Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
DNA replication wikipedia , lookup
DNA profiling wikipedia , lookup
DNA repair protein XRCC4 wikipedia , lookup
DNA polymerase wikipedia , lookup
Zinc finger nuclease wikipedia , lookup
United Kingdom National DNA Database wikipedia , lookup
DNA nanotechnology wikipedia , lookup
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 1 Sequencing the whole genomes of a family (2010 Science 328 636). 98 crossovers in maternal genome 57 crossovers in paternal genome Mutation rate is 1x10-8 per position per haploid genome 70 new mutations in each diploid human genome CpG sites mutate at a rate 11 times higher than other sites 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 Phe Phe Cys Phe Phe WT UUU UUU UGG UUU UUU Phe Phe Trp Phe Phe 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 Met Phe Ser Phe Ser Phe Ser WT AUG UUC AGC UUU AGC UUU AGC Met Phe Ser Phe Ser Phe Ser Mut Mutations that occur in introns are also silent Mutations that occur in non-genic regions are often 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 OR Increased expression of mRNA might result in increased levels of protein 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 8 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 9 Nonsense --- UUU UUU CAG UUU UUU ------- Phe Phe Gln Phe Phe --Nonsense mutation --- UUU UUU UAG UUU UUU ------- Phe Phe STOP 10 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!! 11 Nonsense suppressor 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. Trp AUC ---UAG--- MetAla Phe Phe Trp Phe Phe AAA AUC AAA AAA 12 5’--- UUU UUU UAG UUU UUU -----3’ 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 ? Trp Trp AUG ---UAC--- AUC ---UAG--- Normal tRNA Mutant tRNA 13 14 Generation of mutations Spontaneous mutations Replication induced mutations of DNA Usually base substitutions (Most errors are corrected) Meiotic crossing over can induce mutations Small additions and deletions Environmental 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-rays15 Methods used to study mutations Gross chromosomal changesdeletions, insertions, inversions, translocations Cytology- microscopy- karyotype Point mutations Small deletions, insertions Recombinant DNA technologies 16 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 position 6 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 17 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! 18 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 19 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 20 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 amplification of known sequence --- 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 21 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. 5’ AAAAAAAATTTTAAAAAAA 3’ Will associate with 3’ TTTTTTTTAAAATTTTTTT 5’ This occurs with complementary DNA/DNA, DNA/RNA, RNA/RNA 22 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 23 Restriction Enzymes Enzymes which cut DNA at specific sequences SmaI Analysis revealed that the enzyme recognized and cut the following sequence 5’ 3’ CCAGGG3’ GGTCCC5’ | 5’ CCCGGG3’ 3’ GGGCCC5’ | This sequence is symmetrical. If one rotates it about the axis It reads the same EcoRI | 5’ GAATTC3’ 3’ CTTAAG5’ | 24 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 25 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 26 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. Previously we used specific genes as markers on the chromosome and Map units to indicate distance between the markers(relative distance). Its like using specific landmarks to identify your location along a road. Restriction sites are also used as landmarks along the chromosome. 20Mu 10Mu cy 8 HindIII EcoRI 9 vg SmaI pr 27 Restriction sites serve as landmarks in the DNA with which a physical map of a specific DNA sequence can be created. 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 Human Chimp Gibbon 28 Deletions and additions Normal Globin gene HindIII EcoRI 4 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. 29 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. - Marker EcoRI HindIII Marker EcoRI HindIII EcoRI/HindIII Gel electrophoresis 6 5 4 The rate of migration of a DNA fragment is inversely proportional to its size. Larger the size, slower its movement. 3 2 + 1 2 HindIII EcoRI 5 EcoRI 3 HindIII EcoRI HindIII EcoRI 1 30 Mapping You are given a 20 kb fragment of DNA After trying many enzymes you find that EcoRI and HindIII cut the fragment 20 14 12 HindIII Marker uncut EcoRI HindIII 14kb and 6kb EcoRI 12kb 6kb and 2kb Solve the map 6 14 6 4 2 1 12 2 6 31 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 6 12 2 1 8 H 4 E 6 E 32 2 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 33 Mapping HindIII 14 6 EcoRI 12 6 2 HindIII/EcoRI 8 6 4 2 34 Different 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 35 Mapping HindIII 12 8 EcoRI 12 6 2 HindIII/EcoRI 8 6 4 2 HindIII 12 & 8 12 & 8 36 Mapping EcoRI 12 6 2 PstI 13 7 PstI/EcoRI 12 5 2 1 37 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: 9 6 8 2 1 1 6 E 8 H 4 E 2 38 There is a 3kb deletion in the 4kb HindIII/EcoRI fragment 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 14 14 14 14 12 12 12 12 6 6 6 6 4 4 4 4 2 2 2 2 1 1 1 1 39 Cloning 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 or cloning of 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 | 40 Recombinant DNA A reasonable question is how did we get the 20kb fragment in the first place? 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 | AATT-------------------------------------TTAA AATT-------------------------------------TTAA 41 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 AATT 42 Plasmids Small circular autonomously replicating extrachromosomal DNA Modified plasmids, called cloning vectors Are used by molecular biologists to isolate Large quantities of a given DNA fragment Plasmids used for cloning share three properties Unique restriction site Antibiotic resistance Origin of replication Bacterial genome Plasmid DNA (5000kb) (3kb) 43 Plasmid elements Origin of replication: This is a DNA element that allows the plasmid to be replicated and duplicated in bacteria. Each time the bacterium divides, the plasmid also needs to divide and go with the daughter cells. If a plasmid cannot replicate in bacteria, then it will be lost. 44 Plasmid elements Antibiotic resistance: This allows for the presence of the plasmid to be selectively maintained in a given strain of bacteria Lab bacterial strains are sensitive to antibiotics. When grown on plates with antibiotics, they die. The presence of a plasmid with the antibiotics resistance gene allows these lab strains to grow on plates with the antibiotic. You are therefore selecting for bacterial colonies with the Plasmid 45 Plasmid elements Unique restriction sites: For cloning the plasmid needs too be linearized. Most cloning vectors have unique restriction sites. If the plasmid contains more than one site for a given restriction enzyme, this results in fragmentation of the plasmid Why does this matter? 46 pUC18 pUC18 is a commonly used plasmid: pUC= plasmid University of California Plasmid pBR322 pUC18 pACYC pSC101 replicon pMB1 pMB1 p15A pSC101 copy No 15 500 10 5 47 Ligation PLASMID GENOMIC DNA AATT The EcoRI linearized plasmid DNA is mixed with human EcoRI digested DNA The sticky ends hybridize and anneal and a recombinant plasmid is generated 48 Plasmid propagation- Transformation 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 ligated into a plasmid, unlimited copies of a single gene can be produced.The process of amplifying and isolating the human DNA fragment is called DNA cloning. 49 Inter-species Gene transfer CF gene on a plasmid in E.coli Isolate Plasmid Transfect human cell with CF plasmid Human Cell is cf-/cfIt becomes CF+ after transfection 50