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10
Gene Isolation and Manipulation
WORKING WITH THE FIGURES
1.
Figure 10-1 shows that specific DNA fragments can be synthesized in vitro
prior to cloning. What are two ways to synthesize DNA inserts for recombinant
DNA in vitro?
Answer: DNA can be amplified from genomic sequences in vitro using the
polymerase chain reaction (PCR) or by copying the mRNA sequences into
cDNA using reverse transcriptase.
2.
In Figure 10-4, why is cDNA made only from mRNA and not also from tRNAs
and ribosomal RNAs?
Answer: cDNA is made from mRNA and not from tRNAs or rRNAs because
polyT primers are used to prime the first DNA strand synthesis. Only the
polyadenylated mRNAs will anneal to the primers.
3.
Redraw Figure 10-6 with the goal of adding one EcoRI end and one XhoI end.
Below is the Xhol recognition sequence.
Recognition sequence:
...CTCGAG...
...GAGCTC...
After cut:
...CTCGAG...
...GAGCTC...
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Answer:
4.
Redraw Figure 10-7 so that the cDNA can insert into an XhoI site of a vector
rather than into an EcoRI site as shown.
Answer:
5.
In Figure 10-11, determine approximately how many BAC clones are needed to
provide 1 × coverage of
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a. the yeast genome (12 Mbp).
b. the E. coli genome (4.1 Mbp).
c. the fruit-fly genome (130 Mbp).
Answer:
a. The yeast genome is approximately 12 Mb, if the average insert size is
100-200 kb, then 60-120 BAC clones will give 1X coverage.
b. For E. coli, only 21 to 41 clones will be required.
c. For fruitfly, 650-1300 clones will be required.
6.
In Figure 10-15, why does DNA migrate to the anode (+ pole)?
Answer: DNA moves toward the positive pole in agarose gel electrophoresis
because it is negatively charged.
7.
In Figure 10-18(a), why are DNA fragments of different length and all ending in
an A residue synthesized?
Answer: As DNA is synthesized in the sequencing reaction, the enzyme will
randomly insert either dATP or ddATP across from T residues. If ddATP is
selected, then the chain will terminate, as there is no 3OH available for addition
of the next nucleotide. As a result, the reaction will include fragments that have
terminated at every position where an A is required.
8.
As you will see in Chapter 15, most of the genomes of higher eukaryotes (plants
and animals) are filled with DNA sequences that are present in hundreds, even
thousands, of copies throughout the chromosomes. In the chromosome-walking
procedure shown in Figure 10-20, how would the experimenter know whether
the fragment he or she is using to “walk” to the next BAC or phage is
repetitive? Can repetitive DNA be used in a chromosome walk?
Answer: The researcher would find that the fragment he or she was using as a
probe to identify adjacent sequences would hybridize to many non-overlapping
clones. Such repetitive sequences could not be used in a chromosome walk,
since the repeated sequence would be present in many different genomic
locations and it would be impossible to determine which was the correct one.
9.
Redraw Figure 10-24 to include the positions of the single and double
crossovers.
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Answer:
10. In Figure 10-26, why do only plant cells that have T-DNA inserts in their
chromosomes grow on the agar plates? Do all of the cells of a transgenic plant
grown from one clump of cells contain T-DNA? Justify your answer.
Answer: The T-DNA carries the kanamycin resistance gene; therefore, only
cells that have acquired T-DNA inserts will grow on the agar plates containing
the drug that selects for KanR.
11. In Figure 10-28, what is the difference between extra-chromosomal DNA and
integrated arrays of DNA? Are the latter ectopic? What is distinctive about the
syncitial region that makes it a good place to inject DNA?
Answer: Extrachromosomal arrays are maintained independently of the C.
elegans chromosomes, while the integrated arrays become incorporated into the
genome. The integrated arrays are ectopic, as they do not integrate into the
homologous sequences in their normal chromosomal locus. The syncitial regio
is a good place to inject DNA because there are a large number of nuclei in
shared cytoplasm, any of which can take up the injected DNA. In addition,
these cells will become egg or sperm, so the introduced genes will be passed on
to individuals in the next generation.
BASIC PROBLEMS
12.
From this chapter, make a list of all the examples of
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(a) the hybridization of single-stranded DNAs and
(b) proteins that bind to DNA and then act on it.
