
Termination of protein synthesis
Orientation of the noncoding DNA strand of a gene
inside the nucleus.
on ribosomes.
inside the Golgi apparatus.
on the chromosomes.
only during mitosis.
In DNA replication, new nucleotides are added
(a)
(b)
(c)
(d)
(e)
to the 5' end of the each nascent strand.
to the 3' end of each nascent strand.
to both ends of each nascent strand.
to the 5' end of the continuous strand and the 3' end of discontinuous
strand fragments.
to the 3' end of the continuous strand and the 5' end of discontinuous
strand fragments.
Page 3
2.
RNA primase is responsible for
(a)
(b)
(c)
(d)
(e)
3
The enzyme responsible for the most of the nascent chain synthesis during DNA
replication in E. coli is
(a)
(b)
(c)
(d)
(e)
4
the addition of new nucleotides to the 5' end of a nascent RNA strand
during transcription.
the addition of new nucleotides to the 3' end of a nascent RNA strand
during transcription.
the addition of the 5' leader sequence to eukaryotic mRNA during posttranscriptional processing.
the addition of the 3' poly-A tail to eukaryotic mRNA during posttranscriptional processing.
none of the above.
DNA polymerase I
DNA polymerase II
DNA polymerase III
DNA-directed RNA polymerase
DNA ligase
In an experiment, cells of actively growing E. coli were incubated for several
minutes with radiolabeled deoxyribonucleosides (dNTPs). The DNA was isolated
from the cells, denatured into single strands, and separated according to size by
gradient centrifugation. The radioactivity in the tube corresponded to a single
longer piece of DNA and a heterogeneous mixture of much shorter DNA
fragments. Which of the following statements best explains these results?
(a)
(b)
(c)
(d)
(e)
The longer strand required a template, but the shorter strands did not.
The longer strand had been synthesized from 5'→3', and the shorter
fragments had been synthesized from 3'→5'.
The longer strand had been synthesized from 3'→5', and the shorter
fragments had been synthesized from 5'→3'.
The longer strand represented discontinuous strand synthesis, and the
shorter fragments represented continuous strand synthesis.
The longer strand represented continuous strand synthesis, and the shorter
fragments represented discontinuous strand synthesis.
Page 4
5
Okasazi fragments
(a)
(b)
(c)
(d)
(e)
6
The 3' exonuclease activity of DNA polymerase I in E. coli is responsible for
(a)
(b)
(c)
(d)
(e)
41.
are fragments of nascent DNA formed in discontinuous strand synthesis
during DNA replication.
are fragments of nascent DNA formed in continuous strand synthesis
during DNA replication.
are double-stranded DNA molecules with sticky ends, formed by cleavage
with type II restriction endonucleases.
are fragments of RNA that are spliced out and discarded during posttranslational processing of eukaryotic mRNA.
are fragments of the amino-terminal end of a nascent protein chain,
removed by post-translational proteolytic cleavage in almost all proteins.
removal of the primer.
removal of nucleotides with mismatched bases.
removal of introns.
addition of new nucleotides.
formation of the final phosphodiester bond between two adjacent nascent
chain fragments.
Eukaryotic cells can be grown in a medium containing bromodeoxyuridine
(BrdU), which substitutes for thymidine during DNA replication. A fluorescent
dye that binds to double-stranded DNA makes it possible to recognize
chromosomes or chromatids in which one or both strands of the DNA helix
contain BrdU. The chromosome or chromatid fluoresces brightly when only one
strand of its DNA contains BrdU, but it fluoresces weakly when both strands of
DNA contain BrdU. In an experiment, unlabeled human cells in G1 were
transferred to a medium containing BrdU and colchicine. The cells proceeded
through one round of DNA replication, but the colchicine halted mitosis at
metaphase. The metaphase chromosomes were isolated from the cells and
analyzed under a fluorescence microscope. Assuming that DNA replication in
humans is semiconservative, which of the following do you predict will best
describe the appearance of the metaphase chromosomes?
(a)
(b)
(c)
(d)
(e)
One chromosome in each homologous pair will be brightly fluorescent,
and the other chromosome will not be fluorescent.
