Nucleic Acids
MCD Year 1
Anil Chopra
Contents
Nucleic Acids 1 – Nucleic Acids and Chromosomes ........................................................................ 1
Nucleic Acids 2 – DNA Replication, the Cell Cycle and Mitosis ........................................................ 6
Nucleic Acids 3 - Gene Organisation and Transcription 1 ............................................................... 9
Nucleic Acids 4 – Gene Organisation and Transcription II............................................................. 13
Nucleic Acids 5 - Protein translation and translational modification ............................................ 15
Nucleic Acids 6 – Analysis of DNA ............................................................................................... 19
Nucleic Acids 1 – Nucleic Acids and Chromosomes
Alex Perkins
1. Draw the structure of a nucleotide labelling the sugar, base and phosphates and
explain the difference between a nucleotide and a nucleoside.
Nucleic acids
 DNA and RNA are nucleic acids – molecules of heredity.
 They are macromolecules made up of a large number of nucleotides.
 A nucleotide is composed of a base, a sugar, and a phosphate group.
 A nucleoside is composed of a base and a sugar (no phosphate).
 The sugar in DNA is deoxyribose, the sugar in RNA is ribose.

Nucleotide: deoxyadenosine 5’-triphosphate (dATP)
NH
phosphate
2
N
N
base
H
H
O
_
O
O
P
O
O
O
P
_
O
O
_
H
O C 5ÕH
P
O
N
N
_
O
H 3Õ
HO

sugar – deoxyribose in DNA
1Õ
4Õ
2Õ
H
- ribose in RNA
H
A nucleoside is a nucleotide without the phosphate
2. List the bases found in DNA and RNA and indicate which ones are purines and which
ones are pyrimidines.
DNA and RNA bases:

Pyrimidines:
O
NH2
H
H
N
O
N
N
Cytosine
(C)
H
O
N
Thymine
(T)
DNA only
O
CH3
H
H
O
H
N
N
Uracil
(U)
RNA only
H

Purines:
NH2
O
N
N
N
H
H

H
N
N
H2 N
N
N
Adenine
Guanine
(A)
(G)
The nucleosides: (deoxy)cytidine, (deoxy)thymidine, (deoxy)uridine, (deoxy)adenosine,
(deoxy)guanosine
The nucleotides: deoxyadenosine 5’-triphosphate (dATP); adenosine monophosphate
(AMP) etc.

.
3. Describe a single DNA chain and explain the difference between the 5’ and 3’ ends.
Draw the structure of the double-stranded helix of DNA (not atomic structure)
showing base-pairing, the major and minor grooves, and the directionality of the
chains.
A single DNA chain
5' end
The DNA double helix
NH2
N
N
O
O
N
P OO
5'
N
NH2
O
3'
minor groove
N
H
O
N
O P OO
5'
O
O
O
3'
N
O
O
N
P OO
NH
major groove
H
5'
N
NH2
O
3'
OH
H
3' end





DNA is a long chain of deoxyribose units linked by phosphodiester links.
The phosphate on the 5’ carbon is linked to the –OH on the 3’ carbon along the chain.
On each deoxyribose there is a base.
The chain has two ends. The 5’ end and the 3’ end. It is not symmetrical.
The primary sequence is the linear sequence of the bases. By convention, the nucleotide
sequence is specified in the 5’ to 3’ direction.

The secondary structure of DNA is a right-handed double helix. The two chains in the
helix run in opposite directions.
The deoxyribose and phosphate groups run along the outside of the helix, with the
negative charges outside.
The bases point inwards and the flat planes are perpendicular to the helix.
The two chains are held together by hydrogen bonds between the bases.
The two strands are complementary in their sequence due to the specificity of basepairing. Adenine always pairs with Thymine; Guanine always pairs with Cytosine.




