CSULA Molecular Biology Melting Temperature & Nucleotides Questions

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PRACTICE MIDTERM – BIMM 100
This MIDTERM will be 20 multiple answer choice questions worth 2 points each (40
points total). For full credit, select all answers that are correct for each question. This
Midterm is “open book” (you may use notes, lecture slides, textbooks) but any
communication with another person, live or digitally, will be considered cheating and
result in failure of this class. You are required to have video on in Zoom to get
credit for taking this exam.
1.
Using the below graph showing the relationship between GC content and melting
temperature (Tm), what is the percent Thymine in a DNA sample with a Tm of 95°C?
Percentage of G-C pairs
100
80
60
40
20
0
65 70 75 80 85 90 95 100 105
Tm ( C)
20%
40%
60%
30%
2.
If Integrase created a staggered cut between the nucleotides in bold and underlined,
what is the sequence of the target site direct repeat that would result after integration of
a transposon at this position?
5’ TGATTAGCGCTTAGACTTAGCACGGATCTACTAAGTTCGCGTTAGCTA 3’
3’ ACTAATCGCGAATCTGAATCGTGCCTAGATGATTCAAGCGCAATCGAT 5’
5’ CGGATC 3’
5’ ACGGATCT 3’
3’ CGGATC 5’
5’ TTACGTAG 3’
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PRACTICE MIDTERM – BIMM 100
3.
Which of the following statements are consistent with the results of this DNase I
Protection Assay used to analyze the chromatin structure of the Hemoglobin B (HBB)
gene in different tissues?
Tissue:
DNase I (µg/ml):
adult
0 1 2
fetal
embryonic
0 1 2 0 1 2
HBB
The HBB gene is likely to be associated with Histone 3 acetylated on lysine 9
(H3K9Ac) in adult tissue
The HBB gene is likely to be highly expressed in adult tissue
Heterochromatin Protein 1 (HPI) is likely to be associated with the HBB gene in
embryonic tissue
The HBB gene is likely to be associated with Histone 3 methylated on lysine 9
(H3K9Me) in adult tissue
4.
Which of the following could be used as a probe to detect this Hemoglobin B gene
(HBB) sequence in the Southern Blot part of a DNase I Protection Assay?
5’ ACATTTGCTTCTGACACAAC 3’
3’ TGTAAACGAAGACTGTGTTG 5’
5’ GUUGUGUCAG 3’ radiolabeled with 32P
5’ GTTGTGTCAG 3’ radiolabeled with 32P
5’ GTTGTGTCAG 3’ labeled with a fluorescent tag
5’ ACATTTGCTT 3’ radiolabeled with 32P
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PRACTICE MIDTERM – BIMM 100
5.
If an Alu element has inserted into the Hemoglobin B gene (HBB) and this changes the
levels but not the sequence of the HBB protein, where could it have inserted?
promoter
5’UTR
3’UTR
coding region
6.
If you performed a reaction in a test tube with a Telomerase enzyme containing the
template sequence of 5’ CCCCUAA 3’ and a telomere primer sequence of 5’ AATCCCC
3’, what repeat sequence would be added on to the end of the primer?
None
5’ TTAGGG 3’
5’ AATCCC 3’
5’ CCCTAA 3’
7.
Using single letter abbreviations, what are the first 4 nucleotides of primers you would
need in a PCR reaction to amplify the entire sequence of DNA shown below?
5’ GTAGGCATCATCATCATCATCATCATCATCATCATGAATC 3’
5’ GTAG 3’
5’ GATT 3’
5’ CTAC 3’
5’ TTAG 3’
8.
If a DNA transposon inserted into GENE-A blocks its expression, but normal expression
of GENE-A is completely restored when that DNA transposon hops out of GENE-A,
which mechanisms could have been used to repair GENE-A?
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PRACTICE MIDTERM – BIMM 100
Homologous recombination
End-joining
Nucleotide excision
Mismatch excision
9.
Patient 2
Patient 1
Marker
To screen for Huntington’s Disease, 20 nucleotide primers immediately flanking the 5’
CAG 3’ repeat region in the HD gene were used in PCR reactions with DNA from 2
Patients. Based on these results, which of the following can be concluded? Note that
the Marker lane shows where different sized DNA fragments run in this gel.
350
300
250
200
150
100
50
Patient 1 has 20 repeats in both copies of the HD gene
Patient 2 has 20 repeats in one copy and 100 repeats in the other copy of the HD
gene.
Patient 2 has a higher risk of developing Huntington’s Disease than Patient 1.
Patient 1 has 100 repeats in both copies of the HD gene
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PRACTICE MIDTERM – BIMM 100
10.
Below are results from an experiment designed to test the effect of microbial toxins on
DNA replication in human cells. Which of the following conclusions are consistent with
the results?
TREATMENT
No toxin (control)
Toxin-X
Toxin-Z
RESULT
Complete synthesis of new DNA
Many short unconnected fragments of new DNA are detected
No new DNA strands are detected
Toxin-X inhibits Ligase
Toxin-Z inhibits Primase
Toxin-X inhibits Helicase
Toxin-Z inhibits RNase H
11.
Suspect 3
Suspect 2
Suspect 1
Cigarette
Marker
DNA Fingerprinting was performed on DNA taken from a cigarette left at a crime scene,
as well as from 3 different people suspected of being present at the crime. Which of the
following options are consistent with the results?
