UU UUU S,UUU Card) Idmuuuunffi1ãa(Password) … UUU S,UUU Card) Idmuuuunffi1ãa(Password) Zl
1 Mutations and the code Frameshift mutations A single base-pair deletion or insertion results in a...
-
date post
22-Dec-2015 -
Category
Documents
-
view
221 -
download
0
Transcript of 1 Mutations and the code Frameshift mutations A single base-pair deletion or insertion results in a...
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 AGCMet 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 AGCMet 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
5
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
6
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
7
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 mutagenicImproperly spliced mRNA will result in the intron being Translated
Mutations in tRNA or aminoacyl-tRNA synthase are mutagenic
8
Nonsense mutations
Nonsense mutations alter one codon so that it now encodes for a STOP codon
UUU UUU UGC UUU UUUPhe Phe Cys Phe Phe
UUU UUU UGA UUU UUUPhe Phe STOP
Nonsense mutations insert a stop codon which results in premature termination
Truncated polypeptide usually results in loss of function for polypeptide
9
Nonsense suppressor mutations!
These are the result of a mutation in the anti-codon loop of a specific tRNAIt allows the tRNA to recognize a nonsense codon and base pair with it.
DNA
Gene encoding tRNATRP
Point mutation occurs in the anticodon loopThis allows this tRNA to base pair with a stop codon and ?
10
--- UUU UUU UAG UUU UUU -------- Phe Phe STOP
5’--- UUU UUU UAG UUU UUU -----3’
AUC
Trp
AAA
MetAla
Phe
Phe
--- Phe Phe Trp Phe Phe ---->
A mutant protein that is larger than normal will be synthesized!!
Trp-tRNA has mutationIn anticodonThis allows it to pairwith a stop codon
Nonsense suppressor
11
Nonsense and Nonsense suppressor
--- UUU UUU CAG UUU UUU -------- Phe Phe Gln Phe Phe ---
--- UUU UUU UAG UUU UUU -------- Phe Phe STOP
Nonsense mutation
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.
AUC
Trp
---UAG---
5’--- UUU UUU UAG UUU UUU -----3’AUC
Trp
AAA
MetAla
Phe
Phe
AAA
Phe
AAA
Phe
12
Generation of mutations
Spontaneous mutations
Replication induced mutations of DNA
Usually base substitutionsMost spontaneous errors are corrected
Mutations during meiotic pairingSmall additions and deletions
Environment induced changesExposure 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-
13
Methods used to study mutations
Gross chromosomal changes- deletions, insertions, inversions, translocations
Cytology- microscopy- karyotype
Point mutationsSmall deletions, insertions
Recombinant DNA technologies
14
Recombinant DNA technology
When genes are mutated - proteins are mutated- DISEASE STATES OCCUR
Sickle cell Anemia
Globin2 alpha globin chains2 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 necessaryTake advantage of enzymes and reactions that naturally occur in bacteria
15
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!
16
p53To 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 phenotypeIt usually provides No Information about how the protein generates the phenotype
For p53We 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 informationIs the location in the cell affected
How does it influence the behavior of the cell during division
Organism phenotype
17
Alkaptonuria
Degenerative disease. Darkening of connective tissue, arthritisDarkening of urine
1902 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 metabolismHomogentisic 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 individualsand was caused by lack of enzyme that splits aromatic rings of amino Acids.