4. Describe melting and re-annealing of complementary strands and what is meant by
Watson-Crick base-pairing.
Watson-Crick base pairs
H
N H
N
O
O
N
CH3
N
N
N
H N
N
N
N H
O
N
N
Thymine
(T)
2 hydrogen bonds
 less stable
N H
N
O
Adenine
(A)
H
H N
H
Guanine
(G)
Cytosine
(C)
3 hydrogen bonds
 more stable
Melting and re-annealing
 High temperature and/or low salt concentration causes the two strands to melt or
disassociate.
 If you then lower the temperature or increase the salt concentration, the two melted
strands will re-anneal into a double helix.
 Hybridisation: in a mixture of DNA with different sequences, the complementary strands
will find each other in the mixture.
5. Compare the genomes of E.coli and Homo Sapiens.
The E.coli genome
 E.coli has 4.7 x 106 base pairs in a single circular double-stranded molecule.
 The length of the E.coli DNA is 1.4 mm.
 The DNA in E.coli is tightly packaged –
The human genome
 The human genome (diploid) consists of about 6 x 109 base pairs of DNA.
 The DNA is divided into chromosomes that each contain a linear double-helical DNA
molecule of about 200 x 106 base pairs.
 Prior to cell division, the DNA condenses into discrete chromosomes, visible by
microscopy.
 A diploid cell has 46 chromosomes; 22 pairs of ‘normal’ chromosomes and 2 sex
chromosomes.
6. Draw a diagram illustrating the packaging of DNA into nucleosomes and relate this to
chromosome structure.
Packaging of eukaryotic DNA



The DNA in a diploid human cell is nearly 2 m long. To fit into cells, the DNA is tightly
packaged into chromatin.
Chromatin consists of DNA and proteins.
The lowest level of packaging is the nucleosome, which consists of DNA wrapped around
histone proteins.
A nucleosome
8 histones: 2 each of 2A, 2B, 3 and 4
About 200 base pairs of DNA wrapped around
the histones (core DNA)
Histone 1 between the
nucleosomes

Linker DNA
The nucleosomes form a chain, which pack into a helical array.
Packaging of DNA into a chromosome

As the DNA has already replicated, there are two identical copies and two identical
chromatids (sister chromatids) for each chromosome at metaphase.
Short region of DNA double helix
Beads on a string form of chromatin
30 nm chromatin fiber of packed
nucleosomes
Section of chromosome in extended
form
Condensed section of metaphase
chromosome
Entire metaphase chromosome
consisting of two sister chromatids
Net result: each DNA molecule
has been packaged into a mitotic
chromosome that is 10 000 fold
shorter than its extended length
7. Describe the human karyotype.


A karyotype is an organised profile of someone’s chromosomes.
A diploid human cell has 46 chromosomes
- 22 pairs of ‘normal’ chromosomes (autosomes)
- 2 sex chromosomes (X and Y)
- Sex chromosomes: XX for female; XY for male
G-banded karyotype of a Normal Male Cell

Two homologues of each chromosome
Nucleic Acids 2 – DNA Replication, the Cell Cycle and
Mitosis
Anil Chopra
1. Explain semi-conservative replication.

New copies of DNA made should be identical.

Each daughter cell inherits one older strand and one new strand.

Each strand serves as a template.
2. Describe the reaction catalysed by DNA polymerases.
DNA Helicase: uses ATP to break the hydrogen bonds between the base pairs to
“unzip” the DNA.
DNA Polymerase: synthesises new DNA by adding nucleotides to the 3’
(prime) end of the growing chain.
3. Describe how nucleoside analogs can be used as drugs.
Some nucleoside analogs can be used as drugs because they
terminate chains. E.g.
Acyclovir  used in treatment for Herpes
Cytosine arabinose used in chemotherapy
Dideoxycytosine (ddC) HIV drug
Azidothymidine (AZT)  HIV drug.
4. Describe the functions of the components of the replication complex in prokaryotes including
the terms template, primer, leading strand, lagging strand, Okasaki fragment and replication
fork.

DNA polymerases need a template and a primer.

The replication process begins at discrete points and occurs bi-directionally. IT
ALWAYS OCCURS FROM 5’ TO 3’.

On the leading strand continuous synthesis

On the lagging strand  discontinuous synthesis
1. DNA helicase unwinds helix.
2. When about 1000 bases are exposed, a short RNA primer of about 5-10 bases is
synthesised by primase (an RNA polymerase).
3. DNA polymerase III continually adds deoxynucleotide
phosphates onto the end of the primer to form the chain
extension. This requires ATP.
4. On the lagging strand there are many RNA primers that are
added to various points on the template strand. Theses are
acted on by DNA polymerase III to produce Okazaki
fragments. This also requires ATP. The lagging strand
loops around so that the DNA polymerase III can work on
both strands at the same time.
5. When the Okazaki fragment reaches the RNA primer in
front of it, DNA polymerase I removes the RNA
primer using a 5’ to 3’ exonuclease. The DNA
polymerase I synthesizes the new DNA through the
RNA primer region.
6. Binding proteins bind to the lagging strand in order to prevent local secondary structures
forming.
7. The Okazaki fragments are then joined together by DNA ligase.
5. D
e
s
c
r
i
be how accuracy is
maintained by proofreading and the use
of RNA
primers.
The high
fidelity of DNA replication requires a proofreading mechanism to ensure no mistakes are made.
Mutations (changes in DNA sequence) are very dangerous to the organism. Any errors in replication
cannot be repaired.


