Chapter Eighteen: Recombinant DNA Technology
COMPREHENSION QUESTIONS
1.
List some of the affects and applications of recombinant DNA technology.
Recombinant DNA technology has had profound effects on all fields of biology.
Whole genomes have been sequenced, structures of genes elucidated, the patterns of
molecular evolution studied. Recombinant DNA technology is now used to diagnose
and screen for genetic diseases, and gene therapy is being explored. Recombinant
DNA is used to make pharmaceutical products such as recombinant insulin and
clotting factors. Genetically modified organisms will change the lives of farmers
and improve agricultural productivity and the quality of food and fiber.
2.
What common feature is seen in the sequences recognized by type II restriction
enzymes?
The recognition sequences are palindromic and 4–8 base pairs long.
3.
What role do restriction enzymes play in bacteria? How do bacteria protect their
own DNA from the action of restriction enzymes?
Restriction enzymes cut foreign DNA, such as viral DNA, into fragments. Bacteria
protect their own DNA by modifying bases, usually by methylation, at the
recognition sites.
*4. Explain how gel electrophoresis is used to separate DNA fragments of different
lengths.
Gel electrophoresis acts as a molecular sieve. The gel is an aqueous matrix of
agarose or polyacrylamide. DNA molecules are loaded into a slot or well at one
end of the gel. When an electric field is applied, the negatively charged DNA
molecules migrate toward the positive electrode. Shorter DNA molecules are less
hindered by the agarose or polyacrylamide matrix and migrate faster than longer
DNA molecules, which must wind their way around obstacles and through the pores
in the gel matrix.
*5. After DNA fragments are separated by gel electrophoresis, how can they be
visualized?
DNA molecules can be visualized by staining with a fluorescent dye. Ethidium
bromide intercalates between the stacked bases of the DNA double helix, and the
ethidium bromide-DNA complex fluoresces orange when irradiated with an
ultraviolet light source. Alternatively, they can be visualized by attaching
radioactive or chemical labels to the DNA before it is placed in the gel.
6.
What is the purpose of Southern blotting? How is it carried out?
Southern blotting is used to detect and visualize specific DNA fragments that have a
sequence complementary to a labeled DNA probe. DNA is first cleaved into
fragments with restriction endonucleases. The fragments are separated by size via
gel electrophoresis. These fragments are then denatured and transferred by blotting
224 Chapter Eighteen: Recombinant DNA Technology
onto the surface of a membrane filter. The membrane filter now has single-stranded
DNA fragments bound to its surface, separated by size as in the gel. The filter is
then incubated with a solution containing a denatured, labeled probe DNA. The
probe DNA hybridizes to its complementary DNA on the filter. After washing excess
unbound probes, the labeled probe hybridized to the DNA on the filter can be
detected using the appropriate methods to visualize the label. For radioactively
labeled probes, the bound probe is detected by exposure to X-ray film.
*7. What are the differences between Southern, Northern, and Western blotting?
In all three techniques, macromolecules are separated by size via gel
electrophoresis and are transferred to the surface of a membrane filter. Southern
blotting transfers DNA, Northern blotting transfers RNA, and Western blotting
transfers protein.
*8. Give three important characteristics of cloning vectors.
Cloning vectors should have:
(1) An origin of DNA replication so they can be maintained in a cell.
(2) A gene, such as antibiotic resistance, to select for cells that carry the
vector.
(3) A unique restriction site or series of sites to cut and ligate a foreign DNA
molecule.
9.
Briefly describe four different methods for inserting foreign DNA into plasmids,
giving the strengths and weaknesses of each.
Restriction cloning: Vector and foreign DNA are cut with the same restriction
enzyme, then ligated together with DNA ligase. This is the most straightforward,
simplest method of cloning. The disadvantage is that matching restriction sites may
not be available. Ligation also produces undesirable products, such as the vector
ligating to itself, without the foreign DNA insert.
