Prescott`s Microbiology, 9th Edition Chapter 13 – Bacterial Genome

advertisement
Prescott’s Microbiology, 9th Edition
Chapter 13 – Bacterial Genome Replication and Expression
GUIDELINES FOR ANSWERING THE MICRO INQUIRY QUESTIONS
Figure 13.1 Based on what we now know about proteins, why can we conclude from this
experiment that genetic information was unlikely to be carried by proteins?
Now we know (and they did not back then) that the genetic code is degenerate, thus there are 2-6
nucleotide triplet codons that code for each amino acid. Thus, you cannot go backwards from
the amino acid to the codon, because you don’t know what one it is. In Griffith’s experiment
above, heat-killing the smooth strain denatures the proteins, but the DNA survives (including the
genes that code for the capsule synthesis). This is an example of transformation.
Figure 13.4 To which carbon of ribose (deoxyribose) is each of the following bonded: adenine
and the hydroxyls of adenosine and deoxyadenosine?
In a nucleotide, the purine (including adenine) or pyrimidine base is always covalently attached
to the 1’ (one prime) carbon of the ribose sugar. Adenosine (part of RNA) has two hydroxyl
groups attached to the 2’ and 3’ ribose carbons. Deoxyadenosine (part of DNA) has only on
hydroxyl group, attached to the 3’ carbon. So, one difference between RNA and DNA is ribose
versus deoxyribose, with RNA have a 2’ hydroxyl and DNA having a 2’ hydrogen.
Figure 13.5 How many H bonds are there between adenine and thymine, and between guanine
and cytosine?
There are 2 hydrogen bonds between adenine and thymine and 3 hydrogen bonds between
guanine and cytosine. This is why GC-rich DNA has a slightly higher melting temperature than
AT-rich DNA of the same length.
Figure 13.7 Identify the two other peptide bonds of the tetrapeptide chain.
Shown in yellow are the two other peptide bonds.
Figure 13.10 What provides the energy to fuel this reaction?
Reaction coupling is used to make this reaction favorable. The reaction shown above is coupled
with hydrolysis of the pyrophosphate (PPi) into inorganic phosphate (2 Pi), thus the overall
reaction pair is favorable.
1
© 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in
any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Prescott’s Microbiology, 9th Edition
Figure 13.12 What is the difference between helicase and gyrase? Which is a topoisomerase?
Helicase is an enzyme that unwinds or separates the DNA double helix during replication
whereas DNA gyrase is a topoisomerase that nicks and seals the DNA and relieves supercoiling
that can occur by rapid unwinding.
Figure 13.15 Why can’t DNA polymerase perform this reaction?
During the polymerization reaction, an incoming nucleotide triphosphate is added to the 3’ OH
of a DNA strand (figure 13.10), and a pyrophosphate is cleaved from it. With a nick, there is no
incoming nucleotide triphosphate, nor any phosphate to be removed. The ligase enzyme that
performs this reaction has to have another mechanism to make the reaction favorable, and so
utilized NAD+ or ATP as an energy source.
Figure 13.20 Why is the nontemplate strand called the “coding strand”?
The coding strand, in blue above, is identical in sequence to the mRNA (except Ts are replaced
with Us). The template strand, in purple, is complementary to the coding stand and to the
mRNA.
Figure 13.25 Are the -35 and -10 regions considered “upstream” or “downstream” of the +1
nucleotide?
RNA polymerase is considered to move “down the stream.” So, according to the direction of
transcription (shown as black arrow), any negatively numbered nucleotides (including the -35
and -10 boxes in the sigma factor-binding region of the promoter) are upstream of the +1 site.
The +1 site is the first nucleotide in DNA that is mirrored in the mRNA, so the remainder of the
gene is downstream from the +1 start site. Important note: the +1 start site for transcription is
NOT the first codon (the start site of translation).
Table 13.4 For each of the above codons, determine what is normally encodes. Choose one of
the variants and suggest how it might have evolved.
Looking at Table 13.3, students can determine the common coding for a codon. For example,
AGA commonly codes for Arg. It may be evolutionarily-related to the common stop codon,
UGA.
Figure 13.33 Why is simultaneous transcription and translation impossible in eukaryotes?
Transcription occurs in the nucleus in eukaryotes, while translation occurs in the cytoplasm.
Thus compartmentalization prevents this.
Figure 13.35 What would be the outcome if an aminoacyl-tRNA synthetase added the wrong
amino acid to a tRNA (i.e., the anticodon specified a different amino acid than that added to the
3’ end of the tRNA)?
This question points out the fact that the ribosome does not have any proofreading activity.
Error control is at the level of the synthetase enzymes, which is why they are very specific in
charged tRNAs with certain amino acids. If a tRNA was charged with the incorrect amino acid,
the ribosome would simply incorporate it, thus the protein would have a single amino acid
mutation (not coded in the gene). This has been done in E. coli.
2
© 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in
any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Prescott’s Microbiology, 9th Edition
Figure 13.37 Why would it be impossible for fMet-tRNA to initiate peptide bond formation with
another amino acid? (Hint: Examine figure 13.40 closely.)
The front end of a protein is the amino, or N-terminus, while the back end and last part
synthesized is the carboxy or C-terminus. FMet is the first amino acid, and can form a peptide
bond using its carboxyl end with the second amino acid in the protein. However, another amino
acid cannot be added in front of the fMet because the amino group is not free, it is covered with
the formyl group. Thus all internal Met residues do not have the formyl group.
Figure 13.40 What provides the energy to fuel this reaction?
No extra energy source is required for peptide bond formation because the bond linking an
amino acid to tRNA is high in energy. This high energy was put in during the charging reaction
catalyzed by the amino acyl tRNA synthetase enzymes (figure 13.35), in which ATP is
hydrolyzed.
Figure 13.45 What are two distinguishing features of protein translocation via the Tat system
was compared to that using the Sec system?
The Sec pathway translocates unfolded proteins whereas the Tat pathway translocates folded
proteins. Furthermore, the Tat pathway only translocates proteins that feature twin arginine
residues in their signal sequence.
3
© 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in
any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Download