DNA replication has an error frequency of about 1 change per 109 base pairs.
Before a new nucleotide is added, the previous nucleotide is checked for correct
base-pairing.

In errors with DNA synthesis, the nucleotide with the incorrect
base will not properly fit and bond with its corresponding base pair. The
phosphodiester bond (between then nucleotides) is then hydrolysed by the
exonuclease activity of DNA polymerase III and the correct
base is added.

In errors with RNA primers, the RNA primer is
simply removed and replaced with accurate DNA synthesis.
6. Draw a diagram showing replication of the E.coli chromosome.
The E.coli chromosome is circular. There is a single unique point of
origin (Ori.C) from where the two replication forks occur
simultaneously in opposite directions. The two forks meet at the
other side of the chromosome. Replication of the E.coli chromosome uses different enzymes.
7. Describe the replication of mammalian
chromosomes.
Mammalian chromosomes are long and linear. Replication
originates at various different points along the DNA at
intervals of around 100kilobase pairs and proceeds bidirectionally. It finishes when all the forks have met.
There are at least 5 different DNA polymerases:
Pol a synthesises the primers for both leading and lagging strands
Pol b is equivalent to Pol I in E.coli
Pol d synthesises the leading strand
Pol e synthesises the lagging strand
Pol g replicates mitochondrial DNA
8. Describe the different phases of the cell cycle.
M phase: Mitosis; cell division; 1hr. This is where the 2
chromatids separate.
G1 phase: Gap phase 1 (prior to DNA syntesis); 10 hrs. Each
chromosome is still a one double helix.
S phase: Synthesis of DNA (replication); 9hrs
G2 phase: Gap phase 2 (between DNA synthesis and mitosis); 4 hrs. Each
chromosome is 2 identical chromatids.
G0: cells which have stopped dividing
9. Draw a diagram
showing how the
chromosomes
segregate at
metaphase.
Nucleic Acids 3 - Gene Organisation and Transcription 1
Anil Chopra
1.
Describe the basic differences between DNA and RNA.
 DNA is hereditary and is split up into functional units called genes.
 Genes code for production of functional RNA’s and proteins which give cells their particular
characteristics e.g.
o red blood cells produce haemoglobin
o B-cell lymphocytes produce antibodies.
 The DNA sequence of an organism is its genome.
 Human genome contains between 25 000 and 35 000 genes and this is contained in EVERY CELL.
 46 chromosomes, (22 pairs of autosomal chromosomes and one pair of sex chromosomes X & Y).
 When genes are expressed, they are used in that particular cell. Some ‘housekeeping genes’ are
expressed in all cells. 25% of genes are required for specific cell function.
 When genes are expressed they produce, RNA (ribonucleic acid). This is the same as DNA except
for that:
o It is single stranded.
o The base thymine is replaced with Uralic.
o The pentose sugar is ribose. (not deoxyribose)
o There are 3 different types of it: mRNA, tRNA and rRNA.
2.
Describe what is meant by 'transcription'.
 In the process of transcription, RNA is produced as a template for protein translation.
 This always occurs in the nucleus.
3.
List the major functional classes of RNA and the classes of RNA polymerase involved in
synthesising each of these.
 Gene transcription is carried out by enzymes called “RNA Polymerases”. There are 3 types
o
RNA Polymerase I -Transcribes rRNA genes.
o
RNA Polymerase II - Transcribes genes encoding proteins into mRNA.
o
RNA Polymerase III- Transcribes tRNA and 5S RNA genes.
4.
Describe what is meant by a "gene promoter".
A gene promoter is a DNA sequence at which the transcription initiation complex assembles. This
complex is needed for the elongation reaction of RNA synthesis to occur.
Describe what is meant by a "Transcription factor".
5.
A transcription factor regulates the level of transcription of a given gene. It is also known as a DNA
binding protein. There are 2 types:
 Transcriptional activators – activate gene expression.
 Transcriptional repressors – suppress gene expression.
These act together to bring about changes in expression of genes. External signals such as