Tailing: Vector and foreign DNA fragments are blunt-ended either by filling in
or trimming overhanging sequences after restriction digestion or by cutting with a
blunt-cutting restriction endonuclease. Terminal deoxynucleotidyl transferase is
used to add complementary homopolymeric DNA tails to the 3' ends of the vector
and insert. After annealing of the insert to the vector by the complementary tails,
the molecules are covalently joined with DNA ligase. This method has the
advantage that the vector cannot self-ligate. The disadvantage is that the procedure
destroys the restriction site in the vector, so the insert cannot be cut out. Also, the
added tail sequences may interfere with the function of either the foreign DNA
insert or the vector.
Ligation of blunt ends with T4 DNA ligase: Any two blunt ends of DNA
fragments can be ligated, so the method does not depend on the presence of suitable
restriction sites. The disadvantage is that blunt-end ligations are inefficient, and
vector self-ligation is difficult to control.
Addition of linkers: Small synthetic DNA sequences containing restriction sites
are ligated to blunt ends of DNA fragments. Restriction digestion then generates
sticky ends that can be ligated as in restriction cloning. The advantage is that,
Chapter Eighteen: Recombinant DNA Technology 225
again, the method does not depend on the availability of suitable restriction sites;
any restriction site can be added. The disadvantage is that this involves extra steps
and is more difficult than simple restriction cloning.
10. How are plasmids transferred into bacterial cells?
Plasmids are transferred into bacterial cells by transformation. Most laboratory
strains of E. coli require chemical treatment to render them competent to take up
DNA by transformation. Another common alternative is electroporation, using an
electric discharge to drive DNA into cells.
*11. Briefly explain how an antibiotic-resistance gene and the lacZ gene can be used as
markers to determine which cells contain a particular plasmid.
Many plasmids designed as cloning vectors carry a gene for antibiotic resistance
and the lacZ gene. The lacZ gene on the plasmid has been engineered to contain
multiple unique restriction sites. Foreign DNAs are inserted into one of the unique
restriction sites in the lacZ gene. After transformation, E. coli cells carrying the
plasmid are plated on a medium containing the appropriate antibiotic to select for
cells that carry the plasmid. The medium also contains an inducer for the lac
operon, so the cells express the lacZ gene, and X-gal, a substrate for galactosidase that will turn blue when cleaved by -galactosidase. The colonies
that carry plasmid without foreign DNA inserts will have intact lacZ genes, make
functional -galactosidase, cleave X-gal, and turn blue. Colonies that carry
plasmid with foreign DNA inserts will not make functional -galactosidase because
the lacZ gene is disrupted by the foreign DNA insert. They will remain white. Thus,
cells carrying plasmids with inserts will form white colonies.
12. How are genes inserted into bacteriophage  vectors? What advantages do  vectors
have over plasmids?
 vectors are prepared by removing the nonessential  of the genome from the
center, leaving two “arms.” The two arms are ligated to either side of a foreign
DNA molecule and then packaged into phage particles.  vectors are very efficient
at delivering DNA into E. coli cells. Moreover, they have the capacity to carry
foreign DNA fragments up to 23 kb in length. Finally, because only DNA molecules
40–50 kb can be packaged into a  phage particle, self-ligated vectors or inserts by
themselves will not be packaged or delivered into cells.
*13. What is a cosmid? What are the advantages of using cosmids as gene vectors?
A cosmid is a plasmid vector with a plasmid origin of DNA replication, unique
restriction sites for cloning, and selectable marker genes that also has the lambda
cos site so it can be packaged into  phage particles for efficient delivery into E.
coli cells. It can accommodate DNA fragments up to 44 kb.
14. What are yeast artificial chromosomes and shuttle vectors? When are these cloning
vectors used?
