CHAPTER 17
Application and Experimental Questions
E1. Researchers can isolate a sample of cells, such as skin fibroblasts, and grow them in
the laboratory. This procedure is called a cell culture. A cell culture can be exposed to a
sample of DNA. If the cells are treated with agents that make their membranes permeable
to DNA, the cells may take up the DNA and incorporate the DNA into their
chromosomes. This process is called transformation or transfection. Experimenters have
transformed human skin fibroblasts with methylated DNA and then allowed the
fibroblasts to divide for several cellular generations. The DNA in the daughter cells was
then isolated, and the segment that corresponded to the transformed DNA was examined.
This DNA segment in the daughter cells was also found to be methylated. However, if
the original skin fibroblasts were transformed with unmethylated DNA, the DNA found
in the daughter cells was also unmethylated. Do fibroblasts undergo de novo methylation,
maintenance methylation, or both? Explain your answer.
Answer: These results indicate that the fibroblasts undergo maintenance methylation
because they can replicate and methylate DNA if it has already been methylated.
However, the cells do not undergo de novo methylation, because if the donor DNA was
unmethylated, the DNA in the daughter cells remains unmethylated.
E2.
Restriction enzymes, described in Chapter 19, are enzymes that recognize a
particular DNA sequence and cleave the DNA (along the DNA backbone) at that site.
The restriction enzyme known as NotI recognizes the sequence
5’–GCGGCCGC–3’
3’–CGCCGGCG–5’
However, if the cytosines in this sequence have been methylated, NotI will not cleave the
DNA at this site. For this reason, NotI is commonly used to investigate the methylation
state of CpG islands.
A researcher has studied a gene, which we will call gene T, that is found in corn.
This gene encodes a transporter that is involved in the uptake of phosphate from the soil.
A CpG island is located near the core promoter of gene T. The CpG island has a single
NotI site. The arrangement of gene T is shown here.
[Insert Text Art 17.6]
A SalI restriction site is located upstream from the CpG island, and an EcoRI
restriction site is located near the end of the coding sequence for gene T. The distance
between the SalI and NotI sites is 1,500 bp and the distance between the NotI and EcoRI
sites is 3,800 bp. No other sites for SalI, NotI, or EcoRI are found in this region.
Here is the question. Let’s suppose a researcher has isolated DNA samples from
four different tissues in a corn plant. These include the leaf, the tassel, a section of stem,
and a section of root. The DNA was then digested with all three restriction enzymes,
separated by gel electrophoresis, and then probed with a DNA fragment that is
complementary to the gene T coding sequence. The results are shown here.
[Insert Text Art 17.7]
In which type of tissue is the CpG island methylated? Does this make sense based on the
function of the protein encoded by gene T?
Answer: If the DNA band is 3,800 bp, it means that the site is unmethylated, because it
has been cut by NotI (and NotI cannot cut methylated DNA). If the DNA fragment is
5,300 bp, the DNA is not cut by NotI, so we assume that it is methylated. Now let’s begin
our interpretation of these data with lane 4. When the gene is isolated from root tissue,
the DNA is not methylated because it runs at 3,800 bp. This suggests that gene T is
expressed in root tissue. In the other samples, the DNA runs at 5,300 bp, indicating that
the DNA is methylated. The pattern of methylation seen here is consistent with the
known function of gene T. We would expect it to be expressed in root cells, because it
functions in the uptake of phosphate from the soil. It would be silenced in the other parts
of the plant via methylation.
E3.
A muscle-specific gene was cloned and then subjected to promoter bashing as
described in Solved problem S5. As shown here, six regions, labeled A–F, were deleted,
and then the DNA was transformed into muscle cells.
[Insert Text Art 17.8]
The data shown here are from this experiment.
Percentage of
β-Galactosidase Activity
100
20
330
100
5
15
<1
Region Deleted
None
A
B
C
D
E
F
Explain these results.
Answer: Based on these results, there are enhancers that are located in regions A, D, and
E. When these enhancers are deleted, the level of transcription is decreased. There also
appears to be a silencer in region B, because a deletion of this region increases the rate of
transcription. There do not seem to be any response elements in region C, or at least not
any that function in muscle cells. Region F contains the core promoter, so the deletion of
this region inhibits transcription.
E4.
A gene that is normally expressed in pancreatic cells was cloned and then
subjected to promoter bashing as described in solved problem S5. As shown here, four
regions, labeled A–D, were individually deleted, and then the DNA was transformed into
pancreatic cells or into kidney cells.
