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
is still appropriate because one of the DNA strands is used as a template to make the
RNA. The term
is not appropriate because the RNA made from nonstructural genes does not
code for a polypeptide sequence.
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.
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
) 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
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
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