Yeast artificial chromosomes (YACs) and shuttle vectors are both used as cloning
vectors to replicate and maintain foreign DNA sequences in eukaryotic cells. Yeast
226 Chapter Eighteen: Recombinant DNA Technology
artificial chromosomes have a yeast origin of DNA replication, a centromeric
sequence, and telomeres. They are used to clone fragments of DNA on the order of
1 million bp in length. Shuttle vectors are plasmids that carry both bacterial and
eukaryotic origins of DNA replication, so they can be replicated in either E. coli or
eukaryotic host cells.
*15. How does a genomic library differ from a cDNA library? How is each created?
A genomic library is generated by cloning fragments of chromosomal DNA into a
cloning vector. Chromosomal DNA is randomly fragmented by shearing or by
partial digestion with a restriction enzyme. A cDNA library is made from mRNA
sequences. Cellular mRNAs are isolated and then reverse transcriptase is used to
copy the mRNA sequences to cDNA, which are cloned into plasmid or phage
vectors.
16. How are probes used to screen DNA libraries? Explain how a synthetic probe can
be prepared when the protein product of a gene is known.
The DNA library must first be plated out, either as colonies for plasmid libraries or
as phage plaques on a bacterial lawn for phage libraries. The colonies or plaques
are transferred to membrane filters. A nucleic acid probe can be used to identify
colonies or phage plaques that contain identical or similar sequences by
hybridization. If no cloned DNA probe is available, but the amino acid sequence of
the protein is known, then degenerate synthetic oligonucleotides can be synthesized
that represent all possible coding sequences for a sequence of 7–10 amino acids.
The synthetic oligonucleotides can be end-labeled and used as hybridization probes
to screen a DNA library.
17. Explain how chromosome walking can be used to find a gene.
Chromosome walking is used isolate DNA clones encoding genes defined solely by
mutations, for which no DNA or amino acid sequence is known. Mapping
experiments locate the gene to a chromosome and to a region of the chromosome.
The chromosome walk begins with a neighboring gene that has been previously
cloned. Clones that overlap this initial gene are isolated, then the overlapping
clones are used to isolate additional overlapping clones, until the set of overlapping
clones spans the chromosomal distance between the previously cloned gene and the
unknown gene.
18. Discuss some of the considerations that must go into developing an appropriate
cloning strategy.
In developing a cloning strategy, one must consider what is known about the gene
to be cloned, the size and nature of the gene, and the ultimate purpose for cloning
the gene. What is known about the gene will determine whether a nucleic acid or
antibody probe can be used or whether chromosome walking must be used. Large
genes with many introns will require vectors capable of handling large inserts,
whereas small genes can be cloned in plasmids. The ultimate purpose, whether for
sequencing or expression and in what host cell, will determine the choice of cloning
vector.
Chapter Eighteen: Recombinant DNA Technology 227
*19. Briefly explain how the polymerase chain reaction is used to amplify a specific
DNA sequence. What are some of the limitations of PCR?
First the double-stranded template DNA is denatured by high temperature. Then
synthetic oligonucleotide primers corresponding to the ends of the DNA sequence to
be amplified are annealed to the single-stranded DNA template strands. These
primers are extended by a thermostable DNA polymerase so that the target DNA
sequence is duplicated. These steps are repeated 30 times or more. Each cycle of
denaturation, primer annealing, and extension results in doubling the number of
copies of the target sequence between the primers.
PCR amplification is limited by several factors. One is that sequence of the gene
to be amplified must be known, at least at the ends of the region to be amplified, in
order to synthesize the PCR primers. Another is that the extreme sensitivity of the
technique renders it susceptible to contamination. A third limitation is that the most
common thermostable DNA polymerase used for PCR, Taq DNA polymerase, has a
relatively high error rate. A fourth limitation is that PCR amplification is usually
limited to DNA fragments of up to a few thousand base pairs; optimized DNA
polymerase mixtures and reaction conditions extend the amplifiable length to
around 20 kb.