In every chromosome, each chromatid will be brightly fluorescent.
In every chromosome, one chromatid will be weakly fluorescent, and the
other chromatid will not be fluorescent.
In every chromosome, each chromatid will be weakly fluorescent.
In every chromosome, one chromatid will be weakly fluorescent and one
chromatid will be brightly fluorescent.
Page 5
7
A “shuttle vector” is a recombinant DNA vector capable of replicating in more
than one bacterial species. Which of the following must a shuttle vector have?
(a)
(b)
(c)
(d)
(e)
Pribnow boxes from both bacterial species.
Promoter regions from both bacterial species.
Shine-Dalgarno sequences from both bacterial species.
Origins of replication from both bacterial species.
Ribosomes from both bacterial species.
***********************************************************************
Eukaryotic cells can be grown in a medium containing bromodeoxyuridine
(BrdU), which substitutes for thymidine during DNA replication. A fluorescent
dye that binds to double-stranded DNA makes it possible to recognize
chromosomes or chromatids with BrdU incorporated in their DNA. You can tell
the difference between unlabeled chromatids, chromatids with one strand of the
DNA double helix labeled, and chromatids with both strands of the DNA double
helix labeled. The chromatid fluoresces brightly when only one strand contains
BrdU, but weakly when both strands contain BrdU. The following diagrams
represent the possible fluorescence patterns in metaphase chromosomes. Double
hatch marks represent bright fluorescence, single marks represent weak
fluorescence, and a clear white chromatid represents no fluorescence.
Key:
I.
II.
III.
Neither chromatid is labeled with BrdU. (No fluorescence)
Each chromatid is labeled in one strand of DNA. (Brightly
fluorescent)
Each chromatid is labeled in both strands of DNA. (Weakly
fluorescent)
Page 6
IV.
V.
VI.
One chromatid is labeled in one strand only, and the other
chromatid is labeled in both strands.
One chromatid is unlabeled, and the other chromatid is labeled in
one strand only.
One chromatid is unlabeled, and the other chromatid is labeled in
both strands.
A group of cultured eukaryotic cells, all at the G1 stage and containing unlabeled
chromosomes, were placed in a medium containing BrdU and allowed to grow.
Samples were removed during the first, second, and third mitotic divisions.
Mitosis in each of these samples was arrested at metaphase using colchicine.
Chromosomes were isolated from cells in the samples and examined for
fluorescence. Remember that chromosomes isolated at metaphase are in the
replicated state (a pair of chromatids connected at the centromere). In addition,
each chromatid in a replicated chromosome has one double helix. The two
helices are the products of DNA replication.
See next page for an explanation.
8
IF eukaryotic DNA replication were conservative, then what will be the
appearance of the metaphase chromosomes after the third mitotic division? (The
Roman numerals refer to the diagram above.)
(a)
(b)
(c)
(d)
(e)
9
IF eukaryotic DNA replication were semiconservative, then what will be the
appearance of the metaphase chromosomes after the third mitotic division? (The
Roman numerals refer to the diagram above.)
(a)
(b)
(c)
(d)
(e)
10
1/3 will look like I and 2/3 will look like III.
1/4 will look like II, 1/4 like III, 1/4 like V, and 1/4 like VI.
1/2 will look like III and 1/2 will look like IV.
3/4 will look like II and 1/4 will look like V.
3/4 will look like III and 1/4 will look like VI.
1/3 will look like I and 2/3 will look like III.
1/4 will look like II, 1/4 like III, 1/4 like V, and 1/4 like VI.
1/2 will look like III and 1/2 will look like IV.
3/4 will look like II and 1/4 will look like V.
3/4 will look like III and 1/4 will look like VI.
Based on what you learned about DNA replication in class, what do you predict
about the appearance of the metaphase chromosomes after the third mitotic
division? (The Roman numerals refer to the diagram above.)
(a)
(b)
(c)
(d)
(e)
1/3 will look like I and 2/3 will look like III.
1/4 will look like II, 1/4 like III, 1/4 like V, and 1/4 like VI.
1/2 will look like III and 1/2 will look like IV.
