School of Science
Hamilton Campus
Session 2011 – 12
Trimester 1
Module Code: BIOL09020
PURE AND APPLIED GENETICS
ANSWER SCHEME
Date: January 2012
Time:
Duration:
Answer THREE questions, at least ONE from each section
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BIOL09020
1.
a)
Answer Scheme
January 2012
Discuss the features of the plasmid pUC18/19 (Figure 1) which
make it a good vector for cloning DNA in bacterial cells.
(6 marks)
Answer
Key features:
As a plasmid it can be replicated inside the host cell. Often high
copy number.
Contains multiple unique cloning sites (single restriction sites
to insert fragment).
Selectable markers to enable transformed cells to be identified.
May also include:
Relatively small and easily isolated from the cell. Cells can be
forced to take up the plasmid in vitro.
b)
Describe the steps you would perform to clone a DNA fragment in
E.coli using the vector in figure 1.
Answer
The DNA to be cloned and the plasmid are cut with the same
restriction enzyme. The enzyme should cut uniquely within the
MCS and at either end of the DNA fragment. Enzymes that
produce sticky ends are preferred. The plasmid could be treated
with alkaline phosphatase prior to ligation to reduce the chance
of vector religation. Cut plasmid and fragment are mixed in a
ligation reaction with DNA ligase. The complementary sticky
ends should base pair and ligase forms the phosphodiester
bonds. The ligation mix would be used to transform bacterial
cells, E.coli. Probable method would involve calcium chloride
and heat shock (explanation of principles of technique). Control
with no DNA included. Both the ligation reaction and the
transformation reaction are inefficient resulting untransformed
cells, cells containing plasmid only with no inserted DNA and
cells containing the desired recombinant plasmid. The desired
cells are selected – after transformation cells are plated onto agar
containing ampicillin and X-gal. Ampicillin will kill all
untransformed cells which are not resistant to the antibiotic while
all cells containing the plasmid will survive. Those containing
plasmid only or recombinant plasmid are selected using X-gal.
The lac Z gene in the plasmid codes for  galactosidase which
converts the substrate X-gal into a blue product making those
colonies containing plasmid only blue. In the recombinant
plasmid the foreign DNA fragment is inserted into the lac Z gene
inactivating it therefore it does not produce B gal cannot break
down the substrate and therefore the colonies remain
white/cream. To confirm the insert is present a plasmid miniprep
is used to isolate the plasmid. Cut the isolated plasmid with
restriction enzymes appropriate to show the presence of the
fragment and its orientation. Run on agarose gel and visualise.
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(14 marks)
BIOL09020
Answer Scheme
January 2012
Figure 1
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BIOL09020
Answer Scheme
January 2012
2.
You are studying a gene coding for an enzyme that breaks down
toxins which are harmful to the cell.
a)
Describe how you would use the polymerase chain reaction (PCR)
to amplify the gene from genomic DNA.
(12 marks)
Answer
Answer should include key points as follows:
Basic components are DNA template (Generally double
stranded DNA, no need to be purified, continuous
sequence spanning the region to be amplified), DNA
Polymerase (thermostable), Oligonucleotide primers (Two
synthetic oligonucleotides (20–30bp) that hybridise to
opposite strands of the DNA and flank the DNA of interest
allowing extension to the 3’ end), Deoxynucleotide
triphosphates, Reaction buffer, Magnesium.
PCR is a repeated cycle of 3 steps each cycle the number
of specific DNA templates doubles.
Denaturation (94oC) – Heat is used to break the H bonds
between strands denaturing the DNA allowing it to be used
as a template.
Annealing (40 – 60oC) – the primers anneal to the template,
exact temperature depends on the primer.
Extension (72oC) – DNA Polymerase adds nucleotides to
the 3’end of the primer.
Repeated at least 30 times.
b)
Outline how the basic PCR technique can be modified to determine
if the gene is expressed after exposure to toxins.
(8 marks)
Answer
Reverse transcriptase PCR should be described. RNA is
isolated from the cell and reverse transcribed to cDNA using
the enzyme reverse transcriptase. Oligo dT can be used as the
primer which would produce cDNA from mRNA. The presence
of cDNA specific for the enzyme can be determined by using
cDNA as the template in PCR with primers specific for the gene
of interest. A product will only be produced if the mRNA was
present originally. May suggest the amount of expression could
be determined using qPCR.
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BIOL09020
3.