Answer:
a. The following are examples of hybridization of single-stranded DNAs
discussed in this chapter: use of primers (DNA sequencing, PCR),
hybridization of sticky ends in cloning, probe hybridization (Southern
blotting, Northern blotting, diagnosis of mutations, identification of clones,
and in situ hybridization).
b. The following are examples of proteins that bind to and act on DNA
discussed in this chapter: restriction enzymes, DNA polymerase (Taq
polymerase), reverse transcriptase, and DNA ligase.
13.
Compare and contrast the use of the word recombinant as used in the phrases
(a) “recombinant DNA” and
(b) “recombinant frequency.”
Answer:
a. Recombinant DNA is the term used to describe techniques used to generate
hybrid molecules consisting of DNA from two different sources. In this use,
recombinant refers to the artificial joining of DNA from different sources.
b. Recombinant frequency is used to describe the statistical likelihood that two
loci in the same organism will assort together or separately during meiosis.
Here, recombinant refers to the combinations of loci that are different from
the parent.
14. Why is ligase needed to make recombinant DNA? What would be the
immediate consequence in the cloning process if someone forgot to add it?
Answer: Ligase is an essential enzyme within all cells that seals breaks in the
phosphate-sugar backbone of DNA. During DNA replication it joins Okazaki
fragments to create a continuous strand, and in cloning, it is used to join the
various DNA fragments with the vector. If it was not added, the vector and
cloned DNA would simply fall apart.
15.
In the PCR process, if we assume that each cycle takes 5 minutes, how
manyfold amplification would be accomplished in 1 hour?
Answer: Each cycle takes 5 minutes and doubles the DNA. In 1 hour there
would be 12 cycles, so the DNA would be amplified 212 = 4096-fold
amplification.
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16.
The position of the gene for the protein actin in the haploid fungus Neurospora
is known from the complete genome sequence. If you had a slow-growing
mutant that you suspected of being an actin mutant and you wanted to verify
that it was one, would you
(a) clone the mutant by using convenient restriction sites flanking the actin gene
and then sequence it or
(b) amplify the mutant sequence by using PCR and then sequence it?
Answer: The great advantage of PCR is that fewer procedures are necessary
compared with cloning. However, it requires that the sequence be known, and
that the primers are each present once in the genome and are sufficiently close
(less than 2 kb). If these conditions are met, the primers determine the
specificity of the DNA segment amplified and thus would be most efficient.
17.
You obtain the DNA sequence of a mutant of a 2-kb gene in which you are
interested and it shows base differences at three positions, all in different
codons. One is a silent change, but the other two are missense changes (they
encode new amino acids). How would you demonstrate that these changes are
real mutations and not sequencing errors? (Assume that sequencing is about
99.9 percent accurate.)
Answer: Resequencing the relevant gene should be done, as this will tell you if
the original sequence was correct. You can then use the mutant sequence in a
gene replacement experiment (depending on the organism) to see if the mutant
phenotype is actually the result of the sequence variation you detected.
18.
In a T-DNA transformation of a plant with a transgene from a fungus (not found
in plants), the presumptive transgenic plant does not express the expected
phenotype of the transgene. How would you demonstrate that the transgene is in
fact present? How would you demonstrate that the transgene was expressed?
Answer: You could isolate DNA from the suspected transgenic plant and probe
for the presence of the transgene by Southern hybridization.
19.
How would you produce a mouse that is homozygous for a rat growth- hormone
transgene?
Answer: Inject the rat growth hormone gene (RGH) into fertilized eggs of mice.
These eggs are then implanted into a surrogate mother and any resulting
offspring mated to test their offspring to see if any are transgenic for RGH.
Because RGH will be inherited in a dominant fashion, any large and transgenic
offspring will be heterozygous at this point. The transgenic siblings will need to
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be mated to each other in order to generate mice that are homozygous for the
RGH transgene.
20.
Why was cDNA and not genomic DNA used in the commercial cloning of the
human insulin gene?
Answer: The commercial cloning of insulin was into bacteria. Bacteria are not
capable of processing introns. Genomic DNA would include the introns, while
cDNA is a copy of processed (and thus intron-free) mRNA.
21.