3/4 will look like II and 1/4 will look like V.
3/4 will look like III and 1/4 will look like VI.
Page 7
The BrdU problem is similar to the Meselson-Stahl experiment. At the start of the
experiment, each chromosome (DNA double helix) is unreplicated, and its strands
are unlabeled.. The G1 cells are transferred to BrdU before the first round of
replication begins, and the cells remain in BrdU for the remainder of the experiment.
This means that every new strand formed will be labeled in BrdU. When the
chromosomes are analyzed after the 1st, 2nd, & 3rd round of replication, each pair of
sister chromatids represents the two daughter helices formed during replication in the
S phase.
Let a blue line
|
represent an unlabeled strand, and a red line
|
represent a labeled
strand.
If replication is CONSERVATIVE, the DNA during the experiment will follow this
pattern:
||
Unlabeled, unreplicated helix at the start of the experiment
↓
st
1 round of replication.
The two old strands stay together, and the two new strands join together if replication
is conservative
||•||
All
↓
2nd round of replication.
||•|| ||•||
rd
½
and ½
↓
3 round of replication.
||•|| ||•|| ||•|| ||•||
¼
and ¾
, or ¼ like VI and ¾ like III.
Page 8
If replication is SEMICONSERVATIVE, the DNA during the experiment will follow
this pattern:
||
Unlabeled, unreplicated helix at the start of the experiment
st
↓
1 round of replication.
In semiconservative replication,
each new helix consists of one old strand and one new strand.
||•||
nd
All
↓
2 round of replication.
||•|| ||•||
rd
All
↓
3 round of replication.
||•|| ||•|| ||•|| ||•||
½
and ½
, or ½ like IV and ½ like III.
Since DNA replication is known to be semiconservative, the results expected in the
experiment are ½ like IV and ½ like III.
************************************************************************
Page 9
The following choices are used for questions 11 - 14.
(a)
(b)
(c)
(d)
(e)
Pribnow box
CAAT box
-35 consensus sequence
Shine-Dalgarno sequence
Hogness box
d 11
A sequence in the leader region of mRNA, believed to be responsible for orienting
the mRNA on the small ribosome subunit in the initiation of translation.
a 12
In prokaryotes, a promoter consensus sequence 10 bases upstream from the first
base transcribed.
e 13
In eukaryotes, a promoter consensus sequence 25 bases upstream from the first
base transcribed. Very few books use the term “Hogness box” anymore, using
instead the term “TATA box” You will not use the term “Hogness box” on your
exam.
b 14
In eukaryotes, a promoter consensus sequence 70 bases upstream from the first
base transcribed.
************************************************************************
15
Rho factors
(a)
(b)
(c)
(d)
(e)
16
are responsible for the termination of translation.
are responsible for the termination of transcription in eukaryotes.
are responsible for the termination of transcription in prokaryotes.
are responsible for the initiation of transcription in eukaryotes.
are responsible for the initiation of transcription in prokaryotes.
Sigma factors
(a)
(b)
(c)
(d)
(e)
assist the small ribosome subunit in recognizing mRNA during the
initiation of translation.
assist prokaryotic RNA polymerase in recognizing the origin of replication
during the initiation of replication.
assist prokaryotic RNA polymerase in recognizing the promoter region
during the initiation of transcription.
assist eukaryotic RNA polymerase in recognizing the promoter region
during the initiation of transcription.
assist eukaryotic RNA polymerase in recognizing the origin of replication
during the initiation of replication.
Page 10
17
In post-transcriptional processing of eukaryotic mRNA, the enzyme terminal
transferase is responsible for
(a)
(b)
(c)
(d)
(e)
The addition of a poly-A tail to the 3' end of the mRNA.
The addition of a methylated guanosine cap to the 3' end of the mRNA.
The formation of peptide bond between the amino group of one amino
acid and the carboxyl group of another amino acid.
The addition of a poly-A tail to the 5' end of the mRNA.
The addition of a methylated guanosine cap to the 5' end of the mRNA.
************************************************************************
The nucleotide sequence listed below represents the transcriptional template strand of a
gene.
