Quiz#10Describe elongation stage of translation in prokaryote.Peptide bond formationtRNA Charging: The Second Genetic CodeGenomics:TRANSGENICSmake transgenic miceQuiz#10 Describe elongation stage of translation in prokaryote.3 major steps1. delivery of aa-tRNA to A site (GTP dependent) (efTu)2. PT activity- peptidly transferase forms peptide bond (PT)3. translocation from AP site. Shift to next reaction (Ef-G)Peptide bond formationWhy we say rRNA, instead of protein, plays a major role in peptide bond formation.1. No ribosomal proteins have been identified that have peptidyl transferase (PT) activity.2. Drugs (e.g., Chloramphenicol) that inhibit PT bind to the 23S rRNA, in the PT loop ofDomain V.3. Mutations that provide resistance to the drugs that inhibit PT map to the same PT loop.4. Nearly all (99%) of the protein can be stripped from the 50S subunit, and still get have PTactivity. 5. The X-ray crystal structure of the 50S subunit shows that only RNA chains (PT loop,etc.) are close enough to catalyze a reaction.How will you use experiment to prove this?Puromycin relase assay: Puromycin looks like tyrosine, therefore it can go into the active site If apeptide is released containing puromycin, then puromycin must have gone into the active site.Therefore puroycin can be used to map the active siteIf the 30S subunit is stripped, the reaction still worked. This means that the PT activity must not be inthis subunit, but in the 50S large subunitThe PT activity was mapped to the large (50S) subunit. But It’s composed of protein and rRNA. Howwas it shown that the PT activity was in one of the rRNA components of 50S and not protein? Treat extract with protease to denature proteinsstill get activity. Therefore protein component NOTresponsible for PT activityTreat extract with RNase to denature RNA lose activity. Therefore some RNA component of 50Smust be responsible for PT activityd) Now the question becomes which RNA component (23S and 5S) of the 50S particle contains the PTactive site?Yarus analog looks like transition state so it must be in the active site when they look at the crystalstructure.23S is the only RNA component of 50S that is close enough to the Yarus analog in the crystalstructure. This means that the active site must be there.tRNA Charging: The Second Genetic CodeWhy it could be called the second genetic code?This process is very specific. One Aminoacyl-tRNA synthetase only recognizes one kind of tRNA. Name the major regions of tRNA secondary structure from 5’ to 3’D loopAnticodon loopvariable loop T loop Acceptor stem. What regions are recognized by Aminoacyl-tRNA synthetases mainly?Acceptor stemanticodonThe error rate of charging is very low: about 1 in 10,000. How is this low error achieved?The double-sieve model The enzyme has two sites------activation site and editing site.Only the amino acid which fits into the activation site can be activated and become Aminoacyl-AMP(aa-AMP)The enzyme also has proofreading capability. Once a wrong Aminoacyl-AMP complex is formed, theediting site will hydrolyze the complex before it can be covalently attached to the tRNA Genomics:The Arabidopsis thaliana genome contains approx. ____25,000______________ genes.The Oryza (rice) genome contains approx. ______32,000-56,000_____________ genes.What are some factors that could explain the large difference (>10-fold) in the sizes of the rice and Arabidopsis genomes (in terms of base pairs).• Amounts of highly repeated DNA • Amounts of "Selfish DNA“ (transposons, etc.)• Frequency and sizes of introns• Other intergenic DNA• Genetic redundancy•TRANSGENICSmake transgenic mice1. Describe the characteristics of the knockout (KO) plasmid that must be used carry the gene ofinterest. Two markers—tk1&tk2 makes cells suspeptible to gancyclovirNeo-confer neomycin resistant to cell.Homologous regionsTransgene2. What type of cells are originally transformed with the KO plasmid?Stem cells. Only stem cells have potential to grow into whole animals, which will produce progenyand pass the knockout gene to the next generation.3.How are cells that contain the interrupted gene selected?Select cells which are resistant to neomycin and also resistant to gancyclovir. If homologousrecombination happens, tk1 and tk2 , which confer gancyclovir-sensitive to cells, will not go to thegenome. Only neo gene can be integrated to the genome, so the transformant cell is resistant to bothneomycin and gancyclovir4.What then happens to these cells with the interrupted gene?The interrupted gene will lose its function and be knocked
View Full Document
Unlocking...