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Mutations and the code
Frameshift mutations
A single base-pair deletion or insertion results in a change
in the reading frame
AUG UUU AGC UUU AGC UUU AGC
WT
Met Phe Ser Phe Ser Phe Ser
Delete C
AUG UUU AGU UUA GCU UUA GC
Met Phe Ser Leu Ala Leu
Insert C
AUG UUU AGC CUU UAG CUU UAG C
Met Phe Ser Leu STOP
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Frameshift mutations- Deletion
A single base-pair deletion or insertion results in a change
in the reading frame
AUG UUU AGC UUU AGC UUU AGC
Met Phe Ser Phe Ser Phe Ser
Delete C
Delete GC
Delete AGC
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Frameshift mutations-Insertion
A single base-pair deletion or insertion results in a change
in the reading frame
AUG UUU AGC UUU AGC UUU AGC
Met Phe Ser Phe Ser Phe Ser
Insert C
Insert CC
Insert CCC
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Missense mutations
Missense mutations alters ONE codon so that it encodes
a different amino acid
UUU UUU UGC UUU UUU
WT
UUU UUU UGG UUU UUU
mut
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Consequences of Missense
Mutations
Missense mutations alter one of the many amino acids
that make a protein
Its consequences depend on which amino acid is altered
Conservative mutations:
K to R
Nonconservative mutations:
K to E
Surface Vs buried
Mutations in globular domains Vs un structured tails
Silent mutations
Mutations in non-coding regions
Nonsense mutations
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Silent
Mutations
Silent mutations do not alter the amino acid sequence!
AUG UUU AGC UUU AGC UUU AGC
WT
AUG UUC AGC UUU AGC UUU AGC
Mut
Mutations that occur in introns are also silent
Mutations that occur in non-genic regions are silent
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Mutations in non-protein coding regions
Mutations in the promoter or ribosome binding site are
also mutagenic
Reduced expression of mRNA might result in reduced levels
of proteins
Mutations in splicing junctions may also be mutagenic
Improperly spliced mRNA will result in the intron being
Translated
Mutations in tRNA or aminoacyl-tRNA synthase are mutagenic
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Nonsense mutations
Nonsense mutations alter one codon so that it now encodes
for a STOP codon
UUU UUU UGC UUU UUU
Phe Phe Cys Phe Phe
UUU UUU UGA UUU UUU
Phe Phe STOP
Nonsense mutations insert a stop codon which results in
premature termination
Truncated polypeptide usually results in loss of function
for polypeptide
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Nonsense suppressor mutations!
These are the result of a mutation in the anti-codon loop of
a specific tRNA
It allows the tRNA to recognize a nonsense codon and base
pair with it.
DNA
Gene encoding tRNATRP
Point mutation occurs in the anticodon loop
This allows this tRNA to base pair with a stop codon and ?
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Nonsense suppressor
--- UUU UUU UAG UUU UUU ------- Phe Phe STOP
Trp-tRNA has mutation
In anticodon
This allows it to pair
with a stop codon
MetAla
Phe
Phe
Trp
AAA AUC
5’--- UUU UUU UAG UUU UUU -----3’
--- Phe Phe Trp Phe Phe ---->
A mutant protein that is larger than normal will be synthesized!!
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Nonsense and Nonsense suppressor
--- UUU UUU CAG UUU UUU ------- Phe Phe Gln Phe Phe --Nonsense mutation
--- UUU UUU UAG UUU UUU ------- Phe Phe STOP
Trp
What will happen if an
individual carries both a
nonsense mutation in a gene
and a nonsense suppressor
mutation in the anticodon
loop of one of the trptRNA genes.
AUC
---UAG---
MetAla
Phe
Phe
Trp
Phe
Phe
AAA AUC AAA AAA
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5’--- UUU UUU UAG UUU UUU -----3’
Generation of mutations
Spontaneous mutations
Replication induced mutations of DNA
Usually base substitutions
Most spontaneous errors are corrected
Mutations during meiotic pairing
Small additions and deletions
Environment induced changes
Exposure to physical mutagens - radioactivity or chemicals
Depurination (removal of A or G)
Repair results in random substitution during replication
Deamination (removal of amino group of base) (nitrous acid)
Cytosine--uracil--bp adenine--replication-Oxidation (oxoG)
guanine--oxoguanine--bp adenine--replication -Base analog incorporation during replication BU-T
Intercalating agents
X-rays-
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Methods used to study mutations
Gross chromosomal changesdeletions, insertions, inversions, translocations
Cytology- microscopy- karyotype
Point mutations
Small deletions, insertions
Recombinant DNA technologies
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Recombinant DNA
technology
When genes are mutated - proteins are mutatedDISEASE STATES OCCUR
Sickle cell Anemia
Globin
2 alpha globin chains
2 beta globin chains
Mol wt 16100 daltons xfour = 64650 daltons
Single point mutation in beta-globin
Converts Glu to Val at position6
Need to know mutation
Need to look at genes of individuals
Genes lie buried in 6billion base pairs of DNA
(46 chromosomes).