After Drosophila DNA has been treated with a restriction enzyme, the
fragments are inserted into plasmids and selected as clones in E. coli. With the
use of this “shotgun” technique, every DNA sequence of Drosophila in a
library can be recovered.
a. How would you identify a clone that contains DNA encoding the protein
actin, whose amino acid sequence is known?
b. How would you identify a clone encoding a specific tRNA?
Answer:
a. Because the actin protein sequence is known, a probe could be synthesized
by “guessing” the DNA sequence based on the amino acid sequence. (This
works best if there is a region of amino acids that can be coded with
minimal redundancy.) Alternatively, the gene for actin cloned in another
species can be used as a probe to find the homologous gene in Drosophila.
If an expression vector was used, it might also be possible to detect a clone
coding for actin by screening with actin antibodies.
b. Hybridization using the specific tRNA as a probe could identify a clone
coding for itself.
22.
In any particular transformed eukaryotic cell (say, of Saccharomyces
cerevisiae), how could you tell if the transforming DNA (carried on a circular
bacterial vector)
a. replaced the resident gene of the recipient by double crossing over or single
crossing over?
b. was inserted ectopically?
Answer:
a. The transformed phenotype would map to the same locus. If gene
replacement occurred by a double crossing-over event, the transformed cells
would not contain vector DNA. If a single crossing-over took place, the
entire vector would now be part of the linear Saccharomyces chromosome.
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b. The transformed phenotype would map to a different locus than that of the
auxotroph if the transforming gene was inserted ectopically (i.e., at another
location).
23.
In an electrophoretic gel across which is applied a powerful electrical
alternating pulsed field, the DNA of the haploid fungus Neurospora crassa (n =
7) moves slowly but eventually forms seven bands, which represent DNA
fractions that are of different sizes and hence have moved at different speeds.
These bands are presumed to be the seven chromosomes. How would you show
which band corresponds to which chromosome?
Answer: Size, translocations between known chromosomes, and hybridization
to probes of known location can all be useful in identifying which band on a
PFGE gel corresponds to a particular chromosome.
24.
The protein encoded by the cystic-fibrosis gene is 1480 amino acids long, yet
the gene spans 250 kb. How is this difference possible?
Answer: Conservatively, the amount of DNA necessary to encode this protein
of 445 amino acids is 445  3 = 1335 base pairs. When compared with the
actual amount of DNA used, 60 kb, the gene appears to be roughly 45 times
larger than necessary. This “extra” DNA mostly represents the introns that must
be correctly spliced out of the primary transcript during RNA processing for
correct translation. (There are also comparatively very small amounts of both 5´
and 3´ untranslated regions of the final mRNA that are necessary for correct
translation encoded by this 60-kb of DNA.)
25. In yeast, you have sequenced a piece of wild-type DNA and it clearly contains a
gene, but you do not know what gene it is. Therefore, to investigate further, you
would like to find out its mutant phenotype. How would you use the cloned
wild-type gene to do so? Show your experimental steps clearly.
Answer: The typical procedure is to “knock out” the gene in question and then
see if there is any observable phenotype. One methodology to do this is
described in the companion text. A recombinant vector carrying a selectable
gene within the gene of interest is used to transform yeast cells. Grown under
appropriate conditions, yeast that have incorporated the marker gene will be
selected. Many of these will have the gene of interest disrupted by the selectable
gene. The phenotype of these cells would then be assessed to determine gene
function.
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CHALLENGING PROBLEMS
26.
Prototrophy is often the phenotype selected to detect transformants.
Prototrophic cells are used for donor DNA extraction; then this DNA is cloned
and the clones are added to an auxotrophic recipient culture. Successful
transformants are identified by plating the recipient culture on minimal medium
and looking for colonies. What experimental design would you use to make sure
that a colony that you hope is a transformant is not, in fact,
a. a prototrophic cell that has entered the recipient culture as a contaminant?
b. a revertant (mutation back to prototrophy by a second mutation in the
originally mutated gene) of the auxotrophic mutation?
Answer:
a. To ensure that a colony is not, in fact, a prototrophic contaminant, the
prototrophic line should be sensitive to a drug to which the recipient is
resistant. A simple additional marker would also achieve the same end.
b. Use a nonrevertible auxotroph as the recipient (such as one containing a
deletion.)
27.