*20. Briefly explain in situ hybridization, giving some applications of this technique.
In situ hybridization involves hybridization of radiolabeled or fluorescently labeled
DNA or RNA probes to DNA or RNA molecules that are still in the cell. This
technique can be used to visualize the expression of specific mRNAs in different
cells and tissues and the location of genes on metaphase or polytene chromosomes.
21. What is DNA footprinting?
DNA footprinting is a technique used to determine the binding sites of proteins on
DNA sequences. DNA sequences containing a protein binding site are end-labeled,
and then partially digested with DNase in the presence or absence of the DNA
binding protein. In the absence of the protein, DNase generates a continuous ladder
of fragments generated by DNase cleavage at all possible sites. This ladder of
fragments is visualized by gel electrophoresis and autoradiography. In the presence
of the protein, the binding sites are protected from DNase cleavage, resulting in a
gap in the ladder of DNA fragments. The position of this gap identifies the protein
binding site in the DNA sequence.
22. Briefly explain how site-directed mutagenesis is carried out.
In oligonucleotide-directed mutagenesis, an oligonucleotide containing the desired
mutation in the sequence is synthesized. This mutant oligonucleotide is annealed to
denatured target DNA template and used to direct DNA synthesis. The result is a
double-stranded DNA molecule with a mismatch at the site to be mutated. When
transformed into bacterial cells, bacterial repair enzymes will convert the molecule
to the mutant form about 50% of the time.
228 Chapter Eighteen: Recombinant DNA Technology
*23. What are knockout mice, how are they produced, and for what are they used?
Knockout mice have a target gene disrupted or deleted. First, the target gene is
cloned. The middle portion of the gene is replaced with a selectable marker,
typically the neo gene that confers resistance to G418. This construct is then
introduced back into mouse embryonic stem cells and cells with G418 resistance
are selected. The surviving cells are screened for cells where the chromosomal copy
of the target gene has been replaced with the neo-containing construct by
homologous recombination of the flanking sequences. The embryonic stem cells are
then injected into mouse blastocyst-stage embryos and the chimeric embryos are
transferred to the uterus of a pseudopregnant female mouse. The knockout cells will
participate in the formation of many tissues in the mouse fetus, including germ-line
cells. The chimeric offspring are interbred to produce offspring that are
homozygous for the knockout allele.
The phenotypes of the knockout mice provide information about the function of
the gene.
24. Describe how RFLPs can be used in gene mapping.
RFLPs provide molecular genetic markers that segregate in crosses or pedigrees
like traditional genetic markers, and therefore can be used to map genes. The major
advantage is that RFLPs are highly variable, so that most individuals in the
population are heterozygous. For example, the inheritance of a disease gene in a
human pedigree can be tested for co-inheritance with a set of RFLP markers spaced
several centiMorgans apart on each human chromosome. The genotype of each
individual in the pedigree for the disease gene can be deduced from their
phenotypes and the phenotypes of their progeny, and the RFLP genotypes can be
visualized by Southern blot analysis, probing with each RFLP marker probe.
Statistically significant co-inheritance of the disease allele with an RFLP marker
indicates linkage.
*25. What is DNA fingerprinting? What types of sequences are examined in DNA
fingerprinting?
DNA fingerprinting is the typing of an individual for genetic markers at highly
variable loci. This is useful for forensic investigations, to determine whether the
suspect could have contributed to the evidentiary DNA obtained from blood or
other bodily fluids found at the scene of a crime. Other applications include
paternity testing and the identification of bodily remains.
The RFLP loci traditionally used for DNA fingerprinting are called variable
number of tandem repeat (VNTR) loci; these consist of short tandem repeat
sequences located in introns or spacer regions between genes. The number of
repeat sequences at the locus does not affect the phenotype of the individual in any
way, so these loci are highly variable in the population. More recently, tandem
repeat loci with smaller repeat sequences of just a few nucleotides, called short
tandem repeats (STRs), have been adopted because they can be amplified by PCR.