Molecular analyses necessary
Take advantage of enzymes and reactions that naturally
occur in bacteria
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Why all the Hoopla?
Why all the excitement over recombinant DNA?
It provides a set of techniques that allows us to study
biological processes at the level of individual proteins
in individuals!
It plays an essential role in understanding the genetic basis
of cancer in humans
Recently found that mutations in a single gene called p53
are the most common Genetic lesion in cancers.
More than 50% of cancers contain a mutation in p53
Cells with mutant p53
Chromosomes fragment
Abnormal number of chromosomes
Abnormal cell proliferation!
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p53
To understand the complete biological role of p53 protein
and its mutant phenotype we need to study the gene at
multiple levels:
Genetics- mutant gene- mutant phenotype
Now what?
Genetics will relate specific mutation to specific phenotype
It usually provides No Information about how the protein
generates the phenotype
For p53
We would like to know
The nucleotide sequence of the gene and the mutation that
leads to cancer
When and in which cells the gene is normally expressed
(in which cells is it transcribed)
At the protein level--Amino acid sequence
Three-dimensional structure
Interactions with other proteins
Cellular information
Is the location in the cell affected
How does it influence the behavior of the cell during division
Organism phenotype
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Alkaptonuria
Degenerative disease. Darkening of connective tissue, arthritis
Darkening of urine
1902
Garrod characterized the disorderusing Mendels rules- Autosomal recessive.
Affected individuals had normal parents and normal offspring.
1908
Garrod termed the defect- inborn error of metabolism
Homogentisic acid is secreted in urine of these patients.
This is an aromatic compound and so Garrod suggested that it
was an intermediate that was accumulating in mutant individuals
and was caused by lack of enzyme that splits aromatic rings of
amino Acids.
1958
La Du showed that accumulation of homogentistic acid
is due to absence of enzyme in liver extracts
1994
Seidman mapped gene to chromosome 3 in human
1996
Gene cloned and mutant identified P230S &V300G
2000
Enzyme principally expressed in liver and kidneys
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Basic techniques
--- Nucleic acid hybridization
complementary strands will associate and form double
stranded molecules
--- Restriction Enzymes
These enzymes recognize and cleave DNA at specific
sequences
--- Blotting
Allows analysis of a single sequence in a mixture
--- DNA cloning
This allows the isolation and generation of a large number
of copies of a given DNA sequence
--- DNA sequencing
Determining the array of nucleotides in a DNA molecule
--- PCR
--- Transformation
Stably integrating a piece of DNA into the genome of an
organism
--- Genetic engineering
Altering the DNA sequence of a given piece of DNA
--- Genomics
Analyzing changes in an entire genome
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Nucleic acid
hybridization
Complementary strands of DNA or RNA will specifically associate
DNA is heated to 100C, the hydrogen bonds linking the two
strands are broken
The double helix dissociates into single strands.
As the solution is allowed to cool, strands with complementary
sequences readily re-form double helixes.
This is called Nucleic acid hybridization.
AAAAAAAATTTTAAAAAAA
Will associate with
TTTTTTTTAAAATTTTTTT
This occurs with complementary
DNA/DNA, DNA/RNA, RNA/RNA
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Li-Fraumeni syndrome
This technique is very sensitive and specific.
A single 200 nucleotide sequence when added to a solution
of a million sequences will specifically hybridize with the
ONE complementary sequence
Usefulness
Li-Fraumeni syndrome
Individuals in a family have a propensity to develop tumors
at an early age
Often these families have a deletion in the p53 gene
When this family has a child, they might want to know
if their child has normal p53 or not
Nucleic acid hybridization provides a means to rapidly
determine whether the sequence is present or not
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Restriction
Enzymes
Enzymes which cut DNA at specific sequences
SmaI
Analysis revealed that the enzyme recognized and cut the
following sequence
|
5’ CCCGGG3’
3’ GGGCCC5’
|
This sequence is symmetrical. If one rotates it about the axis
It reads the same
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Linear/Circular DNA
A linear DNA molecule with ONE HindII site will be cut into
two fragments
A circular DNA molecule with ONE HindII site will generate
one DNA fragment
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Restriction sites
SmaI
5’ CCCGGG3’
3’ GGGCCC5’
5’CCC3’
3’GGG5’
5’GGG3’
3’CCC5’
EcoRI is another commonly used restriction enzyme
5’GAATTC3’
3’CTTAAG5’
5’G3’
3’CTTAA5’
5’AATTC3’
3’G5’
Unlike SmaI which produces a blunt end,
EcoRI produces sticky or cohesive ends
These cohesive ends facilitate formation of recombinant
DNA molecules
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Restriction
maps
Restriction maps are descriptions of the number, type and
distances between Restriction sites on a piece of DNA.
Very useful for molecular biologists.
Restriction sites serve as landmarks in the DNA with which a
physical map of a specific DNA sequence can be created.