temperature, hormones, growth factors, voltage e.t.c. all have an effect on transcription factors.
Mutations in transcription factors cause hereditary disorders, and abnormalities in transcription
factor expression are found in many cancers.
Describe with the aid of diagrams the processes involved in transcribing a eukaryotic gene.
6.
The TATA sequence in DNA is the point at which transcription starts. This is done by RNA polymerase
II.
1.
2.
3.
4.
5.
TF IID (containing a TATA binding protein) which partially unwinds the helix
TF IIA and TF IIB bind to the TF IID.
RNA polymerase II binds to TF IIB
TF IIF is already bound to RNA polymerase and is joined by TF IIE, TF IIH, and TF IIJ
This further unwinds the helix to facilitate the transcription by RNA polymerase II.
Transcription factors increase the level of transcription on binding by “bending” the DNA and
modifying chromatin by acylation of histones. They can be used in therapeutic treatment: e.g.
 Aspirin: Aspirin stops the breakdown of a transcription factor which increases the production of
cytokines. This reduces inflammation.
 Half of all lymphoblastic leukaemias have mutated transcription factors.
 Breast cancer: breast cancer cells over express the transcription factor that is the oestrogen
receptor.
Nucleic Acids 4 – Gene Organisation and Transcription II
Anil Chopra
1.
2.
3.
4.
5.
6.
Describe, with the aid of diagrams, the events that take place in pre- mRNA processing.
Define what is meant by a “splice donor site”
Define what is meant by a “splice acceptor site”
Describe the “lariat” intermediate in mRNA splicing
Define the function of the “Spliceosome”
Describe the addition of a “cap” and “poly A tail” to pre-messenger (hn-) RNA.
 When RNA is produced initially from the gene is still not ready to be used in translation yet. It is
known as Pre-mRNA, primary transcript or heterogenous nuclear hn RNA.
 It needs to be processed first. This occurs in the nucleus.
 DNA is formed from
o
Exons – parts of the DNA that will be transcribed in both the pre-mRNA and the final
mRNA.
o
Introns – parts of the DNA that will be transcribed in the pre-mRNA but then edited out
in the final mRNA.
This occurs in a series of steps:
1) The introns that are to be edited out will start with the sequence GU (this is the splice donor
site) and end with the sequence AG (this is the splice acceptor site).
2) Processing requires small ribonuclear proteins. U1 binds to the donor site.
3) U
2
,
U4 and U6 all bind to the intron itself, and U5 binds to the splice acceptor site.
4) Splice donor site sequence is cleaved off and the “G” base from the UG sequence at the
splice donor site curves round to form a phosphodiester bond between itself and an “A”
residue in the intron.
5) The phosphodiester bond at the “G” end of the intron at the spice acceptor site is cleaved
forming a lariat structure.
6) A ligase enzyme attaches all the exons together.
7) A CAP is attached to the end of the mRNA.
8) A poly A tail is added to the end of the mRNA one base at a time. It is added downstream of
the sequence AAUAAA.
7.
With examples, describe how mutations in splice sites feature in human disease.
33% of mutations in inherited diseases occur at splice donor/acceptor sites.
Thalassaemia: this is a disorder in which there is an imbalance in the relative amounts of globulin chains
in red blood cells. -thalassaemia is when there is a deficiency of  chains.
This results in sever anaemia, which then results in increase in iron uptake causing hepatomegaly
(enlargement of the liver), darkening of skin (due to iron-stimulated melanin production),
cardiomyopathy.
Nucleic Acids 5 - Protein translation and translational
modification
Anil Chopra
1. Outline the mechanisms by which ribosomes can translate an mRNA sequence into a
protein sequence.
 From the mRNA produced in transcription and processing (so only the exons are in the
final mRNA), every 3 base pairs codes for 1 amino acid.
 This is known as a codon.
 There are 64 codons that code for 20 different amino acids, therefore some codons code
for the same amino acid, some are START CODONS (Met – AUG) and some are STOP
CODONS.
 Rare amino acids have few codons.
 mRNA consists of the 7MeG CAP at one end, and a poly A tail at the other. In between