Chapter Eighteen: Recombinant DNA Technology 229
26. What is gene therapy?
Gene therapy is the correction of a defective gene by either gene replacement or the
addition of a wild-type copy of the gene. For this to work, enough of the cells of the
critically affected tissues or organs must be transformed with the functional copy of
the gene to restore normal physiology.
27. As the first recombinant DNA experiments were being carried out, there was
concern among some scientists about this research. What were these concerns and
how were they addressed?
The initial concerns were that recombinant DNA technology would lead to the
release of novel pathogens that could attack humans or other species. Even the
release of nonpathogenic organisms could have adverse effects on ecological
systems if they outcompete native species. NIH established a set of guidelines for
experiments involving recombinant DNA technology, specifying both biological and
physical containment levels for experiments at different risk levels. Stringent
approval processes are required for the release of any genetically modified
organism into the field.
APPLICATION QUESTIONS AND PROBLEMS
*28. Suppose that a geneticist discovers a new restriction enzyme in the bacterium
Aeromonas ranidae. This restriction enzyme is the first to be isolated from this
bacterial species. Using the standard convention for abbreviating restriction
enzymes, give this new restriction enzyme a name (for help see footnote to Table
18.2).
The first three letters are taken from the genus and species name, and the Roman
numeral indicates the order in which the enzyme was isolated. Therefore, the
enzyme should be named AraI.
29. How often, on average, would you expect a type II restriction endonuclease to cut a
DNA molecule if the recognition sequence for the enzyme had 5 bp? (Assume that
the four types of bases are equally likely to be found in the DNA and that the bases
in a recognition sequence are independent.) How often would the endonuclease cut
the DNA if the recognition sequence had 8 bp?
Since DNA has four different bases, the frequency of any sequence of n bases is
equal to 1/(4n). A 5-bp recognition sequence will occur with a frequency of 1/(45),
or once every 1024 bp. An 8-bp recognition sequence will occur with a frequency of
1 per 48, or 65,536 bp.
*30. A microbiologist discovers a new type II restriction endonuclease. When DNA is
digested by this enzyme, fragments that average 1,048,500 bp in length are
230 Chapter Eighteen: Recombinant DNA Technology
produced. What is the most likely number of base pairs in the recognition sequence
of this enzyme?
Here, 4n = 1,048,500. So n = 10; a 10-bp recognition sequence is most likely.
31. Will restriction sites for an enzyme that has 4 bp in its restriction site be closer
together, farther apart, or similarly spaced, on average, compared with those of an
enzyme that has 6 bp in its restriction site? Explain your reasoning.
The restriction sites for an enzyme with a 4-bp recognition sequence should be
spaced closer together than the sites for an enzyme with a 6-bp recognition
sequence. The 4-bp recognition sequence will occur with an average frequency of
once every 44 = 256 bp, whereas the 6-bp recognition sequence will occur with an
average frequency of once every 46 = 4096 bp.
*32. About 60% of the base pairs in a human DNA molecule are AT. If the human genome has 3
billion base pairs of DNA, about how many times will the following restriction sites be
present?
(a) BamHI (restriction site = 5'—GGATCC—3')
(b) EcoRI (restriction site = 5'—GAATTC—3')
(c) HaeIII (restriction site = 5'—GGCC—3')
We must first calculate the frequency of each base. Given that AT base pairs
comprise 60% of the DNA, we deduce that the frequency of A is 0.3 and frequency
of T is 0.3. The GC base pairs must comprise 40% of the DNA; therefore, the
frequency of G is 0.2 and the frequency of C is 0.2.
(a) BamH1 GGATCC is then (0.2)(0.2)(0.3)(0.3)(0.2)(0.2) = 0.000144
3,000,000,000(0.000144) = 432,000 times.
