Biol 112 Nov. 2 (C)

Biol 112 Nov. 2 (C) - 02/11/2011 Terms relative to...

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Unformatted text preview: 02/11/2011 Terms relative to mutations: Mutations 1. Wild type: as isolated from nature (superscript +, for example trp+) 2. Genotype: the total DNA sequences in an organism. 3. Genome: the total DNA content in an organism. 4. Phenotype: all of the observable properties (usually DNA sequence is not included) 5. Mutant: organism with a changed genotype (superscript ‐, for example trp‐) 6. Mutation: a change in the genotype. Mutations: origins Bacterial examples Would adding non‐coding DNA to the genome be a mutation? A. No, a mutation is a change in the DNA sequence of the DNA already in the cell B. Yes, adding DNA alters the total DNA sequences and so is a mutation. C. Depends on whether any of the properties of the cell are changed Base pairing 100,000:1 preference for the right base New DNA Template Overall the error rate in DNA synthesis is about one mistake every 109 base pairs (bp) that are replicated. Assume we have a strain of bacterial that has about 106 bp in its genome. And we have a culture of these bacteria with around 109 bacteria per mL. At this cell density, if the cells double once, the number of mutant bacteria per mL that have been created will be approximately : A. B. C. D. E. 10 100 10,000 100,000 1,000,000 Proof reading: 3'‐exonuclease dCTP 5’ A G G 3’ T C C G T C A G T 5’ dGTP DnaQ protein‐part of pol III Base pairing error Mismatch dATP dTTP New DNA Template 5’ A G G T 3’ 3’ T C C G T C A G T 5’ T New DNA 5’ A G G 3’ DNA polymerase Reverses direction, DnaQ removes incorrect nucleotide. Pol III adds tries again Template 3’ T C C G T C A G T 5’ New DNA Template 5’ A G G C 3’ 3’ T C C G T C A G T 5’ 1 02/11/2011 Proof reading: 3'‐exonuclease dnaQ gene product Which of these describe phenotype of a dnaQ‐ cell with reduced the DnaQ activity? 1. The types of mutations caused by errors in selection of the bases have several similar names Base pair substitutions Point mutations A. There would be no change in phenotype from wildtype. B. The cells would not survive, this would be a lethal mutation. C. The cell would have higher levels of DNA polymerase III to replicate the DNA. D. The cell would acquire more mutations than the wild type. 5’ 3’ 2. The other common forms of mutation are insertion and deletion mutations. In these mutations, DNA sequences are added (an insertion) or removed (a deletion) from the DNA. This can happen during replication through “strand slippage” 5’ CGTTTT GCAAAAACCTAG Strand slippage by new strand 5’ 3’ CGTTT Strand slippage GCAAAAACCTAG by template 5’ 5’ C G T T T T 5’ C G T T T GCAAAAACCTAG 3’ GGAAAAACCTAG 5’ Replication continues 3’ 3’ GCAAAAACCTAG 5’ GCAAAAAACCTAG Wild type Insertion of T:A Parent 1st Gen Progeny Exposure to Nitrous acid Deamination C to CD 2nd Gen Progeny G C G CD Base pairing changed 3’ GCAAAAACCTAG 5’ 5’ C G T T T G G A T C 3’ 3’ GCAAACCTAG Wild type 5’ Deletion of 2 T:A Removes the NH2 from C. It would make C more like T Replication Cell division Replication Cell division G C 5’ 5’ 3’ CGTTTTTGGATC Deamination by acid Mutations caused mis‐pairing during replication One example: acid treatment of DNA Parent 5’ AA Replication continues Next replication 5’ C G T T T T T T G G A T C 3’ 3’ 5’ AA GCAAACCTAG Next replication 5’ 3’ CGTTTTTGGATC GCAAACCTAG 5’ C G T T T G G A T C T 5’ C G T T T T T G G A T C 3’ 3’ 5’ G C G C A CD A T Mismatch mutation Wild type Mismatch eliminated Mutation stabilized If DNA were treated so every NH2 were removed from every C and A, replication would make so many mistakes it would be lethal. A CD 2 02/11/2011 Mismatch Repair • DNA