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 ?

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--- 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

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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

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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

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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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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

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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

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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

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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 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

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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

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

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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 axisIt 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

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.

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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

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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.

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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.

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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

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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 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

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Mapping example

Hi Ec Hi/Ec12 12 88 6 6

42 2

Ps Ps/Ec13 127 5

21

Three different enzymesHiEcPs

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MappingHindIII EcoRI HindIII/EcoRI12 12 88 6 6

42 2

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Mapping

EcoRI PstI PstI/EcoRI12 13 126 7 52 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 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:

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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

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

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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

|GAATTCCTTAAG |

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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

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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

AATT

AATT

TTAA

TTAA

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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 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.