(b) EcoRI GAATTC = (0.2)(0.3)(0.3)(0.3)(0.3)(0.2) = 0.000324
3,000,000,000(0.000324) = 972,000 times.
(c) HaeIII GGCC = (0.2)(0.2)(0.2)(0.2) = 0.0016
3,000,000,000(0.0016) = 4,800,000 times
*33. Restriction mapping of a linear piece reveals the following EcoRI restriction sites:
EcoRI site 2
EcoRI site 1
2 kb
4 kb
5 kb
(a) This piece of DNA is cut with EcoRI, the resulting fragments are separated by gel
electrophoresis, and the gel is stained with ethidium bromide. Draw a picture of the
bands that will appear on the gel.
(b) If a mutation that alters EcoRI site 1 occurs in this piece of DNA, how will the banding
pattern on the gel differ from the one you drew in part (a)?
(c) If mutations that alter EcoRI site 1 and 2 occur in this piece of DNA, how will the
banding pattern on the gel differ from the one you drew in part (a)?
Chapter Eighteen: Recombinant DNA Technology 231
(d) If a 1000-bp insertion occurred between the two restriction sites, how would the
banding pattern on the gel differ from the one you drew in part (a)?
(e) If a 500-bp deletion occurred between the two restriction sites, how would the banding
pattern on the gel differ from the one you drew in part (a)?
(a) (b) (c) (d) (e)
11 kb
6 kb
5 kb
4 kb
3.5 kb
2 kb
*34. Which vectors (plasmid,  phage, cosmid) can be used to clone a continuous fragment of
DNA with the following lengths?
(a) 4 kb: plasmid.
(b) 20 kb:  phage.
(c) 35 kb: cosmid.
35. A geneticist uses a plasmid for cloning that has a gene that confers resistance to
penicillin and the lacZ gene. The geneticist inserts a piece of foreign DNA into a
restriction site that is located within the lacZ gene and transforms bacteria with the
plasmid. Explain how the geneticist could identify bacteria that contain a copy of a
plasmid with the foreign DNA.
The geneticist should plate the bacteria on agar medium containing penicillin to
select for cells that have taken up the plasmid. The medium should also have X-gal
and an inducer of the lac operon, such as IPTG or even lactose. Cells that have
taken up a plasmid without foreign DNA will have an intact lacZ gene, produce
functional -galactosidase, and cleave X-gal to make a blue dye. These colonies
will turn blue. In contrast, cells that have taken up a plasmid containing a foreign
DNA inserted into the lacZ gene will be unable to make functional -galactosidase.
These colonies will be white.
*36. Suppose that you have just graduated from college and have started working at a
biotechnology firm. Your first job assignment is to clone the pig gene for the
hormone prolactin. Assume that the pig gene for prolactin has not yet been isolated,
sequenced, or mapped; however, the mouse gene for prolactin has been cloned and
the amino acid sequence of mouse prolactin is known. Briefly explain two different
strategies that you might use to find and clone the pig gene for prolactin.
One strategy would be to use the mouse gene for prolactin as a probe to find the
homologous pig gene from a pig genomic or cDNA library.
A second strategy would be to use the amino acid sequence of mouse prolactin
to design degenerate oligonucleotides as hybridization probes to screen a pig DNA
library.
232 Chapter Eighteen: Recombinant DNA Technology
Yet a third strategy would be to use the amino acid sequence of mouse prolactin
to design a pair of degenerate oligonucleotide PCR primers to PCR amplify the pig
prolactin gene.
37. A genetic engineer wants to isolate a gene from a scorpion that encodes the deadly toxin
found in its stinger, with the ultimate purpose of transferring this gene to bacteria and
producing the toxin for use as a commercial pesticide. Isolating the gene requires a DNA
library. Should the genetic engineer create a genomic library or a cDNA library? Explain
your reasoning.
The engineer should create a cDNA library. Bacteria cannot splice introns. If the engineer
wants to express the toxin in bacteria, then she needs a cDNA sequence that has been
reverse transcribed from mRNA, and therefore has no intron sequences.