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Sequence Divergence
The restriction map is also a reflection of the nucleotide sequence
arrangement of a gene
By comparing maps we can surmise differences in the sequence
between species
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Deletions and
additions
Normal Globin gene
HindIII
EcoRI
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Globin gene from a patient
4
HindIII
3
EcoRI
5
HindIII
EcoRI
EcoRI
3
8
HindIII
5
EcoRI
EcoRI
3
With restriction maps, the relationship between genes can be
determined without having to actually sequence the genes.
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Gel electrophoresis
Agarose gel electrophoresis
The length of the DNA can
be accurately determined by
allowing the charged DNA to
run through an agarose gel.
DNA moves towards the
Positive electrode.
The rate of migration of a
DNA fragment is inversely
proportional to its size.
Larger the size, slower its
movement.
2
HindIII
EcoRI
5
EcoRI
3
HindIII
EcoRI
HindIII
EcoRI
1
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Mapping
Marker
EcoRI
HindIII
You are given a 20 kb fragment of DNA
After trying many enzymes you find
That EcoRI and HindIII cut the fragment
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HindIII 14kb and 6kb
EcoRI 12kb 6kb and 2kb
Solve the map
12
6
4
2
1
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Mapping
Since HindIII cut the 20kb fragment once, in which of the
three EcoRI fragments. Does it cut?
Marker
EcoRI
HindIII
EcoRI+HindIII
A double digest with both enzymes will provide the answer
Fragments of 8kb, 6kb, 4kb and 2kb
The double digest does not alter the
size of the 6kb and 2kb fragments
The 12kb fragment is lost. Also 8+4=12
14
12
6
4
8
4
2
1
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Mapping
How are these fragments ordered?
Marker
EcoRI
HindIII
EcoRI+ HindIII
The HindIII single digest tells us that they must be ordered so
that One side adds up to 6kb and the other side adds up to 14kb
14
12
6
4
2
1
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Mapping
HindIII
14
6
EcoRI
12
6
2
HindIII/EcoRI
8
6
4
2
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Mapping example
Hi
12
8
Ec
12
6
2
Hi/Ec
8
6
4
2
Ps
13
7
Ps/Ec
12
5
2
1
Three different enzymes
Hi
Ec
Ps
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Mapping
HindIII
12
8
EcoRI
12
6
2
HindIII/EcoRI
8
6
4
2
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Mapping
EcoRI
12
6
2
PstI
13
7
PstI/EcoRI
12
5
2
1
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4
8
Mapping
deletions
Say you isolated this DNA from a region coding for the globin
gene, from a normal Patient and one suffering from thalassemia.
The fragment was 17kb rather than 20kb in the patient with
Thalassemia!
The restriction patterns were as following:
HindIII
14
3
EcoRI
9
6
2
Double
8
6
2
1
With similar reasoning as described above, the following
map is produced:
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Mapping
Marker
2kb
2kb+HindIII
Marker
6kb
6kb+HindIII
Marker
EcoRI
HindIII
EcoRI+
HindIII
Marker
12kb
12kb+HindIII
Often maps are more complex and difficult to analyze using
single and double digests alone.
To simplify the analyses, you can isolate each EcoRI band
From the gel and then digest with HindIII
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14
14
14
12
12
12
12
6
6
6
6
4
4
4
4
2
2
2
2
1
1
1
1
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Recombinant DNA
A reasonable question is how did we get the 20kb fragment in
the first place?
Also how did we obtain the p53 probe
To understand the origin of the fragment we must address
the issue of:
The construction of Recombinant DNA molecules
Recombinant DNA is generated through cutting and pasting of
DNA to produce novel sequence arrangements
Restriction enzymes such as EcoRI produce staggered cuts
leaving short single-stranded tails at the ends of the fragment.
These “cohesive or sticky” ends allow joining of different DNA
fragments
When a piece of DNA is cut with EcoRI,
you get
|
GAATTC
CTTAAG
|
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AATT--------------------- AATT-----------------------------------------TTAA ---------------------TTAA
Plasmids
Plasmids are naturally occurring circular pieces of DNA in E. coli
The plasmid DNA is circular and usually has one EcoRI site.
It is cut with EcoRI to give a linear plasmid DNA molecule
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Ligation
PLASMID
GENOMIC DNA
The EcoRI linearized plasmid DNA is mixed with human EcoRI
digested DNA
The sticky ends hybridize and anneal and a recombinant
plasmid is generated
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Plasmid
propagation
The plasmid DNA can replicate in bacteria
and therefore many copies of the
plasmid will be made. The human DNA
fragment in the plasmid will also multiply
along with the plasmid DNA.
Normally a gene is present as 2 copies
in a cell. If the gene is 3000bp long
there are 6x103 bp in a total of 6x109 bp
of the human genome
Once cloned into a plasmid, unlimited copies
of a single gene can be produced.The
process of amplifying and isolating the
human DNA fragment is called cloning.
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