C12 - C1. A. tRNA genes encode tRNA molecules, and rRNA...

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C1. A. tRNA genes encode tRNA molecules, and rRNA genes encode the rRNAs found in ribosomes. There are also genes for the RNAs found in snRNPs, etc. B. The term template strand is still appropriate because one of the DNA strands is used as a template to make the RNA. The term coding strand is not appropriate because the RNA made from nonstructural genes does not code for a polypeptide sequence. C. Yes. C2. The formation of the open complex, and the release of sigma factor. C3. A consensus sequence is the most common nucleotide sequence that is found within a group of related sequences. An example is the –35 and –10 consensus sequence found in bacterial promoters. At –35, it is TTGACA, but it can differ by one or two nucleotides and still function efficiently as a promoter. In the consensus sequences within bacterial promoters, the –35 site is primarily for recognition by sigma factor. The –10 site, also known as the Pribnow box, is the site where the DNA will begin to unwind to allow transcription to occur. C4. GGCATTGTCA C5. Mutations that make a sequence more like the consensus sequence are likely to be up promoter mutations while mutations that cause the promoter to deviate from the consensus sequence are likely to be down promoter mutations. Also, in the –10 region, AT pairs are favored over GC pairs, because the role of this region is to form the open complex. AT pairs are more easily separated because they form only two hydrogen bonds compared to GC pairs, which form three hydrogen bonds. A. Up promoter B. Down promoter C. Up promoter C6. The most highly conserved positions are the first, second, and sixth. In general, when promoter sequences are conserved, they are more likely to be important for binding. That explains why changes are not found at these positions; if a mutation altered a conserved position, the promoter would probably not work very well. By comparison, changes are occasionally tolerated at the fourth position and frequently at the third and fifth positions. The positions that tolerate changes are less important for binding by sigma factor. C7. In Figure 12.5, each α− helix (labeled 2 and 3 ) appears to occupy a region that is approximately one-half of a complete turn of the double helix. Since a complete turn of a DNA double helix involves 10 bp, each a helix appears to be bound to about 5 bp in the major groove of the DNA; 5 bp would span a linear distance of approximately 1.7 nm (as described in Chapter 9, Figure 9.17). If we divide 1.7 nm by 0.15 nm/amino acid, we obtain a value of 11.3 amino acids per helix. This would create about 3.15 turns of an a helix; there are three complete turns per helix, for a total of six complete turns for both helices 2 and 3. C8. This will not affect transcription. However, it will affect translation by preventing the initiation of polypeptide
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This note was uploaded on 04/07/2009 for the course BIO 325 taught by Professor Saxena during the Spring '08 term at University of Texas at Austin.

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C12 - C1. A. tRNA genes encode tRNA molecules, and rRNA...

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