Answer Scheme
January 2012
Two types of genes, oncogenes and tumour suppressor genes have
been shown to play a role in the development of cancer. Discuss the
normal function(s) of these genes in the cell and examples of
mutations that can lead to cancer.
(20 marks)
Answer
Answer might include the following points.
Oncogenes are cancer causing genes ie. expression results in
cancer. These are mutated versions of the normal gene (protooncogene) genetically dominant; one copy can change the cells
behaviour. They are cellular genes whose function is to
promote normal growth and division of cells. Expression of
these genes is tightly controlled in a normal cell. Changes to
the genes may increase their expression resulting in increased
cell division. Oncogenes may be cell surface receptors, growth
factors, molecules involved in transmembrane signalling,
second messengers or nuclear proteins.
Both qualitative and quantitative effects in protein expression
should be discussed with examples.
Burkitts lymphoma is an example of a quantitative change
translocation alters transcription of the myc oncogene on
chromosome 8 to chromosome 14, 22 and 2 in regions where
there are immunoglobulin loci. These genes are actively
transcribed. There are two possibilities a repressor region is
destroyed or myc is influenced by the immunoglobulin
enhancer both result in increased expression. C-myc is a DNA
binding protein. Qualitative change example could be EGF
receptor and erbB. ErbB is a truncated EGF receptor which is
permanently switched on. There are many other examples that
could be used.
Tumour suppressor genes are genes whose normal function is
to inhibit the cell cycle and cell division. The gene coding for
p53 is mutated in the majority of cancers implying that the loss
of the normal gene product is associated with malignancy. p53
plays a central role in controlling the cell cycle and loss of
function leads to uncontrolled cell growth. It is a transcription
factor, which activates the expression of cellular genes whose
products cause the cell cycle to arrest in G1. Loss of p53
means these genes are not activated and the inhibition in G1 is
removed.
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BIOL09020
Answer Scheme
January 2012
4.
a)
Compare and contrast the content and organisation of the genome
in prokaryotes and eukaryotes.
(10 marks)
Answer
A typical 50kb segment of human genomic DNA may
contain:
One gene coding for a protein. Genes coding for proteins
contain introns and exons (on average 9 per gene). During
transcription the entire gene is copied to pre mRNA but
introns are subsequently removed by splicing to produce
mRNA which is used to direct protein synthesis.
Gene segments
Pseudogene – non-functional copy of a gene. May have
been a gene inactivated by mutation which has accumulated
further mutations or derived from mRNA. A DNA copy of the
mRNA is made by reverse transcriptase. The copy is then
inserted into the genome but is inactive as it lacks introns
and upstream sequences.
Interspersed repeat sequences that occur in several places
throughout the genome.
Microsatellites – sequences that are repeated several times.
50% non coding, non-repetitive single copy DNA with no
known function
Bacterial DNA is much smaller and generally circular. It is
much more compact, with far more genes in a smaller space
and very few repetitive sequences. Very little non-coding
DNA ~ 11%. No introns. Genes are often very close,
arranged in operons where several genes are transcribed as
a single mRNA from one promoter. Sense sequences are on
both strands. Bacteria often contain plasmids. Plasmids are
small extra-chromosomal DNA that contain genes. The
genes give bacteria additional properties e.g. antibiotic
resistance but are dispensable.
b)
Outline why the ends of linear chromosomes get shorter during
successive rounds of replication in eukaryotes.
(4 marks)
Answer
The chromosome becomes shorter once the last primer is
removed from the end. DNA polymerase has no 3’ end to add
on to complete the gap. Diagrams might help to explain this.
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BIOL09020
c)
Answer Scheme
January 2012
Discuss the effect telomere shortening has on the cell and describe
the mechanism some cells have developed to overcome.
(6 marks)
Answer
The chromosome shortens every time the cell divides.
Shortening limits the number of cell divisions as eventually
genes become damaged and cells can no longer divide. The
presence of active telomerase prevents shortening. Telomerase
provides a short molecule of RNA to allow extension of the
3’end. There is then sufficient length to prime the 5’ end once
the primer is removed the 5’end is the correct size. (important
in stem cells and cancer cells).
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BIOL09020
5.
Answer Scheme
January 2012
Discuss the events that occur at the level of transcription when
E.coli is grown on a mixture of glucose and lactose.
(20 marks)
Answer
E. coli will use glucose as a carbon source for growth first.
Glucose levels will decline while lactose remains constant.