100
90
80
70
60
50
Suspect 1 left DNA on the cigarette
Suspect 2 left DNA on the cigarette
Suspect 3 left DNA on the cigarette
Suspects 1 and 2 are twins
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PRACTICE MIDTERM – BIMM 100
12.
For the below sequence, where the +1 site is in bold underline and the +10 and -10
sites are also labeled, what are the first 3 nucleotides of the RNA transcribed from this
sequence?
+1
+10
-10
GAGCGACATAATACGATTAT 3’
5‘
5’ AUA 3’
5’ UAU 3’
5’ AAU 3’
5’ UUA 3’
13.
To identify transcriptional regulatory elements, reporters containing 100 base pairs of
Promoter Region DNA were tested in an expression system, where higher numbers
indicate higher expression levels. The black boxes indicate the region mutated in each
reporter. Which of the following can be concluded from these results?
Mutant
number
-100
Promoter Region
-80
-60
-40
-20
Reporter gene
expression
+1
5
1
5
2
10
3
5
4
0
5
5
The -40 to -20 region contains a Positive regulatory element
The -70 to -60 region contains a Negative regulatory element
The -60 to -40 region contains a Negative regulatory element
The -70 to -60 region contains a Positive regulatory element
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PRACTICE MIDTERM – BIMM 100
14.
Below are the results of mutational analysis using reporters containing 100 base pairs of
Promoter Region DNA in an expression system, where higher numbers indicate higher
expression levels. The black boxes indicate the region mutated in each reporter. Based
on these results, what is the most likely type of core promoter element in this
sequence?
Mutant
number
-100
Promoter Region
-80
-60
-40
-20
Reporter gene
expression
+1
5
1
5
2
10
3
5
4
0
5
5
TATA box
Initiator
CpG Island
Enhancer
15.
Which of the below conditions would result in no expression from the lac operon?
High lactose, low glucose, mutation in lac Repressor so that it can bind the
Operator and lactose at the same time
No lactose, high cAMP
High lactose, high glucose, mutation in sigma 70 protein that increases its affinity
for binding the Promoter
No lactose, deletion of the Repressor binding site in the Operator sequence, low
glucose
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PRACTICE MIDTERM – BIMM 100
16.
Protein A + B
Protein B
Protein A
No protein
base
pairs
150
Marker
To study the binding of Proteins A and B to the GENE-N promoter region, you perform
DNase I Footprinting using the -150 to +1 region of GENE-N with the -150 position
radiolabeled. Based on these results, which of the following can be concluded?
–
125
100
75
50
25
+
-150 -125 -100 -75 -50 -25 +1
GENE-N promoter region
Protein B binds position -50 to -25
Protein A alone cannot not bind GENE-N promoter region
Protein B binds about 25 base pairs of DNA in the GENE-N promoter region
Protein B binds position -100 to -125
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PRACTICE MIDTERM – BIMM 100
17.
Protein A + B
Protein B
Protein A
No Protein
Based on the results of the below EMSA (Gel Shift) assay, using radiolabeled GENE-N
promoter DNA as the probe and the indicated Proteins, which of the following can be
concluded?
GENE-N
promoter DNA
Protein A requires Protein B to bind the GENE-N promoter DNA
Protein B requires Protein A to bind the GENE-N promoter DNA
Protein A + B results in the highest level of transcription
Protein B inhibits Protein A from binding the GENE-N promoter DNA
18.
Below are the results of an In Vivo Reporter Assay to test if Element N in the GENE-N
promoter regulates expression in response to viral infection. Based on these results,
where higher numbers indicate higher levels of reporter gene expression and + means
the Element is present and – means the Element is absent, which of the following can
be concluded?
Hours after viral infection
0 6 12 18
+ Element N: 0 5 10 10
– Element N: 5 5
5
5
Before viral infection, Repressor proteins inhibit expression of the reporter
through Element N
Element N is needed to respond to viral infection
At 12 hours after infection, Repressor proteins decrease expression of the
reporter through Element N
Before viral infection, Activator proteins promote expression of the reporter
through Element N
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PRACTICE MIDTERM – BIMM 100
19.
To analyze the association of specific proteins with the GENE-N promoter in viral
infected cells, you perform a chromatin immunoprecipitation (ChIP) assay using
antibodies (a-) for the proteins listed to the right of the results shown below. H3 =
Histone 3, TBP = Tata Binding Protein. Which of the following conclusions are
consistent with these results?
Hours after viral infection
0
6
12
18
a -acetylated
H3 (K9)
a -methylated
H3 (K9)
a -Protein A
a -Protein B
a-TBP
INPUT DNA
control
Expression of GENE-N would first be detected at 12 hours after viral infection
GENE-N is repressed before viral infection
After viral infection, Protein B associates with GENE-N before RNA Polymerase
II
Protein A is Heterochromatin Protein 1 (HP1)
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PRACTICE MIDTERM – BIMM 100
20.
Which of the following could explain why the human Myoglobin gene is expressed in
cardiomyocyte cells but not in hepatocyte cells?
Different transcriptional regulatory proteins are present in cardiomyocyte and
hepatocyte cells
Different histone modifications are present in the chromatin associated with the
Myoglobin gene in cardiomyocyte and hepatocyte cells
Different promoters are present in cardiomyocyte and hepatocyte cells
The Myoglobin gene is in cardiomyocyte cells but not in hepatocyte cells

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