1958 La Du showed that accumulation of homogentistic acidis 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
18
Basic techniques
--- Nucleic acid hybridizationcomplementary strands will associate and form double stranded molecules
--- Restriction EnzymesThese enzymes recognize and cleave DNA at specific sequences
--- BlottingAllows analysis of a single sequence in a mixture
--- DNA cloningThis allows the isolation and generation of a large number of copies of a given DNA sequence
--- DNA sequencingDetermining the array of nucleotides in a DNA molecule
--- PCR
--- TransformationStably integrating a piece of DNA into the genome of an organism
--- Genetic engineeringAltering the DNA sequence of a given piece of DNA
--- GenomicsAnalyzing changes in an entire genome
19
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 brokenThe 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 complementaryDNA/DNA, DNA/RNA, RNA/RNA
20
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
UsefulnessLi-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
21
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 axisIt reads the same
22
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
23
Restriction sites
5’ CCCGGG3’3’ GGGCCC5’
5’CCC3’ 5’GGG3’3’GGG5’ 3’CCC5’
EcoRI is another commonly used restriction enzyme
5’GAATTC3’3’CTTAAG5’
5’G3’ 5’AATTC3’3’CTTAA5’ 3’G5’
Unlike SmaI which produces a blunt end, EcoRI produces sticky or cohesive endsThese cohesive ends facilitate formation of recombinant DNA molecules
SmaI
24
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.
25
Sequence Divergence
The restriction map is also a reflection of the nucleotide sequence arrangement of a geneBy comparing maps we can surmise differences in the sequence between species
26
Deletions and additions
EcoR
I
Hin
dII
I
EcoR
I
Hin
dII
I
EcoR
I
3 5 8 4
Normal Globin gene
Globin gene from a patient
EcoR
I
Hin
dII
I
EcoR
I
Hin
dII
I
EcoR
I
3 5 3 4
With restriction maps, the relationship between genes can be determined without having to actually sequence the genes.
27
Gel electrophoresis
EcoR
I
Hin
dII
I
EcoR
I
Hin
dII
I
EcoR
I
1 3 5 2
Hin
dII
I
EcoR
I
Agarose gel electrophoresisThe length of the DNA can be accurately determined byallowing the charged DNA torun through an agarose gel.
DNA moves towards the Positive electrode.
The rate of migration of aDNA fragment is inverselyproportional to its size.Larger the size, slower itsmovement.
28
Mapping
You are given a 20 kb fragment of DNAAfter trying many enzymes you findThat EcoRI and HindIII cut the fragment
HindIII 14kb and 6kb
EcoRI 12kb 6kb and 2kb
Solve the map
1
2
4
6
14
Mark
er
EcoR
IH
ind
III
12
29
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 fragmentsThe 12kb fragment is lost. Also 8+4=12
1
2
4
6
14
Mark
er
EcoR
IH
ind
III
12
EcoR
I+H
ind
III
8
4
30
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
1
2
4
6
14
Mark
er
EcoR
IH
ind
III
12
EcoR
I+
Hin
dII
I
31
Mapping
HindIII EcoRI HindIII/EcoRI14 12 86 6 6
42 2
32
Mapping example
Hi Ec Hi/Ec12 12 88 6 6
42 2
Ps Ps/Ec13 127 5
21
Three different enzymesHiEcPs
33
MappingHindIII EcoRI HindIII/EcoRI12 12 88 6 6
42 2
34
Mapping
EcoRI PstI PstI/EcoRI12 13 126 7 52 2
1
4 8
35
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 withThalassemia!
The restriction patterns were as following:
HindIII EcoRI Double14 9 83 6 6
2 21
With similar reasoning as described above, the following map is produced:
36
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
1
2
4
6
14
Mark
er
EcoR
IH
ind
III
12
EcoR
I+H
ind
III
1
2
4
6
14
Mark
er
12
1
2
4
6
14
12
1
2
4
6
14
12
Mark
er
Mark
er
12
kb
12
kb
+H
ind
III
6kb
6kb
+H
ind
III
2kb
2kb
+H
ind
III
37
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
|GAATTCCTTAAG |
38
Plasmids
AATT--------------------- ---------------------TTAA
AATT--------------------- ---------------------TTAA
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
39
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
AATT
AATT
TTAA
TTAA
40
Plasmid propagation
The plasmid DNA can replicate in bacteria and therefore many copies of the plasmid will be made. The human DNAfragment in the plasmid will also multiplyalong 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.