[Insert Text Art 15.9]
The data shown here are from this experiment.
Cell Type
Percentage of
Region Deleted
Transformed
β-Galactosidase Activity
None
Pancreatic
100
A
Pancreatic
5
B
Pancreatic
100
C
Pancreatic
100
D
Pancreatic
<1
None
Kidney
<1
A
Kidney
<1
B
Kidney
100
C
Kidney
<1
D
Kidney
<1
If we assume that the upstream region has one silencer and one enhancer, answer the
following questions:
A. Where are the silencer and enhancer located?
B. Why don’t we detect the presence of the silencer in the pancreatic cells?
C. Why isn’t this gene normally expressed in kidney cells?
Answer:
A.
Based on the transformation of kidney cells, a silencer is in region B. Based on
the transformation of pancreatic cells, an enhancer is in region A.
B.
The pancreatic cells must not express the repressor protein that binds to the
silencer in region B.
C.
The kidney cells must express a repressor protein that binds to region B and
represses transcription. Kidney cells express the downstream gene only if the silencer is
removed. As mentioned in part B, this repressor is not expressed in pancreatic cells.
E5.
A gel retardation assay can be used to determine if a protein binds to a segment of
DNA. When a segment of DNA is bound by a protein, its mobility will be retarded and
the DNA band will appear higher in the gel. In the gel retardation assay shown here, a
cloned gene fragment that is 750 bp in length contains a regulatory element that is
recognized by a transcription factor called protein X. Previous experiments have shown
that the presence of hormone X results in transcriptional activation by protein X. The
results of a gel retardation assay are shown here.
[Insert Text Art 17.10]
Explain the action of hormone X.
Answer: The results indicate that protein X binds to the DNA fragment and retards its
mobility (lanes 3 and 4). However, the hormone is not required for DNA binding.
Because we already know that the hormone is needed for transcriptional activation, it
must play some other role. Perhaps the hormone activates a signaling pathway that leads
to the phosphorylation of the transcription factor, and phosphorylation is necessary for
translational activation.
E6.
Explain how the data of Fire and Mellow suggested that double-stranded RNA is
responsible for the silencing of the mex-3 gene.
Answer: When they injected just antisense mex-3 RNA, they observed lower but
detectable levels of mex-3 mRNA. However, the injection of both sense and antisense
RNA, which would form a double-stranded structure, resulted in a complete loss of mex3 mRNA in the cells. In this way, double-stranded RNA would silence the expression of
the mex-3 gene.
E7.
Chapter 19 describes a blotting method known as Northern blotting, in which a
short segment of cloned DNA is used as a probe to detect RNA that is transcribed from a
particular gene. The DNA probe, which is radioactive, is complementary to the RNA that
the researcher wishes to detect. After the radioactive probe DNA binds to the RNA
within a blot of a gel, the RNA is visualized as a dark (radioactive) band on an X-ray
film. The method of Northern blotting can be used to determine the amount of a
particular RNA transcribed in a given cell type. If one type of cell produces twice as
much of a particular mRNA compared to another cell, the band appears twice as intense.
For this question, a researcher has a DNA probe that is complementary to the
ferritin mRNA. This probe can be used to specifically detect the amount of ferritin
mRNA on a gel. A researcher began with two flasks of human skin cells. One flask
contained a very low concentration of iron, and the other flask had a high concentration
of iron. The mRNA was isolated from these cells and then subjected to Northern blotting,
using a probe that is complementary to the ferritin mRNA. The sample loaded in lane 1
was from the cells grown in a low concentration of iron, and the sample in lane 2 was
from the cells grown in a high concentration of iron. Three Northern blots are shown
here, but only one of them is correct. Based on your understanding of ferritin mRNA
regulation, which blot (a, b, or c) would be your expected result? Explain. Which blot (a,
b, or c) would be your expected result if the gel had been probed with a DNA segment
that is complementary to the transferrin receptor mRNA?
[Insert Text Art 17.11]
Answer: The gel shown in (a) is correct for ferritin mRNA regulation. The presence and
absence of iron does not affect the amount of mRNA; it affects the ability of the mRNA
to be translated. The gel shown in (b) is correct for transferrin receptor mRNA. In the
absence of iron, the IRP binds to the IRE in the mRNA and stabilizes it, so the mRNA
level is high. In the presence of iron, the iron binds to IRP and IRP is released from the
IRE, and the transferrin receptor mRNA is degraded.