Bacterial DNA replication

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Dortha McCoy
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1 Bacterial DNA replication Summary: What problems do these proteins solve? Tyr OH attacks PO4 and forms a covalent intermediate Structural changes in the protein open the gap by 20 Å! 1

2 Summary: What problems do these proteins solve? Helicase Polymerase Core ssdna binding Remove +sc at fork (swivel) Decatenation Ligase Function Primase Primer removal Clamp loader Sliding clamp DnaB Primase (DnaG) pol I s 5-3 exo pol III (α, ε, θ subunits) γ complex β SSB gyrase topo IV E. coli DNA ligase SV40 (simian virus 40) T antigen pol α primase FEN 1 (also RNaseH) pol δ, ε RF-C PCNA RF-A topo I or topo II topo II DNA ligase I other model systems include bacteriophage T4 and yeast The ends of (linear) eukaryotic chromosomes cannot be replicated by the replisome. Not enough nucleotides for primase to start last lagging strand fragment Chromosome ends shorten every generation! 2

3 Telomere shortening signals trouble! Telomere binding proteins (TBPs) 1. Telomere shortening releases telomere binding proteins (TBPs) 2. Further shortening affects expression of telomereshortening sensitive genes 3. Further shortening leads to DNA damage and mutations. Telomerase replicates the ends (telomeres) Telomere ssdna Telomerase extends the leading strand! Synthesis is in the 5 -->3 direction. Telomerase is a ribonucleoprotein (RNP). The enzyme contains RNA and proteins. The RNA templates DNA synthesis. The proteins include the telomerase reverse transcriptase TERT. 3

4 Telomerase cycles at the telomeres TERT protein Telomere ssdna TER RNA template Telomerase extends a chromosome 3 overhang 4

Conserved structures in TER and TERT Core secondary structures shared in ciliate and vertebrate telomerase RNAs (TERs). (Sequences highly variable.

5 Conserved structures in TER and TERT Core secondary structures shared in ciliate and vertebrate telomerase RNAs (TERs). (Sequences highly variable.) nucleotides 1000s of nucleotides 1300 nucleotides TERT protein sequences conserved Starting and stopping summary 1. DNA replication is controlled at the initiation step. 2. DNA replication starts at specific sites in E. coli and yeast. 3. In E. coli, DnaA recognizes OriC and promotes loading of the DnaB helicase by DnaC (helicase loader) 4. DnaA and DnaC reactions are coupled to ATP hydrolysis. 5. Bacterial chromosomes are circular, and termination occurs opposite OriC. 6. In E. coli, the helicase inhibitor protein, Tus, binds 10 ter DNA sites to trap the replisome at the end. 7. Eukaryotic chromosomes are linear, and the chromosome ends cannot be replicated by the replisome. 8. Telomerase extends the leading strand at the end. 9. Telomerase is a ribonucleoprotein (RNP) with RNA (template) and reverse-transcriptase subunits. 5

6 DNA methods summary 1. Restriction enzymes cut at specific DNA sites. (N) 2. Vectors allow genes to be cloned and proteins expressed. (N) 3. Gel electrophoresis separates DNA on the basis of size. 4. DNAs can be synthesized (up to ~100 bases commercially). (N) 5. PCR amplifies any target DNA sequence. (N) 6. Genes and genomes can be sequenced by chain termination. (N) 7. Oligonucleotides can be used to change bases by site- directed mutagenesis. (N) 8. Southern blotting detects sequences by hybridization. 9. Microarrays detect gene expression patterns over the genome. 10. Genes can be knocked out (deleted) or replaced in prokaryotes and eukaryotes. (N) Restriction enzymes cut DNA at specific sites 6

Restriction enzymes cut DNA at specific sites 3 types of ends: 5 overhang, blunt and 3 overhang Cognate methyl transferases protect host genome from digestion.

7 Restriction enzymes cut DNA at specific sites 3 types of ends: 5 overhang, blunt and 3 overhang Cognate methyl transferases protect host genome from digestion. Restriction-modification systems degrade foreign DNA. Average frequency of restriction sites in random DNA sequences Site size Average frequency 1/256 (1/4 x 1/4 x 1/4 x 1/4) 1/4,096 1/65,536 The average occurrence of each sequence = 1/4 n, where n = the site length and all bases are equally represented 7

8 Lots of different recognition sites known Core four bases Flanking bases None A----T C----G G----C T----A A simple cloning procedure 1. Cut insert and vector DNA with a restriction enzyme 2. Mix and join ends with DNA ligase. The ends should match for efficient ligation. 8

Cloning without DNA ligase Ligation-independent cloning Gateway cloning 1. Prepare open vector and insert with the same long sticky ends No dt in template T T + Pol I Klenow fragment + datp 2.

