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Formation of mRNAs RNA processing.docx

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Department
Biochemistry
Course
Biochemistry 2280A
Professor
Eric Ball
Semester
Fall

Description
Formation of mRNAs RNA processing Eukaryotic cells Primary transcript becomes a mRNA for export out of the nucleus and translation 3 steps in RNA processing  5’ capping  3 ‘ polyadenylation  Splicing Each requires spec enzymes 5`cap  5 end first nucleotide added phosphate then a 7 methyl guanosine through a 5’-5’ linkage 5’ capping importance  Required for mRNA export form nucleus  Req for translation of mRNA into protein (translation initation) 3`polyadenylation  Happens while translation is happening  Downstream the aaaaa seq signals a cleavage from the actual RNA ending and after 30 base pairs poly a tail to be added on Importance  Helps protect 3 end of mRNA from degradation  Indicates that 3 end of mRNA is intact and therefore is important for export out of nucleus and for translation RNA splicing  Coding called exons non coding called introns  In eukaryotic cells protein encoding seq are interrupted by one or more noncoding deq called introns  Almost all mammalian cells have introns  Exons are expressed  Introns are interrupted to remember this stuff Example human beta globin gene  Full gene is 200,000 pairs but most is thrown out to make globin gene  Why have so much garbage  Reason is cause its not the introns can sometimes be used to actually make other mRNA proteins Why splicing important  Differential splicing  Signal RNA can be spliced in diff ways to create related but distinct proteins  Splicing enhances the coding capacity of the genome  Ex one long DNA strand creates many different tropomyosin that are related but diff Splice mechanism  Needs a 5’ splice junction  3’ splice junction  Branch point it has an Adenine that is very important to identify it  Introns are removed in two consecutive transesterification reactions  Attack by the 2’ OH of the adenine at the branch point with the 5’ splice junction  OH is now on the 3’ end of the exon  The 3’ on the 5; junction excon now attacks the 3’splice junction  3’OH of the free 5’ junction attacks the 3’ splice junc
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