1. A mutation that inactivates the regulatory gene of a repressible operon in an E. coli cell
would result in
A) inactivation of RNA polymerase by alteration of its active site.
B) continuous translation of the mRNA because of alteration of its structure.
C) continuous transcription of the structural gene controlled by that regulator.
D) irreversible binding of the repressor to the operator.
E) complete inhibition of transcription of the structural gene controlled by that
regulator.
2. There is a mutation in the repressor that results in a molecule known as a superrepressor because it represses the lac operon permanently. Which of these would
characterize such a
mutant?
A) It cannot make a functional repressor.
B) It cannot bind to the inducer.
C) It makes molecules that bind to one another.
D) It makes a repressor that binds CAP.
E) It cannot bind to the operator.
3. Transcription of the structural genes in an inducible operon
A) starts when the pathway's substrate is present.
B) stops when the pathway's product is present.
C) occurs continuously in the cell.
D) does not result in the production of enzymes.
E) starts when the pathway's product is present.
4. Gene expression might be altered at the level of post-transcriptional processing in
eukaryotes rather than prokaryotes because of which of the following?
A) Prokaryotic genes are expressed as mRNA, which is more stable in the cell.
B) Eukaryotic coded polypeptides often require cleaving of signal sequences
before
localization.
C) Eukaryotic exons may be spliced in alternative patterns.
D) Eukaryotic mRNAs get 5' caps and 3' tails.
E) Prokaryotes use ribosomes of different structure and size.
Questions 5&6: A researcher found a method she could use to manipulate and quantify
phosphorylation and methylation in embryonic cells in culture.
5. One of her colleagues suggested she try increased methylation of C nucleotides in a
mammalian system. Which of the following results would she most likely see?
A) increased chromatin condensation
B) inactivation of the selected genes
C) decreased chromatin condensation
D) abnormalities of mouse embryos
E) decreased binding of transcription factors
6. Within a cell, the amount of protein made using a given mRNA molecule depends
partly on
A) the degree of DNA methylation.
B) the presence of certain transcription factors.
C) the rate at which the mRNA is degraded.
D) the types of ribosomes present in the cytoplasm.
E) the number of introns present in the mRNA.
7. Assume that you are trying to insert a gene into a plasmid. Someone gives you a
preparation of genomic DNA that has been cut with restriction enzyme X. The gene you
wish to insert has sites on both ends for cutting by restriction enzyme Y. You have a
plasmid with a single site for Y, but not for X. Your strategy should be to
A) cut the plasmid twice with restriction enzyme Y and ligate the two
fragments onto the ends of the DNA fragments cut with restriction
enzyme X.
B) cut the plasmid with restriction enzyme X and insert the fragments cut
with restriction enzyme Y into the plasmid.
C) cut the plasmid with restriction enzyme X and then insert the gene
into the plasmid.
D) cut the DNA again with restriction enzyme Y and insert these
fragments into the plasmid cut with the same enzyme.
E) insert the fragments cut with restriction enzyme X directly into the
plasmid without cutting the plasmid.
8. A principal problem with inserting an unmodified mammalian gene into a BAC, and
then getting that gene expressed in bacteria, is that
A) bacteria translate polycistronic messages only.
B) bacteria cannot remove eukaryotic introns.
C) prokaryotes use a different genetic code from that of eukaryotes.
D) bacterial RNA polymerase cannot make RNA complementary to
mammalian DNA.
E) bacterial DNA is not found in a membrane-bounded nucleus and is
therefore incompatible with mammalian DNA.
9. Which of the following best describes the complete sequence of steps occurring during
every cycle of PCR?
1. The primers hybridize to the target DNA.
2. The mixture is heated to a high temperature to denature the double-stranded
target DNA.
3. Fresh DNA polymerase is added.
4. DNA polymerase extends the primers to make a copy of the target DNA.
A) 3, 4, 2
B) 2, 1, 4
C) 2, 3, 4
D) 1, 3, 2, 4
E) 3, 4, 1, 2
10. Of the three modes of gene regulation shown in the figure, which is the fastest in
response time?
A.
B.
C.
D.
post-translational control
transcriptional control
translational control
All three are equally fast.
11. Of the three modes of gene regulation shown in the figure above, which is the most
efficient in resource use?
A.
B.
C.
D.
post-translational control
transcriptional control
translational control
All three are equally efficient.
12. At which of the following stages does transcriptional control occur?
DNA →a mRNA →b protein →c activated protein
A. a
B. b
C. c
D. all three are correct
13. Although the expression of most genes is tightly regulated, some genes are expressed
at roughly constant rates (i.e., constitutively). Which of the following genes would you
predict to be constitutively expressed?
A. genes involved in the degradation of tryptophan
B. genes involved in the transport of the sugar maltose
C. genes involved in the degradation of arabinose (a sugar)
D. genes involved in the biosynthesis of the amino acid tryptophan
E. genes that code for ribosomal RNAs
14. Your genome is 99.9 percent identical to your classmate’s genome. Considering
the relationship between the genome and the proteome, what is the most likely
relationship between your proteome and that of your classmate?