9 Cloning without DNA ligase Ligation-independent cloning Gateway cloning 1. Prepare open vector and insert with the same long sticky ends No dt in template T T + Pol I Klenow fragment + datp 2. Mix and let the ends anneal. A T T A A T T A 1. Prepare an insert flanked by sites for a site-specific DNA recombinase. 2. Mix insert with the closed vector containing the recipient recombination site and recombinase enzyme. E Transform the nicked plasmid. The plasmid is repaired in vivo. 3. (Have lunch.) Transform. Vectors allow DNA sequences to be cloned - 1 Ori + selectable marker + cloning site (polylinker) Phage λ for cloning big (7-25 kb) DNA pieces 9

10 Vectors allow DNA sequences to be cloned - 2 Shuttle vectors: move genes between organisms Expression vectors: Make your favorite protein Reporter genes: β-gal, GFP... Vectors allow DNA sequences to be cloned -3 Transient transfection: eukaryotes Stable transduction 10

Gel electrophoresis separates DNA on the basis of size Agarose: big fragments (>300 bp) Acrylamide: smaller fragments, higher resolution Mobility

11 Gel electrophoresis separates DNA on the basis of size Agarose: big fragments (>300 bp) Acrylamide: smaller fragments, higher resolution Mobility proportional to log MW. Chemical DNA synthesis Sequential rounds of coupling, oxidation and deprotection of the 5 OH build up the oligonucleotide

12 Chemical DNA synthesis Sequential rounds of coupling, oxidation and deprotection of the 5 OH build up the oligonucleotide. 3 5 Frontiers in DNA synthesis Currently: nucleotides routine (Assemble 5 kb) 10,000 = largest. Primer set for the human genome (30,000 genes) ~ $10 4 Goal 1: Make yeast chromosome 3: 300 kb without errors! (Jeff Boeke) Goal 2: Assemble 16 X 10 6 w/o errors for ~$1000 (George Church) 12

13 DNA sequencing by partial chain termination ddntps terminate the chain DNA sequencing by partial chain termination ddntps terminate the chain Small amount of ddgtp + excess dgtp partially terminates chains at Cs in the template 13

DNA sequencing by partial chain termination 1.

Separating the mixture of products by size reveals the

14 DNA sequencing by partial chain termination 1. All fragments start at the primer 2. All fragments ending in a particular base have a different length and a different color tag 3. Separating the mixture of products by size reveals the sequence. Two strategies for genome sequencing Hierarchical Shotgun Sequencing Sequencing 14

Site-directed mutagenesis 1. Anneal divergent mutagenic primers. 2.

15 PCR (Polymerase Chain Reaction): isolate and amplify any DNA sequence Copies: N cycles amplifies the target sequence 2 N -fold. Site-directed mutagenesis 1. Anneal divergent mutagenic primers. 2. Replicate entire plasmid with a DNA pol lacking 5 -->3 exonuclease. 3. Select against parental strands. 15

Check insertion site by Southern blotting Southern blotting detects DNA sequences by hybridization 1. Digest DNA using restriction enzyme(s) 2.

16 Gene replacement in mice -- make donor cells 1. Insert drug markers into genome of ES cells 2. Select to enrich for homologous recombinants Neo r confers resistance to G-418. tk HSV confers sensitivity to ganciclovir. Check insertion site by Southern blotting Southern blotting detects DNA sequences by hybridization 1. Digest DNA using restriction enzyme(s) 2. Run gel 3. Transfer DNA from gel to (nitrocellulose) paper. 4. Denature DNA, hybridize probe DNA, and wash off excess probe. 5. Detect the probe on the paper. E.g. by autoradiography. Northern blotting detects RNA on the gel. 16

Microarrays detect expressed genes by hybridization Each spot has a different synthetic

Label cdnas with red fluorophore in one condition and green fluorophore in another reference

Relative to the reference, Red=enriched, yellow = =, green = depleted.

expresses extra copies of the growth hormone gene? 1. Inject ES cells into early embryos, 2.

17 Microarrays detect expressed genes by hybridization Each spot has a different synthetic oligonucleotide complementary to a specific gene. 1. Label cdnas with red fluorophore in one condition and green fluorophore in another reference condition. 2. Mix red and green DNA and hybridize to a microarray. 3. Relative to the reference, Red=enriched, yellow = =, green = depleted. Gene replacement in mice -- germline incorporation Transgenic mice express a new gene Which mouse expresses extra copies of the growth hormone gene? 1. Inject ES cells into early embryos, 2. Transfer embryos to foster mother, 3. Breed chimeric mice and screen for progeny with mutant germ line, 4. Screen progeny DNA for mutation, 5. Mate heterozygotes (X+/X-), 6. Screen progeny DNA for KO genotype (X-/X-). Entire process takes a year. 17

Gene replacement in plants -- engineered crops E.g. Golden rice synthesizes β-carotene DNA methods summary 1. Restriction enzymes cut at specific DNA sites. (N) 2.

18 Gene replacement in plants -- engineered crops E.g. Golden rice synthesizes β-carotene DNA methods summary 1. Restriction enzymes cut at specific DNA sites. (N) 2. Vectors allow genes to be cloned and proteins expressed. (N) 3. Gel electrophoresis separates DNA on the basis of size. 4. DNAs can be synthesized (up to ~100 bases commercially). (N) 5. PCR amplifies any target DNA sequence. (N) 6. Genes and genomes can be sequenced by chain termination. (N) 7. Oligonucleotides can be used to change bases by site- directed mutagenesis. (N) 8. Southern blotting detects sequences by hybridization. 9. Microarrays detect gene expression patterns over the genome. 10. Genes can be knocked out (deleted) or replaced in prokaryotes and eukaryotes. (N) 18

Molecular Cell Biology - Problem Drill 11: Recombinant DNA

Molecular Cell Biology - Problem Drill 11: Recombinant DNA Question No. 1 of 10 1. Which of the following statements about the sources of DNA used for molecular cloning is correct? Question #1 (A) cdna