CHAPTER 21
Application and Experimental Questions
E1.
Would the following methods be described as linkage, cytogenetic, or physical
mapping?
A.
Fluorescence in situ hybridization (FISH)
B.
Conducting dihybrid crosses to compute map distances
C.
Chromosome walking
D.
Examination of polytene chromosomes in Drosophila
E.
Use of RFLPs in crosses
F.
Using BACs and cosmids to construct a contig
Answer:
A.
Cytogenetic mapping
B.
Linkage mapping
C.
Physical mapping
D.
Cytogenetic mapping
E.
Linkage mapping
F.
Physical mapping
E2.
In an in situ hybridization experiment, what is the relationship between the
sequence of the probe DNA and the site on the chromosomal DNA where the probe
binds?
Answer: They are complementary to each other.
E3.
Describe the technique of in situ hybridization. Explain how it can be used to map
genes.
Answer: In situ hybridization is a cytological method of mapping. A probe that is
complementary to a chromosomal sequence is used to locate the gene microscopically
within a mixture of many different chromosomes. Therefore, it can be used to
cytologically map the location of a gene sequence. When more than one probe is used,
the order of genes along a particular chromosome can be determined.
E4.
The cells from a malignant tumor were subjected to in situ hybridization using a
probe that recognizes a unique sequence on chromosome 14. The probe was detected
only once in each of these cells. Explain these results and speculate on their significance
with regard to the malignant characteristics of these cells.
Answer: Because normal cells contain two copies of chromosome 14, one would expect
that a probe would bind to complementary DNA sequences on both of these
chromosomes. If a probe recognized only one of two chromosomes, this means that one
of the copies of chromosome 14 has been lost, or it has suffered a deletion in the region
where the probe binds. With regard to cancer, the loss of this genetic material may be
related to the uncontrollable cell growth.
E5.
Figure 21.2 describes the technique of FISH. Why is it necessary to “fix” the cells
(and the chromosomes inside of them) to the slides? What does it mean to fix them? Why
is it necessary to denature the chromosomal DNA?
Answer: The term fixing refers to procedures that chemically freeze cells and prevent
degradation. After fixation has occurred, the contents within the cells do not change their
morphology. In a sense, they are frozen in place. For a FISH experiment, this keeps all
the chromosomes within one cell in the vicinity of each other; they cannot float around
the slide and get mixed up with chromosomes from other cells. Therefore, when we see a
group of chromosomes in a FISH experiment, this group of chromosomes comes from a
single cell.
It is necessary to denature the chromosomal DNA so that the probe can bind to it.
The probe is a segment of DNA that is complementary to the DNA of interest. The
strands of chromosomal DNA must be separated (i.e., denatured) so that the probe can
bind to complementary sequences.
E6.
Explain how the use of DNA probes with different fluorescence emission
wavelengths can be used in a single FISH experiment to map the locations of two or more
genes. This method is called chromosome painting. Explain why this is an appropriate
term.
Answer: After the cells and chromosomes have been fixed to the slide, it is possible to
add two or more different probes that recognize different sequences (i.e., different sites)
within the genome. Each probe has a different fluorescence emission wavelength.
Usually, a researcher will use computer imagery that recognizes the wavelength of each
probe and then assigns that probe a bright color. The color seen by the researcher is not
the actual color emitted by the probe; it is a secondary color assigned by the computer. In
a sense, the probes, with the aid of a computer, are “painting” the regions of the
chromosomes that are recognized by a probe. An example of chromosome painting is
shown in Figure 21.3. In this example, human chromosome 5 is painted with six different
colors.
E7.
A researcher is interested in a gene that is found on human chromosome 21.
Describe the expected results of a FISH experiment using a probe that is complementary
to this gene. How many spots would you see if the probe was used on a sample from a
normal individual versus an individual with Down syndrome?
Answer: If the sample was from an unaffected individual, two spots (one on each copy of
chromosome 21) would be observed. Three spots would be observed if the sample was
from a person with Down syndrome, because the person has three copies of chromosome
21.
