School of Science
Paisley Campus
Session 2011 – 12
Trimester 2
Module Code:BIOL09020
PURE AND APPLIED GENETICS
ANSWER SCHEME
Date: May 2012
Time:
Duration:
Answer THREE questions, at least ONE from each Section
Page 1 of 10
Continued overleaf
BIOL09020
ANSWER SCHEME
May 2012
Section A
Answer at least one question
1.
a)
Discuss the features of the plasmid pBR322 (Figure 1)
which makes 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 containing BamHI restriction sites at either end
in E.coli using the vector in figure 1.
(14 marks)
Answer
The DNA to be cloned and the plasmid are cut with the
restriction enzyme BamHI. The enzyme should cut uniquely
within the tetracyclin resistant gene opening the plasmid and
at either end of the DNA fragment. 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. Ampicillin will kill all untransformed cells which
are not resistant to the antibiotic while all cells containing the
plasmid will survive.
Page 2 of 10
Continued overleaf
BIOL09020
ANSWER SCHEME
May 2012
Those containing plasmid only or recombinant plasmid are
selected using tetracyclin. Cells containing plasmid only will
be resistant to tetracyclin while in the recombinant plasmid
the foreign DNA fragment is inserted into the tetracyclin
resistant gene inactivating it therefore it does confer
resistance to tetracyclin. A copy is made of the colonies on
the agar and Amp plate and those that are killed by tet are
selected. 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.
Figure 1
Page 3 of 10
Continued overleaf
BIOL09020
2.
ANSWER SCHEME
May 2012
Write an account of the effects of genomic mutations with
reference to specific diseases. (20 marks)
(20 marks)
Answer
The answer should include a description of the different
types of mutation and examples of specific diseases. The
answer is fairly open but may cover some of the points
below.
A mutation is a change in a short region of DNA. This can
be a change in one nucleotide, point mutation, or insertion
or deletion of a few nucleotides, frameshift. Mutations can
arise from replication errors or environmental agents such
as chemicals or radiation. Some mutations are silent maybe
if they occur in non-coding DNA while others if they occur
in coding DNA or control regions have a significant effect.
A frameshift mutation should be explained (including
diagrams) and its effect on the cell. Examples might include
Cystic Fibrosis and loss of phenylalanine often by a
deletion of nucleotides.
Replication slippage and trinucleotide repeat expansion
disease could be discussed using specific examples e.g.
Huntington’s disease.
Point mutations change the sequence of a codon; this can
be synonymous – code for the same amino acid, nonsynonymous – code for a different amino acid. The effect of
this depends on the function of the specific amino acid. It
may have no effect, reduce activity of the protein or if the
change is to a key amino acid e.g. at an active site complete
loss of function, nonsense – code for a stop codon which
results in a short protein usually non-functional unless it
occurs near the end, readthrough – a stop codon is
changed to an amino acid producing a longer protein. Most
proteins might tolerate this but it could affect folding and
therefore function.
Mutations out with the coding region can have a variety of
effects including inactivation of promoters or regulatory
sequences e.g. if the mutation is to the sequence
recognised by a DNA binding protein. Mutations at the
intron exon boundary might affect splicing e.g. in β
thalassemia.
Mutations in germ cells and somatic cells. Recessive
mutations and loss of function e.g. cystic fibrosis or gain of
function which may confer an abnormal activity on a protein
but often affects the regulatory region resulting in altered
expression of the protein. An example might be a mutation
in the regulatory region of a gene coding for a protein with a
role in cell growth leads to over expression and cancer.
Page 4 of 10
Continued overleaf
BIOL09020
3.
a)
ANSWER SCHEME
May 2012
A DNA library is a useful resource for a scientist. Describe
the method(s) that would be used to create
(8 marks)
i) genomic library and
ii) cDNA library.
Answer
i) Genomic library is a collection of clones, which
between them contain the DNA of an entire organism.
The genomic DNA is isolated, cut with restriction
enzymes, the fragments are ligated into vectors and host
cells are transformed with the ligation mix. In theory
each transformed cell contains a vector with a piece of
the genome and between them they contain the entire
genome.
ii) cDNA is synthesized from mRNA by reverse
transcriptase. mRNA is used as the template by reverse
transcriptase adding to a primer e.g. oligo dT creating a
cDNA/mRNA hybrid. RNA is destroyed by alkali and the
synthesized cDNA strands can be used as a template for
the synthesis of the second strand. cDNA fragments are
then ligated into vectors and used to transform cells.
b)
Compare and contrast the content of the two libraries.
(6 marks)
Answer
Genomic library is larger representing the entire genome
coding and non coding containing at least one copy of
each piece of DNA. cDNA library only represents the
coding portion of DNA no non coding including introns.
It will only represent the genes that are expressed in the
cell the library has been made from. It may contain
multiple copies of highly expressed genes.
Page 5 of 10
Continued overleaf
BIOL09020
c)
ANSWER SCHEME
May 2012
Describe a method to select a specific fragment of DNA
with a known sequence from a DNA library.
(6 marks)
Answer
A probe is required to isolate a specific DNA sequence;
this can be a single stranded DNA sequence, RNA, gene
from another species, cDNA for a genomic library which
is complementary to the sequence of interest. The cells
of the library are transferred to membrane, lysed, the
DNA is made single stranded by treatment with alkali
and bound to the membrane using heat or UV light. The
membrane is incubated with the labelled probe which
will bind to the membrane. Non specific binding is
removed by washing. The probe will bind specifically to
the complementary DNA sequence via H bonding
between bases. The label allows the correct cell to be
identified.