*38. A protein has the following amino acid sequence:
Met-Tyr-Asn-Val-Arg-Val-Tyr-Lys-Ala-Lys-Trp-Leu-Ile-His-Thr-Pro
You wish to make a set of probes to screen a cDNA library for the sequence that
encodes this protein. Your probes should be at least 18 nucleotides in length.
(a) Which amino acids in the protein should be used to construct the probes so that
the least degeneracy results? (Consult the genetic code in Figure 15.12.)
A probe of 18 nucleotides must be based on 6 amino acids. The 6 amino acid
stretch with the least degeneracy is Val-Tyr-Lys-Ala-Lys-Trp. This sequence
avoids the amino acids arg and leu, which have 6 codons each.
(b) How many different probes must be synthesized to be certain that you will find
the correct cDNA sequence that specifies the protein?
Val and Ala have four codons each, Tyr and Lys have two codons each, and Trp
has one codon. Therefore, there are 4 × 2 × 2 × 4 × 2 × 1 possible sequences,
or 128.
*39. A gene in mice is discovered that is similar to a gene in yeast. How might it be
determined whether this gene is essential for development in mice?
This gene must first be cloned, possibly by using the yeast gene as a probe to screen
a mouse genomic DNA library. The cloned gene is then engineered to replace a
substantial portion of the protein coding sequence with the neo gene. This construct
is then introduced into mouse embryonic stem cells. After transfer to the uterus of a
pseudopregnant mouse, the progeny are tested for the presence of the knockout
allele. Progeny with the knockout allele are interbred. If the gene is essential for
embryonic development, no homozygous knockout mice will be born. The arrested
or spontaneously aborted fetuses can then be examined to determine how
development has gone awry in fetuses that are homozygous for the knockout allele.
*40. A hypothetical disorder called G syndrome is an autosomal dominant disease
characterized by visual, skeletal, and cardiovascular defects. The disorder appears
in middle age. Because the symptoms of the disorder are variable, the disorder is
difficult to diagnose. Early diagnosis is important, however, because the
cardiovascular symptoms can be treated if the disorder is recognized early. The
gene for G syndrome is known to reside on chromosome 7, and it is closely linked
Chapter Eighteen: Recombinant DNA Technology 233
to two RFLPs on the same chromosome, one at the A locus and one at the C locus.
The genes at the G, A, and C loci are very close together, and there is little crossing
over between them. The following RFLP alleles are found at the A and C loci:
A locus: A1, A2, A3, A4
C locus: C1, C2, C3
Sally, shown in the following pedigree, is concerned that she might have G
syndrome. Her deceased mother had G syndrome, and she has a brother with the
disorder. A geneticist genotypes Sally and her immediate family for the A and C
loci and obtains the genotypes shown on the pedigree.
A1 A1
C1 C3
Sally
A1 A3
C2 C3
A1 A2
C2 C3
A1 A2
C1 C2
(a) Assume that there is no crossing over between the A, C, and G loci. Does Sally
carry the gene that causes G syndrome? Explain why or why not.
First, we must determine the RFLP genotype of Sally’s mother. Based on the
RFLP genotypes of Sally and her siblings and their father, we deduce that
Sally’s mother must have had A2A3 and C2C2. The linkage relationships of
these chromosomes are A1C1 and A1C3 from the father and A2C2 and A3C2
from the mother. The mother passed on an A2C2 to Sally’s brother with G
syndrome; therefore the G syndrome allele must be linked with A2C2. Since
Sally inherited the A2C2 chromosome from her mother, she must also have
inherited the G syndrome allele, assuming that no crossover occurred between
the A, C, and G loci.
(b) Draw the arrangement of the A, C, and G alleles on the chromosomes for all
members of the family.