these there is an untranslated region at both the 3’ and the 5’ end.
AAAAAn
7MeG
5’cap 5’UTR
coding region
3’UTR polyA
2. Describe the role of aminoacyl tRNAs in ensuring the fidelity of the genetic code.
The processed mRNA leaves the nucleus and attaches to
ribosomes where translation occurs. Here tRNA (transfer
RNA) molecules transport amino acids to the ribosome:
 There is only one tRNA molecule per amino acid and they are
covalently linked by aminoacyl tRNA synthetases using ATP.
 In order to ensure fidelity of the genetic code there are 20 types of
aminoacyl tRNA synthetases so that each amino
acid links with its specific tRNA molecule.
 At one end of the tRNA molecule, there is an
anticodon containing the complementary base
pairs which forms an antiparallel relationship with the mRNA.
3. State how a ribosome recognises the start and end of a
sequence to be translated.
Translation:
1) Ribosomes consist of 2 subunits, 40S and 60S. These
subunits dissociate/split.
2) Preinitiation complex is formed which contains Met-tRNA+eIFs+40S subunit. (Met
is always the first amino acid)
3) This then binds to the mRNA.
4) T
h
e
larger 60S subunit attaches to the mRNA in the same way. (with the tRNA and eIFs
attached)
5) GTP  GDP ensures correct base pairing.
6) The second tRNA molecule holding the next amino acid in the chin binds to the
mRNA molecule immediately adjacent to the previous one i.e. in frame.
7) The two amino acids are
joined together with a
peptide bond peptidyl
transferase on the 60S
subunit.
8) The first of tRNA
molecule dissociates
and the second
tRNA is now in the peptidyl (P) site.
9) Elongation Factors (EFs) use the energy of GTP to enhance the efficiency and
accuracy of translation by providing “pauses” (e.g. GTP hydrolysis) that allow
incorrect base pairs to dissociate.
10) This continues until it reaches the stop codon.
11) When this is recognised, release factors bind to the empty A site and the peptide
chain is released.
12) The ribosome then releases from the mRNA.
4. Explain why some antibiotics inhibit protein synthesis in prokaryotes but not
eukaryotes.
Antibiotics are used to inhibit the action of prokaryotes. They do this by interfering with
stages of their replication or protein synthesis as both of these are complex procedures
requiring many steps. E.g.:
• Streptomycin
Inhibits initiation
• Tetracycline
Inhibits tRNA binding
• Erythromycin
Inhibits translocation
• Chloramphenicol
Inhibits peptide transfer
• Puromycin
Terminates elongation
• Cycloheximide
Inhibits peptide transfer (in eukaryotes)
5. Identify the features of a newly-synthesised protein that are required for it to enter the
secretory pathway.
As cells contain many compartments, there needs a mechanism to transfer proteins across
membranes:
 The first 20-24 amino acids are a signal sequence.
 This is recognised and bound to by a signal recognition particle SRP.
 This SRP binds to receptors on the surface of the ER (or other compartment).
 The amino acid chain is then fed through to the RER lumen where the signal sequence is
cleaved of and the amino acid chain is folded.
6. Give examples of the ways in which newly-synthesised proteins can be posttranslationally modified.
After synthesis, the proteins undergo post-translational modification which
involves various things including:
• Proteolytic cleavage (e.g. insulin -> A and B chains)
• Disulphide bond formation (e.g. insulin)
• Addition of carbohydrate (Glycosylation)
• Addition of phosphate
(Phosphorylation)
• Addition of lipid groups
(Prenylation, Acylation)
Nucleic Acids 6 – Analysis of DNA
Anil Chopra
1.
2.
Explain the term hybridisation, used for binding of a probe to a nucleic acid.
Describe the reactions carried out by restriction enzymes (restriction endonucleases)
and explain their usefulness in analysis of DNA.
DNA Cloning: is a method of selectively amplifying DNA sequences of interest to generate
homogenous DNA populations. There are 2 types:
–
–
Cell-based DNA cloning (in vivo)
Cell-free DNA cloning (in vitro) - Polymerise Chain Reaction
Cell Based DNA Cloning
1)
The target DNA and a replicon (e.g. plasmid, yeast artificial chromosome) are
cut by restriction endonucleases.