A cloned fragment of DNA was sequenced by using the dideoxy chaintermination method. A part of the autoradiogram of the sequencing gel is
represented here.
a. Deduce the nucleotide sequence of the DNA nucleotide chain synthesized
from the primer. Label the 5′ and 3′ ends.
b. Deduce the nucleotide sequence of the DNA nucleotide chain used as the
template strand. Label the 5′ and 3′ ends.
c. Write out the nucleotide sequence of the DNA double helix (label the 5′ and
3′ ends).
d. How many of the six reading frames are “open” as far as you can tell?
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Answer:
a. The gel can be read from the bottom to the top in a 5´-to-3´ direction. The
sequence is
5´ T T C G A A A G G T G A C C C C T G G A C C T T T A G A 3´
b. By complementarity, the template was
3´ A A G C T T T C C A C T G G G G A C C T G G A A A T C T 5´
c. The double helix is
5´ T T C G A A A G G T G A C C C C T G G A C C T T T A G A 3´
3´ A A G C T T T C C A C T G G G G A C C T G G A A A T C T 5´
d. Open reading frames have no stop codons. There are three frames for each
strand, for a total of six possible reading frames. For the strand read from
the gel, the transcript would be
5´ U C U A A A G G U C C A G G G G U C A C C U U U C G A A 3´
And for the template strand
5´ U U C G A A A G G U G A C C C C U G G A C C U U U A G A 3´
Stop codons are in bold and underlined. The two stop codons in the mRNA
read from the template strand are both in the same reading frame. There are
a total of four open reading frames of the six possible.
28.
The cDNA clone for the human gene encoding tyrosinase was radioactively
labeled and used in a Southern analysis of EcoRI-digested genomic DNA of
wild-type mice. Three mouse fragments were found to be radioactive (were
bound by the probe). When albino mice were used in this Southern analysis, no
genomic fragments bound to the probe. Explain these results in relation to the
nature of the wild-type and mutant mouse alleles.
Answer: The region of DNA that encodes tyrosinase in “normal” mouse
genomic DNA contains two EcoRI sites. Thus, after EcoRI digestion, three
different-sized fragments hybridize to the cDNA clone. When genomic DNA
from certain albino mice is subjected to similar analysis, there are no DNA
fragments that contain complementary sequences to the same cDNA. This
indicates that these mice lack the ability to produce tyrosinase because the DNA
that encodes the enzyme must be deleted.
310 Chapter Ten
29.
Transgenic tobacco plants were obtained in which the vector Ti plasmid was
designed to insert the gene of interest plus an adjacent kanamycin-resistance
gene. The inheritance of chromosomal insertion was followed by testing
progeny for kanamycin resistance. Two plants typified the results obtained
generally. When plant 1 was backcrossed with wild-type tobacco, 50 percent of
the progeny were kanamycin resistant and 50 percent were sensitive. When
plant 2 was backcrossed with the wild type, 75 percent of the progeny were
kanamycin resistant and 25 percent were sensitive. What must have been the
difference between the two transgenic plants? What would you predict about the
situation regarding the gene of interest?
Answer: Plant 1 shows the typical inheritance for a dominant gene that is
heterozygous. Assuming kanamycin resistance is dominant to kanamycin
sensitivity, the cross can be outlined as follows:

kanS/kanS

1/2 kanR/kanS
1/2 kanS/kanS
kanR/kanS
This would suggest that the gene of interest would be inserted once into the
genome.
Plant 2 shows a 3:1 ratio in the progeny of the backcross. This suggests that
there have been two unlinked insertions of the kanR gene and presumably the
gene of interest as well.

kanS1/kanS1 ; kanS2/kanS2

1/4 kanR1/kanS1 ; kanR2/kanS2
1/4 kanR1/kanS1 ; kanS2/kanS2
1/4 kanS1/kanS1 ; kanR2/kanS2
1/4 kanS1/kanS1 ; kanS2/kanS2
kanR1/kanS1 ; kanR2/kanS2
30.
A cystic-fibrosis mutation in a certain pedigree is due to a single nucleotide-pair
change. This change destroys an EcoRI restriction site normally found in this
position. How would you use this information in counseling members of this
family about their likelihood of being carriers? State the precise experiments
needed. Assume that you find that a woman in this family is a carrier, and it
transpires that she is married to an unrelated man who also is a heterozygote for
cystic fibrosis, but, in his case, it is a different mutation in the same gene. How
would you counsel this couple about the risks of a child’s having cystic
fibrosis?
