Prescott’s Microbiology, 9th Edition
Chapter 17 – Recombinant DNA Technology
GUIDELINES FOR ANSWERING THE MICRO INQUIRY QUESTIONS
Figure 17.1 Which of the DNA molecules shown are recombinant?
Students should first define recombinant. Then they can accurately determine that the product of
step 3 (vector with insert) is a recombinant DNA molecule, and the bacteria made by step 4 is a
recombinant organism.
Figure 17.3 Examine the uncut piece of DNA shown in the upper half of this figure. Where would
an exonuclease cleave?
Exonucleases cleave from the ends.
Table 17.2 Which of the above enzymes yield blunt ends? Which yield sticky ends with a 5’
overhang? What about a 3’ overhang?
The enzymes with blunt products is AluI. Enzymes with sticky 5’ overhang products include all
the other enzymes. There are no enzymes pictured with 3’ overhangs.
Figure 17.4 Why might two DNA fragments inadvertently be cloned into a single vactor when
using this cloning strategy?
Two fragments of DNA from the organism could anneal together, making a concatemer, which
would still have the two appropriate ends to be accepted into the vector, because both sites
(target DNA and vector) were cut with the same enzyme. Note the fragment could fit into the
vector in either direction as well. Ask students how to avoid this (directional cloning – cutting
ends with two different enzymes).
Figure 17.5 Why must introns be removed from eukaryotic DNA before it can be expressed in a
bacterium?
Bacteria do not contain the genes or proteins for the spliceosome, and thus cannot splice RNA.
So the introns would not be removed and would be translated, but they do not code functional
protein.
Figure 17.6 In a linear piece of DNA is cut with a restriction enzyme for which it has four
recognition sites, how many bands would be visible on a gel if the products of digestion were
electrophoresed?
The drawing will be helpful to the students. Note this assumes that products will not be the same
size (and thus form only one band on the gel). Five products would be visible with a linear target
and four with a circular target.
Figure 17.8 Why, after three cycles, are the vast majority of amplified DNA molecules (i.e., PCR
products) the size defined by the distance between the forward and reverse primers?
After three cycles, the ratio is two molecules of hybrid one strand of original target DNA with
one strand product DNA, along with six molecules of pure product DNA, or 75% product DNA.
Figure 17.10 Which plasmid is a shuttle vector? Why?
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© 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
Shuttle vectors can ‘shuttle’ between at least two hosts, and so can be replicated in at least two
different cell types. Plasmid Yep24 can be replicated in the prokaryote E. coli or the eukaryote
Saccharomyces cerevisiae. This allows easy amplification in E. coli for use in experimentation in
the yeast.
Figure 17.11 What would you conclude if you obtained only blue colonies after cloning into a
vector that enables blue/white screening of transformants? Why might such a result occur?
Blue transformants have the plasmid without an insert, because of expression of the lacZ gene.
White transformants have the insert disrupting the lacZ gene. So, only blue colonies means none
have an insert in the plasmid, thus unsuccessful cloning but successful transformation. This
might be caused by a faulty ligase enzyme or incompatible ends on insert and plasmid.
Figure 17.12 In what ways does the BAC shown here differ from the plasmid pUC19 shown in
figure 17.10?
pUC encodes resistance to ampicillin and enables blue/white cloning selection. pBeloBAC11
encodes resistance to chloramphenicol and also enables blue/white cloning selection. This BAC
differs from pUC19 in that it contains genes that ensure a replication complex will be formed
(repE), as well as genes for proper partitioning of one newly replicated BAC to each daughter
cell (sopA, sopB, and sopC). The primary functional difference is the size of the inserts accepted,
with pUC accepting inserts up to 10-15 kb and pBelo accepting up to 300 kb.
Figure 17.13 Why are long fragments (e.g., 20,000 bp) of genomic DNA often desired when
constructing a genomic library?
Genes in eukaryotes can be quite large, and so large fragments have a greater statistical change
of containing a complete gene for expression.
Figure 17.15 What special considerations are necessary if one is constructing a translational
fusion of a protein known to be exported?
The export mechanism needs to be known (end of chapter 13). That will determine is the fusion
protein will be exported or not. For example, the sec system relies on a signal peptide, so the
fusion should be exported normally. However, the tat pathway exports folded proteins, and since
the fusion is larger than the normal protein itself, export may be limited or blocked.
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© 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.