Restriction endonucleases (type II) work by cleaving DNA at specific
sequences.

These are usually short and pallendromic (the same backward as they
are forward).

Can produce “blunt ends”
or
“sticky ends”.

Generally, the longer the sequence cut, the less frequently it appears in
the DNA.
2)
3)
4)
The DNA fragments are mixed and joined by a DNA ligase.
The newly formed recombinant DNA is introduced into bacteria/yeast cells.
The replicon contains an antibiotic resistance marker therefore only those with
the replicon to survive when treated with antibiotics.
5)
The cell culture is expanded and the recombinant DNA is isolated.
3. Describe how DNA fragments can be separated on the basis of size.
Separation of DNA fragments by Electrophoresis
As DNA is a negatively charged molecule, it will move towards the positive anode if it is in
an electrolyte mixture. Smaller fragments of DNA will move faster than larger ones and so it
is easy to isolate DNA which has been modified or added to. This process is called
electrophoresis. This can then go on to be hybridised.
Hybridisation
If one wants to find out where a specific DNA sequence is in a chromosome or on a gene, a
hybridisation assay is done:
1) A labelled nucleic acid probe is mixed in amongst unlabelled nucleotides (DNA, RNA or
oligonucleotides). Probes can be made in various ways
a.
DNA Probes can either be synthesised by:
i.
Nick translation , this is where a chunk of DNA is
removed (“nicked”) first by DNase and the rest by exonuclease.
Then a certain DNA polymerase synthesises that
radioactively label the new strand. This becomes
the probe.
Random primed labelling, this is where
the DNA is denatured and random
oligonucleotides are added. The DNA polymerase uses the
oligonucleotides as primers to synthesise the new radioactively
labelled probe.
b.
RNA probes are made by
using a radioactively labelled probe in DNA plasmids
and then adding RNA polymerase, which produced
radioactively labelled RNA probes.
ii.
c.
Oligonucleotides are labelled with a radioactive
ATP molecule which is added to the 5’ by an enzyme
called Polynucleotide kinase.
2) B
o
t
h
t
h
e
t
a
r
get DNA and the labelled DNA probes are denatured (either by heating or mixing with some
solution) and bound to a solid support e.g. nylon or nitrocellulose membrane. This readily binds
single-stranded nucleic acid (e.g. denatured DNA) and then hybridised with a solution of
(radioactively or fluorescently) labelled (*) probe.
3) They then hybridise in that some of the labelled primers mix in to the target DNA. These can be
recognised by photographic film and therefore isolated.
4. Explain the concept of stringency of hybridisation, and the factors that contribute to stringency.

The DNA is denatured by heating until all the hydrogen bonds holding
the two chains together are broken. This can depend on the length of the strand (longer chain=more
bonds), the nature of the bonds (G-C has one more bond than A-T), and environment (certain atoms
such as Na+ stabilise the bonds and certain molecules such as urea destabilise them).

The melting temperature is taken as the mid point of the transition
between double and single stranded DNA. (Around 87O C)

This is then allowed to cool down at around 25oC where hybridisation
occurs. (i.e. annealing of the DNA strands)
Hybridisation stringency (i.e. the power to distinguish between related sequences) increases with:
- Increase in temperature
- Decrease in Na+ concentration
5. Explain how the polymerase chain reaction (PCR) is used to amplify small amounts of DNA for
subsequent analysis.
Polymerase Chain Reaction
This is a process by which specific target DNA can be amplified (copied many times) in a collection of
DNA sequences.



2 primers are made – one for each strand that is to be copied.
The DNA is heated till it is denatured and so it is single stranded.
A special enzyme
known as Thermostable Thermophilus
aquaticus DNA polymerase along with dNTPs
extend 5’-3’ from the primers and generate new
strands.

Denature 94oC

Anneal 50-60oC

Extend 72oC

This cycle is repeated many
times.
6. Describe in general terms the way in which PCR primers would be selected
to amplify a given DNA sequence.

They are usually about 20 nucleotides
in length.

They should not involve too many repeated nucleotides as these form
hairpins.

The percentage of C-G bonds along with their length should give you a
rough estimate of their melting temperature.

Complementary base pairs at either 3’ ends should be avoided in order
to prevent primer dimers.
7. Describe the principles behind the standard method of obtaining the sequence of a piece of DNA.
1) Obtain the template strand you wish to sequence, and then add a DNA primer complementary to
the 3’ end.
2) With DNA polymerase, synthesise the complementary strand.
3) Then use this newly formed complementary strand to synthesise the original DNA using
fluorescent or radioactively marked nucleotides in 4 different reactions. In each of these reactions,
add a small amount of ddNTPs (dideoxyribonucleoside triphosphate).
 These come with 4 different bases attached to them and therefore will halt the reaction only
after that corresponding base – however many times that may come up.
 Add each one to one reaction.
 You will then, in each reaction, get DNA chains of varying lengths. e.g. in the “G” reaction (where
you added ddGTP), you may get…
o ddGATGCG
o ddGCGTAATGCG
o ddGCTGATCGCGTAATGCG
4) This halts the production of the fluorescently marked strands at different points.
5) Therefore the strands are all different lengths, which means that in electrophoresis, they will all
appear at different levels.
6) The mixtures of the four different reactions are simply put in 4 columns which can be read out
from top to bottom to give you the sequence – this can be done manually if the nucleotides
have been radioactively marked, or automatically if they have been fluorescently marked.