During the decline the bacteria show exponential growth.
Growth is followed by a lag after which lactose levels decline
with a second exponential growth suggesting lactose is used
as a carbon source. When the lactose is depleted the bacteria
enter a stationary phase.
In the absence of lactose the lactose repressor is bound to the
operator preventing binding of RNA Polymerase and
transcription of the operon. The repressor occasionally
detaches allowing a few molecules to be transcribed. When the
bacteria encounter lactose the low level of enzymes allow it to
be transported into the cell and metabolised to glucose and
galactose. An intermediate in this reaction is allolactose, which
binds to the repressor, changes its conformation so it can no
longer bind to the operator. RNA Polymerase can then bind to
the promoter and transcribe the 3 genes. Transcription is still
low without the activation of CAP.
Catabolite activator Protein
This protein binds to sites upstream of operons including the
lactose operon and increases the efficiency of transcription
initiation.
When E.coli is grown in the presence of glucose as well as
lactose the glucose acts to override the lac operon. Glucose
controls the activity of adenylate cyclase an enzyme that
converts ATP to cAMP. Glucose inhibits the enzyme decreasing
cAMP levels. The CAP can bind to DNA only in the presence of
cAMP therefore in the presence of glucose binding is inhibited
and the lac operon is only transcribed at a very low level.
Diagrams showing the arrangement of the operon may be
useful.
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BIOL09020
6.
Answer Scheme
January 2012
Three types of RNA are involved in Translation (Protein synthesis)
discuss the role and interaction of all three.
(20 marks)
Answer
This is an example of what might be included but alternatives
are acceptable.
The function and role of each RNA molecule should be
described in detail.
mRNA carries the code to be translated copied from DNA. It is
read 5’ to 3’ in groups of 3 nucleotides, codon.
rRNA and proteins make up the ribosome, site of protein
synthesis. Ribosomes consist of two subunits, small and large
which are separate in the cytoplasm when not involved in
protein synthesis. The small subunit and IF3 attach to mRNA.
Attachment is to the ribosome-binding site (Shine Dalgarno
sequence). This is a sequence on the mRNA upstream of the
start codon which is complementary to a sequence on the 3’
end of 16s rRNA. Attachment positions the small subunit over
the initiation codon, AUG. The structure of tRNA should be
drawn or described. Two key areas highlighted – anticodon –
triplet of nucleotides that base pair with the codon on mRNA,
specific interaction involving base pairing. A diagram could be
used to show pairing and direction of RNA’s. Wobble could be
described. The codon- anticodon interaction brings the correct
tRNA.
- Amino acid acceptor stem, 3’ end has CCA where amino acid
is attached. Addition of amino acid is catalysed by aminoacyl
tRNA synthetases. Reaction can be described, isoaccepting
tRNAs. The enzymes have high fidelity for their tRNA
structurally similar amino acids can be edited. Some amino
acids are attached and modified. The key is to have the correct
amino acid attached.
When the large subunit attaches to form the complete ribosome
there is space for 2 tRNA molecules plus exit site in the cavity
between the subunits. The small subunit has the codonanticodon interaction and the large subunit has the aminoacyl
end. The initiator tRNA occupies site P, peptidyl site. The
second site A, aminoacyl has the second codon exposed and
the tRNA complementary to the codon is brought to the site by
elongation factors, which ensures the correct amino acid is
attached. Contacts between tRNA, mRNA and 16sRNA ensure
correct tRNA is accepted, correct base pairs must be formed
between all 3 nucleotides.
A peptide bond is now formed between the amino acids held by
the 2 tRNA’s. Peptidyl transferase catalyses this reaction and
the release of the amino acid from the tRNA in site P. In
bacteria this activity is part of the 23s rRNA and is therefore a
ribozyme. The reaction requires energy.
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BIOL09020
Answer Scheme
January 2012
All the following occur at once:
 Ribosome moves along 3 nucleotides to expose third codon in
site A.
 The tRNA carrying the dipeptide moves from site A to P.
 The empty tRNA in site P moves to a third exit site in bacteria or is
ejected in eukaryotes
Energy is required. Electron microscopy suggests that the 2 ribosomal
subunits rotate slightly in opposite directions creating a space enabling
the ribosome to slide along the mRNA. The empty A site is filled with the
appropriate tRNA and the elongation cycle continues to the end of the
open reading frame. Diagrams would be good.
END OF QUESTION PAPER
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