End of Section A
Please turn over for Section B
Page 6 of 10
Continued overleaf
BIOL09020
ANSWER SCHEME
May 2012
Section B
Answer at least one question
4.
Outline the mechanisms of DNA replication and repair.
(20 marks)
Answer
Semi Conservative mechanism including details of
Meselson and Stahl experiment. Origin of replication,
formation of replication bubble. Outline the enzymes and
proteins involved and their role (DNA Polymerase, primase,
helicase, ligase etc.) Leading and lagging strand synthesis.
Termination.
Proofreading by DNA Polymerase which has a 3’-5’
exonuclease activity which detects a mismatched base pair
and removes it.
General Excision repair – Proteins scan the DNA ,recognise
mismatch, identify incorrect strand, cut DNA either side of
error, remove strand (may require helicase), fill gap with
DNA Polymerase III and seal with DNA ligase. Direct
mechanism may include repair of thymine dimmers by
Photolyase.
5.
A protein, G26, has been isolated from cells. The presence of
this protein in cells results in the increased expression of gene
Z. Structural studies have determined that this protein has a zinc
finger motif.
a) Discuss possible role of the protein, G26, in the cell.
(4 marks)
Answer
The zinc binding motif is commonly found in DNA binding
proteins. The protein probably recognises and binds to a
specific DNA sequence. As it enhances expression of Gene
Z it is likely to bind upstream of this gene enhancing
transcription.
Page 7 of 10
Continued overleaf
BIOL09020
ANSWER SCHEME
May 2012
b) Describe experiments you would perform to gain more
information about this protein, G26 and its interaction with
DNA.
(16 marks)
Answer
The answer should discuss methods for studying DNA
binding proteins. This should include a description of the
methods and the information that could be gained from the
experiment. Methods could include gel retardation,
Footprinting, modification interference, purifying the
protein, studying structure by X-ray crystallography and
NMR.
Gel retardation experiment. Based on the fact that naked
DNA and DNA with a protein attached can be separated on
agarose gels. The section of DNA thought to contain a
binding site is digested with restriction enzymes then the
digest mixed with an extract of nuclear proteins (the pure
protein will be used if it is available) or not. Banding
patterns are then compared. Diagrams could be used to
show how the fragment bound to protein is held back in the
gel. This tells us the general location of a protein-binding.
DNase I footprinting is then performed on the identified
fragment. DNase footprinting gives a more localised region
of binding. The fragment is labelled, incubated with nuclear
extract or not and digested with DNase I under limiting
conditions. In the absence of nuclear extract each copy of
the fragment should be cut once this results in a ladder of
bands differing by one nucleotide. In the presence of
nuclear extract there is a gap in the ladder, which
corresponds to the region the protein, has bound and
protected the DNA from digestion. This identifies the region
on the DNA where the protein binds.
Modification interference identifies nucleotides which are
key for protein binding. Individual nucleotides are modified
and the effect on binding is monitored.
Purifying the protein – the DNA binding protein can be
purified by passing a nuclear extract through a column on
which the DNA sequence immobilsed.
X-ray crystallography and NMR spectroscopy allow the
structure of the protein and its interaction with DNA to be
analysed.
Page 8 of 10
Continued overleaf
BIOL09020
6.
ANSWER SCHEME
May 2012
a) Human genomic DNA is tightly packaged as chromatin in
the nucleus of the cell. Discuss the evidence and
proposals on how gene expression can be influenced by
the structure of chromatin.
(10 marks)
Answer
For genes to be expressed they have to be accessible to the
binding of proteins such as enhancers or transcription
initiation complex. Chromatin structure may influence
expression by the degree of packaging and the position of
the nucleosome. Experiments treating chromatin with
DNAse I indicated they were sensitive areas around genes
being expressed suggesting these areas were more open
allowing access to the DNase I. These are called functional
domains and their boundaries are marked by insulators.
Insulators are sequences 1-2 kb in length that are thought
to modify chromatin packaging and prevent genes within a
domain being influenced by regulatory molecules in an
adjacent domain. Experiments where genes were inserted
into DNA either with or without insulators showed that in
the presence of insulators gene expression was always
close to 100% compared to genes inserted without
insulators where expression was variable (diagrams would
help illustrate some of these ideas). This suggests that
insulators modify the degree of packaging allowing the
genes to be more accessible to proteins influencing
transcription.
The position of the nucleosome also influences gene
expression. Experiments have shown that acetylation of
lysine amino acids on the nucleosome reduce its affinity for
DNA. Enzymes that catalyse this reaction have been found
in cells. Three models have been proposed for movement of
nucleosomes – remodelling, sliding and transfer.
Page 9 of 10
Continued overleaf
BIOL09020
ANSWER SCHEME
May 2012
b) Discuss the events that occur at the level of transcription
in E.coli in response to the levels of tryptophan.
Answer
The tryptophan operon codes for a set of Five genes (trpA,
trpB, trpC, trpD, and trpE) involved in the synthesis of the
amino acid tryptophan. Another gene (trpR) produces an
inactive repressor protein. Accumulation of the end product
(tryptophan) represses synthesis of the enzymes.
Tryptophan binds to the inactive repressor protein at an
allosteric site. Binding of tryptophan changes the
conformation and the repressor + tryptophan complex
binds to the operator, repressing the operon. Tryptophan
can accumulate due to internal production or from external
sources.
END OF QUESTION PAPER
Page 10 of 10
(10 marks)