Nucleic AcidsMCD Year 1

Anil Chopra

ContentsNucleic 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 ChromosomesAlex 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)

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:

__

H

O

H

HO

H

C H

HO

1Õ

2Õ3Õ

4Õ

5Õ

2

N

N N

N

NH

H

H

PO

O

O

PO

O

O

PO

O

O

_

_

sugar – deoxyribose in DNA

- ribose in RNA

base phosphate

Cytosine(C)

Uracil(U)

RNA only

Thymine(T)

DNA only

N

N

O

O

H

H

HN

N

O

O

CH3

H

HN

N

NH2

O H

H

Purines:

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.

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.

Adenine

(A)

Guanine

(G)

N

N N

N

N H 2

H

H

N

N N

O

H2 N

H

minor groove

major groove

The DNA double helix

-

O

O

H

H

H

N

HN

N

N

O

N 2N

O

O

O

PO O

PO O

O

PO O

O

N

NN

N

NH2

O

H

O

N

NH2

O

5'

3'

3'

3'

5'

5'

OH

-

-

5' end

3' end

A single DNA chain

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 base-

pairing. 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.

Adenine(A)

Thymine(T)

Guanine(G)

Cytosine(C)

3 hydrogen bonds more stable

2 hydrogen bonds less stable

Watson-Crick base pairs

NN

N

O

HH

NN

N

N O

NH

H

HN

N

N

N N

NN

O

O

CH3

H

HH

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 bacterium is only 3 m long.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.

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.

Linker DNA

About 200 base pairs of DNA wrapped aroundthe histones (core DNA)

Histone 1 between thenucleosomes

8 histones: 2 each of 2A, 2B, 3 and 4

A nucleosome

Short region of DNA double helix

Beads on a string form of chromatin

30 nm chromatin fiber of packednucleosomes

Section of chromosome in extendedform

Condensed section of metaphasechromosome

Entire metaphase chromosomeconsisting of two sister chromatids

Net result: each DNA moleculehas been packaged into a mitoticchromosome that is 10 000 foldshorter 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 MitosisAnil 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. Describ

e how accuracy is maintained by proof-reading and the use of RNA primers.

The high fidelity of DNA replication requires a proof-reading 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 bi-directionally. 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 1Anil 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 typeso 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.

5. Describe what is meant by a "Transcription factor".

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.

6. Describe with the aid of diagrams the processes involved in transcribing a eukaryotic gene.

The TATA sequence in DNA is the point at which transcription starts. This is done by RNA polymerase II.

1. TF IID (containing a TATA binding protein) which partially unwinds the helix2. TF IIA and TF IIB bind to the TF IID.3. RNA polymerase II binds to TF IIB4. TF IIF is already bound to RNA polymerase and is joined by TF IIE, TF IIH, and TF IIJ5. 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 IIAnil Chopra

1. Describe, with the aid of diagrams, the events that take place in pre- mRNA processing. 2. Define what is meant by a “splice donor site”3. Define what is meant by a “splice acceptor site”4. Describe the “lariat” intermediate in mRNA splicing5. Define the function of the “Spliceosome”6. 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) U2, 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 modificationAnil 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.

5’cap 5’UTR coding region 3’UTR polyA

7MeG AAAAAn

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) The

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.1) When this is recognised, release factors bind to the empty A site and the peptide

chain is released.

11) 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 post-translationally 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 DNAAnil Chopra

1. Explain the term hybridisation, used for binding of a probe to a nucleic acid. 2. 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) The DNA fragments are mixed and joined by a DNA ligase.

3) The newly formed recombinant DNA is introduced into bacteria/yeast cells.

4) 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.

ii. 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.

c. Oligonucleotides are labelled with a radioactive ATP molecule which is added to the 5’ by an enzyme called Polynucleotide kinase.

2) Both

the target 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 ddGATGCGo ddGCGTAATGCGo 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.