Mutations and the code Frameshift mutations

1
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
2
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
3
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
4
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
5--- UUU UUU UAG UUU UUU -----3
--- Phe Phe Trp Phe Phe ----gt 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
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 trp-tRNA genes.
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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 changes- deletions,
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 mutated-
DISEASE 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 Interaction
s 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
• Garrod characterized the disorder-
• using 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

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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

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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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The method
Isolate DNA Patient
Isolate DNA normal individual
(The probe)
Fragment DNA, Heat to denature Add radiolabeled
ssDNA(p53 probe) Gradually and slowly cool
solution
Radiolabeled p53 probe associates with DNA in
normal individual
If patient is deficient for p53 gene
Radiolabeled p53 probe is unable to associates
with DNA in patient
Add enzyme (nuclease) that specifically
degrades ssDNA molecules. dsDNA remains degraded
No radiolabel present in dsDNA (because p53 probe
could not anneal)
Radiolabel present in dsDNA
Extremely sensitive p53 is 3000bp, human genome
is 3000,000,000 bp Technique detects one sequence
of 3000bp from 1 million sequences of 3000bp
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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
5CCC3 5GGG3 3GGG5 3CCC5
EcoRI is another commonly used restriction enzyme
5G3 5AATTC3 3CTTAA5 3G5
5GAATTC3 3CTTAAG5
Unlike SmaI which produces a blunt end, EcoRI
produces sticky or cohesive ends These cohesive
ends facilitate formation of recombinant DNA
molecules
25
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.
26
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
3
5
8
4
EcoRI
EcoRI
EcoRI
HindIII
HindIII
Globin gene from a patient
3
5
3
4
EcoRI
EcoRI
EcoRI
HindIII
HindIII
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.
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Mapping
You are given a 20 kb fragment of DNA After
trying many enzymes you find That EcoRI and
HindIII cut the fragment
HindIII 14kb and 6kb EcoRI 12kb 6kb and
2kb Solve the map
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Mapping
Since HindIII cut the 20kb fragment once, in
which of the three EcoRI fragments. Does it
cut? 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
8412
EcoRIHindIII
Marker
HindIII
EcoRI
14
12
8
6
4
4
2
1
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Mapping
How are these fragments ordered?
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
EcoRI HindIII
Marker
HindIII
EcoRI
14
12
6
4
2
1
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Mapping
HindIII EcoRI HindIII/EcoRI 14 12 8 6 6 6
4 2 2
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Mapping example
Hi Ec Hi/Ec 12 12 8 8 6 6 4 2 2
Ps Ps/Ec 13 12 7 5 2 1
Three different enzymes Hi Ec Ps
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Mapping
HindIII EcoRI HindIII/EcoRI 12 12 8 8 6 6
4 2 2
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Mapping
EcoRI PstI PstI/EcoRI 12 13 12 6 7 5 2 2
1
4
8
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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 EcoRI Double
• 14 9 8
• 3 6 6
• 2 2
• 1

With similar reasoning as described above, the
following map is produced
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Mapping
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
2kbHindIII
12kbHindIII
6kbHindIII
EcoRI HindIII
Marker
Marker
Marker
HindIII
Marker
12kb
2kb
EcoRI
6kb
14
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
GAATTC CTTAAG
When a piece of DNA is cut with EcoRI, you get
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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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Triticale
Early attempts to cross wheat and rye produced
only sterile offspring. It was not until new
techniques were developed that it was possible
to produce fertile hybrids The genome of wheat
and rye (these are from different species) were
combined to form a new species! These allowed
wheat and rye plants to overcome the species
barriers and form fertile offspring. Once
these fertile plants were obtained they were used
in Crosses with wheat and rye plants to develop
genetically novel plants with combinations of
traits from wheat and rye parents. These plants
were hailed as a superfood that would solve the
problem of world hunger!
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Cotton
A species of bacteria produces a potent natural
pesticide This toxin is suitable for organic
farming The gene necessary for producing the
toxin was isolated. It was introduced into the
genome of cotton plants The plant now produced
-toxin eliminating the need for pesticide. This
would reduce the harmful effects of pesticides on
humans!!!!
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Tomato
Ethylene gas released by fruit accelerates the
ripening process Prevention of ethylene
production would block the fruit from ripening
prematurely and spoiling on the way to the
market. The ethylene biosynthetic pathway is as
follows Precursor-----gtACC------gtethylene
ACC synthase
ACC oxidase
Mutagenesis of these enzymes allows slow
ripening!!!!