E8. What is a contig? Explain how you would determine that two clones in a contig are
overlapping.
Answer: A contig is a collection of clones that contain overlapping segments of DNA
that span a particular region of a chromosome. To determine if two clones are
overlapping, one could conduct a Southern blotting experiment. In this approach, one of
the clones is used as a probe. If it is overlapping with the second clone, it will bind to it in
a Southern blot. Therefore, the second clone is run on a gel and the first clone is used as a
probe. If the band corresponding to the second clone is labeled, this means that the two
clones are overlapping.
E9.
Contigs are often made using BAC or cosmid vectors. What are the advantages
and disadvantages of these two types of vector? Which type of contig would you make
first, a BAC or cosmid contig? Explain.
Answer: A BAC vector can contain extremely large pieces of DNA, so it is used as a first
step to align the segments of DNA in a physical mapping study. However, it is difficult to
work with them in subcloning and DNA sequencing experiments. Cosmids, by
comparison, contain smaller segments of the genome. The locations of cosmids can be
determined by hybridizing them to BACs. The cosmids can then be used for subcloning
and DNA sequencing.
E10. Describe the molecular features of a BAC cloning vector. What is the primary
advantage of a BAC compared to plasmid or viral vectors?
Answer: BAC cloning vectors have the replication properties of a bacterial chromosome
and the cloning properties of a plasmid. To replicate like a chromosome, the BAC vector
contains an origin of replication from an F factor. Therefore, in a bacterial cell, a BAC
can behave as a chromosome. Like a plasmid, BACs also contain selectable markers and
convenient cloning sites for the insertion of large segments of DNA. The primary
advantage is the ability to clone very large pieces of DNA.
E11. In general terms, what is a polymorphism? Explain the molecular basis for a
restriction fragment length polymorphism (RFLP). How is an RFLP detected
experimentally? Why are RFLPs useful in physical mapping studies? How can they be
used to clone a particular gene?
Answer: A polymorphism refers to genetic variation at a particular locus within a
population. If the polymorphism occurs within gene sequences, this is allelic variation. A
polymorphism can also occur within genetic markers such as RFLPs. The molecular basis
for an RFLP is that two distinct individuals will have variation in their DNA sequences
and some of the variation may affect the relative locations of restriction enzyme sites.
Because this occurs relatively frequently between unrelated individuals, many RFLPs can
be identified. They can be detected by restriction digestion and gel electrophoresis and
then Southern blotting. They are useful in mapping studies because it is relatively easy to
find many of them along a chromosome where they can serve as points of reference in
genetic maps. In this regard, they can be used in gene cloning as a starting point for a
chromosomal walk.
E12. A woman has been married to two different men and produced five children. This
group is analyzed with regard to three different STSs: STS-1 is 146 and 122 bp; STS-2 is
102 and 88 bp, and STS-3 is 188 and 204 bp. The mother is homozygous for all three
STSs: STS-1 = 122, STS-2 = 88, and STS-3 = 188. Father 1 is homozygous for STS-1 =
122 and STS-2 = 102, and heterozygous for STS-3 = 188/204. Father 2 is heterozygous
for STS-1 = 122/146, STS-2 = 88/102, and homozygous for STS-3 = 204. The five
children have the following results:
[Insert Text Art 21.4]
Which children can you definitely assign to father 1 and father 2?
Answer: Child 1 and child 3 belong to father 2; child 2 and child 4 belong to father 1;
child 5 could belong to either father.
E13. An experimenter used primers to nine different STSs to test their presence along
five different BAC clones. The results are shown here.
Alignment of STSs and BACs
STSs
1
2
3
4
5
6
7
8
9
BACs
1
–
–
–
–
–
+
+
–
+
2
+
–
–
–
+
–
–
+
–
3
–
–
–
+
–
+
–
–
–
4
–
+
+
–
–
–
–
+
–
5
–
–
+
–
–
–
–
–
+
Make a contig map that describes the alignment of the five BACs.