A. More than 99.9 percent identical
B. 99.9 percent identical
C. Less than 99.9 percent identical
15. The sequencing of the human genome has allowed scientists to
A. understand regulatory sequences that are important for gene expression.
B. locate genes that cause disease.
C. understand evolutionary relationships by comparing human genes to genes
in other organisms.
D. investigate gene families and their origins.
E. All of the above
16. Proteomics has been used to compare
A. DNA sequences between closely related species.
B. gene expression during embryonic development.
C. protein sequences between closely related species.
D. shotgun cloned sequences.
E. prokaryotic genomes.
17. Which of the following statements comparing the human genome to that of
invertebrates like C. elegans and D. melanogaster is true?
A. A larger fraction of DNA in the human genome is for coding proteins.
B. The human genome is more than ten times larger.
C. The average human gene codes for more proteins than the average
invertebrate gene.
D. Both a and b
E. Both b and c
Short Answer:
1. Draw a plasmid. Include the labels in the table below.
2. Match terms in Column A with the appropriate term in Column B. Some
responses in Column A have only one answer, some have multiple, some have
none.
Column A
Marker gene: GFP, Tetracycline, Ampicillin
cDNA: in some cases inserted gene is GFP to
indicate expression as study of promoter
control
Origin of Replication: none in Column B
Restriction Enzyme Site: EcoR1, BamH1
Column B
1. EcoR1
2. Bam H1
3. Tetracycline resistance
4. Ampicillin resistance
5. GFP
3. It is understood that gene expression can be modified by remodeling
chromatin structure. Explain why histone modifications can cause epigenetic
inheritance patterns. (2 marks)
By increasing positive charge on histone proteins (by methylation of histones), DNA
which is negatively charged wraps around the histones more tightly reducing the
ability for DNA to be transcribed. The attraction between DNA and histones can be
reduced if histone proteins are acetylated (addition of acetyl groups that have
negative charges). Acetylation of histones loosens the attraction of DNA to the
histones (loosens the nucleosome) allowing for transcription of DNA.
4. Fill-in the following table to summarize your understanding of the control
over the Lac Operon.
Conditions
Outside of Cell
Conditions Inside the Cell
Adenylyl Cyclase
(active or not
active)
cAMP Levels
(high or low)
CAP
(active or not
active)
High Glucose
Not active
Low
Not active
Transcription of
Lac Operon
(frequent or
infrequent)
Infrequent
Low Glucose
Active
High
Active
Frequent
5. Why biologists choose to isolate mRNA rather than DNA from the pancreas?
can get cDNA from mature mRNA to insert into bacterium to produce huge
quantities of E. coli.
6. What kind of techniques can be used to study gene expression during
development? (2 marks) cDNA library can be constructed by reverse
transcription of mRNA in a particular tissue
7. What is a genomic library? Explain how its created. (4 marks) genomic
library is a collection of DNA fragments that together comprise an organism.
Construct by 1- cut genomic DNA with RE, 2- insert fragments into vectors, 3recombinant vectors taken up by host bacterial cells, 4- each produces a
colony of recombinant cells
8. The gel in the figure shows the RFLP analysis of DNA samples obtained from
a crime scene. Blood stains on a suspect’s shirt (B) were analyzed and
compared with blood from the victim (V) and from the suspect (S). Are the
blood stains on the shirt from the victim or from the suspect? Explain. If you
examine the lane that contains the blood stains, you can find bands that do
not match a band in either the victim’s or the suspect’s lane. As a result, the
blood that contains this band could not have come from the suspect.
a. Investigators find an additional blood sample in the garage, but
they are not sure whether it is related to the crime. After
undergoing electrophoresis, the gel in the figure below shows
the results of DNA fingerprinting of blood samples
obtained from the victim (V), her husband (H), the
suspect (S), and the blood from the garage (B2). Is
the blood from the garage related to the crime?
Explain your answer. The blood from the garage
does not match the suspect, but it has many bands
in common with both the victim and her husband.
The blood from the garage most likely belongs to
one of their children and therefore is either not
related to the crime or may implicate one of the
children in the crime.
9. What is dideoxy method for DNA sequencing? (4 marks)
It’s a method of sequencing that includes a proportion of
dideoxyribonucleotides in the reaction mixture to allow
their addition to the growing strand but don’t have an
available 3’OH and the polymerization ends at each
addition of a dideoxyribonucleotide.
With a different colour fluorescent tag on each of the four
different dideoxyribonucleotides, the terminal nucleotide
can be known by the colour of fluorescence. The
collection of strands (some closer to completion of the
non-template, and some were terminated before) are run
on an electrophoresis gel. The shortest strands run the
furthest. The sequence of the non-template strand can be
read from the gel starting with the shortest fragments
that have run the longest distance on the gel.
This method works well when the appropriate
concentrations of both deoxyribonucleotides and
dideoxyribonucleotides are such that the non-template
strand can extend completely in some case, with enough
dideoxyribonucleotides to cause termination regularly with every basepair.
10. How are open reading frames recognized in a genomic sequence? What kind
of information can be derived from an open reading frame? (2 marks) open
reading frames can be recognized by the start and stop codons for translation
and by intron consensus sequences.