polymerase leaves a mismatched pair in the sequence. • Mismatched bases do not hydrogen bond properly and can be detected. • Mismatched bases are corrected AFTER DNA synthesis is complete. Mismatch repair (mislabeled in Course Notes as Nucleotide Excision Repair) 5’ 3’ CGTTTTTGGATC 3’ GCAAAAGCCTAG Excision of nucleotides near mismatch by ‘repair enzymes’ 5’ 5’ 3’ CGTTTTTGGATC DNA pol I fills the gap 3’ GCA AG 5’ 5’ 3’ CGTTTTTGGATC 3’ GCA AG 5’ DNA ligase seals up 5’ 3’ CGTTTTTGGATC 3’ Mismatch repair 5’ 3’ CGTTTTTGGATC 3’ GCAAAAGCCTAG 5’ How do the repair enzymes know which strand to remove? A. B. C. D. GCAAAAACCTAG 5’ Dealing with UV radiation The light from the sun includes radiation in the energy range called “ultraviolet” This energy range can cause changes to the DNA. In particular where there are pyrimidines next to one another on the DNA (e.g) 5’ GAGATTGCACA 3’ 3’ CTCTAACGTGT 5’ Covalent bonds can be formed. The enzymes can tell which one makes the correct protein product. The enzymes may not know and it may be random. The old strand of DNA (the parental) is marked. The new strand of DNA (the daughter strand) is marked. Photoreactivation An enzyme called DNA Photolyase specifically binds pyrimidine dimers. It absorbs light and uses the absorbed energy to break the bonds between the adjacent pyrimidines. Photoreactivation If you shine UV light on some bacteria and then divide the cells. You then grow half of the cells in the dark and half of the cells in the light. You will observe: A. B. C. D. The number of mutants in the dark and light will be the same. The number of mutants in the dark grown cells will be higher. The number of mutants in the light grown cells will be higher. The cells in both cases will die so the experiment can’t be done. 3 02/11/2011 Figure 14-18 NUCLEOTIDE EXCISION REPAIR 1. Enzymes detect an Nucleotide Excision Repair irregularity in DNA structure and cut the damaged strand. • Repair system (proteins) that recognizes distortions in DNA double helix. Damaged bases 2. An enzyme excises e.g. thymine dimers nucleotides on the damaged strand. • Uses: – a helicase to unwind DNA – Endonuclease to cut DNA in 2 places – DNA pol I to fill the gap – DNA ligase to seal. 5 3 3. DNA polymerase fills in the gap in the 5 3 direction. 3 5 4. DNA ligase links the new and old nucleotides. Repaired damage Dealing with UV damage during replication the “sloppier copier”. The sloppier copier. This would be ‘the end’ for the cell. But it has a mechanism to deal with the situation. First a regulatory protein (RecA) detects the large stretch of single strand DNA. DNA polymerase III and it’s associated proteins cannot replicate past a thymine dimer (or several other types of damage). So if the replication fork gets to a dimer before it is repaired, replication on the strand with the dimer stops. But the other strand can continue for a while. The sloppier copier. Effect of mutation on a protein‐if it is in the coding region RecA activates a form of DNA polymerase (DNA polymerase V). This is an enzyme that has much lower “fidelity”. It does not demand precise base pairing to continue synthesis, so it replicate past the dimer. The region around the dimer will have a much higher rate of mutations—but at least the DNA will be copied. After the dimer has been past, DNA pol III will replace DNA pol V and replication will resume. 5’…ATCGTATACGGACGT…3’ 3’…TAGCATATGCCTGCA…5 TAC ATG Codon AAC TTG TAG ATC TAT ATA UAC AAC UAG UAU Amino acid Tyr Asn Stop Tyr Mutation name ‐ Missense Nonsense Silent Protein Wild type Depends Incomplete Wild type High rate of errors Cell phenotype wild type depends on environment wild type 4 ...
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