Answer:
Deduced Outcome
STSs
4
6
7
9
3
2
8 5 1
BAC
3
<—————>
1
<————————>
5
<—————>
4
<——————>
2
<———————>
E14. In the Human Genome Project, researchers have collected linkage data from many
crosses in which the male was heterozygous for markers and many crosses where the
female was heterozygous for markers. The distance between the same two markers,
computed in map units or centiMorgans, is different between males and females. In other
words, the linkage maps for human males and females are not the same. Propose an
explanation for this discrepancy. Do you think the sizes of chromosomes (excluding the
Y chromosome) in human males and females are different? How could physical mapping
resolve this discrepancy?
Answer: An explanation is that the rate of recombination between homologous
chromosomes is different during oogenesis compared to spermatogenesis. Physical
mapping measures the actual distance (in bp) between markers. The physical mapping of
chromosomes in males and females reveals that they are the same lengths. Therefore, the
sizes of chromosomes in males and females are the same. The differences obtained in
linkage maps are due to differences in the rates of recombination during oogenesis versus
spermatogenesis.
E15. Take a look at solved problem S4. Let’s suppose a male is heterozygous for two
polymorphic sequence-tagged sites. STS-1 exists in two sizes: 211 bp and 289 bp. STS-2
also exists in two sizes: 115 bp and 422 bp. A sample of sperm was collected from this
man, and individual sperm were placed into 30 separate tubes. Into each of the 30 tubes
were added the primers that amplify STS-1 and STS-2, and then the samples were
subjected to PCR. The following results were obtained:
[Insert Text Art 21.5]
A.
What is the arrangement of these two sequence-tagged sites in this
individual?
B.
What is the linkage distance between STS-1 and STS-2?
C.
Could this approach of analyzing a population of sperm be applied to
RFLPs?
Answer:
A.
One homologue contains the STS-1 that is 289 bp and STS-2 that is 422 bp, while
the other homologue contains STS-1 that is 211 bp and STS-2 that is 115 bp. This is
based on the observation that 28 of the sperm have either the 289 bp and 422 bp bands or
the 211 bp and 115 bp bands.
B.
There are two recombinant sperm; see lanes 12 and 18. Because there are two
recombinant sperm out of a total of thirty:
2
 100
30
= 6.7 mu
Map distance =
C.
In theory, this method could be used. However, there is not enough DNA in one
sperm to carry out an RFLP analysis unless the DNA is amplified by PCR.
E16. Compared to a conventional plasmid, what additional sequences are required in a
YAC vector so it can behave like an artificial chromosome? Describe the importance of
each required sequence.
Answer: Besides a selectable marker and an origin that will replicate in E. coli, YAC
vectors also require two telomere sequences, a centromere sequence, and an ARS
sequence. The telomeres are needed to prevent the shortening of the artificial
chromosome from the ends. The centromere sequence is needed for the proper
segregation of the artificial chromosome during meiosis and mitosis. The ARS sequence
is the yeast equivalent of an origin of replication, which is needed so the YAC DNA can
be replicated.
E17. When conducting physical mapping studies, place the following methods in their
most logical order:
A.
Clone large fragments of DNA to make a BAC library.
B.
Determine the DNA sequence of subclones from a cosmid library.
C.
Subclone BAC fragments to make a cosmid library.
D.
Subclone cosmid fragments for DNA sequencing.
Answer: The proper order is A, C, D, B.
1. Clone large fragments of DNA to make a BAC library.
2. Subclone BAC fragments to make a cosmid library.
3. Subclone cosmid fragments for DNA sequencing.
4. Determine the DNA sequence of subclones from a cosmid library.
E18. Four cosmid clones, which we will call cosmid A, B, C, and D, were subjected to
a Southern blot in pairwise combinations. The insert size of each cosmid was also
analyzed. The following results were obtained:
Cosmid
Insert Size (bp)
Hybridized to?
A
6,000
C
B
2,200
C, D
C
11,500
A, B, D
D
7,000
B, C
Draw a map that shows the order of the inserts within these four cosmids.
Answer: Note that the insert of cosmid B is contained completely within the insert of
cosmid C.