Father: A1 C1 g, A1 C3 g
Mother: A2 C2 G, A3 C2 g
Sally’s unaffected brother: A1 C3 g, A3 C2 g
Sally’s affected brother: A1 C3 g, A2 C2 G
Sally: A1 C1 g, A2 C2 G
CHALLENGE QUESTIONS
41. Suppose that you are hired by a biotechnology firm to produce a strain of giant fruit
flies, by using recombinant DNA technology, so that genetics students will not be
forced to strain their eyes when looking at tiny flies. You go to the library and learn
that growth in fruit flies is normally inhibited by a hormone called shorty substance
P (SSP). You decide that you can produce giant fruit flies if you can somehow turn
234 Chapter Eighteen: Recombinant DNA Technology
off the production of SSP. SSP is synthesized from a compound called XSP in a
single-step reaction catalyzed by the enzyme runtase:
XSP———————>SSP
runtase
A researcher has already isolated cDNA for runtase and has sequenced it, but the
location of the runtase gene in the Drosophila genome is unknown.
In attempting to devise a strategy for turning off the production of SSP and
producing giant flies by using standard recombinant DNA techniques, you discover
that deleting, inactivating, or otherwise mutating this DNA sequence in Drosophila
turns out to be extremely difficult. Therefore you must restrict your genetic
engineering to gene augmentation (adding new genes to cells). Describe the
methods that you will use to turn off SSP and produce giant flies by using
recombinant DNA technology.
One possible solution is to create a gene for expression of oligonucleotides in the
cell that will act like oligonucleotide drugs. Given the DNA sequence of runtase,
one could fuse a short portion of the cDNA for runtase in reverse orientation to a
strong promoter, so that short antisense RNA molecules are produced in
abundance. These antisense RNA molecules will hybridize with runtase mRNA and
prevent its expression. Another route is to design a ribozyme whose targeting
sequences are complementary to runtase mRNA. A DNA construct that would
express a ribozyme can be transformed into the cell. The expressed ribozyme would
then selectively target and cleave runtase mRNA.
42. A rare form of polydactyly (extra fingers and toes) in humans is due to an X-linked
recessive gene, whose chromosomal location is unknown. Suppose a geneticist
studies the family whose pedigree is shown here. She isolates DNA from each
member of this family, cuts the DNA with a restriction enzyme, separates the
resulting fragments by gel electrophoresis, and transfers the DNA to nitrocellulose
by Southern blotting. She then hybridizes the nitrocellulose with a cloned DNA
sequence that comes from the X chromosome. The pattern of bands that appears on
the autoradiograph is shown below each person in the pedigree.
Chapter Eighteen: Recombinant DNA Technology 235
(a) For each person in the pedigree, give his or her genotype for RFLPs revealed by
the probe. (Remember that males are hemizygous for X-linked genes, and
females can be homozygous or heterozygous.)
The probe reveals three distinct RFLP alleles in this pedigree. I will designate
these types as indicated on the pedigree diagram on the previous page: A =
single high band, B = two intermediate bands, C = three bands (one
intermediate plus two low bands). Father = CY; mother = AB; children from
left to right: AY, BY, BC, BY, AC, BY, AY (Y indicates Y chromosome).
(b) Is there evidence for close linkage between the probe sequence and the X-linked
gene for polydactyly? Explain your reasoning.
Yes, but with low confidence. All three sons who inherited the B allele of the
RFLP locus also have polydactyly. Conversely, two sons who did not inherit the
B allele do not have polydactyly. Therefore, the observed recombination
frequency between the RFLP locus and the polydactyly locus is less than 1/5, or
20%. However, these data are far too limited to make a statistically reliable
estimate of recombination frequency between these loci.
(c) How many of the daughters in the pedigree are likely to be carriers of X-linked
polydactyly? Explain your reasoning.
Only one daughter inherited the RFLP B allele, so only one daughter is likely to
be a carrier of X-linked polydactyly.