Template DNA strand
18
Which of the following is the nontemplate DNA strand complementary to the
template strand?
(a)
(b)
(c)
(d)
(e)
19
5' CTACGTAGTTGAAGACAT 3'
3' CTACGTAGTTGAAGACAT 5'
5' AUGUCUUCAACUACGUAG 3'
5' ATGTCTTCAACTACGTAG 3'
3' ATGTCTTCAACTACGTAG 5'
Which of the following is the mRNA transcribed from the template DNA (assume
that there is no intron splicing or other processing).
(a)
(b)
(c)
(d)
(e)
20
3' TACAGAAGTTGATGCATC 5'
3' AUGUCUUCAACUACGUAG 5'
5' AUGUCUUCAACUACGUAG 3'
3' UACAGAAGUUGAUGCAUC 5'
5' UACAGAAGUUGAUGCAUC 3'
5' GAUUCUACUUCAGACGAU 3'
Which of the following is the peptide that is produced when the mRNA is
translated? (The standard 3-letter abbreviations for amino acids are used.)
(a)
(Amino end) Tyr Arg Ser (Carboxyl end)
(b)
(Amino end) Asp Ser Thr Ser Asp Asp (Carboxyl end)
(c)
(Carboxyl end) Asp Ser Thr Ser Asp Asp (Amino end)
(d)
(Carboxyl end) Met Ser Ser Thr Thr (Amino end)
(e)
(Amino end) Met Ser Ser Thr Thr (Carboxyl end)
***********************************************************************
Page 11
21
Which of the following best describes the action of an aminoacyl tRNA
synthetase enzyme?
(a)
(b)
(c)
(d)
(e)
22`
Which of the following is used directly as an energy source by enzymes in the
initiation and elongation stages of translation?
(a)
(b)
(c)
(d)
(e)
23
It attaches the amino end of an amino acid to the 5' end of a tRNA
molecule.
It attaches the carboxyl end of an amino acid to the 5' end of a tRNA
molecule.
It attaches the carboxyl end of an amino acid to the 3' end of a tRNA
molecule.
It attaches the amino end of an amino acid to the 3' end of a tRNA
molecule.
It attaches the anticodon of a tRNA to the codon on the mRNA.
ATP
CTP
GTP
UTP
lactose
A certain mutation in E. coli is a chain-termination (or nonsense) mutation.
Compared to the wild type, one of the nucleotide bases in the mutant is substituted
for another so that a sense codon UCG in the coding region is changed into the
stop codon UAG. This means that translation terminates prematurely, resulting in
a protein that is too short to carry out its enyzmatic functions.
It is possible to isolate another E. coli mutant gene called a nonsense suppressor
mutation. If cloned into the strain with the chain-termination mutation, the
nonsense suppressor causes the normal wild phenotype to return to the strain.
However, the nonsense suppressor gene does not encode the same protein as the
chain termination gene. What does the nonsense suppressor encode?
(a)
(b)
(c)
(d)
(e)
The nonsense suppressor encodes a repressor protein that blocks
transcription of the chain-termination gene.
The nonsense suppressor encodes the aminoacyl tRNA synthetase specific
for tRNA with the anticodon AGC.
The nonsense suppressor encodes a tRNA molecule with the anticodon
AGC.
The nonsense suppressor encodes a tRNA molecule with the anticodon
AUC.
The nonsense suppressor encodes a Rho factor protein.
Page 12
24
To which site on the ribosome does the next aminoacyl tRNA bind during the
elongation phase of translation?
(a)
(b)
(c)
(d)
(e)
1.
The 5' site
The 3' site
The A site
The P site
The amino site
What are the special features found at the 5' end and the 3' end of mature
eukaryotic mRNA? (4 pt)
At the 5’ end, there is a methylated guanosine cap attached via a 5’ -> 5’
triphophate linkage. It is added during posttranscriptional processing by a
capping enzyme.
At the 3’ end is a poly-A tail consisting of about 100 – 200 consecutive
adenosine nucleotides. It is added during post-transcriptional processing by a
terminal transferase enzyme.
2.
The nucleotide sequence listed below represents the nontemplate strand of a gene.