E19. What is an STS (sequence-tagged site)? How are STSs generated experimentally?
What are the uses of STSs? Explain how a microsatellite can produce a polymorphic
STS.
Answer: A sequence-tagged site (STS) is a segment of DNA, usually quite short (e.g.,
100 to 400 bp in length), that serves as a unique site in the genome. STSs are identified
using primers in a PCR reaction. STSs serve as molecular markers in genetic mapping
studies. Sometimes the region within an STS may contain a microsatellite. A
microsatellite is a short DNA segment that is variable in length, usually due to a short
repetitive sequence. When a microsatellite is within an STS, the length of the STS will
vary among different individuals or even the same individual may be heterozygous for
the STS. This makes the STS polymorphic. Polymorphic STSs can be used in linkage
analysis, because their transmission can be followed in family pedigrees and through
crosses of experimental organisms.
E20. A human gene, which we will call gene X, is located on chromosome 11 and is
found as a normal allele and a recessive disease-causing allele. The location of gene X
has been approximated on the map shown here that contains four STSs, labeled STS-1,
STS-2, STS-3, and STS-4.
STS-1
STS-2
STS-3 Gene X
STS-4
A.
Explain the general strategy of positional cloning.
B.
If you applied the approach of positional cloning to clone gene X, where
would you begin? As you progressed in your cloning efforts, how would you know if you
were walking toward gene X or away from gene X?
C.
How would you know you had reached gene X? (Keep in mind that gene
X exists as a normal allele and a disease-causing allele.)
Answer:
A.
The general strategy is shown in Figure 21.12. The researcher begins at a certain
location and then walks toward the gene of interest. You begin with a clone that has a
marker that is known to map relatively close to the gene of interest. A piece of DNA at
the end of the insert is subcloned and then used in a Southern blot to identify an adjacent
clone in a cosmid DNA library. This is the first “step.” The end of this clone is subcloned
to make the next step. And so on. Eventually, after many steps, you will arrive at your
gene of interest.
B.
In this example, you would begin at STS-3. If you walked a few steps and
happened upon STS-2, you would know that you were walking in the wrong direction.
C.
This is a difficult aspect of chromosome walking. Basically, you would walk
toward gene X using DNA from a normal individual and DNA from an individual with a
mutant gene X. When you have found a site where the sequences are different between
the normal and mutant individual, you may have found gene X. You would eventually
have to confirm this by analyzing the DNA sequence of this region and determining that
it encodes a functional gene.
E21. Describe how you would clone a gene by positional cloning. Explain how a
(previously made) contig would make this task much easier.
Answer: The first piece of information you would start with is the location of a gene or
marker that is known from previous mapping studies to be close to the gene of interest.
You would begin with a clone containing this gene (or marker) and follow the procedure
of chromosome walking to eventually reach the gene of interest. A contig would make
this much easier because you would not have to conduct a series of subcloning
experiments to reach your gene. Instead, you could simply analyze the members of the
contig.
E22. Discuss the general differences between hierarchical shotgun sequencing versus
whole genome shotgun sequencing.
Answer: In the hierarchical approach, the genome is mapped to create a YAC or BAC
contig for each chromosome. The inserts within each contig are then subjected to shotgun
sequencing. By comparison, in the whole genome shotgun sequencing approach, the
mapping step is eliminated. Instead, the whole genome is chopped up into small pieces,
which are sequenced from both ends. This latter method is quicker because it eliminates
the mapping step, which can be quite time consuming.
E23. Discuss the advantages of next-generation sequencing techologies.
Answer: The overall advantage of next-generation sequencing is that a large amount of
DNA sequence can be obtained in a short period of time, and at a lower cost. One
common innovation is that the need to subclone fragments of DNA into vectors is no
longer necessary. Also, some of the methods involve the parallel analysis of an enormous
number of samples simultaneously.
E24. What is meant by “sequence by synthesis”?
Answer: Sequence by synthesis is a form of next-generation DNA sequencing in which
the base sequence is determined as the DNA strand is being made. Pyrosequencing is an
example.