Guideline answers Jan 2013
Model Answer
Section A
1 a) Describe in detail the steps you would perform to clone the DNA
fragment below (Figure 2) in E.coli using the vector in Figure 1 (Nb. The
MCS in the vector contains unique restriction sites for BamH1, EcoR1
and Pst1).
Figure 1
DNA Fragment
Figure 2
All steps should be described.
Cut both vector and DNA with restriction enzyme either BamHI only
(should treat vector with alkaline phosphatase to reduce vector
relegation) or BamHI and PstI (to achieve directional cloning).
DNA Ligase to ligate fragment into vector outlining possible
Transformation of E.coli with ligation mix using calcium
chloride/heat shock or alternative. Possible outcomes and
description of selection method.
b) Figure 3 below shows the results of a real time PCR experiment using
SYBR green.
Model Answer
i) Describe the function of SYBR green in the reaction
SYBR green binds to double stranded DNA emitting fluorescence.
It can therefore be used to monitor PCR by the increase in
fluorescence after each cycle (doubling of DNA).
ii) Estimate the Ct value for each of the four samples
Ct = 18, 21, 25 and 28 – read from the baseline.
iii) From the Ct values discuss briefly what this tells you about the samples.
The higher the Ct value the less the amount of starting template.
Figure 3
Section A Continued
2 a) Oncogenes and tumour suppressor genes have been shown to play a
role in the development of cancer. Outline the normal function(s) of
these genes in the cell and using one example of each discuss how they
can contribute to the development of cancer.
Oncogene – gene that codes for a protein with a role in cell division
e.g. growth factor, GF receptor, intracellular signalling molecule,
transcription factor. Mutation in one allele results in uncontrolled
growth. A specific example should be described.
Model Answer
Tumour suppressor gene – protein which normally inhibits the cell
cycle. Requires a mutation in both alleles to produce nonb) functional protein. A specific example should be described.
DNA libraries are useful resources for scientists. Compare and contrast
the two types of library, genomic and cDNA.
Will need to include a description of the differences. Genomic
library is a collection of clones, which between them contain the
DNA of an entire organism.
A cDNA library is synthesized from mRNA by reverse transcriptase.
cDNA collection is smaller representative only of coding DNA and
then only genes that are expressed in the cell the mRNA is isolated
from – no upstream or downstream sequences or introns. Genes
that are highly expressed will be well represented in the library.
gDNA is the entire genome found in every cell.
3 a) Select one genetic disease. Describe the mutation or chromosomal
abnormality, the disease and the method of diagnosis.
Can select any mutation or chromosomal abnormality. The
mutation should be described and the effect on the protein and/or
disease caused. The test used in labs to detect the disease should
be described e.g. FISH for chromosomal abnormalities, PCR or
sequencing for many mutations.
b) Expressing a human recombinant protein in bacteria raises several
issues due to differences in the transcription and translation mechanism
of the two organisms. Outline two issues and potential solutions.
Two issues should be selected the problem and potential solutions
described such as; inability of bacteria to remove introns solved by
using cDNA or synthesised using amino acid sequence, difference
in promoters solved by transcribing from bacterial promoter within
vector; fusion proteins.
Section B
4 a) Describe the experiment of Meselson and Stahl and how this contributed
to our understanding of DNA replication.
Proved the mechanism of semi conservative replication. E.coli
were grown in N-15 therefore all DNA was heavier density as seen
by density centrifugation. Medium was replaced by normal N-14
and cells allowed one division. All DNA now appeared to be
medium density suggesting a mix of parental and daughter. On
Model Answer
second and subsequent divisions a lighter N-14/N-14 appeared and
increased. Diagrams can be used.
Discuss the effect telomere shortening has on the cell and describe the
mechanism some cells have developed to overcome this shortening.
Shortening of the telomere limits the number of cell divisions. The
enzyme, telomerase can restore the telomere. Contains an RNA
sequence and reverse transcriptase activity used to extend the 3’
end of the telomere. There is now room for primase and DNA
polymerase to extend the 5’ end.
Describe the process of the elongation step in translation outlining the
role of the three main types of RNA.
Elongation events are similar in prokaryotes and eukaryotes. When
the large subunit attaches to form the complete ribosome there is
space for 2 tRNA molecules in the cavity between the subunits. The
small subunit has the codon-anticodon 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 EF-Tu (eEF-1 in euks), 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 from the hydrolysis of GTP held by EF-Tu.
EF-Tu is inactivated and ejected from the system to be regenerated
by EF-Ts.
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
Model Answer
continues to the end of the open reading frame. Diagrams would be
5 Discuss the main differences in transcription between prokaryotes and
In prokaryotes RNA Polymerase (via sigma subunit) binds to
promoter sequence at -35. DNA is opened at -10 (A/T rich).
Transcription starts at +1. First few ribonucleotides are added the
sigma leaves. This allows RNA Polymerase to continue. DNA is
opened about 12-14 bp. RNA polymerase uses the 3’ to 5’ strand as a
template. RNA transiently base pairs with the template DNA.
Termination occurs after a hairpin or rho protein.
In eukaryotes note differences in promoter sequence at -25 and
initiation sequence. Binding of TBP to promoter followed by
number of transcription factors and RNA Polymerase to create an
initiation complex. Note the difficulty in accessing the genome due
to histones and packaging. Several elongation factors involved
although similar mechanism. The processing of mRNA should
cover 5’ capping, polyadenylation and intron splicing.
6 The diagram below (Figure 4) monitors four parameters of the growth of
E. coli on a medium containing both glucose and lactose.
a) Describe what the results in Figure 4 show for each parameter.
Brief description of graph. E. coli uses glucose as a carbon source
for growth first. Glucose levels decline immediately 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 now used
as a carbon source. This is confirmed by the increase in beta
galactosidase activity which is required for lactose metabolism.
When the lactose is depleted the bacteria enter a stationary phase.
b) Discuss the events that are occurring at the molecular level in these cells.
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
Model Answer
Polymerase can then binds to the promoter and transcribe the 3
Catabolite activator Protein
This protein binds to sites upstream of operons including the
lactose operon and increases the efficiency of transcription
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 not transcribed.
Figure 4

Jan 2013a answer guide 1