Please give: (a) the nucleotide sequence of the template strand, (b) the nucleotide
sequence of the mRNA formed (assuming no introns are present), and (c) the
amino acid sequence of the peptide chain encoded by this sequence. Be sure to
include which ends are 5', 3', amino, or carboxyl. (16 pt)
nontemplate DNA strand
5' CGATGTCTTCAACTACGTAG 3'
template DNA strand:
3’GCTACAGAAGTTGATGCATC 5’
mRNA:
5’CGAUGUCUUCAACUACGUAG 3’
Peptide chain:
(amino end) met - ser - ser - thr - thr (carboxyl end)
(starting at an initiation codon)
3.
Each of the following is related to or involved in DNA replication in the
bacterium Escherichia coli. Please briefly describe or state the role of each.
Note: for enzymes, you should state each separate enzymatic activity, and state its
purpose. (2 pt each)
(a)
DNA ligase Forms the final phosphodoester bond between two fragments of a
nascent chain during DNA replication.
(b)
Primase: Creates short RNA primers that are needed by DNA polymerase,
because DNA polymerase cannot initiate the de novo synthesis of a new chain.
(c)
DNA polymerase I: This enzyme has 5’ -> 3’ polymerase activity capable of
elongating a nascent strand. It also has 3’ exonuclease activity that serves to remove
mismatched bases (proofreading function). It also has 5’ exonuclease activity that is
responsible for primer excision. Its major role in vivo is primer excision and
proofreading.
Page 13
(d)
DNA polymerase III: This enzyme has 5’-> 3’ poylmerase activity and also 3’
exonuclease (proofreading) activity, but no 5’ exonuclease activity. It is responsible for
most of the nascent chain elongation during replication in E. coli.
(e)
Okasaki fragments: These are short fragments of nascent strand DNA
representing the discontinuously synthesized strand (lagging strand).
6.
In an experiment, cells of actively growing E. coli were incubated for several
minutes with radiolabeled deoxyribonucleosides (dNTPs). The DNA was isolated
from the cells, denatured into single strands, and separated according to size by
gradient centrifugation. The radioactivity in the tube corresponded to a single
longer piece of DNA and a heterogeneous mixture of much shorter DNA
fragments. Briefly explain these results. In your answer, you should use big words
like “discontinuous.” (4 pt)
The longer strand represented continuous strand synthesis, and the shorter
fragments represented discontinuous strand synthesis (Okasaki fragments)..
5.
Eukaryotic cells can be grown in a medium containing bromodeoxyuridine
(BrdU), which substitutes for thymidine during DNA replication. A fluorescent
dye that binds to double-stranded DNA makes it possible to recognize
chromosomes or chromatids with BrdU incorporated in their DNA. You can tell
the difference between unlabeled chromatids, chromatids with one strand of the
DNA double helix labeled, and chromatids with both strands of the DNA double
helix labeled. The chromatid fluoresces brightly when only one strand contains
BrdU, but weakly when both strands contain BrdU. The following diagrams
represent the fluorescence of metaphase chromosomes.
Page 14
A group of cultured eukaryotic cells, all at the G1 stage, were placed in a medium
containing BrdU and allowed to grow. Samples were removed during the first,
second, and third mitotic divisions. Mitosis in each of these samples was arrested
at metaphase using colchicine. Chromosomes were isolated from cells in the
samples and examined for fluorescence.
What will be the appearance of the metaphase chromosomes isolated in the first,
second, and third mitotic divisions after transferring the cells to the BrdU
medium? Answer this question by sketching the chromosomes, as shown in the
diagram. Use double hatch marks to indicate bright fluorescence, and single hatch
marks for weak fluorescence. (6 pt)
Here are two hints.
Hint 1: Chromosomes isolated at metaphase are in the replicated state (a pair of
chromatids connected at the centromere)
Hint 2: Each chromatid in a replicated chromosome has one double helix. The
two helices are the products of DNA replication.
See pgs. 8 & 9 for an explanation. This question asks you to predict what the
appearance should be, so you should state the results for SEMICONSERVATIVE
replication.
Page 15