Answer: Assuming that the DNA from this region is cloned, it could be used as
a probe to detect this RFLP on Southern blots. DNA from individuals within
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this pedigree would be isolated (typically from blood samples containing white
blood cells) and restricted with EcoRI, and Southern blots would be performed.
Individuals with this mutant CF allele would have one band that would be larger
(owing to the missing EcoRI site) when compared with wild type. Individuals
that inherited this larger EcoRI fragment would, at minimum, be carriers for
cystic fibrosis. In the specific case discussed in this problem, a woman that is
heterozygous for this specific allele marries a man that is heterozygous for a
different mutated CF allele. Just knowing that both are heterozygous, it is
possible to predict that there is a 25 percent chance of their child’s having CF.
However, because the mother’s allele is detectable on a Southern blot, it would
be possible to test whether the fetus inherited this allele. DNA from the fetus
(through either CVS or amniocentesis) could be isolated and tested for this
specific EcoRI fragment. If the fetus did not inherit this allele, there would be a
0 percent chance of its having CF. On the other hand, if the fetus inherited this
allele, there would be a 50 percent chance the child will have CF.
31.
Bacterial glucuronidase converts a colorless substance called X-Gluc into a
bright blue indigo pigment. The gene for glucuronidase also works in plants if
given a plant promoter region. How would you use this gene as a reporter gene
to find the tissues in which a plant gene that you have just cloned is normally
active? (Assume that X-Gluc is easily taken up by the plant tissues.)
Answer: The promoter and control regions of the plant gene of interest must be
cloned and joined in the correct orientation with the glucuronidase gene. This
places the reporter gene under the same transcriptional control as the gene of
interest. The companion text discusses the methodology used to create
transgenic plants. Transform plant cells with the reporter gene construct, and as
discussed in the text, grow into transgenic plants. The glucuronidase gene will
now be expressed in the same developmental pattern as the gene of interest and
its expression can easily be monitored by bathing the plant in an X-Gluc
solution and assaying for the blue reaction product.
32.
The plant Arabidopsis thaliana was transformed by using the Ti plasmid into
which a kanamycin-resistance gene had been inserted in the T-DNA region.
Two kanamycin-resistant colonies (A and B) were selected, and plants were
regenerated from them. The plants were allowed to self-pollinate, and the
results were as follows:
Plant A selfed → 3/4 progeny resistant to kanamycin
1/4 progeny sensitive to kanamycin
Plant B selfed → 5/16 progeny resistant to kanamycin
1/16 progeny sensitive to kanamycin
a. Draw the relevant plant chromosomes in both plants.
b. Explain the two different ratios.
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Answer:
a. and b. During Ti plasmid transformation, the kanamycin gene will insert
randomly into the plant chromosomes. Colony A, when selfed, has 3/4
kanamycin-resistant progeny, and colony B, when selfed, has 15/16
kanamycin-resistant progeny. This suggests that there was a single insertion
into one chromosome in colony A and two independent insertions on
separate chromosomes in colony B. This can be schematically represented
by showing a single insertion within one of the pair of chromosome “A” for
colony A.
Chromosomes “A”
T-DNA
T-DNA
kanamycinR
and two independent insertions into one of each of the pairs of
chromosomes “B” and “C” for colony B.
Chromosomes “B”
T-DNA
T-DNA
kanamycin R
Chromosomes “C”
T-DNA
T-DNA
kanamycin R
Genetically, this can be represented as
Colony A
Colony B
kanRA/kanSA
kanRB/kanSB ; kanRC/kanSC
When these are selfed
kanRA/kanSA  kanRA/kanSA

1/4 kanRA/kanRA
1/2 kanRA/kanSA
1/4 kanSA/kanSA
kanRB/kanSB ; kanRC/kanSC  kanRB/kanSB ; kanRC/kanSC

9/16 kanRB/– ; kanRC/–
3/16 kanRB/– ; kanSC/kanSC
3/16 kanSB/kanSB ; kanRC/–
1/16 kanSB/kanSB ; kanSC/kanSC